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Student's Solutions Guide
to accompany
.:..·
~'
··Discrete· .
:
...·.Mathematics·
.
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.
and Its
Applications
SEVENTH EDITION
Prepared by
Jerrold Grossman .
Student's Solutions Guide
to accompany
Discrete Mathematics
and Its Applications
Seventh Edition
Kenneth H. Rosen
Monmouth University
(and formerly AT&T Laboratories)
Prepared by
Jerrold W. Grossman
Oakland University
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Learn
•
Succeed"
The McGrow·Hill Companies
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learn
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Student's Solutions Guide to accompany
DISCRETE MATHEMATICS AND ITS APPLICATIONS, SEVENTH EDITION
KENNETH H. ROSEN
Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Compames, Inc., 1221 Avenue of the Americas, New
York, NY 10020. Copyright© 2012 and 2007 by The McGraw-Hill Companies, Inc. All rights reserved.
Pnnted in the United States of America.
No part of this publication may be reproduced or distributed many form or by any means, or stored ma database or retrieval system,
without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network or other electronic
storage or transm1ss10n, or broadcast for distance leammg.
1234567890QDB~DB10987654321
ISBN: 978-0-07-735350-6
MHID: 0-07-735350-1
www.mhhe.com
Preface
This Student's Solutions Guide for Discrete Mathematics and Its Applications, seventh edition,
contains several useful and important study aids.
• SOLUTIONS TO ODD-NUMBERED EXERCISES
The bulk of this work consists of solutions to all the odd-numbered exercises in
the text. These are not just answers, but complete, worked-out solutions, showing how
the principles discussed in the text and examples are applied to the problems. I have also
added bits of wisdom, insights, warnings about errors to avoid, and extra comments that
go beyond the question as posed. Furthermore, at the beginning of each section you will
find some general words of advice and hints on approaching the exercises in that section.
• REFERENCES FOR CHAPTER REVIEWS
Exact page references, theorem and example references, and answers are provided as a
guide for all the chapter review questions in the text. This will make reviewing for
tests and quizzes particularly easy.
• A GUIDE TO WRITING PROOFS
Near the end of this book is a section on writing proofs, a skill that most students find difficult to master. Proofs are introduced formally in Chapter 1 (and proofs by
mathematical induction are studied in Chapter 5), but exercises throughout the text ask
for proofs of propositions. Reading this section when studying Sections 1.6-1.8, and then
periodically thereafter, will be rewarded, as your proof-writing ability grows.
• REFERENCES AND ADVICE ON THE WRITING PROJECTS
Near the end of this book you will find some general advice on the Writing Projects
given at the end of each chapter. There is a discussion of various resources available in the
mathematical sciences (such as Mathematical Reviews and the World Wide Web), tips on
doing library research, and some suggestions on how to write well. In addition, there
is a rather extensive bibliography of books and articles that will be useful when
researching these projects. We also provide specific hints and suggestions for each project,
with pointers to the references; these can be found in the solutions section of this manual,
at the end of each chapter.
• SAMPLE CHAPTER TESTS
Near the end of this book you will find a sequence of 13 chapter tests, comparable
to what might be given in a course. You can take these sample tests in a simulated test
setting as practice for the real thing. Complete solutions are provided, of course.
• PROBLEM-SOLVING TIPS AND LIST OF COMMON MISTAKES
People beginning any endeavor tend to make the same kinds of mistakes. This is
especially true in the study of mathematics. I have included a detailed list of common
misconceptions that students of discrete mathematics often have and the kinds of errors
they tend to make. Specific examples are given. It will be useful for you to review this list
from time to time, to make sure that you are not falling into these common traps. Also
included in this section is general advice on solving problems in mathematics, which you
will find helpful as you tackle the exercises.
• CRIB SHEETS
Finally, I have prepared a set of 13 single-page "crib sheets," one for each chapter
of the book. They provide a quick summary of all the important concepts, definitions, and
lll
theorems in the chapter. There are at least three ways to use these. First, they can be used
as a reference source by someone who wants to brush up on the material quickly or reveal
gaps in old knowledge. Second, they provide an excellent review sheet for studying for tests
and quizzes, especially useful for glancing over in the last few minutes. And third, a copy
of this page (augmented by your own notes in the margins) is ideal in those cases where an
instructor allows the students to come to a test with notes.
Several comments about the solutions in this volume are in order. In many cases, more
than one solution to an exercise is presented, and sometimes the solutions presented here are
not the same as the answers given in the back of the text. Indeed, there is rarely only one way
to solve a problem in mathematics. You may well come upon still other valid ways to arrive at
the correct answers. Even if you have solved a problem completely, you will find that reviewing
the solutions presented here will be valuable, since there is insight to be gained from seeing how
someone else handles a problem that you have just solved.
Exercises often ask that answers be justified or verified, or they ask you to show or prove
a particular statement. In all these cases your solution should be a proof, i.e., a mathematical
argument based on the rules of logic. Such a proof needs to be complete, convincing, and correct.
Read your proof after finishing it. Ask yourself whether you would understand and believe it if
it were presented to you by your instructor.
Although I cannot personally discuss with you my philosophy on learning discrete mathematics by solving exercises, let me include a few general words of advice. The best way to learn
mathematics is by solving problems, and it is crucial that you first try to work these exercises
independently. Consequently, do not use this Guide as a crutch. Do not look at the solution
(or even the answer) to a problem before you have worked on it yourself. Resist the temptation
to consult the solution as soon as the going gets rough. Make a real effort to work the problem
completely on your own-preferably to the point of writing down a complete solution-before
checking your work with the solutions presented here. If you have not been able to solve a problem and have reached the point where you feel it necessary to look at the answer or solution,
try reading it only casually, looking for a hint as to how you might proceed; then try working
on the exercise again, armed with this added information. As a last resort, study the solution
in detail and make sure you could explain it to a fellow student.
I want to thank Jerry Grossman for his extensive advice and assistance in the preparation of
this entire Guide, Paul Lorczak, Suzanne Zeitman, and especially Georgia Mederer for doublechecking the solutions, Ron Marash for preparing the advice on writing proofs, and students
at Monmouth College and Oakland University for their input on preliminary versions of these
solutions.
A tremendous amount of effort has been devoted to ensuring the accuracy of these solutions,
but it is possible that a few scattered errors remain. I would appreciate hearing about all that
you find, be they typographical or mathematical. You can reach me using the Reporting of
Errata link on the companion website's Information Center at www.mhhe.com/rosen.
One final note: In addition to this Guide, you will find the companion website created
for Discrete Mathematics and Its Applications an invaluable resource. Included here are a
Web Resources Guide with links to external websites keyed to the textbook, numerous Extra
Examples to reinforce important topics, Interactive Demonstration Applets for exploring key
algorithms, Self Assessment question banks to gauge your understanding of core concepts, and
many other helpful resources. See the section titled "The Companion Website" on page xvi of
the textbook for more details. The address is www.mhhe.com/rosen.
Kenneth H. Rosen
iv
Contents
Preface
iii
CHAPTER 1
The Foundations: Logic and Proofs
1.1
Propositional Logic
1
8
1.2
Applications of Propositional Logic
1.3
Propositional Equivalences
11
1.4
Predicates and Quantifiers
17
1.5
Nested Quantifiers
23
1.6
Rules of Inference
32
1. 7
Introduction to Proofs
36
Proof Methods and Strategy
40
1.8
Guide to Review Questions for Chapter 1
44
Supplementary Exercises for Chapter 1
45
48
Writing Projects for Chapter 1
CHAPTER 2
Basic Structures: Sets, Functions,
Sequences, Sums, and Matrices
2.1
Sets
50
2.2
Set Operations
53
Functions
59
2 .3
68
2.4
Sequences and Summations
Cardinality of Sets
75
2.5
2.6
Matrices
79
83
Guide to Review Questions for Chapter 2
Supplementary Exercises for Chapter 2
84
Writing Projects for Chapter 2
87
CHAPTER 3
Algorithms
3.1
Algorithms
88
3.2
The Growth of Functions
97
3.3
Complexity of Algorithms
103
Guide to Review Questions for Chapter 3
107
Supplementary Exercises for Chapter 3
108
112
Writing Projects for Chapter 3
CHAPTER 4
Number Theory and Cryptography
4.1
Divisibility and Modular Arithmetic
113
116
4.2
Integer Representations and Algorithms
4.3
Primes and Greatest Common Divisors
122
130
4.4
Solving Congruences
4.5
Applications of Congruences
137
Cryptography
140
4.6
Guide to Review Questions for Chapter 4
142
Supplementary Exercises for Chapter 4
143
Writing Projects for Chapter 4
147
v
1
50
88
113
CHAPTER 5
CHAPTER 6
CHAPTER 7
CHAPTER 8
Induction and Recursion
5.1
Mathematical Induction
149
Strong Induction and Well-Ordering
161
5.2
5.3
Recursive Definitions and Structural Induction
5.4
Recursive Algorithms
176
Program Correctness
182
5.5
Guide to Review Questions for Chapter 5
183
Supplementary Exercises for Chapter 5
185
Writing Projects for Chapter 5
195
Counting
6.1
The Basics of Counting
197
The Pigeonhole Principle
206
6.2
6.3
Permutations and Combinations
211
Binomial Coefficients and Identities
6.4
216
6.5
Generalized Permutations and Combinations
6.6
Generating Permutations and Combinations
Guide to Review Questions for Chapter 6
230
Supplementary Exercises for Chapter 6
231
237
Writing Projects for Chapter 6
Discrete Probability
7.1
An Introduction to Discrete Probability
Probability Theory
242
7.2
7.3
Bayes' Theorem
247
7.4
Expected Value and Variance
250
Guide to Review Questions for Chapter 7
255
Supplementary Exercises for Chapter 7
256
261
Writing Projects for Chapter 7
149
167
197
220
227
239
239
Advanced Counting Techniques
8.1
Applications of Recurrence Relations
262
Solving Linear Recurrence Relations
272
8.2
8.3
Divide-and-Conquer Algorithms and Recurrence Relations
8.4
Generating Functions
286
8.5
Inclusion-Exclusion
298
8.6
Applications of Inclusion-Exclusion
300
Guide to Review Questions for Chapter 8
304
Supplementary Exercises for Chapter 8
305
Writing Projects for Chapter 8
310
CHAPTER 9
9.1
9.2
9.3
9.4
Relations
Relations and Their Properties
312
n-ary Relations and Their Applications
Representing Relations
322
Closures of Relations
325
Vl
262
282
312
320
9.5
Equivalence Relations
329
Partial Orderings
337
9.6
345
Guide to Review Questions for Chapter 9
Supplementary Exercises for Chapter 9
347
351
Writing Projects for Chapter 9
CHAPTER 10
Graphs
10.1
Graphs and Graph Models
352
10.2
Graph Terminology and Special Types of Graphs
355
10.3
Representing Graphs and Graph Isomorphism
361
10.4
Connectivity
368
10.5
Euler and Hamilton Paths
375
10.6
Shortest-Path Problems
381
10.7
Planar Graphs
385
10.8
Graph Coloring
389
Guide to Review Questions for Chapter 10
393
Supplementary Exercises for Chapter 10
395
Writing Projects for Chapter 10
401
352
Trees
CHAPTER 11
11.1
Introduction to Trees
403
11.2
Applications of Trees
408
11.3
Tree Traversal
417
Spanning Trees
421
11.4
11.5
Minimum Spanning Trees
427
Guide to Review Questions for Chapter 11
Supplementary Exercises for Chapter 11
Writing Projects for Chapter 11
434
403
429
430
CHAPTER 12
Boolean Algebra
12.1
Boolean Functions
436
12.2
Representing Boolean Functions
440
12.3
Logic Gates
443
12.4
Minimization of Circuits
445
Guide to Review Questions for Chapter 12
453
Supplementary Exercises for Chapter 12
454
Writing Projects for Chapter 12
456
436
CHAPTER 13
Modeling Computation
13.1
Languages and Grammars
457
Finite-State Machines with Output
464
13.2
469
13.3
Finite-State Machines with No Output
Language Recognition
474
13.4
Turing Machines
478
13.5
482
Guide to Review Questions for Chapter 13
483
Supplementary Exercises for Chapter 13
Writing Projects for Chapter 13
487
457
Vll
APPENDIXES
Appendix 1
Appendix 2
Appendix 3
489
Axioms for the Real Numbers
and the Positive Integers
489
Exponential and Logarithmic Functions
Pseudocode
491
491
A Guide to Proof-Writing
492
References and Advice on Writing Projects
503
Sample Chapter Tests with Solutions
511
Common Mistakes in Discrete Mathematics
Solving Problems in Discrete Mathematics
List of Common Mistakes
541
540
540
551
Crib Sheets
Vlll
Section 1.1
Propositional Logic
1
CHAPTERl
The Foundations: Logic and Proofs
SECTION 1.1
Propositional Logic
Manipulating propositions and constructing truth tables are straightforward. A truth table is constructed by
finding the truth values of compound propositions from the inside out; see the solution to Exercise 31, for
instance. This exercise set also introduces fuzzy logic.
1. Propositions must have clearly defined truth values, so a proposition must be a declarative sentence with no
free variables.
a) This is a true proposition.
b) This is a false proposition (Tallahassee is the capital).
c) This is a true proposition.
d) This is a false proposition.
e) This is not a proposition (it contains a variable; the truth value depends on the value assigned to x).
f) This is not a proposition, since it does not assert anything.
3. a) Mei does not have an MP3 player.
b) There is pollution in New Jersey.
c) 2+1#3
d) It is not the case that the summer in Maine is hot and sunny. In other words, the summer in Maine is not
hot and sunny, which means that it is not hot or it is not sunny. It is not correct to negate this by saying
"The summer in Maine is not hot and not sunny." [For this part (and in a similar vein for part (b)) we need
to assume that there are well-defined notions of hot and sunny; otherwise this would not be a proposition
because of not having a definite truth value.]
5. a) Steve does not have more than 100 GB free disk space on his laptop. (Alternatively: Steve has less than
or equal to 100 GB free disk space on his laptop.)
b) Zach does not block e-mails and texts from Jennifer. (Alternatively, and more precisely: Zach does not
block e-mails from Jennifer, or he does not block texts from Jennifer. Note that negating an "and" statement
produces an "or" statement. It would not be correct to say that Zach does not block e-mails from Jennifer,
and he does not block texts from Jennifer. That is a stronger statement than just the negation of the given
statement.)
c) 7·11·13#999.
d) Diane did not ride her bike 100 miles on Sunday.
7. a) This is false, because Acme's revenue was larger.
b) Both parts of this conjunction are true, so the statement is true.
c) The second part of this disjunction is true, so the statement is true.
d) The hypothesis of this conditional statement is false and the conclusion is true, so by the truth-table
definition this is a true statement. (Either of those conditions would have been enough to make the statement
true.)
2
Chapter 1
The Foundations: Logic and Proofs
e) Both parts of this biconditional statement are true, so by the truth-table definition this is a true statement.
9. This is pretty straightforward, using the normal words for the logical operators.
a) Sharks have not been spotted near the shore.
b) Swimming at the New Jersey shore is allowed, and sharks have been spotted near the shore.
c) Swimming at the New Jersey shore is not allowed, or sharks have been spotted near the shore.
d) If swimming at the New Jersey shore is allowed, then sharks have not been spotted near the shore.
e) If sharks have not been spotted near the shore, then swimming at the New Jersey shore is allowed.
f) If swimming at the New Jersey shore is not allowed, then sharks have not been spotted near the shore.
g) Swimming at the New Jersey shore is allowed if and only if sharks have not been spotted near the shore.
h) Swimming at the New Jersey shore is not allowed, and either swimming at the New Jersey shore is allowed
or sharks have not been spotted near the shore. Note that we were able to incorporate the parentheses by
using the word "either" in the second half of the sentence.
11. a) Here we have the conjunction p A q.
b) Here we have a conjunction of p with the negation of q, namely p A --.q. Note that "but" logically means
the same thing as "and."
c) Again this is a conjunction: --.p A --.q.
d) Here we have a disjunction, p V q. Note that V is the inclusive or, so the "(or both)" part of the English
sentence is automatically included.
e) This sentence is a conditional statement, p -> q.
f) This is a conjunction of propositions, both of which are compound: (p V q) A (p-> --.q).
g) This is the biconditional p
f-+
q.
13. a) This is just the negation of p, so we write --.p.
b) This is a conjunction ("but" means "and"): p A --.q.
c) The position of the word "if" tells us which is the antecedent and which is the consequence: p
d) -ip - t --.q
e) The sufficient condition is the antecedent: p
->
->
q.
q.
f)qA--.p
g) ''Whenever" means "if": q
-> p.
15. a) "But" is a logical synonym for "and" (although it often suggests that the second part of the sentence is
likely to be unexpected). So this is r A --.p.
b) Because of the agreement about precedence, we do not need parentheses in this expression: --.p A q A r.
c) The outermost structure here is the conditional statement, and the conclusion part of the conditional
statement is itself a biconditional: r -> ( q f-+ --.p) .
d) This is similar to part (b): --.q A --.p A r.
e) This one is a little tricky. The statement that the condition is necessary is a conditional statement in one
direction, and the statement that this condition is not sufficient is the negation of the conditional statement in
the other direction. Thus we have the structure (safe -> conditions) A--.( conditions -> safe). Fleshing this out
gives our answer: (q -> (--.r A --.p)) A --.( (--.r A --.p) -> q). There are some logically equivalent correct answers
as well.
f) We just need to remember that "whenever" means "if" in logic: (p A r)
->
--.q.
17. In each case, we simply need to determine the truth value of the hypothesis and the conclusion, and then use
Section 1.1
Propositional Logic
3
the definition of the truth value of the conditional statement. The conditional statement is true in every case
except when the hypothesis (the "if" part) is true and the conclusion (the ''then" part) is false.
a) Since the hypothesis is true and the conclusion is false, this conditional statement is false.
b) Since the hypothesis is false and the conclusion is true, this conditional statement is true.
c) Since the hypothesis is false and the conclusion is false, this conditional statement is true. Note that the
conditional statement is false in both part (b) and part ( c); as long as the hypothesis is false, we need look
no further to conclude that the conditional statement is true.
d) Since the hypothesis is false, this conditional statement is true.
19. a) Presumably the diner gets to choose only one of these beverages, so this is an exclusive or.
b) This is probably meant to be inclusive, so that long passwords with many digits are acceptable.
c) This is surely meant to be inclusive. If a student has had both of the prerequisites, so much the better.
d) At first glance one might argue that no one would pay with both currencies simultaneously, so it would
seem reasonable to call this an exclusive or. There certainly could be cases, however, in which the patron
would pay a portion of the bill in dollars and the remainder in euros. Therefore, an inclusive or seems better.
21. a) If this is an inclusive or, then it is allowable to take discrete mathematics if you have had calculus or
computer science or both. If this is an exclusive or, then a person who had both courses would not be allowed
to take discrete mathematics-only someone who had taken exactly one of the prerequisites would be allowed
in. Clearly the former interpretation is intended; if anything, the person who has had both calculus and
computer science is even better prepared for discrete mathematics.
b) If this is an inclusive or, then you can take the rebate, or you can sign up for the low-interest loan, or you
can demand both of these incentives. If this is an exclusive or, then you will receive one of the incentives but
not both. Since both of these deals are expensive for the dealer or manufacturer, surely the exclusive or was
intended.
c) If this is an inclusive or, you can order two items from column A (and none from B), or three items from
column B (and none from A), or five items (two from A and three from B). If this is an exclusive or, which it
surely is here, then you get your choice of the two A items or the three B items, but not both.
d) If this is an inclusive or, then lots of snow, or extreme cold, or a combination of the two will close school.
If this is an exclusive or, then one form of bad weather would close school but if both of them happened then
school would meet. This latter interpretation is clearly absurd, so the inclusive or is intended.
23. a) If the wind blows from the northeast, then it snows. ["Whenever" means ''if."]
b) If it stays warm for a week, then the apple trees will bloom. [Sometimes word order is flexible in English,
as it is here. Other times it is not-"The man bit the dog" does not have the same meaning as "The dog bit
the man."]
c) If the Pistons win the championship, then they beat the Lakers.
d) If you get to the top of Long's Peak, then you must have walked eight miles. [The necessary condition is
the conclusion.]
e) If you are world famous, then you will get tenure as a professor. [The sufficient condition is the antecedent.]
f) If you drive more than 400 miles, then you will need to buy gasoline. [The word ''then" is sometimes
omitted in English sentences, but it is still understood.]
g) If your guarantee is good, then you must have bought your CD player less than 90 days ago. [Note that
"only if" does not mean "if"; the clause following the "only if" is the conclusion, not the antecedent.]
h) If the water is not too cold, then Jan will go swimming. [Note that "unless" really means "if not." It also
can be taken to mean "or."]
Chapter 1
4
The Foundations: Logic and Proofs
25. In each case there will be two statements. It is being asserted that the first one holds true if and only if the
second one does. The order doesn't matter, but often one order is more colloquial English.
a) You buy an ice cream cone if and only if it is hot outside.
b) You win the contest if and only if you hold the only winning ticket.
c) You get promoted if and only if you have connections.
d) Your mind will decay if and only if you watch television.
e) The train runs late if and only if it is a day I take the train.
27. Many forms of the answers for this exercise are possible.
a) One form of the converse that reads well in English is "I will ski tomorrow only if it snows today." We
could state the contrapositive as ''If I don't ski tomorrow, then it will not have snowed today." The inverse is
"If it does not snow today, then I will not ski tomorrow."
b) The proposition as stated can be rendered "If there is going to be a quiz, then I will come to class." The
converse is "If I come to class, then there will be a quiz." (Or, perhaps even better, "I come to class only if
there's going to be a quiz.") The contrapositive is "If I don't come to class, then there won't be a quiz." The
inverse is "If there is not going to be a quiz, then I don't come to class.''
c) There is a variable ("a positive integer") in this sentence, so technically it is not a proposition. Nevertheless,
we can treat sentences such as this in the same way we treat propositions. Its converse is "A positive integer
is a prime if it has no divisors other than 1 and itself." (Note that this can be false, since the number 1
satisfies the hypothesis but not the conclusion.) The contrapositive of the original proposition is "If a positive
integer has a divisor other than 1 and itself, then it is not prime." (We are simplifying a bit here, replacing
''does not have no divisors" by "has a divisor." Note that this is always true, assuming that we are talking
about positive divisors.) The inverse is "If a positive integer is not prime, then it has a divisor other than 1
and itself."
29. A truth table will need 2n rows if there are n variables.
a)2 1 =2
b)2 4 =16
d)2 4 =16
c)2 6 =64
31. To construct the truth table for a compound proposition, we work from the inside out. In each case, we will
show the intermediate steps. In part ( d), for example, we first construct the truth table for p V q, then the
truth table for p /\ q, and finally combine them to get the truth table for (p V q) -+ (p /\ q). For parts (a)
and (b) we have the following table (column three for part (a), column four for part (b)).
P
'P
p/\-ip
pV-ip
T
F
F
T
F
F
T
T
For part ( c) we have the following table.
v -iq
p
q
-iq
T
T
F
F
T
F
T
F
F
T
F
T
T
T
F
T
p
(p
v -iq)
-+
q
T
F
T
F
For part ( d) we have the following table.
p
q
pVq
p /\ q
T
T
F
F
T
F
T
F
T
T
T
F
T
F
F
F
(p
v q)
-+
T
F
F
T
(p /\ q)
5
Propositional Logic
Section 1.1
For part ( e) we have the following table. This time we have omitted the column explicitly showing the
negations of p and q. Note that this true proposition is telling us that a conditional statement and its
contrapositive always have the same truth value.
p
q
p-+ q
T
T
F
F
T
F
T
F
T
F
T
T
•q
-+
(p-+ q)
'P
f--7
T
F
T
T
(•q
-+
•p)
T
T
T
T
For part ( f) we have the following table. The fact that this proposition is not always true tells us that knowing
a conditional statement in one direction does not tell us that the conditional statement is true in the other
direction.
p
q
T
T
F
F
T
F
T
F
p
--'>
q
q-+ p
T
F
T
T
(p-+ q)
T
T
F
T
-+
(q-+ p)
T
T
F
T
33. To construct the truth table for a compound proposition, we work from the inside out. In each case, we will
show the intermediate steps. In part (a), for example, we first construct the truth table for p V q, then the
truth table for p EB q, and finally combine them to get the truth table for (p V q) -+ (p EB q). For parts (a),
(b), and ( c) we have the following table (column five for part (a), column seven for part (b), column eight
for part ( c)).
(pV q) EB (p/\ q)
p q pVq
(pV q) -+ (p EB q)
p /\ q
(p EB q) -+ (p /\ q)
pEB q
T
T
F
F
T
F
T
F
F
T
T
F
T
T
T
F
F
T
T
T
T
F
F
T
T
F
F
F
F
T
T
F
For part ( d) we have the following table.
p
q
'P
p f--7 q
---
•P ,_. q
(p ,_. q) EB (•P ,_. q)
T
T
F
F
T
F
T
F
F
F
T
T
T
F
F
T
F
T
T
F
T
T
T
T
For part ( e) we need eight rows in our truth table, because we have three variables.
p f--7 q
(p ,_. q) EB (•p ,_. •r)
p q r
•P ,_. •r
•r
'P
T
F
F
F
T
F
F
F
T
T
F
T
T
T
F
T
T
For part (f) we have the following table.
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
T
F
F
F
F
T
T
F
T
F
T
F
T
F
F
T
F
T
F
F
T
F
T
p
q
•q
p EB q
p EB •q
T
T
F
F
T
F
T
F
F
T
F
T
F
T
T
F
T
F
F
T
T
T
F
F
T
T
F
(p EB q)
-+
T
F
F
T
(p EB •q)
The Foundations: Logic and Proofs
Chapter 1
6
35. The techniques are the same as in Exercises 31-34. For parts (a) and (b) we have the following table (column
four for part (a), column six for part (b)).
p
q
•q
T
T
F
F
T
F
T
F
F
T
F
T
p
---+
•q
-.p
'P
F
F
T
T
F
T
T
T
+-+
q
F
T
T
F
For parts (c) and (d) we have the following table (columns six and seven, respectively).
p
q
T
T
F
F
T
F
T
F
p
---+
q
-.p
'P
F
F
T
T
T
F
T
T
---+
(p---+ q)
q
v (•p---+ q)
(p---+ q) /\ (•p---+ q)
T
T
T
T
T
T
T
F
T
F
T
F
For parts ( e) and ( f) we have the following table (this time we have not explicitly shown the columns for
negation). Column five shows the answer for part (e), and column seven shows the answer for part (f).
p
q
T
T
F
F
T
F
T
F
p
+-+
T
F
F
T
q
'P
+-+
(p
q
+-+
q)
v (•p +-+ q)
•P
T
T
T
T
F
T
T
F
+-+
(•p
•q
T
F
F
T
+-+
-.q)
+-+
(p
+-+
q)
T
T
T
T
37. The techniques are the same as in Exercises 31-36, except that there are now three variables and therefore
eight rows. For part (a), we have
p
q
r
•q
•q Vr
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
F
F
T
T
F
F
T
T
T
F
T
T
T
F
T
T
p
q
r
-.p
q ---+ r
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
F
F
F
F
T
T
T
T
T
F
T
T
T
F
T
T
p
---+ ( •q
V r)
T
F
T
T
T
T
T
T
For part (b), we have
Parts ( c) and ( d) we can combine into a single table.
-.p
---+ ( q ---+
T
T
T
T
T
F
T
T
r)
Section 1.1
Propositional Logic
7
p
q
r
p ____, q
•p
'P ____, r
(p ____, q) V (•p ____, r)
(p ____, q) A (•p ____, r)
T
T
T
T
F
F
T
T
T
T
T
F
F
F
F
F
T
F
T
F
T
F
T
T
T
T
F
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
F
F
F
F
F
T
T
F
F
F
T
F
F
F
T
F
T
F
For part ( e) we have
p
q
r
T
T
T
T
F
T
T
F
F
T
T
T
F
F
F
F
F
p
F
T
F
T
•q
T
T
F
F
F
T
T
F
F
F
F
T
T
T
T
T
F
39. This time the truth table needs 24
p q r
F
F
F
F
F
F
F
q
F
F
Finally, for part ( f) we have
p q r
•P
T T T
F
T T F
F
T F T
F
T F F
F
F T T
T
F T F
T
F F T
T
F F F
T
T
T
T
T
T
T
T
T
F
<,--+
•p
•q
<,--+
F
F
T
T
T
T
T
T
T
F
F
T
T
T
•q
q ,,._, r
(•P ,,__, •q) ,,__, (q ,,__, r)
T
T
F
F
F
T
T
T
F
F
F
T
T
T
F
F
= 16 rows. Note the systematic order in which we list the possibilities.
s
p
<,--+
T
T
T
F
F
T
F
F
T
T
F
F
F
T
T
T
T
F
T
F
T
F
T
F
F
F
F
F
T
F
T
T
F
F
F
F
T
F
T
F
F
F
T
T
F
T
T
T
T
F
T
T
(p ,,._, q) V (•q ,,._, r)
T
T
F
F
T
T
T
T
F
F
F
F
F
•q ,,._, r
F
F
T
T
T
T
F
q
r ,,._, s
(p ,,._, q) ,,._, (r ,,._, s)
T
T
F
F
F
F
T
T
T
F
T
T
F
F
F
F
F
F
T
T
F
F
F
F
F
T
T
T
T
T
F
F
F
F
T
T
T
T
T
T
F
F
41. The first clause (p V q V r) is true if and only if at least one of p, q, and r is true. The second clause
( •p V •q V •r) is true if and only if at least one of the three variables is false. Therefore both clauses are true,
and therefore the entire statement is true, if and only if there is at least one T and one F among the truth
values of the variables, in other words, that they don't all have the same truth value.
Chapter 1
8
The Foundations: Logic and Proofs
= 111 1111; bitwise AND= 000 0000; bitwise XOR = 1111111
OR = 1111 1010; bitwise AND= 1010 0000; bitwise XOR = 0101 1010
43. a) bitwise OR
b) bitwise
c) bitwise OR= 10 01111001; bitwise AND= 00 0100 0000; bitwise XOR= 10 00111001
d) bitwise OR= 1111111111; bitwise AND= 00 0000 0000; bitwise XOR= 1111111111
45. For "Fred is not happy," the truth value is 1 - 0.8
= 0.2.
For "John is not happy,'' the truth value is 1- 0.4 = 0.6.
47. For "Fred is happy, or John is happy,'' the truth value is max(0.8, 0.4) = 0.8.
For ''Fred is not happy, or John is not happy," the truth value is max(0.2, 0.6)
Exercise 45).
=
0.6 (using the result of
49. One great problem-solving strategy to try with problems like this, when the parameter is large ( 100 statements
here) is to lower the parameter. Look at a simpler problem, with just two or three statements, and see if you
can figure out what's going on. That was the approach used to discover the solution presented here.
a) Some number of these statements are true, so in fact exactly one of the statements must be true and the
other 99 of them must be false. That is what the 99th statement is saying, so it is true and the rest are false.
b) The 10oth statement cannot be true, since it is asserting that all the statements are false. Therefore it must
be false. That makes the first statement true. Now if the 99th statement were true, then we would conclude
that statements 2 through 100 were false, which contradicts the truth of statement 99. So statement 99
must be false. That means that statement 2 is true. We continue in this way and conclude that statements
1 through 50 are all true and statements 51 through 100 are all false.
c) If there are an odd number of statements, then we'd run into a contradiction when we got to the middle. If
there were just three statements, for example, then statement 3 would have to be false, making statement 1
true, and now the truth of statement 2 would imply its falsity and its falsity would imply its truth. Therefore
this situation cannot occur with three (or any odd number of) statements. It is a logical paradox, showing
that in fact these are not statements after all.
SECTION 1.2
Applications of Propositional Logic
Applications of propositional logic abound in computer science, puzzles, and everyday life. For example, much
of the operation of our legal system is based on conditional statements. Boolean searches are increasingly
important in using the Web (see Exercises 13-14 for example).
1. Recall that '' q unless •p" is another way to state p _, q. In this problem, •P is a, so p is •a; and q is •e.
Therefore the statement here is •a _, •e. This could also be stated equivalently as e _, a (if you can edit,
then you must be an administrator).
3. Recall that p only if q means p _, q. In this case, if you can graduate then you must have fulfilled the three
listed requirements. Therefore the statement is g _, (r /\ (•m) /\ (•b)). Notice that in everyday life one might
actually say "You can graduate if you do these things," but logically that is not what the rules really say.
5. This is similar to Exercise 3. If you are eligible to be President, then you must satisfy the requirements:
e _, (a/\ (b V p) /\ r). Notice that it is only the requirement of being native-born that can be overridden by
having parents who were citizens, so b V p is grouped as one of the three conditions.
Section 1.2
Applications of Propositional Logic
9
7. a) Since "whenever" means "if," we have q---+ p.
b) Since "but" means "and," we have q /\ 'P.
c) This sentence is saying the same thing as the sentence in part (a), so the answer is the same: q---+ p.
d) Again, we recall that "when" means "if" in logic: •q---+ •p.
9. Let m, n, k, and i represent the propositions ''The system is in multiuser state," "The system is operating
normally," "The kernel is functioning,'' and "The system is in interrupt mode," respectively. Then we want
to make the following expressions simultaneously true by our choice of truth values for m, n, k, and i:
m ,,_.. n,
n
---+
k,
-,kV i,
-im
---+
i,
-,i
In order for this to happen, clearly i must be false. In order for •m ---+ i to be true when i is false, the
hypothesis •m must be false, so m must be true. Since we want m ,,_.. n to be true, this implies that n must
also be true. Since we want n ---+ k to be true, we must therefore have k true. But now if k is true and i is
false, then the third specification, •kV i is false. Therefore we conclude that this system is not consistent.
11. Let s be "The router can send packets to the edge system"; let a be "The router supports the new address
space"; let r be "The latest software release is installed." Then we are told s ---+ a, a ---+ r, r ---+ s, and
•a. Since a is false, the first conditional statement tells us that s must be false. From that we deduce from
the third conditional statement that r must be false. If indeed all three propositions are false, then all four
specifications are true, so they are consistent.
13. This is similar to Example 6, about universities in New Mexico. To search for beaches in New Jersey. we could
enter NEW AND JERSEY AND BEACHES. If we enter (JERSEY AND BEACHES) NOT NEW, then
we'll get websites about beaches on the isle of Jersey, except for sites that happen to use the word "new" in a
different context (e.g., a recently opened beach there). If we were sure that the word "isle" was in the name
of the location, then of course we could enter ISLE AND JERSEY AND BEACHES.
15. There are many correct answers to this problem, but all involve some sort of double layering, or combining a
question about the kind of person being addressed with a question about the information being sought. One
solution is to ask this question: "If I were to ask you whether the right branch leads to the ruins, would you
say 'yes'?" If the villager is a truth-teller, then of course he will reply "yes" if and only if the right branch
leads to the ruins. Now let us see what the liar says. If the right branch leads to the ruins, then he would say
"no" if asked whether the right branch leads to the ruins. Therefore, the truthful answer to your convoluted
question is "no." Since he always lies, he will reply "yes." On the other hand, if the right branch does not lead
to the ruins, then he would say "yes'' if asked whether the right branch leads to the ruins; and so the truthful
answer to your question is "yes"; therefore he will reply "no." Note that in both cases, he gives the same
answer to your question as the truth-teller; namely, he says "yes" if and only if the right branch leads to the
ruins. A more detailed discussion can be found in Martin Gardner's Scientific American Book of Mathematical
Puzzles and Diversions (Simon and Schuster, 1959), p. 25; reprinted as Hexafiexagons and Other Mathematical
Diversions: The First Scientific American Book of Puzzles and Games (University of Chicago Press, 1988).
17. The question was "Does everyone want coffee?" If the first professor did not want coffee, then he would
know that the answer to the hostess's question was "no." Therefore we-and the hostess and the remaining
professors-know that the first professor does want coffee. The same argument applies to the second professor,
so she, too, must want coffee. The third professor can now answer the question. Because she said "no,'' we
conclude that she does not want coffee. Therefore the hostess knows to bring coffee to the first two professors
but not to the third.
10
Chapter 1
The Foundations: Logic and Proofs
19. If A is a knight, then he is telling the truth, in which case B must be a knave. Since B said nothing, that
is certainly possible. If A is a knave, then he is lying, which means that his statement that at least one of
them is a knave is false; hence they are both knights. That is a contradiction. So we can conclude that A is
a knight and B is a knave.
21. If A is a knight, then he is telling the truth, in which case B must be a knight as well, since A is not a knave.
(If p V q and •p are both true, then q must be true.) Since B said nothing, that is certainly possible. If A
is a knave, then his statement is patently true, but that is a contradiction to the behavior of knaves. So we
can conclude that A is a knight and B is a knight.
23. If A is a knight, then he should be telling the truth, but he is asserting that he is a knave. So that cannot
be. If A is a knave, then in order for his statement to be false, B must be a knight. So we can conclude that
A is a knave and B is a knight.
25. Neither the knight nor the knave would say that he is the knave, so B must be the spy. Therefore C is lying
and must be the knave, and A is therefore the knight (and told the truth).
27. We know that B is not the knight, because if he were, then his assertion that A is telling the truth would
mean that there were two knights. Clearly C is not the knight, because he claims he is the spy. Therefore
A is the knight. That means that B was telling the truth, so he must be the spy. And C is the knave, who
falsely asserts that he is the spy.
29. We can tell nothing here; each of the six permutations is possible. The knight will always say that he is the
knight; the knave will always lie, so he might also say that he is the knight; and the spy may lie and say that
he is the knight.
31. If there were a solution, then whoever is the knave here is speaking the truth when he says that he is not the
spy. Because knaves always lie, we get a contradiction. Therefore there are no solutions.
33. Because of the first piece of information that Steve has, let's assume first that Fred is not the highest paid.
Then Janice is. Therefore Janice is not the lowest paid, so by the second piece of information that Steve has,
Maggie is the highest paid. But that is a contradiction. Therefore we know that Fred is the highest paid.
Next let's assume that Janice is not the lowest paid. Then our second fact implies that Maggie is the highest
paid. But that contradicts the fact that Fred is the highest paid. Therefore we know that Janice is the lowest
paid. So it appears that the only hope of a consistent set of facts is to have Fred paid the most, Maggie next,
and Janice the least. (We have just seen that any other assumption leads to a contradiction.) This assumption
does not contradict either of our two facts, since in both cases, the hypothesis is false.
35. Let's use the letters B, C, G, and H for the statements that the butler, cook, gardner, and handyman are
telling the truth, respectively. We can then write each fact as a true proposition: B -+ C; •(CA G), which
is equivalent to ·CV --,Q (see the discussion of De Morgan's law in Section 1.3); •( •G A ·H), which is
equivalent to G V H; and H -+ ·C. Suppose that B is true. Then it follows from the first of our propositions
that C must also be true. This tells us (using the second proposition) that G must be false, whence the third
proposition makes H true. But now the fourth proposition is violated. Therefore we conclude that B cannot
be true. If fact, the argument we have just given also proves that C cannot be true. Therefore we know that
the butler and the cook are lying. This much already makes the first, second, and fourth propositions true,
regardless of the truth of G or H. Thus either the gardner or the handyman could be lying or telling the
truth; all we know (from the third proposition) is that at least one of them is telling the truth.
Section 1.3
Propositional Equivalences
11
37. If the first sign were true, then the second sign would also be true. In that case, we could not have one true
sign and one false sign. Rather, the second sign is true and the first is false; there is a lady in the second room
and a tiger in the first room.
39. The given conditions imply that there cannot be two honest senators. Therefore, since we are told that there
is at least one honest senator, there must be exactly 49 corrupt senators.
41. a) The output of the OR gate is q V •r. Therefore the output of the AND gate is p /\ (q V •r). Therefore the
output of this circuit is •(P /\ (q V •r)).
b) The output of the top AND gate is (•p) /\ (•q). The output of the bottom AND gate is p /\ r. Therefore
the output of this circuit is ( (•p) /\ (•q)) V (p /\ r).
43. We have the inputs come in from the left, in some cases passing through an inverter to form their negations.
Certain pairs of them enter OR gates, and the outputs of these and other negated inputs enter AND gates.
The outputs of these AND gates enter the final OR gate.
SECTION 1.3
Propositional Equivalences
The solutions to Exercises 1-10 are routine; we use truth tables to show that a proposition is a tautology or
that two propositions are equivalent. The reader should do more than this, however; think about what the
equivalence is saying. See Exercise 11 for this approach. Some important topics not covered in the text are
introduced in this exercise set, including the notion of the dual of a proposition, disjunctive normal form
for propositions, functional completeness, satisfiability, and two other logical connectives, NAND and
NOR. Much of this material foreshadows the study of Boolean algebra in Chapter 12.
1. First we construct the following truth tables, for the propositions we are asked to deal with.
pVF
pVT
p /\ p
p
p/\ T
pt\ F
pVp
T
T
T
T
F
T
F
F
T
F
F
F
The first equivalence, p /\ T
p, is valid because the second column p /\ T is identical to the first column p.
Similarly, part (b) comes from looking at columns three and one. Since column four is a column of F's, and
column five is a column of T's, part ( c) and part ( d) hold. Finally, the last two parts follow from the fact
that the last two columns are identical to the first column.
T
F
=
Chapter 1
12
The Foundations: Logic and Proofs
3. We construct the following truth tables.
p
q
pVq
qVp
p/\ q
q /\p
T
T
T
T
T T
T
T
F
T F
F
F T
T
T
F
F
F
F
F
F F
F
Part (a) follows from the fact that the third and fourth columns are identical; part (b) follows from the fact
that the fifth and sixth columns are identical.
5. We construct the following truth table and note that the fifth and eighth columns are identical.
p
q
r
qVr
p/\(qVr)
p /\ q
p /\ r
(p/\q) V (p/\r)
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
T
F
T
T
T
F
T
T
T
F
F
F
F
F
T
T
F
F
F
F
F
F
T
F
T
F
F
F
F
F
T
T
T
F
F
F
F
F
7. De Morgan's laws tell us that to negate a conjunction we form the disjunction of the negations, and to negate
a disjunction we form the conjunction of the negations.
a) This is the conjunction "Jan is rich, and Jan is happy." So the negation is "Jan is not rich, or Jan is not
happy."
b) This is the disjunction '"Carlos will bicycle tomorrow, or Carlos will run tomorrow." So the negation is
''Carlos will not bicycle tomorrow, and Carlos will not run tomorrow." We could also render this as ''Carlos
will neither bicycle nor run tomorrow.''
c) This is the disjunction ''Mei walks to class, or Mei takes the bus to class." So the negation is ''Mei does
not walk to class, and Mei does not take the bus to class." (Maybe she gets a ride with a friend.) We could
also render this as ''Mei neither walks nor takes the bus to class."
d) This is the conjunction ''Ibrahim is smart, and Ibrahim is hard working." So the negation is "Ibrahim is
not smart, or Ibrahim is not hard working.''
9. We construct a truth table for each conditional statement and note that the relevant column contains only
T's. For parts (a) and (b) we have the following table (column four for part (a), column six for part (b)).
E____!l_
T
T
F
T
F
T
F F
p /\ q
(p /\ q)
T
F
F
F
--+
p
T
T
T
T
p
vq
T
T
T
F
p
--+
(p
v q)
T
T
T
T
For parts ( c) and ( d) we have the following table (columns five and seven, respectively).
P
q
TT
TF
F T
F F
'P
P-+ q
F
F
T
F
T
T
T
T
'P
-+
(p-+ q)
T
T
T
T
P /\ q
T
F
F
F
(p /\ q)
-+
(p--+ q)
T
T
T
T
For parts (e) and (f) we have the following table. Column five shows the answer for part (e), and column
seven shows the answer for part ( f).
Section 1.3
13
Propositional Equivalences
p
p
q
T
T
T
F
T
F
F
F
T
F
T
T
---'>
q
•(p
---'>
q)
•(p
---'>
F
q)
---'>
p
•(p
•q
---'>
q)
T
T
T
F
T
T
T
F
F
T
T
F
T
T
T
---'>
-,q
11. Here is one approach: Recall that the only way a conditional statement can be false is for the hypothesis to be
true and the conclusion to be false; hence it is sufficient to show that the conclusion must be true whenever the
hypothesis is true. An alternative approach that works for some of these tautologies is to use the equivalences
given in this section and prove these "algebraically." We will demonstrate this second method in some of the
solutions.
a) If the hypothesis is true, then by the definition of /\ we know that p is true. Hence the conclusion is also
true. For an algebraic proof, we exhibit the following string of equivalences, each one following from one of the
laws in this section: (p /\ q) ---+ p = -, (p /\ q) V p = (•p V •q) V p = (•q V •p) V p = •q V (•P V p) = •q V T
T.
The first logical equivalence is the first equivalence in Table 7 (with p /\ q playing the role of p, and p playing
the role of q ); the second is De Morgan's law; the third is the commutative law; the fourth is the associative
law; the fifth is the negation law (with the commutative law); and the sixth is the domination law.
=
b) If the hypothesis p is true, then by the definition of V, the conclusion p V q must also be true.
c) If the hypothesis is true, then p must be false; hence the conclusion p ---+ q is true, since its hypothesis is
false. Symbolically we have •p---+ (p---+ q) = ••p V (•P V q) = p V (•p V q) = (p V •p) V q =TV q = T.
d) If the hypothesis is true, then by the definition of /\ we know that q must be true. This makes the
conclusion p ---+ q true, since its conclusion is true.
e) If the hypothesis is true, then p
what we wanted to show.
---+
q must be false. But this can happen only if p is true, which is precisely
f) If the hypothesis is true, then p
what we wanted to show.
---+
q must be false. But this can happen only if q is false, which is precisely
13. We first construct truth tables and verify that in each case the two propositions give identical columns. The
fact that the fourth column is identical to the first column proves part (a), and the fact that the sixth column
is identical to the first column proves part {b).
p
q
p /\ q
T T
T F
F T
F F
Alternately, we can argue as follows.
T
F
F
F
p
v (p /\ q)
T
T
F
F
p
vq
T
T
T
F
p /\ (p
v q)
T
T
F
F
a) If p is true, then p V (p /\ q) is true, since the first proposition in the disjunction is true. On the
hand, if p is false, then both parts of the disjunction are false. Hence p V (p /\ q) always has the same
value as p does, so the two propositions are logically equivalent.
b) If p is false, then p /\ (p V q) is false, since the first proposition in the conjunction is false. On the
hand, if p is true, then both parts of the conjunction are true. Hence p /\ (p V q) always has the same
value as p does, so the two propositions are logically equivalent.
other
truth
other
truth
15. We need to determine whether we can find an assignment of truth values to p and q to make this proposition
false. Let us try to find one. The only way that a conditional statement can be false is for the hypothesis to
be true and the conclusion to be false. Hence we must make •P false, which means we must make p true.
Furthermore, in order for the hypothesis to be true, we will need to make q false, so that the first part of
the conjunction will be true. But now with p true and q false, the second part of the conjunction is false.
14
Chapter 1
The Foundations: Logic and Proofs
Therefore the entire hypothesis is false, so this assignment will not yield a false conditional statement. Since
we have argued that no assignment of truth values can make this proposition false, we have proved that this
proposition is a tautology. (An alternative approach would be to construct a truth table and see that its final
column had only T's in it.) This tautology is telling us that if we know that a conditional statement is true,
and that its conclusion is false, then we can conclude that its antecedent is also false.
17. The proposition -.(p
f--7
q) is true when p and q do not have the same truth values, which means that p and
q have different truth values (either p is true and q is false, or vice versa). These are exactly the cases in
which p f--7 -.q is true. Therefore these two expressions are true in exactly the same instances, and therefore
are logically equivalent.
19. The proposition -.p
f--7 q is true when -.p and q have the same truth values, which means that p and q have
different truth values (either p is true and q is false, or vice versa). By the same reasoning, these are exactly
the cases in which p f--7 --iq is true. Therefore these two expressions are true in exactly the same instances,
and therefore are logically equivalent.
21. This is essentially the same as Exercise 17. The proposition -i(p
f--7
q) is true when p
f--7
q is false. Since
p f--7 q is true when p and q have the same truth value, it is false when p and q have different truth values
(either p is true and q is false, or vice versa). These are precisely the cases in which -.p f--7 q is true.
23. We'll determine exactly which rows of the truth table will have F as their entries. In order for (p _, r) I\ ( q _, r)
to be false, we must have at least one of the two conditional statements false, which happens exactly when r
is false and at least one of p and q is true. But these are precisely the cases in which p V q is true and r is
false, which is precisely when (p V q) _, r is false. Since the two propositions are false in exactly the same
situations, they are logically equivalent.
25. We'll determine exactly which rows of the truth table will have Fas their entries. In order for (p _, r)V(q _, r)
to be false, we must have both of the two conditional statements false, which happens exactly when r is false
and both p and q are true. But this is precisely the case in which p I\ q is true and r is false, which is precisely
when (p I\ q) _, r is false. Since the two propositions are false in exactly the same situations, they are logically
equivalent.
27. This fact was observed in Section 1.1 when the biconditional was first defined. Each of these is true precisely
when p and q have the same truth values.
29. We will show that if p _, q and q _, r are both true, then p _, r is true. Thus we want to show that if p is
true, then so is r. Given that p and p _, q are both true, we conclude that q is true; from that and q _, r
we conclude that r is true, as desired. This can also be done with a truth table.
31. To show that these are not logically equivalent, we need only find one assignment of truth values to p, q, and
r for which the truth values of (p _, q) _, r and p _, (q _, r) differ. One such assignment is F for all three.
Then (p _, q) _, r is false and p _, (q _, r) is true.
33. To show that these are not logically equivalent, we need only find one assignment of truth values to p, q, r,
and s for which the truth values of (p _, q) _, (r _, s) and (p _, r) _, (q _, s) differ. Let us try to make the
first one false. That means we have to make r _, s false, so we want r to be true and s to be false. If we
let p and q be false, then each of the other three simple conditional statements (p _, q, p _, r, and q _, s)
will be true. Then (p-t q) _, (r _, s) will be T-t F, which is false; but (p-t r) _, (q _, s) will be T-t T,
which is true.
Section 1.3
Propositional Equivalences
15
35. We apply the rules stated in the preamble.
a) pV-.qV-.r
b) (pVqVr)/\s
c) (p/\T)V(q/\F)
37. If we apply the operation for forming the dual twice to a proposition, then every symbol returns to what it
originally was. The /\ changes to the V, then changes back to the /\. Similarly the V changes to the /\, then
back to the V. The same thing happens with the T and the F. Thus the dual of the dual of a proposition s,
namely (s*)*, is equal to the original proposition s.
39. Let p and q be two compound propositions involving only the operators /\, V, and -, ; we can also allow them
to involve the constants T and F. We want to show that if p and q are logically equivalent, then p* and q*
are logically equivalent. The trick is to look at 'P and -.q. They are certainly logically equivalent if p and
q are. Now if p is a conjunction, say r /\ s, then 'P is logically equivalent, by De Morgan's law, to -.r V -.s;
a similar statement applies if p is a disjunction. If r and/or s are themselves compound propositions, then
we apply De Morgan's laws again to "push" the negation symbol -, deeper inside the formula, changing /\ to
V and V to /\. We repeat this process until all the negation signs have been "pushed in" as far as possible
and are now attached to the atomic (i.e., not compound) propositions in the compound propositions p and q.
Call these atomic propositions p 1 , p 2 , etc. Now in this process De Morgan's laws have forced us to change
each /\ to V and each V to /\. Furthermore, if there are any constants T or F in the propositions, then they
will be changed to their opposite when the negation operation is applied: -.T is the same as F, and -.F is the
same as T. In summary, 'P and -.q look just like p* and q*, except that each atomic proposition p, within
them is replaced by its negation. Now we agreed that 'P -.q; this means that for every possible assignment
of truth values to the atomic propositions p 1 , P2, etc., the truth values of 'P and -.q are the same. But
assigning T to Pi is the same as assigning F to 'Pi , and assigning F to p, is the same as assigning T to 'Pi .
Thus, for every possible assignment of truth values to the atomic propositions, the truth values of p* and q*
are the same. This is precisely what we wanted to prove.
=
41. There are three ways in which exactly two of p, q, and r can be true. We write down these three possibilities
as conjunctions and join them by V to obtain the answer: (p /\ q /\ -.r) V (p /\ -.q /\ r) V (-.p /\ q /\ r). See
Exercise 42 for a more general result.
43. Given a compound proposition p, we can construct its truth table and then, by Exercise 42, write down a
proposition q in disjunctive normal form that is logically equivalent to p. Since q involves only -, , /\, and
V, this shows that -, , /\, and V form a functionally complete collection of logical operators.
45. Given a compound proposition p, we can, by Exercise 43, write down a proposition q that is logically equivalent
to p and uses only ' , /\, and V. Now by De Morgan's law we can get rid of all the /\ 's by replacing each
occurrence of P1 /\ P2 /\ · · · /\ Pn with the equivalent proposition -.(-.pi V 'P2 V · · · V 'Pn).
47. The proposition -.(p /\ q) is true when either p or q, or both, are false, and is false when both p and q are
true; since this was the definition of p I q, the two are logically equivalent.
49. The proposition -.(p V q) is true when both p and q are false, and is false otherwise; since this was the
definition of p l q, the two are logically equivalent.
51. A straightforward approach, using the results of Exercise 50, parts (a) and (b), is as follows: (p ----+ q) =
(-.p V q) =((pl p) V q) =(((pl p) l q) l ((pl p) l q)). If we allow the constant F in our expression, then a
simpler answer is Fl ((Fl p) l q).
53. This is clear from the definition, in which p and q play a symmetric role.
16
Chapter 1
The Foundations: Logic and Proofs
55. A truth table for a compound proposition involving p and q has four lines, one for each of the following
combinations of truth values for p and q: TT, TF, FT, and FF. Now each line of the truth table for the
compound proposition can be either T or F. Thus there are two possibilities for the first line; for each of those
there are two possibilities for the second line, giving 2 · 2 = 4 possibilities for the first two lines; for each of
those there are two possibilities for the third line, giving 4 · 2 = 8 possibilities for the first three lines; and
finally for each of those, there are two possibilities for the fourth line, giving 8 · 2 = 16 possibilities altogether.
This sort of counting will be studied extensively in Chapter 6.
57. Let do, me, and in stand for the propositions "The directory database is opened,'' "The monitor is put
in a closed state," and "The system is in its initial state,'' respectively. Then the given statement reads
-iin ___, (do ___, me). By the third line of Table 7 (twice), this is equivalent to in V (-ido V me). In words, this
says that it must always be true that either the system is in its initial state, or the data base is not opened, or
the monitor is put in a closed state. Another way to render this would be to say that if the database is open,
then either the system is in its initial state or the monitor is put in a closed state.
59. Disjunctions are easy to make true, since we just have to make sure that at least one of the things being
"or-ed" is true. In this problem, we notice that 'P occurs in four of the disjunctions, so we can satisfy all of
them by making p false. Three of the remaining disjunctions contain r, so if we let r be true, those will be
taken care of. That leaves only p V --.q V s and q V --.r V --.s , and we can satisfy both of those by making q and
s both true. This assignment, then, makes all nine of the disjunctions true.
61. a) With a little trial and error we discover that setting p = F and q = F produces (FVT) /\ (TVF) /\ (TVT),
which has the value T. So this compound proposition is satisfiable. (Note that this is the only satisfying
truth assignment.)
b) We claim that there is no satisfying truth assignment here. No matter what the truth values of
p and q
might be, the four implications become T ___, T, T ___, F, F ___, T, and F ___, F, in some order. Exactly one
of these is false, so their conjunction is false.
c) This compound proposition is not satisfiable. In order for the first clause, p +-+ q, to be true, p and q must
have the same truth value. In order for the second clause, (--.p) +-+ q, to be true, p and q must have opposite
truth values. These two conditions are incompatible, so there is no satisfying truth assignment.
63. This is done in exactly the same manner as was described in the text for a 9 x 9 Sudoku puzzle, with the
variables indexed from 1 to 4, instead of from 1 to 9, and with a similar change for the propositions for the
2
X
2 blocks: /\~=O /\~=O /\!=l
v;=l V~=l p(2r + i, 2s + j, n).
65. We just repeat the discussion in the text, with the roles of the rows and columns interchanged: To assert that
column j contains the number n, we form v;= 1 p(i,j,n). To assert that column j contains all 9 numbers,
we form the conjunction of these disjunctions over all nine possible values of n, giving us /\~=l Vi=l p( i, j, n).
To assert that every column contains every number, we take the conjunction of /\~=l
1 p(i,j,n) over all
nine columns. This gives us /\~= 1 /\~= 1 v;= 1 p( i, j, n).
Vi=
Section 1.4
Predicates and Quantifiers
17
SECTION 1.4 Predicates and Quantifiers
The reader may find quantifiers hard to understand at first. Predicate logic (the study of propositions with
quantifiers) is one level of abstraction higher than propositional logic (the study of propositions without
quantifiers). Careful attention to this material will aid you in thinking more clearly, not only in mathematics
but in other areas as well, from computer science to politics. Keep in mind exactly what the quantifiers mean:
Yx means "for all x" or "for every x," and ::Jx means "there exists an x such that" or "for some x." It is
good practice to read every such sentence aloud, paying attention to English grammar as well as meaning. It
is very important to understand how the negations of quantified statements are formed, and why this method
is correct; it is just common sense, really.
The word "any" in mathematical statements can be ambiguous, so it is best to avoid using it. In negative
contexts it almost always means "some" (existential quantifier), as in the statement "You will be suspended
from school if you are found guilty of violating any of the plagiarism rules" (you don't have to violate all the
rules to get into trouble-breaking one is sufficient). In positive contexts, however, it can mean either "some"
(existential quantifier) or "every" (universal quantifier), depending on context. For example, in the sentence
"The fraternity will be put on probation if any of its members is found intoxicated," the use is existential
(one drunk brother is enough to cause the sanction); but in the sentence "Any member of the sorority will
be happy to lead you on a tour of the house," the use is universal (every member is able to be the guide).
Another interesting example is an exercise in a mathematics textbook that asks you to show that "the sum
of any two odd numbers is even." The author clearly intends the universal interpretation here-you need to
show that the sum of two odd numbers is always even. If you interpreted the question existentially, you might
say, "Look, 3 + 5 = 8, so I've shown it is true--you said I could do it for any numbers, and those are the ones
I chose.''
1. a) T, since 0 :::; 4
b) T, since 4 :::; 4
c) F, since 6
i
4
3. a) This is true.
b) This is false, since Lansing, not Detroit, is the capital.
c) This is false (but Q(Boston, Massachusetts) is true).
d) This is false, since Albany, not New York, is the capital.
5. a) There is a student who spends more than five hours every weekday in class.
b) Every student spends more than five hours every weekday in class.
c) There is a student who does not spend more than five hours every weekday in class.
d) No student spends more than five hours every weekday in class. (Or, equivalently, every student spends
less than or equal to five hours every weekday in class.)
7. a) This statement is that for every x, if x is a comedian, then x is funny. In English, this is most simply
stated, "Every comedian is funny."
b) This statement is that for every x in the domain (universe of discourse), x is a comedian and x is funny.
In English, this is most simply stated, "Every person is a funny comedian." Note that this is not the sort
of thing one wants to say. It really makes no sense and doesn't say anything about the existence of boring
comedians; it's surely false, because there exist lots of x for which C(x) is false. This illustrates the fact that
you rarely want to use conjunctions with universal quantifiers.
c) This statement is that there exists an x in the domain such that if x is a comedian then x is funny. In
English, this might be rendered, "There exists a person such that ifs/he is a comedian, then s/he is funny."
Note that this is not the sort of thing one wants to say. It really makes no sense and doesn't say anything about
the existence of funny comedians; it's surely true, because tbere exist lots of x for which C(x) is false (recall
Chapter 1
18
The Foundations: Logic and Proofs
the definition of the truth value of p -+ q). This illustrates the fact that you rarely want to use conditional
statements with existential quantifiers.
d) This statement is that there exists an x in the domain such that x is a comedian and x is funny. In
English, this might be rendered, ''There exists a funny comedian" or "Some comedians are funny" or "Some
funny people are comedians."
9. a) We assume that this sentence is asserting that the same person has both talents. Therefore we can write
3x(P(x) /\ Q(x)).
b) Since "but" really means the same thing as "and" logically, this is 3x(P(x) /\ •Q(x))
c) This time we are making a universal statement: Vx(P(x) V Q(x))
d) This sentence is asserting the nonexistence of anyone with either talent, so we could write it as -,::Jx(P(x) V
Q(x)). Alternatively, we can think of this as asserting that everyone fails to have either of these talents, and
we obtain the logically equivalent answer 'v'x•(P(x) V Q(x)). Failing to have either talent is equivalent to
having neither talent (by De Morgan's law), so we can also write this as Vx((•P(x)) /\ (•Q(x)). Note that it
would not be correct to write Vx((•P(x)) V (•Q(x)) nor to write 'v'x•(P(x) /\ Q(x)).
11. a) T, since 0
= 02
d) F, since -1-=/:- (-1) 2
b) T, since 1
= 12
e) T (let x
=
1)
c) F, since 2 f 22
f) F (let x = 2 )
13. a) Since adding 1 to a number makes it larger, this is true.
b) Since 2 · 0 = 3 · 0, this is true.
c) This statement is true, since 0 = -0.
d) This is true for the nonnegative integers but not for the negative integers. For example, 3(-2) 'f:_ 4(-2).
Therefore the universally quantified statement is false.
15. Recall that the integers include the positive and negative integers and 0.
a) This is the well-known true fact that the square of a real number cannot be negative.
b) There are two real numbers that satisfy n 2 = 2, namely
±J2, but there do not exist
any integers with
this property, so the statement is false.
c) If n is a positive integer, then n 2 ;:::: n is certainly true; it's also true for n = 0; and it's trivially true if n
is negative. Therefore the universally quantified statement is true.
d) Squares can never be negative; therefore this statement is false.
17. Existential quantifiers are like disjunctions, and universal quantifiers are like conjunctions. See Examples 11
and 16.
a) We want to assert that P(x) is true for some x in the universe, so either P(O) is true or P(l) is true
or P(2) is true or P(3) is true or P(4) is true. Thus the answer is P(O) V P(l) V P(2) V P(3) V P(4). The
other parts of this exercise are similar. Note that by De Morgan's laws, the expression in part (c) is logically
equivalent to the expression in part (f), and the expression in part (d) is logically equivalent to the expression
in part (e).
b) P(O) /\ P(l) /\ P(2) /\ P(3) /\ P(4)
c) •P(O) V ·P(l) V ·P(2) V ·P(3) V ·P(4)
d) ·P(O) /\ ·P(l) /\ ·P(2) /\ ·P(3) /\ ·P(4)
e) This is just the negation of part (a): •(P(O) V P(l) V P(2) V P(3) V P(4))
f) This is just the negation of part (b): •(P(O) /\ P(l) /\ P(2) /\ P(3) /\ P(4))
Section 1.4
Predicates and Quantifiers
19
19. Existential quantifiers are like disjunctions, and universal quantifiers are like conjunctions. See Examples 11
and 16.
a) We want to assert that P(x) is true for some x in the universe, so either P(l) is true or P(2) is true or
P(3) is true or P(4) is true or P(5) is true. Thus the answer is P(l) V P(2) V P(3) V P(4) V P(5).
b) P(l) /\ P(2) /\ P(3) /\ P( 4) /\ P(5)
c) This is just the negation of part (a): •(P(l) V P(2) V P(3) V P(4) V P(5))
d) This is just the negation of part (b): •(P(l) /\ P(2) /\ P(3) /\ P(4) /\ P(5))
e) The formal translation is as follows: (((1 =I- 3) -+ P(l)) /\ ((2 =I- 3) -+ P(2)) /\ ((3 =I- 3) -+ P(3)) /\
((4 =I- 3) -+ P(4)) /\ ((5 =I- 3) -+ P(5))) V (•P(l) V -.P(2) V -.P(3) V -.P(4) V ·P(5)). However, since the
hypothesis x =I- 3 is false when x is 3 and true when x is anything other than 3, we have more simply
(P(l) /\ P(2) /\ P(4) /\ P(5)) V (•P(l) v-.P(2) v-.P(3) V •P(4) V •P(5)). Thinking about it a little more, we
note that this statement is always true, since if the first part is not true, then the second part must be true.
21. a) One would hope that if we take the domain to be the students in your class, then the statement is true. If
we take the domain to be all students in the world, then the statement is clearly false, because some of them
are studying only other subjects.
b) If we take the domain to be United States Senators, then the statement is true. If we take the domain to
be college football players, then the statement is false, because some of them are younger than 21.
c) If the domain consists of just Princes William and Harry of Great Britain (sons of the late Princess Diana),
then the statement is true. It is also true if the domain consists of just one person (everyone has the same
mother as him- or herself). If the domain consists of all the grandchildren of Queen Elizabeth II of Great
Britain (of whom William and Harry are just two), then the statement is false.
d) If the domain consists of Bill Clinton and George W. Bush, then this statement is true because they do not
have the same grandmother. If the domain consists of all residents of the United States, then the statement
is false, because there are many instances of siblings and first cousins, who have at least one grandmother in
common.
23. In order to do the translation the second way, we let C(x) be the propositional function "x is in your class."
Note that for the second way, we always want to use conditional statements with universal quantifiers and
conjunctions with existential quantifiers.
a) Let H (x) be "x can speak Hindi." Then we have 3x H (x) the first way, or 3x (C (x) /\ H (x)) the second
way.
b) Let F(x) be "x is friendly." Then we have 't:/x F(x) the first way, or 't:/x(C(x)-+ F(x)) the second way.
c) Let B(x) be "x was born in California." Then we have 3x-.B(x) the first way, or 3x(C(x) /\ -.B(x)) the
second way.
d) Let M(x) be "x has been in a movie." Then we have 3xM(x) the first way, or 3x(C(x) /\ M(x)) the
second way.
e) This is saying that everyone has failed to take the course. So the answer here is 't:/x -.L(x) the first way, or
't:/x(C(x)-+ -.L(x)) the second way, where L(x) is "x has taken a course in logic programming."
25. Let P(x) be "x is perfect"; let F(x) be "x is your friend"; and let the domain (universe of discourse) be all
people.
a) This means that everyone has the property of being not perfect: 't:/x •P(x). Alternatively, we can write
this as --,::Jx P(x), which says that there does not exist a perfect person.
b) This is just the negation of "Everyone is perfect":
--,\:/x P( x).
c) If someone is your friend, then that person is perfect: 't:/x(F(x)
statement with universal quantifiers.
-+
P(x)). Note the use of conditional
20
Chapter 1
The Foundations: Logic and Proofs
d) We do not have to rule out your having more than one perfect friend. Thus we have simply :Jx(F(x)/\P(x)).
Note the use of conjunction with existential quantifiers.
e) The expression is Vx(F(x) /\ P(x)). Note that here we did use a conjunction with the universal quantifier,
but the sentence is not natural (who could claim this?). We could also have split this up into two quantified
statements and written (Vx F(x)) /\(\Ix P(x)).
f) This is a disjunction. The expression is (--NxF(x)) V (:Jx....,P(x)).
27. In all of these, we will let Y(x) be the propositional function that x is in your school or class, as appropriate.
a) If we let V(x) be "x has lived in Vietnam," then we have :lx V(x) if the universe is just your schoolmates,
or :Jx(Y(x) /\ V(x)) if the universe is all people. If we let D(x, y) mean that person x has lived in country y,
then we can rewrite this last one as :Jx(Y(x) /\ D(x, Vietnam)).
b) Ifwe let H(x) be "x can speak Hindi," then we have :Jx-,H(x) ifthe universe is just your schoolmates,
or 3x(Y(x) /\ -iH(x)) if the universe is all people. If we let S(x, y) mean that person x can speak language y,
then we can rewrite this last one as 3x(Y(x) /\ -,S(x, Hindi)).
c) If we let J(x), P(x), and C(x) be the propositional functions asserting x 's knowledge of Java, Prolog, and
C++, respectively, then we have :Jx(J(x) /\P(x) /\C(x)) if the universe is just your schoolmates, or :Jx(Y(x) /\
J(x) /\ P(x) /\ C(x)) if the universe is all people. If we let K(x, y) mean that person x knows programming
language y, then we can rewrite this last one as 3x(Y(x) /\ K(x, Java)/\ K(x, Prolog) /\ K(x, C++ )) .
d) If we let T(x) be "x enjoys Thai food," then we have VxT(x) if the universe is just your classmates, or
Vx(Y(x) -+ T(x)) if the universe is all people. If we let E(x, y) mean that person x enjoys food of type y,
then we can rewrite this last one as Vx(Y(x) -+ E(x, Thai)).
e) If we let H (x) be "x plays hockey,'' then we have 3x -,H (x) if the universe is just your classmates, or
:Jx(Y(x) /\ -iH(x)) if the universe is all people. If we let P(x, y) mean that person x plays game y, then we
can rewrite this last one as :Jx(Y(x) /\ -iP(x, hockey)).
29. Our domain (universe of discourse) here is all propositions. Let T(x) mean that x is a tautology and C(x)
mean that x is a contradiction. Since a contingency is just a proposition that is neither a tautology nor a
contradiction, we do not need a separate predicate for being a contingency.
a) This one is just the assertion that tautologies exist: :Jx T(x).
b) Although the word "all" or "every" does not appear here, this sentence is really expressing a universal
meaning, that the negation of a contradiction is always a tautology. So we want to say that if x is a
contradiction, then -ix is a tautology. Thus we have Vx(C(x) -+ T(-,x)). Note the rare use of a logical
symbol (negation) applied to a variable ( x); this is purely a coincidence in this exercise because the universe
happens itself to be propositions.
c) The words "can be" are expressing an existential idea-that there exist two contingencies whose disjunction
is a tautology. Thus we have :Jx:Jy( -iT(x) /\ -iC(x) /\ -iT(y) /\ -iC(y) /\ T(x Vy)). The same final comment as
in part (b) applies here. Also note the explanation about contingencies in the preamble.
d) As in part (b ), this is the universal assertion that whenever x and y are tautologies, then so is x /\ y; thus
we have Vx\ly((T(x) /\ T(y)) -+ T(x /\ y)).
31. In each case we just have to list all the possibilities, joining them with V if the quantifier is :J, and joining
them with /\ if the quantifier is V.
a) Q(O, 0,0) /\ Q(O, 1, 0)
b) Q(O, 1, 1) V Q(l, 1, 1) V Q(2, 1, 1)
c) -iQ(O, 0, 0) V -iQ(O, 0, 1)
d) -iQ(O, 0, 1) V -iQ(l, 0, 1) V -iQ(2, 0, 1)
33. In each case we need to specify some predicates and identify the domain (universe of discourse).
a) Let T(x) be the predicate that x can learn new tricks, and let the domain be old dogs. Our original
statement is :Jx T(x). Its negation is -,::Jx T(x), which we must to rewrite in the required manner as Vx -iT(x).
Section 1.4
Predicates and Quantifiers
21
In English this reads "Every old dog is unable to learn new tricks" or "All old dogs can't learn new tricks."
(Note that this does not say that not all old dogs can learn new tricks-it is saying something stronger than
that.) More colloquially, we can say "No old dogs can learn new tricks."
b) Let C(x) be the predicate that x knows calculus, and let the domain be rabbits. Our original statement
is --.3x C(x). Its negation is, of course, simply 3x C(x). In English this reads "There is a rabbit that knows
calculus.''
c) Let F(x) be the predicate that x can fly, and let the domain be birds. Our original statement is l::/x F(x).
Its negation is --.1::/x F(x) (i.e., not all birds can fly), which we must to rewrite in the required manner as
3x --.F (x) . In English this reads "There is a bird who cannot fly."
d) Let T(x) be the predicate that x can talk, and let the domain be dogs. Our original statement is --.3xT(x).
Its negation is, of course, simply 3xT(x). In English this reads "There is a dog that talks."
e) Let F(x) and R(x) be the predicates that x knows French and knows Russian, respectively, and let the
domain be people in this class. Our original statement is --.3x(F(x) /\ R(x)). Its negation is, of course, simply
3x(F(x) /\ R(x)). In English this reads "There is someone in this class who knows French and Russian."
35. a) As we saw in Example 13, this is true, so there is no counterexample.
b) Since 0 is neither greater than nor less than 0, this is a counterexample.
c) This proposition says that 1 is the only integer-that every integer equals 1. If is obviously false, and any
other integer, such as -111749, provides a counterexample.
37. In each case we need to make up predicates. The answers are certainly not unique and depend on the choice
of predicate, among other things.
a) l::/x((F(x,25000) V S(x,25)) _, E(x)), where E(x) is "Person x qualifies as an elite flyer in a given year,"
F(x, y) is "Person x flies more than y miles in a given year,'' and S(x, y) is "Person x takes more than y
flights in a given year"
b) l::/x(((M(x)/\T(x, 3))V(•M(x)/\T(x, 3.5))) _, Q(x)), where Q(x) is "Person x qualifies for the marathon,"
l'vf(x) is "Person x is a man,'' and T(x, y) is "Person x has run the marathon in less than y hours"
c) M _, ((H(60) V (H(45) /\ T)) /\ l::/yG(B,y)), where Mis the proposition ''The student received a masters
degree," H(x) is "The student took at least x course hours," T is the proposition "The student wrote a
thesis," and G(x, y) is "The person got grade x or higher in his course y"
d) 3x ((T(x, 21) /\ G(x, 4.0)), where T(x, y) is "Person x took more than y credit hours" and G(x,p) is
"Person x earned grade point average p" (we assume that we are talking about one given semester)
39. In each case we pretty much just write what we see.
a) If there is a printer that is both out of service and busy, then some job has been lost.
b) If every printer is busy, then there is a job in the queue.
c) If there is a job that is both queued and lost, then some printer is out of service.
d) If every printer is busy and every job is queued, then some job is lost.
41. In each case we need to make up predicates. The answers are certainly not unique and depend on the choice
of predicate, among other things.
a) (3x F(x, 10)) _, 3x S(x), where F(x, y) is "Disk x has more than y kilobytes of free space,'' and S(x) is
"Mail message x can be saved"
b) (3xA(x)) _, l::/x(Q(x) _, T(x)), where A(x) is ''Alert xis active," Q(x) is "Message xis queued," and
T(x) is "Message x is transmitted"
c) l::/x((x -1- main console)_, T(x)), where T(x) is "The diagnostic monitor tracks the status of system x"
d) l::/x(•L(x) _, B(x)), where L(x) is "The host of the conference call put participant x on a special list''
and B(x) is "Participant x was billed"
22
Chapter 1
The Foundations: Logic and Proofs
43. A conditional statement is true if the hypothesis is false. Thus it is very easy for the second of these propositions
to be true-just have P(x) be something that is not always true, such as "The integer x is a multiple of 2.''
On the other hand, it is certainly not always true that if a number is a multiple of 2, then it is also a multiple
of 4, so if we let Q(x) be "The integer xis a multiple of 4,'' then \ix(P(:r) _, Q(x)) will be false. Thus these
two propositions can have different truth values. Of course, for some choices of P and Q, they will have the
same truth values, such as when P and Q are true all the time.
45. Both are true precisely when at least one of P(x) and Q(x) is true for at least one value of x in the domain
(universe of discourse).
47. We can establish these equivalences by arguing that one side is true if and only if the other side is true. For
both parts, we will look at the two cases: either A is true or A is false.
a) Suppose that A is true. Then the left-hand side is logically equivalent to \:/xP(x), since the conjunction
of any proposition with a true proposition has the same truth value as that proposition. By similar reasoning
the right-hand side is equivalent to \ixP(x). Therefore the two propositions are logically equivalent in this
case; each one is true precisely when P(x) is true for every x. On the other hand, suppose that A is false.
Then the left-hand side is certainly false. Furthermore, for every x, P(x) A A is false, so the right-hand side
is false as well. Thus in all cases, the two propositions have the same truth value.
b) This problem is similar to part (a). If A is true, then both sides are logically equivalent to 3xP( x). If A
is false, then both sides are false.
49. We can establish these equivalences by arguing that one side is true if and only if the other side is true. For
both parts, we will look at the two cases: either A is true or A is false.
a) Suppose that A is true. Then for each x, P(x) _,A is true, because a conditional statement with a true
conclusion is always true; therefore the left-hand side is always true in this case. By similar reasoning the
right-hand side is always true in this case (here we used the fact that the domain is nonempty). Therefore the
two propositions are logically equivalent when A is true. On the other hand, suppose that A is false. There
are two sub cases. If P( x) is false for every x, then P( x) _, A is vacuously true (a conditional statement
with a false hypothesis is true), so the left-hand side is vacuously true. The same reasoning shows that the
right-hand side is also true, because in this subcase 3xP(x) is false. For the second subcase, suppose that
P(x) is true for some x. Then for that x, P(x) _,A is false (a conditional statement with a true hypothesis
and false conclusion is false), so the left-hand side is false. The right-hand side is also false, because in this
subcase :3xP(.r) is true but A is false. Thus in all cases, the two propositions have the same truth value.
b) This problem is similar to part (a). If A is true, then both sides are trivially true, because the conditional
statements have true conclusions. If A is false, then there are two subcases. If P( x) is false for some x,
then P(x) _,A is vacuously true for that x (a conditional statement with a false hypothesis is true), so the
left-hand side is true. The same reasoning shows that the right-hand side is true, because in this subcase
\ixP(x) is false. For the second subcase, suppose that P(x) is true for every x. Then for every x, P(x) _,A
is false (a conditional statement with a true hypothesis and false conclusion is false), so the left-hand side is
false (there is no x making the conditional statement true). The right-hand side is also false, because it is a
conditional statement with a true hypothesis and a false conclusion. Thus in all cases, the two propositions
have the same truth value.
51. We can show that these are not logically equivalent by giving an example in which one is true and the other is
false. Let P(x) be the statement "x is odd" applied to positive integers. Similarly let Q(x) be ''x is even."
Then since there exist odd numbers and there exist even numbers, the statement 3xP(x) A 3xQ(x) is true.
On the other hand, no number is both odd and even, so 3x(P(x) A Q(x)) is false.
Section 1.5
Nested Quantifiers
23
53. a) This is certainly true: if there is a unique x satisfying P(x), then there certainly is an x satisfying P(x).
b) Unless the domain (universe of discourse) has fewer than two items in it, the truth of the hypothesis
implies that there is more than one x such that P(x) holds. Therefore this proposition need not be true. (For
example, let P(x) be the proposition x 2 2:: 0 in the context of the real numbers. The hypothesis is true, but
there is not a unique x for which x 2 2:: 0.)
c) This is true: if there is an x (unique or not) such that P(x) is false, then we can conclude that it is not
the case that P(x) holds for all x.
55. A Prolog query returns a yes/no answer if there are no variables in the query, and it returns all values that
make the query true if there are.
a)
b)
c)
d)
One of the facts was that Chan was the instructor of Math 273, so the response is yes.
None of the facts was that Patel was the instructor of CS 301, so the response is no.
Prolog returns the names of the people enrolled in CS 301, namely juana and kiko.
Prolog returns the names of the courses Kiko is enrolled in, namely math273 and cs301.
e) Prolog returns the names of the students enrolled in courses which Grossman is the instructor for (which
is just CS 301), namely juana and kiko.
57. Following the idea and syntax of Example 28, we have the following rule: sibling(X, Y) :- mother(M,X),
mother(M,Y), father(F,X), father(F,Y). Note that we used the comma to mean "and"; X and Y must
have the same mother and the same father in order to be (full) siblings.
59. a) This is the statement that every person who is a professor is not ignorant. In other words, for every person,
if that person is a professor, then that person is not ignorant. In symbols: Vx(P(x) ---+ •Q(x)). This is not
the only possible answer. We could equivalently think of the statement as asserting that there does not exist
an ignorant professor: --,:Jx(P(x) /\ Q(x)).
b) Every person who is ignorant is vain: Vx(Q(x)---+ R(x)).
c) This is similar to part (a): Vx(P(x)---+ ·R(x)).
d) The conclusion (part ( c)) does not follow. There may well be vain professors, since the premises do not
rule out the possibility that there are vain people besides the ignorant ones.
61. a) This is asserting that every person who is a baby is necessarily not logical: Vx(P(x)---+ •Q(x)).
b)
c)
d)
e)
If a person can manage a crocodile, then that person is not despised: Vx(R(x)---+ •S(x)).
Every person who is not logical is necessarily despised: Vx(•Q(x)---+ S(x)).
Every person who is a baby cannot manage a crocodile: Vx(P(x)---+ ·R(x)).
The conclusion follows. Suppose that x is a baby. Then by the first premise, x is illogical, and hence,
by the third premise, x is despised. But the second premise says that if x could manage a crocodile, then
x would not be despised. Therefore x cannot manage a crocodile. Thus we have proved that babies cannot
manage crocodiles.
SECTION 1.5
Nested Quantifiers
Nested quantifiers are one of the most difficult things for students to understand. The theoretical definition of
limit in calculus, for example, is hard to comprehend because it has three levels of nested quantifiers. Study
the examples in this section carefully before attempting the exercises, and make sure that you understand the
solutions to the exercises you have difficulty with. Practice enough of these until you feel comfortable. The
effort will be rewarded in such areas as computer programming and advanced mathematics courses.
Chapter 1
24
The Foundations: Logic and Proofs
1. a) For every real number x there exists a real number y such that x is less than y. Basically, this is asserting
that there is no largest real number-for any real number you care to name, there is a larger one.
b) For every real number x and real number y, if x and y are both nonnegative, then their product is
nonnegative. Or, more simply, the product of nonnegative real numbers is nonnegative.
c) For every real number x and real number y, there exists a real number z such that xy = z. Or, more
simply, the real numbers are closed under multiplication. (Some authors would include the uniqueness of z
as part of the meaning of the word closed.)
3. It is useful to keep in mind that x and y can be the same person, so sending messages to oneself counts in
this problem.
a) Formally, this says that there exist students x and y such that x has sent a message to y. In other words,
there is some student in your class who has sent a message to some student in your class.
b) This is similar to part (a) except that x has sent a message to everyone, not just to at least one person.
So this says there is some student in your class who has sent a message to every student in your class.
c) Note that this is not the same as part (b). Here we have that for every x there exists a y such that x has
sent a message to y. In other words, every student in your class has sent a message to at least one student in
your class.
d) Note that this is not the same as part (c), since the order of quantifiers has changed. In part (c), y
could depend on x; in other words, the recipient of the messages could vary from sender to sender. Here the
existential quantification on y comes first, so it's the same recipient for all the messages. The meaning is that
there is a student in your class who has been sent a message by every student in your class.
e) This is similar to part ( c), with the role of sender and recipient reversed: every student in your class has
been sent a message from at least one student in your class. Again, note that the sender can depend on the
recipient.
f) Every student in the class has sent a message to every student in the class.
5. a) This simply says that Sarah Smith has visited www. att. com.
b) To say that an x exists such that x has visited www. imdb. org is just to say that someone (i.e., at least
one person) has visited www. imdb. org.
c) This is similar to part (b). Jose Orez has visited some website.
d) This is asserting that a y exists that both of these students has visited. In other words, Ashok Puri and
Cindy Yoon have both visited the same website.
e) When there are two quantifiers of opposite types, the sentence gets more complicated. This is saying that
there is a person ( y) other than David Belcher who has visited all the websites that David has visited (i.e., for
every website z, if David has visited z, then so has this person). Note that it is not saying that this person has
visited only websites that David has visited (that would be the converse conditional statement)-this person
may have visited other sites as well.
f) Here the existence of two people is being asserted; they are said to be unequal, and for every website z,
one of these people has visited z if and only if the other one has. In plain English, there are two different
people who have visited exactly the same websites.
7. a) Abdallah Hussein does not like Japanese cuisine.
b) Note that this is the conjunction of two separate quantified statements. Some student at your school likes
Korean cuisine, and everyone at your school likes Mexican cuisine.
c) There is some cuisine that either Monique Arsenault or Jay Johnson likes.
d) Formally this says that for every x and z, there exists a y such that if x and z are not equal, then it is
not the case that both x and z like y. In simple English, this says that for every pair of distinct students at
Section 1.5
Nested Quantifiers
25
your school, there is some cuisine that at least one them does not like.
e) There are two students at your school who have exactly the same tastes (i.e., they like exactly the same
cuisines).
f) For every pair of students at your school, there is some cuisine about which they have the same opinion
(either they both like it or they both do not like it).
9. We need to be careful to put the lover first and the lovee second as arguments in the propositional function L.
a) 'VxL(x, Jerry)
b) Note that the "somebody" being loved depends on the person doing the loving, so we have to put the
universal quantifier first: 'Vx3yL(x, y).
c) In this case, one lovee works for all lovers, so we have to put the existential quantifier first: 3y'VxL(x, y).
d) We could think of this as saying that there does not exist anyone who loves everybody ( -i3x'VyL(x, y) ),
or we could think of it as saying that for each person, we can find a person whom he or she does not love
('Vx3y-iL(x, y). These two expressions are logically equivalent.
e) 3x-iL(Lydia, x)
f) We are asserting the existence of an individual such that everybody fails to love that person: 3x'Vy-iL(y, x).
g) In Exercises 52-54 of Section 1.4, we worked with a notation for the existence of a unique object satisfying
a certain condition. Employing that device, we could write this as 3!x'VyL(y, x). In Exercise 52 of the present
section we will discover a way to avoid this notation in general. What we have to say is that the x asserted
here exists, and that every z satisfying this condition (of being loved by everybody) must equal x. Thus we
obtain 3x('VyL(y, x) /\ 'Vz(('VwL(w, z)) -> z = x)). Note that we could have used y as the bound variable
where we used w; since the scope of the first use of y had ended before we came to this point in the formula,
reusing y as the bound variable would cause no ambiguity.
h) We want to assert the existence of two distinct people, whom we will call x and y, whom Lynn loves, as
well as make the statement that everyone whom Lynn loves must be either x or y: 3x3y(x =/=- y/\L(Lynn,x)/\
L(Lynn, y) /\ 'Vz(L(Lynn, z)-> (z = x V z = y))).
i) 'VxL(x, x) (Note that nothing in our earlier answers ruled out the possibility that variables or constants
with different names might be equal to each other. For example, in part (a), x could equal Jerry, so that
statement includes as a special case the assertion that Jerry loves himself. Similarly, in part (h), the two
people whom Lynn loves either could be two people other than Lynn (in which case we know that Lynn does
not love herself), or could be Lynn herself and one other person.)
j) This is asserting that the one and only one person who is loved by the person being discussed is in fact that
person: 3x'Vy(L(x, y) +-+ x = y).
11. a) We might want to assert that Lois is a student and Michaels is a faculty member, but the sentence doesn't
really say that, so the simple answer is just A(Lois, Professor Michaels).
b) To say that every student (as opposed to every person) has done this, we need to restrict our universally
quantified variable to being a student. The easiest way to do this is to make the assertion being quantified
a conditional statement. As a general rule of thumb, use conditional statements with universal quantifiers
and conjunctions with existential quantifiers (see part (d), for example). Thus our answer is 'Vx(S(x) ->
A(x,Professor Gross)).
c) This is similar to part (b): 'Vx(F(x)-> (A(x,Professor Miller) V A(Professor Miller,x))). Note the need
for parentheses in these answers.
d) There is a student such that for every faculty member, that student has not asked that faculty member a
question. Note how we need to include the Sand F predicates: 3x(S(x) /\ 'Vy(F(y)-> -,A(x, y))). We could
also write this as 3x(S(x) /\ -i3y(F(y) /\ A(x, y))).
e) This is very similar to part (d), with the role of the players reversed: 3x(F(x) /\ 'Vy(S(y)-> -iA(y,x))).
26
Chapter 1
The Foundations: Logic and Proofs
f) This is a little ambiguous in English. If the statement is that there is a very inquisitive student, one
who has gone around and asked a question of every professor, then this is similar to part ( d), without the
negation: 3x(S(x) /\ \:/y(F(y) ---> A(x, y))). On the other hand, the statement might be intended as asserting
simply that for every professor, there exists some student who has asked that professor a question. In other
words, the questioner might depend on the questionee. Note how the meaning changes with the change in
order of quantifiers. Under the second interpretation the answer is \fy(F(y)---> 3x(S(x) /\ A(x, y))). The first
interpretation is probably the intended one.
g) This is pretty straightforward, except that we have to rule out the possibility that the askee is the same
as the asker. Our sentence needs to say that there exists a faculty member such that for every other faculty
member, the first has asked the second a question: 3x(F(x) /\ \fy((F(y) /\ y
-:f x)--->
A(x, y))).
h) There is a student such that every faculty member has failed to ask him a question: 3x(S(x) /\ \fy(F(y)--->
--.A(y, x))).
13. Be careful to put in parentheses where needed; otherwise your answer can be either ambiguous or wrong.
a) Clearly this is simply --.M(Chou, Koko).
b) We can give two answers, which are equivalent by De Morgan's law: --.(M(Arlene, Sarah)VT(Arlene, Sarah))
or --.M(Arlene, Sarah)/\ --.T(Arlene, Sarah).
c) Clearly this is simply --.M(Deborah, Jose).
d) Note that this statement includes the assertion that Ken has sent himself a message: \fx M(x, Ken).
e) We can write this in two equivalent ways, depending on whether we want to say that everyone has
failed to phone Nina or to say that there does not exist someone who has phoned her: \fx --.T(x, Nina) or
--.3x T(x, Nina).
f) This is almost identical to part (d): Vx(T(x,Avi) V M(x,Avi)).
g) To get the "else" in there, we have to make sure that y is different from x in our answer: 3x\fy(y
M(x, y)).
h) This is almost identical to part (g): 3x\fy(y :f- x---> (M(x, y) V T(x, y))).
#x
i) We need to assert the existence of two distinct people who have sent e-mail both ways: 3x3y(x
#
--->
y /\
M(x, y) /\ M(y, x)).
j) Only one variable is needed: 3x M(x, x).
k) This poor soul (x in our expression) did not receive a message or a phone call (i.e., did not receive a
message and did not receive a phone call) from any person y other than possibly himself: 3x\fy(x -:f y --->
(--.M(y,x) f\--.T(y,x))).
1) Here y is "another student": \:/x3y(x -:f y /\ (M(y, x) V T(y, x))).
rn) This is almost identical to part (i): 3x3y(x -:f y /\ M(x, y) /\ T(y, x)).
n) Note how the "everyone else" means someone different from both x and y in our expression (and note that
there are four possibilities for how each such person z might be contacted): 3x3y(x # y /\ Vz((z # x /\ z :fy)---> (M(x, z) V M(y, z) V T(x, z) V T(y, z)))).
15. The answers presented here are not the only ones possible; other answers can be obtained using different
predicates and different variables, or by varying the domain (universe of discourse).
a) \:/xN(x, discrete mathematics), where N(x, y) is "x needs a course in y" and the domain for x is computer
science students and the domain for y is academic subjects
b) 3xO(x,personal computer), where O(x,y) is "x owns y," and the domain for xis students in this class
c) \:/x3yP(x,y), where P(x,y) is "x has taken y"; x ranges over students in this class, and y ranges over
computer science courses
d) 3x3yP(x, y), with the environment of part (c) (i.e., the same definition of P and the same domain)
Section 1.5
Nested Quantifiers
27
e) Vx\lyP(x,y), where P(x,y) is "x has been in y"; x ranges over students in this class, and y ranges over
buildings on campus
f) 3x3y\lz(P(z,y)--+ Q(x,z)), where P(z,y) is ''z is in y" and Q(x,z) is "x has been in z"; x ranges over
students in this class, y ranges over buildings on campus, and z ranges over rooms
g) Vx\ly3z(P(z, y) /\ Q(x, z)), with the environment of part (f)
17. a) We need to rule out the possibility that the user has access to another mailbox different from the one that
is guaranteed: Vu3m(A(u, m) /\ Vn(n -1- m --+ ·A(u, n))), where A(u, m) means that user u has access to
mailbox m.
b) 3p\le(H(e)--+ S(p,running))-+ S(kernel,working correctly), where H(e) means that error condition e is
in effect and S(x,y) means that the status of xis y. Obviously there are other ways to express this with
different choices of predicates. Note that "only if" is the converse of "if,'' so the kernel's working properly is
the conclusion, not the hypothesis.
c) Vu\ls(E(s, .edu)--+ A(u,s)), where E(s,x) means that websites has extension x, and A(u,s) means that
user u can access website s
d) This is tricky, because we have to interpret the English sentence first, and different interpretations would
lead to different answers. We will assume that the specification is that there exist two distinct systems such
that they monitor every remote server, and no other system has the property of monitoring every remote
system. Thus our answer is 3x3y(x -1- y /\ Vz((Vs M(z, s)) f-7 (z = x V z = y))), where M(a, b) means that
system a monitors remote server b. Note that the last part of our expression serves two purposes-it says
that x and y do monitor all servers, and it says that no other system does. There are at least two other
interpretations of this sentence, which would lead to different legitimate answers.
19. a) Vx\ly((x < 0) /\ (y < 0)--+ (x + y < 0))
b) What does "necessarily" mean in this context? The best explanation is to assert that a certain universal
conditional statement is not true. So we have ---,\fx\ly((x > 0) /\ (y > 0)--+ (x - y > 0)). Note that we do
not want to put the negation symbol inside (it is not true that the difference of two positive integers is never
positive), nor do we want to negate just the conclusion (it is not true that the sum is always nonpositive). We
could rewrite our solution by passing the negation inside, obtaining 3x3y((x > 0) /\ (y > 0) /\ (x - y:::; 0)).
+ y 2 2: (x + y) 2 )
Vx\ly (lxyl = lxllYI)
c) Vx\ly (x 2
d)
21. Vx3a3b3dd ((x
> 0) --+ x = a 2 +b 2 +c 2 +d 2 ), where the domain (universe of discourse) consists of all integers
23. a) Vx\ly((x < 0) /\ (y < 0)--+ (xy > 0))
b)Vx(x-x=O)
c) To say that there are exactly two objects that meet some condition, we must have two existentially quantified
variables to represent the two objects, we must say that they are different, and then we must say that an
object meets the conditions if and only if it is one of those two. In this case we have Vx3a3b (a -1- b /\ Vc(c 2
Xf-7 (c=aVc=b))).
d) \Ix ((x < 0) --+ 3y (x
0
= y 2 ))
=
where the domain (universe of discourse) consists of all real numbers
25. a) This says that there exists a real number x such that for every real number y, the product xy equals y.
That is, there is a multiplicative identity for the real numbers. This is a true statement, since x = 1 is the
identity.
b) The product of two negative real numbers is always a positive real number.
c) There exist real numbers x and y such that x 2 exceeds y but x is less than y. This is true, since we can
take x
= 2 and
y
= 3, for instance.
28
Chapter 1
The Foundations: Logic and Proofs
d) This says that for every pair of real numbers x and y , there exists a real number z that is their sum. In
other words, the real numbers are closed under the operation of addition, another true fact. (Some authors
would include the uniqueness of z as part of the meaning of the word closed.)
27. Recall that the integers include the positive and negative integers and 0.
a) The import of this statement is that no matter how large n might be, we can always find an integer m
bigger than n 2 . This is certainly true; for example, we could always take m = n 2 + 1.
b) This statement is asserting that there is an n that is smaller than the square of every integer; note that
n is not allowed to depend on m, since the existential quantifier comes first. This statement is true, since
we could take, for instance, n = -3, and then n would be less than every square, since squares are always
greater than or equal to 0.
c) Note the order of quantifiers: m here is allowed to depend on n. Since we can take m = -n, this statement
is true (additive inverses exist for the integers).
d) Here one n must work for all m. Clearly n = 1 does the trick, so the statement is true.
e) The statement is that the equation n 2 + m 2 = 5 has a solution over the integers. This is true; in fact there
are eight solutions, namely n = ±1, m = ±2, and vice versa.
f) The statement is that the equation n 2 + m 2
= 6 has a solution over the integers. There are only a small
finite number of cases to try, since if lml or lnl were bigger than 2 then the left-hand side would be bigger
than 6. A few minutes reflection shows that in fact there is no solution, so the existential statement is false.
g) The statement is that the system of equations {n + m = 4, n - m = 1} has a solution over the integers.
By algebra we see that there is a unique solution to this system, namely n = 2 ~, m = 1 ~ . Since there do not
exist integers that make the equations true, the statement is false.
h) The statement is that the system of equations {n + m = 4, n - m = 2} has a solution over the integers.
By algebra we see that there is indeed an integral solution to this system, namely n = 3, m = 1. Therefore
the statement is true.
i) This statement says that the average of two integers is always an integer. If we take m = 1 and n = 2,
for example, then the only p for which p = ( m + n) /2 is p =
which is not an integer. Therefore the
statement is false.
q,
29. a) P(l, 1) /\ P(l, 2) /\ P(l, 3) /\ P(2, 1) /\ P(2, 2) /\ P(2, 3) /\ P(3, 1) /\ P(3, 2) /\ P(3, 3)
b) P(l, 1) V P(l, 2) V P(l, 3) V P(2, 1) V P(2, 2) V P(2, 3) V P(3, 1) V P(3, 2) V P(3, 3)
c) (P(l, 1) /\ P(l, 2) /\ P(l, 3)) V (P(2, 1) /\ P(2, 2) /\ P(2, 3)) V (P(3, 1) /\ P(3, 2) /\ P(3, 3))
d) (P(l, 1) V P(2, 1) V P(3, 1)) /\ (P(l, 2) V P(2, 2) V P(3, 2)) /\ (P(l, 3) V P(2, 3) V P(3, 3))
Note the crucial difference between parts ( c) and ( d).
31. As we push the negation symbol toward the inside, each quantifier it passes must change its type. For logical
connectives we either use De Morgan's laws or recall that •(p----+ q) p /\ --.q.
=
a)
--.\fx3y\fz T(x, y, z)
=3x--.3y\fz T(x, y, z)
=3x\fy---,\fzT(x,y,z)
= 3x\fy3z--.T(x, y, z)
b)
--.(\fx3y P(x, y) V \fx3y Q(x, y))
=--.\fx3y P(x, y) /\ --.\fx3y Q(x, y)
=3x--.3yP(x,y) /\ 3x--.3yQ(x,y)
= 3x\fy--.P(x, y) /\ 3x\fy---, Q(x, y)
c)
Section 1.5
29
Nested Quantifiers
-,'tfx3y(P(x, y) /\ 3z R(x, y, z))
=3x-,3y(P(x, y) /\ 3z R(x, y, z))
=3x'tly •(P(x, y) /\ 3z R(x, y, z))
= 3x't/y(•P(x, y) V -,3z R(x, y, z))
= 3x't/y(•P(x,y) V'tfz-,R(x,y,z))
d)
•'tlx3y(P(x,y)-+ Q(x,y))
=3x-,3y(P(x,y)-+ Q(x,y))
=3x'Vy•(P(x,y)-+ Q(x,y))
= 3x't/y(P(x, y) /\ •Q(x, y))
33. We need to use the transformations shown in Table 2 of Section 1.4, replacing -,'tj by 3-,, and replacing -,3
by \;/-,. In other words, we push all the negation symbols inside the quantifiers, changing the sense of the
quantifiers as we do so, because of the equivalences in Table 2 of Section 1.4. In addition, we need to use De
Morgan's laws (Section 1.3) to change the negation of a conjunction to the disjunction of the negations and
to change the negation of a disjunction to the conjunction of the negations. We also use the double negation
law.
a) 3x3y •P(x, y)
b) 3y'tlx•P(x,y)
c) We can think of this in two steps. First we transform the expression into the equivalent expression
3y3x•(P(x,y) V Q(x,y)), and then we use De Morgan's law to rewrite this as 3y3x(•P(x,y) /\ •Q(x,y)).
d) First we apply De Morgan's law to write this as a disjunction: (•3x3y•P(x,y)) V (•V'x't/yQ(x,y)). Then
we push the negation inside the quantifiers, and note that the two negations in front of P then cancel out
(••P(x,y)
P(x,y)). So our final answer is (V'x't/yP(x,y)) V (3x3y•Q(x,y)).
e) First we push the negation inside the outer universal quantifier, then apply De Morgan's law, and finally push it inside the inner quantifiers: 3x-,(3y't/zP(x,y,z) /\ 3z't/yP(x,y,z)) = 3x(•3y't/zP(x,y,z) V
-,3z't/yP(x,y,z)) = 3x(V'y3z-,P(x,y,z) VV'z3y•P(x,y,z)).
=
35. If the domain (universe of discourse) has at least four members, then no matter what values are assigned to x,
y, and z, there will always be another member of the domain, different from those three, that we can assign
to w to make the statement true. Thus we can use a domain such as United States Senators. On the other
hand, for any domain with three or fewer members, if we assign all the members to x, y, and z (repeating
some if necessary), then there will be nothing left to assign to w to make the statement true. For this we can
use a domain such as your biological parents.
37. In each case we need to specify some predicates and identify the domain (universe of discourse).
a) To get into the spirit of the problem, we should let T(x,y) be the predicate that x has taken y, where
x ranges over students in this class and y ranges over mathematics classes at this school. Then our original
statement is V'x3y3z(y =f. z /\ T(x,y) /\ T(x,z) /\ V'w(T(x,w) -+ (w = y V w = z))). Here y and z are
the two math classes that x has taken, and our statement says that these are different and that if x has
taken any math class w, then w is one of these two. We form the negation by using Table 2 of Section 1.4
and De Morgan's law to push the negation symbol that we place before the entire expression inwards, to
achieve 3x't/y't/z(y = z V •T(x, y) V •T(x, z) V 3w(T(x, w) /\ w =f. y /\ w =f. z)). This can also be expressed as
3x't/y't/z(y =f. z-+ (•T(x,y) V •T(x,z) V 3w(T(x,w) /\ w =f. y /\ w =f. z))). Note that we formed the negation
of a conditional statement by asserting that the hypothesis was true and the conclusion was false. In simple
English, this last statement reads "There is someone in this class for whom no matter which two distinct math
courses you consider, these are not the two and only two math courses this person has taken."
b) Let V (x, y) be the predicate that x has visited y, where x ranges over people and y ranges over countries.
The statement seems to be asserting that the person identified here has visited country y if and only if
Chapter 1
30
The Foundations: Logic and Proofs
y is not Libya. So we can write this symbolically as :3x\iy(V(x, y) +---+ y =I- Libya). One way to form the
negation of P +---+ Q is to write P +---+ -,Q; this can be seen by looking at truth tables. Thus the negation
is \ix:3y(V(x, y) +---+ y = Libya). Note that there are two ways for a biconditional to be true; therefore in
English this reads "For every person there is a country such that either that country is Libya and the person
has visited it, or else that country is not Libya and the person has not visited it." More simply, ''For every
person, either that person has visited Libya or else that person has failed to visit some country other than
Libya." If we are willing to keep the negation in front of the quantifier in English, then of course we could
just say "There is nobody who has visited every country except Libya," but that would not be in the spirit of
the exercise.
c) Let C ( x, y) be the predicate that x has climbed y, where x ranges over people and y ranges over mountains
in the Himalayas. Our statement is -,::Jx\iy C(x, y). Its negation is, of course, simply :3x\iy C(x, y). In English
this reads "Someone has climbed every mountain in the Himalayas."
d) There are different ways to approach this, depending on how many variables we want to introduce. Let
M(x, y, z) be the predicate that x has been in movie z with y, where the domains for x and y are movie
actors, and for z is movies. The statement then reads: \ix((:3z lvl(x, Kevin Bacon, z))v(:3y:3zi:3z 2 (M(x, y, z1)/\
M(y, Kevin Bacon, z 2 )))). The negation is formed in the usual manner: :3x((Vz -,Af(x, Kevin Bacon, z)) /\
(\iy\iz 1\iz 2 (-,M(x, y, z 1) V -,Af(y, Kevin Bacon, z 2 ) ))) . In simple English this means that there is someone who
has neither been in a movie with Kevin Bacon nor been in a movie with someone who has been in a movie
with Kevin Bacon.
39. a) Since the square of a number and its additive inverse are the same, we have many counterexamples, such
as x = 2 and y = - 2.
b) This statement is saying that every number has a square root. If x is negative (like x = -4), or, since we
are working in the domain of the integers, x is not a perfect square (like x = 6 ), then the equation y 2 = x
has no solution.
c) Since negative numbers are not larger than positive numbers, we can take something like x = 17 and
y = -1 for our counterexample.
41. We simply want to say that a certain equation holds for all real numbers: \ix\iy\iz((x · y) · z = x · (y · z)).
43. We want to say that for each pair of coefficients (the m and the b in the expression mx + b ), as long as m
is not 0, there is a unique x making that expression equal to 0. So we write \im\ib(m =I- 0-+ :3x(mx + b =
O/\ \iw(mw + b = 0-+ w
= x))).
Notice that the uniqueness is expressed by the last part of our proposition.
45. This statement says that every number has a multiplicative inverse.
a) In the universe of nonzero real numbers, this is certainly true. In each case we let y
b) Integers usually don't have inverses that are integers. If we let x
= 3,
= 1/ x.
then no integer y satisfies xy
=
1.
Thus in this setting, the statement is false.
c) As in part (a) this is true, since 1/x is positive when x is positive.
47. We use the equivalences explained in Table 2 of Section 1.4, twice:
-,::Jx\iyP(x, y)
=\fx-,\fyP(x, y) =\ix::Jy-,P(x, y)
49. a) We prove this by arguing that whenever the first proposition is true, so is the second; and that whenever the
second proposition is true, so is the first. So suppose that \ixP(x) /\ :3xQ(x) is true. In particular, P always
holds, and there is some object, call it y, in the domain (universe of discourse) that makes Q true. Now to
show that the second proposition is true, suppose that x is any object in the domain. By our assumptions,
P(x) is true. Furthermore, Q(y) is true for the particular y we mentioned above. Therefore P(x) /\ Q(y) is
Section 1.5
Nested Quantifiers
31
true for this x and y. Since x was arbitrary, we have showed that \/x3y(P(x) /\ Q(y)) is true, as desired.
Conversely, suppose that the second proposition is true. Letting x be any member of the domain allows us to
assert that there exists a y such that P(x) /\ Q(y) is true, and therefore Q(y) is true. Thus by the definition
of existential quantifiers, 3xQ(x) is true. Furthermore, our hypothesis tells us in particular that \/xP(x) is
true. Therefore the first proposition, \/xP(x) /\ 3xQ(x) is true.
b) This is similar to part (a). Suppose that \/xP(x)V3xQ(x) is true. Thus either P always holds, or there is
some object, call it y, in the domain that makes Q true. In the first case it follows that P(x) V Q(y) is true
for all x, and so we can conclude that \/x3y(P(x) V Q(y)) is true (it does not matter in this case whether Q(y)
is true or not). In the second case, Q(y) is true for this particular y, and so P(x) V Q(y) is true regardless of
what x is. Again, it follows that \/x3y(P(x) V Q(y)) is true. Conversely, suppose that the second proposition
is true. If P( x) is true for all x, then the first proposition must be true. If not, then P( x) fails for some x,
but for this x there must be a y such that P(x) V Q(y) is true; hence Q(y) must be true. Therefore 3yQ(y)
holds, and thus the first proposition is true.
51. This will essentially be a proof by (structural) mathematical induction (see Sections 5.1-5.3), where we show
how a long expression can be put into prenex normal form if the subexpressions in it can be put into prenex
normal form. First we invoke the result of Exercise 45 from Section 1.3 to assume without loss of generality
that our given proposition uses only the logical connectives V and --, . Then every proposition must either be a
single propositional variable (like P), the disjunction of two propositions, the negation of a proposition, or the
universal or existential quantification of a predicate. (There is a small technical point that we are sliding over
here; disjunction and negation need to be defined for predicates as well as for propositions, since otherwise we
would not be able to write down such things as \/x(P(x) /\ Q(x)). We assume that all that we have done for
propositions applies to predicates as well.)
Certainly every proposition that involves no quantifiers is already in prenex normal form; this is the base
case of our induction. Next suppose that our proposition is of the form QxP(x), where Q is a quantifier.
Then P(x) is a shorter expression than the given proposition, so (by the inductive hypothesis) we can put it
into prenex form, with all of its quantifiers coming at the beginning. Then Qx followed by this prenex form
is again in prenex form and is equivalent to the original proposition. Next suppose that our proposition is of
the form --,p. Again, we can invoke the inductive hypothesis and assume that P is already in prenex form,
with all of its quantifiers coming at its front. We now slide the negation symbol past all the quantifiers, using
the equivalences in Table 2 of Section 1.4. For example, --,\fx3yR(x, y) becomes 3x\fy--,R(x, y), which is in
prenex form.
Finally, suppose that our given proposition is a disjunction of two propositions, P V Q, each of which can
(again by the inductive hypothesis) be assumed to be in prenex normal form, with their quantifiers at the front.
There are several cases. If only one of P and Q has quantifiers, then we invoke the result of Exercise 46 of
Section 1.4 to bring the quantifier in front of both. We then apply our process to what remains. For example,
P V \/xQ(x) is equivalent to \/x(P V Q(x)), and then P V Q(x) is put into prenex form. Another case is that
the proposition might look like 3xR(x) V 3xS(x). In this case, by Exercise 45 of Section 1.4, the proposition
is equivalent to 3x(R(x) V S(x)). Once again, by the inductive hypothesis we can then put R(x) V S(x) into
prenex form, and so the entire proposition can be put into prenex form. Similarly, using Exercise 48 of the
present section we can transform \/xR(x) V\/xS(x) into the equivalent \/x\/y(R(x) VS(y)); putting R(x) V S(y)
into prenex form then brings the entire proposition into prenex form. Finally, if the proposition is of the form
\/xR(x) V 3xQ(x), then we invoke Exercise 49b of the present section and apply the same construction.
Note that this proof actually gives us the process for finding the proposition in prenex form equivalent to
the given proposition-we just work from the inside out, dealing with one logical operation or quantifier at a
Chapter 1
32
The Foundations: Logic and Proofs
time. Here is an example:
't:/xP(x) V -dx(Q(x) V 't:/yR(x, y))
='t:/xP(x) V -dx't:/y(Q(x) V R(x, y))
='t:/xP(x) V Vx:ly•(Q(x) V R(x, y))
:::::: 't:/x't:/z(P(x) V 3y•(Q(z) V R(z, y)))
:::::: 't:/x't:/z:Jy(P(x) V •(Q(z) V R(z, y)))
SECTION 1.6
Rules of Inference
This section lays the groundwork for understanding proofs. You are asked to understand the logical rules of
inference behind valid arguments, and you are asked to construct some highly stylized proofs using these rules.
The proofs will become more informal in the next section and throughout the remainder of this book (and
your mathematical studies).
1. This is modus ponens. The first statement is p ____, q, where p is "Socrates is human" and q is "Socrates is
mortal." The second statement is p. The third is q. Modus ponens is valid. We can therefore conclude that
the conclusion of the argument (third statement) is true, because the hypotheses (the first two statements)
are true.
3. a) This is the addition rule. We are concluding from p that p V q must be true, where p is "Alice is a
mathematics major" and q is ''Alice is a computer science major."
b) This is the simplification rule. We are concluding from p A q that p must be true, where p is "Jerry is a
mathematics major" and q is "Jerry is a computer science major."
c) This is modus ponens. We are concluding from p ____, q and p that q must be true, where p is "it is rainy"
and q is ''the pool will be closed.''
d) This is modus tollens. We are concluding from p ____, q and •q that •p must be true, where p is "it will
snow today'' and q is ''the university will close today."
e) This is hypothetical syllogism. We are concluding from p ____, q and q ____, r that p ____, r must be true, where
p is "I will go swimming," q is ''I will stay in the sun too long," and r is "I will sunburn."
5. Let w be the proposition "Randy works hard," let d be the proposition "Randy is a dull boy," and let j
be the proposition "Randy will get the job." We are given premises w, w ____, d, and d ____, •j. We want to
conclude •j. We set up the proof in two columns, with reasons, as in Example 6.
Step
1. w
2. w----; d
3. d
4. d----; · j
5. · j
Reason
Hypothesis
Hypothesis
Modus ponens using (2) and (3)
Hypothesis
Modus ponens using (3) and (4)
7. First we use universal instantiation to conclude from "For all x, if x is a man, then x is mortal" the special
case of interest, "If Socrates is a man, then Socrates is mortal." Then we use modus ponens to conclude that
Socrates is mortal.
Section 1.6
Rules of Inference
33
9. a) Because it was sunny on Tuesday, we assume that it did not rain or snow on Tuesday (otherwise we cannot
do anything with this problem). If we use modus tollens on the universal instantiation of the given conditional
statement applied to Tuesday, then we conclude that I did not take Tuesday off. If we now apply disjunctive
syllogism to the disjunction in light of this conclusion, we see that I took Thursday off. Now use modus
ponens on the universal instantiation of the given conditional statement applied to Thursday; we conclude
that it rained or snowed on Thursday. One more application of disjunctive syllogism tells us that it rained on
Thursday.
b) Using modus tollens we conclude two things-that I did not eat spicy food and that it did not thunder.
Therefore by the conjunction rule of inference (Table 1), we conclude "I did not eat spicy food and it did not
thunder."
c) By disjunctive syllogism from the first two hypotheses we conclude that I am clever. The third hypothesis
gives us no useful information.
d) We can apply universal instantiation to the conditional statement and conclude that if Ralph (respectively,
Ann) is a CS major, then he (she) has a PC. Now modus tollens tells us that Ralph is not a CS major. There
are no conclusions to be drawn about Ann.
e) The first two conditional statements can be phrased as "If x is good for corporations, then x is good for
the U.S." and "If x is good for the U.S., then x is good for you." If we now apply universal instantiation
with x being "for you to buy lots of stuff,'' then we can conclude using modus ponens twice that for you to
buy lots of stuff is good for the U.S. and is good for you.
f) The given conditional statement is "For all x, if x is a rodent, then x gnaws its food." We can form the
universal instantiation of this with x being a mouse, a rabbit, and a bat. Then modus ponens allows us to
conclude that mice gnaw their food; and modus tollens allows us to conclude that rabbits are not rodents. We
can conclude nothing about bats.
11. We are asked to show that whenever p 1 , p 2 , ... , Pn are true, then q--+ r must be true, given that we know
that whenever P1 , P2 , ... , Pn and q are true, then r must be true. So suppose that p 1 , p 2 , ... , Pn are true.
We want to establish that q --+ r is true. If q is false, then we are done, vacuously. Otherwise, q is true, so
by the validity of the given argument form, we know that r is true.
13. In each case we set up the proof in two columns, with reasons, as in Example 6.
a) Let c(x) be ''x is in this class," let j(x) be "x knows how to write programs in JAVA," and let h(x) be
"x can get a high paying job." We are given premises c(Doug), j(Doug), and Vx(j(x)
to conclude 3x(c(x) /\ h(x)).
Step
1. Vx(j(x)--+ h(x))
2. j (Doug) --+ h(Doug)
3. j(Doug)
4. h(Doug)
5. c(Doug)
6. c(Doug) /\ h(Doug)
7. 3x(c(x) /\ h(x))
--+
h(x)), and we want
Reason
Hypothesis
Universal instantiation using (1)
Hypothesis
Modus ponens using (2) and (3)
Hypothesis
Conjunction using (4) and (5)
Existential generalization using (6)
b) Let c(x) be "x is in this class," let w(x) be "x enjoys whale watching," and let p(x) be "x cares about
ocean pollution." We are given premises 3x(c(x) /\ w(x)) and Vx(w(x) --+ p(x)), and we want to conclude
3x (c( x) /\ p( x)) . In our proof, y represents an unspecified particular person.
Chapter 1
34
Step
1. 3x(c(x) /\ w(x))
2.
3.
4.
5.
6.
7.
8.
9.
c(y)Aw(y)
w(y)
c(y)
\lx(w(x) _.., p(x))
w(y) _.., p(y)
p(y)
c(y) /\ p(y)
3x(c(x) /\p(x))
The Foundations: Logic and Proofs
Reason
Hypothesis
Existential instantiation using ( 1)
Simplification using (2)
Simplification using (2)
Hypothesis
Universal instantiation using (5)
Modus ponens using (3) and (6)
Conjunction using (4) and (7)
Existential generalization using (8)
c) Let c(x) be "xis in this class," let p(x) be "x owns a PC," and let w(x) be "x can use a word processing
program." We are given premises c(Zeke), \lx(c(x) _.., p(x)), and \lx(p(x) _.., w(x)), and we want to conclude
w(Zeke).
Step
1. \lx(c(x) _, p(x))
2. c(Zeke) _.., p(Zeke)
3. c(Zeke)
4. p(Zeke)
5. \lx(p(x) _, w(x))
6. p(Zeke) _, w(Zeke)
Reason
7. w(Zeke)
Modus ponens using (4) and (6)
Hypothesis
Universal instantiation using (1)
Hypothesis
Modus ponens using (2) and (3)
Hypothesis
Universal instantiation using (5)
d) Let j(x) be "x is in New Jersey," let f(x) be "x lives within fifty miles of the ocean,'' and let s(x) be "x
has seen the ocean." We are given premises Vx(j(x) _, f(x)) and 3x(j(x) /\ •s(x)), and we want to conclude
3x(f (x) /\ •s( x)) . In our proof, y represents an unspecified particular person.
Step
1. 3x(j(x) /\ •s(x))
2. j(y) /\ •s(y)
3. j(y)
4. \lx(j(x) _, f(x))
5. j(y) _, f(y)
6. f(y)
7. •s(y)
8. f(y) /\ •s(y)
9. 3x(f(x) /\ •s(x))
Reason
Hypothesis
Existential instantiation using (1)
Simplification using (2)
Hypothesis
Universal instantiation using (4)
Modus ponens using (3) and (5)
Simplification using (2)
Conjunction using (6) and (7)
Existential generalization using (8)
15. a) This is correct, using universal instantiation and modus ponens.
b) This is invalid. After applying universal instantiation, it contains the fallacy of affirming the conclusion.
c) This is invalid. After applying universal instantiation, it contains the fallacy of denying the hypothesis.
d) This is valid by universal instantiation and modus tollens.
17. We know that some x exists that makes H (x) true, but we cannot conclude that Lola is one such x. Maybe
only Suzanne is happy and everyone else is not happy. Then 3x H(x) is true, but H(Lola) is false.
19. a) This is the fallacy of affirming the conclusion, since it has the form "p _.., q and q implies p."
b) This reasoning is valid; it is modus tollens.
c) This is the fallacy of denying the hypothesis, since it has the form "p _, q and •P implies •q ."
Section 1.6
Rules of Inference
35
21. Let us give an argument justifying the conclusion. By the second premise, there is some lion that does not
drink coffee. Let us call him Leo. Thus we know that Leo is a lion and that Leo does not drink coffee. By
simplification this allows us to assert each of these statements separately. The first premise says that all lions
are fierce; in particular, if Leo is a lion, then Leo is fierce. By modus ponens, we can conclude that Leo is
fierce. Thus we conclude that Leo is fierce and Leo does not drink coffee. By the definition of the existential
quantifier, this tells us that there exist fierce creatures that do not drink coffee; in other words, that some
fierce creatures do not drink coffee.
23. The error occurs in step (5), because we cannot assume, as is being done here, that the c that makes P true
is the same as the c that makes Q true.
25. We are given the premises Vx(P(x) ---+ Q(x)) and •Q(a). We want to show •P(a). Suppose, to the contrary,
that •P(a) is not true. Then P(a) is true. Therefore by universal modus ponens, we have Q(a). But this
contradicts the given premise •Q(a). Therefore our supposition must have been wrong, and so •P(a) is true,
as desired.
27. We can set this up in two-column format.
Step
1. 'v'x(P(x) /\ R(x))
2. P(a) /\ R(a)
3. P(a)
4. 'v'x(P(x)---+ (Q(x) /\ S(x)))
5. Q(a) /\ S(a)
6. S(a)
7. R(a)
8. R(a)/\S(a)
9. 'v'x(R(x) /\ S(x))
Reason
Premise
Universal instantiation using (1)
Simplification using (2)
Premise
Universal modus ponens using (3) and (4)
Simplification using (5)
Simplification using (2)
Conjunction using (7) and (6)
Universal generalization using (5)
29. We can set this up in two-column format. The proof is rather long but straightforward if we go one step at a
time.
Step
Reason
1. 3x•P(x)
Premise
Existential instantiation using ( 1)
2. ·P(c)
Premise
3. 'v'x(P(x) V Q(x))
4. P(c) V Q(c)
Universal instantiation using (3)
5. Q(c)
Disjunctive syllogism using (4) and (2)
Premise
6. 'v'x(•Q(x) V S(x))
Universal instantiation using (6)
7. •Q(c) V S(c)
8. S(c)
Disjunctive syllogism using (5) and (7), since ••Q(c) Q(c)
Premise
9. 'v'x(R(x)---+ •S(x))
Universal instantiation using (9)
10. R(c)---+ •S(c)
=
11. ·R(c)
12. 3x·R(x)
Modus tollens using (8) and (10), since ••S(c)
Existential generalization using ( 11)
= S(c)
31. Let p be "It is raining"; let q be "Yvette has her umbrella"; let r be "Yvette gets wet." Then our assumptions
are •p V q, •q V •r, and p V •r. Using resolution on the first two assumptions gives us •p V •r. Using
resolution on this and the third assumption gives us •r, so Yvette does not get wet.
36
Chapter 1
The Foundations: Logic and Proofs
33. Assume that this proposition is satisfiable. Using resolution on the first two clauses allows us to conclude
q V q; in other words, we know that q has to be true. Using resolution on the last two clauses allows us to
conclude •q V •q; in other words, we know that •q has to be true. This is a contradiction. So this proposition
is not satisfiable.
35. This argument is valid. We argue by contradiction. Assume that Superman does exist. Then he is not
impotent, and he is not malevolent (this follows from the fourth sentence). Therefore by (the contrapositives
of) the two parts of the second sentence, we conclude that he is able to prevent evil, and he is willing to
prevent evil. By the first sentence, we therefore know that Superman does prevent evil. This contradicts the
third sentence. Since we have arrived at a contradiction, our original assumption must have been false, so we
conclude finally that Superman does not exist.
SECTION 1.7 Introduction to Proofs
This introduction applies jointly to this section and the next (1.8).
Learning to construct good mathematical proofs takes years. There is no algorithm for constructing the
proof of a true proposition (there is actually a deep theorem in mathematical logic that says this). Instead, the
construction of a valid proof is an art, honed after much practice. There are two problems for the beginning
student-figuring out the key ideas in a problem (what is it that really makes the proposition true?) and
writing down the proof in acceptable mathematical language.
Here are some general things to keep in mind in constructing proofs. First, of course, you need to find out
exactly what is going on-why the proposition is true. This can take anywhere from ten seconds (for a really
simple proposition) to a lifetime (some mathematicians have spent their entire careers trying to prove certain
conjectures). For a typical student at this level, tackling a typical problem, the median might be somewhere
around 15 minutes. This time should be spent looking at examples, making tentative assumptions, breaking
the problem down into cases, perhaps looking at analogous but simpler problems, and in general bringing all
of your mathematical intuition and training to bear.
It is often easiest to give a proof by contradiction, since you get to assume the most (all the hypotheses
as well as the negation of the conclusion), and all you have to do is to derive a contradiction. Another thing
to try early in attacking a problem is to separate the proposition into several cases; proof by cases is a valid
technique, if you make sure to include all the possibilities. In proving propositions, all the rules of inference
are at your disposal, as well as axioms and previously proved results. Ask yourself what definitions, axioms,
or other theorems might be relevant to the problem at hand. The importance of constantly returning to the
definitions cannot be overstated!
Once you think you see what is involved, you need to write down the proof. In doing so, pay attention both
to content (does each statement follow logically? are you making any fallacious arguments? are you leaving out
any cases or using hidden assumptions?) and to style. There are certain conventions in mathematical proofs,
and you need to follow them. For example, you must use complete sentences and say exactly what you mean.
(An equation is a complete sentence, with "equals" as the verb; however, a good proof will usually have more
English words than mathematical symbols in it.) The point of a proof is to convince the reader that your line
of argument is sound, and that therefore the proposition under discussion is true; put yourself in the reader's
shoes, and ask yourself whether you are convinced.
Most of the proofs called for in this exercise set are not extremely difficult. Nevertheless, expect to have
a fairly rough time constructing proofs that look like those presented in this solutions manual, the textbook,
or other mathematics textbooks. The more proofs you write, utilizing the different methods discussed in this
Section 1. 7
Introduction to Proofs
37
section, the better you will become at it. As a bonus, your ability to construct and respond to nonmathematical
arguments (politics, religion, or whatever) will be enhanced. Good luck!
1. We must show that whenever we have two odd integers, their sum is even. Suppose that a and b are two
odd integers. Then there exist integers s and t such that a = 2s + 1 and b = 2t + 1. Adding, we obtain
a+ b = (2s + 1) + (2t + 1) = 2(s + t + 1). Since this represents a+ b as 2 times the integer s + t + 1, we
conclude that a + b is even, as desired.
3. We need to prove the following assertion for an arbitrary integer n: "If n is even, then n 2 is even." Suppose
that n is even. Then n = 2k for some integer k. Therefore n 2 = (2k ) 2 = 4k 2 = 2(2k 2 ). Since we have
written n 2 as 2 times an integer, we conclude that n 2 is even.
5. We can give a direct proof. Suppose that m + n is even. Then m + n = 2s for some integer s. Suppose that
n + p is even. Then n + p = 2t for some integer t. If we add these [this step is inspired by the fact that we
want to look at m + p], we get m + p + 2n = 2s + 2t. Subtracting 2n from both sides and factoring, we have
m + p = 2s + 2t - 2n = 2(s + t - n). Since we have written m + p as 2 times an integer, we conclude that
m + p is even, as desired.
7. The difference of two squares can be factored: a 2 - b2 = (a+ b)(a - b). If we can arrange for our given odd
integer to equal a + b and for a - b to equal 1, then we will be done. But we can do this by letting a and
b be the integers that straddle n/2. For example, if n = 11, then we take a = 6 and b = 5. Specifically, if
n = 2k + 1, then we let a = k + 1 and b = k. Here, then, is our proof. Since n is odd, we can write n = 2k + 1
for some integer k . Then (k + 1) 2 - k 2 = k 2 + 2k + 1 - k 2 = 2k + 1 = n. This expresses n as the difference
of two squares.
9. The proposition to be proved here is as follows: If r is a rational number and i is an irrational number, then
s = r + i is an irrational number. So suppose that r is rational, i is irrational, and s is rational. Then by
Example 7 the sum of the rational numbers s and -r must be rational. (Indeed, ifs= a/b and r = c/d,
where a, b, c, and d are integers, with b =/=- 0 and d =/=- 0, then by algebra we see that s + (-r) = (ad-be)/ (bd) ,
so that patently s + (-r) is a rational number.) But s + (-r) = r + i - r = i, forcing us to the conclusion
that i is rational. This contradicts our hypothesis that i is irrational. Therefore the assumption that s was
rational was incorrect, and we conclude, as desired, that s is irrational.
11. To disprove this proposition it is enough to find a counterexample, since the proposition has an implied universal quantification. We know from Example 10 that J2 is irrational. If we take the product of the irrational
number J2 and the irrational number J2, then we obtain the rational number 2. This counterexample
refutes the proposition.
13. We give an proof by contraposition. The contrapositive of this statement is "If l/x is rational, then x is
rational" so we give a direct proof of this contrapositive. Note that since 1/x exists, we know that x =/=- 0. If
l/x is rational, then by definition 1/x = p/q for some integers p and q with q =/=- O. Since 1/x cannot be 0
(if it were, then we'd have the contradiction 1 = x · 0 by multiplying both sides by x), we know that p
=/=-
0.
Now x = 1/(1/x) = l/(p/q) = q/p by the usual rules of algebra and arithmetic. Hence x can be written as
the quotient of two integers with the denominator nonzero. Thus by definition, x is rational.
15. We will prove the contrapositive (that if it is not true that x 2: 1 or y 2: 1, then it is not true that x + y 2: 2),
using a direct argument. Assume that it is not true that x 2: 1 or y 2: 1. Then (by De Morgan's law) x < 1
and y < 1. Adding these two inequalities, we obtain x
proof is complete.
+y <
2. This is the negation of x
+ y 2:
2, and our
38
Chapter 1
The Foundations: Logic and Proofs
17. a) We must prove the contrapositive: If n is odd, then n 3 + 5 is even. Assume that n is odd. Then we can
write n = 2k+l for some integer k. Then n 3 +5 = (2k+1) 3 +5 = 8k 3 +12k 2 +6k+6 = 2(4k 3 +6k 2 +3k+3).
Thus n 3 + 5 is two times some integer, so it is even.
b) Suppose that n 3 + 5 is odd and that n is odd. Since n is odd, and the product of odd numbers is odd,
in two steps we see that n 3 is odd. But then subtracting we conclude that 5, being the difference of the two
odd numbers n 3 + 5 and n 3 , is even. This is not true. Therefore our supposition was wrong, and the proof
by contradiction is complete.
19. The proposition we are trying to prove is "If 0 is a positive integer greater than 1, then 02 > 0 ." Our
proof is a vacuous one, exactly as in Example 5. Since the hypothesis is false, the conditional statement is
automatically true.
21. The proposition we are trying to prove is "If a and b are positive real numbers, then (a+ b) 1 2'. a 1 + b1 ."
Our proof is a direct one. By the definition of exponentiation, any real number to the power 1 is itself. Hence
(a+ b) 1 = a+ b = a 1 + b1 . Finally, by the addition rule, we can conclude from (a+ b) 1 = a 1 + b1 that
(a+ b) 1 2 a 1 + b1 (the latter being the disjunction of (a+ b) 1 = a 1 + b1 and (a+ b) 1 > a 1 + b1 ). One might
also say that this is a trivial proof, since we did not use the hypothesis that a and b are positive (although of
course we used the hypothesis that they are numbers).
23. We give a proof by contradiction. If there were nine or fewer days on each day of the week, this would account
for at most 9 · 7 = 63 days. But we chose 64 days. This contradiction shows that at least ten of the days
must be on the same day of the week.
25. One way to prove this is to use the rational root test from high school algebra: Every rational number that
satisfies a polynomial with integer coefficients is of the form p / q, where p is a factor of the constant term of
the polynomial, and q is a factor of the coefficient of the leading term. In this case, both the constant and the
leading coefficient are 1, so the only possible values for p and q are ±1. Therefore the only possible rational
roots are ±1/(±1), which means that 1 and -1 are the only possible rational roots. Clearly neither of them
is a root, so there are no rational roots.
Alternatively, we can follow the hint. Suppose by way of contradiction that a/b is a rational root, where
a and b are integers and this fraction is in lowest terms (that is, a and b have no common divisor greater
than 1 ). Plug this proposed root into the equation to obtain a 3 /b3 + a/b + 1 = 0. Multiply through by b3 to
obtain a 3 + ab 2 + b3 = 0. If a and b are both odd, then the left-hand side is the sum of three odd numbers
and therefore must be odd. If a is odd and b is even, then the left-hand side is odd+ even+ even, which is
again odd. Similarly, if a is even and b is odd, then the left-hand side is even+ even+ odd, which is again
odd. Because the fraction a/b is in simplest terms, it cannot happen that both a and b are even. Thus in
all cases, the left-hand side is odd, and therefore cannot equal 0. This contradiction shows that no such root
exists.
27. We must prove two conditional statements. First, we assume that n is odd and show that 5n + 6 is odd (this
is a direct proof). By assumption, n = 2k + 1 for some integer k. Then 5n + 6=5(2k+1) + 6 = lOk + 11 =
2(5k + 5) + 1. Since we have written 5n + 6 as 2 times an integer plus 1, we have showed that 5n + 6 is
odd, as desired. Now we give an proof by contraposition of the converse. Suppose that n is not odd-in other
words, that n is even. Then n = 2k for some integer k. Then 5n + 6 = lOk + 6 = 2(5k + 3). Since we
have written 5n + 6 as 2 times an integer, we have showed that 5n + 6 is even. This completes the proof by
contraposition of this conditional statement.
Section 1.7
Introduction to Proofs
39
29. This proposition is true. We give a proof by contradiction. Suppose that m is neither 1 nor -1. Then mn
has a factor (namely lml) larger than 1. On the other hand, mn = 1, and 1 clearly has no such factor.
Therefore we conclude that m = 1 or m = -1. It is then immediate that n = 1 in the first case and n = -1
in the second case, since mn = 1 implies that n = 1/m.
31. Perhaps the best way to do this is to prove that all of them are equivalent to x being even, which one can
discover easily enough by trying a few small values of x. If x is even, then x = 2k for some integer k.
Therefore 3x + 2 = 3 · 2k + 2 = 6k + 2 = 2(3k + 1), which is even, since it has been written in the form 2t,
where t = 3k+l. Similarly, x+5 = 2k+5 = 2k+4+1=2(k+2)+1, so x+5 is odd; and x 2 = (2k) 2 = 2(2k 2),
so x 2 is even. For the converses, we will use a proof by contraposition. So assume that x is not even; thus x
is odd and we can write x = 2k + 1 for some integer k. Then 3x + 2=3(2k+1) + 2 = 6k + 5 = 2(3k + 2) + 1,
which is odd (i.e., not even), since it has been written in the form 2t + 1, where t = 3k + 2. Similarly,
x+5 = 2k+1+5 = 2(k+3), so x+5 is even (i.e., not odd). That x 2 is odd was already proved in Example 1.
This completes the proof.
33. It is easiest to give proofs by contraposition of (i) --+ (ii), (ii) --+ (i), (i) --+ (iii) , and (iii) --+ (i). For the
first of these, suppose that 3x + 2 is rational, namely equal to p / q for some integers p and q with q -=I- 0.
Then we can write x = ( (p/ q) - 2) /3 = (p - 2q) / (3q), where 3q -=I- 0. This shows that x is rational. For
the second conditional statement, suppose that x is rational, namely equal to p/q for some integers p and q
with q -=I- 0. Then we can write 3x + 2 = (3p + 2q) / q, where q -=I- 0. This shows that 3x + 2 is rational. For
the third conditional statement, suppose that x/2 is rational, namely equal to p/q for some integers p and q
with q -=I- 0 . Then we can write x = 2p / q, where q -=I- 0 . This shows that x is rational. And for the fourth
conditional statement, suppose that xis rational, namely equal to p/q for some integers p and q with q -=I- 0.
Then we can write x/2 = p/(2q), where 2q -=I- 0. This shows that x/2 is rational.
35. The steps are valid for obtaining possible solutions to the equations. If the given equation is true, then we
can conclude that x = 1 or x = 6, since the truth of each equation implies the truth of the next equation.
However, the steps are not all reversible; in particular, the squaring step is not reversible. Therefore the
possible answers must be checked in the original equation. We know that no other solutions are possible, but
we do not know that these two numbers are in fact solutions. If we plug in x = 1 we get the true statement
2 = 2; but if we plug in x = 6 we get the false statement 3 = -3. Therefore x = 1 is the one and only
solution of -./x + 3 = 3 - x.
37. Suppose that we have proved p 1 --+ p 4 --+ p2--+ p5--+ p3--+ p 1 . Imagine these conditional statements arranged
around a circle. Then to prove that each one of these propositions (say Pi) implies each of the others (say p 1 ),
we just have to follow the circle, starting at Pi, until we come to Pj , using hypothetical syllogism repeatedly.
39. We can give a very satisfying proof by contradiction here. Suppose instead that all of the numbers a 1 , a 2 ,
... , an are less than their average, which we can denote by A. In symbols, we have a, < A for all i. If we
add these n inequalities, we see that
a1 + a2 + · · · + an < nA .
By definition,
A = a1 + a2 + · · · + an .
n
The two displayed formulae clearly contradict each other, however: they imply that nA < nA. Thus our
assumption must have been incorrect, and we conclude that at least one of the numbers a, is greater than or
equal to their average.
40
Chapter 1
The Foundations: Logic and Proofs
41. We can prove that these four statements are equivalent in a circular way: (i) --> (ii) --> (iii) --> (iv) --> ( i).
For the first, we want to show that if n is even, then n + 1 is odd. Assume that n is even. Then n = 2k
for some integer k. Thus n + 1 = 2k + 1, so by definition n + 1 is odd. This completes the first proof.
Next we give a direct proof of (ii) --> (iii). Suppose that n + 1 is odd, say n + 1 = 2k + 1. Then
3n + 1 = 2n + (n + 1) = 2n + 2k + 1 = 2(n + k) + 1. Since this shows that 3n + 1 is 2 times an integer plus 1,
we conclude that 3n+ 1 is odd, as desired. For the next proof, suppose that 3n+ 1 is odd, say 3n+ 1=2k+1.
Then 3n = (3n + 1) - 1 = (2k + 1) - 1 = 2k. Therefore by definition 3n is even. Finally, we must prove that
if 3n is even, then n is even. We will do this using a proof by contraposition. Suppose that n is not even.
Then n is odd, so we can write n = 2k + 1 for some integer k. Then 3n = 3(2k + 1) = 6k + 3 = 2(3k + 1) + 1.
This exhibits 3n as 2 times an integer plus 1, so 3n is odd, completing the proof by contraposition.
SECTION 1.8
Proof Methods and Strategy
The preamble to the solutions for Section 1. 7 applies here as well, so you might want to reread it at this time.
In addition, the section near the back of this Guide, entitled "A Guide to Proof- Writing," provides an excellent
tutorial, with many additional examples. Don't forget to take advantage of the many additional resources on
the website for this text, as well.
If you are majoring in mathematics, then proofs are the bread and butter of your fleld. Most likely you
will take a course devoted entirely to learning how to read and write proofs, using one of the many textbooks
available on this subject. For a review of many of them (as well as reviews of hundreds of mathematics books),
see this site provided by the Mathematical Association of America: mathdl.maa.org/MathDL/19/.
1. We give an exhaustive proof-just check the entire domain. For n = 1 we have 12 + 1 = 2 2: 2 = 2 1 . For
n = 2 we have 22 + 1 = 5 2 4 = 22 . For n = 3 we have 32 + 1 = 10 2 8 = 23 • For n = 4 we have
42 + 1 = 17 2: 16 = 24 . Notice that for n 2 5, the inequality is no longer true.
3. Following the hint, we consider the two cases determined by the relative sizes of x and y. First suppose that
x 2 y. Then by definition max(x, y) = x and min(x, y) = y. Therefore in this case max(x, y) + min(x, y) =
.r + y, exactly as desired. For the second (and final) case, suppose that x < y. Then max(x,y) = y and
min(x, y) = x. Therefore in this case max(x, y) + min(x, y) = y + x = x + y, again the desired conclusion.
Hence in all cases, the equality holds.
5. Because Ix - YI = IY - xi, the values of x and y are interchangeable. Therefore, without loss of generality,
we can assume that x 2 y. In this case, Ix - YI = x - y, so the first expression gives us
x+y-(x-y)
2
and indeed y is the smaller. Similarly, the second expression gives us
x+y+(x-y)
x+y+x-y
2
2
2x
=
2 =x,
and indeed x is the larger.
7. There are several cases to consider. If x and y are both nonnegative, then lxl + IYI = x+y = lx+yl. Similarly,
if both are negative, then lxl + IYI = (-x) + (-y) = -(x + y) = Ix+ YI, since x + y is negative in this case.
The complication (and strict inequality) comes if one of the variables is nonnegative and the other is negative.
By the symmetry of the roles of x and y here (strictly speaking, by the commutativity of addition), we can
assume without loss of generality that it is x that is nonnegative and y that is negative. So we have x 2: 0
and y < 0.
Section 1.8
Proof Methods and Strategy
41
Now there are two subcases to consider within this case, depending on the relative sizes of the nonnegative
numbers x and -y. First suppose that x 2 -y. Then x + y 2 0. Therefore Ix + YI = x + y, and this
quantity is a nonnegative number smaller than x (since y is negative). On the other hand lxl + IYI = x + IYI
is a positive number bigger than x. Therefore we have Ix+ YI < x < lxl + IYI, as desired.
Finally, consider the possibility that x < -y. Then Ix+ YI= -(x + y) = (-x) + (-y) is a positive
number less than or equal to -y (since -x is nonpositive). On the other hand lxl + IYI = lxl + (-y) is a
positive number greater than or equal to -y. Therefore we have Ix+ YI :S: -y :S: lxl + IYI, as desired.
9. We want to find consecutive squares that are far apart. If n is large enough, then (n + 1) 2 will be much
bigger than n 2 , and that will do it. Let's take n = 100. Then 100 2 = 10000 and 101 2 = 10201, so the 201
consecutive numbers 10001, 10002, ... , 10200 are not perfect squares. The first 100 of these will satisfy the
requirements of this exercise. Our proof was constructive, since we actually exhibited the numbers.
11. We try some small numbers and discover that 8
=
23 and 9
= 32 . In fact, this is the only solution, but the
proof of this fact is not trivial.
13. One way to solve this is the following nonconstructive proof. Let x = 2 (rational) and y = J2 (irrational).
If xY = 2v'2 is irrational, we are done. If not, then let x = 2v'2 and y = J2/ 4; x is rational by assumption,
and y is irrational (if it were rational, then J2 would be rational). But now xY = (2v'2)v'2/ 4 = 2v'2·(VZ)/ 4 =
2 1 / 2 = J2, which is irrational, as desired.
15. a) This statement asserts the existence of x with a certain property. If we let y = x, then we see that P(x)
is true. If y is anything other than x, then P(y) is not true. Thus x is the unique element that makes P
true.
b) The first clause here says that there is an element that makes P true. The second clause says that whenever
two elements both make P true, they are in fact the same element. Together this says that P is satisfied by
exactly one element.
c) This statement asserts the existence of an x that makes P true and has the further property that whenever
we find an element that makes P true, that element is x. In other words, x is the unique element that makes P
true. Note that this is essentially the same as the definition given in the text, except that the final conditional
statement has been replaced by its contrapositive.
17. The equation la - cl = lb- cl is equivalent to the disjunction of two equations: a - c = b- c or a - c = -b + c.
The first of these is equivalent to a = b, which contradicts the assumptions made in this problem, so the
original equation is equivalent to a - c = -b + c. By adding b + c to both sides and dividing by 2, we see
that this equation is equivalent to c = (a + b) /2. Thus there is a unique solution. Furthermore, this c is an
integer, because the sum of the odd integers a and b is even.
19. We are being asked to solve n = (k - 2) + (k + 3) for k. Using the usual, reversible, rules of algebra, we see
that this equation is equivalent to k = (n - 1) /2. In other words, this is the one and only value of k that
makes our equation true. Since n is odd, n - 1 is even, so k is an integer.
21. If x is itself an integer, then we can take n = x and
E = 0. No other solution is possible in this case, since
if the integer n is greater than x, then n is at least x + 1, which would make E 2 1. If x is not an integer,
then round it up to the next integer, and call that integer n. We let E = n - x. Clearly 0 :S: E < 1, this is the
only E that will work with this n, and n cannot be any larger, since E is constrained to be less than 1.
Chapter 1
42
The Foundations: Logic and Proofs
23. If x = 5 and y = 8, then the harmonic mean is 2·5·8/(5+8) ~ 6.15, and the geometric mean is J5-8 ~ 6.32.
If x = 10 and y = 100, then the harmonic mean is 2·10·100/(10+100) ~ 18.18, and the geometric mean
is v'lO · 100 ~ 31.62. We conjecture that the harmonic mean of x and y is always less than their geometric
mean if x and y are distinct positive real numbers (clearly if x = y then both means are this common value).
So we want to verify the inequality 2xy/(x + y) < JXY. Multiplying both sides by (x + y)/(2v'xfj) gives us
the equivalent inequality JXY < (:r + y)/2, which is proved in Example 14.
25. The key point here is that the parity (oddness or evenness) of the sum of the numbers written on the board
never changes. If j and k are both even or both odd, then their sum and their difference are both even, and
we are replacing the even sum j + k by the even difference IJ - kl , leaving the parity of the total unchanged. If
j and k have different parities, then erasing them changes the parity of the total, but their difference Jj - kl is
odd, so adding this difference restores the parity of the total. Therefore the integer we end up with at the end
of the process must have the same parity as 1+2 + · · · + (2n). It is easy to compute this sum. If we add the
first and last terms we get 2n + 1; if we add the second and next-to-last terms we get 2 + (2n - 1) = 2n + 1;
and so on. In all we get n sums of 2n + 1, so the total sum is n(2n
two odd numbers and therefore is odd, as desired.
+ 1).
If n is odd, this is the product of
27. Without loss of generality we can assume that n is nonnegative, since the fourth power of an integer and the
fourth power of its negative are the same. To get a handle on the last digit of n, we can divide n by 10,
obtaining a quotient k and remainder l, whence n = lOk + l, and l is an integer between 0 and 9, inclusive.
Then we compute n 4 in each of these ten cases. We get the following values, where ?? is some integer that is
a multiple of 10, whose exact value we do not care about.
(10k + 0) 4 = 10000k4 = 10000k4
(10k + 1)
4
(10k + 2) 4
4
+0
3
=
10000k + ?? . k + ?? . k 2 + ?? . k + 1
=
10000k4 + ?? . k 3 + ?? . k 2 + ?? . k + 16
(10k + 3) 4 = 10000k 4 + ?? . k 3 + ?? . k 2 + ?? . k + 81
(10k + 4) 4 = 10000k 4 +??. k 3 + ?? . k 2 +??. k + 256
(10k + 5) 4 = 10000k4 + ?? . k 3 + ?? . k 2 + ?? . k + 625
(10k + 6) 4 = 10000k 4 + ?? . k 3 + ?? . k 2 + ?? . k + 1296
+ 7) 4 = 10000k4 + ?? . k 3 + ?? . k 2 + ?? . k + 2401
(10k + 8) 4 = 10000k 4 + ?? . k 3 + ?? . k 2 + ?? . k + 4096
(10k + 9) 4 = 10000k 4 + ?? . k 3 + ?? . k 2 + ?? . k + 6561
(lOk
Since each coefficient indicated by ?? is a multiple of 10, the corresponding term has no effect on the ones
digit of the answer. Therefore the ones digits are 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, respectively, so it is always
a 0 , 1 , 5 , or 6 .
29. Because n 3
n2
+ n3
31. Since 54
=
> 100 for all n > 4, we need only note that n =
1, n
= 2, n = 3, and n = 4 do not satisfy
100.
= 625,
for there to be positive integer solutions to this equation both x and y must be less than 5.
This means that each of x 4 and y 4 is at most 44 = 256, so their sum is at most 512 and cannot be 625.
33. We give a proof by contraposition. Assume that it is not the case that a ::; ijn or b ::; ijn or c ::; ijn. Then
it must be true that a > ifi1 and b > -ifii, and c > ijn. Multiplying these inequalities of positive numbers
together we obtain abc < ( ijri,) 3 = n, which implies the negation of our hypothesis that n = abc.
Section 1.8
Proof Methods and Strategy
43
35. The idea is to find a small irrational number to add to the smaller of the two given rational numbers. Because
we know that J2 is irrational, we can use a small multiple of v'2. Here is our proof: By finding a common
denominator, we can assume that the given rational numbers are a/b and c/b, where b is a positive integer
and a and c are integers with a < c. In particular, (a+ 1)/b:::; c/b. Thus x = (a+ ~J2)/b is between the
two given rational numbers, because 0 < J2 < 2. Furthermore, x is irrational, because if x were rational,
then 2(bx - a) = v'2 would be as well, in violation of Example 10 in Section 1. 7.
37. a) Without loss of generality, we may assume that the x sequence is already sorted into nondecreasing order,
since we can relabel the indices. There are only a finite number of possible orderings for the y sequence, so if we
can show that we can increase the sum (or at least keep it the same) whenever we find Yi and y1 that are out of
order (i.e., i < j but Yi > y 1 ) by switching them, then we will have shown that the sum is largest when the y
sequence is in nondecreasing order. Indeed, if we perform the swap, then we have added XiYJ +x1 yi to the sum
and subtracted XiYi +x1 y1 . The net effect, then, is to have added XiYJ +x1 yi -XiYi -x1 y 1 = (x 1 -xi)(Yi -yj),
which is nonnegative by our ordering assumptions.
b) This is similar to part (a). Again we assume that the x sequence is already sorted into nondecreasing
order. If the y sequence is not in nonincreasing order, then Yi < y1 for some i < j. By swapping Yi and Yj
we increase the sum by XiYJ + x 1 yi - XiYi - x 1 y1 = (x 1 - xi)(Yi - y1 ), which is nonpositive by our ordering
assumptions.
39. In each case we just have to keep applying the function f until we reach 1, where J(x) = 3x + 1 if x is odd
and f(x) = x/2 if x is even.
a) f(6) = 3, f(3) = 10, f(lO) = 5, f(5) = 16, f(16) = 8, f(8) = 4, f(4) = 2, f(2) = 1. We abbreviate this
to 6 _.., 3 _.., 10 _.., 5 _.., 16 _.., 8 _.., 4 _.., 2 _.., 1.
b) 7 _.., 22 _.., 11 _.., 34 _.., 17 _.., 52 _.., 26 _.., 13 _.., 40 _.., 20 _.., 10 _.., 5 _.., 16 _.., 8 _.., 4 _.., 2 _.., 1
c) 17 _.., 52 _.., 26 _.., 13 _.., 40 _.., 20 _.., 10 _.., 5 _.., 16 _.., 8 _.., 4-* 2 _.., 1
d) 21 _.., 64 _.., 32 _.., 16 _.., 8 _.., 4 _.., 2 _.., 1
41. We give a constructive proof. Without loss of generality, we can assume that the upper left and upper right
corners of the board are removed. We can place three dominoes horizontally to fill the remaining portion of
the first row, and we can place four dominoes horizontally in each of the other seven rows to fill them.
43. The number of squares in a rectangular board is the product of the number of squares in each row and the
number of squares in each column. We are given that this number is even, so there is either an even number
of squares in each row or an even number of squares in each column. In the former case, we can tile the board
in the obvious way by placing the dominoes horizontally, and in the latter case, we can tile the board in the
obvious way by placing the dominoes vertically.
45. We follow the suggested labeling scheme. Clearly we can rotate the board if necessary to make the removed
squares be 1 and 16. Square 2 must be covered by a domino. If that domino is placed to cover squares 2
and 6, then the following domino placements are forced in succession: 5-9, 13-14, and 10-11, at which point
there is no way to cover square 15. Otherwise, square 2 must be covered by a domino placed at 2-3. Then the
following domino placements are forced: 4-8, 11-12, 6-7, 5-9, and 10-14, and again there is no way to cover
square 15.
47. Remove the two black squares adjacent to one of the white corners, and remove two white squares other
than that corner. Then no domino can cover that white corner, because neither of the squares adjacent to it
remains.
Chapter 1
44
The Foundations: Logic and Proofs
49. a) It is not hard to find the five patterns:
1
EE
4
5
qpLfb
b) It is clear that the pattern labeled 1 and the pattern labeled 2 will tile the checkerboard. It is harder to
find the tiling for patterns 3 and 4, but a little experimentation shows that it is possible.
It remains to argue that pattern 5 cannot tile the checkerboard. Label the squares from 1 to 64, one row
at a time from the top, from left to right in each row. Thus square 1 is the upper left corner, and square 64 is
the lower right. Suppose we did have a tiling. By symmetry and without loss of generality, we may suppose
that the tile is positioned in the upper left corner, covering squares 1, 2, 10, and 11. This forces a tile to be
adjacent to it on the right, covering squares 3, 4, 12. and 13. Continue in this manner and we are forced to
have a tile covering squares 6, 7, 15, and 16. This makes it impossible to cover square 8. Thus no tiling is
possible.
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 1
1. a) See p. 3.
b) This is not a boring course.
2. a) See pp. 4, 6, and 9.
b) Disjunction: 'TU go to the movies tonight or I'll finish my discrete mathematics homework." Conjunction:
"I'll go to the movies tonight and I'll finish my discrete mathematics homework." Exclusive or: "I'll go to the
movies tonight or I'll finish my discrete mathematics homework, but not both." Conditional statement: "If
I'll go to the movies tonight, then I'll finish my discrete mathematics homework." Biconditional: "I'll go to
the movies tonight if and only if I'll finish my discrete mathematics homework."
b) Seep. 8.
c) Converse: "If I go for a walk in the woods tomorrow, then it will be sunny." Contrapositive: "If I don't go
for a walk in the woods tomorrow, then it will not be sunny.''
3. a) See p. 6.
4. a) See p. 25.
b) using truth tables; symbolically, using identities in Tables 6-8 in Section 1.3; by giving a valid argument
about the possible truth values of the propositional variables involved
c) Use the fact that r ---; -iq
=
-,r
V -iq, or use truth tables.
Supplementary Exercises
45
5. a) Each line of the truth table corresponds to exactly one combination of truth values for the n atomic
propositions involved. We can write down a conjunction that is true precisely in this case, namely the
conjunction of all the atomic propositions that are true and the negations of all the atomic propositions that
are false. If we do this for each line of the truth table for which the value of the compound proposition is
to be true, and take the disjunction of the resulting propositions, then we have the desired proposition in its
disjunctive normal form. See Exercise 42 in Section 1.3.
b) See Exercise 43 in Section 1.3.
c) See Exercises 50 and 52 in Section 1.3.
6. See pp. 40 and 42. The negation of VxP(x) is 3x-,P(x), and the negation of 3xP(x) is Vx•P(x).
7. a) In the second, x can depend on y. In the first, the same x must "work" for every y.
b) See Example 4 in Section 1.5.
8. See pp. 69-70. This is a valid argument because it uses the valid rule of inference called modus tollens.
9. This is a valid argument because it uses the universal modus ponens rule of inference. Therefore if the premises
are true, the conclusion must be true.
10. a) See pp. 82, 83, and 86.
b) For a direct proof, the hypothesis implies that n = 2k for some k, whence n + 4 = 2(k + 2), so n + 4
is even. For a proof by contraposition, suppose that n + 4 is odd; hence n + 4 = 2k + 1 for some k.
Then n = 2(k - 2) + 1, so n is odd, hence not even. For a proof by contradiction, assume that n = 2k
and n + 4 = 21+1 for some k and l. Subtracting gives 4 = 2(1 - k) + 1, which means that 4 is odd, a
contradiction.
11. a) See p. 87.
b) Suppose that 3n + 2 is odd, so that 3n + 2 = 2k + 1 for some k. Multiply both sides by 3 and subtract 1,
obtaining 9n+5 = 6k+2 = 2(3k+ 1). This shows that 9n+5 is even. We prove the converse by contraposition.
Suppose that 3n + 2 is not odd, i.e., that it is even. Then 3n + 2 = 2k for some k. Multiply both sides by 3
and subtract 1, obtaining 9n + 5 = 6k - 1 = 2(3k - 1) + 1. This shows that 9n + 5 is odd.
12. No-we could add to these p2
-->
p3 and p 1
-->
p 4 , for example.
13. a) Find a counterexample, i.e., an object c such that P(c) is false.
b) n = 1 is a counterexample.
14. See p. 96.
15. See p. 99.
16. See Example 4 in Section 1.8.
SUPPLEMENTARY EXERCISES FOR CHAPTER 1
1. a) q--> p (note that "only if" does not mean "if")
b) q /\ p
c) •q V •P (assuming inclusive use of the English word "or" is intended by the speaker)
d) q '"'"'p (this is another way to say "if and only if" in English words)
3. We could use truth tables, but we can also argue as follows.
a) Since q is false but the conditional statement p
b) The disjunction says that either
-->
q is true, we must conclude that p is also false.
p or q is true. Since p is given to be false, it follows that q must be true.
46
Chapter 1
The Foundations: Logic and Proofs
5. The inverse of p -+ q is •p -+ •q. Therefore the converse of the inverse is •q -+ •p. Note that this is the same
as the contrapositive of the original statement. The converse of p-+ q is q-+ p. Therefore the converse of
the converse is p -+ q, which was the original statement. The contrapositive of p -+ q is •q -+ •p. Therefore
the converse of the contrapositive is •p -+ •q, which is the same as the inverse of the original statement.
7. The straightforward approach is to use disjunctive normal form. There are four cases in which exactly three of
the variables are true. The desired proposition is (p/\q/\r/\•s) V (p/\q/\•r/\s) V (p/\•q/\r/\s) V (•p/\q/\r/\s).
9. Translating these statements into symbols, using the obvious letters, we have •t -+ •g, •g -+ •q, r -+ q, and
•t /\ r. Assume the statements are consistent. The fourth statement tells us that •t must be true. Therefore
by modus ponens with the first statement, we know that •g is true, hence (from the second statement) that •q
is true. Also, the fourth statement tells us that r must be true, and so again modus ponens (third statement)
makes q true. This is a contraction: q /\ •q. Thus the statements are inconsistent.
11. We make a table of the eight possibilities for p, q, and r, showing the truth values of the four propositions.
p .q r
•(p-+ (q /\ r))
pV •q
(p/\r)V(q-+p)
•r
T
T
T
T
T
T
F
T
F
T
F
T
T
T
F
T
T
T
T
F
T
T
T
T
T
F
T
F F
F
F T T
F
F
F
F
T
F
F T F
F F T
F
T
F
T
F F F
F
T
T
T
If we look at the first row of the table, we see that if the student rejects the first proposition, accepts the second,
rejects the third, and accepts the fourth, then the resulting commitments are consistent, because the second
and fourth propositions and the negations of the first and third propositions are all true in this case in which p,
q, and r are all true. Similarly, looking at the sixth row of the table, where p and r are false but q is true, we
see that a student who accepts the third proposition and rejects the other three also wins. Scanning the entire
table, we see that the winning answers are reject-accept-reject-accept, accept-accept-accept-accept, acceptaccept-reject-accept, reject-reject-reject-reject, reject-reject-accept-reject, and reject-accept-accept-accept.
13. Aaron must be a knave, because a knight would never make the false statement that all of them are knaves.
If Bohan is a knight, then he would be speaking the truth if Crystal is a knight, so that is one possibility. On
the other hand, Bohan might be a knave, in which case his statement is already false, regardless of Crystal's
identity. In this case, if Crystal were also a knave, then Aaron would have told the truth, which is impossible.
So there are two possibilities for the ordered triple (Aaron, Bohan, Crystal), namely (knave, knight, knight)
and (knave, knave, knight).
15. We are told that exactly one of these people committed the crime, and exactly one (the guilty party) is
a knight. We look at the three cases to determine who the knight is. If Amy were the knight, then her
protestations of innocence would be true, but that cannot be, since we know that the knight is guilty. If Claire
were the knight, then her statement that Brenda is not a normal is true; and since Brenda cannot be the
knight in this situation, Brenda must be a knave. That means that Brenda is lying when she says that Amy
was telling the truth; therefore Amy is lying. This means that Amy is guilty, but that cannot be, since Amy
isn't the knight. So Brenda must be the knight. Amy is an innocent normal who is telling the truth when she
says she is innocent; Brenda is telling the truth when she says that Amy is telling the truth; and Claire is a
normal who is telling the truth when she says that Brenda is not a normal. So Brenda committed the crime.
Supplementary Exercises
47
17. The definition of valid argument is an argument in which the truth of all the premises forces the truth of
conclusion. In this example, the two premises can never be true simultaneously, because they are contradictory,
irrespective of the true status of the tooth fairy. Therefore it is (vacuously) true that whenever both of the
premises are true, the conclusion is also true (irrespective of your luck at finding gold at the end of the
rainbow). Because the premises are not both true, we cannot conclude that the conclusion is true.
19. This is done in exactly the same manner as was described in the text for a 9 x 9 Sudoku puzzle (Section 1.3),
with the variables indexed from 1 to 16, instead of from 1 to 9, and with a similar change for the propositions
for the 4 x 4 blocks:
/\;=0 /\;=0 /\~6=1 v;=l v~=l p( 4r + i, 4s + j, n).
21. a) F, since 4 does not di vi de 5
b) T, since 2 di vi des 4
c) F, by the counterexample in part (a)
d) T, since 1 divides every positive integer
e) F, since no number is a multiple of all positive integers (No matter what positive integer n one chooses, if
we take m = n + 1, then P( m, n) is false, since n + 1 does not divide n.)
f) T, since 1 divides every positive integer
23. The given statement tells us that there are exactly two elements in the domain. Therefore if we let the domain
be anything with size other than 2 the statement will be false.
25. For each person we want to assert the existence of two different people who are that person's parents. The
most elegant way to do so is Vx3y3z(y =I z /\ Vw(P(w, x) <---+ (w = y V w = z))). Note that we are saying that
w is a parent of x if and only if w is one of the two people whose existence we asserted.
27. To express the statement that exactly n members of the domain satisfy P, we need to use n existential
quantifiers, express the fact that these n variables all satisfy P and are all different, and express the fact that
every other member of the domain that satisfies P must be one of these.
a) This is a special case, however. To say that there are no values of x that make P true we can simply write
--,::JxP(x) or Vx--,P(x).
b) This is the same as Exercise 52 in Section 1.5, because 31 xP(x) is the same as 3!xP(x). Thus we can
write 3x(P(x) /\ Vy(P(y)---> y
= x)).
c) Following the discussion above, we write :3x1:3x2(P(x1) /\P(x 2 ) /\x 1 =I x2 /\ Vy(P(y)---> (y = x1 Vy= x2))).
d) We expand the previous answer to one more variable: :3x 1:3x 2:3x 3 (P(x 1) /\ P(x 2 ) /\ P(x 3 ) /\ x 1 =I x 2 /\ x 1 =I
X3 /\ X2 =fa X3 /\ Vy(P(y)---> (y = X1 Vy= X2 Vy= x3))).
29. Suppose that 3x(P(x)---> Q(x)) is true. Then for some x, either Q(x) is true or P(x) is false. If Q(x) is
true for some x, then the conditional statement VxP(x)---> 3xQ(x) is true (having true conclusion). If P(x)
is false for some x, then again the conditional statement VxP(x) ---> 3xQ(x) is true (having false hypothesis).
Conversely, suppose that 3x(P(x)---> Q(x)) is false. That means that for every x, the conditional statement
P(x) ---> Q(x) is false, or, in other words, P(x) is true and Q(x) is false. The latter statement implies that
3xQ(x) is false. Thus VxP(x) ---> 3xQ(x) has a true hypothesis and a false conclusion and is therefore false.
31. No. For each x there may be just one y making P(x, y) true, so that the second proposition will not be true.
For example, let P( x, y) be x + y = 0, where the domain (universe of discourse) is the integers. Then the
first proposition is true, since for each x there exists a y, namely -x, such that P(x, y) holds. On the other
hand, there is no one x such that x + y = 0 for every y.
33. Let T(s,c,d) be the statement that students has taken class c in department d. Then, with the domains
(universes of discourse) being the students in this class, the courses at this university, and the departments in
the school of mathematical sciences, the given statement is VsVd::lcT(s, c, d).
Chapter 1
48
The Foundations: Logic and Proofs
35. Let T(x, y) mean that student x has taken class y, where the domain is all students in this class. We want to
say that there exists exactly one student for whom there exists exactly one class that this student has taken.
So we can write simply :3!x:3!y T(x. y). To do this without quantifiers, we need to expand the uniqueness
quantifier using Exercise 52 in Section 1.5. Doing so, we have :3x\fz((:3y\fw(T(z,w) +-+ w = y)) +-+ z = x).
37. By universal instantiation we have P(a)
Q(a) and Q(a)
•Q(a). and again by modus tollens we conclude •P(a).
ft
-+
R(a). By modus tollens we then conclude
fi
is rational, then x is rational, assuming throughout that x 2'. 0.
= p / q is rational, q -j. 0. Then x = ( fi) 2 = p 2 / q2 is also rational ( q2 is again nonzero).
39. We give a proof by contraposition that if
Suppose that
-+
41. We can give a constructive proof by letting m = 10500
+ 1.
Then m 2 = (10 500
+ 1) 2 > (10 500 ) 2 =
10 1000.
43. The first three positive cubes are 1, 8, and 27. If we want to find a number that cannot be written as the
sum of eight cubes, we would look for a number that is 7 more than a small multiple of 8. Indeed, 23 will
do. \Ve can use two S's but then would have to use seven l's to reach 23, a total of nine numbers. Clearly
no smaller collection will do. This counterexample disproves the statement.
45. The first three positive fifth powers are 1, 32, and 243. If we want to find a number that cannot be written as
the sum of 36 fifth powers, we would look for a number that is 31 more than a small multiple of 32. Indeed,
7 · 32 - 1 = 223 will do. We can use six 32's but then would have to use 31 l's to reach 223, a total of 37
numbers. Clearly no smaller collection will do. This counterexample disproves the statement.
WRITING PROJECTS FOR CHAPTER 1
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. An excellent website for this is www. paradoxes. co. uk. It includes a bibliography.
2. Search your library's on-line catalog for a book with the word fuzzy in the title. You might find [BaGo],
[DuPr], [Ka], [Ko3], or [McFr], for example.
3. Look for this article (available on the Web): Marques-Silva, J. (2008) Practical Applications of Boolean
Satisfiability. Also try a book published by the American Mathematical Society, the proceedings of a workshop
on this topic: Satisfiability Problem: Theory and Applications.
4. A Web search for ''solving sudoku" should get you more than enough sources.
5. Even if you can't find a set, you may find some articles about it in materials for high school students and
teachers, such as old issues of Mathematics Teacher. published by the National Council of Teachers of Mathematics. This journal, and possibly even copies of the game, may exist in the education library at your school
(if there is one). The company that currently produces it has a website: wffnproof. com.
6. Martin Gardner and others have written some books that annotate Carroll's writings quite extensively. Lewis
Carroll has become a cult figure in certain circles. See also [Cal], [Ca2], and [Ca3], for original material.
7. A textbook on logic programming and/or the language PROLOG, such as [Ho2] or [Sal], would be a logical
place to start. Many bookstores have huge computer science sections, so that source should not be ignored.
Writing Projects
49
8. A course on computational logic at Stanford in 2005-2006 had a Web page with class notes:
logic. stanford. edu/classes/cs157 /2005fall/cs157 .html. Enderton's book on logic [En] would be a
possible choice for background information.
9. There are books on this subject, such as [Du].
10. A place to start might be a recent article on this topic in Science [Re]. As always, a Web search will also turn
up more information.
11. There is an excellent article on this by Keith Devlin, writing for the Mathematical Association of America
(MAA); see www.maa.org/devlin/devlin_05_03 .html.
12. The well-known on-line encyclopedia made up of articles by contributors is usually quite good, with accurate information and useful links and cross-references. See their article on Chomp: en. wikipedia. org/wiki/Chomp.
13. The references given in the text are the obvious place to start. The mathematics education field has bought
into P6lya's ideas, especially as they relate to problem-solving. See what the National Council of Teachers of
Mathematics (www.nctm.org) has to say about it.
14. The classic work in this field is [GrSh].
Chapter 2
50
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
CHAPTER2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
SECTION 2.1
Sets
This exercise set (note that this is a "set" in the mathematical sense!) reinforces the concepts introduced in
this section-set description, subset and containment, cardinality, power set, and Cartesian product. A few of
the exercises (mostly some of the even-numbered ones) are a bit subtle. Keep in mind the distinction between
"is an element of" and "is a subset of." Similarly, there is a big difference between 0 and { 0}. In dealing
with sets, as in most of mathematics, it is extremely important to say exactly what you mean.
1. a) {1,-1}
b) {1,2,3,4,5,6,7,8,9,10,11}
c) {O, 1, 4, 9, 16, 25, 36, 49, 64, 81}
d) 0 ( v'2 is not an integer)
3. Recall that one set is a subset of another set if every element of the first set is also an element of the second.
We just have to apply this definition here.
a) Because every nonstop flight is a flight, every element in the first set is also an element of the second, so
the first set is a subset of the second. Because there are flights that do have intermediate stops (say, from
New York to Atlanta with a stop in Detroit), the second set is not a subset of the first.
b) Because there is certainly at least one person who speaks English but not Chinese, and vice versa, neither
set is a subset of the other.
c) Every flying squirrel is a living creature that can fly, so the first set is a subset of the second. Because
birds, for instance, are leaving creatures that can fly but are not flying squirrels, the second set is not a subset
of the first.
5. a) Yes; order and repetition do not matter.
b) No; the first set has one element, and the second has two elements.
c) No; the first set has no elements, and the second has one element (namely the empty set).
7. a) Since 2 is an integer greater than 1, 2 is an element of this set.
b) Since 2 is not a perfect square ( 12 < 2, but n 2 > 2 for n > 1 ), 2 is not an element of this set.
c) This set has two elements, and as we can clearly see, one of those elements is 2.
d) This set has two elements, and as we can clearly see, neither of those elements is 2. Both of the elements
of this set are sets; 2 is a number, not a set.
e) This set has two elements, and as we can clearly see, neither of those elements is 2. Both of the elements
of this set are sets; 2 is a number, not a set.
f) This set has just one element, namely the set { {2}}. So 2 is not an element of this set. Note that {2} is
not an element either, since {2} =f. { {2}}.
9. a) This is false, since the empty set has no elements.
b) This is false. The set on the right has only one element, namely the number 0, not the empty set.
c) This is false. In fact, the empty set has no proper subsets.
Section 2.1
Sets
51
d) This is true. Every element of the set on the left is, vacuously, an element of the set on the right; and the
set on the right contains an element, namely 0, that is not in the set on the left.
e) This is false. The set on the right has only one element, namely the number 0, not the set containing the
number 0.
f) This is false. For one set to be a proper subset of another, the two sets cannot be equal.
g) This is true. Every set is a subset of itself.
11. a) T (in fact x is the only element)
b) T (every set is a subset of itself)
d) T (in fact, {x} is the only element)
c) F (the only element of {x} is a letter, not a set)
e) T (the empty set is a subset of every set)
f) F (the only element of {x} is a letter, not a set)
13. The four months whose names don't contain the letter R form a subset of the set of twelve months, as shown
here.
January
February
March
Apnl
September
November
0 ctober
December
15. We put the subsets inside the supersets. We also put dots in certain regions to indicate that those regions are
not empty (required by the fact that these are proper subset relations). Thus the answer is as shown.
u
1 7. We need to show that every element of A is also an element of C. Let x E A. Then since A ~ B, we can
conclude that x E B. Furthermore, since B ~ C, the fact that x E B implies that x E C, as we wished to
show.
19. The cardinality of a set is the number of elements it has. The number of elements in its elements is irrelevant.
a) 1
b) 1
21. a) {0,{a}}
c) 2
d) 3
b) {0, {a}, {b}, {a,b}}
c) {0,{0},{{0}},{0,{0}}}
23. a) Since the set we are working with has 3 elements, the power set has 23
=8
elements.
b) Since the set we are working with has 4 elements, the power set has 24 = 16 elements.
c) The power set of the empty set has 2° = 1 element. The power set of this set therefore has 21
In particular, it is { 0, { 0}}. (See Example 14.)
= 2 elements.
25. We need to prove two things, because this is an "if and only if" statement. First, let us prove the "if" part.
We are given that A ~ B. We want to prove that the power set of A is a subset of the power set of B, which
means that if C ~ A then C ~ B. But this follows directly from Exercise 17. For the "only if" part, we are
given that the power set of A is a subset of the power set of B. We must show that every element of A is also
an element of B. So suppose a is an arbitrary element of A. Then {a} is a subset of A, so it is an element
of the power set of A. Since the power set of A is a subset of the power set of B, it follows that {a} is an
element of the power set of B, which means that {a} is a subset of B. But this means that the element of
{a}, namely a, is an element of B, as desired.
52
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
27. In each case we need to list all the ordered pairs, and there are 4 x 2 = 8 of them.
a) {(a,y), (a,z), (b,y), (b,z), (c,y), (c,z), (d,y), (d,z)}
b) {(y,a), (y,b),(y,c), (y,d),(z,a), (z,b),(z,c), (z,d)}
29. This is the set of triples (a, b, c), where a is an airline and b and c are cities. For example, (American,
Rochester Hills Michigan, Middletown New Jersey) is an element of this Cartesian product. A useful subset of
this set is the set of triples (a, b, c) for which a flies between band c. For example, (Delta, Detroit, New York)
is in this subset, but the triple mentioned earlier is not.
31. By definition, 0 x A consists of all pairs (x, a) such that x E 0 and a E A. Since there are no elements
x E
0 , there are no such pairs, so 0 x A = 0 . Similar reasoning shows that A x 0 = 0.
33. Recall that A 2 = A x A. Therefore here we must list all the ordered pairs of elements of A.
a) {(0,0), (0,1),(0,3),(1,0), (1,1),(1,3), (3,0),(3,1),(3,3)}
b) {(1,1),(1,2),(1,a),(1,b),(2,1), (2,2),(2,a),(2,b),(a,1),(a,2),(a,a),(a,b),(~ 1), (~2), (b,a),
(~b)}
35. The Cartesian product Ax B has mn elements. (This problem foreshadows the general discussion of counting
in Chapter 5.) To see that this answer is correct, note that for each a E A there are n different elements b E B
with which to form the pair (a, b). Since there are m different elements of A, each leading to n different
pairs, there must be mn pairs altogether.
37. The Cartesian product Ax A has m 2 elements. (This problem foreshadows the general discussion of counting
in Chapter 6.) To see that this answer is correct, note that for each a E A there are m different elements
b EA (including a itself) with which to form the pair (a,b). Since there are m different elements of A, each
leading to m different pairs, there must be m 2 pairs altogether. Similarly, for each of the m 2 choices of a and
b, there are m choices of c with which to form the triple (a, b, c) in A 3 , so A 3 has m 3 elements. Continuing
in this way, we see that An has m n elements.
39. The only difference between Ax Bx C and (Ax B) x C is parentheses, so for all practical purposes one can
think of them as essentially the same thing. By Definition 9, the elements of A x B x C consist of 3-tuples
(a, b, c), where a E A, b E B, and c E C. By Definition 8, the elements of (Ax B) x C consist of ordered
pairs (p, c), where p E A x B and c E C, so the typical element of (A x B) x C looks like ((a, b), c). A
3-tuple is a different creature from a 2-tuple, even if the 3-tuple and the 2-tuple in this case convey exactly
the same information. To be more precise, there is a natural one-to-one correspondence between Ax B x C
and (Ax B) x C given by (a, b, c) +---+ ((a, b), c).
41. a) Every real number has its square not equal to -1. Alternatively, the square of a real number is never -1.
This is true, since squares of real numbers are always nonnegative.
b) There exists an integer whose square is 2. This is false, since the only two numbers whose squares are 2,
namely J2 and -J2, are not integers.
c) The square of every integer is positive. This is almost true, but not quite, since 02 f 0.
d) There is a real number equal to its own square. This is true, since x = 1 (as well as x = 0) fits the bill.
43. In each case we want the set of all values of x in the domain (the set of integers) that satisfy the given equation
or inequality.
a) The only integers whose squares are less than 3 are the integers whose absolute values are less than 2. So
the truth set is {x E Z I x 2 < 3} = {-1,0,1}.
Section 2.2
53
Set Operations
b) All negative integers satisfy this inequality, as do all nonnegative integers other than 0 and 1. So the truth
set is {x E Z x 2 > x} = Z- {O, 1} = { ... ,-2,-1,2,3,4, ... }.
c) The only real number satisfying this equation is x = -1/2. Because this value is not in our domain, the
truth set is empty: {x E Z 2x + 1 = O} = 0.
J
J
45. First we prove the statement mentioned in the hint. The "if" part is immediate from the definition of equality.
The "only if" part is rather subtle. We want to show that if {{a},{a,b}} = {{c},{c,d}}, then a= c and
b = d. First consider the case in which a I- b. Then { {a}, {a, b}} has exactly two elements, both of which
are sets; exactly one of them contains one element, and exactly one of them contains two elements. Thus
{{c}, {c, d}} must have the same property; hence c cannot equal d, and so {c} is the element containing
one element. Hence {a} = {c}, and so a = c. Also in this case the two-element elements {a, b} and {c, d}
must be equal, and since b I- a = c, we must have b = d. The other possibility is that a = b. Then
{{a},{a,b}} ={{a}}, a set with one element. Hence {{c},{c,d}} must also have only one element, which
can only happen when c = d and the set is { {c}} . It then follows that a = c, and hence b = d, as well.
Now there is really nothing else to prove. The property that we want ordered pairs to have is precisely
the one that we just proved is satisfied by this definition. Furthermore, if we look at the proof, then it is
clear how to "recover" both a and b from { {a}, {a, b}} . If this set has two elements, then a is the unique
element in the one-element element of this set, and b is the unique member of the two-element element of this
set other than a. If this set has only one element, then a and b are both equal to the unique element of the
unique element of this set.
47. We can do this recursively, using the idea from Section 5.4 of reducing a problem to a smaller instance of the
same problem. Suppose that the elements of the set in question are listed: A= {a 1 , a 2 , a 3 , ... , a,,}. First we
will write down all the subsets that do not involve a,,. This is just the same problem we are talking about all
over again, but with a smaller set---one with just n - 1 elements. We do this by the process we are currently
describing. Then we write these same subsets down again, but this time adjoin a,, to each one. Each subset
of A will have been written down, then-first all those that do not include a,,, and then all those that do.
For example, using this procedure the subsets of {p, d, q} would be listed in the order 0, {p} , {d} ,
{p,d}, {q}, {p,q}, {d,q}, {p,d,q}.
An alternative solution is given in the answer key in the back of the textbook.
SECTION 2.2 Set Operations
Most of the exercises involving operations on sets can be done fairly routinely by following the deEnitions.
It is important to understand what it means for two sets to be equal and how to prove that two given sets
are equal-using membership tables, using the definition to reduce the problem to logic, or showing that
each is a subset of the other; see, for example, Exercises 5-24. It is often helpful when looking at operations
on sets to draw the Venn diagram, even if you are not asked to do so. The symmetric difference is a
fairly important set operation not discussed in the section; it is developed in Exercises 32-43. Two other new
concepts, multisets and fuzzy sets, are also introduced in this set of exercises.
Chapter 2
54
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
1. a) the set of students who live within one mile of school and walk to class (only students who do both of these
things are in the intersection)
b) the set of students who either live within one mile of school or walk to class (or, it goes without saying,
both)
c) the set of students who live within one mile of school but do not walk to class
d) the set of students who live more than a mile from school but nevertheless walk to class
3. a) We include all numbers that are in one or both of the sets, obtaining {O, 1, 2, 3, 4, 5, 6}.
b) There is only one number in both of these sets, so the answer is { 3}.
c) The set of numbers in A but not in B is {1, 2, 4, 5}.
d) The set of numbers in B but not in A is {O, 6}.
5. By definition A is the set of elements of the universal set that are not in A. Not being in A means being
in A. Thus A is the same set as A. We can give this proof in symbols as follows:
A = { x I •X E A} = { x I ••X E A} = { x I x E A} = A.
7. These identities are straightforward applications of the definitions and are most easily stated using set-builder
notation. Recall that T means the proposition that is always true, and F means the proposition that is
always false.
a) AU U = { x I x E A V x E U} = { x I x E A V T} = { x I T} = U
b) An 0 = { x Ix EA/\ x E 0} = { x Ix EA /\ F} = { x IF}= 0
9. a) We must show that every element (of the universal set) is in AU A. This is clear, since every element is
either in A (and hence in that union) or else not in A (and hence in that union).
b) We must show that no element is in An A. This is clear, since An
and not in A at the same time, obviously an impossibility.
A consists of elements that are in A
11. These follow directly from the corresponding properties for the logical operations or and and.
a) AU B = { x I x E A V x E B} = { x I x E B V x E A} = B U A
b) An B = { x Ix EA/\ x EB}= { x Ix EB /\ x EA}= B n A
13. We will show that these two sets are equal by showing that each is a subset of the other. Suppose x E
An (AU B). Then x EA and x EAU B by the definition of intersection. Since x EA, we have proved that
the left-hand side is a subset of the right-hand side. Conversely, let x E A. Then by the definition of union,
x E AU B as well. Since both of these are true, x E An (AU B) by the definition of intersection, and we
have shown that the right-hand side is a subset of the left-hand side.
15. This exercise asks for a proof of one of De Morgan's laws for sets. The primary way to show that two sets are
equal is to show that each is a subset of the other. In other words, to show that X = Y, we must show that
whenever x E X, it follows that x E Y, and that whenever x E Y, it follows that x E X. Exercises 5- 7 could
also have been done this way, but it was easier in those cases to reduce the problems to the corresponding
problems of logic. Here, too, we can reduce the problem to logic and invoke De Morgan's law for logic, but
this problem requests specific proof techniques.
a) This proof is similar to the proof of the dual property, given in Example 10. Suppose x E AU B. Then
x rf_ AU B. which means that x is in neither A nor B. In other words, x rf- A and x rf_ B. This is equivalent
to saying that x E A and x E B. Therefore x E An B' as desired. Conversely, if x E An B' then x E A
and x E B. This means x rf- A and x rf- B, so x cannot be in the union of A and B . Since x rf- A U B, we
conclude that x E A U B, as desired.
Section 2.2
55
Set Operations
b) The following membership table gives the desired equality, since columns four and seven are identical.
A
E
1
1
1
0
0
0
AUE
--1
1
1
0
1
0
AUE
--0
0
0
1
A
E
AnE
--0
0
0
1
-
0
0
1
1
0
1
0
1
17. This exercise asks for a proof of a generalization of one of De Morgan's laws for sets from two sets to three.
a) This proof is similar to the proof of the two-set property, given in Example 10. Suppose x E An B n C.
Then x ~ An B n C, which means that x fails to be in at least one of these three sets. In other words, x ~ A
or x r:f. E or x r:f. C. This is equivalent to saying that x E A or x E E or x E C. Therefore x E AU E U C, as
desired. Conversely, if x E AUE UC, then x E A or x E E or x E C. This means x r:f. A or x r:f. E or x r:f. C,
so x cannot be in the intersection of A, E and C. Since x r:f. A n E n C, we conclude that x E A n E n C,
as desired.
b) The following membership table gives the desired equality, since columns five and nine are identical.
A
E
c
AnEnC
AnEnC
0
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
1
0
0
1
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
1
1
A
E
c
AUE UC
0
0
0
0
1
1
1
1
0
0
0
1
0
1
0
1
0
0
1
1
1
0
0
1
1
1
1
1
1
1
1
1
19. a) This is clear, since both of these sets are precisely { x I x E A /\ x r:f. E}.
b) One approach here is to use the distributive law; see the answer section for that approach. Alternatively,
we can argue directly as follows. Suppose x E (An E) U (An E). Then we know that either x E An E or
x E An B (or both). If either case, this forces x E A. Thus we have shown that the left-hand side is a subset
of the right-hand side. For the opposite direction, suppose x E A. There are two cases: x E E and x r:f. E.
In the former case, x is then an element of An E and therefore also an element of (An E) U (An E). In the
latter cases, x E E and therefore x is an element of An E and therefore also an element of (An E) U (An E).
21. There are many ways to prove identities such as the one given here. One way is to reduce them to logical
identities (some of the associative and distributive laws for V and /\). Alternately, we could argue in each
case that the left-hand side is a subset of the right-hand side and vice versa. Another method would be to
construct membership tables (they will have eight rows in order to cover all the possibilities). Here we choose
the second method. First we show that every element of the left-hand side must be in the right-hand side
as well. If x E AU (EU C), then x must be either in A or in EU C (or both). In the former case, we
can conclude that x E AUE and thus x E (AUE) UC, by the definition of union. In the latter case, x
must be either in E or in C (or both). In the former subcase, we can conclude that x E AU B and thus
x E (AUE) UC, by the definition of union; in the latter subcase, we can conclude that x E (AU B) UC,
again using the definition of union. The argument in the other direction is practically identical, with the roles
of A, E, and C switched around.
23. There are many ways to prove identities such as the one given here. One way is to reduce them to logical
identities (some of the associative and distributive laws for V and /\ ). Alternately, we could argue in each
case that the left-hand side is a subset of the right-hand side and vice versa. Another method would be to
construct membership tables (they will have eight rows in order to cover all the possibilities). Here we choose
Chapter 2
56
Basic Structures: Sets, Functions, Sequences, Sums, and l\fatrices
the third method. We construct the following membership table and note that the fifth and eighth columns
are identical.
A B
BnC
Au (B n C)
AuB
Aue
(Au B) n (AU C)
c
1
1
1
1
0
0
0
0
1
1
0
0
1
0
1
0
1
0
0
0
1
1
1
0
1
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
1
1
1
0
1
0
0
0
0
0
0
25. These are straightforward applications of the definitions.
a) The set of elements common to all three sets is {4, 6}.
b) The set of elements in at least one of the three sets is {O, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
c) The set of elements in C and at the same time in at least one of A and B is {4, 5, 6, 8, 10}.
d) The set of elements either in C or in both A and B (or in both of these) is {O, 2, 4, 5, 6, 7, 8, 9, 10}.
27. a) In the figure we have shaded the A set with horizontal bars (including the double-shaded portion, which
includes both horizontal and vertical bars), and we have shaded the set B- C with vertical bars (that portion
inside B but outside C. The intersection is where these overlap-the double-shaded portion (shaped like an
arrowhead).
b) In the figure we have shaded the set A
nB
with horizontal bars (including the double-shaded portion,
which includes both horizontal and vertical bars), and we have shaded the set A n C with vertical bars. The
union is the entire region that has any shading at all (shaped like a tilted mustache).
c) In the figure we have shaded the set An B with horizontal bars (including the double-shaded portion,
which includes both horizontal and vertical bars), and we have shaded the set An C with vertical bars. The
union is the entire region that has any shading at all (everything inside A except the triangular middle portion
where all three sets overlap) portion (shaped like an arrowhead).
(a)
(c)
(b)
29. a) If B adds nothing new to A, then we can conclude that all the elements of B were already in A. In other
words, B <:;; A.
b) In this case, all the elements of A are forced to be in B as well, so we conclude that A <:;; B.
c) This equality holds precisely when none of the elements of A are in B (if there were any such elements, then
A - B would not contain all the elements of A). Thus we conclude that A and B are disjoint (A n B = 0).
d) We can conclude nothing about A and B in this case, since this equality always holds.
e) Every element in A- B must be in A, and every element in B - A must not be in A. Since no item can
be in A and not be in A at the same time, there are no elements in both A - B and B - A. Thus the only
way for these two sets to be equal is if both of them are the empty set. This means that every element of A
must be in B, and every element of B must be in A. Thus we conclude that A = B.
31. This is the set-theoretic version of the contrapositive law for logic, which says that p
to --.q -+ --.p. We argue as follows.
A<:;; B
=Vx(x E A-+ x E B) =Vx(x tf. B-+ x tf. A) =Vx(x E B
-+
x EA)
-+
=B <:;;A
q is logically equivalent
Section 2.2
Set Operations
57
33. Clearly this will be the set of students majoring in computer science or mathematics but not both.
35. This is just a restatement of the definition. An element is in (AU B) - (An B) if it is in the union (i.e., in
either A or B), but not in the intersection (i.e., not in both A and B).
37. We will use the result of Exercise 36 as well as some obvious identities (some of which are in Exercises
a) A EB A= (A - A) U (A - A)= 0 u 0 = 0
c) A EB U =(A - U) u (U - A) = 0 u A= A
6~10).
b) A EB 0 =(A - 0) u (0 - A)= Au 0 =A
d) A EB A= (A - A) u (A - A)= Au A= U
39. We can conclude that B = 0. To see this, suppose that B contains some element b. If b E A, then b is
excluded from A EB B, so A EB B cannot equal A. On the other hand, if b tf: A, then b must be in A EB B,
so again A EB B cannot equal A. Thus in either case, A EB B -/=- A. We conclude that B cannot have any
elements.
41. Yes. To show that A= B, we need to show that x E A implies x E B and conversely. By symmetry, it will
be enough to show one direction of this. So assume that A EB C = B EB C, and let x E A be given. There
are two cases to consider, depending on whether x E C. If x E C, then by definition we can conclude that
x t/: A EB C. Therefore x tf: B EB C. Now if x were not in B, then x would be in B EB C (since x E C by
assumption). Since this is not true, we conclude that x E B, as desired. For the other case, assume that
x t/: C. Then x E A EB C. Therefore x E B EB C as well. Again, if x were not in B, then it could not be in
B EB C (since x t/: C by assumption). Once again we conclude that x E B, and the proof is complete.
43. Yes. Both sides equal the set of elements that occur in an odd number of the sets A, B, C, and D.
45. We give a proof by contradiction. If A U B is finite, then it has n elements for some natural number n. But
A already has more than n elements, because it is infinite, and A U B has all the elements that A has, so
A U B has more than n elements. This contradiction shows that A U B must be infinite.
47. a) The union of these sets is the set of elements that appear in at least one of them. In this case the sets
are "increasing": Ai ~ A2 ~ • · · ~ An. Therefore every element in any of the sets is in An, so the union is
An= {1,2, ... ,n}.
b) The intersection of these sets is the set of elements that appear in all of them. Since Ai = {1} , only the
number 1 has a chance to be in the intersection. In fact 1 is in the intersection, since it is in all of the sets.
Therefore the intersection is Ai = {1}.
49. a) Here the sets are increasing. A bit string of length not exceeding 1 is also a bit string of length not
exceeding 2, so Ai ~ Az. Similarly, Az ~ A3 ~ A4 ~ · · · ~ An. Therefore the union of the sets Ai, Az, ... ,
An is just An itself.
b) Since Ai is a subset of all the Ai 's, the intersection is Ai , the set of all nonempty bit strings of length
not exceeding 1, namely {O, 1}.
51. a) As i increases, the sets get larger: Ai C A2 C A 3 · · ·. All the sets are subsets of the set of integers,
and every integer is included eventually, so LJ:i Ai = Z. Because Ai is a subset of each of the others,
n:i Ai= Ai= {-1, 0, 1}.
b) All the sets are subsets of the set of integers, and every nonzero integer is in exactly one of the sets, so
LJ:i Ai = Z - {O}. Each pair of these sets are disjoint, so no element is common to all of the sets. Therefore
n:iAi = 0.
58
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
c) This is similar to part (a), the only difference being that here we are working with real numbers. Therefore
1 Ai = R (the set of all real numbers), and
1 Ai = A1 = [-1, 1] (the interval of all real numbers
between -1 and 1, inclusive).
LJ:
n:
d) This time the sets are getting smaller as i increases: · · · C A 3 C A2 C A 1 . Because A 1 includes all
LJ:
the others,
1 Ai = A1 = [1, oo) (all real numbers greater than or equal to 1). Every number eventually
gets excluded as i increases, so
1 A, = 0. Notice that oo is not a real number, so we cannot write
1 Ai = { oo}.
n:
n:
53. The ith digit in the string indicates whether the ith number in the universal set (in this case the number i)
is in the set in question.
a) {1,2,3,4,7,8,9,10}
b) {2,4,5,6,7}
c) {1, 10}
55. We are given two bit strings, representing two sets. We want to represent the set of elements that are in the
first set but not the second. Thus the bit in the ith position of the bit string for the difference is 1 if the ith
bit of the first string is 1 and the ith bit of the second string is 0, and is 0 otherwise.
57. We represent the sets by bit strings of length 26, using alphabetical order. Thus
A is represented by 11 1110 0000 0000 0000 0000 0000,
B is represented by 01 1100 1000 0000 0100 0101 0000,
To
a)
b)
c)
d)
C is represented by 00 1010 0010 0000 1000 0010 0111, and
D is represented by 00 0110 0110 0001 1000 0110 0110.
find the desired sets, we apply the indicated bitwise operations to these strings.
11 1110 0000 0000 0000 0000 0000 V 01 1100 1000 0000 0100 0101 0000 =
111110 1000 0000 0100 0101 0000, which represents the set {a,b,c,d,e,g,p,t,v}
11 1110 0000 0000 0000 0000 0000 /\ 011100 1000 0000 0100 0101 0000 =
01 1100 0000 0000 0000 0000 0000, which represents the set {b, c, d}
(11 1110 0000 0000 0000 0000 0000 V 00 0110 0110 0001 1000 0110 0110) /\
(01 1100 1000 0000 0100 0101 0000 v 00 1010 0010 0000 1000 0010 0111) =
11 1110 0110 0001 1000 0110 0110 /\ 01 1110 1010 0000 1100 0111 0111 =
011110 0010 0000 1000 0110 0110, which represents the set {b,c,d,e,i,o,t,u,x,y}
11 1110 0000 0000 0000 0000 0000 v 01 1100 1000 0000 0100 0101 0000 v
00 1010 0010 0000 1000 0010 0111 v 00 0110 0110 0001 1000 0110 0110 =
11 1110 1110 0001 1100 0111 0111, which represents the set
{a,b,c,d,e,g,h,i,n,o,p,t,u,v,x,y,z}
59. We simply adjoin the set itself to the list of its elements.
a) {1,2,3,{l,2,3}}
b) {0}
c) {0,{0}}
d) {0,{0},{0,{0}}}
61. a) The multiplicity of a in the union is the maximum of 3 and 2, the multiplicities of a in A and B.
Since the maximum is 3, we find that a occurs with multiplicity 3 in the union. Working similarly with b,
c (which appears with multiplicity 0 in B), and d (which appears with multiplicity 0 in A), we find that
AU B = {3 ·a, 3 · b, 1 · c, 4 · d}.
b) This is similar to part (a), with "maximum" replaced by "minimum." Thus An B = {2 ·a, 2 · b}. (In
particular, c and d appear with multiplicity 0-i.e., do not appear-in the intersection.)
c) In this case we subtract multiplicities, but never go below 0. Thus the answer is { 1 · a, 1 · c}.
d) Similar to part (c) (subtraction in the opposite order); the answer is {1 · b,4 · d}.
e) We add multiplicities here, to get {5 · a, 5 · b, 1 · c, 4 · d}.
Section 2.3
Functions
59
63. Assume that the universal set contains just Alice, Brian, Fred, Oscar, and Rita. We subtract the degrees of
membership from 1 to obtain the complement. Thus F is {0.4 Alice, 0.1 Brian, 0.6 Fred, 0.9 Oscar, 0.5 Rita},
and R is {0.6 Alice, 0.2 Brian, 0.8 Fred, 0.1 Oscar, 0.3 Rita}.
65. Taking the minimums, we obtain {0.4 Alice, 0.8 Brian, 0.2 Fred, 0.1 Oscar, 0.5 Rita} for F n R.
SECTION 2.3
Functions
The importance of understanding what a function is cannot be overemphasized-functions permeate all of
mathematics and computer science. This exercise set enables you to make sure you understand functions
and their properties. Exercise 33 is a particularly good benchmark to test your full comprehension of the
abstractions involved. The definitions play a crucial role in doing proofs about functions. To prove that a
function f : A ---> B is one-to-one, you need to show that x 1 -j. x 2 ---> f(xi) -j. f(x2) for all x 1 , x2 E A. To
prove that such a function is onto, you need to show that \::/y EB 3x EA (f(x) = y).
1. a) The expression I/x is meaningless for x
is no rule at all. In other words,
f
= 0,
which is one of the elements in the domain; thus the "rule"
(0) is not defined.
b) Things like Hare undefined (or, at best, are complex numbers).
c) The "rule" for f is ambiguous. We must have f(x) defined uniquely, but here there are two values
associated with every x, the positive square root and the negative square root of x 2 + 1 .
3. a) This is not a function, because there may be no zero bit in S, or there may be more than one zero bit
in S. Thus there may be no value for f (S) or more than one. In either case this violates the definition of a
function, since f(S) must have a unique value.
b) This is a function from the set of bit strings to the set of integers, since the number of 1 bits is always a
clearly defined nonnegative integer.
c) This definition does not tell what to do with a nonempty string consisting of all O's. Thus, for example,
f(OOO) is undefined. Therefore this is not a function.
5. In each case we want to find the domain (the set on which the function operates, which is implicitly stated in
the problem) and the range (the set of possible output values).
a) Clearly the domain is the set of all bit strings. The range is Z; the function evaluated at a string with n
l's and no O's is n, and the function evaluated at a string with n O's and no l's is -n.
b) Again the domain is clearly the set of all bit strings. Since there can be any natural number of O's in a bit
string, the value of the function can be 0, 2, 4, .... Therefore the range is the set of even natural numbers.
c) Again the domain is the set of all bit strings. Since the number of leftover bits can be any whole number
from 0 to 7 (if it were more, then we could form another byte), the range is {O, 1, 2, 3, 4, 5, 6, 7}.
d) As the problem states, the domain is the set of positive integers. Only perfect squares can be function
values, and clearly every positive perfect square is possible. Therefore the range is {l, 4, 9, 16, ... }.
7. In each case, the domain is the set of possible inputs for which the function is defined, and the range is the
set of all possible outputs on these inputs.
a) The domain is z+ x z+, since we are told that the function operates on pairs of positive integers (the word
"pair" in mathematics is usually understood to mean ordered pair). Since the maximum is again a positive
integer, and all positive integers are possible maximums (by letting the two elements of the pair be the same),
the range is z+ .
Chapter 2
60
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
z+ . Since the decimal representation of an integer has to have at least one
digit, at most nine digits do not appear, and of course the number of missing digits could be any number less
than 9. Thus the range is {O, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
b) We are told that the domain is
c) We are told that the domain is the set of bit strings. The block 11 could appear no times, or it could
appear any positive number of times, so the range is N.
d) We are told that the domain is the set of bit strings. Since the first 1 can be anywhere in the string, its
position can be 1, 2, 3, .... If the bit string contains no l's, the value is 0 by definition. Therefore the range
is N.
9. The floor function rounds down and the ceiling function rounds up.
a)l
b)O
c)O
d)-1
e)3
f)-1
g)l~+l~lJ=l~+2J=l2~J=2
h) lH~JJ = l~. 2J = llJ = 1
11. We need to determine whether the range is all of {a, b, c, d}. It is for the function in part (a), but not for the
other two functions.
13. a) This function is onto, since every integer is 1 less than some integer. In particular, f(x + 1) = x.
b) This function is not onto. Since n 2 + 1 is always positive, the range cannot include any negative integers.
c) This function is not onto, since the integer 2, for example, is not in the range. In other words, 2 is not the
cube of any integer.
d) This function is onto. If we want to obtain the value x, then we simply need to start with 2x, since
f(2x) = l2x/2l = lxl = x for all x E Z.
15. An onto function is one whose range is the entire codomain. Thus we must determine whether we can write
every integer in the form given by the rule for f in each case.
a) Given any integer n, we have f(O, n) = n, so the function is onto.
b) Clearly the range contains no negative integers, so the function is not onto.
c) Given any integer m, we have f(m, 25) = m, so the function is onto. (We could have used any constant
in place of 25 in this argument.)
d) Clearly the range contains no negative integers, so the function is not onto.
e) Given any integer m, we have f(m, 0) = m, so the function is onto.
17. a) This may be one-to-one (if there are no shared offices) or not (if there are shared offices).
b) Again, this depends on the policy of the school. Assuming that no bus gets more than one teacher
chaperone, this function will be one-to-one.
c) This is most likely not one-to-one, because two teachers might have the same salary. On the other hand,
it may happen that everyone's salary is different, in which case it is one-to-one.
d) This is clearly a one-to-one function, because social security numbers are unique--no two people can have
the same social security number.
19. a) The codomain could be the set of all offices in the school. This is probably not onto; for example, the
secretaries probably have offices, and those offices are not in the range of this function (because the secretaries
are not in the domain).
b) The codomain could be the set of busses going on the field trip. This is probably an onto function, as long
as school policy is that every bus has to have a teacher chaperone.
c) The codomain could be all positive real numbers (with the understanding that the number represents the
salary in dollars). The function is clearly not onto; no teacher has a salary of y'2 dollars or 10 10 dollars.
Section 2.3
Functions
61
d) The codomain could be all strings of nine digits. This function is not onto, because the social security
number of everyone who is not a teacher at the school is not in the range of the function.
21. Obviously there are an infinite number of correct answers to each part. The problem asked for a "formula."
Parts (a) and ( c) seem harder here, since we somehow have to fold the negative integers into the positive
ones without overlap. Therefore we probably want to treat the negative integers differently from the positive
integers. One way to do this with a formula is to make it a two-part formula. If one objects that this is not
"a formula,'' we can counter as follows. Consider the function g( x) = L2x J/2x. Clearly if x 2: 0, then 2x is a
positive integer, so g( x) = 2x /2x = 1. If x < 0, then 2x is a number between 0 and 1, so g( x) = 0/2x = 0.
Ifwe want to define a function that has the value fi(x) when x 2: 0 and h(x) when x < 0, then we can use
the formula g(x) · fi(x) + (1- g(x)) · h(x).
a) We could map the positive integers (and 0) into the positive multiples of 3, say, and the negative integers
into numbers that are 1 greater than a multiple of 3, in a one-to-one manner. This will give us a function
that leaves some elements out of the range. So let us define our function as follows:
f(x) = {
3x+3
3lxl + 1
if x 2: 0
if x < 0
The values of f on the inputs 0 through 4 are then 3, 6, 9, 12, 15; and the values on the inputs -1 to -4 are
4, 7, 10, 13. Clearly this function is one-to-one, but it is not onto since, for example, 2 is not in the range.
b) This is easier. We can just take f(x) =
lxl + 1.
It is clearly onto, but f(n) and f(-n) have the same
value for every positive integer n, so f is not one-to-one.
c) This is similar to part (a), except that we have to be careful to hit all values. Mapping the nonnegative
integers to the odds and the negative integers to the evens will do the trick:
f(x) = {
~~11
if x 2: 0
if x < 0
d) Here we can use a trivial example like f(x) = 17 or a simple nontrivial one like f(x) = x 2
+ 1.
Clearly
these are neither one-to-one nor onto.
23. a) One way to determine whether a function is a bijection is to try to construct its inverse. This function
is a bijection, since its inverse (obtained by solving y = 2x + 1 for x) is the function g(y) = (y - 1) /2.
Alternatively, we can argue directly. To show that the function is one-to-one, note that if 2x + 1 = 2x' + 1,
then x = x'. To show that the function is onto, note that 2( (y - 1) /2) + 1 = y, so every number is in the
range.
b) This function is not a bijection, since its range is the set ofreal numbers greater than or equal to 1 (which
is sometimes written [1, oo)), not all of R. (It is not injective either.)
c) This function is a bijection, since it has an inverse function, namely the function f(y) = y 1 13 (obtained by
solving y = x 3 for x).
d) This function is not a bijection. It is easy to see that it is not injective, since x and -x have the same
image, for all real numbers x. A little work shows that the range is only { y I 0.5 :S: y < 1} = [0.5, 1).
25. The key here is that larger denominators make smaller fractions, and smaller denominators make larger
fractions. We have two things to prove, since this is an "if and only if" statement. First, suppose that f is
strictly decreasing. This means that f(x) > f(y) whenever x < y. To show that g is strictly increasing,
suppose that x < y. Then g(x) = 1/ f(x) < 1/ f(y) = g(y). Conversely, suppose that g is strictly increasing.
This means that g( x) < g(y) whenever x < y. To show that f is strictly decreasing, suppose that x < y.
Then f(x) = l/g(x) > l/g(y) = f(y).
62
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
27. a) Let f be the given strictly decreasing function from R to itself. We need to show that f(a) = f(b) implies
a= b for all a, b E R. We give an indirect proof by proving the contrapositive: if a i- b, then f(a) i- f(b).
There are two cases. Suppose a < b; then because f is strictly decreasing, it follows that f(a) > f(b).
Similarly, if a> b, then f(a) < f(b). Thus in either case, f(a) i- f(b).
b) We need to make the function decreasing, but not strictly decreasing, so, for example, we could take
the trivial function f (x) = 17. If we want the range to be all of R, we could define
f(x) = -x- l for x < -1; f(x) = 0 for -1:::; x:::; 1; and f(x) = -x+ 1forx>1.
f in parts this way:
29. The function is not one-to-one (for example, f(2) = 2 = f(-2) ), so it is not invertible. On the restricted
domain, the function is the identity function from the set of nonnegative real numbers to itself, f (x) = x, so
it is one-to-one and onto and therefore invertible; in fact, it is its own inverse.
31. In each case, we need to compute the values of f(x) for each x ES.
a) Note that f(±2) = l(±2) 2 /3J = l4/3J = 1, J(±l) = l(±1) 2 /3J = ll/3J = 0, f(O) = l0 2 /3J = lOJ = 0,
and f (3) = l 32 /3 J = l 3J = 3. Therefore f (S) = {O, 1, 3}.
b) In addition to the values we computed above, we note that f(4) = 5 and f(5) = 8. Therefore f(S) =
{0,1,3,5,8}.
c) Note this time also that f(7)
= 16 and f(ll) = 40, so f(S) = {0,8, 16,40}.
d) {f(2),f(6),f(10),f(l4)} = {1,12,33,65}
33. In both cases, we can argue directly from the definitions.
a) Assume that both f and g are one-to-one. We need to show that fog is one-to-one. This means that
we need to show that if x and y are two distinct elements of A, then f(g(x)) i- f(g(y)). First, since g is
one-to-one, the definition tells us that g(x) i- g(y). Second, since now g(x) and g(y) are distinct elements
of B, and since f is one-to-one, we conclude that f(g(x)) i-f(g(y)), as desired.
b) Assume that both f and g are onto. We need to show that f o g is onto. This means that we need to
show that if z is any element of C, then there is some element x E A such that f(g(x)) = z. First, since
f is onto, we can conclude that there is an element y E B such that f (y) = z. Second, since g is onto and
y E B, we can conclude that there is an element x EA such that g(x) = y. Putting these together, we have
z = f(y) = f(g(x)), as desired.
35. To establish the setting here, let us suppose that g : A ___, B and f : B ___, C. Then f o g: A ___, C. We are
told that f and fog are onto. Thus all of C gets "hit" by the images of elements of B; in fact, each element
in C gets hit by an element from A under the composition f o g. But this does not seem to tell us anything
about the elements of B getting hit by the images of elements of A. Indeed, there is no reason that they
must. For a simple counterexample, suppose that A= {a}, B = {b 1 ,b2 }, and C = {c}. Let g(a) = b1 , and
let f(b 1 ) = c and f(b 2 ) = c. Then clearly f and fog are onto, but g is not, since b2 is not in its range.
37. We just perform the indicated operations on the defining expressions. Thus f + g is the function whose value
at x is (x 2 +1) + (x + 2), or, more simply, (f + g)(x) = x 2 + x + 3. Similarly Jg is the function whose value
at x is (x 2 + l)(x + 2); in other words, (f g)(x) = x 3 + 2x 2 + x + 2.
39. We simply solve the equation y = ax + b for x. This gives x = (y - b) /a, which is well-defined since a i- 0.
Thus the inverse is 1- 1 (y) = (y-b)/a. To check that our work is correct, we must show that f of- 1 (y) = y for
all y E R and that 1- 1 o f(x) = x for all x E R. Both of these are straightforward algebraic manipulations.
For the first, we have f o 1- 1 (y) = J(f- 1 (y)) = J((y- b)/a) = a((y- b)/a) + b = y. The second is similar.
Functions
Section 2.3
63
41. a) Let us arrange for 8 and T to be nonempty sets that have empty intersection. Then the left-hand side will
be f(0), which is the empty set. If we can make the right-hand side nonempty, then we will be done. We can
make the right-hand side nonempty by making the codomain consist of just one element, so that J(8) and
f(T) will both be the set consisting of that one element. The simplest example is as follows. Let A = {1, 2}
and B = {3}. Let f be the unique function from A to B (namely /(1) = /(2) = 3 ). Let S = {1} and
T = {2}. Then !(8 n T) = /(0) = 0, which is a proper subset of /(8) n f (T) = {3} n {3} = {3}.
b) Assume that f is one-to-one. We must show that every element of l (8) n l (T) is an element of l (8 n T).
Let y E B be an element of 1(8) n l(T). Then y E 1(8), so y = f(xi) for some x 1 E 8. Similarly, y E l (T),
so y = l (x2) for some X2 ET. Because l is one-to-one, it follows that x 1 = x 2 . This element is therefore in
8 n T' so y E l (8 n T).
43. a) We want to find the set of all numbers whose floor is 0. Since all numbers from 0 to 1 (including 0 but
not 1) round down to 0, we conclude that g- 1 ( {O}) = { x I 0 ~ x < 1} = [O, 1).
b) This is similar to part (a). All numbers from -1 to 2 (including -1 but not 2) round down to -1, 0,
or l; we conclude that g- 1 ({-1,0, 1}) = {x I -1~x<2} = [-1,2).
c) Since g(x) is always an integer, there are no values of x such that g(x) is strictly between 0 and 1. Thus
the inverse image in this case is the empty set.
45. Note that the complementation here is with respect to the relevant universal set. Thus S = B - 8 and
l- 1 (8) =A- 1- 1 (8). There are two things to prove in order to show that these two sets are equal: that the
left-hand side of the equation is a subset of the right-hand side, and that the right-hand side is a subset of the
left-hand side. First let x E 1- 1 (S). This means that l (x) ES, or equivalently that l (x) 8. Therefore by
definition of inverse image, x 1- 1 (8), so x E 1- 1 (8). For the other direction, assume that x E 1- 1 (8).
Then x
l- 1 (8). By definition this means that l(x)
8, which means that l(x) ES. Therefore by
definition, x E l- 1 (S).
tt
tt
tt
tt
47. There are three cases. Define the "fractional part" of x to be l(x) = x- lxJ. Clearly l(x) is always between
0 and 1 (inclusive at 0, exclusive at 1), and x = lxJ + l(x). If l(x) is less than~, then x - ~will have a
value slightly less than lx J ' so when we round up, we get lx J. In other words, in this case x - ~ l = lx J '
and indeed that is the integer closest to x. If l(x) is greater than~' then x - ~will have a value slightly
greater than lx J , so when we round up, we get lx J + 1. In other words, in this case Ix - ~ l = lx J + 1,
and indeed that is the integer closest to x in this case. Finally, if the fractional part is exactly ~ , then x is
midway between two integers, and x - ~ l = lx J' which is the smaller of these two integers.
r
r
49. We can write the real number x as l x J + t, where t is a real number satisfying 0 ~ t < 1. Since t = x - l x J ,
we have 0 ~ x - lx J < 1. The first two inequalities, x - 1 < lx J and lx J ~ x, follow algebraically. For the
other two inequalities, we can write x = x l - E, where again 0 ~ E < 1. Then 0 ~ x l - x < 1, and again
the desired inequalities follow by easy algebra.
r
r
51. a) One direction (the "only if" part) is obvious: If x < n, then since lxJ ~ x it follows that lxJ < n. We
will prove the other direction (the "if" part) indirectly (we will prove its contrapositive). Suppose that x 2: n.
Then "the greatest integer not exceeding x" must be at least n, since n is an integer not exceeding x. That
is, lx J 2: n.
b) One direction (the "only if" part) is obvious: If n < x, then since x ~ Ixl it follows that n < fx l . We
will prove the other direction (the "if" part) indirectly (we will prove its contrapositive). Suppose that n 2: x.
Then "the smallest integer not less than x" must be no greater than n, since n is an integer not less than x.
That is, lxl ~ n.
Chapter 2
64
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
53. If n is even, then n = 2k for some integer k. Thus l n/2 J = l k J = k = n/2. If n is odd, then n = 2k + 1 for
some integer k. Thus Ln/2J = Lk + ~J = k = (n -1)/2.
55. Without loss of generality we can assume that x 2 0, since the equation to be proved is equivalent to the
same equation with -x substituted for x. Then the left-hand side is f-x l by definition, and the right-hand
side is - l x J. Thus this problem reduces to Exercise 54. Its proof is straightforward. Write x as n + E, where
n is a natural number and E is a real number satisfying 0 s E < 1. Then clearly -xl = -n - El = -n and
-LxJ =-Ln+Ej =-n as well.
r
r
57. In some sense this question is its own answer-the number of integers strictly between a and b is the number
of integers strictly between a and b. Presumably we seek an expression involving a, b, and the floor and/or
ceiling function to answer this question. If we round a down and round b up to integers, then we will be
looking at the smallest and largest integers just outside the range of integers we want to count, respectively.
These values are of course la J and fb l , respectively. Then the answer is fbl - la J - 1 (just think of counting
all the integers between these two values, excluding both ends-if a row of fenceposts one foot apart extends
for k feet, then there are k - 1 fenceposts not counting the end posts). Note that this even works when, for
example, a= 0.3 and b = 0.7.
59. Since a byte is eight bits, all we are asking for in each case is fn/81 , where n is the number of bits.
a) f7/8l = 1
b) f17/8l = 3
d) f28800/8l = 3600
c) f1001/8l = 126
61. In each case we need to divide the number of bytes (octets) by 1500 and round up. In other words, the answer
r
is n/15001, where n is the number of bytes.
a) f150,000/1500l = 100
d) 45,300,000;15001 = 30,200
r
b) f384,000/1500l = 256
c) fl,544,000/15001 = 1030
63. The graph will look exactly like the graph of the function f (x) = lx J, shown in Figure lOa, except that the
picture will be compressed by a factor of 2 in the horizontal direction, since x has been replaced by 2x.
-2
-1
-1
_,
_,
-2
-3
65. This is a step function, with values changing only at the integers. We note the pattern that f(x) jumps by 1
when x passes through an odd integer (because of the lxJ term), and by 2 when x passes through an even
integer (an additional jump caused by the l x /2 J term). The result is as shown.
-3 -2 -1
-1
-2
_,
-3
-4
_,
-5
1
2
3
4
Section 2.3
65
Functions
67. a) The graph will look exactly like the graph of the function f(x) = lxJ, shown in Figure lOa, except that
the picture will be shifted to the left by ~ unit, since x has been replaced by x + ~ = x - (- ~).
3
_,
2
_,
1 _,
-4
- 2 _ , -1
_,
4
-2
_,
-3
b) The graph will look exactly like the graph of the function f(x) = l2xj, shown in the solution to Exercise 63,
except that the picture will be shifted to the left by ~ unit, since x has been replaced by x + ~. Alternatively,
we can note that f(x) can be rewritten as l2xj + 1, so the graph is the graph shown in the solution to
Exercise 63 shifted up one unit.
-2
-1
_,
_,
-1
-2
_,
-3
c) The graph will look exactly like the graph of the function f(x) = j x l, shown in Figure lOb, except that
the x-axis is stretched by a factor of 3. Thus we can use the same picture and just relabel the x-axis.
-12 -9 -6 -3
~
-1
~
3
6
9 12
-2
~
-3
d) The graph is a step version of the usual hyperbola y = 1/x. Note that x = 0 is not in the domain. The
graph can be drawn by first plotting the points at which 1/x is an integer (x = 1, ±~, ±~, ... ) and then
filling in the horizontal segments, making sure to note that they go to the right (for example, if x is a little
bigger than ~, then 1/x is a little less than 2, so f(x) = 2, since we are rounding up here). Note that
f(x) = 1 for x 2 1, and f(x) = 0 for x < -1.
eo
4
3
_,
-
~
-1
~
-1
_,
eo
-2
-3
e) The key thing to note is that since we can pull integers outside the floor and ceiling function (identity (4)
in Table 1), we can write f (x) more simply as l x J + j x l · When x is an integer, this is just 2x. When x is
between two integers, however, this has the value of the integer between the two integers 2 l x J and 2 j x l . The
graph is therefore as shown here.
66
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
4
•
~
•
-2
-1
-1
•
-2
-3
~
•
-4
f} The basic shape is the parabola, y = x 2 . In particular, for x an even integer, l(x) = x 2 , since the terms
inside the floor and ceiling function symbols are integers. However, because of these step functions, the curve
is broken into steps. At even integers other than x = 0 there are isolated points in the graph. Also, the graph
takes jumps at all the integer and half-integer values outside the range -2 < x < ~ (where in fact l(x) = 0).
The portion of the graph near the origin is shown here.
_,
•
-2
4
-1
-1
-2
-3
-4
g) Despite the complicated-looking formula, this is really quite similar to part (a}; in fact, we'll see that
it's identical! First note that the expression inside the outer ceiling function symbols is always going to be a
half-integer; therefore we can tell exactly what its rounded-up value will be, namely l x - ~ J+ 1. Furthermore,
since identity (4) of Table 1 allows us to move the 1 inside the floor function symbols, we have l(x) = lx+ ~J.
Therefore this is the same function as in part (a).
69. We simply need to solve the equation y = x 3 +1 for x. This is easily done by algebra: x = (y-1) 113 . Therefore
the inverse function is given by the rule 1- 1 (y) = (y-1) 113 (or, equivalently, by the rule 1- 1 (x) = (x -1) 113 ,
since the variable in the definition is just a dummy variable).
71. We can prove all of these identities by showing that the left-hand side is equal to the right-hand side for all
possible values of x. In each instance (except part ( c), in which there are only two cases), there are four cases
to consider, depending on whether x is in A and/or B.
a) If x is in both A and B, then fAnB(x) = 1; and the right-hand side is 1 · 1 = 1 as well. Otherwise
x tJ_ A n B , so the left-hand side is 0, and the right-hand side is either 0 · 1 or 1 · 0 or 0 · 0, all of which are
also 0.
b} If xis in both A and B, then lAuB(x) = 1; and the right-hand side is 1+1-1·1=1 as well. If xis
in A but not B, then x E AU B, so the left-hand side is still 1, and the right-hand side is 1 + 0 - 1 · 0 = 1,
as desired. The case in which x is in B but not A is similar. Finally, if x is in neither A nor B, then the
left-hand side is 0, and the right-hand side is 0 + 0 - 0 · 0 = 0 as well.
c) If x E A, then x tJ_ A, so f-x( x) = 0. The right-hand side equals 1 - 1 = 0 in this case, as well. On the
other hand, if x tJ_ A, then x E A, so the left-hand side is 1, and the right-hand side is 1 - 0 = 1 as well.
d} If x is in both A and B, then x tJ_ A EBB, so 1AEFJB(x) = 0. The right-hand side is 1+1 - 2 · 1·1 = 0
as well. Next, if x E A but x tJ_ B, then x E A EBB, so the left-hand side is 1. The right-hand side is
1 + 0 - 2 · 1 · 0 = 1 as well. The case x E B /\ x tJ_ A is similar. Finally, if x is in neither A nor B, then
x tJ_ A EBB, so the left-hand side is 0; and the right-hand side is also 0 + 0 - 2 · 0 · 0 = 0.
Section 2.3
Functions
67
73. a) This is true. Since lxJ is already an integer, flxJl = lxJ.
b) A little experimentation shows that this is not always true. To disprove it we need only produce a
counterexample, such as x = ~. In that case the left-hand side is l 1J = 1, while the right-hand side is
2. 0 =
o.
c) This is true. We prove it by cases. If x is an integer, then by identity (4b) in Table 1, we know that
fx + y l = x + fy l , and it follows that the difference is 0. Similarly, if y is an integer. The remaining case
is that x = n + E and y = m + 8, where n and m are integers and E and 8 are positive real numbers
less than 1 . Then x + y will be greater than m + n but less than m + n + 2, so fx + y l will be either
m + n + 1 or m + n + 2. Therefore the given expression will be either (n + 1) + (m + 1) - (m + n + 1) = 1 or
(n + 1) + (m + 1) - (m + n + 2) = 0, as desired.
d) This is clearly false, as we can find with a little experimentation. Take, for example, x
Then the left-hand side is f3/10l = 1, but the right-hand side is 1 · 3 = 3.
= 1/10 and
y
= 3.
e) Again a little trial and error will produce a counterexample. Take x = 1/2. Then the left-hand side is 1
while the right-hand side is 0.
75. a) If x is a positive integer, then the two sides are identical. So suppose that x = n 2 + m + E, where n is
the largest perfect square integer less than x, m is a nonnegative integer, and 0 < E < 1. For example, if
x = 13.2, then n = 3, m = 4, and E = 0.2. Then both fa and
= Jn 2 + m are between n and n + 1.
Therefore both sides of the equation equal n.
VIXJ
b) If x is a positive integer, then the two sides are identical. So suppose that x = n 2
- m - E, where n is
the smallest perfect square integer greater than x, m is a nonnegative integer, and E is a real number with
0 < E < 1. For example, if x = 13.2, then n = 4, m = 2, and E = 0.8. Then both fa and
= Jn 2 -- m
are between n - 1 and n. Therefore both sides of the equation equal n.
v1JXT
77. In each case we easily read the domain and codomain from the notation. The domain of definition is obtained
by determining for which values the definition makes sense. The function is total if the domain of definition
is the entire domain (so that there are no values for which the partial function is undefined).
a) The domain is Z and the codomain is R. Since division is possible by every nonzero number, the domain
of definition is all the nonzero integers; {0} is the set of values for which f is undefined. (It is not total.)
b) The domain and codomain are both given to be Z. Since the definition makes sense for all integers, this
is a total function, whose domain of definition is also Z; the set of values for which f is undefined is 0.
c) By inspection, the domain is the Cartesian product Z x Z , and the co domain is Q . Since fractions cannot
have a 0 in the denominator, we must exclude the "slice" of Z x Z in which the second coordinate is 0. Thus
the domain of definition is Z x (Z - {O}), and the function is undefined for all values in Z x {O}. It is not a
total function.
d) The domain is given to be Z x Z and the codomain is given to be Z. Since the definition makes sense
for all pairs of integers, this is a total function, whose domain of definition is also Z x Z; the set of values for
which f is undefined is 0.
e) Again the domain and codomain are Z x Z and Z, respectively. Since the definition is only stated for
those pairs in which the first coordinate exceeds the second, the domain of definition is { (m, n) I m > n}, and
therefore the set of values for which the function is undefined is { (m, n) I m ::; n}. It is not a total function.
79. a) By definition, to say that S has cardinality m is to say that S has exactly m distinct elements. Therefore
we can imagine enumerating them, as a child counts objects: assign the first object to 1, the second to 2, and
so on. This provides the one-to-one correspondence.
b) By part (a), there is a bijection f from S to {1,2, ... ,m} and a bijection
This tells us that g-
1
from T to {1,2, ... ,m}.
is a bijection from {1, 2, ... , m} to T. Then the composition g- 1 of is the desired
g
68
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
bijection from S to T.
SECTION 2.4
Sequences and Summations
This exercise set contains a lot of routine practice with the concept of and notation for sequences. It also
discusses telescoping sums; the product notation, corresponding to the summation notation discussed in
the section; and the factorial function, which occurs repeatedly in subsequent chapters. There are also a
few challenging exercises on more complicated sequences and sums.
1. a)
a0
=2·(-3) 0 +5°=2·1+1=3
b)
c) a4 = 2 · (-3) 4 + 54 = 2 · 81+625 = 787
a1
=2·(-3) 1 +5 1 =2·(-3)+5=-1
d) a 5 = 2 · (-3) 5 + 55 = 2 · (-243) + 3125 = 2639
3. In each case we simply evaluate the given function at n = 0, 1, 2, 3.
a) ao = 2° + 1 = 2,
b) ao = 11 = 1,
a1
a1
= 21 + 1 = 3,
= 22 = 4,
a2
a2
= 22 + 1 = 5, a3 = 23 + 1 = 9
= 33 = 27, a3 = 44 = 256
c) ao = L0/2J = 0, a1 = Ll/2J = 0, a2 = L2/2J = 1, a3 = L3/2j = 1
d) ao = L0/2j + I0/21 = 0 + 0 = 0, ai = Ll/2j + ll/21 = 0 + 1 = 1, az = L2/2j + 12/21
a3 = L3/2J + 13/21=1+2 = 3. Note that Ln/2J + ln/21 always equals n.
1+1 = 2,
5. In each case we just follow the instructions.
a) 2,5,8,11,14,17,20,23,26,29
b) 1,1,1,2,2,2,3,3,3,4
c) 1,1,3,3,5,5,7,7,9,9
d) This requires a bit of routine calculation. For example, the fifth term is 5! - 25 = 120 - 32 = 88. The first
ten terms are -1, -2, -2, 8, 88, 656, 4912, 40064, 362368, 3627776.
e) 3,6,12,24,48,96,192,384,768,1536
f) 2,4,6,10,16,26,42,68,110,178
g) For n = 1, the binary expansion is 1, which has one bit, so the first term of the sequence is 1. For n = 2,
the binary expansion is 10, which has two bits, so the second term of the sequence is 2. Continuing in this
way we see that the first ten terms are 1,2,2,3,3,3,3,4,4,4. Note that the sequence has one 1, two 2's,
four 3's, eight 4's, as so on, with 2k-l copies of k.
h) The English word for 1 is "one" which has three letters, so the first term is 3. This makes a good
brain-teaser; give someone the sequence and ask her or him to find the pattern. The first ten terms are
3,3,5,4,4,3,5,5,4,3.
7. One pattern is that each term is twice the preceding term. A formula for this would be that the nth term
is 2n-l. Another pattern is that we obtain the next term by adding increasing values to the previous term.
Thus to move from the first term to the second we add 1; to move from the second to the third we add 2 ;
then add 3, and so on. So the sequence would start out 1, 2, 4, 7, 11, 16, 22, .... We could also have trivial
answers such as the rule that the first three terms are 1, 2, 4 and all the rest are 17 (so the sequence is
1, 2, 4, 17, 17, 17, ... ), or that the terms simply repeat 1, 2, 4, 1, 2, 4, 1, 2, 4, .... Here is another pattern: Take
n points on the unit circle, and connect each of them to all the others by line segments. The inside of the
circle will be divided into a number of regions. What is the largest this number can be? Call that value an.
If there is one point, then there are no lines and therefore just the one original region inside the circle; thus
a 1 = 1. If n = 2, then the one chord divides the interior into two parts, so a 2 = 2. Three points give us a
triangle, and that makes four regions (the inside of the triangle and the three pieces outside the triangle), so
a 3 = 4. Careful drawing shows that the sequence starts out 1, 2, 4, 8, 16, 31. That's right: 31, not 32.
Section 2.4
Sequences and Summations
69
Creative students may well find other rules or patterns with various degrees of appeal.
9. We need to compute the terms of the sequence one at a time, since each term is dependent upon one or more
of the previous terms.
a) We are given ao = 2. Then by the recurrence relation an = 6an-1 we see (by letting n = 1) that a 1 =
6ao = 6·2 = 12. Similarly a 2 = 6a 1 = 6·12 = 72, then a3 = 6a 2 = 6·72 = 432, and a 4 = 6a 3 = 6·432 = 2592.
b) a1 = 2 (given), a2 = ai = 22 = 4, a3 =a~= 42 = 16, a 4 =a~= 16 2 = 256, a5 =a~= 256 2 = 65536
c) This time each term depends on the two previous terms. We are given a 0 = 1 and a 1 = 2. To compute
= 2 in the recurrence relation, obtaining a 2 = a 1 + 3a0 = 2 + 3 · 1 = 5. Then we have
a3 = a2 + 3a1 = 5 + 3 · 2 = 11 and a4 = a3 + 3a2 = 11 + 3 · 5 = 26.
a2 we let n
d) ao = 1 (given), a1 = 1 (given), a 2 = 2a1+2 2ao = 2 · 1+4 · 1 = 6, a 3 = 3a2 + 32a 1 =3·6+9 · 1 = 27,
a4 = 4a3 + 4 2a2 = 4 · 27 + 16 · 6 = 204
e) We are given ao = 1, ai = 2, and a 2 = 0. Then a 3 = a 2 + a 0 = 0 + 1 = 1 and a 4 = a 3 + a 1 = 1 + 2 = 3.
11. a) We simply plug in n = 0, n = 1, n = 2, n = 3, and n = 4. Thus we have a 0 = 2° + 5 · 3° = 1 + 5 · 1 = 6,
a1 = 21 + 5 · 3I = 2 + 5 · 3 = 17, a 2 = 22 + 5 · 32 = 4 + 5 · 9 = 49, a 3 = 23 + 5 · 33 = S + 5 · 27 = 143, and
a4 = 24 + 5 · 34 = 16 + 5 · Sl = 421.
b) Using our data from part (a), we see that 49 = 5 · 17 - 6 · 6, 143 = 5 · 49 - 6 · 17, and 421 = 5 · 143 - 6 · 49.
c) This is algebra. The messiest part is factoring out a large power of 2 and a large power of 3. If we substitute
n - 1 for n in the definition we have an-1 = 2n-l + 5 · 3n-I; similarly an_ 2 = 2n- 2 + 5 · 3n- 2 . We start with
the right-hand side of our desired identity:
5an-I - 6an-2 = 5(2n-l + 5 · 3n-l) - 6(2n- 2 + 5 · 3n- 2)
= 2n- 2(10 - 6) + 3n- 2(75 - 30)
= 2n- 2 · 4 + 3n- 2 · 9 · 5
= 2n
+ 3n . 5 =
an
13. In each case we have to substitute the given equation for an into the recurrence relation an =San-I -16an_ 2
and see if we get a true statement. Remember to make the appropriate substitutions for n (either n - 1 or
n - 2) on the right-hand side. What we are really doing here is performing the inductive step in a proof by
mathematical induction: if the formula is correct for an-I (and also for an- 2 , etc., in some cases), then the
formula is also correct for an .
a) Plugging an
Therefore an
=
0 into the equation an
=
8an-1 - 16an_ 2 , we obtain the true statement that 0 = 0.
= 0 is a solution of the recurrence relation.
b) Plugging an = 1 into the equation an =San-I -16an-2, we obtain the false statement 1 = S· 1-16· 1 = -S.
Therefore an = 1 is not a solution.
c) Plugging an= 2n into the equation an= 8an-l -16an_ 2 , we obtain the statement 2n = s.2n- 1 -15.2n- 2 .
By algebra, the right-hand side equals 2n- 2 (8 · 2 - 16) = 0. Since this is not equal to the left-hand side, we
conclude that an = 2n is not a solution.
d) Plugging an= 4n into the equation an= San-I -16an-2, we obtain the statement 4n = s.4n-l _ 15.4n- 2 .
By algebra, the right-hand side equals 4n- 2 (S · 4-16) = 4n- 2 • 16 = 4n- 2 · 42 = 4n. Since this is the left-hand
side, we conclude that an = 4n is a solution.
e) Plugging an = n4n into the equation an= 8an-l -16an_ 2 , we obtain the statement n4n = 8(n-1)4n-l l6(n-2)4n- 2 . By algebra, the right-hand side equals 4n- 2 (S(n-1)·4-16(n-2)) = 4n- 2 (32n-32-16n+32) =
4n- 2(l6n) = 4n- 2 · 42 n = n4n. Since this is the left-hand side, we conclude that an= n4n is a solution.
Chapter 2
70
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
f) Plugging a,, = 2 · 4n + 3n4" into the equation an = 8an-l - 16a,,_ 2 , we obtain the statement 2 · 4n +
3n4" = 8(2 · 4n-l + 3(n - 1)4n-l) - 16(2 · 4n- 2 + 3(n - 2)4n- 2). By algebra, the right-hand side equals
4"- 2 (8 · 2 ·4+8 ·3(n-1) · 4-16 · 2-16 · 3(n-2)) = 4n- 2(64+96n-96-32-48n+96) = 4n- 2(48n+32) =
4n- 2 ·4 2(3n+2) = (2+3n)4n. Since this is the same as the left-hand side, we conclude that an= 2.4n +3n4n
is a solution.
g) Plugging an= (-4)n into the equation a,, = 8an_ 1 -l6a 11 _2, we obtain the statement (-4)" = 8·(-4)"- 1 16· (-4)n-2. Byalgebratheright-handsideequals (-4)"- 2(8·(-4)-16) = (-4)n- 2(-48) = -3(-4)n. Since
this is not equal to the left-hand side, we conclude that an= (-4)n is not a solution.
h) Plugging a,, = n 2 4n into the equation an = 8a11 _1
2
- l6a 11 _ 2 , we obtain the statement n 4n = 8(n 1
2
2
2
2
2
2
1) 4n- -16(n-2) 4n- . By algebra, the right-hand side equals 4n- (8(n-1) ·4-16(n-2) ) = 4n- 2 (32(n 2 2n + 1) -16(n 2 - 4n+ 4)) = 4n- 2(32n 2 - 64n + 32 -16n 2 + 64n- 64) = 4"- 2 (16n 2 - 32) = 4n- 2 · 42 (n 2 - 2) =
4"(n 2 - 2). Since this is not equal to the left-hand side, we conclude that an= n 24n is not a solution.
15. In each case we have to plug the purported solution into the right-hand side of the recurrence relation and see
if it simplifies to the left-hand side. The algebra can get tedious, and it is easy to make a mistake.
a) We have
an-l + 2an-2 + 2n - 9 = -(n - 1) + 2 + 2(-(n - 2) + 2) + 2n - 9
= -n+2 =an.
b) We have
an-l + 2an-2 + 2n - 9 = 5(-1)"- 1 - (n - 1) + 2 + 2(5(-1) 11 - 2 - (n - 2) + 2) + 2n - 9
= 5(-1)"- 2(-1+2) - n + 2 =a,,.
Note that we had to factor out ( -1) 11 -
2
and that this is the same as ( -1 )" since ( -1 ) 2
= 1.
c) We have
an-l + 2a,,_2 + 2n - 9 = 3(-l)n-l + 2n-l - (n - 1) + 2 + 2(3(-1)"- 2 + 2n- 2 - (n - 2) + 2) + 2n - 9
= 3(-1)"- 2(-1+2) + 2n- 2(2 + 2) - n + 2 =a,,.
Note that we had to factor out 211 - 2 and that 2n- 2 · 4 = 2".
d) We have
an-l + 2an-2 + 2n - 9 = 7 · 2n-l - (n - 1) + 2 + 2(7 · 2"- 2 - (n - 2) + 2) + 2n - 9
= 2n- 2(7 · 2 + 2 · 7) - n + 2 =an.
17. In the iterative approach, we write a,, in terms of a,,_ 1 , then write an-l in terms of an-2 (using the recurrence
relation with n - 1 plugged in for n), and so on. When we reach the end of this procedure, we use the given
initial value of a 0 . This will give us an explicit formula for the answer or it will give us a finite series, which
we then sum to obtain an explicit formula for the answer.
a) We write
an = 3an-l
2
= 3(3a 11 -2) = 3 an-2
2
3
= 3 (3an-3) = 3 an-3
Note that we figured out the last line by following the pattern that had developed in the first few lines.
Therefore the answer is an = 2 · 3" .
Section 2.4
Sequences and Summations
71
b) We write
an= 2 +an-I
= 2+
(2 + an-2)
= (2 + 2) + an-2 = (2 · 2) + an-2
= (2 · 2) + (2 + an-3) = (3 · 2) + an-3
= (n · 2) + an-n = (n · 2) + ao = (n · 2) + 3 = 2n + 3.
Again we figured out the last line by following the pattern that had developed in the first few lines. Therefore
the answer is an = 2n + 3.
c) We write (note that it is more convenient to put the an-I at the end)
an= n +an-I
= n + ((n - 1) + an-2) = (n + (n - 1)) + an-2
= (n + (n - 1)) + ((n - 2) + an-3) = (n + (n - 1) + (n - 2)) + an-3
=
(n + (n - 1) + (n - 2) + · · · + (n - (n -1))) + an-n
= (n + (n - 1) + (n - 2) + · · · + 1) + ao
n(n + 1)
2
_n 2 +n+2
1
+ 2
Therefore the answer is an= (n 2 + n + 2)/2. The formula used to obtain the last line-for the sum of the
first n positive integers~is given in Table 2.
d) We write
an = 3 + 2n +an-I
= 3 + 2n + (3 + 2(n - 1) + an-2) = (2 · 3 + 2n + 2(n - 1)) + an-2
= (2 · 3 + 2n + 2(n - 1)) + (3 + 2(n - 2) + an-3)
= (3 · 3 + 2n + 2(n - 1) + 2(n - 2)) + an-3
= (n · 3 + 2n + 2(n - 1) + 2(n - 2) + · · · + 2(n - (n - 1))) + an-n
= (n · 3 + 2n + 2(n - 1) + 2(n - 2) + · · · + 2 · 1) + a0
= 3n + 2 · n(n + l) + 4 = n 2 + 4n + 4.
2
Therefore the answer is an
integers from Table 2.
e) We write
= n 2 + 4n + 4. Again we used the formula for the sum of the first n positive
an= -1+2an-I
= -1+2(-1+2an-2) = -3 + 4an-2
= -3 + 4(-1+2an-3) = -7 + 8an-3
= -7 + 8(-1+2an-4) = -15 + l6an-4
= -15 + 16(-1+2an-5) = -31+32an-5
= -(2n - 1) + 2nan-n = -2n + 1 + 2n · 1 = 1.
This time it was somewhat harder to figure out the pattern developing in the coefficients, but it became clear
after we carried out the computation far enough. The answer, namely that an = 1 for all n, it is clear in
retrospect, after we found it, since 2 · 1 - 1 = 1.
Chapter 2
72
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
f) We write
an= 1+3an-l
2
= 1+3(1+3an-2) = (1 + 3) + 3 an-2
2
2
3
= (1 + 3) + 3 (1 + 3an-3) = (1 + 3 + 3 ) + 3 an-3
= (l+3+3 2 +···+3n-l)+3nan-n
=
1 + 3 + 32 + · · · + 3n-l + 3n
3n+l - 1
(a geometric series)
3-1
3n+l -1
2
Thus the answer is an = (3n+l - 1) /2.
g) We write
an= nan-l
=
n(n - l)an-2
= n(n - l)(n - 2)an-3 = n(n - l)(n - 2)(n - 3)an-4
=
n(n - l)(n - 2)(n - 3) · · · (n - (n - 1)) an-n
= n(n - l)(n - 2)(n - 3) · · · 1 · ao
= n! · 5 = 5n!.
h) We write
2
= 2n(2(n - l)an-2) = 2 (n(n - l))an-2
= 22(n(n -1)) (2(n - 2)an-3) = 23 (n(n - l)(n - 2))an-3
= 2nn(n - l)(n - 2)(n - 3) · · · (n - (n - l))an-n
= 2nn(n - l)(n - 2)(n - 3) · · · 1 · ao
19. a) Since the number of bacteria triples every hour, the recurrence relation should say that the number of
bacteria after n hours is 3 times the number of bacteria after n - 1 hours. Letting bn denote the number of
bacteria after n hours, this statement translates into the recurrence relation bn = 3bn-l.
b) The given statement is the initial condition b0 = 100 (the number of bacteria at the beginning is the number
of bacteria after no hours have elapsed). We solve the recurrence relation by iteration: bn = 3bn-l = 32bn_ 2 =
· · · = 3nbn-n = 3nbo. Letting n = 10 and knowing that b0 = 100, we see that b10 = 310 · 100 = 5,904,900.
21. a) Let Cn be the number of cars produced in the first n months. The initial condition could be taken to be
c0 = 0 (no cars are made in the first 0 months). Since n cars are made in the nth month, and since Cn- l
cars are made in the first n - 1 months, we see that Cn = Cn-l + n.
b) The number of cars produced in the first year is c 12 . To compute this we will solve the recurrence relation
and initial condition, then plug in n = 12 (alternately, we could just compute the terms c1 , c2 , ... , c 12
Section 2.4
Sequences and Summations
73
directly from the definition). We proceed by iteration exactly as we did in Exercise 17c:
+ Cn-1
= n + ((n -1) + Cn-2) = (n + (n - 1)) + Cn-2
= (n + (n -1)) + ((n - 2) + Cn-3) = (n + (n -1) + (n - 2)) + Cn-3
Cn =
=
n
(n
+ (n -
1)
+ (n -
2)
+ · · · + (n -
(n - 1)))
+ c,,_n
= (n + (n - 1) + (n - 2) + · · · + 1) +co
n(n+l)
n 2 +n
2
+O= - 2 Therefore the number of cars produced in the first year is (12 2
+ 12) /2
=
78.
c) We found the formula in our solution to part (b).
23. Each month our account accrues some interest that must be paid. Since the balance the previous month
is B(k - 1), the amount of interest we owe is (0.07 /12)B(k - 1). After paying this interest, the rest of
the $100 payment we make each month goes toward reducing the principal. Therefore we have B(k) =
B(k- l) - (100- (0.07/12)B(k-1)). This can be simplified to B(k) = (1 + (0.07/12))B(k-1) -100. The
initial condition is B(O) = 5000. If one calculates this as k goes from 0 to 60, we see the balance gradually
decrease and finally become negative when k = 60 (i.e., after five years).
25. In some sense there are no right answers here. The solutions stated are the most appealing patterns that the
author has found.
a) It looks as if we have one 1 and one 0, then two of each, then three of each, and so on, increasing the
number of repetitions by one each time. Thus we need three more 1's (and then four O's) to continue the
sequence.
b) A pattern here is that the positive integers are listed in increasing order, with each even number repeated.
Thus the next three terms are 9, 10, 10.
c) The terms in the odd locations (first, third, fifth, etc.) are just the successive terms in the geometric
sequence that starts with 1 and has ratio 2, and the terms in the even locations are all 0. The nth term is
0 if n is even and is 2 m + ~ for some positive integer m-this would imply that n :::; m 2 + m + ~ and n > m 2 + m, an
impossibility. Therefore {yin}=
{Jn+~},
and we are done.
An alternative solution is provided in the answer section of the text.
29. a) 2+3+4+5+6=20
b) 1-2+4-8+16=11
c) 3+3+···+3=10·3=30
d) This series "telescopes": each term cancels part of the term before it (see also Exercise 35). The sum is
(2 - 1) + (4 - 2) + (8 - 4) + ... + (512 - 256) = -1+512 = 511.
31. We use the formula for the sum of a geometric progression: L~=O arJ = a(rn+I - l)/(r - 1).
a) Here a= 3, r = 2, and n = 8, so the sum is 3(2 9 - 1)/(2 - 1) = 1533.
b) Here a = 1, r = 2, and n = 8. The sum taken over all the values of j from 0 to n is, by the formula,
(2 9 - 1)/(2 - 1) = 511. However, our sum starts at j = 1, so we must subtract out the term that isn't there,
namely 2°. Hence the answer is 511 - 1 = 510.
c) Again we have to subtract the missing terms, so the sum is ((-3) 9
4921 - 1 - (-3) = 4923.
d) 2((-3) 9
-
-
1)/((-3) - 1) - (-3) 0
-
(-3)I =
1)/((-3) - 1) = 9842
33. The easiest way to do these sums, since the number of terms is reasonably small, is just to write out the
summands explicitly. Note that the inside index (j) runs through all of its values for each value of the outside
index ( i ).
a) (1+1) + (1+2) + (1+3) + (2 + 1) + (2 + 2) + (2 + 3) = 21
b) (0 + 3 + 6 + 9) + (2 + 5 + 8 + 11) + (4 + 7 + 10 + 13) = 78
c) (1+1+1) + (2 + 2 + 2) + (3 + 3 + 3) = 18
d) (0 + 0 + 0) + (1 + 2 + 3) + (2 + 4 + 6) = 18
35. If we just write out what the sum means, we see that parts of successive terms cancel, leaving only two terms:
n
L(a1
-
a1 -I) = aI - ao
+ a2 -
aI
+ a3 -
J=I
37. a) We use the hint, where
ak
= k2 :
n
n
k=I
k=I
a2 +···+an-I - an-2 +an - an-I =an - ao
Section 2.5
Cardinality of Sets
75
b) We can use the distributive law to rewrite
terms of the sum we want, S
=
L~=l k:
n
Now we solve for S, obtaining S
2::~= 1 (2k -1) (which we know from part (a) equals n 2 ) in
n
n
= (n 2 + n)/2, which is usually expressed as n(n + 1)/2.
39. This exercise is like Example 23. From Table 2 we know that 2::~~1 k
99 · 100/2
= 4950.
Therefore the desired sum is 20100 - 4950
=
= 200 · 201/2 = 20100,
and 2::~~ 1 k
=
15150.
41. If we write down the first few terms of this sum we notice a pattern. It starts (1 + 1+1) + (2 + 2 + 2 +
2 + 2) + (3 + 3 + 3 + 3 + 3 + 3 + 3) + · · ·. There are three l's, then five 2'ss, then seven 3'ss, and so on;
in general there are (i + 1) 2 - i 2 = 2i + 1 copies of i. So we need to sum i(2i + 1) for an appropriate
range of values for i. We must find this range. It gets a little messy at the end if m is such that the
sequence stops before a complete range of the last value is present. Let n = LvmJ - 1. Then there are
n + 1 blocks, and (n + 1) 2 - 1 is where the next-to-last block ends. The sum of those complete blocks is
I:~=l i(2i + 1) = I:~= 1 2i 2 + i
=
n(n + 1)(2n + 1)/3 + n(n + 1)/2. The remaining terms in our summation
all have the value n + 1 and the number of them present is m - ((n + 1) 2 - 1). Our final answer is therefore
n(n + 1)(2n + 1)/3 + n(n + 1)/2 + (n + l)(m - (n + 1) 2 + 1).
43. a) 0 (anything times 0 is 0)
b) 5 . 6 . 7 . 8 = 1680
c) Each factor is either 1 or -1, so the product is either 1 or -1. To see which it is, we need to determine
how many of the factors are -1. Clearly there are 50 such factors, namely when i
( -1 ) 50 = 1, the product is 1.
=
1, 3, 5, ... , 99. Since
d) 2 . 2 ... 2 = 210 = 1024
45. O! + 1! + 2! + 3! + 4!
SECTION 2.5
= 1+
1+ 1·2 + 1·2 ·3 + 1·2 ·3·4
=
1 + 1 + 2 + 6 + 24
= 34
Cardinality of Sets
Don't feel bad if you find this section confusing. When Cantor started talking about sizes of infinity in the
nineteenth century, many mathematicians thought he made no sense. The basic rule to keep in mind is that
if an infinite set can be given in a list, then it is countable. It is not always easy to find the right list. Various
indirect means are also available for showing that an infinite set is countable, such as showing that the set is
a subset of a countable set, or showing that it is the union of a countable collection of countable sets. Proving
sets uncountable usually requires some sort of diagonal argument, although in fact Cantor's first proof of the
uncountability of the real numbers used a different approach (a nice summary can be found in an article in
the American Mathematical Monthly, 117:7 (2010), 633~637).
76
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
1. a) The negative integers are countably infinite. Each negative integer can be paired with its absolute value
to give the desired one-to-one correspondence: 1 +--> -1, 2 +--> -2, 3 +--> -3, and so on.
b) The even integers are countably infinite. We can list the set of even integers in the order 0, 2, -2, 4, -4,
6, -6, .... and pair them with the positive integers listed in their natural order. Thus 1 +--> 0, 2 +--> 2, 3 +--> -2,
4 +--> 4, and so on. There is no need to give a formula for this correspondence-the discussion given is quite
sufficient; but it is not hard to see that we are pairing the positive integer n with the even integer f(n), where
f(n) = n if n is even and f(n) = 1 - n if n is odd.
c) This set is again countably infinite. We can list its elements in the order 99, 98, 97, .... A formula for a
correspondence with the set of positive integers is given by f (n) = 100 - n. For example, the positive integer
117 is paired with -17.
d) The proof that the set of real numbers between 0 and 1 is not countable (Example 5) can easily be modified
to show that the set of real numbers between 0 and 1/2 is not countable. We need to let the digit d, be
something like 2 if d,, "I 2 and 3 otherwise. The number thus constructed will be a real number between 0
and 1/2 that is not in the list.
e) This set is finite; it has cardinality 999,999.999.
f) This set is countably infinite, exactly as in part (b); the only difference is that there we are looking at
the multiples of 2 and here we are looking at the multiples of 7. The correspondence is given by pairing the
positive integer n with 7n/2 if n is even and -7(n - 1)/2 if n is odd: 0, 7, -7, 14, -14, 21, -21, ....
3. a) The bit strings not containing 0 are just the bit strings consisting of all l's, so this set is {.\ 1, 11, 111,
1111, ... } , where ,\ denotes the empty string (the string of length 0). Thus this set is countably infinite,
where the correspondence matches the positive integer n with the string of n - 1 l's.
b) This is a subset of the set of rational numbers, so it is countable (see Exercise 16). To find a correspondence,
we can just follow the path in Example 4, but omit fractions in the top three rows (as well as continuing to
omit those fractions that are duplicates of rational numbers already encountered).
c) This set is uncountable, as can be shown by applying the diagonal argument of Example 5.
d) This set is uncountable, as can be shown by applying the diagonal argument of Example 5.
5. Suppose m new guests arrive at the fully occupied hotel. If we move the guest in Room 1 to Room m
+
1,
the guest in Room 2 to Room m + 2, and so on, then rooms with numbers from 1 to m become vacant. The
new guests can then occupy these rooms.
7. We can use the guests in the even-numbered rooms to occupy the original rooms, and the guests in the oddnumbered rooms to occupy the rooms in the second building. Specifically, for each positive integer n, put the
guest currently in Room 2n into Room n, and the guest currently in Room 2n - 1 into Room n of the new
building.
9. There is more than one way to do this. Here is one method. First spread out the original guests so that the
gaps between occupied rooms get larger and larger. Specifically, keep the first guest (i.e., the one currently in
Room 1) in Room 1; leave Room 2 vacant; put the second guest (the one currently in Room 2) into Room 3;
leave Rooms 4 and 5 vacant; put the third guest into Room 6; leave Rooms 7, 8, and 9 vacant, and so on.
Have the guests from the first bus fill the first free room in each gap (Rooms 2, 4, 7, 11, and so on). After
this is done, once again the gaps between occupied rooms get larger and larger. (The unoccupied rooms are
now 5. 8, 9, 12. 13. 14, and so on.) So we can repeat the process with the second busload. We continue in
this manner for the countable infinity of busloads. An alternative approach is given in the answer key.
11. In each case, we can make the intersection what we want it to be, and then put additional elements into A
and into B (with no overlap) to make them uncountable.
Section 2.5
Cardinality of Sets
77
a) The simplest solution would be to make An B = 0. So, for example, take A to be the interval (1, 2) of
real numbers, and take B to be the interval (3, 4).
b) Take the example from part (a) and adjoin the positive integers. Thus, let A
=
(1, 2) U z+ and let
B = (3, 4) U z+.
c) Let A= (1,3) and B
=
(2,4).
13. Suppose that A is countable. This means either that A is finite or that there exists a one-to-one correspondence
f from A to z+. In the former case, there is a one-to-one function g from A to a subset of z+ (the range of
g is the first n positive integers, where IAI = n ). In either case, we have met the requirements of Definition 2,
so IAI : : ; jz+j. Conversely, suppose that IAI : : ; jz+j. By definition, this means that there is a one-to-one
function g from A to z+, so A has the same cardinality as a subset of z+ (namely, the range of g ). Now
by Exercise 16 we conclude that A is countable.
15. This is just the contrapositive of Exercise 16 and so follows directly from it. In more detail, suppose that B
were countable, say with elements b1 , b2 , .... Then since As:;; B, we can list the elements of A using the order
in which they appear in this listing of B . Therefore A is countable, contradicting the hypothesis. Thus B is
not countable.
17. Yes. We need to look at this from the other direction, by noting that A= (An B) U (A - B). We are given
that B is countable, so its subset An B is also countable (Exercise 16). If A - B were also countable, then,
since the union of two countable sets is countable (Theorem 1), we would conclude that A is countable. But
we are given that A is not countable. Therefore our assumption that A - B is countable is wrong, and we
conclude that A - B is uncountable. (This is an example of a proof by contradiction.)
19. By what we are given, we know that there are bijections f from A to B and g from C to D. Then we can
define a bijection from Ax C to Bx D by sending (a,c) to (f(a),g(c)). This is clearly one-to-one and onto,
so we have shown that A x C and B x D have the same cardinality.
IAI : : ; IBI is that there is a one-to-one function f : A _, B. Similarly, we are given a
one-to-one function g : B _, C. By Exercise 33 in Section 2.3, the composition go f : A_, C is one-to-one.
Therefore by definition IAI : : ; ICI.
21. The definition of
23. This proof implicitly uses an assumption called the Axiom of Choice. Define a sequence a 1 , a 2 , a 3 , . . . of
elements of A as follows. First, a 1 is any element of A. Once we have selected a 1 , a 2 , a 3 , ... , ak, let ak+l
be any element of A - {a 1, a2, ... , ak}. Such an element must exist because A is infinite. The resulting set
{a 1, a2, a 3 , ... } is the desired countably infinite subset of A.
25. The set of finite strings of characters over a finite alphabet is countably infinite, because we can list these
strings in alphabetical order by length. For example, if the alphabet is {a, b, c}, then our list is ,\, a, b, c,
aa , ab, ac, ba, bb, be, ca, cb, cc, aaa, aab, . . . . (See also Exercise 29.) Therefore the infinite set S can be
identified with an infinite subset of this countable set, which by Exercise 16 is also countably infinite.
27. Since empty sets do not contribute any elements to unions, we can assume that none of the sets in our
given countable collection of countable sets is the empty set. If there are no sets in the collection, then
the union is empty and therefore countable. Otherwise let the countable sets be A 1 , A 2 , . . . . (If there
are only a finite number k of them, then we can still assume that they form an infinite sequence by taking
Ak+l = Ak+ 2 = · · · = A 1 . ) Since each set A, is countable and nonempty, we can list its elements in a sequence
as a, 1 , ai 2 , ... ; again, if the set is finite we can list its elements and then list a,1 repeatedly to assure an
Chapter 2
78
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
infinite sequence. Now we just need a systematic way to put all the elements a 21 into a sequence. We do
this by listing first all the elements a21 in which i + j = 2 (there is only one such pair, (1, 1) ), then all the
elements in which i + j = 3 (there are only two such pairs, (1, 2) and (2, 1) ), and so on; except that we do
not list any element that we have already listed. So, assuming that these elements are distinct, our list starts
a 11 , a12 , a 21 , a 13, a 22 , a 31 , a14, . . . . (If any of these terms duplicates a previous term, then it is simply
omitted.) The result of this process will be either an infinite sequence or a finite sequence containing all the
elements of the union of the sets A 2 • Thus that union is countable.
29. There are only a finite number of bit strings of each finite length, so we can list all the bit strings by listing first
those of length 0, then those of length 1, etc. The listing might be >., 0, 1, 00, 01, 10, 11, 000, 001, .... (Recall
that >. denotes the empty string.) Actually this is a special case of Exercise 27: the set of all bit strings is the
union of a countable number of countable (actually finite) sets, namely the sets of bit strings of length n for
n = 0, 1, 2, ....
31. A little experimentation with this function shows the pattern:
f(l, 1)
f(l, 2)
j(l, 3)
f(l, 4)
f (l, 5)
f(l, 6)
=1
=2
=4
=
f(2, 1)
f(2, 2)
f(2, 3)
f(2, 4)
f (2, 5)
f (2, 6)
7
= 11
=
f (l, 7) =
16
22
=3
= 5
=8
= 12
= 17
= 23
=6
=9
= 13
f(3, 4) = 18
f (3, 5) = 24
= 10
= 14
= 19
f(4, 4) = 25
f(3, 1)
f(3, 2)
f(3, 3)
f(4, 1)
f(4, 2)
f(4, 3)
f(5, 1) = 15
f(5, 2) = 20
f(5, 3) = 26
f(6, 1) = 21
f(6, 2) = 27
We see by looking at the diagonals of this table that the function takes on successive values as m+n increases.
When m + n = 2, f(m, n) = 1. When m + n = 3, f(m, n) takes on the values 2 and 3. When m + n = 4,
f (m, n) takes on the values 4, 5, and 6. And so on. It is clear from the formula that the range of values the
function takes on for a fixed value of m + n, say m + n = x, is (x- 2 ~(x-l) + 1 through (x- 2 ~(x-l) + (x - 1),
since m can assume the values 1, 2, 3, ... , (x - 1) under these conditions, and the first term in the formula is
a fixed positive integer when m + n is fixed. To show that this function is one-to-one and onto, we merely
need to show that the range of values for x + 1 picks up precisely where the range of values for x left off, i.e.,
that f(x - 1, 1) + 1 = j(l, x). We compute:
f(x - 1, 1) + 1 = (x - 2)2(x - 1)
+ (x -
1)
+1=
x2 - ; + 2 = (x
~ l)x + 1 =
f(l, x)
33. It suffices to find one-to-one functions f : (0, 1) --+ [O, 1] and g : [O, 1] --+ (0, 1). We can obviously use the
function f(x) = x in the first case. For the second, we can just compress [O, 1] into, say, [~,~];the increasing
linear function g(x) = (x + 1)/3 will do that. It then follows from the Schroder-Bernstein theorem that
l(O, 1)1
=
l[O, l]I.
35. We can follow the hint or argue as follows, which really amounts to the same thing. (See the answer key for
a proof using bit strings.) Suppose there were such a one-to-one correspondence f from z+ to the power
set of z+ (the set of all subsets of z+ ). Thus, for each x E z+, f(x) is a subset of z+. We will derive a
contradiction by showing that f is not onto; we do this by finding an element not in its range. To this end,
let A = { x I x tJ_ f(x)}. We claim that A is not in the range of f. If it were, then A = f(x 0 ) for some
x 0 E z+ . Let us look at whether x 0 E A or not. On the one hand, if x 0 E A, then by the definition of A,
it must be true that x 0 tJ_ f (x 0 ), which means that x 0 tJ_ A; that is a contradiction. On the other hand, if
if x 0 tJ_ A, then by the definition of A, it must be true that x 0 E f(x 0 ), which means that x 0 EA, again a
contradiction. Therefore no such one-to-one correspondence exists.
Section 2.6
79
Matrices
37. We argued in the solution to Exercise 29 that the set of all strings of symbols from the alphabet {O, 1} is
countable, since there are only a finite number of bit strings of each length. There was nothing special about
the alphabet {O, 1} in that argument. For any finite alphabet (for example, the alphabet consisting of all
upper and lower case letters, numerals, and punctuation and other mathematical marks typically used in a
programming language), there are only a finite number of strings of length 1 (namely the number of symbols
in the alphabet), only a finite number of strings of length 2 (namely, the square of this number), and so
on. Therefore, using the result of Exercise 27, we conclude that there are only countably many strings from
any given finite alphabet. Now the set of all computer programs in a particular language is just a subset of
the set of all strings over that alphabet (some strings are meaningless jumbles of symbols that are not valid
programs), so by Exercise 16, this set, too, is countable.
39. In Exercise 37 we saw that there are only a countable number of computer programs, so there are only a
countable number of computable functions. In Exercise 38 we saw that there are an uncountable number
of functions. Hence not all functions are computable. Indeed, in some sense, since uncountable sets are so
much bigger than countable sets, almost all functions are not computable! This is not really so surprising; in
real life we deal with only a small handful of useful functions, and these are computable. Note that this is
a nonconstructive proof-we have not exhibited even one noncomputable function, merely argued that they
have to exist. Actually finding one is much harder, but it can be done. For example, the following function
is not computable. Let T be the function from the set of positive integers to {O, 1} defined by letting T(n)
be 0 if the number 0 is in the range of the function computed by the nth computer program (where we list
them in alphabetical order by length) and letting T(n) = 1 otherwise.
SECTION 2.6 Matrices
In addition to routine exercises with matrix calculations, there are several exercises here asking for proofs
of various properties of matrix operations. In most cases the proofs follow immediately from the definitions
of the matrix operations and properties of operations on the set from which the entries in the matrices are
drawn. Also, the important notion of the (multiplicative) inverse of a matrix is examined in Exercises 18~21.
Keep in mind that some matrix operations are performed "entrywise," whereas others operate on whole rows
or columns at a time. Exercise 29 foreshadows material in Section 9.4.
1. a) Since A has 3 rows and 4 columns, its size is 3 x 4.
b) The thfrd column of A is the 3 x 1 matdx [
i] ·
c) The second row of A is the 1 x 4 matrix [2 0 4 6] .
d) This is the element in the third row, second column, namely 1.
e) The transpose of A is the 4 x 3 matrix [ :
~
!]·
3 6 7
3. a) We use the definition of matrix multiplication to obtain the four entries in the product AB. The (1, l)th
entry is the sum anb 11 + ai2b 21 = 2 · 0 + 1·1 = 1. Similarly, the (1, 2)th entry is the sum anb12 + a12b22 =
2·4+1·3 = 11; (2, l)th entry is the sum a 21 b11 +a 22 b21 =3·0 + 2 · 1 = 2; and (2,2)th entry is the sum
a 21 bi2 + a 22 b22 = 3 · 4 + 2 · 3 = 18. Therefore the answer is [;
i~]
.
b) The calculation is similar. Again, to get the (i,j)th entry of the product, we need to add up all the
products a,kbk1 . You can visualize "lifting" the ith row from the first factor (A) and placing it on top of the
Chapter 2
80
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
lh column from the second factor ( B), multiplying the pairs of numbers that lie on top of each other, and
taking the sum. Here we have
rn
-2 -1]
Yi [;
2
0
=
[1·3+(-1)·1
0·3+1·1
2·3+3·1
1·(-2)+(-1)·0 1·(-1)+(-1)·2]
0. (-2) + 1. 0
0·(-1)+1·2
2. (-2) + 3. 0
2·(-1)+3·2
U~: Tl
c) The calculation is similar to the previous parts:
[
34 -3]
-1 [-1 3 2 -2]
0
-1
-2
5
0
-1
4
-3
[4·(-l)+(-3)·0
3·(-1)+(-1)·0
0·(-1)+(-2)·0
(-1)·(-1)+5·0
15 -4
-3
-3 10
2
l
2 -8
0
1 -8 18
[-4
-~3
4·3+(-3)·(-1)
3·3+(-1)·(-1)
0·3+(-2)·(-1)
(-1)·3+5·(-1)
l
4. 2 + (-3). 4 4. (-2) + (-3) . (-3)
3·2+(-1)·4 3. (-2) + (-1). (-3)
0. 2 + (-2). 4 0. (-2) + (-2) . (-3)
(-1)·2+5·4 (-1). (-2) + 5. (-3)
l
5. First we need to observe that A = [ai1 ] must be a 2 x 2 matrix; it must have two rows since the matrix it
is being multiplied by on the left has two columns, and it must have two columns since the answer obtained
has two columns. If we write out what the matrix multiplication means, then we obtain the following system
of linear equations:
2a11 + 3a21 = 3
2a12 + 3a22 = 0
la11 + 4a21 = 1
la12 + 4a22 = 2
Solving these equations by elimination of variables (or other means-it's really two systems of two equations
each in two unknowns), we obtain a 11 = 9/5, a12 = -6/5, a21 = -1/5, a22 = 4/5. As a check we compute
that, indeed,
2 3] [ 9/5
[1 4
-1/5
-6/5] = [3 OJ
4/5
1 2 .
7. Since the (i,j)th entry of 0 +A is the sum of the (i,j)th entry of 0 (namely 0) and the (i,j)th entry of A,
this entry is the same as the (i,j)th entry of A. Therefore by the definition of matrix equality, 0 +A= A.
A similar argument shows that A + 0 = A.
9. We simply look at the (i,j)th entries of each side. The (i,j)th entry of the left-hand side is ai1 + (bi 1 + ci1 ).
The (i, j)th entry of the right-hand side is (aij + bi1 ) + c,1 . By the associativity law for real number addition,
these are equal. The conclusion follows.
11. In order for AB to be defined, the number of columns of A
BA to be defined, the number of columns of B must equal
integers m and n, it must be the case that A is an m x n
to say this is to say that A must have the same size as Bt
must equal the number of rows of B. In order for
the number of rows of A. Thus for some positive
matrix and B is an n x m matrix. Another way
(and/or vice versa).
Section 2.6
81
Matrices
13. Let us begin with the left-hand side and find its (i, j)th entry. First we need to find the entries of BC. By
k
definition, the (q, j)th entry of BC is
E
bqrCr1 . (See Section 2.4 for the meaning of summation notation.
r=l
This symbolism is a shorthand way of writing bq1C1j + bq2c21 + · · · + bqkCkJ .) Therefore the (i,j)th entry of
k
p
A(BC) is
E
a,q (
q=l
E
bqrCrj).
r=l
By distributing multiplication over addition (for real numbers), we can move
p
the term
inside the inner summation, to obtain
k
E E
aiqbqrcr1 . (We are also implicitly using associativity
q=l r=l
of multiplication of real numbers here, to avoid putting parentheses in the product aiqbqrCrj . )
a,q
k
A similar analysis with the right-hand side shows that the (i, j)th entry there is equal to
k
p
I: (I:
r=l q=l
a,qbqr )cr1
p
E E a,qbqrCrj.
Now by the commutativity of addition, the order of summation (whether we sum over
r=l q=l
r first and then q, or over q first and then r) does not matter, so these two expressions are equal, and the
proof is complete.
15. Let us begin by computing An for the first few values of n.
A1 =
[~
3] A4 [10 4]1 ' A5 [10 5]1 .
2]
1] , A
2
3
1
=[1o 1 ' A
=[1
o1'
It seems clear from this pattern, then, that An
-- [01
nl]
=
=
. (A proof of this fact could be given using
mathematical induction, discussed in Section 5.1.)
17. a) The (i,j)th entry of (A+ B)t is the (j, i)th entry of A+ B, namely a1 , + b1 ,. On the other hand, the
(i,j)th entry of At+ Bt is the sum of the (i,j)th entries of At and Bt, which are the (j, i)th entries of A
and B, again aji + b1 i. Hence (A+ B)t =At+ Bt.
n
b) The (i,j)th entry of (AB)t is the (j,i)th entry of AB, namely
E
a 1 kbki.
k=l
of summation notation. This symbolism is a shorthand way of writing
(SeeSection2.4forthemeaning
baj 1 bii
+
n
a 12 b 2 ,
+ ··· +
a 1 nbni .)
On
E bk,aJk (since the (i,k)th entry of Bt is bki and the (k,j)th
k=l
entry of At is ajk). By the commutativity of multiplication of real numbers, these two values are the same,
so the matrices are equal.
the other hand, the (i,j)th entry of BtAt is
19. All we have to do is form the products AA- 1 and A- 1A, using the purported A- 1 , and see that both of
them are the 2 x 2 identity matrix. It is easy to see that the upper left and lower right entries in each case
are (ad - be)/ (ad - be) = 1, and the upper right and lower left entries are all 0.
r
= I, where I is the n x n identity matrix. Since matrix multiplication is
associative, we can write this product as
21. We must show that An (A -
1
An ((A- 1r) = A(A ... (A(AA- 1)A- 1) ... A- 1)A-l.
By dropping each AA- 1 = I from the center as it is obtained, this product reduces to I. Similarly
((A- 1)n)An =I. Therefore by definition (An)- 1 = (A- 1
(A more formal proof requires mathematical induction; see Section 5.1.)
r.
23. By definition, the (i,j)th entry of A+ At is ai1 + a 1 ,. Similarly, the (j, i)th entry of A+ At is a 1 i + a, 1 .
By the commutativity of addition, these are equal, so A+ At is symmetric by the definition of symmetric
matrices.
Chapter 2
82
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
25. Using the idea in Exercise 24, we see that the given system can be expressed as AX = B, where A is the
coefficient matrix, X is an n x 1 matrix with Xi the entry in its ith row, and B is the n x 1 matrix of
right-hand sides. Specifically we have
~~~ :nd[ t r infrs~t]-. ~:::f:: can find x
1
:::e
-1
-1
But Exercise 18 tells us
simply by computing A -1 B.
3
We should plug in x 1 = 1, x 2 = -1, and x 3 = -2 to see that these do indeed form the solution.
27. These routine exercises simply require application of the appropriate definitions. Parts (a) and (b) are entrywise operations, whereas the operation 8 in part (c) is similar to matrix multiplication (the (i,j)th entry of
A 8 B depends on the ith row of A and the lh column of B ).
a) AV B
l
~ ~ ~ i
[
29. Note that Al 2l means A 8 A, and Al 3l means A 8 A 8 A. We just apply the definition.
c)
A v A l2l V A l3l
=
[ 1~
o~ o~l
31. These are immediate from the commutativity of the corresponding logical operations on variables.
a) AV B = [a, 1 V bi1 ] = [bi 1 V a,1 ] = B VA
b) B /\A= [b, 1 /\ a, 1 ] = [ai 1 /\ b, 1 ] =A/\ B
33. These are immediate from the distributivity of the corresponding logical operations on variables.
a) AV (B /\ C) = [ai 1 V (b, 1
b) A/\ (B V C) = [ai 1
/\
/\
ci1 )]
(bi 1 V ci1 )]
= [(a, 1 V bi1 ) /\ (a, 1 V ci1 )] =(AV B) /\(AV C)
= [(ai1 /\ bi1 ) V (ai 1 /\ ci1 )] =(A/\ B) V (A/\ C)
35. The proof is identical to the proof in Exercise 13, except that real number multiplication is replaced by /\,
and real number addition is replaced by V. Briefly, in symbols, A 8 (B 8 C)
p
k
[V V aiqAbqrAcr1 ]
q=l r=l
k
=
p
k
k
q=l
r=l
p
[V V aiq/\bqr/\cr1 ] = [V (V
r=l q=l
p
= [ V aiq /\ ( V bqr /\ Crj)] =
r=l q=l
a,q/\bqr) /\cr1]
= (A8B)8C.
83
Review Questions
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 2
1. Seep. 119. To prove that A is a subset of B we need to show that an arbitrarily chosen element x of A must
also be an element of B .
2. The empty set is the set with no elements. It satisfies the definition of being a subset of every set vacuously.
3. a) See p. 121.
b) See p. 128.
4. a) See p. 121.
b) always
5. a) See pp. 127 and 128 and the preamble to Exercise 32 in Section 2.2.
b) union: integers that are odd or positive; intersection: odd positive integers; difference: even positive
integers; symmetric difference: even positive integers together with odd negative integers
6. a) A= B =(A<:;;; B /\ B <:;;;A)= Vx(x EA+-+ x EB)
b) See pp. 129-132.
c) An B n C =An (Bu C) =(An B) u (An C) =(A- B) u (A - C); use Venn diagrams
7. Underlying each set identity is a logical equivalence. See, for instance, Example 11 in Section 2.2.
8. a) See p. 139.
b) Z, Z,
9. a) See p. 141.
f) f(n) = 42548
10. a) Seep. 145:
z+ = N - {O}
b) See p. 143.
1- 1 (b) =a= f(a)
=
c) f(n) = n
d) f(n)
=
2n
e) f(n) = fn/21
b
b) when it is one-to-one and onto
c) yes-itself
11. a) See p. 149.
b) integers
12. Hint: subtract 5 from each term and look at the resulting sequence.
13. The formula depends on the initial condition, namely the value of a0 . After that, each term is 5 less than the
preceding term, so an = ao - 5n.
14. See p. 164.
15. Set up a one-to-one correspondence between the set of positive integers and the set of all odd integers, such
as 1 +-+ 1, 2 +-+ -1, 3 +-+ 3, 4 +-+ -3, 5 +-+ 5, 6 +-+ -5, and so on.
16. See Example 5 in Section 2.5.
17. See page 179. For AB to be defined, the number of columns of A must equal the number of rows of B.
18. See Example 4 in Section 2.6.
84
Chapter 2
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
SUPPLEMENTARY EXERCISES FOR CHAPTER 2
1. a)
A=
the set of words that are not in A
b) A n B = the set of words that are in both A and B
c) A - B = the set of words that are in A but not B
d) An B =(AU B) = the set of words that are in neither A nor B
e) A EBB= the set of words that are in A or B but not both (can also be written as (A - B) U (B - A) or
as (AUB)-(AnB))
3. Yes. We must show that every element of A is also an element of B. So suppose a is an arbitrary element
of A. Then {a} is a subset of A, so it is an element of the power set of A. Since the power set of A is a
subset of the power set of B, it follows that {a} is an element of the power set of B, which means that {a}
is a subset of B. But this means that the element of {a}, namely a, is an element of B, as desired.
5. We will show that each side is a subset of the other. First suppose x E A - (A - B). Then x E A and
x tf- A - B. Now the only way for x not to be in A - B, given that it is in A, is for it to be in B. Thus we
have that x is in both A and B , so x E A n B. For the other direction, let x E A n B . Then x E A and
x E B. It follows that x tf- A - B, and so x is in A - (A - B).
7. We need only provide a counterexample to show that (A - B) - C is not necessarily equal to A - (B - C).
Let A = C = {l}, and let B = 0. Then (A - B) - C = ({1} - 0) - {l} = {l} - {l} = 0, whereas
A- (B-C) = {1}- (0- {1}) = {1}-0 = {1}.
9. This is not necessarily true. For a counterexample, let A = B = {1, 2}, let C = 0, and let D = {1}. Then
(A- B) - (C - D) = 0 - 0 = 0, but (A - C) - (B - D) = {1, 2} - {2} = {1}.
11. a) Since 0 ~ An B ~ A~ AU B ~ U, we have the order 101 ::; IA n Bl ::; IAI ::; IA U Bl ::; IUI.
~ AEBB ~ AUB. Also recall that IAUBI = IAI + IBl-IAnBI, so that IAUBI is always less
than or equal to IAI + IBI. Putting this all together, we have 101 ::; IA- Bl ::; IA EB Bl ::; IA U Bl ::; IAI + IBI.
b) Note that A-B
13. a) Yes, f is one-to-one, since each element of the domain {1, 2, 3, 4} is sent by f to a different element of the
codomain. No, g is not one-to-one, since g sends the two different elements a and d of the domain to the
same element, 2.
b) Yes, f is onto, since every element in the codomain {a, b, c, d} is the image under f of some element in
the domain {1,2,3,4}. In other words, the range off is the entire codomain. No, g is not onto, since the
element 4 in the codomain is not in the range of g (is not the image under g of any element of the domain
{a, b, c, d} ).
c) Certainly f has an inverse, since it is one-to-one and onto. Its inverse is the function from {a, b, c, d} to
{1,2,3,4} that sends a to 3, sends b to 4, sends c to 2, and sends d to 1. (Each element in {a,b,c,d} gets
sent by 1- 1 to the element in {l,2,3,4} that gets sent to it by f .) Since g is not one-to-one and onto, it
has no inverse.
15. If f is one-to-one, then f provides a bijection between S and f (S), so they have the same cardinality by
definition. If f is not one-to-one, then there exist elements x and y in S such that f(x) = f(y). Let
S = {x,y}. Then ISi = 2 but lf(S)I = 1. Note that we do not need the hypothesis that A and B are
finite.
Supplementary Exercises
85
17. The key is to look at sets with just one element. On these sets, the induced functions act just like the original
functions. So let x be an arbitrary element of A. Then {x} E P(A), and S1({x}) = {l(y) I y E {x}} =
{f(x)}. By the same reasoning, S9 ({x}) = {g(x)}. Since S1 = S9 , we can conclude that {f(x)} = {g(x)},
and so necessarily l(x) = g(x).
19. This is certainly true if either x or y is an integer, since then this equation is equivalent to the identity (4a)
in Table 1 of Section 2.3. Otherwise, write x and y in terms of their integer and fractional parts: x = n + E
and y = m + 8, where n = Lx J , 0 < E < 1, m = LY J , and 0 < 6 < l. If 8 + E < 1, then the equation is true,
since both sides equal m + n; if 8 + E 2: 1, then the equation is false, since the left-hand side equals m + n,
but the right-hand side equals m + n + 1. In summary, the equation is true if and only if either at least one
of x and y is an integer or the sum of the fractional parts of x and y is less than 1. (Note that the second
condition in the disjunction subsumes the first.)
+ f and y = m + fj, where n = lx J ,
0 ::; E < 1, m = lY J , and 0 ::; 8 < 1. If 8 = E = 0, then both sides equal n + m. If E = 0 but 8 > 0, then
the left-hand side equals n + m + 1, but the right-hand side equals n + m. If E > 0, then the right-hand side
equals n + m + 1, so the two sides will be equal if and only if E + fj::; 1 (otherwise the left-hand side would
be n + m + 2 ). In summary, the equation is true if and only if either both x and y are integers, or x is not
an integer but the sum of the fractional parts of x and y is less than or equal to 1 .
21. Write x and y in terms of their integer and fractional parts: x = n
23. If xis an integer, then clearly LxJ + Lm-xJ = x+m-x = m. Otherwise, write x in terms of its integer and
fractional parts: x = n+E, where n = LxJ and 0 < E < l. In this case LxJ + Lm-xJ = Ln+cJ + Lm-n-Ej =
n + m - n - 1 = m - 1 , because we had to round m - n - E down to the next smaller integer.
i.
25. Write n = 2k + 1 for some integer k. Then n 2 = 4k 2 + 4k + 1, so n 2 /4 = k 2 + k +
Therefore In 2 /41
2
2
2
2
2
k + k + 1 . But we also have (n + 3) / 4 = (4k + 4k + 1 + 3) / 4 = (4k + 4k + 4) / 4 = k + k + 1 .
27. Let us write x = n + (r / m) + E, where n is an integer, r is an nonnegative integer less than m, and E is
a real number with 0::; E < 1/m. In other words, we are peeling off the integer part of x (i.e., n = lxJ)
and the whole multiples of 1/m beyond that. Then the left-hand side is lnm + r + mEj =nm+ r. On
the right-hand side, the terms LxJ through Lx + (m - r - 1)/mJ are all just n, and the remaining terms,
if any, from Lx + (m - r)/mJ through Lx + (m - 1)/mJ, are all n + l. Therefore the right-hand side is
(m - r)n + r(n + 1) =nm+ r as well.
29. This product telescopes. The numerator in the fraction for k cancels the denominator in the fraction for
k + 1. So all that remains of the product is the numerator for k = 100 and the denominator for k = 1, namely
101/1=101.
31. There is no good way to determine a nice rule for this kind of problem. One just has to look at the sequence
and see what seems to be happening. In this sequence, we notice that 10 = 2 · 5, 39 = 3 · 13, 172 = 4 · 43,
and 885 = 5 · 177. We then also notice that 3 = 1 · 3 for the second and third terms. So each odd-indexed
term (assuming that we call the first term a 1 ) comes from the term before it, by multiplying by successively
larger integers. In symbols, this says that a 2 n+l = n·a 2 n for all n > 0. Then we notice that the even-indexed
terms are obtained in a similar way by adding: a 2 n = n + a 2 n- 1 for all n > 0. So the next four terms are
a13
= 6 · 891 = 5346,
a14
= 7 + 5346 = 5353,
a 15
= 7 · 5353 = 37471. and
a 14
= 8 + 37471=37479.
33. If such a function 1 exists, then S equals the union of a countable number of countable sets, namely
1- 1 (2) U · · ·. It follows from Exercise 27 in Section 2.5 that S is countable.
1- 1 ( 1) U
Chapter 2
86
Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
35. Because there is a one-to-one correspondence between R and the open interval (0, 1) (given by f(x) =
2 arctan(x)/7r, it suffices to shows that l(O, 1) x (O. l)I = l(O, l)j. We use the Schroder-Bernstein theorem; it
suffices to find injective functions f from (0,1) to (0,1) x (0,1) and g from (0,1) x (0,1) to (0,1). For f
we can just use f(x) = (x, ~);it is clearly injective. For g we follow the hint. Suppose (x, y) E (0, 1) x (0, 1),
and represent x and y with their decimal expansions, never choosing the expansion of any number that ends
in an infinite string of 9s (we can avoid that by having all finite decimals end in an infinite string of Os). Let
x = O.x 1x2x3 ... and y = O.y1Y2Y3 ... be these expansions. Let g(x, y) be the decimal expansion obtained
by interweaving these two strings, namely O.X1Y1X2Y2X3y3 .... Because we can recover x and y from g(x, y)
(namely by taking every other digit starting with the first or second decimal digit, respectively), it follows
that g is one-to-one, and our proof is complete.
37. Let us begin by computing An for the first few values of n.
A1 = [ 0
-1
l]
0
'
A2=
[-1
0
0]
-1
Since A 4 = I, the pattern will repeat from here: A 5
Thus for all n 2 0 we have
'
=
A3= [O
1
-1]
[l OJ
0 .A=o1·
4
A 4 A = IA = A, A 6
-1 ]
0
,
=
= A 3, and
A2 , A7
A 4n +4 = [
1 0]
0
1
so on.
.
39. (The notation cl means the identity matrix I with each entry multiplied by the real number c; thus this
matrix consists of e's along the main diagonal and O's elsewhere.) Let A
= [: ~].
We will determine
what these entries have to be by using the fact that AB = BA for a few judiciously chosen matrices B. First
let B
0
=w
= [~
~] . Then
and u
= x.
AB = [ ~ : ] , and BA = [ ~
Next choose B
matrix A must be in the form [ ~
= [~
~] . Since these two must be equal, we know that
~] . Then we get [ ~ ~]
~] , which is just
u
[
~ ~] , whence v = 0.
Therefore the
times the identity matrix, as desired.
41. a) The (i,j)th entry of A 8 0 is by definition the Boolean sum (V) of some Boolean products (A) of the
form a,k A 0. Since the latter always equals 0, every entry is 0, so A 8 0 = 0. Similarly 0 8 A consists of
entries that are all 0, so it, too, equals 0.
b) Since V operates entrywise, the statements that A V 0 = A and 0 V A = A follow from the facts that
a,J V 0 = a,1 and 0 V a,j = a,J .
c) Since A operates entrywise, the statements that A A 0 = 0 and 0 A A = 0 follow from the facts that
ai1 A 0 = 0 and 0 A a,1 = 0.
Writing Projects
87
WRITING PROJECTS FOR CHAPTER 2
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. A classic source here is [Wil]. It gives a very readable account of many philosophical issues in the foundations
of mathematics, including the topic for this essay.
2. Our list of references mentions several history of mathematics books, such as [Bo4] and [Ev3]. You should
also browse the shelves in your library, around QA 21.
3. Go to the Encyclopedia's website, oeis. org.
4. A Web search should turn up some useful references here, including an article in Science News Online. It gets
its name from the fact that a graph describing it looks like the output of an electrocardiogram.
5. A Web search for this phrase will turn up much information.
6. A classic source here is [Wil]. It gives a very readable account of many mathematical and philosophical issues
in the foundations of mathematics, including the topic for this essay. Of course a Web search will turn up lots
of useful material, as well. Your essay should delve into the generalized continuum hypothesis for the higher
orders of infinity: 2Nn = ~n+l .
88
Chapter 3
Algorithms
CHAPTER3
Algorithms
SECTION 3.1
Algorithms
J\fany of the exercises here are actually miniature programming assignments. Since this is not a book on
programming. we have glossed over some of the finer points. For example, there are (at least) two ways to
pass variables to procedures-by value and by reference. In the former case the original values of the arguments
are not changed. In the latter case they are. In most cases we will assume that arguments are passed by
reference. None of these exercises are tricky; they just give the reader a chance to become familiar with
algorithms written in pseudocode. The reader should refer to Appendix 3 for more details of the pseudocode
being used here.
1. Initially max is set equal to the first element of the list, namely 1. The for loop then begins, with i set equal
to 2. Immediately i (namely 2) is compared to n, which equals 10 for this sequence (the entire input is
known to the computer, including the value of n ). Since 2 < 10. the statement in the loop is executed. This
is an if. .. then statement, so first the comparison in the if part is made: max (which equals 1) is compared
to a, = a 2 = 8. Since the condition is true, namely 1 < 8, the then part of the statement is executed, so
max is assigned the value 8.
The only statement in the for loop has now been executed, so the loop variable i is incremented (from 2
to 3 ), and we repeat the process. First we check again to verify that i is still less than n (namely 3 < 10 ),
and then we execute the if. .. then statement in the body of the loop. This time, too, the condition is satisfied,
since max = 8 is less than a 3 = 12. Therefore the assignment statement max := a, is executed, and max
receives the value 12.
Next the loop variable is incremented again, so that now i = 4. After a comparison to determine that
4 < 10, the if. .. then statement is executed. This time the condition fails, since max = 12 is not less than
a4 = 9. Therefore the then part of the statement is not executed. Having finished with this pass through
the loop, we increment i again, to 5. This pass through the loop, as well as the next pass through, behave
exactly as the previous pass, since the condition max < a, continues to fail. On the sixth pass through the
loop, however, with i = 7, we find again that max < a,, namely 12 < 14. Therefore max is assigned the
value 14.
After three more uneventful passes through the loop (with i = 8, 9, and 10), we finally increment i to
11. At this point, when the comparison of i with n is made, we find that i is no longer less than or equal
to n, so no further passes through the loop are made. Instead, control passes beyond the loop. In this case
there are no statements beyond the loop, so execution halts. Note that when execution halts, max has the
value 14 (which is the correct maximum of the list), and i has the value 11. (Actually in many programming
languages, the value of i after the loop has terminated in this way is undefined.)
3. We will call the procedure AddEmUp. Its input is a list of integers, just as was the case for Algorithm l.
Indeed, we can just mimic the structure of Algorithm l. We assume that the list is not empty (an assumption
made in Algorithm 1 as well).
Section 3.1
89
Algorithms
procedure AddEmUp(a 1 , a 2 , ... , an: integers)
sum:= a1
for i := 2 to n
sum:= sum+ a,
return sum
{ sum is the sum of all the elements in the list}
5. We need to go through the list and find cases when one element is equal to the following element. However,
in order to avoid listing the values that occur more than once more than once, we need to skip over repeated
duplicates after we have found one duplicate. The following algorithm will do it. (If we wanted to "return"
the answer, we would form c 1 , c2 , ... , ck into a list and return that list.)
procedure duplicates(a 1 , a 2 , ... , an: integers in nondecreasing order)
k := 0 {this counts the duplicates}
j := 2
while j Sn
if a1 = a1 _ 1 then
k := k+ 1
Ck
:= a1
while j S n and a1 =
j := j
Ck
+1
j := j + 1
{ c1, c2, ... , Ck is the desired list}
7. We need to go through the list and record the index of the last even integer seen.
procedure last even location(a 1 , a2, ... , an: integers)
k := 0
for i := 1 ton
if a, is even then k := i
return k { k is the desired location (or 0 if there are no evens)}
9. We just need to look at the list forward and backward simultaneously, going about half-way through it.
procedure palindrome check(a 1 a2 ... an : string)
answer:= true
fori:=lto ln/2J
if a,# an+I-i then answer:= false
return answer { answer is true if and only if string is a palindrome}
11. We cannot simply write x := y followed by y := x, because then the two variables will have the same value,
and the original value of x will be lost. Thus there is no way to accomplish this task with just two assignment
statements. Three are necessary, and sufficient, as the following code shows. The idea is that we need to save
temporarily the original value of x.
temp:= x
x := y
y :=temp
13. We will not give these answers in quite the detail we used in Exercise 1.
a) Note that n = 8 and x = 9. Initially i is set equal to 1 . The while loop is executed as long as i S 8
and the ith element of the list is not equal to 9. Thus on the first pass we check that 1 S 8 and that 9 # 1
(since a 1 = 1 ), and therefore perform the statement i := i + 1. At this point i = 2. We check that 2 S 8 and
9 # 3, and therefore again increment i, this time to 3. This process continues until i = 7. At that point the
condition "i S 8 and 9 # a," is false, since a1 = 9. Therefore the body of the loop is not executed (so i is
still equal to 7), and control passes beyond the loop.
Chapter 3
90
Algorithms
The next statement is the if... then statement. The condition is satisfied, since 7 ::::; 8, so the statement
location := i is executed, and location receives the value 7. The else clause is not executed. This completes
the procedure, so location has the correct value, namely 7, which indicates the location of the element x
(namely 9) in the list: 9 is the seventh element.
b) Initially i is set equal to 1 and j is set equal to 8. Since i < j at this point, the steps of the while
loop are executed. First m is set equal to l(l + 8)/2J = 4. Then since x (which equals 9) is greater than
a 4 (which equals 5), the statement i := m + 1 is executed, so i now has the value 5. At this point the first
iteration through the loop is finished, and the search has been narrowed to the sequence a 5 , ... , a 8 .
In the next pass through the loop (there is another pass since i
< j is still true), m becomes l(5 + 8)/2J
6. Since again x > am , we reset i to be m + 1 , which is 7. The loop is now repeated with i = 7 and j
This time m becomes 7, so the test x > am (i.e., 9 > 9) fails; thus j := m is executed, so now j = 7.
=
= 8.
At this point i f:. j, so there are no more iterations of the loop. Instead control passes to the statement
beyond the loop. Since the condition x = a, is true, location is set to 7, as it should be, and the algorithm
is finished.
15. We need to find where x goes, then slide the rest of the list down to make room for x, then put x into the
space created. In the procedure that follows, we employ the trick of temporarily tacking x + 1 onto the end of
the list, so that the while loop will always terminate. Also note that the indexing in the for loop is slightly
tricky since we need to work from the end of the list toward the front.
procedure insert(x, ai, a2, ... , an : integers)
{the list is in order: a 1 ::::; a2 ::::; · · · ::::; an }
an+l := X + 1
i := 1
while x >a,
i := i + 1 {the loop ends when i is the index for x}
for j := 0 to n - i {shove the rest of the list to the right}
an- 1 +1 := an-J
a,:= x
{ x has been inserted into the correct spot in the list, now of length n
+ 1}
17. This algorithm is similar to max, except that we need to keep track of the location of the maximum value, as
well as the maximum value itself. Note that we need a strict inequality in the test max a2 then interchange a 1 and a 2
if a2 > a3 then interchange a2 and a3
if a 1 > a2 then interchange a 1 and a 2
{the first three elements are now in nondecreasing order}
23. For notation, assume that f : A ___, B, where A is the set consisting of the distinct integers a 1 , a 2 , ... , an,
and B is the set consisting of the distinct integers b1 , b2 , ... , bm . All n + m + 1 of these entities (the elements
of A, the elements of B, and the function f) are the input to the algorithm. We set up an array called hit
(indexed by the integers) to keep track of which elements of B are the images of elements of A; thus hit(b,)
equals 0 until we find an aJ such that f (a1 ) = b,, at which time we set hit(b,) equal to 1. Simultaneously
we keep track of how many hits we have made (i.e., how many times we changed some hit(b,) from 0 to 1 ).
If at the end we have made m hits, then f is onto; otherwise it is not. Note that we record the output as a
logical value assigned to the variable that has the name of the procedure. This is a common practice in some
programming languages.
Chapter 3
92
Algorithms
procedure onto(!: function, a1, a2, ... , an, b1, b2, ... , bm : integers)
for i := 1 tom
hzt(b,) := 0 {no one has been hit yet}
count := 0 {there have been no hits yet}
for j := 1 ton
if hit(f(a1 )) = 0 then {a new hit!}
hit(f(a1 )) := 1
count := count + 1
if count = m then return true else return false
{ f is onto if and only if there have been m hits}
25. This algorithm is straightforward.
procedure count ones(a1a 2 ... an : bit string)
count:= 0 {no l's yet}
for i := 1 to n
if a, = 1 then count := count + 1
return count { count contains the number of l's}
27. We start with the pseudocode for binary search given in the text and modify it. In particular, we need to
compute two middle subscripts (one third of the way through the list and two thirds of the way through) and
compare x with two elements in the list. Furthermore, we need special handling of the case when there are
two elements left to be considered. The following pseudocode is reasonably straightforward.
procedure ternary search(x: integer, a 1 , a 2 , ... , an : increasing integers)
i := 1
j := n
while i < j - 1
l := l(i + j)/3J
u := l2(i + j)/3J
if x > au then i := u + 1
else if x > az then
i := l + 1
j := u
else j := l
if x =a, then location:= i
else if x = a1 then location := j
else location := 0
return location
{ location is the subscript of the term equal to x ( 0 if not found)}
29. The following algorithm will find the first mode in the sequence. At each point in the execution of this
algorithm, modecount is the number of occurrences of the element found to occur most often so far (which is
called mode). Whenever a more frequently occurring element is found (the main inner loop), modecount and
mode are updated.
Section 3.1
Algorithms
93
procedure find a mode(a 1 , a 2 , •.. , an: nondecreasing integers)
modecount := 0
i := 1
while i::; n
value:= a,
count:= 1
while i ::; n and a, = value
count := count+ 1
i := i
if count
+1
> modecount then
modecount := count
mode := value
return mode
{ mode is the first value occurring most often, namely modecount times}
31. The following algorithm goes through the terms of the sequence one by one, and, for each term, compares it
to all previous terms. If it finds a match, then it stores the subscript of that term in location and terminates
the search. If no match is ever found, then location is set to 0.
procedure find duplicate(a 1 , a 2 , ... , an : integers)
location := 0 {no match found yet}
i := 2
while i ::; n and location = 0
j := 1
while j < i and location= 0
if a, = a1 then location := i
else j := j + 1
i := i + 1
return location
{ location is the subscript of the first value that repeats a previous value in the sequence
and is 0 if there is no such value}
33. The following algorithm goes through the terms of the sequence one by one, and, for each term, checks whether
it is less than the immediately preceding term. If it finds such a term, then it stores the subscript of that term
in location and terminates the search. If no term satisfies this condition, then location is set to 0.
procedure find decrease(a 1 , a 2 , ... , an : positive integers)
location := 0 {no match found yet}
i := 2
while i ::; n and location = 0
if a, < a,_ 1 then location := i
else i := i + 1
return location
{ location is the subscript of the first value that is less than the immediately preceding one
and is 0 if there is no such value}
35. There are four passes through the list. On the first pass, the 3 and the 1 are interchanged first, then the
next two comparisons produce no interchanges, and finally the last comparison results in the interchange of
the 7 and the 4. Thus after one pass the list reads 1, 3, 5, 4, 7. During the next pass, the 5 and the 4 are
interchanged, yielding 1, 3, 4, 5, 7. There are two more passes, but no further interchanges are made, since the
list is now in order.
37. We need to add a Boolean variable to indicate whether any interchanges were made during a pass. Initially
this variable, which we will call stilLinterchanging, is set to true. If no interchanges were made, then we can
94
Chapter 3
Algorithms
quit. To do this neatly, we turn the outermost loop into a while loop that is executed as long as i
stilLinterchanging is true. Thus our pseudocode is as follows.
< n and
procedure betterbubblesort( ai, . .. , an)
i := 1
stilLinterchanging :=true
while i < n and stilLinterchanging
stilLinterchanging := false
for j := 1 to n - i
if a1 > a1 + 1 then
stilLinterchanging := true
interchange a1 and a1 + 1
i := i + 1
{ a 1 •... , an is in nondecreasing order}
39. We start with 3, 1, 5, 7, 4. The first step inserts 1 correctly into the sorted list 3, producing 1, 3, 5, 7, 4. Next
5 is inserted into 1, 3, and the list still reads 1, 3, 5, 7, 4, as it does after the 7 is inserted into 1, 3, 5. Finally,
the 4 is inserted, and we obtain the sorted list 1, 3, 4, 5, 7. At each insertion, the element to be inserted is
compared with the elements already sorted, starting from the beginning, until its correct spot is found, and
then the previously sorted elements beyond that spot are each moved one position toward the back of the list.
41. We assume that when the least element is found at each stage, it is interchanged with the element in the
position it wants to occupy.
a) The smallest element is 1, so it is interchanged with the 3 at the beginning of the list, yielding 1, 5, 4, 3, 2.
Next, the smallest element among the remaining elements in the list (the second through fifth positions) is 2,
so it is interchanged with the 5 in position 2, yielding 1, 2, 4, 3, 5. One more pass gives us 1, 2, 3, 4, 5. At this
point we find the fourth smallest element among the fourth and fifth positions, namely 4, and interchange it
with itself, again yielding 1, 2, 3, 4, 5. This completes the sort.
b) The process is similar to part (a). We just show the status at the end of each of the four passes: 1, 4, 3,
2,5; 1,2,3,4.5; 1.2,3,4,5; 1,2,3,4,5.
c) Again there are four passes. but all interchanges result in the list remaining as it is.
43. We carry out the linear search algorithm given as Algorithm 2 in this section, except that we replace x =I- a,
by x < a,. and we replace the else clause with else location := n + 1. The cursor skips past all elements in
the list less than x, the new element we are trying to insert, and ends up in the correct position for the new
element.
45. We are counting just the comparisons of the numbers in the list, not any comparisons needed for the bookkeeping in the for loop. The second element in the list must be compared with the first and compared with
itself (in other words, when j = 2 in Algorithm 5, i takes the values 1 and 2 before we drop out of the
while loop). The third element must be compared with the first two and itself, since it exceeds them both.
We continue in this way, until finally the nth element must be compared with all the elements. So the total
number of comparisons is 2 + 3 + 4 + · · · + n, which can be written as (n 2 + n - 2)/2. This is the worst case
for insertion sort in terms of number of comparisons; see Example 6 in Section 3.3. On the other hand, no
movements of elements are required, since each new element is already in its correct position.
47. There are two kinds of steps-the searching and the inserting. We assume the answer to Exercise 44, which
is to use Algorithm 3 but replace the final check with if x < a, then location := i else location := i + 1.
So the first step is to find the location for 2 in the list 3, and we insert it in front of the 3, so the list now
reads 2. 3, 4, 5. 1, 6. This took one comparison. Next we use binary search to find the location for the 4, and
Section 3.1
Algorithms
95
we see, after comparing it to the 2 and then the 3, that it comes after the 3, so we insert it there, leaving
still 2, 3, 4, 5, 1, 6. Next we use binary search to find the location for the 5, and we see, after comparing it
to the 3 and then the 4, that it comes after the 4, so we insert it there, leaving still 2, 3, 4, 5, 1, 6. Next we
use binary search to find the location for the 1, and we see, after comparing it to the 3 and then the 2 and
then the 2 again, that it comes before the 2, so we insert it there, leaving 1, 2, 3, 4, 5, 6. Finally we use binary
search to find the location for the 6, and we see, after comparing it to the 3 and then the 4 and then the 5,
that it comes after the 5, so we insert it there, giving the final answer 1, 2, 3, 4, 5, 6. Note that this took 11
comparisons in all.
49. We combine the search technique of Algorithm 3, as modified in Exercises 44 and 47, with the insertion part
of Algorithm 5.
procedure binary insertion sort(a1, a2, ... , an : real numbers with n 2: 2)
for j := 2 ton
{binary search for insertion location i }
left:= 1
right:= j - 1
while left < right
middle:= L(left + right)/2J
if a 1 > amiddle then left := middle + 1
else right := middle
if aj < azeft then i := left else i := left + 1
{insert a1 in location i by moving ai through a1 _ 1 toward back of list}
m :=a1
for k := 0 to j - i - 1
aj-k
ai
:= a 1 -k-1
:=m
{ ai, a2, ... , an are sorted}
51. If the elements are in close to the correct order, then we would usually find the correct spot for the next item
to be inserted near the upper end of the list of already-sorted elements. Hence the variation from Exercise 50,
which starts comparing at that end, would be best.
53. In each case we use as many quarters as we can, then as many dimes to achieve the remaining amount, then
as many nickels, then as many pennies.
a) The algorithm uses the maximum number of quarters, two, leaving 1 cent. It then uses the maximum
number of dimes (none) and nickels (none), before using one penny.
b) two quarters, leaving 19 cents, then one dime, leaving 9 cents, then one nickel, leaving 4 cents, then four
pennies
c) three quarters, leaving 1 cent, then one penny
d) two quarters, leaving 10 cents, then one dime
55. In each case we uses as many quarters as we can, then as many dimes to achieve the remaining amount, then
as many pennies.
a) The algorithm uses the maximum number of quarters, two, leaving 1 cent. It then uses the maximum
number of dimes (none), before using one penny. Since the answer to Exercise 53a used no nickels anyway,
the greedy algorithm here certainly used the fewest coins possible.
b) The algorithm uses two quarters, leaving 19 cents, then one dime, leaving 9 cents, then nine pennies. The
greedy algorithm thus uses 12 coins. Since there are no nickels available, we must either use nine pennies or
else use only one quarter and four pennies, along with four dimes to reach the needed total of 69 cents. This
uses only nine coins, so the greedy algorithm here did not achieve the optimum.
96
Chapter 3
Algorithms
c) The algorithm uses three quarters, leaving 1 cent, then one penny. Since the answer to Exercise 53c used
no nickels anyway, the greedy algorithm here certainly used the fewest coins possible.
d) The algorithm uses two quarters, leaving 10 cents, then one dime. Since the answer to Exercise 53c used
no nickels anyway, the greedy algorithm here certainly used the fewest coins possible.
57. We first sort the talks by finishing times and get the following list: 9:00-9:45, 9:30-10:00, 9:50-10:15, 10:1010:25, 10:00-10:30, 10:15-10:45, 10:30-10:55, 10:30-11:00, 11:00-11:15, 10:55-11:25, 10:45-11:30. We then go
through the list in order and schedule every talk that is compatible with the talks already scheduled. So first
we schedule the 9:00-9:45 talk, then the first one on our list that starts after that talk is finished, namely the
9:50-10:15 talk. Next comes the 10:15-10:45 talk, and then the 11:00-11:15 talk. There are no more talks
that start after this one ends, so we are done, having scheduled four talks.
59. a) We propose the following greedy algorithm. Order the talks by starting time. Number the lecture halls 1,
2, 3, and so on. For each talk, assign it to lowest numbered lecture hall that is currently available. So, for
example, using the talks from Exercise 57, we would assign the 9:00-9:45 talk to lecture hall 1, the 9:30-10:00
talk to lecture hall 2, the 9:50-10:15 talk to lecture hall 1, the 10:00-10:30 talk to lecture hall 2, the 10:1010:25 talk to lecture hall 3, the 10: 15-10:45 talk to lecture hall 1, the 10:30-10:55 talk to lecture hall 2, the
10:30-11:00 talk to lecture hall 3, the 10:45-11:30 talk to lecture hall 1, the 10:55-11:25 talk to lecture hall 2,
and the 11:00-11:15 talk to lecture hall 3. Therefore three halls were sufficient.
b) This algorithm is optimal, because if it uses n lecture halls, then at the point the nth hall was first assigned,
it had to be used (otherwise a lower-numbered hall would have been assigned), which means that n talks were
going on simultaneously (this talk just assigned and the n - 1 talks currently in halls 1 through n - 1 ).
61. In the algorithm presented here, the input consists, for each man, of a list of all women in his preference order,
and for each woman, of a list of all men in her preference order. At the risk of being sexist, we will let the men
be the suitors and the women the suitees (although obviously we could reverse these roles). The procedure
needs to have data structures (lists) to keep track, for each man, of his status (rejected or not) and the list of
women who have rejected him, and, for each woman, of the men currently on her proposal list.
procedure stable(A11, M2, ... , lvf8 , W1, W2, ... , W 8 ,: preference lists)
for i := 1 to s
mark man i as rejected
for i := 1 to s
set man i's rejection list to be empty
for j := 1 to s
set woman j's proposal list to be empty
while rejected men remain
for i := 1 to s
if man i is marked rejected then add i to the proposal list
for the woman j who ranks highest on his preference list
but does not appear on his rejection list, and mark i as not rejected
for j := 1 to s
if woman j's proposal list is nonempty then remove from
j's proposal list all men i except the man io who ranks highest
on her preference list, and for each such man i mark him
as rejected and add j to his rejection list
for j := 1 to s
match j with the one man on j's proposal list
{This matching is stable.}
63. Suppose the assignment is not stable. Then there is a man m and a woman w such that m prefers w to the
woman (call her w') with whom he is matched, and w prefers m to the man with whom she is matched. But
Section 3.2
The Growth of Functions
97
m must have proposed to w before he proposed to w' , since he prefers the former. And since m did not end
up matched with w, she must have rejected him. Since women reject a suitor only when they get a better
proposal, and they eventually get matched with a pending suitor, the woman with whom w is matched must
be better in her eyes than m, contradicting our original assumption. Therefore the matching is stable.
65. The algorithm is simply to run the two programs on their inputs concurrently and wait for one to halt. This
algorithm will terminate by the conditions of the problem, and we'll have the desired answer.
SECTION 3.2
The Growth of Functions
The big-0 notation is used extensively in computer science and other areas. Think of it as a crude ruler for
measuring functions in terms of how fast they grow. The idea is to treat all functions that are more or less
the same as one function-one mark on this ruler. Thus, for example, all linear functions are simply thought
of as O(n). Although technically the big-0 notation gives an upper bound on the growth of a function, in
practice we choose the smallest big-0 estimate that applies. (This is made more rigorous with the big-8
notation, also discussed in this section.) In essence, one finds best big-0 estimates by discarding lower order
terms and multiplicative constants. Furthermore, one usually chooses the simplest possible representative of
the big-0 (or big-8) class (for example, writing 0( n 2 ) rather than 0(3n 2 + 5) ). A related concept, used in
combinatorics and applied mathematics, is the little-o notation, dealt with in Exercises 61-69.
1. Note that the choices of witnesses C and k are not unique.
a) Yes, since 1101::::; lxl for all x > 10. The witnesses are C = 1 and k = 10.
b) Yes, since l3x + 71 ::::; l4xl = 4lxl for all x > 7. The witnesses are C = 4 and k = 7.
c) No. There is no constant C such that lx 2 + x + 1 I :::; Clxl for all sufficiently large x. To see this, suppose
this inequality held for all sufficiently large positive values of x. Then we would have x 2 :::; Cx, which would
imply that x ::::; C for all sufficiently large x, an obvious impossibility.
d) Yes. This follows from the fact that log x < x for all x > 1 (which in turn follows from the fact that
x < 2x, which can be formally proved by mathematical induction-see Section 5.1). Therefore l5logxl::::; 5lxl
for all x > 1. The witnesses are C = 5 and k = 1.
lx J : : ;
e) Yes. This follows from the fact that
and k = 0.
x. Thus
f) Yes. This follows from the fact that lx/21::::; (x/2)
The witnesses are C = 1 and k = 2.
Ilx JI : : ;
+ 1.
lxl for all x
> 0. The witnesses are C
Thus llx/211:::; l(x/2)
+ ll:::;
3. We need to put some bounds on the lower order terms. If x > 9 then we have x 4
x4
+ x 4 + x 4 + x 4 = 4x4 •
Therefore x 4
+ 9x3 + 4x + 7 is
O(x 4 ), taking witnesses C
=
1
lxl for all x > 2.
+ 9x 3 + 4x + 7 <
= 4 and k = 9.
5. We use long division to rewrite this function:
x2
+1
x+l
x
2
2
1+2 = x - 1
x+l
x+l
-
+ _2_
x+l
= x _ 1 + _2_
x+l
Now this is certainly less than x as long as x > 1, so our function is O(x). The witnesses are C = 1 and
k = 1.
98
Chapter 3
Algorithms
7. a) Since log x grows more slowly than x, x 2 log x grows more slowly than x 3 , so the first term dominates.
Therefore this function is O(x 3 ) but not O(xn) for any n < 3. More precisely, 2x3 +x 2 log x :::; 2x 3 +x 3 = 3x 3
for all x, so we have witnesses C
=3
and k
= 0.
b) We know that log x grows so much more slowly than x that every power of log x grows more slowly
than x (see Exercise 58). Thus the first term dominates, and the best estimate is O(x 3 ). More precisely,
(log x ) 4 < x 3 for all x > 1, so 3x 3 + (log x ) 4 :::; 3x3 + x 3 = 4x 3 for all x, so we have witnesses C = 4 and
k = 1.
c) By long division, we see that f(x) = x +lower order terms. Therefore this function is O(x), so n
fact, f(x)
=x
+ x~l
:::;
= 1. In
> 1, so the witnesses can be taken to be C = 2 and k = 1.
this quotient has the form f(x) = 1 +lower order terms. Therefore this function
= 0. Since 5 log x < x 4 for x > 1, we have f (x) :::; 2x 4 / x 4 = 2, so we can take as
2x for all x
d) Again by long division,
is 0(1). In other words, n
witnesses C = 2 and k = 1.
9. On the one hand we have x 2 + 4x + 17:::; x 2 + x 2 + x 2 = 3x 2 :::; 3x3 for all x > 17, so x 2 + 4x + 17 is O(x 3 ),
with witnesses C = 3 and k = 17. On the other hand, if x 3 were O(x 2 + 4x + 17), then we would have
x 3 :::; C(x 2 + 4x + 17) :::; 3Cx 2 for all sufficiently large x. But this says that x :::; 3C, clearly impossible for
the constant C to satisfy for all large x. Therefore x 3 is not O(x 2 + 4x + 17).
11. For the first part we have 3x 4 + 1 :::; 4x 4 = 8lx 4 /21 for all x > 1; we have witnesses C
second part we have x 4 /2:::; 3x4 :::; 1 · J3x 4 + lJ for all x; witnesses are C = 1, k = 0.
= 8, k =
1. For the
> 0. In terms of witnesses we have
C = 1 and k = 0. On the other hand, if 3n were 0(2n), then we would have 3n :::; C · 2n for all sufficiently
large n. This is equivalent to C ::::: ( ~) n, which is clearly impossible, since ( ~) n grows without bound as n
increases.
13. To show that 2n is 0(3n) it is enough to note that 2n :::; 3n for all n
15. A function f is 0(1) if lf(x)J :::; C for all sufficiently large x. In other words,
is bounded for all x > k (where k is some constant).
f is 0(1) if its absolute value
17. Let C1, C2, ki, and k2 be numbers such that lf(x)I :::; C1lg(x)I for all x > ki and lg(x)I :::; C2lh(x)I
for all x > k 2 . Let C = C1 C2 and let k be the larger of k1 and k2. Then for all x > k we have
lf(x)J:::; C1lg(x)I:::; C1C2lh(x)I = Clh(x)I, which is precisely what we needed to show.
19. Because 2n+i = 2 · 2n, clearly it is 0(2n) (take C = 2). To see that 22n is not 0(2n), look at the ratio:
22 n /2n = 2n. Because this is unbounded, there is no constant C such that 22 n :::; C · 2n for all sufficiently
large n.
21. The order is lOOOlogn, yin, nlogn, n 2/1000000, 2n, 3n, 2n!. Thateachisbig-0 ofthenextisclear.
23. Because n log n is 0( n 3 12) but n 3 12 is not 0( n log n), the first algorithm uses fewer operations for large n.
In fact, if we solve n log n < n 3 12 , we get n > 4. In other words, the first algorithm uses fewer operations for
all n > 4.
25. a) The significant terms here are the n 2 being multiplied by the n; thus this function is O(n 3 ).
b) Since log n is smaller than n, the significant term in the first factor is n 2 . Therefore the entire function
is O(n 5 ).
c) For the first factor we note that 2n < n! for n 2: 4, so the significant term is n!. For the second factor,
the significant term is n 3 . Therefore this function is O(n3 n!).
Section 3.2
The Growth of Functions
99
27. a) First we note that log(n 2 +1) and logn are in the same big-0 class, since logn 2 = 2logn. Therefore the
second term here dominates the first, and the simplest good answer would be just 0( n 2 log n).
b) The first term is in the same big-0 class as O(n 2(logn) 2), while the second is in a slightly smaller class,
0( n 2 log n). (In each case, we can throw away the smaller order terms, since they are dominated by the terms
we are keeping-this is the essence of doing big-0 estimates.) Therefore the answer is O(n 2(1ogn) 2).
c) The only issue here is whether 2n or n 2 is the faster-growing, and clearly it is the former. Therefore the
best big-0 estimate we can give is O(n 2").
29. We can use the following rule of thumb to determine what simple big- 8 function to use: throw away all the
lower order terms (those that don't grow as fast as other terms) and all constant coefficients.
a) This function is 8(x), so it is not 8(x 2), since x 2 grows faster than x. To be precise, x 2 is not 0(17x+ll).
For the same reason, this function is not O(x 2).
b) This function is 8(x 2); we can ignore the "+ 1000'' since it is a lower order term. Of course, since J(x)
is 8(x 2), it is also O(x 2).
c) This function grows more slowly than x 2 , since logx grows more slowly than x. Therefore f(x) is not
8(x 2) or O(x 2).
d) This function grows faster than x 2 . Therefore f(x) is not 8(x 2), but it is D(x 2 ).
e) Exponential functions (with base larger than 1) grow faster than all polynomials, so this function is not
O(x 2) and therefore not 8(x 2). But it is O(x 2).
f) For large values of x, this is quite close to x 2 , since both factors are quite close to x. Certainly l x J · fxl
is always between x 2/2 and 2x 2 , for x > 2. Therefore this function is 8(x 2) and hence also O(x 2).
31. If J(x) is 8(g(x)), then lf(x)I::; C2lg(x)I and lg(x)I::; c1 1lf(x)I for all x > k. Thus f(x) is O(g(x)) and
g(x) is O(f(x)). Conversely, suppose that f(x) is O(g(x)) and g(x) is O(f(x)). Then (with appropriate
choice of variable names) we may assume that IJ(x)I ::; C 2lg(x)I and lg(x)I ::; Clf(x)I for all x > k. (The
k here will be the larger of the two k's involved in the hypotheses.) If C > 0 then we can take C 1 = c~ 1
to obtain the desired inequalities in "f(x) is G(g(x)) ." If C ::; 0, then g(x) = 0 for all x > k, and hence
by the first inequality f(x) = 0 for all x > k; thus we have f(x) = g(x) for all x > k, and we can take
C1=C2=1.
33. The definition of "f(x) is 8(g(x))" is that f(x) is both O(g(x)) and O(g(x)). That means that there are
positive constants C1, k1, C2, and k2 such that IJ(x)I ::; C2lg(x)I for all x > k2 and lf(x)I ?: C1lg(x)I for
all x > k1 . That is practically the same as the statement in this exercise. We need only note that we can
take k to be the larger of k1 and k2 if we want to prove the "only if" direction, and we can take k1 = k2 = k
if we want to prove the "if" direction.
35. In the following picture, the wavy line is the graph of the function f. For simplicity we assume that the graph
of g is a straight line through the origin. Then the graphs of C1 g and C2 g are also straight lines through the
origin, as drawn here. The fact that f(x) is G(g(x)) is shown by the fact that for x > k the graph off is
confined to the shaded wedge-shaped space between these latter two lines (see Exercise 33). (We assume that
g(x) is positive for positive x, so that lg(x)I is the same as g(x).)
Chapter 3
100
Algorithms
y
37. Looking at the definition tells us that if f(x) is 8(1) then lf(x)I has to be bounded between two positive
constants. In other words, f(x) can't get too large (either positive or negative), and it can't get too close
to 0.
39. We are given that lf(x)I ::::; Cl9(x)I for all x > k. Hence lr(x)I = lf(x)ln ::::; cn19n(x)I for all x
x) is 0(9n (x)) (take the constants in the definition to be en and k).
r(
> k,
so
41. Since the functions are given to be increasing and unbounded, we may assume that they both take on values
greater than 1 for all sufficiently large x. The hypothesis can then be written as f(x)::::; C9(x) for all x
> k.
If we take the logarithm of both sides, then we obtain log J(x) ::::; log C +log 9(x). Finally, this latter expression
is less than 2log9(x) for large enough x, since log9(x) is growing without bound. Note that the "converse"
implication does not hold; even though log(3n) is O(log(2n)) (both functions are linear), it is not true that
3" is 0(2n).
43. By definition there are positive constants C 1 , C{, C2, C~, k1, k{, k2, and k~ such that fi(x) 2 C1l9(x)I
for all x > ki, fi(x)::::; Cil9(x)I for all x > ki, h(x) ~ C2l9(x)I for all x > k2, and h(x)::::; C~l9(x)I for all
x > k~. We are able to omit the absolute value signs on the f(x) 's since we are told that they are positive; we
are also told here that the 9(x) 's are positive, but we do not need that. Adding the first and third inequalities
we obtain fi(x) + h(x) 2 (C1 + C2)l9(x)I for all x > max(k1, k2); and similarly with the second and fourth
inequalities we know Ji (x) + h(x) ::::;
definition of being 8(g(x)).
(Ci +
q)l9(x)I for all x
> max(ki, k~).
Thus
Ji (x) +
h(x) meets the
If the f's can take on negative values, then this is no longer true. For example, let fi (x) = x 2 + x, let
h(x) = -x 2 + x, and let g(x) = x 2 . Then each fi(x) is 8(9(x)), but the sum is 2x, which is not 8(g(x)).
45. This is not true. It is similar to Exercise 43, and essentially the same counterexample suffices. Let Ji (x) =
x 2 +2x, h(x) = x 2 +x, and 9(x) = x 2 . Then clearly Ji (x) and h(x) are both 8(9(x)), but (11 - h)(x) = x
is not.
47. The key here is that if a function is to be big-0 of another, then the appropriate inequality has to hold for all
large inputs. Suppose we let f (x) = x 2 for even x and x for odd x. Similarly, we let 9( x) = x 2 for odd x
and x for even x. Then clearly neither inequality lf(x)I::::; Clg(x)I nor J9(x)I::::; Clf(x)I holds for all x, since
for even x the first function is much bigger than the second, while for odd x the second is much bigger than
the first.
49. We are given that there are positive constants C1, Ci, C2, q, k1,
ki,
k2, and k~ such that lf1(x)I
2
C1lg1(x)I for all x > k1, lfi(x)I::::; Cil91(x)I for all x > ki, lh(x)I ~ C2l92(x)I for all x > k2, and
Ih (x) I ::::; q Jg 2(x) I for all x > k~. Since h and 9 2 never take on the value 0, we can rewrite the last two
of these inequalities as ll/ h(x)I ::::; (1/C2)ll/g2(x)I and 11/ h(x)I ~ (1/crn1/g2(x)J. Now we multiply the
first inequality and the rewritten fourth inequality to obtain lfi(x)/ h(x)I ~ (Ci/q)lg1(x)/g2(x)I for all
x > max(k 1 ,k~). Working with the other two inequalities gives us lfi(x)/h(x)I::::; (Ci/C2)lg1(x)/92(x)I for
all x > max( ki, k2) . Together these tell us that Ji/ h is big- 8 of gi/ 9 2 .
Section 3.2
The Growth of Functions
101
51. We just make the analogous change in the definition of big-8 that was made in the definition of big-0: there
exist positive constants C1, C2, ki, k2, ki, k~ such that lf(x,y)I :S C1lg(x,y)I for all x > ki and y > k2,
and lf(x, y)I 2: C2lg(x, y)I for all x > ki and y > k~.
53. For all values of x and y greater than 1, each term of the expression inside parentheses is less than x 2 y,
so the entire expression inside parentheses is less than 3x 2 y. Therefore our function is less than 27x 6 y 3 for
all x > 1 and y > 1. By definition this shows that it is big-0 of x 6 y 3 . Specifically, we take C = 27 and
ki = k2 = 1 in the definition.
55. For all positive values of x and y, we know that lxy J :::; xy by definition (since the floor function value cannot
exceed the argument). Thus lxyj is O(xy) from the definition, taking C = 1 and k1 = k 2 = 0. In fact, lxyj
is also O(xy) (and therefore 8(xy) ); this is easy to see since Lxyj 2: (x - 1)(y - 1) 2: ( ~x) ( ~y) = i-xy for all
x and y greater than 2.
57. Because d < c, clearly nd < nc for all n 2: 2; therefore nd is O(nc). To see that nd is not O(nc), look at
the ratio: nd /nc = nd-c. Because this is unbounded (the exponent is positive), there is no constant C such
that nd :::; enc for all sufficiently large n.
59. If f and g are positive-valued functions for which we can compute the limit of the ratio f(x)/g(x) as x __, oo,
then the value of that limit will tell us about the asymptotic behavior of those functions relative to each other.
Indeed, if the limit is a finite number C, then for large enough x, we have f (x) < (C + 1)g( x) , so f (n) is
O(g(n)). If the limit is oo, then clearly f(n) is not O(g(n)).
In the case at hand, limx_, 00 xd /bx
=0
(apply L'H6pital's rule
Idl
times), and similarly limx_, 00 bx jxd
=
oo. The desired conclusions follow.
61. All that we need to do is determine whether the ratio of the two functions approaches 0 as x approaches
infinity.
.
x2
a ) 1im X---+(X)
X3
=
l'
1
1m X--+cx::>
X
=0
1
· l' s ru1e £or t h e second equa1.ity )
b) ll·m x log x -_ 11·m log x -- 11·m - -- -- 0 (using L'H'op1ta
X--->00
x2
X--->00
x
X--->00 x ln 2
2
c) lim x
X--->00 2X
d) lim
X---+00
x2
= lim
X--->00
~lx
=
2x n 2
+x +1 =
X2
lim
X--->00
lim (1
X---+(X)
2
(1 )
2X n 2 2
+ -X1 + 2X1 ) =
63. The picture shows the graph of y
= x2
=
0 (with two applications of L'H6pital's rule)
1 =/= 0
increasing quite rapidly and y
ratio is hard to see on the picture; it rises to about y
= 0.53
= x log x
at about x
=
increasing less rapidly. The
2.7 and then slowly decreases
toward 0. The limit as x __, oo of (x log x) / x 2 is in fact 0.
x2
y
30
20
x log x
10
/x log x)/x 2
2
3
4
5
6
Chapter 3
102
65. No. As one example, take f(x) = x- 2 and g(x)
Algorithms
= x- 1 . Then f(x) is o(g(x)), since lim x- 2 /x- 1
2
1
= lim 2.r- -x- = 2°=1 f' O.
x~oo
2
lim 1/x = 0. On the other hand lim (2.r- ;2xX---->OC•
:r~,'.X)
1
)
X---1'CX";
67. a) Since the limit of f(x)/g(x) is 0 (as x----+ oo ), so too is the limit of lf(x)l/lg(x)I. In particular, for x large
enough, this ratio is certainly less than 1. In other words lf(x)I:::; lg(x)I for sufficiently large x, which meets
the definition of "f(x) is O(g(.r,))."
b) We can simply let f (:r) = g( x) be any function with positive values. Then the limit of their ratio is 1,
not 0, so f(x) is not o(g(x)), but certainly f(x) is O(g(x)).
69. This follows immediately from Exercise 67a (whereby we can conclude that h(x) is O(g(x))) and Corollary 1
to Theorem 2.
71. What we want to show is equivalent to the statement that log(nn) is at most a constant times log(n!), which
in turn is equivalent to the statement that nn is at most a constant power of n! (because of the fact that
ClogA = log(Ac)~see Appendix 2). We will show that in fact nn:::; (n!) 2 for all n > 1. To do this, let us
write (n!) 2 as (n · l) · ( (n - l) · 2)- ( (n - 2) · 3) · · · (2 · (n -1)) · (1 · n). Now clearly each product pair (i + 1) · (n- i)
is at least as big as n (indeed, the ones near the middle are significantly bigger than n ). Therefore the entire
product is at least as big as nn, as desired.
73. For n = 5 we compute that log5! ::::o 6.9 and (5log5)/4 ::::o 2.9, so the inequality holds (it actually holds for
all n > 1). Therefore we can assume that n ;:::: 6. Since n! is the product of all the integers from n down
to 1, we certainly have n! > n(n - l)(n - 2) · · · In/21 (since at least the term 2 is missing). Note that there
are more than n/2 terms in this product, and each term is at least as big as n/2. Therefore the product is
greater than (n/2)(n/ 2 J. Taking the log of both sides of the inequality, we have
n)n/2
n
n
n
logn! >log (
=
2 1og 2 = 2 (logn-1) > (nlogn)/4,
2
since n > 4 implies log n - 1 > (log n) /2.
75. In each case we need to evaluate the limit of f (x) / g( x) as x ----+ oo. If it equals 1, then f and g are asymptotic;
otherwise (including the case in which the limit does not exist) they are not. Most of these are straightforward
applications of algebra, elementary notions about limits, or L'H6pital's rule.
a) f (x) = log(x 2 + 1) ;:::: log(x 2 ) = 2 log .r. Therefore f (.r) / g(:r) ;:::: 2 for all x. Thus the limit is not 1 (in fact,
of course, it's 2), so f and g are not asymptotic.
b) By the algebraic rules for exponents, f(x)/g(x) = 2- 4 = 1/16. Therefore the limit of the ratio is 1/16,
not 1, so f and g are not asymptotic.
22'
"
2
c) lim - 2 = lim 22 -x . As :r gets large, the exponent grows without bound, so this limit is oo. Thus f
x---+oo 2x
:r------roo
and g are not asymptotic.
2x 2 +x+l
d) lim
= lim 2 1 -:r. As :r gets large, the exponent grows in the negative direction without bound,
2
.r---+CX) 2.r + 2 x
.T---+00
so this limit is 0. Thus f and g are not asymptotic.
Section 3.3
Complexity of Algorithms
SECTION 3.3
103
Complexity of Algorithms
Some of these exercises involve analysis of algorithms, as was done in the examples in this section. These
are a matter of carefully counting the operations of interest, usually in the worst case. Some of the others
are algebra exercises that display the results of the analysis in real terms-the number of years of computer
time, for example, required to solve a large problem. Horner's method for evaluating a polynomial, given
in Exercise 14, is a nice trick to know. It is extremely handy for polynomial evaluation on a pocket calculator
(especially if the calculator is so cheap that it does not use the usual precedence rules).
1. The statement t := t + ij is executed just 12 times, so the number of operations is 0(1). (Specifically, there
are just 24 additions or multiplications.)
3. The nesting of the loops implies that the assignment statement is executed roughly n 2 /2 times. Therefore
the number of operations is 0( n 2 ) .
5. Assuming that the algorithm given to find the smallest element of a list is identical to Algorithm 1 in Section 3.1,
except that the inequality is reversed (and the name max replaced by the name min), the analysis will be
identical to the analysis given in Example 1 in the current section. In particular, there will be 2n - 1
comparisons needed, counting the bookkeeping for the loop.
7. The linear search would find this element after at most 9 comparisons ( 4 to determine that we have not yet
finished with the while loop, 4 more to determine if we have located the desired element yet, and 1 to set the
value of location). Binary search, according to Example 3, will take 2 log 32 + 2 = 2 · 5 + 2 = 12 comparisons.
Since 9 < 12, the linear search will be faster, in terms of comparisons.
9. The algorithm simply scans the bits one at a time. Thus clearly O(n) comparisons are required (perhaps one
for bookkeeping and one for looking at the ith bit, for each i from 1 to n ).
11. a) We can express the suggested algorithm in pseudocode as follows. Notice that the Boolean variable disjoint
is set to true at the beginning of the comparison of sets S, and S1 , and becomes false if and when we find
an element common to those two sets. If disjoint is never set to false, then we have found a disjoint pair, and
answer is set to true. This process is repeated for each pair of sets (controlled by the outer two loops).
procedure disjointpair(S1 , S 2 , ... , Sn : subsets of {1, 2, ... , n})
answer:= false
for i := 1 ton
for j := i + 1 ton
disjoint := true
fork:= 1 ton
if k E Si and k E S1 then disjoint:= false
if disjoint then answer := true
return answer
b) The three nested loops imply that the elementhood test needs to be applied O(n 3 ) times.
13. a) Here we have n = 2, a0 = 1, a 1 = 1, a 2 = 3, and c = 2. Initially, we set power equal to 1 and y equal
to 1. The first time through the for loop (with i = 1 ) , power becomes 2 and so y becomes 1 + 1 · 2 = 3.
The second and final time through the loop, power becomes 2 · 2 = 4 and y becomes 3 + 3 · 4 = 15. Thus
the value of the polynomial at x = 2 is 15.
b) Each pass through the loop requires two multiplications and one addition. Therefore there are a total of
2n multiplications and n additions in all.
104
Chapter 3
Algorithms
15. This is an exercise in algebra, numerical analysis (for some of the parts), and using a calculator. Since each bit
operation requires 10- 9 seconds, we want to know for what value of n there will be at most 10 9 bit operations
required. Thus we need to set the expression equal to 109 , solve for n, and round down if necessary.
a) Solving logn = 109 , we get (recalling that "log'' means logarithm base 2) n = 2109 . By taking log 10 of
both sides, we find that this number is approximately equal to 103oo,ooo,ooo. Obviously we do not want to
write out the answer explicitly!
b) Clearly n = 109 .
c) Solving n log n = 109 is not trivial. There is no good formula for solving such transcendental equations.
An algorithm that works well with a calculator is to rewrite the equation as n = 109 /log n, enter a random
starting value, say n = 2, and repeatedly calculate a new value of n. Thus we would obtain, in succession,
n = 109 /log2 = 109 , n = 109 /log(10 9 ):::::; 33,447,777.3, n = 109 /log(33,447,777.3):::::; 40,007,350.14, n =
109 /log( 40,007,350.14) :::::; 39,598,061.08, and so on. After a few more iterations, the numbers stabilize at
approximately 39,620,077. 73, so the answer is 39,620,077.
d) Solving n 2 = 109 gives n = 104 ·5 , which is 31,622 when rounded down.
e) Solving 2n = 109 gives n = log(10 9 ):::::; 29.9. Rounding down gives the answer, 29.
f) The quickest way to find the largest value of n such that n! :::; 109 is simply to try a few values of n. We
find that 12! :::::; 4.8 x 108 while 13! :::::; 6.2 x 109 , so the answer is 12.
17. If each bit operation takes 10- 12 second, then we can carry out 10 12 bit operations per second, and therefore
60 · 10 12 bit operations per minute. Therefore in each case we want to solve the equation f(n) = 60 · 10 12 for
n and round down to an integer. Obviously a calculator will come in handy here.
a) If loglogn
= 60 · 10 12 , then n = 22
12
60
60 10 12
,
which is an unfathomably huge number.
1012
b) If log n = 60 · 10 , then n = 2
, which is still an unfathomably huge number.
5
2
12
c) If (log n ) = 60 · 10 , then log n = V6Q · 106 , so n = L2v'60·l0 J , which is still an extremely large number
(it has over 2 million digits).
d) If l,000,000n = 60 · 10 12 , then n = 60 · 106 = 60,000,000.
e) If n 2 = 60 · 10 12 , then n
f) If 2n = 60 · 10 12 , then n
the base 2.)
2
= lV66 · 106 J = 7,745,966.
= Llog(60 · 10 12 )J = Llog 60 + 12log10J = 45. (Remember, we are taking log to
g) If 2n = 60 · 10 12 , then n
to the base 2.)
= h/log(60 · 10 12 )J = l yflog 60 + 12log10J = 6. (Remember, we are taking log
19. In each case, we just multiply the number of seconds per operation by the number of operations (namely 250 ).
To convert seconds to minutes, we divide by 60; to convert minutes to hours, we divide by 60 again. To
convert hours to days, we divide by 24; to convert days to years, we divide by 365 ~ .
a) 250 x 10- 6
b)
250
c)
250
10- 9
= 1,125,899,907 seconds:::::; 36 years
= 1,125,899.907 seconds:::::; 13 days
x
x 10- 12 = 1,125.899907 seconds:::::; 19 minutes
21. In each case we want to compare the function evaluated at n
+1
to the function evaluated at n. The most
desirable form of the comparison (subtraction or division) will vary.
a) Notice that log( n + 1) - log n = log "~ 1 . If n is at all large, the fraction in this expression is approximately
equal to 1, and therefore the expression is approximately equal to 0. In other words, hardly any extra time is
required. For example, in going from n = 10 to n = 11, the number of extra milliseconds is log 11/10:::::; 0.14.
b) lOO(n + 1) - lOOn = 100. One hundred extra milliseconds are required, independent of n.
c) Because (n + 1) 2 - n 2 = 2n + 1, we conclude that 2n + 1 additional milliseconds are needed for the larger
problem.
Section 3.3
Complexity of Algorithms
d) Because (n + 1) 3
105
n 3 = 3n 2 + 3n + 1, we conclude that 3n 2 + 3n + 1 additional milliseconds are needed
-
for the larger problem.
e) This time it makes more sense to use a ratio comparison, rather than a difference comparison. Because
2n+i /2n = 2, we see that twice as much time is required for the larger problem.
2
f) Because 2 107.
c) n 3 is not less than a constant times n 2 + 18n
n > 107.
+ 107,
since their ratio exceeds n 3 / (3n 2 )
= n/3
for all
4. (log n ) 3 , fo,, lOOn + 101, n 3 /1000000, 2nn 2 , 3n, n!
5. a) For the sum, take the largest term; for the product multiply the factors together.
b) g(n) = nn- 2 2n = (2n)n /n 2
6. a) the largest, average, and smallest number of comparisons used by the algorithm before it stops, among all
lists of n integers
b) all are n - 1
7. a) See pp. 194-196.
b) See p. 220.
c) No-it depends on the lists involved. (However, the worst case complexity for binary search is always
better than that for linear search for lists of any given size except for very short lists.)
8. a) See pp. 196-197.
b) On the first pass, the 5 bubbles down to the end, producing 2 4 1 3 5. On the next pass, the 4 bubbles
down to the end, producing 213 4 5. On the next pass, the 1 and the 2 are swapped. No further changes are
made on the fourth pass.
c) O(n 2 ); see Example 5 in Section 3.3
9. a) See pp. 197-198.
b) On the first pass, the 5 is inserted into its correct position relative to the 2, producing 2 5 14 3. On the
next pass, the 1 is inserted into its correct position relative to 2 5, producing 1 2 5 4 3. On the next pass, the
4 is inserted into its correct position relative to 1 2 5, producing 1 2 4 5 3. On the final pass, the 3 is inserted,
producing the sorted list.
c) O(n 2 ); see Example 6 in Section 3.
10. a) See p. 198.
b) See Example 6 and Theorem 1 in Section 3.1.
c) See Exercise 55b in Section 3.1.
11. Seep. 226 for "tractable.'' A solvable problem is simply one that can be solved by an algorithm. The halting
problem is proved on pp. 201-202 to be unsolvable.
Chapter 3
108
Algorithms
SUPPLEMENTARY EXERCISES FOR CHAPTER 3
1. a) This algorithm will be identical to the algorithm first largest for Exercise 17 of Section 3.1, except that we
want to change the value of location each time we find another element in the list that is equal to the current
value of max. Therefore we simply change the strict less than ( <) in the comparison max < a, to a less than
or equal to (:::; ), rendering the fifth line of that procedure ''if max :::; a, then."
b) The number of comparisons used by this algorithm can be computed as follows. There are n - 1 passes
through the for loop, each one requiring a comparison of max with ai. In addition, n comparisons are needed
for bookkeeping for the loop (comparison of i with n, as i assumes the values 2, 3, ... , n + 1). Therefore
2n - 1 comparisons are needed altogether, which is O(n).
3. a) We will try to write an algorithm sophisticated enough to avoid unnecessary checking. The answer-true
or false-will be placed in a variable called answer.
procedure zeros(a 1 a2 ... an : bit string)
i := 1
answer :=false {no pair of zeros found yet}
while i < n and -ianswer
if a, = 1 then i := i + 1
else if a,+1 = 1 then i := i + 2
else answer:= true
return answer
{ answer was set to true if and only if there were a pair of consecutive zeros}
b) The number of comparisons depends on whether a pair of O's is found and also depends on the pattern of
increments of the looping variable i. Without getting into the intricate details of exactly which is the worst
case, we note that at worst there are approximately n passes through the loop, each requiring one comparison
of a, with 1 (there may be two comparisons on some passes, but then there will be fewer passes). In addition,
n bookkeeping comparisons of i with n are needed (we are ignoring the testing of the logical variable answer).
Thus a total of approximately 2n comparisons are used, which is 0( n).
5. a) and b). We have a variable min to keep track of the minimum as well as a variable max to keep track of
the maximum.
procedure smallest and largest(a 1 , a 2 , ... , an: integers)
min:= ai
max:= ai
for i := 2 ton
if a, < min then min := a,
if a,> max then max:= a,
{ min is the smallest integer among the input, and max is the largest}
c) There are two comparisons for each iteration of the loop, and there are n - 1 iterations, so there are 2n - 2
comparisons in all.
7. We think of ourselves as observers as some algorithm for solving this problem is executed. We do not care
what the algorithm's strategy is, but we view it along the following lines, in effect taking notes as to what is
happening and what we know as it proceeds. Before any comparisons are done, there is a possibility that each
element could be the maximum and a possibility that it could be the minimum. This means that there are
2n different possibilities, and 2n - 2 of them have to be eliminated through comparisons of elements, since
we need to find the unique maximum and the unique minimum. We classify comparisons of two elements as
"nonvirgin" or "virgin," depending on whether or not both elements being compared have been in any previous
comparison. A virgin comparison, between two elements that have not yet been involved in any comparison,
eliminates the possibility that the larger one is the minimum and that the smaller one is the maximum; thus
Supplementary Exercises
109
each virgin comparison eliminates two possibilities, but it clearly cannot do more. A nonvirgin comparison,
one involving at least one element that has been compared before, must be between two elements that are still
in the running to be the maximum or two elements that are still in the running to be the minimum, and at least
one of these elements must not be in the running for the other category. For example, we might be comparing
x and y, where all we know is that x has been eliminated as the minimum. If we find that x > y in this case,
then only one possibility has been ruled out-we now know that y is not the maximum. Thus in the worst
case, a nonvirgin comparison eliminates only one possibility. (The cases of other nonvirgin comparisons are
similar.) Now there are at most l n/2 J comparisons of elements that have never been compared before, each
removing two possibilities; they remove 2ln/2J possibilities altogether. Therefore we need 2n - 2 - 2ln/2J
more comparisons that, as we have argued, can remove only one possibility each, in order to find the answers in
the worst case, since 2n -2 possibilities have to be eliminated. This gives us a total of 2n - 2- 2 ln/2 J + l n/2 J
comparisons in all. But 2n-2-2ln/2J+ln/2j = 2n-2-ln/2j = 2n-2+f-n/2l = f2n-n/2l-2 = f3n/2l-2,
as desired.
Note that this gives us a lower bound on the number of comparisons used in an algorithm to find the
minimum and the maximum. On the other hand, Exercise 6 gave us an upper bound of the same size. Thus
the algorithm in Exercise 6 is the most efficient algorithm possible for solving this problem.
9. The following uses the brute-force method. The sum of two terms of the sequence ai, a 2 , ... , an is given by
ai + a1 , where i < j. If we loop through all pairs of such sums and check for equality, we can output two pairs
whenever we find equal sums. A pseudocode implementation for this process follows. Because of the nested
loops, the complexity is O(n 4 ).
procedure equal sums(a1, a2, ... , an)
for i := 1 ton
for j := i + 1 to n {since we want i < j}
fork:= 1 ton
for l : = k + 1 to n {since we want k < l }
if ai + a1 = ak + a1 and (i,j)-/:- (k, l) then output these pairs
11. After the comparison and possible exchange of adjacent elements on the first pass, from front to back, the
list is 3, 1, 4, 5, 2, 6, where the 6 is known to be in its correct position. After the comparison and possible
exchange of adjacent elements on the second pass, from back to front, the list is 1, 3, 2, 4, 5, 6, where the 6
and the 1 are known to be in their correct positions. After the next pass, the result is 1, 2, 3, 4, 5, 6. One
more pass finds nothing to exchange, and the algorithm terminates.
13. There are possibly as many as n - 1 passes through the list (or parts of it-it depends on the particular way
it is implemented), and each pass uses O(n) comparisons. Thus there are O(n 2 ) comparisons in all.
15. Since log n < n, we have (n log n + n 2 ) 3 :::; ( n 2 + n 2 ) 3 :::; (2n 2 ) 3 = 8n6 for all n
(nlogn+n 2 ) 3 is O(n6 ),withwitnesses C=8 and k=O.
> 0. This proves that
17. In the first factor the x 2 term dominates the other term, since (Iogx) 3 is O(x). Therefore by Theorem 2 in
Section 3.2, this term is O(x 2 ). Similarly, in the second factor, the 2x term dominates. Thus by Theorem 3
of Section 3.2, the product is O(x 2 2x).
19. Let us look at the ratio
n!
n·(n-l)·(n-2)···3·2·1
n n-l n-2
3 2 1
= 2. -2-. -2- ... 2. 2. 2
2n
2 · 2 · 2· · ·2 · 2 · 2
Each of the fractions in the final expression is greater than 1 except the last one, so the entire expression is
at least (n/2)/2 = n/4. Since n!/2n increases without bound as n increases, n! cannot be bounded by a
constant times 2n, which tells us that n! is not 0(2n).
Chapter 3
110
Algorithms
21. Each of these functions is of the same order as n 2 , so all pairs are of the same order. One way to see this is to
think about ''throwing away lower order terms.'' Notice in particular that log 2n = n, and that the n 3 terms
cancel in the fourth function.
23. We know that exponential functions grow faster than polynomial functions, so such a value of n must exist. If
we take logs of both sides, then we obtain the equivalent inequality 2 100 log( n) < n, or n/ log n > 2 100 . This
tells us that n has to be very large. In fact, n = 2 100 is not large enough, because for that value, the left-hand
side is smaller by a factor of 100. If we go a little bigger, however, then the inequality will be satisfied. For
example, if n = 2 107 , then we have
___!:!:_
logn
= 2107 = 128 . 2100 > 2100 .
107
107
25. Clearly nn is growing the fastest, so it belongs at the end of our list. Next, l.OOOln is an exponential function,
so it is bigger (in big-0 terms) than any of the others not yet considered. Next, note that J1og 2 n < log 2 n
for large n, so taking 2 to the power of these two expressions shows that 2vflog 2 n is O(n). Therefore, the
competition for third and fourth places from the right in our list comes down to n i.oool and n(log n) 1001 , and
since all positive powers of n grow faster than any power of logn (see Exercise 58 in Section 3.2), the former
wins third place. Finally, to see that (log n )2 is the slowest-growing of these functions, compare it to 2vflog 2 n
by taking the base-2 log of both and noting that 2 log log n grows more slowly than yflog n. So the required
list is (logn) 2 , 2V10g2 n, n(logn) 1001 , nl.OOOl, l.OOOln, nn.
27. We want the functions to play leap-frog, with first one much bigger, then the other. Something like this will do
the trick: Let J(n) = n 2ln/ 2J+i. Thus, f(l) = 11 , J(2) = 23 , f(3) = 3 3 , f(4) = 4 5 , J(5) = 55 , f(6) = 6 7 ,
and so on. Similarly, let g(n) = n 2 rn/ 2l. Thus, g(l) = 12 , g(2) = 22 , g(3) = 34 , g(4) = 44 , g(5) = 56 ,
g(6) = 66 , and so on. Then for even n we have J(n)/g(n) = n, and for odd n we have g(n)/J(n) = n.
Because both ratios are unbounded, neither function is big- 0 of the other.
29. a) We loop through all pairs (i, j) with i < j and check whether a,+ a1
all values of k).
= ak for some
k (by looping through
procedure brute(a 1 , a2, ... , an : integers)
for i := 1 to n - 1
for j := i + 1 to n
fork:= 1 ton
if a,+ a1 = ak then return true else return false
b) Because of the loop within a loop within a loop, clearly the time complexity is O(n 3).
31. There are six possible matchings. We need to determine which ones are stable. The matching [(m 1, wi),
( m 2, w 2), (m 3, w3)] is not stable, because m 3 and w 2 prefer each other over their current mate. The matching
[(m 1, w 1), (m 2, w3), (m 3, w 2)] is stable; although m 1 would prefer W3, she ranks him lowest, and men m2
and m 3 got their first picks. The matching [(m 1, w 2), (m 2, w1), (m 3, w3)] is stable; although m 1 would prefer
w 1 or w3 , they both rank him lowest, and m 2 and m 3 will not break their respective matches because
their potential girlfriends got their first choices. The matching [(m 1,w 2),(m 2,w 3),(m3,w 1)] is not stable,
because m 3 and w 3 prefer each other over their current mate. The matching [(m 1, w3), (m 2, w 1), (m 3, w2)]
is not stable, because m 2 and w3 will run off together. Finally, [(m 1,w 3),(m2,w 2),(m3,w 1)] is not stable,
again because m 3 and w3 will run off together. To summarize, the stable matchings are [(m 1, w1). (m2, w3),
(m 3, w 2 )] and [(m 1, w 2), (m 2, w1), (m 3, w3)]. Reading off this list, we see that the valid partners are as follows:
for m 1 : w1 and w 2 ; for m 2 : w1 and w3 ; for m3: w2 and W3; for w1 : m1 and m2; for w2: m1 and m3; and
for w3 : m2 and m3 .
Supplementary Exercises
111
33. We just take the definition of male optimal and female pessimal and interchange the sexes. Thus, a matching
in which each woman is assigned her valid partner ranking highest on her preference list is called female
optimal, and a matching in which each man is assigned his valid partner ranking lowest on his preference list
is called male pessimal.
35. a) We just rewrite the preamble to Exercise 60 in Section 3.1 with the appropriate modifications: Suppose
we have s men m 1 , m 2 , ... , ms and t women w 1 , w 2 , ... , Wt. We wish to match each person with a person
of the opposite gender, to the extent possible (namely, min(s, t) marriages). Furthermore suppose that each
person ranks, in order of preference, with no ties, the people of the opposite gender. We say that a matching
of people of opposite genders to form couples is stable if we cannot find a man m and a woman w who
are not assigned to each other such that m prefers w over his assigned partner (or lack of a partner) and w
prefers m to her assigned partner (or lack of a partner). (We assume that each person prefers any mate to
being unmatched.)
b) The simplest way to do this is to create Js - tJ fictitious people (men or women, whichever is in shorter
supply) so that the number of men and the number of women become the same, and to put these fictitious
people at the bottom of the preference lists of the people of opposite gender. The preference lists for the
fictitious people can be specified arbitrarily. We then just run the algorithm as before.
c) Because being unmatched is least desirable, it follows immediately from the proof that the original algorithm produces a stable matching (Exercise 63 in Section 3.1) that the modified algorithm produces a stable
matching.
37. If we schedule the jobs in the order shown, then Job 3 will finish at time 20, Job 1 will finish at time 45,
Job 4 will finish at time 50, Job 2 will finish at time 65, and Job 5 will finish at time 75. Comparing these to
the deadlines, we see that only Job 2 is late, and it is late by 5 minutes (we'll call the time units minutes for
convenience). A similar calculation for the other order shows that Job 1 is 10 minutes late, and Job 2 is 15
minutes late, so the maximum lateness is 15.
39. Consider the first situation in Exercise 37. We saw there that it is possible to achieve a maximum lateness of
5. If we schedule the jobs in order of increasing time required, then Job 1 will be scheduled last and finish at
time 75. This will give it a lateness of 25, which gives a maximum lateness worse than the previous schedule.
(Using the algorithm from Exercise 40 gives a maximum lateness of only 5.)
41. a) We want to maximize the mass of the contents of the knapsack. So we can try each possible subset of the
available items to see which ones can fit into the knapsack (i.e., have total mass not exceeding W) and thereby
find a subset having the maximum mass. In detail, then, examine each of the 2n subsets S of { 1, 2, ... , n},
and for each subset compute the total mass of the corresponding items (the sum of w 1 for all j E S). Keep
track of the subset giving the largest such sum that is less than or equal to W, and return that subset as
the output of the algorithm. Note that this is quite inefficient, because the number of subsets to examine is
exponential. In fact, there is no known efficient algorithm for solving this problem.
b) Essentially we find the solution here by inspection. Among the 32 subsets of items we try (from the empty
set to the set of all five items), we stumble upon the subset consisting of the food pack and the portable stove,
which has total mass equal to the capacity, 18.
43. a) The makespan is always at least as large as the load on the processor assigned to do the lengthiest job, which
must obviously be at least max1 = 1 , 2 ,... ,n t1 . Therefore the minimum makespan L* satisfies this inequality.
b) The total amount of time the processors need to spend working on the jobs (the total load) is 2:7= 1 t 1 .
Therefore the average load per processor is ~ :L;=l t1 . The maximum load cannot be any smaller than the
average, so the makespan is always at least this large. It follows that the minimum makespan L * is at least
this large, as we were asked to prove.
112
Chapter 3
Algorithms
45. The algorithm will assign job 1 to processor 1 (one of the processors with smallest load at that point, namely 0),
assign job 2 to processor 2 (for the same reason), and assign job 3 to processor 3. The loads are 3, 5, and 4 at
this point. Then job 4 gets assigned to processor 1 because it has the lightest load; the loads are now 10, 5,
4. Finally, job 5 gets assigned to processor 3, giving it load 12, so the makespan is 12. Notice that we can do
better by assigning job 1 and job 4 to processor 1 (load 10), job 2 and job 3 to processor 2 (load 9), and job 5
to processor 3 (load 8), for a makespan of 10. Notice, too, that this latter solution is best possible, because
to achieve a makespan of 9, all three processors would have to have a load of 9, and this clearly cannot be
achieved.
WRITING PROJECTS FOR CHAPTER 3
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. Algorithms are as much a subject of computer science as they are a subject of mathematics. Thus sources
on the history of computer science (or perhaps even verbose introductory programming or comprehensive
computer science textbooks) might be a good place to look. See also [Ha2].
2. The original is [Ba2]. Good history of mathematics books would be a place to follow up.
3. See [Mel], or do a Web search.
4. Knuth's volume 3 [Kn] is the bible for sorting algorithms.
5. Look up Moore's law.
6. A modern book on algorithm design, such as [CoLe] is the obvious best source.
7. The Association for Computing Machinery (ACM) is the organization that bestows this prestigious award.
8. This is a vast and extremely important subject in computer science today. One basic book on the subject is
[Gi]; see also an essay in [De2].
9. See the comments for Writing Project 8, above. One basic algorithm to think about doing in parallel is
sorting. Imagine a group of school-children asked to arrange themselves in a row by height, and any child can
determine whether another child is shorter or taller than him/herself.
10. Many websites discuss NP-complete problems, including www.claymath.org/millennium/P _vs_NP, which discusses the $1,000,000 prize the Clay Institute has offered for finding an efficient algorithm for such problems
or showing that none exists. The classic book reference is [GaJo].
11. See [GaJo] or a website about NP-complete problems.
Section 4.1
Divisibility and Modular Arithmetic
113
CHAPTER4
Number Theory and Cryptography
SECTION 4.1
Divisibility and Modular Arithmetic
Number theory is playing an increasingly important role in computer science. This section and these exercises
just scratch the surface of what is relevant. Many of these exercises are simply a matter of applying definitions.
It is sometimes hard for a beginning student to remember that in order to prove something about a concept
(such as modular arithmetic), it is usually necessary to invoke the definition! Exercises 34-44 hint at the
rich structure that modular arithmetic has (sometimes resembling real number arithmetic more than integer
arithmetic). In many contexts in mathematics and computer science, modular arithmetic is more relevant and
convenient than ordinary integer arithmetic.
1. a) yes, since 68 = 17 · 4
c) yes, since 357 = 17 · 21
b) no, remainder= 16
d) no, remainder= 15
3. If a I b, then we know that b = at for some integer t. Therefore be = a( tc), so by definition a I be.
5. The given conditions imply that there are integers s and t such that a = bs and b = at. Combining these, we
obtain a = ats; since a =/=- 0, we conclude that st = 1. Now the only way for this to happen is for s = t = 1
or s = t = -1. Therefore either a= b or a= -b.
7. The given condition means that be= (ac)t for some integer t. Since c =/=- 0, we can divide both sides by c to
obtain b = at. This is the definition of a Ib, as desired.
9. In each case we need to find (the unique integers) q and r such that a= dq + r and 0:::; r
d are the given integers. In each case q = La/ dJ.
< d, where a and
b) -111 = 11 · ( -11) + 10, so q = -11 and r = 10
c) 789=23·34+7,so q=34 and r=7
d) 1001=13·77+0,so q=77 and r=O
e) 0 = 19 · 0 + 0 , so q = 0 and r = 0
f) 3 = 5 · 0 + 3, so q = 0 and r = 3
g) -1 = 3 · (-1) + 2, so q = -1 and r = 2
h) 4 = 1 · 4 + 0, so q = 4 and r = 0
a) 19 = 7 · 2 + 5, so q = 2 and r = 5
11. We are doing arithmetic modulo 12 for this exercise.
a) Because 11+80 mod 12
= 7,
the clock reads 7:00.
b) Because 12 - 40 mod 12 = -28 mod 12 = -28 + 36 mod 12 = 8, the clock reads 8:00.
c) Because 6 + 100 mod 12 = 10, the clock reads 10:00.
13. In each case we merely have to compute the expression on the right mod 13. This means dividing it by 13
and taking the (nonnegative) remainder.
a) 9 · 4 mod 13 = 36 mod 13
= 10
b) 11·9 mod 13
= 99 mod
13 = 8
c) 4 + 9 mod 13 = 13 mod 13 = 0
d) 2 · 4 + 3 · 9 mod 13 = 35 mod 13
2
2
e) 4 + 9 mod 13 = 97 mod 13 = 6
f) 4 3 - 93 mod 13 = -665 mod 13 = 11 (because -665 = -52 · 13 + 11)
=9
Chapter 4
114
Number Theory and Cryptography
15. The given condition, that a mod m = b mod m, means that a and b have the same remainder when divided
by m. In symbols, a = qi m + r and b = q2m + r for some integers qi , q2, and r. Subtracting these two
equations gives us a - b = (qi - q2)m, which says that m divides (is a factor of) a - b. This is precisely the
definition of a= b (mod m).
17. The quotient n/k lies between two consecutive integers, say b-1 and b, possibly equal to b. In symbols, there
exists a positive integer b such that b - 1 < n/k::; b. In particular, fn/kl = b. Also, since n/k > b- 1, we
have n > k(b-1), and so (since everything is an integer) n-1 2: k(b-1). This means that (n -1)/k 2: b-1,
so L(n-1)/kj 2'.b-1. Ontheotherhand, L(n-1)/kj :S(n-1)/k fm/2l
Note that if m is even, then we can, alternatively, take f(m/2) = -m/2.
21. For these problems, we need to perform the division (as in Exercise 9) and report the remainder.
a) 13 = 3 · 4 + 1, so 13 mod 3 = 1
b) -97 = 11 · ( -9) + 2, so -97 mod 11 = 2
c) 155=19·8+3,so 155mod19=3
d) -221=23·(-10)+9,so -221mod23=9
23. Recall that a div m and a mod m are the integer quotient and remainder when a is divided by m.
a) Because 228 = 1 · 119 + 109, we have 228 div 119 = 1 and 228 mod 119 = 109.
b) Because 9009 = 40 · 223 + 89, we have 9009 div 223 = 40 and 9009 mod 223 = 89.
c) Because -10101 = -31 · 333 + 222, we have -10101 div 333 = -31 and -10101 mod 333 = 222. (Note
that 10101 -;- 333 is 30 ~~~, so without the negative dividend we would get a different absolute quotient and
different remainder. But we have to round the negative quotient here, -30 ~~~ , down to -31 in order for the
remainder to be nonnegative.)
d) Because - 765432 = - 21 · 38271 + 38259, we have - 765432 div 38271 = - 21 and - 765432 mod 38271 =
38259.
25. a) Because -15 already satisfies the inequality, the answer is -15.
b) Because 24 is too large to satisfy the inequality, we subtract 31 and obtain the answer is - 7.
c) Because 99 is too smaU to satisfy the inequality, we add 41 and obtain the answer is 140.
27. We just need to start at -1 and repeatedly subtract or add 25 until we exceed the desired range. Thus the
negative values we seek are -1, -26, -51, and -76, and the positive values are 24, 49, 74, and 99.
29. For these problems, we need to divide by 17 and see whether the remainder equals 5. Remember that the
quotient can be negative, but the remainder r must satisfy 0 ::; r < 17.
a) 80 = 17 · 4 + 12 , so 80
"I-
5 (mod 17)
c) -29 = 17 · (-2) + 5, so -29
b) 103 = 17 · 6 + 1, so 103
= 5 (mod 17)
"I-
5 (mod 17)
d) -122 = 17·(-8)+14, so -122-=/=- 5 (mod 17)
Section 4.1
Divisibility and Modular Arithmetic
115
31.
= 13, so the answer is 13.
457 · 182 = 20 · 21 = 420 = 6.
a) Working modulo 23, we have -133
b) Working modulo 23, we have
+ 261 =
128
= (3 2 mod 32) 3 mod 15 = 93 mod 15 = 729 mod 15 = 9
b) (34 mod 17) 2 mod 11 = (81 mod 17) 2 mod 11 = 13 2 mod 11 = 22 mod 11 = 4
c) (19 3 mod 23) 2 mod 31 = ((-4) 3 mod 23) 2 mod 31 = (-64 mod 23) 2 mod 31 = 52 mod 31 = 25
d) (89 3 mod 79) 4 mod 26 = (10 3 mod 79) 4 mod 26 = (1000 mod 79) 4 mod 26 = 52 4 mod 26
04 mod 26 = 0
33. a) (99 2 mod 32) 3 mod 15
35. The hypothesis a= b (mod m) means that m l(a - b). Since we are given that n Im, Theorem l(iii) implies
that n l(a - b). Therefore a= b (mod n), as desired.
37. a) To show that this conditional statement does not necessarily hold, we need to find an example in which
ac =be (mod m), but a¢. b (mod m). Let m = 4 and c = 2 (what is important in constructing this example
is that m and c have a nontrivial common factor). Let a = 0 and b = 2 . Then ac = 0 and be = 4, so
ac =be (mod 4), but 0 ¢. 2 (mod 4).
b) To show that this conditional statement does not necessarily hold, we need to find an example in which
a= b (mod m) and c d (mod m), but ac ¢. bd (mod m). If we try a few randomly chosen positive integers,
we will soon find one. Let m = 5, a= 3, b = 3, c = 1, and d = 6. Then ac = 3 and bd = 729 4 (mod 5),
so 3 1 ¢. 36 (mod 5), even though 3 3 (mod 5) and 1 6 (mod 5).
=
=
=
=
39. By Exercise 38 the sum of two squares must be either 0 + 0, 0 + 1, or 1 + 1, modulo 4. Therefore the sum
cannot be 3 modulo 4, which means that it cannot be of the form 4k
+ 3.
41. There are at least two ways to prove this. One way is to invoke Theorem 5 repeatedly. Since a= b (mod m),
Theorem 5 implies that a· a = b · b (mod m), i.e., a 2 = b2 (mod m). Invoking Theorem 5 again, since
a= b (mod m) and a 2 = b2 (mod m), we obtain a 3 = b3 (mod m). After k - 1 applications of this process,
we obtain ak
bk (mod m), as desired. (This is really a proof by mathematical induction, a topic to be
considered formally in Chapter 5.)
=
Alternately, we can argue directly, using the algebraic identity ak-bk = (a-b)(ak-l +ak- 2 b+· · ·+abk- 2 +
bk-l). Specifically, the hypothesis that a= b (mod m) means that m l(a - b). Therefore by Theorem l(ii),
m divides the right-hand side of this identity, so m l(ak - bk). This means precisely that ak =bk (mod m).
43. The closure property states that a ·m b E Zm whenever a, b E Zm. Recall that Zm = {O, 1, 2, ... , m - 1}
and that a ·m b is defined to be (a · b) mod m. But this last expression will by definition be an integer in
the desired range. To see that multiplication is associative, we must show that (a ·m b) ·m c =a ·m (b ·m c).
This is equivalent to
((a· b mod m) · c) mod m =(a· (b · c mod m)) mod m.
This is true, because both sides equal (a· b · c) mod m (multiplication of integers is associative). Similarly,
multiplication in Zm is commutative because multiplication in Z is commutative, and 1 is the multiplicative
identity for Zm because 1 is the multiplicative identity for Z.
45. We will use + and · for these operations to save space and improve the appearance of the table. Notice
that we really can get by with a little more than half of this table if we observe that these operations are
commutative; then we would need to list a
+ b and
a · b only for a ::;: b.
Chapter 4
116
Number Theory and Cryptography
0+0=0
0+1=1
0+2=2
0+3=3
0+4=4
1+0=1
1+1=2
1+2=3
1+3=4
1+4=0
2+0=2
2+1=3
2+2=4
2+3=0
2+4=1
3+0=3
3+1=4
3+2=0
3+3=1
3+4=2
4+0=4
4+1=0
4+2=1
4+3=2
4+4=3
0 ·O = 0
0·1=0
0. 2 = 0
O· 3 = 0
0. 4 = 0
1·0 = 0
1·1=1
1·2 = 2
1·3 = 3
1·4 = 4
2. 0 = 0
2 · 1=2
2. 2 = 4
2. 3 = 1
2. 4 = 3
3 ·O = 0
3·1=3
3. 2 = 1
3. 3 = 4
3. 4 = 2
4. 0 = 0
4·1=4
4. 2 = 3
4. 3 = 2
4. 4 = 1
47. If d = 1, then f(a) =a and g(a) = 0. Therefore f is clearly one-to-one and onto, and g is neither. If
d > 1, then f is still onto, because f (db) = b for any desired b E Z, but it is clearly not one-to-one, because
f (0) = f (1) = 0. Furthermore, g is clearly not onto, because its range is just {O, 1, 2, ... , d - 1}, and it is not
one-to-one because g(O) = g( d) = 0.
SECTION 4.2
Integer Representations and Algorithms
In addition to having some routine calculation exercises, this exercise set introduces other forms of representing
integers. These are balanced ternary expansion, Cantor expansion, binary coded decimal (or BCD)
representation, and one's and two's complement representations. Each has practical and/or theoretical
importance in mathematics or computer science. If all else fails, one can carry out an algorithm by "playing
computer" and mechanically fallowing the pseudocode step by step.
1. We divide repeatedly by 2, noting the remainders. The remainders are then arranged from right to left to
obtain the binary representation of the given number.
a) We begin by dividing 231 by 2, obtaining a quotient of 115 and a remainder of 1. Therefore a 0 = 1. Next
115/2 = 57, remainder 1. Therefore a 1 = 1. Similarly 57 /2 = 28, remainder 1. Therefore a 2 = 1. Then
28/2 = 14, remainder 0, so a 3 = 0. Similarly a 4 = 0, after we divide 14 by 2, obtaining 7 with remainder 0.
Three more divisions yield quotients of 3, 1, and 0, with remainders of 1, 1, and 1, respectively, so a 5 = a 6 =
a 7 = 1. Putting all this together, we see that the binary representation is (a 7a 6a 5a4a 3a 2a 1a 0 )2 = (1110 0111)2.
As a check we can compute that 2° + 2 1 + 22 + 25 + 26 + 27 = 231.
b) Following the same procedure as in part (a), we obtain successive remainders 0, 0, 1, 0, 1, 1, 0, 1, 1,
0, 0, 0, 1. Therefore 4532 = (1 0001 1011 0100)2.
c) By the same method we obtain 97644
=
(1 0111 1101 0110 1100)2.
3. a) (1 1111)2 = 24 + 23 + 22 + 2 1 + 2° = 16 + 8 + 4 + 2 + 1
that (1 1111) 2 = (10 0000)2 - 1 = 25 - 1 = 31.
b) (10 0000 0001)2 = 29 + 2°
= 31.
An easier way to get the answer is to note
= 513
c) (1 0101 0101)2 = 2 + 2 + 24 + 22 + 2° = 256 + 64 + 16 + 4 + 1 = 341
8
6
d) (110 1001 0001 0000)2 = 2 14 + 2 13 + 2 11 + 28 + 24 = 16384 + 8192 + 2048 + 256 + 16 = 26896
Section 4.2
Integer Representations and Algorithms
117
5. In each case we follow the idea given in Example 7, converting each octal digit to its binary equivalent
(including leading O's where necessary). Note that by convention we then group the binary digits into groups
of fours, starting at the right.
a) Since (5) 8
= (101)2, (7) 8 = (lll)z, and (2) 8 = (010)2, we have (572) 8 = (1 0111 1010)2.
b) We concatenate 1, 110, 000, and 100 to obtain (11 1000 0100)2.
c) (1 0001 0011)2
d) (101 0000 1111)2
7. Following Example 7, we simply write the binary equivalents of each digit: (A)16
(C)16 = (1100)2, (D)16 = (1101)2, (E)16 = (1110)2, and (F) 16
of four binary digits is just for readability by humans.
= (1010)2, (B) 16 = (1011)2,
= (1111)2. Note that the blocking by groups
a) (80E)16 = (1000 0000 1110)2
b) (135AB)16 = (0001 0011 0101 1010 1011)2
c) (ABBA) 16 = (1010 1011 1011 1010)2
d) (DEFACED) 16
= (1101 1110 1111 1010 1100 1110 1101) 2
9. Following Example 7, we simply write the binary equivalents of each digit. Since (A) 16 = (1010)2, (B) 16 =
(1011)2, (C)16 = (1100)2, (D)16 = (1101)2, (E)16 = (1110)2, and (F)15 = (1111)2, we see that (ABCDEF) 16
= (101010111100110111101111)2. Following the convention shown in Exercise 3 of grouping binary digits by
fours, we can write this in a more readable form as 1010 1011 1100 1101 1110 1111.
11. Following Example 7, we simply write the hexadecimal equivalents of each group of four binary digits. Thus
we have (1011 0111 1011)2 = (B7B)16.
13. We adopt a notation that will help with the explanation. Adding up to three leading O's if necessary, write
the binary expansion as ( ... b23b22b21b20b13b12bnb10bo3bo2bo1booh. The value of this numeral is boo+ 2bo1 +
4b02 + 8b03 + 24 b10 + 25 bn + 26 b12 + 27 b13 + 28 b20 + 29 b21 + 210 b22 + 2nb23 + · · ·, which we can rewrite as
boo+ 2bo1+4bo2 + 8bo3 + (b10 + 2b11+4b12 + 8b13) · 24 + (b20 + 2b21+4b22 + 8b23) · 28 + · · ·. Now (b,3b,2b,1b,oh
translates into the hexadecimal digit h, . So our number is ho+ h 1 · 24 + h 2 · 28 + · · · = ho + h 1 · 16 + h2 · 16 2 + · · · ,
which is the hexadecimal expansion ( ... h1h 1 ho)l6.
15. We adopt a notation that will help with the explanation. Adding up to two leading O's if necessary, write
the binary expansion as ( ... b22b21b20b12b11b10bo2bo1booh. The value of this numeral is b00 + 2bo1 + 4bo2 +
23b10 + 24 bn + 25 b12 + 26 b20 + 27 b21 + 28 b22 + · · ·, which we can rewrite as boo+ 2bo 1 + 4bo2 + (b10 + 2bn +
4b12) · 23 + (b20 + 2b21 + 4b22) · 26 + · · ·. Now (b, 2biib,o)z translates into the octal digit h,. So our number is
ho+ h1 · 23 + h2 · 26 + · · · = ho+ h 1 · 8 + h2 · 82 + · · ·, which is the octal expansion (... h 1h 1 h 0 ) 8 .
17. In each case we follow the method of Example 7, blocking by threes instead of fours. We replace each octal
digit of the given numeral by its 3-digit binary equivalent and string the digits together. The first digit is
(7)s = (111)2, the next is (3)s = (011)2, and so on, so we obtain (1 1101 1100 1010 1101 0001)2. For the
other direction, we split the given binary numeral into blocks of three digits, adding initial O's to fill it out:
001 010 111 011. Then we replace each block by its octal equivalent, obtaining the answer (1273) 8 .
19. Since we have procedures for converting both octal and hexadecimal to and from binary (Example 7), to
convert from octal to hexadecimal, we first convert from octal to binary and then convert from binary to
hexadecimal.
21. We can just add and multiply using the grade-school algorithms, working with these very simple addition and
multiplication tables: 0 + 0 = 0, 0 + 1 = 1 + 0 = 1, 1 + 1 = 10, which means that we "carry" the 1 into the
Chapter 4
118
Number Theory and Cryptography
next column; 0 · 0 = 0 · 1 = 1 · 0 = 0, 1 · 1 = 1. See Examples 8 and 10. Note that we can check our work
by converting everything to decimal numerals (the check is shown in parentheses below). For convenience, we
leave off the ''2" subscripts throughout.
a) 100 0111+1110111=1011 1110 (decimal: 71+119 = 190)
100 0111·1110111=10 0001 0000 0001 (decimal: 71·119 = 8449)
b) 11101111+10111101=110101100 (decimal: 239+189=428)
1110llll·10111101=101100000111 0011 (decimal: 239·189 = 45,171)
c) 10 10101010+111110000=100 1001 1010 (decimal: 682 + 496 = 1178)
10 1010 1010 · 1 11110000=10100101001 0110 0000 (decimal: 682 · 496 = 338,272)
d) 10 0000 0001+11 1111 1111=110 0000 0000 (decimal: 513 + 1023 = 1536)
10 0000 0001·1111111111=1000 0000 000111111111 (decimal: 513 · 1023 = 524,799)
23. We can just add and multiply using the grade-school algorithms (working column by column starting at the
right), using the addition and multiplication tables in base eight (for example, 5 + 6 = 13 and 5 · 6 = 36).
When a digit-by-digit answer is too large to fit (i.e., greater than 7). we '"carry" into the next column. Note
that we can check our work by converting everything to decimal numerals (the check is shown in parentheses
below). For convenience. we leave off the "8'' subscripts throughout.
a) 763 + 147 = 1132 (decimal: 499 + 103 = 602)
763 · 147 = 144,305 (decimal: 499 · 103 = 51,397)
b) 6001 + 272 = 6273 (decimal: 3073 + 186 = 3259)
6001 · 272 = 2,134,272 (decimal: 3073 · 186 = 571,578)
c) 1111+777 = 2110 (decimal: 585 + 511=1096)
1111·777 = 1,107,667 (decimal: 585 · 511=298,935)
d) 54321+3456 = 57,777 (decimal: 22,737 + 1838 = 24,575)
54321 · 3456 = 237,326.216 (decimal: 22,737 · 1838 = 41,790,606)
25. In effect, this algorithm computes 7 mod 645, 72 mod 645, 74 mod 645, 78 mod 645, 7 16 mod 645,
... , and then multiplies (modulo 645) the required values. Since 644 = (1010000100)2, we need to multiply
together 74 mod 645, 7128 mod 645, and 7512 mod 645, reducing modulo 645 at each step. We compute by
repeatedly squaring: 72 mod 645 = 49, 74 mod 645 = 49 2 mod 645 = 2401 mod 645 = 466, 78 mod 645 =
466 2 mod 645 = 217156 mod 645 = 436, 716 mod 645 = 436 2 mod 645 = 190096 mod 645 = 466. At this
point we see a pattern with period 2, so we have 7 32 mod 645 = 436, 764 mod 645 = 466, 7128 mod 645 =
436, 7256 mod 645 = 466, and 7512 mod 645 = 436. Thus our final answer will be the product of 466, 436,
and 436, reduced modulo 645. We compute these one at a time: 466 · 436 mod 645 = 203176 mod 645 = 1,
and 1 · 436 mod 645 = 436. So 7644 mod 645 = 436. A computer algebra system will verify this; use the
command "7 &- 644 mod 645;" in Maple, for example. The ampersand here tells Maple to use modular
exponentiation, rather than first computing the integer 7644 , which has over 500 digits, although it could
certainly handle this if asked. The point is that modular exponentiation is much faster and avoids having to
deal with such large numbers.
27. In effect, this algorithm computes 3 mod 99, 32 mod 99, 34 mod 99, 38 mod 99, 3 16 mod 99, ... ,
and then multiplies (modulo 99) the required values. Since 2003 = (11111010011)2, we need to multiply
together 3 mod 99, 32 mod 99, 316 mod 99, 364 mod 99, 3128 mod 99, 3256 mod 99, 3512 mod 99,
and 3 1024 mod 99, reducing modulo 99 at each step. We compute by repeatedly squaring: 32 mod 99 = 9,
34 mod 99 = 81, 38 mod 99 = 81 2 mod 99 = 6561 mod 99 = 27, 316 mod 99 = 27 2 mod 99 =
729 mod 99 = 36, 332 mod 99 = 36 2 mod 99 = 1296 mod 99 = 9, and then the pattern repeats, so
364 mod 99 = 81, 3128 mod 99 = 27, 3256 mod 99 = 36, 3512 mod 99 = 9, and 31024 mod 99 = 81. Thus
Section 4.2
Integer Representations and Algorithms
119
our final answer will be the product of 3, 9, 36, 81, 27, 36, 9, and 81. We compute these one at a time
modulo 99: 3 · 9 is 27, 27 · 36 is 81, 81 · 81 is 27, 27 · 27 is 36, 36 · 36 is 9, 9 · 9 is 81, and finally 81 · 81 is
27. So 32003 mod 99 = 27.
29. The binary expansion of an integer represents the integer as a sum of distinct powers of 2. For example,
since 21 = (1 0101)2, we have 21 = 24 + 22 + 2°. Since binary expansions are unique, each integer can be so
represented uniquely.
31. Let the decimal expansion of the integer a be given by a = (an-lan- 2 ... a 1 a 0 )i 0 . Thus a = 10n- 1an-l +
10n- 2an-2 + · · · + l0a1 + ao. Since 10
1 (mod 3), we have a
an-I + an-2 + · · · + ai + ao (mod 3).
Therefore a= 0 (mod 3) if and only if the sum of the digits is congruent to 0 (mod 3). Since being divisible
by 3 is the same as being congruent to 0 (mod 3), we have proved that a positive integer is divisible by 3 if
and only if the sum of its decimal digits is divisible by 3.
=
=
33. Let the binary expansion of the positive integer a be given by a = (an_ 1an_ 2 ... a 1 a 0 )2. Thus a = a 0 +
2a1 + 22 a2 + · · · + 2n- 1an-l. Since 22
1 (mod 3), we see that 2k
1 (mod 3) when k is even, and
2k = 2 = -1 (mod 3) when k is odd. Therefore we have a= a0 - a 1 + a 2 - a 3 +···±an-I (mod 3). Thus
a = 0 (mod 3) if and only if the sum of the binary digits in the even-numbered positions minus the sum of
the binary digits in the odd-numbered positions is congruent to 0 modulo 3. Since being divisible by 3 is the
same as being congruent to 0 (mod 3), our proof is complete.
=
=
35. a) Since the leading bit is a 1, this represents a negative number. The binary expansion of the absolute value
of this number is the complement of the rest of the expansion, namely the complement of 1001, or 0110.
Since (0110)2 = 6, the answer is -6.
b) Since the leading bit is a 0, this represents a positive number, namely the number whose binary expansion
is the rest of this string, 1101. Since (1101)2=13, the answer is 13.
c) The answer is the negative of the complement of 0001, namely -(1110) 2 = -14.
d) -(0000)2 = O; note that 0 has two different representations, 0000 and 1111
37. We must assume that the sum actually represents a number in the appropriate range. Assume that n bits
are being used, so that numbers strictly between -2n-l and 2n-l can be represented. The answer is almost,
but not quite, that to obtain the one's complement representation of the sum of two numbers, we simply add
the two strings representing these numbers using Algorithm 3. Instead, after performing this operation, there
may be a carry out of the left-most column; in such a case, we then add 1 more to the answer. For example,
suppose that n = 4; then numbers from -7 to 7 can be represented. To add -5 and 3, we add 1010 and
0011, obtaining 1101; there was no carry out of the left-most column. Since 1101 is the one's complement
representation of -2, we have the correct answer. On the other hand, to add -4 and -3, we add 1011
and 1100, obtaining 1 0111. The 1 that was carried out of the left-most column is instead added to 0111,
yielding 1000, which is the one's complement representation of - 7. A proof that this method works entails
considering the various cases determined by the signs and magnitudes of the addends.
2
39. If m is positive (or 0), then the leading bit (an-I) is 0, so the formula reads simply m = 2=~:0 a,2', which
is clearly correct, since this is the binary expansion of m. (See Section 2.4 for the meaning of summation
notation. This symbolism is a shorthand way of writing a0 + 2a 1 + 4a 2 + · · · + 2n- 2 an_ 2 .) Now suppose that
m is negative. The one's complement expansion for m has its leading bit equal to 1. By the definition of one's
complement, we can think of obtaining the remaining n - 1 bits by subtracting -m, written in binary, from
111 ... 1 (with n - 1 1 's), since subtracting a bit from 1 is the same thing as complementing it. Equivalently,
if we view the bit string (an-2an-l ... a0 ) as a binary number, then it represents (2n-l - 1) - (-m). In
Chapter 4
120
symbols, this says that (2n-l -
Number Theory and Cryptography
1) - (-m) = '2:~:02 ai2'. Solving for m gives us the equation we are trying
to prove (since an-1 = 1 ) .
41. Following the definition, if the first bit is a 0, then we just evaluate the binary expansion. If the first bit is a 1,
then we find what number x is represented by the remaining four bits in binary; the answer is then -(2 4 -x).
a) Since the first bit is a 1. and the remaining bits represent the number 9, this string represents the number
-(2 4 - 9) = -7.
b) Since the first bit is a 0 and this is just the binary expansion of 13, the answer is 13.
c) Since the first bit is a 1. and the remaining bits represent the number 1, this string represents the number
-(2 4 -1) = -15.
d) Since the first bit is a 1, and the remaining bits represent the number 15, this string represents the number
-(2 4 -15) = -1. Note that 10000 would represent -(2 4 - 0) = -16, so in fact we can represent one extra
negative number than positive number with this notation.
43. The nice thing about two's complement arithmetic is that we can just work as if it were all in base 2, since
-x (where x is positive) is represented by 2" - x; in other words, modulo 2", negative numbers represent
themselves. However, if overflow occurs, then we must recognize an error. Let us look at some examples,
where n = 5 (i.e., we use five bits to represent numbers between -15 and 15 ). To add 5 + 7, we write
00101 + 00111 = 01100 in base 2, which gives us the correct answer, 12. However, if we try to add 13 + 7
we obtain 01101+00111 = 10100, which represents -12, rather than 20, so we report an overflow error. (Of
course these two numbers are congruent modulo 32.) Similarly, for 5 + (-7), we write 00101+11001=11110
in base 2, and 11110 is the two's complement representation of -2, the right answer. For (-5) + (-7), we
write 11011+11001=110100 in base 2; if we ignore the extra 1 in the left-most column (which doesn't exist),
then this is the two's complement representation of -12, again the right answer. To summarize, to obtain
the two's complement representation of the sum of two integers given in two's complement representation,
add them as if they were binary integers, and ignore any carry out of the left-most column. However, if the
left-most digits of the two addends agree and the left-most digit of the answer is different from their common
value, then an overflow has occurred, and the answer is not valid.
45. If m is positive (or 0), then the leading bit ( an-l) is 0, so the formula reads simply m = 2=~:0 ai2i, which
is clearly correct, since this is the binary expansion of m. (See Section 2.4 for the meaning of summation
notation. This symbolism is a shorthand way of writing a 0 + 2a 1 + 4a 2 + · · · + 2n- 2 an_ 2 .) Now suppose that
m is negative. The two's complement expansion for m has its leading bit equal to 1. By the definition of
two's complement, the remaining n - 1 bits are the binary expansion of 2n-l - (-m). In symbols, this says
2
that 2n-l - ( -m) = 2=~:0 a, 2' . Solving for m gives us the equation we are trying to prove (since an-l = 1).
2
47. Clearly we need 4n digits, four for each digit of the decimal representation.
49. To find the Cantor expansion, we will work from left to right. Thus the first step will be to find the largest
number n whose factorial is still less than or equal to the given positive integer x. Then we determine the
digits in the expansion, starting with an and ending with a 1 .
Section 4.2
Integer Representations and Algorithms
121
procedure Cantor(x: positive integer)
n := 1; factorial := 1
while (n + 1) ·factorial ::::; x
n := n + 1
factorial := factorial · n
{at this point we know that there are n digits in the expansion}
y := x {this is just so we do not destroy the original input}
while n > 0
an:= LY/factorialJ
y := y - an . factorial
factorial := factorial/n
n := n -1
{we are done: x = ann! + an-i(n -1)! + · · · + a22! +ail!}
51. Note that n = 5. Initially the carry is c = 0, and we start the for loop with j = 0. Since a0 = 1 and b0 = 0,
we set d to be L(l + 0 + 0)/2J = O; then so= 1+0 + 0- 2 · 0, which equals 1, and finally c = 0. At the end
of the first pass, then, the right-most digit of the answer has been determined (it's a 1), and there is a carry
of 0 into the next column.
Now j = 1, and we compute d to be L(ai +bi+ c)/2J = L(l + 1+0)/2J = l; whereupon si becomes
1 + 1 + 0 - 2 · 1 = 0, and c is set to 1. Thus far we have determined that the last two bits of the answer are
01 (from left to right), and there is a carry of 1 into the next column.
The next three passes through the loop are similar. As a result of the pass when j = 2 we set d = 1,
s2 = 0, and then c = 1. When j = 3, we obtain d = 1, s 3 = 0, and then c = 1. Finally, when j = 4, we
obtain d = 1, S4 = 1, and then c = 1. At this point the loop is terminated, and when we execute the final
step, s 5 = 1. Thus the answer is 11 0001.
53. We will assume that the answer is not negative, since otherwise we would need something like the one's
complement representation. The algorithm is similar to the algorithm for addition, except that we need to
borrow instead of carry. Rather than trying to incorporate the two cases (borrow or no borrow) into one, as
was done in the algorithm for addition, we will use an if. .. then statement to treat the cases separately. The
notation is the usual one: a= (an-i ... aiao)z and b = (bn-i ... bibo)z
procedure subtract( a, b: nonnegative integers)
borrow:= 0
for j := 0 to n - 1
if a1 - borrow 2:: b1 then
Sj := a1 - borrow - bj
borrow:= 0
else
Sj := a1 + 2 - borrow - b1
borrow:= 1
{assuming a 2:: b, we have a - b = (sn-iSn-2 ... siso)z}
55. To determine which of two integers (we assume they are nonnegative), given in binary as a = (an-i ... aiao)z
and b = (bn-i ... bibo)z, is larger, we need to compare digits from the most significant end ( i = n - 1) to the
least ( i = 0), stopping if and when we find a difference. For variety here we record the answer as a character
string; in most applications it would probably be better to set compare to one of three code values (such as
-1, 1, and 0) to indicate which of the three possibilities held.
122
Chapter 4
Number Theory and Cryptography
procedure compare( a, b: nonnegative integers)
i := n -1
while i > 0 and a, =bi
i := i - 1
if a; > b, then answer:= "a> b"
else if ai < b, then answer:= "a< b"
else answer:= "a= b"
return answer
57. There is one division for each pass through the while loop. Also, each pass generates one digit in the base b
expansion. Thus the number of divisions equals the number of digits in the base b expansion of n. This is
just l logb n J + 1 (for example, numbers from 10 to 99, inclusive, have common logarithms in the interval
[1,2)). Therefore exactly llogbnJ + 1 divisions are required, and this is O(logn). (We are counting only the
actual division operation in the statement q := lq/bJ. If we also count the implied division in the statement
ak := q mod b, then there are twice as many as we computed here. The big- 0 estimate is the same, of
course.)
59. The only time-consuming part of the algorithm is the while loop, which is iterated q times. The work done
inside is a subtraction of integers no bigger than a, which has log a bits. The results now follows from
Example 9.
SECTION 4.3
Primes and Greatest Common Divisors
The prime numbers are the building blocks for the natural numbers in terms of multiplication, just as the
elements (like carbon, oxygen, or uranium) are the building blocks of all matter. Just as we can put two
hydrogen atoms and one oxygen atom together to form water, every composite natural number is uniquely
constructed by multiplying together prime numbers. Analyzing numbers in terms of their prime factorizations
allows us to solve many problems, such as finding greatest common divisors. Prime numbers have fascinated
people for millennia, and many easy-to-state questions about them remain unanswered. Students interested
in pursuing these topics more should definitely consider taking a course in number theory.
1. In each case we can just use trial division up to the square root of the number being tested.
a) Since 21 = 3 · 7, we know that 21 is not prime.
b) Since 2 ,(29, 3 ,(29, and 5 ,(29, we know that 29 is prime. We needed to check for prime divisors only up
to J2§, which is less than 6.
c) Since 2 ,( 71, 3 ,( 71 , 5 ,( 71, and 7 ,( 71, we know that 71 is prime.
d} Since 2 ,(97, 3 ,(97, 5 ,(97, and 7 ,(97, we know that 97 is prime.
= 3 · 37, we know that 111 is not prime.
143 = 11 · 13, we know that 143 is not prime.
e) Since 111
f) Since
3. In each case we can use trial division, starting with the smallest prime and increasing to the next prime once
we find that a given prime no longer is a divisor of what is left. A calculator comes in handy. Alternatively,
one could use a factor tree.
a) We note that 2 is a factor of 88, and the quotient upon division by 2 is 44. We divide by 2 again, and then
again, leaving a quotient of 11. Since 11 is prime, we are done, and we have found the prime factorization:
88 = 23 · 11.
b) 126 = 2 . 63 = 2 . 3 . 21 = 2 . 3 . 3 . 7 = 2 . 32 . 7
= 3 . 243 = 3 . 3 . 81 = 3 . 3 . 3 . 27 = 3 . 3 . 3 . 3 . 9 = 3 . 3 . 3 . 3 . 3 . 3 = 36
1001 = 7. 143 = 7. 11 . 13
c) 729
d}
Section 4.3
Primes and Greatest Common Divisors
e) 1111=ll·101 (we know that 101 is prime because we have already tried all prime factors less than
123
v'IOI)
f) 909090 = 2·454545 = 2·3·151515 = 2·3·3·50505 = 2·3·3·3·16835 = 2·3·3·3·5·3367= 2·3·3·3·5·7·481 =
2 . 3 . 3 . 3 . 5 . 7 . 13 . 37 = 2 . 3 3 . 5 . 7 . 13 . 37
5. 10! = 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 = 2 · 3 · 2 2 · 5 · (2 · 3) · 7 · 23 . 32 · (2. 5) = 28 . 34 . 52 . 7
7. The input is an integer n greater than 1. We try dividing it by all integers from 2 to yn, and if we find one
that leaves no remainder then we know that n is not prime. The pseudocode below accomplishes this.
procedure primetester(n: integer greater than 1)
isprime :=true
d := 2
while isprime and d :::; yn
if n mod d = 0 then isprime :=false
else d := d + 1
return isprime
9. We use what we know about factoring from algebra. In particular, we know that am+ 1 = (a+ l)(am-l am- 2 + am- 3 + · · · - 1). (Notice that this works if and only if m is odd, because the final sign has to be a
plus sign.) Because a and m are both greater than 1, we know that 1 < a+ 1 < am+ 1. This provides a
factoring of am+ 1 into two proper factors, so am+ 1 is composite.
11. We give a proof by contradiction. Suppose that in fact log 2 3 is the rational number p / q, where p and q are
integers. Since log 2 3 > 0, we can assume that p and q are positive. Translating the equation log 2 3 = p / q
into its exponential equivalent, we obtain 3 = 2p/q. Raising both sides to the qth power yields 3q = 2P. Now
this is a violation of the Fundamental Theorem of Arithmetic, since it gives two different prime factorizations
of the same number. Hence our assumption (that log 2 3 is rational) must be wrong, and we conclude that
log 2 3 is irrational.
13. This is simply an existence statement. To prove that it is true, we need only exhibit the primes. Indeed, 3,
5, and 7 satisfy the conditions. (Actually, this is the only example, and a harder problem is to prove that
there are no others.)
15. The prime factors of 30 are 2, 3, and 5. Thus we are looking for positive integers less than 30 that have none
of these as prime factors. Since the smallest prime number other than these is 7, and 72 is already greater
than 30, in fact only primes (and the number 1) will satisfy this condition. Therefore the answer is 1, 7, 11,
13, 17, 19, 23, and 29.
17. a) Since gcd(ll, 15) = 1, gcd(ll, 19) = 1, and gcd(15, 19) = 1, these three numbers are pairwise relatively
prime.
b) Since gcd(15,21) = 3 > 1, these three numbers are not pairwise relatively prime.
c) Since gcd(l2, 17) = 1, gcd(12, 31) = 1, gcd(12, 37) = 1, gcd(l 7, 31) = 1, gcd(l 7, 37) = 1, and gcd(31,
37) = 1, these four numbers are pairwise relatively prime. (Indeed, the last three are primes, and the prime
factors of the first are 2 and 3.)
d) Again, since no two of 7, 8, 9, and 11 have a common factor greater than 1, this set is pairwise relatively
prime.
19. The identity shown in the hint is valid, as can be readily seen by multiplying out the right-hand side (all the
terms cancel-telescope-except for 2ab and -1). We will prove the assertion by proving its contrapositive.
Suppose that n is not prime. Then by definition n = ab for some integers a and b each greater than 1 . Since
a > 1, 2a - 1, the first factor in the suggested identity, is greater than 1. Clearly the second factor is greater
than 1. Thus 2n - 1 = 2ab - 1 is the product of two integers each greater than 1, so it is not prime.
124
Chapter 4
Number Theory and Cryptography
21. We compute (n) here by enumerating the set of positive integers less than n that are relatively prime to n.
a) ¢(4) = l{l,3}1=2
b) ¢(10) = l{l,3, 7,9}1=4
c) ¢(13) = l{l, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}1 = 12
23. All the positive integers less than or equal to pk (and there are clearly pk of them) are less than pk and
relatively prime to pk unless they are a multiple of p. Since the fraction l/p of them are multiples of p, we
have (pk)= pk (l - 1/p) =pk - pk-1.
25. To find the greatest common divisor of two numbers whose prime factorizations are given, we just need to
take the smaller exponent for each prime.
a) The first number has no prime factors of 2, so the gcd has no 2's. Since the first number has seven factors
of 3, but the second number has only five, the gcd has five factors of 3. Similarly the gcd has a factor of 53 .
So the gcd is 35 . 53 .
b) These numbers have no common prime factors, so the gcd is 1 .
e) These numbers have no common prime factors, so the gcd is 1 .
c) 23 17
d) 41. 43. 53
f) The gcd of any positive integer and 0 is that integer, so the answer is 1111.
27. To find the least common multiple of two numbers whose prime factorizations are given, we just need to take
the larger exponent for each prime.
a) The first number has no prime factors of 2 but the second number has 11 of them, so the lcm has 11
factors of 2. Since the first number has seven factors of 3 and the second number has five, the lcm has seven
factors of 3. Similarly the lcm has a factor of 59 and a factor of 73 . So the lcm is 211 · 37 · 59 · 73 .
b) These numbers have no common prime factors, so the lcm is their product, 29 · 37 · 55 · 73 · 11 · 13 · 17.
c) 23 31
d) 41·43 · 53
e) 212 · 3 13 · 5 17 · 721 , as in part (b)
f) It makes no sense to ask for a positive multiple of 0, so this question has no answer. Least common multiples
are defined only for positive integers.
29. First we find the prime factorizations: 92928 = 28 · 3 · 11 2 and 123552 = 25 · 33 · 11 · 13. Therefore
gcd(92928, 123552) = 25 · 3 · 11 = 1056 and lcm(92928, 123552) = 28 · 33 · 11 2 · 13 = 10872576. The requested products are (2 5 · 3 · 11) · (2 8 · 33 · 11 2 · 13) and (2 8 · 3 · 11 2 ) · (2 5 · 33 · 11 · 13), both of which are
213 . 34 . 11 3 . 13 = 11,481,440,256.
31. The important observation to make here is that the smaller of any two numbers plus the larger of the two
numbers is always equal to the sum of the two numbers. Since the exponent of the prime p in gcd( a, b) is the
smaller of the exponents of p in a and in b, and since the exponent of the prime p in lcm(a, b) is the larger
of the exponents of p in a and in b, the exponent of p in gcd(a, b)lcm(a, b) is the sum of the smaller and
the larger of these two values. Therefore by the observation, it equals the sum of the two values themselves,
which is clearly equal to the exponent of p in ab. Since this is true for every prime p, we conclude that
gcd(a, b)lcm(a, b) and ab have the same prime factorizations and are therefore equal.
33. a) By Lemma 1, gcd(12, 18) is the same as the gcd of the smaller of these two numbers ( 12) and the remainder
when the larger ( 18) is divided by the smaller. In this case the remainder is 6, so gcd(12, 18) = gcd(12, 6).
Now gcd(12, 6) is the same as the gcd of the smaller of these two numbers ( 6) and the remainder when the
larger (12) is divided by the smaller, namely 0. This gives gcd(12,6) = gcd(6,0). But gcd(x,O) = x for all
positive integers, so gcd(6,0) = 6. Thus the answer is 6. In brief (the form we will use for the remaining
parts), gcd(12, 18) = gcd(l2,6) = gcd(6,0) = 6.
b) gcd(lll,201) = gcd(lll,90) = gcd(90,21) = gcd(21,6) = gcd(6,3) = gcd(3,0) = 3
Section 4.3
125
Primes and Greatest Common Divisors
c) gcd(lOOl, 1331) = gcd(lOOl, 330) = gcd(330, 11) = gcd(ll, 0) = 11
d) gcd(12345, 54321) = gcd(12345, 4941) = gcd(4941, 2463) = gcd(2463, 15) = gcd(15, 3) = gcd(3, 0) = 3
e) gcd(lOOO, 5040) = gcd(lOOO, 40) = gcd( 40, 0) = 40
f) gcd(9888, 6060) = gcd(6060, 3828) = gcd(3828, 2232) = gcd(2232, 1596) = gcd(1596, 636) = gcd(636, 324)
= gcd(324, 312) = gcd(312, 12) = gcd(12, 0) = 12
35. In carrying out the Euclidean algorithm on this data, we divide successively by 55, 34, 21, 13, 8, 5, 3, 2,
and 1, so nine divisions are required.
37. One can compute gcd(2a - 1, 2b - 1) using the Euclidean algorithm. Let us look at what happens when we
do so. If b = 1, then the answer is just 1, which is the same as 2gcd(a,b) - 1 in this case. Otherwise, we
reduce the problem to computing gcd(2b - 1, (2a - 1) mod (2b - 1)). Now from Exercise 36 we know that
this second argument equals 2a mod b - 1 . Therefore the exponents involved in the continuing calculation are
b and a mod b-exactly the same quantities that are involved in computing gcd( a, b) ! It follows that when
the process terminates, the answer must be 2gcd(a,b) - 1, as desired.
39. a) This first one is easy to do by inspection. Clearly 10 and 11 are relatively prime, so their greatest common
divisor is 1, and 1 = 11 - 10 = ( -1) · 10 + 1 · 11.
b) In order to find the coefficients s and t such that 21s + 44t = gcd(21, 44), we carry out the steps of the
Euclidean algorithm.
44 = 2. 21+2
21=10. 2+1
Then we work up from the bottom, expressing the greatest common divisor (which we have just seen to be 1)
in terms of the numbers involved in the algorithm, namely 44, 21, and 2. In particular, the last equation
tells us that 1 = 21 - 10 · 2, so that we have expressed the gcd as a linear combination of 21 and 2. But now
the first equation tells us that 2 = 44 - 2 · 21; we plug this into our previous equation and obtain
1 = 21 - 10 . (44 - 2 . 21) = 21 . 21 - 10 . 44.
Thus we have expressed 1 as a linear combination (with integer coefficients) of 21 and 44, namely gcd( 21, 44) =
21·21+(-10)·44.
c) Again, we carry out the Euclidean algorithm. Since 48 = 1·36+ 12, and 12 I 36, we know that gcd(36, 48) =
12. From the equation shown here, we can immediately write 12 = ( -1) · 36 + 48.
d) The calculation of the greatest common divisor takes several steps:
55 = 1·34 + 21
34 = 1. 21+13
21=1. 13 + 8
13
= 1·8 + 5
8=1·5+3
5=1·3+2
3=1·2+1
Then we need to work our way back up, successively plugging in for the remainders determined in this
Chapter 4
126
Number Theory and Cryptography
calculation:
1=3-2
= 3 - (5 - 3) = 2. 3 - 5
= 2. (8 - 5) - 5 = 2. 8 - 3. 5
= 2 . 8 - 3 . (13 - 8) = 5 . 8 - 3 . 13
= 5 . (21 - 13) - 3 . 13 = 5 . 21 - 8 . 13
= 5. 21 - 8. (34 - 21) = 13. 21 - 8. 34
= 13. (55 - 34) - 8. 34 = 13. 55 - 21 . 34
e) Here are the two calculations-down to the gcd using the Euclidean algorithm, and then back up by
substitution until we have expressed the gcd as the desired linear combination of the original numbers.
213 = 1. 117 + 96
117 = 1 . 96 + 21
96 = 4. 21+12
21=1. 12 + 9
12 = 1·9 + 3
Since 3 I9, we have gcd(ll 7, 213) = 3.
3 = 12 - 9
= 12 - (21 - 12) = 2 . 12 - 21
= 2. (96 - 4. 21) - 21 = 2. 96 - 9. 21
= 2 . 96 - 9 . ( 117 - 96) = 11 . 96 - 9 . 117
= 11. (213 - 117) - 9. 117 = 11. 213 - 20. 117
f) Clearly gcd(O, 223) = 223, so we can write 223 = s · 0 + 1 · 223 for any integer s.
g) Here are the two calculations-down to the gcd using the Euclidean algorithm, and then back up by
substitution until we have expressed the gcd as the desired linear combination of the original numbers.
2347 = 19. 123 + 10
123 = 12 . 10 + 3
10=3·3+1
Thus the greatest common divisor is 1.
1=10-3·3
= 10 - 3. (123 - 12. 10) = 37. 10 - 3. 123
= 37. (2347 - 19. 123) - 3. 123 = 37. 2347 - 706. 123
h) Here are the two calculations-down to the gcd using the Euclidean algorithm, and then back up by
substitution until we have expressed the gcd as the desired linear combination of the original numbers.
4666 = 3454 + 1212
3454 = 2 . 1212 + 1030
1212 = 1030 + 182
1030 = 5 . 182 + 120
182 = 120 + 62
120 = 62 + 58
62
=
58 + 4
58
=
14. 4 + 2
Section 4.3
Primes and Greatest Common Divisors
127
Since 2 I4, the greatest common divisor is 2.
2 = 58 -14. 4
= 58 - 14. (62 - 58) = 15. 58 - 14. 62
= 15 . (120 - 62) - 14. 62 = 15 . 120 - 29 . 62
= 15 . 120 - 29. (182 - 120) = 44. 120 - 29 . 182
= 44 . (1030 - 5 . 182) - 29 . 182 = 44. 1030 - 249 . 182
= 44 . 1030 - 249 . (1212 - 1030) = 293 . 1030 - 249 . 1212
= 293. (3454 - 2 . 1212) - 249 . 1212 = 293. 3454 - 835 . 1212
= 293. 3454 - 835. (4666 - 3454) = 1128. 3454 - 835 . 4666
i) Here are the two calculations-down to the gcd using the Euclidean algorithm, and then back up by
substitution until we have expressed the gcd as the desired linear combination of the original numbers.
11111=9999+1112
9999 = 8 . 1112 + 1103
1112 = 1103 + 9
1103 = 122 . 9 + 5
9=5+4
5=4+1
Thus 1 is the greatest common divisor.
1=5-4
= 5 - (9 - 5) = 2 . 5 - 9
= 2 . (1103 - 122 . 9) - 9 = 2 . 1103 - 245 . 9
= 2. 1103 - 245. (1112 - 1103) = 247. 1103 - 245. 1112
= 247. (9999 - 8. 1112) - 245. 1112 = 247. 9999 - 2221 . 1112
= 247. 9999 - 2221. (11111 - 9999) = 2468. 9999 - 2221. 11111
41. When we apply the Euclidean algorithm we obtain the following quotients and remainders: Qi = 0, r 2 = 26,
Q2 = 3, r 3 = 13, q3 = 2. Note that n = 3. Thus we compute the successive s's and t's as follows, using the
given recurrences:
s2 = so - Qi si = 1 - 0 · 0 = 1 ,
t2 = to - Qi ti = 0 - 0 · 1 = 0
t3 = ti - Q2t2 = 1 - 3 . 0 = 1
S3 = Si - Q2S2 = 0 - 3 · 1 = -3,
Thus we have s3a + t3b = (-3) · 26 + 1 · 91 = 13, which is gcd(26, 91).
43. When we apply the Euclidean algorithm we obtain the following quotients and remainders: Qi = 1, r 2 = 55,
Q2=l, r3=34, q3=l, r4=21, q4=l, rs=l3, Qs=l, r5=8, q5=l, r1=5, q7=l, rs=3, Qs=l,
rg = 2, q9 = 1, r 10 = 1, q 10 = 2. Note that n = 10. Thus we compute the successive s's and t's as follows,
using the given recurrences:
S2 =So - QiSi = 1 - 1·0 = 1,
t2 =to - Qiti = 0 - 1·1 = -1
t3 =ti - q2t2 = 1 - 1. (-1) = 2
S3 =Si - Q2S2 = 0 - 1·1 = -1,
t4 = t2 - q3t3 = -1 - 1 . 2 = -3
S4 = S2 - Q3S3 = 1 - 1 · ( -1) = 2,
Ss = S3 - Q4S4 = -1 - 1 · 2 = -3,
ts = t3 - q4t4 = 2 - 1 · ( -3) = 5
t5 = t4 - qsts = -3 - 1 . 5 = -8
S5 = S4 - QsSs = 2 - 1 · ( -3) = 5,
s7 = ss - q5s5 = -3 - 1 · 5 = -8,
t1 = ts - q5t5 = 5 - 1 · ( -8) = 13
128
Chapter 4
83 = 85 - q187 = 5 - 1 · ( -8) = 13,
89 = 87 - q 8 8 8 = -8 - 1 · 13 = -21,
810 = 83 - qg8g = 13 - 1 · ( -21) = 34,
Thus we have
810a
+ tiob =
Number Theory and Cryptography
q7 t1 = -8 - 1·13 = -21
tg = t1 - q8t 8 = 13 - 1 · (-21) = 34
tio =ts - q9 t 9 = -21 - 1 · 34 = -55
ts
= t5 -
34 · 144 + (-55) · 89 = 1, which is gcd(144, 89).
45. We start with the pseudocode for the Euclidean algorithm (Algorithm 1) and add variables to keep track of
the 8 and t values. We need three of them, since the new 8 depends on the previous two s's, and similarly
for t. We also need to keep track of q.
procedure extended Euclidean( a, b: positive integers)
x :=a
y := b
oldolds := 1
olds:= 0
oldoldt := 0
oldt := 1
while y "I 0
q := x div y
r := x mod y
x := y
y := r
8 := oldolds - q · olds
t := oldoldt - q · oldt
oldolds := olds
oldoldt := oldt
olds := 8
oldt := t
{ gcd( a, b) is x, and the Bezout coefficients are given by ( oldold8 )a + (oldoldt )b = x }
47. Obviously there are no definitive answers to these problems, but we present below a reasonable and satisfying
rule for forming the sequence in each case.
a) There are l's in the prime locations and O's elsewhere. In other words, the nth term of the sequence is 1
if n is a prime number and 0 otherwise.
b) The suspicious 2's occurring every other term and the appearance of the 11 and 13 lead us to discover
that the nth term is the smallest prime factor of n (and is 1 when n = 1 ) .
c) The nth term is the number of positive divisors of n. For example, the twelfth term is 6, since 12 has the
positive divisors 1, 2, 3, 4, 6, and 12. A tip-off to get us going in the right direction is that there are 2's in
the prime locations.
d) Perhaps the composer of the problem had something else in mind, but one rule here is that the nth term is
0 if and only if n has a repeated prime factor; the l's occur at locations for which n is "square-free" (has no
factor, other than 1, that is a perfect square). For example, 12 has the square 22 , so the twelfth term is 0.
e) We note that all the terms (after the first one) are primes. This leads us to guess that the nth term is the
largest prime less than or equal to n (and is 1 when n = 1).
f) Each term comes from the one before it by multiplying by a certain number. The multipliers are 2, 3,
5, 7, 11, 13, 17, 19, and 23~the primes. So the rule seems to be that we obtain the next term from the
nth term by multiplying by the nth prime number (and we start at 1). In other words, the nth term is the
product of the smallest n - 1 prime numbers.
49. Consider the product n(n + l)(n + 2) for some integer n. Since every second integer is even (divisible by 2),
this product is divisible by 2. Since every third integer is divisible by 3, this product is divisible by 3.
Therefore this product has both 2 and 3 in its prime factorization and is therefore divisible by 2 · 3 = 6.
Section 4.3
Primes and Greatest Common Divisors
129
51. It is hard to know how to get started on this problem. To some extent, mathematics is an experimental
science, so it would probably be a good idea to compute n 2 - 79n + 1601 for several positive integer values
of n to get a feel for what is happening. Using a computer, or at least a calculator, would be helpful. If
we plug in n = 1, 2, 3, 4, and 5, then we get the values 1523, 1447, 1373, 1301, and 1231, all of which are
prime. This may lead us to believe that the proposition is true, but it gives us no clue as to how to prove it.
Indeed, it seems as if it would be very hard to prove that this expression always produces a prime number,
since being prime means the absence of nontrivial factors, and nothing in the expression seems to be very
helpful in proving such a negative assertion. (The fact that we cannot factor it algebraically is irrelevant-in
fact, if it factored algebraically, then it would essentially never be prime.) Perhaps we should try some more
integers. If we do so, we find a lot more prime numbers, but we are still skeptical. Well, perhaps there is some
way to arrange that this expression will have a factor. How about 1601? Well, yes! If we let n = 1601, then
all three terms will have 1601 as a common factor, so that 1601 is a factor of the entire expression. In fact,
1601 2 - 79 · 1601 + 1601 = 1601 · 1523. So we have found a counterexample after all, and the proposition is
false. Note that this was not a problem in which we could proceed in a calm, calculated way from problem
to solution. Mathematics is often like that-lots of false leads and approaches that get us nowhere, and then
suddenly a burst of insight that solves the problem. (The smallest n for which this expression is not prime is
n = 80; this gives the value 1681 = 41 · 41.)
53. Here is one way to find a composite term in the sequence. If we set k = 1, then we get a + b. That number
is greater than 1, but it may not be composite. So let's increase k by a+ b, which will have the effect of
adding a multiple of a+ b to our previous answer, and we will therefore get a composite number, because
a+ b will be a nontrivial factor of it. So setting k =a+ b + 1 should work. Indeed, with that choice we have
ak + b = a(a + b + 1) + b = a 2 +ab+ a+ b, which factors nicely as (a+ l)(a + b). Since a and b are both
positive integers, both factors are greater than 1, and we have our composite number.
55. Recall that the proof that there are infinitely many primes starts by assuming that there are only finitely
many primes P1 , P2 , ... , Pn , and forming the number P1P2 · · · Pn + 1 . This number is either prime or has
a prime factor different from each of the primes p 1 , p 2 , ... , Pn; this shows that there are infinitely many
primes. So, let us suppose that there are only finitely many primes of the form 4k + 3, namely q1 , q2, ... , qn,
where q1 = 3, q2 = 7, and so on.
What number can we form that is not divisible by any of these primes, but that must be divisible by a
prime of the form 4k + 3? We might consider the number 4q1 q2 · · · qn + 3. Unfortunately, this number is not
prime, as it is is divisible by 3 (because q1 = 3). Instead we consider the number Q = 4q 1 q2 · · ·qn -1. Note
that Q is of the form 4k + 3 (where k = q1 q2 · · · qn - 1). If Q is prime, then we have found a prime of the
desired form different from all those listed. If Q is not prime, then Q has at least one prime factor not in
the list q1 , q2 , ... , qn , because the remainder when Q is divided by q1 is q1 - 1 , and q1 - 1 =I- 0. Therefore
q1 AQ for j = 1, 2, ... , n. Because all odd primes are either of the form 4k + 1 or of the form 4k + 3, and the
product of primes of the form 4k + 1 is also of this form (because (4k + 1) (4m + 1) = 4( 4km + k + m) + 1),
there must be a factor of Q of the form 4k + 3 different from the primes we listed. This complete the proof.
57. We need to show that this function is one-to-one and onto. In other words, if we are given a positive integer
x, we must show that there is exactly one positive rational number m/n (written in lowest terms) such that
K(m/n) = x. To do this, we factor x into its prime factorization and then read off them and n such that
K (m / n) = x. The primes that occur to even powers are the primes that occur in the prime factorization
of m, with the exponents being half the corresponding exponents in x; and the primes that occur to odd
powers are the primes that occur in the prime factorization of n, with the exponents being half of one more
than the exponents in x. Since this uniquely determines m and n, there is one and only one fraction, in
Chapter 4
130
Number Theory and Cryptography
lowest terms, that maps to x under K.
SECTION 4.4
Solving Congruences
Many of these exercises are reasonably straightforward calculations, but the amount of arithmetic involved in
some of them can be formidable. Look at the worked out examples in the text if you need help getting the
hang of it. The theoretical exercises, such as #18 and #19 give you a good taste of the kinds of proofs in an
elementary number theory course.
1. We simply need to show that 15 · 7
=1 (mod 26), or in other words, that 15 · 7 -
1 is divisible by 26. But
this quantity is 104, which is 26 · 4.
3. We want to find an integer k such that 4k is 1 greater than a multiple of 9. We compute 4 · 1 = 4 = 0 · 9 + 4,
4. 2 = 8 = 0. 9 + 8, 4. 3 = 12 = 1·9 + 3, 4. 4 = 16 = 1·9 + 7, 4. 5 = 20 = 2. 9 + 2, 4. 6 = 24 = 2. 9 + 6,
4 · 7 = 28 = 3 · 9 + 1. Therefore an inverse of 4 modulo 9 is 7.
5. a) Following the procedure of Example 2, we carry out the Euclidean algorithm to find gcd( 4, 9):
9=2·4+1
4= 4·1
Then we work backwards to rewrite the gcd (the last nonzero remainder, which is 1 here) in terms of 4 and 9:
1=9-2·4
Therefore the Bezout coefficients of 9 and 4 are 1 and -2, respectively. The coefficient of 4 is our desired
answer, namely -2, which is the same as 7 modulo 9. Note that this agrees with our answer in Exercise 3.
b) We proceed as above:
141=7.19 + 8
19 = 2. 8 + 3
8=2·3+2
3=1·2+1
2=2·1
Then we work backwards to rewrite the gcd (the last nonzero remainder, which is 1 here) in terms of 141
and 19:
1=3-1·2
= 3 - 1 . (8 - 2 . 3) = 3 . 3 - 1 . 8
= 3 . (19 - 2 . 8) - 1 . 8 = 3 . 19 - 7. 8
= 3. 19 - 7. (141 - 7. 19) = (-7). 141+52. 19
Therefore the Bezout coefficient of 19 is 52, and that is an inverse of 19 modulo 141.
c) We proceed as above:
89 = 1·55 + 34
55 = 1·34 + 21
34=1·21+13
21=1·13 + 8
13 = 1·8 + 5
8=1·5+3
5=1·3+2
3=1·2+1
2 = 1·2
Section 4.4
Solving Congruences
131
Then we work backwards to rewrite the gcd (the last nonzero remainder, which is 1 here) in terms of 89
and 55:
1=3-1·2
= 3 - 1 . (5 - 1 . 3) = 2 . 3 - 1 . 5
= 2 . (8 - 1 . 5) - 1 . 5 = 2 . 8 - 3 . 5
= 2. 8 - 3. (13 - 1 . 8) = 5. 8 - 3. 13
= 5 . (21 - 1 . 13) - 3 . 13 = 5 . 21 - 8 . 13
= 5. 21 - 8. (34 - 1. 21) = 13. 21 - 8. 34
= 13. (55 - 1 . 34) - 8. 34 = 13. 55 - 21 . 34
= 13. 55 - 21 . (89 - 1 . 55) = 34. 55 - 21 . 89
Therefore the Bezout coefficient of 55 is 34, and that is an inverse of 55 modulo 89.
d) We proceed as above:
232 = 2 . 89 + 54
89 = 1·54 + 35
54 = 1. 35 + 19
35
= 1. 19 + 16
19 = 1·16 + 3
16=5.3+1
3= 3·1
Then we work backwards to rewrite the gcd (the last nonzero remainder, which is 1 here) in terms of 232
and 89:
1=16 - 5. 3
= 16 - 5. (19 - 1 . 16) = 6. 16 - 5. 19
= 6 . (35 - 1 . 19) - 5 . 19 = 6 . 35 - 11 . 19
= 6. 35 - 11. (54 - 1. 35) = 17. 35 - 11. 54
= 17. (89 - 1. 54) - 11. 54 = 17. 89 - 28. 54
= 17. 89 - 28 . (232 - 2 . 89) = 73 . 89 - 28 . 232
Therefore the Bezout coefficient of 89 is 73, and that is an inverse of 89 modulo 232.
7. We follow the hint. Suppose that we had two inverses of a modulo m, say b and c. In symbols, we would
have ba 1 (mod m) and ca 1 (mod m) . The first congruence says that m divides ba - 1 , and the second
says that m divides ca - 1. Therefore m divides the difference (ba -1) - (ca -1) = ba - ca. (The difference
of two multiples of m is a multiple of m.) Thus ba =ca (mod m). It follows immediately from Theorem 7
in Section 4.3 (the roles of a, b, and c need to be permuted) that b = c (mod m), which is what we wanted
to prove.
=
=
9. In Exercise 5a we found that an inverse of 4 modulo 9 is 7. Therefore we multiply both sides of this equation
by 7, obtaining x 35 8 (mod 9). As a check, we compute 4 · 8 = 32 5 (mod 9).
= =
=
11. Our answers are not unique, of course-anything in the same congruence class works just as well.
a) In Exercise 5b we found that an inverse of 19 modulo 141 is 52. Therefore we multiply both sides of this
equation by 52, obtaining x
208 67 (mod 141). As a check, we compute 19 · 67 = 1273 4 (mod 141).
b) In Exercise 5c we found that an inverse of 55 modulo 89 is 34. Therefore we multiply both sides of this
equation by 34, obtaining x
1156
88 (mod 89). As a check, we compute 55 · 88
55 · ( -1) = -55
34 (mod 89).
= =
= =
=
=
=
Chapter 4
132
Number Theory and Cryptography
c) In Exercise 5d we found that an inverse of 89 modulo 232 is 73. Therefore we multiply both sides of this
equation by 73, obtaining x
146 (mod 232). As a check, we compute 89 · 146 = 12994 2 (mod 232).
=
=
13. We follow the hint. Adding 6 to both sides gives the equivalent congruence 15x 2 + 19x + 6
=0
(mod 11),
because 5 + 6 = 11 = 0 (mod 11). This factors as (5x + 3)(3x + 2) = 0 (mod 11). Because there are
no non-zero divisors of 0 working modulo 11, we conclude that the solutions are precisely the solutions of
5x + 3 0 (mod 11) and 3x + 2
0 (mod 11). We solve these by the method of Example 3. By inspection
(trial-and-error) or working it out through the Euclidean algorithm and back-substituting, we find that an
=
=
inverse of 5 modulo 11is9, and multiplying both sides of 5x+3
= =
= 0 (mod 11) by 9 yields x+27 = 0 (mod 11),
= =
so x
-27 6 (mod 11). Similarly, an inverse of 3 modulo 11 is 4, and we get x
-8 3 (mod 11). So the
solution set is {3, 6} (and anything congruent to these modulo 11 ). Plugging these values into the original
equation to check, we have 15·32 +19 · 3 + 6 = 198 0 (mod 11) and 15·62 +19 · 6 + 6 = 660 0 (mod 11).
=
=
15. The hypothesis tells us that m divides ac- be, which is the product (a- b)c. Let m' be m/ gcd(c, m). Then
m' is a factor of m, so certainly m' J(a - b)c. Now since all the common factors of m and c were divided
out of m to get m', we know that m' is relatively prime to c. It follows from Lemma 2 in Section 4.3 that
m' I a - b. But this means that a= b (mod m'), exactly what we were trying to prove.
17. We want to find numbers x such that x 2 = 1 (mod p), in other words, such that p divides x 2 - 1. Factoring
this expression, we see that we are seeking numbers x such that p I( x + 1) (x - 1) . By Lemma 3 in Section 4.3,
this can only happen if p I x + 1 or p I x - 1. But these two congruences are equivalent to the statements
x
-1 (mod p) and x
1 (mod p) .
=
=
19. a) If two of these integers were congruent modulo p, say ia and ja, where 1 ::::; i < j < p, then we would
have p Ija - ia, or p I(j - i)a. By Lemma 2 (or Lemma 3) in Section 4.3, since a is not divisible by p, p must
divide j - i. But this is impossible, since j - i is a positive integer less than p. Therefore no two of these
integers are congruent modulo p.
b) By part (a), since no two of a, 2a, ... , (p - l)a are congruent modulo p, each must be congruent to a
different number from 1 to p- 1. Therefore if we multiply them all together, we will obtain the same product,
modulo p, as if we had multiplied all the numbers from 1 to p - 1. In symbols,
=1 · 2 · 3 · · · (p -
1) (mod p).
1
The left-hand side of this congruence is clearly (p - 1)! · aP- , and the right-hand side is just (p - l)!, as
desired.
a· 2a · 3a · · · (p - l)a
c) Wilson's theorem says that (p - 1)! is congruent to -1 modulo p. Therefore the congruence in part (b)
says that (-1) · ap-l
-1 (mod p). Multiplying both sides by -1, we see that ap-l = 1 (mod p), as desired.
Note that we already assumed the hypothesis that p ,./'a in part (a).
=
p I a, then both sides of aP =a (mod p) are 0 modulo p, so the congruence holds. If not, then we just
multiply the result obtained in part ( c) by a.
d) If
21. Since 2, 3, 5, and 11 are pairwise relatively prime, we can use the Chinese remainder theorem. The answer
will be unique modulo 2 · 3 · 5 · 11 = 330. Using the notation in the text, we have a 1 = 1, m 1 = 2,
a2 = 2, m2 = 3, a3 = 3, m3 = 5, a4 = 4, m4 = 11, m = 330, Ali= 330/2 = 165, M2 = 330/3 = 110,
1\13 = 330/5 = 66, l'vl4 = 330/11=30. Then we need to find inverses y, of M, modulo m, for i = 1,2,3,4.
This can be done by inspection (trial and error), since the moduli here are so small, or systematically using
the Euclidean algorithm, as in Exercise 5; we find that y 1 = 1, y 2 = 2, y 3 = 1, and y4 = 7 (for this last one,
30 8 (mod 11), so we want to solve 8y4 = 1 (mod 11), and we observe that 8 · 7 = 56 = 1 (mod 11) ). Thus
our solution is x = 1 · 165 · 1 + 2 · 110 · 2 + 3 · 66 · 1 + 4 · 30 · 7 = 1643 = 323 (mod 330). So the solutions are
all integers of the form 323 + 330k, where k is an integer.
=
Section 4.4
Solving Congruences
133
23. By definition, the first congruence can be written as x
= 3t + 2
where t is an integer. Substituting this
expression for x into the second congruence tells us that 3t + 2 = 1 (mod 4), which can easily be solved
to show that t
1 (mod 4). From this we can write t = 4u + 1 for some integer u. Thus x = 3t + 2 =
3(4u + 1) + 2 = 12u + 5. We plug this into the third congruence to obtain 12u + 5 = 3 (mod 5), which we
easily solve to give u 4 (mod 5). Hence u = 5v + 4, and so x = 12u + 5 = 12(5v + 4) + 5 = 60v + 53. We
check our answer by confirming that 53 = 2 (mod 3), 53 = 1 (mod 4), and 53 = 3 (mod 5).
=
=
25. We simply translate the steps of the calculation given in the proof of Theorem 2 into pseudocode. Of course,
hidden in line 7 below is a multi-step process of finding inverses in modular arithmetic, which can be accomplished by using the Euclidean algorithm and back-substituting, as in Example 2. The last loop reduces the
answer x to its simplest form modulo m. All solutions are then of the form x + mk, where m is the product
of the moduli and k is an integer.
procedure chinese( m 1 , m2, ... , mn : relatively prime positive integers; a 1 , a 2 , ... , an : integers)
m:= 1
fork:= 1 ton
m :=m·mk
fork:= 1 ton
Mk:= m/mk
Yk := Mi; 1 mod mk
x := 0
fork:= 1 ton
x := x + akMkYk
while x 2: m
x := x-m
return x {the smallest solution to the system { x ak (mod mk), k = 1, 2, ... , n } }
=
27. We cannot apply the Chinese remainder theorem directly, since the moduli are not pairwise relatively prime.
However, we can, using the Chinese remainder theorem, translate these congruences into a set of congruences
that together are equivalent to the given congruence. Since we want x = 4 (mod 12), we must have x = 4 =
1 (mod 3) and x = 4 = 0 (mod 4). Similarly, from the third congruence we must have x = 1 (mod 3) and
x = 2 (mod 7). Since the first congruence is consistent with the requirement that x = 1 (mod 3), we see that
our system is equivalent to the system x
7 (mod 9), x
0 (mod 4), x
2 (mod 7). These can be solved
using the Chinese remainder theorem (see Exercise 21 or Example 5) to yield x
16 (mod 252). Therefore
the solutions are all integers of the form 16 + 252k, where k is an integer.
=
=
=
=
29. We will argue for the truth of this statement using the Fundamental Theorem of Arithmetic. What we must
show is that m1 m 2 · · · mn Ia - b. Look at the prime factorization of both sides of this proposition. Suppose
that p is a prime appearing in the prime factorization of the left-hand side. Then p ImJ for some j. Since
the m, 's are relatively prime, p does not appear as a factor in any of the other m, 's. Now we know from the
hypothesis that mJ Ia - b. Therefore a - b contains the factor p in its prime factorization, and p must appear
to a power at least as large as the power to which it appears in m 1 . But what we have just shown is that
each prime power pr in the prime factorization of the left-hand side also appears in the prime factorization of
the right-hand side. Therefore the left-hand side does, indeed, divide the right-hand side.
31. We are asked to solve the simultaneous congruences x = 1 (mod 2) and x = 1 (mod 3). The solution will be
unique modulo 2 · 3 = 6. By inspection we see that the answer is simply that x
1 (mod 6). The solution
set is {... ,-ll,-5,1,7,13, ... }.
=
33. Fermat's little theorem tells us that 712 = 1 (mod 13). Note that 121
712-10 . 7 = (712)10. 7 = 110 . 7 = 7 (mod 13).
10 · 12
+ 1.
Therefore 7121
Chapter 4
134
35. Fermat's little theorem tells us that under the given conditions
2
a· aP- = aP-l = 1 (mod p). This is precisely the definition that
ap-l
aP- 2
Number Theory and Cryptography
= 1 (mod p). Therefore aP- 2
is an inverse of a modulo p.
. a
=
37. a) We calculate 2340 = (2 10 ) 34 = 134 = 1 (mod 11), since Fermat's little theorem says that 210 = 1 (mod 11).
b) We calculate 2340 = (2·5 ) 68 =32 68 =1 68 =1(mod31), since 32=1(mod31).
c) Since 11 and 31 are relatively prime, and 11·31 = 341, it follows from the first two parts and Exercise 29
that 2340 = 1 (mod 341).
39. a) By Fermat's little theorem we know that 56 = 1 (mod 7); therefore 5 1998 = (5 6) 333 = 1333 = 1 (mod 7),
and so 52003 = 55 · 5 1998 = 3125 · 1 = 3 (mod 7). So 52003 mod 7 = 3. Similarly, 5 10 = 1 (mod 11);
therefore 52000 = (5 10 ) 200 = 1200 = 1 (mod 11), and so 52003 = 53 · 52000 = 125 · 1 = 4 (mod 11). So
1 166
1 (mod 13), and so
52003 mod 11 = 4. Finally, 5 12 = 1 (mod 13); therefore 5 1992 = (5 12 ) 166
2003
1992
11
2003
5
= 5 ·5
48,828, 125 · 1 8 (mod 13). So 5
mod 13 = 8.
=
=
=
=
b) We now apply the Chinese remainder theorem to the results of part (a), as in Example 5. Let m =
7 · 11 · 13 = 1001, llfi = m/7 = 143, 1112 = m/11 = 91, and M 3 = m/13 = 77. We see that 5 is an
inverse of 143 modulo 7. since 143
3 (mod 7), and 3 · 5 = 15
1 (mod 7). Similarly, 4 is an inverse
of 91 modulo 11, and 12 is an inverse of 77 modulo 13. (An algorithm to compute inverses-if we don't
want to find them by inspection as we've done here-is illustrated in Example 2.) Therefore the answer is
(3 · 143 · 5 + 4 · 91·4 + 8 · 77 · 12) mod 1001=10993 mod 1001 = 983.
=
=
41. Let q be a (necessarily odd) prime dividing 2P - 1. By Fermat's little theorem, we know that q I 2q-l - 1.
Then from Exercise 37 in Section 4.3 we know that gcd(2P- l, 2q-l -1) = 2gcd(p.q-l) -1. Since q is a common
divisor of 2P - 1 and 2q-l - 1, we know that gcd(2P - 1, 2q-l - 1) > 1. Hence gcd(p, q - 1) = p, since the
only other possibility, namely gcd(p, q - 1) = 1, would give us gcd(2P - 1, 2q-l - 1) = 1. Hence p I q - 1, and
therefore there is a positive integer m such that q - 1 = mp. Since q is odd, m must be even, say m = 2k,
and so every prime divisor of 2P - 1 is of the form 2kp + 1. F\lfthermore, products of numbers of this form
are also of this form, since (2k 1p + 1)(2k2p + 1) = 4k 1k2p 2 + 2k 1p + 2k 2p + 1 = 2(2k1k2p + k 1 + k3)p + 1.
Therefore all divisors of 2P - 1 are of this form.
43. To decide whether 2 11 - 1 = 204 7 is prime, we need only look for a prime factor not exceeding J2647 ~ 45.
By Exercise 41 every such prime divisor must be of the form 22k + 1. The only candidate is therefore 23. In
fact 2047 = 23 · 89, so we conclude that 2047 is not prime.
We can take the same approach for 217 - 1 = 131,071, but we need either computer algebra software or
patience with a calculator. By Exercise 41 every prime divisor of 217 - 1 must be of the form 34k + 1, so we
need to try all such divisors (or at least those that are not obviously nonprime) up to .Jl31,071
means up to k = 10. No number of this form divides 131,071, so we conclude that it is prime.
~
362, which
45. First note that 2047 = 23·89, so 2047 is composite. To apply Miller's test, we write 2047-1 = 2046 = 2·1023,
so s = 1 and t = 1023. We must show that either 21023 = 1 (mod 204 7) or 21023 = -1 (mod 204 7) . To
compute, we write 21023 = (2 11 ) 93 = 2048 93
193 = 1 (mod 2047), as desired. (We could also compute this
using the modular exponentiation algorithm given in Section 4.2-see Example 12 in that section.)
=
47. We factor 2821 = 7·13·31. We must show that this number meets the definition of Carmichael number, namely
that b2820
1 (mod 2821) for all b relatively prime to 2821. Note that if gcd(b, 2821) = 1, then gcd(b, 7) =
gcd(b, 13) = gcd(b, 31) = 1. Using Fermat's little theorem we find that b6 = 1 (mod 7), b12 = 1 (mod 13),
and b30
1 (mod 31). It follows that b2820 = (b6) 470 = 1 (mod 7), b2820 = (b 12 ) 235 = 1 (mod 13),
2820
and b
= (b30 )94
1 (mod 31). By Exercise 29 (or the Chinese remainder theorem) it follows that
2820
b
1 (mod 2821), as desired.
=
=
=
=
Section 4.4
Solving Congruences
135
49. a) If we multiply out this expression, we get n = 1296m3 + 396m2 + 36m + 1. Clearly 6m In - 1, 12m In - 1,
and 18m In - 1. Therefore, the conditions of Exercise 48 are met, and we conclude that n is a Carmichael
number.
b) Letting
m = 51 gives n = 172,947,529. We note that 6m + 1 = 307, 12m + 1 = 613, and 18m + 1 = 919
are all prime.
51. It is straightforward to calculate the remainders when the integers from 0 to 14 are divided by 3 and by 5.
For example, the remainders when 10 is divided by 3 and 5 are 1 and 0, respectively, so we represent 10 by
the pair (1, 0). The exercise is simply asking us to tabulate these remainders, as in Example 7.
0 = (0, 0)
1 = (1,1)
3 = (0, 3)
2 = (2, 2)
5
4=(1,4)
=
(2, 0)
6=(0,1)
7 = (1,2)
8
=
(2, 3)
9 = (0,4)
10 = (1,0)
11 = (2, 1)
12 = (0, 2)
13=(1,3)
14 = (2,4)
53. The method of solving a system of congruences such as this is given in the proof of Theorem 2. Here we have
m1 = 99, m2 = 98, m3 = 97, and m4 = 95, so that m = 99 · 98 · 97 · 95 = 89403930. We compute the values
Mk = m/mk and obtain Mi = 903070, M2 = 912285, M 3 = 921690, and M 4 = 941094. Next we need to
find the inverses Yk of Mk modulo mk. To do this we first replace each Mk by its remainder modulo mk (to
make the arithmetic easier), and then apply the technique shown in Example 2. For k = 1 we want to find
the inverse of 903070 modulo 99, which is the same as the inverse of 903070 mod 99, namely 91. To do this
we apply the Euclidean algorithm to express 1 as a linear combination of 91 and 99.
99 = 91+8
91=11·8 + 3
8=2·3+2
3=2+1
:.1=3-2
= 3 - (8 - 2 . 3) = 3 . 3 - 8
= 3 . (91 - 11 . 8) - 8 = 3 . 91 - 34. 8
= 3. 91 - 34. (99 - 91) = 37. 91 - 34. 99
We therefore conclude that the inverse of 91 modulo 99 is 37, so we have y 1 = 37. Similar calculations show
that Y2 = 33, y3 = 24, and y4 = 4. Continuing with the procedure outlined in the proof of Theorem 2, we
now form the sum of the products akMkYk, and this will be our solution. We have
65. 903070. 37 + 2. 912285. 33 + 51. 921690. 24 + 10. 941094. 4 = 3397886480.
We want our answer reduced modulo m, so we divide by 89403930 and take the remainder, obtaining 537140.
(All of these calculations are not difficult using a scientific calculator.) Finally, let us check our answer:
537140 mod 99 = 65, 537140 mod 98 = 2, 537140 mod 97 = 51, 537140 mod 95 = 10.
55. For the first question we seek an exponent n such that 2n = 5 (mod 19). For the second we want 2n =
6 (mod 19). There is no known efficient algorithm for finding these exponents, so we might as well just
start computing powers of 2 modulo 19. In each case, we just need to multiply the previous result by 2,
working modulo 19. We have 22 = 4 (mod 19), 23 = 2 · 4 = 8 (mod 19), 24 = 2 · 8 = 16 (mod 19),
25 = 2·16 = 32 = 13 (mod 19), 26 = 2 · 25 = 2 · 13 = 26 = 7 (mod 19), 27 = 2 · 7 = 14 (mod 19), 28 = 2·14 =
28 = 9 (mod 19), 29 = 2·9 = 18 (mod 19), 210 =2·18 = 36 = 17 (mod 19), 211 =2·17 = 34 = 15 (mod 19),
212 = 2 · 15 = 30 = 11(mod19), 213 = 2 · 11 = 22 = 3 (mod 19), 214 = 2 · 3 = 6 (mod 19). Finally! So
we conclude that the discrete logarithm of 6 to the base 2 modulo 19 is 14. Continuing the calculation, we
have 215 = 2 · 6 = 12 (mod 19), 216 = 2 · 12 = 24 = 5 (mod 19). So the discrete logarithm of 5 to the base 2
modulo 19 is 16.
136
Chapter 4
Number Theory and Cryptography
57. A computer algebra system such as Maple facilitates the modular arithmetic calculations. We repeatedly
multiply by 3 and reduce modulo 17. We get 3° = 1 (mod 17), 31 = 3 (mod 17), 32 = 9 (mod 17),
33 = 27 = 10 (mod 17) , and so on. Thus log 3 1 = 0, log 3 3 = 1, log 3 9 = 2, log 3 10 = 3, and so on. If we
collect the data and present them in order of increasing argument, we get the required table. (Of course log 3 0
does not exist.)
log 3 1 = 0
log 3 9
log 3 2 = 14
log 3 3
= 1 log3 4 = 12 log3 5 = 5 log3 6 = 15 log3 7 = 11 log3 8 = 10
= 2 log3 10 = 3 log3 11 = 7 log3 12 = 13 log 3 13 = 4 log 3 14 = 9 log 3 15 = 6 log 3 16 = 8
=
59. We need to prove that if the congruence x 2
a (mod p) has any solutions at all, then it has exactly
two solutions. So let us assume that s is a solution. Clearly -s is a solution as well, since (-s) 2 = s 2 .
Furthermore, -s =fas (mod p), since if it were, we would have 2s = 0 (mod p), which means that p I 2s. Since
p is an odd prime, that means that p Is, so that s 0 (mod p). Therefore a= 0 (mod p), contradicting the
conditions of the problem.
=
It remains to prove that there cannot be more than two incongruent solutions. Suppose that s is one
solution and that t is a second solution. We have s 2 = t 2 (mod p). This means that p I s 2 - t 2 , that is,
p I(s + t) ( s - t) . Since p is prime, Lemma 3 in Section 4. 3 guarantees that p I s - t or p I s + t. This means that
t = s (mod p) or t = -s (mod p). Therefore any solution t must be either the first solution or its negative.
In other words, there are at most two solutions.
61. There is really almost nothing to prove here. The value (~) depends only on whether or not a is a quadratic
residue modulo p, i.e., whether or not the equivalence x 2 =a (mod p) has a solution. Obviously, this depends
only on the equivalence class of a modulo p.
=
=(;)
63. By Exercise 62 we know that (~) (~)
a(p-lJ/ 2 b(p-lJ/ 2 = (ab)(p-l)/ 2
(mod p). Since the only values
either side of this equivalence can take on are ±1, being congruent modulo p is the same as being equal.
65. We follow the hint. Working modulo 5, we want to solve x 2 = 4. It is easy to see that there are exactly
two solutions modulo 5, namely x = 2 and x = 3. Similarly there are only the solutions x = 1 and x = 6
2 or 3 (mod 5) and x
1
modulo 7. Therefore we want to find values of x modulo 5 · 7 = 35 such that x
or 6 (mod 7). We can do this by applying the Chinese remainder theorem (as in Example 5) four times, for
the four combinations of these values. For example, to solve x = 2 (mod 5) and x = 1 (mod 7), we find that
m = 35, Mi = 7, M2 = 5, y 1 = 3, Y2 = 3, so x
2 · 7 · 3 + 1 · 5 · 3 = 57 22 (mod 35). Doing the similar
calculation with the other three possibilities yields the other three solutions modulo 35: x = 8, x = 13, and
x = 27.
=
=
=
=
67. To compute logra (modp), we need to solve re= a (modp) fore. The brute force approach is just to
compute re mod p for e = 0, 1, 2, ... , p - 2 until we get the answer a. This requires about p iterations, each
of which can be done with O(logp) bit operations, since we need only multiply the previous value by r and
find the remainder upon division by p. At worst, we require all p iterations; on average, only half that many.
In either case, the time complexity is O(plogp), which is prohibitively large if pis, say, a 200-digit number.
Section 4.5
Applications of Congruences
SECTION 4.5
137
Applications of Congruences
The great British number theorist G. H. Hardy (1877-1947) once said, "I have never done anything 'useful.'
No discovery of mine has made, or is likely to make, directly or indirectly, for good or ill, the least difference to
the amenity of the world." He was wrong. Number theory has many applications, especially in cryptography
(see Section 4.6). In the present section we saw applications to hashing functions (important for storing
large amounts of information and being able to retrieve it efficiently), pseudorandom numbers (important for
computer simulations), and check digits (important in our technological world). Hardy would be appalled!
The exercises in this section are mostly routine.
1. We are simply asked to compute k mod 97 for each value of k. We do this by dividing the given number by
97 and taking the remainder, which can be found either by multiplying the decimal remainder by 97, or by
subtracting 97 times the quotient from k. (See the solution to Exercise 3 below for details.)
a) 034567981 mod 97 = 91
b) 183211232 mod 97 = 57
c) 220195744 mod 97 = 21
d) 987255335 mod 97 = 5
3. a) We need to compute k mod 31 in each case. A good way to do this on a calculator is as follows. Enter k
and divide by 31. The result will be a number with an integer part and a decimal fractional part. Subtract off
the integer part, leaving a decimal fraction between 0 and 1. This is the remainder expressed as a decimal.
To find out what whole number remainder that really represents, multiply by 31. The answer will be a whole
number (or nearly so-it may require rounding, say from 4.9999 or 5.0001 to 5), and that number is k mod 31.
(ii) 918 mod 31 = 19
(iii) 007 mod 31 = 7
(i) 317 mod 31 = 7
(iv) 100 mod 31 = 7
(v) 111 mod 31 = 18
(vi) 310 mod 31 = 0
b) Take the next available space, where the next space is computed by adding 1 to the space number and
pretending that 30 + 1 = 0.
5. We apply the formula with n = 0 to obtain x 1 = (3 · x 0 + 2) mod 13 = (3 · 1 + 2) mod 13 = 5. Then
X2 = (3 · x1 + 2) mod 13 = (3 · 5 + 2) mod 13 = 17 mod 13 = 4.
Continuing in this way we have
X3 = (3 · 4 + 2) mod 13 = 1. Because this is the same as x 0 , the sequence repeats from here on out. So the
sequence is 1, 5, 4, 1, 5, 4, 1, 5, 4, ....
7. We compute until the sequence begins to repeat:
x 1 = 3 · 2 mod 11 = 6
x2
= 3 · 6 mod 11 = 7
X3
= 3 · 7 mod 11 = 10
x 4 = 3 · 10 mod 11 = 8
x5
Since
X5
= 3 · 8 mod 11 = 2
= xo, the sequence repeats forever: 2, 6, 7, 10, 8, 2, 6, 7, 10, 8, ....
9. We follow the instructions. Because 2357 2 = 5555449 = 05555449, the middle four digits are 5554, so 5554
is our second pseudorandom number. Next 5554 2 = 30846916, so our third pseudorandom number is 8469.
Repeating the same procedure leads to the following five terms: 7239, 4031, 2489, 1951, 8064.
11. We are told to apply the formula Xn+l = x~ mod 7, starting with x 0 = 2. Thus x 1 = 23 mod 7 = 1,
x 2 = 13 mod 7 = 1, and our sequence never gets off the ground! The sequence generated here is 2, 1, 1, 1, ....
13. A correctly transmitted bit string must have an even number of 1's. Therefore we can be sure that there is
an error in (d), but because the other three strings have an even number of 1 's, we cannot detect an error
in any of them. (Of course that doesn't mean that there is no error, because it is possible that two bits were
transmitted incorrectly, in which case the sum modulo 2 does not change.)
Chapter 4
138
Number Theory and Cryptography
15. Let d be the check digit. Then we know that 1·0+2·0+3·7+4·1+5·1+6·9+7·8+8·8+9·1+10·d = 0 (mod 11).
=
This simplifies to 213+10·d 0 (mod 11). But 213
to 4 - d = 0 (mod 11) , or d = 4.
=4 (mod 11), and 10 =-1(mod11), so this is equivalent
17. The ISBN is 0073383090. To check its validity we compute, as in Example 6, 1·0+2·0 + 3 · 7 + 4 · 3 + 5 ·
3 + 6 · 8 + 7 · 3 + 8 · 0 + 9 · 9 + 10 · 0 = 198. Because this is congruent to 0 modulo 11, the check digit was
computed correctly.
19. To determine whether an 11-digit number is a valid USPS money order identification number, we need to
verify that the sum of the first ten digits reduced modulo 9 gives the last digit.
a) 7 + 4 + 0 + 5 + 1 + 4 + 8 + 9 + 6 + 2 mod 9 = 46 mod 9 = 1 -1- 3, so this is not a valid number.
b) 8 + 8 + 3 + 8 + 2 + 0 + 1 + 3 + 4 + 4 mod 9
number.
=
41 mod 9
=
5, which is the last digit, so this is a valid
c) 5 + 6 + 1+5+2+2 + 4 + 0 + 7 + 8 mod 9 = 40 mod 9 = 4, which is the last digit, so this is a valid
number.
d) 6 + 6 + 6 + 0 + 6 + 6 + 3 + 1 + 1 + 7 mod 9 = 42 mod 9 = 6 -1- 8, so this is not a valid number.
21. In each case, we know that xn = x1 + x2 + X3 + X4 + xs + X5 + x1 + xs + Xg + X10 mod 9. (See the preamble
to Exercise 18.) This is equivalent to x 11 = x 1 + x 2 + X3 + x 4 + x 5 + x 6 + x 7 + x 8 + x 9 + x 10 (mod 9), with
0 ::; x 11 :S 8. Therefore we will get an equation modulo 9 involving the unknown Q for each of these valid
postal money order identification numbers.
a) 8 = 4 + 9 + 3 + 2 + 1 + 2 + Q + 0 + 6 + 8 (mod 9), which is equivalent to 8 = Q + 35 = Q + 8 (mod 9).
Therefore Q = 0 (mod 9). There are two single-digit numbers Q that makes this true: Q = 0 and Q = 9, so
it is impossible to know for sure what the smudged digit was.
=
=
b) 8 8 + 5 + 0 + Q + 9 + 1+0 + 3 + 8 + 5 (mod 9), which is equivalent to 8 Q + 39 = Q + 3 (mod 9).
The only single-digit number Q that makes this true is Q = 5, so the smudged digit must have been a 5.
c) 4 = 2 + Q + 9 + 4 + 1+0 + 0 + 7 + 7 + 3 (mod 9), which is equivalent to 4 = Q + 33 = Q + 6 (mod 9).
The only single-digit number Q that makes this true is Q = 7, so the smudged digit must have been a 7.
d) 1=6+6 + 6 + 8 + 7 + Q + 0 + 3 + 2 + 0 (mod 9), which is equivalent to 1=Q+38 = Q + 2 (mod 9).
The only single-digit number Q that makes this true is Q = 8, so the smudged digit must have been an 8.
23. Because the first ten digits are added, any transposition error involving them will go undetected-the sum of
the first ten digits will be the same for the transposed number as it is for the correct number. Suppose the
last digit is transposed with another digit; without loss of generality, we can assume it's the tenth digit and
that x 10 -1- x 11 . Then the correct equation will be
X11
=X1 + X2 +
X3
+
X4
+ X5 +
X5
+
X7
+
Xs
+
Xg
+
X10
(mod 9)
but the equation resulting from the error will read
X10
=X1 + X2 +
X3
+
X4
+
X5
+
X5
+
X7
+
Xs
+
Xg
+ X11 (mod 9) .
=
Subtracting these two equations, we see that the erroneous equation will be true if and only if x 11 - x10
x 10 - x 11 (mod 9). This is equivalent to 2x 11
2x 10 (mod 9), which, because 2 is relatively prime to 9, is
equivalent to x 11 = x 10 (mod 9), which is false. This tells us that the check equation will fail. Therefore we
conclude that transposition errors involving the eleventh digits are detected.
=
25. From Example 5. we know that a valid UPC code must satisfy the equation
Section 4.5
Applications of Congruences
139
Therefore in each case we simply need to compute the left-hand side of this equation modulo 10 and see
whether or not we get 0 as the answer.
a) 3 · 0 + 3 + 3 · 6 + 0 + 3 · 0 + 0 + 3 · 2 + 9 + 3 · 1+4 + 3 · 5 + 2 = 60
= 0 (mod 10), so this is a valid code.
b) 3 · 0 + 1 + 3 · 2 + 3 + 3 · 4 + 5 + 3 · 6 + 7 + 3 · 8 + 9 + 3 · 0 + 3 = 88 "!- 0 (mod 10) , so this is not a valid code.
c) 3 · 7 + 8 + 3 · 2 + 4 + 3 · 2 + 1+3 · 8 + 4 + 3 · 3 + 0 + 3 · 1+4 = 90
d) 3 · 7 + 2 + 3 · 6 + 4 + 3 · 1+2+3·1+7 + 3 · 5 + 4 + 3 · 2 + 5 = 90
=0 (mod 10), so this is a valid code.
=0 (mod 10), so this is a valid code.
27. The digits with even subscripts appear in the formula with coefficient 1, whereas those with odd subscripts
appear with coefficient 3. Therefore if two digits whose positions have the same parity (both odd or both
even) are switched, then th~ sum will be unchanged and such an error cannot be detected. If two digits whose
parities are different are transposed, say an x in an odd position and a y in an even position, then the new
sum will differ from the old sum by (x + 3y) - (3x + y), which equals 2(y- x). As long as the two transposed
digits do not differ by 5, the sum will therefore be different modulo 10; if they do differ by 5, then the
sum will be the same modulo 10. We conclude that transposition errors will be detected if and only if the
transposed digits are an odd number of positions apart (in particular transposing neighboring digits) and do
not differ by 5 .
29. In each case we need to compute a 1 a 2 ... al4 mod 7 and see if we get a 15 . This may be inconvenient on a
calculator with only 12 digits of precision, but one can always divide it out by hand (or, better, use computer
algebra software).
a) 10133334178901 = 7 · 1447619168414 + 3. Therefore 10133334178901 mod 7 = 3 = a 15 , so this is a valid
airline ticket number. (In Maple we could just type 10133334178901 mod 7 and get the response 3.)
b) 00786234277044 mod 7· = 6
-/=-
5 = a 15 , so this is not a valid number.
c) 11327343888253 mod 7 = 1 = a 15 , so this is a valid number.
d) 00012234732287 mod 7 = 1 = a 15 , so this is a valid number.
31. Let's solve a more general problem by ignoring the word "consecutive." First we look at the case in which the
transposition does not involve the check digit itself. Suppose the erroneous number formed by the first 14 digits
occurs when a, is interchanged with aJ , where 1 ~ i < j ~ 14. Because of our decimal place-value numeration
system, before the switch, a, was contributing a, -10 14 -z to the value of the number, and aJ was contributing
aJ · 10 14 -J. Therefore this change has increased the 14-digit number by (aJ -a,)10 14 - ' +(a, -aJ)10 14 -J, which
equals (aJ - a,)(10 14 -z - 10 14 -J). In order for this to still check, this last expression must be equivalent to 0
modulo 7. Obviously this will happen if ai and aJ differ by 7, but it will also happen if (10 14 - 1 -10 14 -J) is a
multiple of 7. A bit of calculation shows that this will happen if and only if j - i = 6 or 12. Thus we cannot
detect the error if the columns in which the transposition occurs are 6 or 12 apart or the transposed digits
differ by 7. Finally, if the digit a 15 is transposed with the digit a,, where 1 ~ i ~ 14, then a 1 a 2 ••• a 14 mod 7
has gone up by (a 15 -a,)10 14 -z and the check digit has gone up by a, - a 15 , so we will not be able to detect this
error ifand only if (a 15 -a,)10 14 - i =a, -a 15 (mod 7), which is equivalent to (a 15 -a,)(10 14 -'+ 1) = 0 (mod 7).
Because 10 14 -'' + 1 = 0 (mod 7) if and only if i = 5 or 11, we conclude that we cannot detect the transposition
error if it interchanges the check digit with a 5 or a 11 or interchanges it with a digit differing from it by 7.
(Of course, the check digit must be 0 through 6, so an error that puts a 7, 8, or 9 in the last position can
also be detected.)
Because transposing consecutive digits is not transposing digits whose positions differ by the quantities
mentioned above, we can detect all transposition errors of consecutive digits unless the digits differ by 7.
33. In each case we will compute 3d1 +4d 2 + 5d3 + 6d4 + 7d5 + 8d 6 + 9d7 mod 11. If this matches the digit given
for ds, then the ISSN is valid, and conversely.
Chapter 4
140
Number Theory and Cryptography
a) 3 · 1 + 4 · 0 + 5 · 5 + 6 · 9 + 7 · 1 + 8 · 0 + 9 · 2 mod 11 = 107 mod 11 = 8. Because ds = 7 =/= 8, this number
is not valid.
b) 3 · 0 + 4 · 0 + 5 · 0 + 6 · 2 + 7 · 9 + 8 · 8 + 9 · 9 mod 11 = 220 mod 11 = 0. Because d8 = 0, this number is
valid.
c) 3 · 1+4 · 5 + 5 · 3 + 6 · 0 + 7 · 8 + 8 · 6 + 9 · 6 mod 11=196 mod 11 = 9. Because d8 = 9, this number is
valid.
d) 3 · 1+4 · 0 + 5 · 0 + 6 · 7 + 7 · 1 + 8 · 2 + 9 · 0 mod 11 = 68 mod 11 = 2. Because d 8 is "X" (representing
10 modulo 11 ), this number is not valid.
35. By subtracting d8 from both sides and noting that -1
=10 (mod 11) , we see that the checking congruence is
equivalent to 3d 1 +4d 2 + 5d3 + 6d4 + 7d 5 + 8d5 + 9d1 + lOds = 0 (mod 11) . It is now easy to see that transposing
adjacent digits x and y (where x is on the left) causes the left-hand side to increase by x and decrease by y,
for a net change of x - y. Because x =/'. y (mod 11) , the congruence will no longer hold. Therefore errors of
this type are always detected.
SECTION 4.6
Cryptography
In addition to exercises about the topics covered in this section, this exercise set introduces the Vigenere cipher
(Exercises 18-22) and a protocol for key exchange (Exercise 33). There is a nice website for encoding and
decoding with the affine cipher (far any function of the farm f(p) = ap + b ), which you can use to check your
answers: www.shodor.org/interactivate/activities/CaesarCipher/. A website for the Vigenere cipher
can be found here: isl ab. oregonstate. edu/koc/ ece575/02Proj ect/Mun+Lee/VigenereCipher. html
1. a) We need to replace each letter by the letter three places later in the alphabet. Thus D becomes G, 0
becomes R, and so on. The resulting message is GR QRW SDVV JR.
b) We need to replace each letter by the letter 13 places later in the alphabet. Thus D becomes Q, 0 becomes
B (we cycle, with A following Z), and so on. The resulting message is QB ABG CNFF TB.
c) This one is a little harder, so it is probably easiest to work with the numbers. For D we have p = 3 because
D is the fourth letter of the alphabet. Then 3 · 3 + 7 mod 26 = 16, so the encrypted letter is the 17th letter,
or Q (remember that we start the sequence at 0). Our original message has the following numerical values:
3-14 13-14-19 15-0-18-18 6-14. Multiplying each of these numbers by 3, adding 7, and reducing modulo 26
gives us 16-23 20-23-12 0- 7-9-9 25-23. Translating back into letters we have QX UXM AHJJ ZX.
3. In each case, we translate the letters to numbers from 0 to 25, then apply the function, then translate back. (See
the solution for Exercise le above for details.) In each case, the numerical message is 22-0-19-2-7 24-14-20-17
18-19-4-15.
a) Adding 14 to each number modulo 26 yields 10-14-7-16-21 12-2-8-5 6-7-18-3. Translating back into letters
yields KOHQV MCIF GHSD.
b) Multiplying each number by 14, adding 21, and reducing modulo 26 yields 17-21-1-23-15 19-9-15-25
13-1-25-23. Translating back into letters yields RVBXP TJPZ NBZX.
c) Multiplying each number by - 7, adding 1, and reducing modulo 26 yields 3-1-24-13-4 15-7-17-12 5-24-25-0.
Translating back into letters yields DBYNE PHRM FYZA.
5. a) We need to undo the encryption operation, which was to choose the letter that occurred ten places later in
the alphabet. Therefore we need to go backwards 10 places (or, what amounts to the same thing, forward 16
places). For example, the C decodes as S. Doing this for each letter, as in Exercise 1, gives us SURRENDER
NOW.
b) BE MY FRIEND
c) TIME FOR FUN
Section 4.6
Cryptography
141
7. We need to play detective. First note that the two-letter word DY occurs twice. Because this was a shift cipher,
we know that the first letter of this word occurs five places beyond the second letter in the alphabet. One of
those letters has to be a vowel. This makes it very likely that the word is either UP or TO, corresponding to
k = 9 or k = 10, respectively. Since TO is a more common word, let us assume k = 10. To decrypt, we shift
each letter in the encrypted message backward 10 places (or forward 16 places) in the alphabet, obtaining
TO SLEEP PERCHANCE TO DREAM (from Hamlet).
9. Following the same strategy as in Exercise 7, we try to figure out k from the fact that MW is a two-letter
word in the encrypted text. What fits best is IS, with k = 4. If we apply that to the three-letter word, we get
ANY, which seems quite promising. We now decode the entire message: ANY SUFFICIENTLY ADVANCED
TECHNOLOGY IS INDISTINGUISHABLE FROM MAGIC.
11. We want to solve the congruence c = 15p + 13 (mod 26) for p. To do that we will need an inverse of
15 modulo 26, which we can obtain using the Euclidean algorithm or by trial and error. It is 7, because
7 · 15 = 105 = 4 · 26 + 1. Therefore we have p = 7(c - 13) mod 26 = 7c - 91 mod 26 = 7c + 13 mod 26.
13. Because the most common letters are E and T, in that order, and the numerical values of E, T, Z, and J are
=
=
=
4, 19, 25, and 9, respectively, we will assume that f( 4)
25 and f(19)
9. This means that 4a + b 25
and 19a + b 9, where we work modulo 26, of course. Subtracting the two equations gives 15a 10, which
simplifies to 3a 2 (because 5 is not a factor of 26, we can divide both sides by 5). We can find an inverse
of 3 modulo 26 using the Euclidean algorithm or by trial and error. It is 9, because 3 · 9 = 27 = 26 + 1.
Therefore a 9 · 2 = 18. Plugging this into 4a + b 25 yields b 25 - 4a = 25 - 72 5. We therefore guess
that the encryption function is f(p) = 18p + 5 mod 26. As a check, we see that f(4) = 25 and f(19) = 9.
=
=
=
=
=
=
=
Because 0'(1) = 3, we put the third letter
first; because 0'(2) = 1, we put the first letter second; and so on. This gives us BEWA REOF MART IANS,
presumably meant to be BEWARE OF MARTIANS.
15. We permute each block of four by undoing the permutation
O".
17. Presumably the message was translated letter by letter, such as by a shift cipher or affine cipher. (Other,
nonlinear, bijections on Z 26 are also possible.)
19. The numerical version of the encrypted text is 14-8-10-24-22-21-7-1-23. If we subtract the values for the
key HOTHOTHOT, namely 7-14-19-7-14-19-7-14-19 and reduce modulo 26, we obtain 7-20-17-17-8-2-0-13-4,
which translates to HURRICANE.
21. If l is the distance between the beginnings of the string that occurs several times, then it may be likely that
the length of the key string is a factor of l. Thus if we have several such values of l, we can find their greatest
common divisor and assume that the length of the key string is a factor of this gcd.
23. Suppose that we know n = pq and (p - 1) (q - 1), and we wish to find p and q. Here is how we do so.
Expanding (p - 1)( q - 1) algebraically we obtain pq - p - q + 1 = n - p - q + 1. Thus we know the value
of n - p - q + 1, and so we can easily calculate the value of p + q (since we know n). But we also know
the value of pq, namely n. This gives us two simultaneous equations in two unknowns, and we can solve
them using the quadratic formula. Here is an example. Suppose that we want to factor n = 341, and we
are told that (p - 1) (q - 1) = 300. We want to find p and q. Following the argument just outlined, we
know that p + q = 341 + 1 - 300 = 42. Plugging q = 42 - p into pq = 341 we obtain p( 42 - p) = 341,
or p 2 - 42p + 341 = 0. The quadratic formula then tells us that p = (42 + .J42 2 - 4 · 341) /2 = 31, and so
the factors are 31 and 42 - 31 = 11. Note that absolutely no trial divisions were involved here-it was just
straight calculation.
Chapter 4
142
Number Theory and Cryptography
25. First we translate UPLOAD into numbers: 2015 1114 0003. For each of these numbers, which we might call M,
we need to compute C =Me mod n = M 17 mod 3233. Note that n = 53 · 61 = 3233 and that gcd(e, (p 1) (q - 1)) = gcd(l 7, 52 · 60) = 1, as it should be. A computational aid tells us that 2015 17 mod 3233 = 2545,
1114 17 mod 3233 = 2757, and 0003 17 mod 3233 = 1211. Therefore the encrypted message is 2545 2757
1211.
27. This problem requires a great amount of calculation. Ideally, one should do it using a computer algebra
package, such as Mathematica or Maple. Let us follow the procedure outlined in Example 9. It was computed
there that the inverse of e = 13 modulo n = 43·59 is d = 937. We need to compute 0667937 mod 2537 = 1808,
1947937 mod 2537 = 1121, and 0671 937 mod 2537 = 0417. (These calculations can in principle be done with
a calculator, using the fast modular exponentiation algorithm, but it would probably take the better part of
an hour and be prone to transcription errors.) Thus the original message is 1808 1121 0417, which is easily
translated into letters as SILVER.
29. We follow the steps given in the text, with p = 23, a= 5, k 1 = 8, and k 2 = 5. Using Maple, we verify that
5 is a primitive root modulo 23, by noticing that 5k as k runs from 0 to 21 produce distinct values (and of
course 522 mod 23 = 1 ). We find that 58 mod 23 = 16. So in Step (2), Alice sends 16 to Bob. Similarly, in
Step (3), Bob sends 55 mod 23 = 20 to Alice. In Step (4) Alice computes 20 8 mod 23 = 6, and in Step (5)
Bob computes 165 mod 23 = 6. These are the same, of course, and thus 6 is the shared key.
31. See Example 10 for the procedure. First Alice translates her message into numbers: 1804 1111 0421 0417
2419 0708 1306. She then applies her decryption transformation sending each block x to x 1183 mod 2867.
(We should verify with Maple that 7 · 1183 mod (60 · 46) = 1.) Using Maple, we see that the blocks become
1804 1183 mod 2867 = 2186, 1111 1183 mod 2867 = 2087, 0421 1183 mod 2867 = 1279, 0417 1183 mod 2867 =
1251, 2419 1183 mod 2867 = 0326, 0708 1183 mod 2867 = 0816, and 1306 1183 mod 2867 = 1948. If her friends
apply Alice's encryption transformation to 2186 2087 1279 1251 0326 0816 1948, they will obtain the numbers
of her original message.
33. Cathy knows the shared key kAlice,Bob, but because she transmitted it to Alice encrypted, no one else knows
it at the time Alice receives it. Alice can decrypt the first part of Cathy's message to find out what the key
is. When Alice sends the second part of Cathy's message, which consists of kAlice.Bob encrypted with Bob's
key, on to Bob, Bob can decrypt it to find the shared key, but it remains hidden from everyone else.
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 4
1. Dividing 210 by 17 gives a quotient of 12 and a remainder of 6, which are the respective requested values.
2. a) 7 a - b
J
c) (10a
+ 13) -
b) 0
=-7; -1 =-8; 3 =17 =-11
(-4b
+ 20)
=
3(a - b)
+ 7(a + b - 1); note that 7 divides both terms
3. See Theorem 5 in Section 4.1.
4. See Example 5 in Section 4.2.
5. Octal: 154533; hexadecimal: D95B
6. 1110 1000 0110 and 1010 0000 1110 1011
7. See p. 258.
8. a) See Example 4 in Section 4.3 and the preceding paragraph on p. 258.
b) 11 2 . 23 . 29
Supplementary Exercises
143
9. a) See p. 265.
b) find all the common factors (not a good algorithm unless the numbers are really small); find the prime
factorization of each integer (works well if the numbers aren't too big and therefore can be easily factored);
use the Euclidean algorithm (really the best method)
c) 1 (use the Euclidean algorithm)
d) 23 35 55 73
10. a) Use the Euclidean algorithm; see Example 17 in Section 4.3.
b) 7 = 5 . 119 - 7. 84
11. a)
aa = 1 (mod m)
b) Express 1 as sa + tm (see Review Question 10). Then s is the inverse of a modulo m.
c) 11
12. a) Multiply each side by the inverse of a modulo m.
13. a) See p. 278.
b) { 17 + l40k I k
E
b) { 10 + 19k I k
E
z}
z}
14. No; n could be a pseudoprime such as 341.
15. 9200
= 918 . 92 . 9180 = 918 . 92 . (918)20
=1 . 81 . 120 = 81 = 5 (mod 19)
16. See Example 6 in Section 4.5.
17. NCCYRF NAQ BENATRF
18. a) See p. 298.
b) The amount of shift, k, is kept secret. It is needed both to encode and to decode messages.
c) Although the key for decoding, d, is kept secret, the keys for encoding, n and e, are published.
19. See pp. 299-301.
20. See p. 302.
SUPPLEMENTARY EXERCISES FOR CHAPTER 4
= 46518, we can conclude that the number of miles is congruent to 46,518 modulo
100,000. So it is 46518+ lOOOOOk for some natural number k (the number of miles driven cannot be negative).
Thus the actual number of miles driven is 46,518 or 146,518, or 246,518, or ....
1. Because 89697 - 43179
3. Obviously there are an infinite number of possible answers. The numbers congruent to 5 modulo 17 include
5, 22, 39, 56, ... , as well as -12, -29, -46, ....
=
5. From the hypothesis ac
be (mod m) we know that ac - be = km for some integer k. Divide both sides
by c to obtain the equation a - b = (km)/c. Now the left-hand side is an integer, and so the right-hand
side must be an integer as well. In other words, cJkm. Letting d = gcd(m,c), we write c =de. Then
the way that c divides km is that d Im and e I k (since no factor of e divides m/d). Thus our equation
reduces to a - b = (k/e)(m/d), where both factors on the right are integers. By definition, this means that
a= b (mod m/d).
7. We give an indirect proof. If n is odd, then n = 2k + 1 for some integer k. Therefore n 2 +1 = (2k + 1) 2 +1 =
4k 2 + 4k + 2 = 2 (mod 4). But perfect squares of even numbers are congruent to 0 modulo 4 (because
(2m) 2 = 4m 2 ), and perfect squares of odd numbers are odd, so n 2 + 1 is not a perfect square.
Chapter 4
144
Number Theory and Cryptography
9. The contribution of the digit in the kth column from the right in the binary expansion of a positive integer is
2k, where we consider the units digit to be 0 columns from the right. Therefore for k 2: 3, the contribution
is divisible by 23 = 8. Any sequence of three digits other than 000 in the three right-most columns will
contribute between 1 and 7, causing the number not to be divisible by 8. Thus a positive integer written in
binary is divisible by 8 if and only if its last three digits are 000.
11. We assume that someone has chosen a positive integer less than 2n, which we are to guess. We ask the person
to write the number in binary, using leading O's if necessary to make it n bits long. We then ask "Is the first
bit a 1 ?", ·'Is the second bit a 1 ?", "Is the third bit a 1 ?", and so on. After we know the answers to these n
questions, we will know the number, because we will know its binary expansion.
13. Without loss of generality, we may assume that the given integer is positive (since n I a if and only if n 1(-a),
and the case a= 0 is trivial). Let the decimal expansion of the integer a be given by a= (an-ian_ 2 ... a 1a 0 )i 0 .
Thus a= 10n-lan-l + 10n- 2an-2 + · · · + lOa1 + ao. Since
a 1 + a 0 (mod 9). Therefore a
0 (mod 9) if and only if
Since being divisible by 9 is the same as being congruent
divisible by 9 if and only if the sum of its decimal digits is
=
15. Note that Qn
a factor of n!
of Qn. Thus
factors of Qn
10 = 1 (mod 9), we have a= an-1 + an-2 + · · · +
the sum of the digits is congruent to 0 (mod 9).
to 0 (mod 9), we have proved that an integer is
divisible by 9.
2: 2. By the Fundamental Theorem of Arithmetic, Qn has a prime factorization. Because k is
for all k ::;: n, when Qn is divided by k the remainder will be 1; therefore k is not a factor
Qn must have a prime factor greater than n. (We have shown that in fact all of the prime
will be greater than rz.)
17. The numbers whose decimal expansion ends in 1 are precisely those integers in the arithmetic progression
lOk + 1. Dirichlet's theorem guarantees that this sequence includes infinitely many primes. The first few are
11, 31, 41, 61, 71, 101, 131, ....
19. Note that even numbers greater than 2 are composite and that 9 is composite. Therefore every odd number
greater than 11 is the sum of two composite numbers (13 = 4 + 9, 15 = 6 + 9, 17 = 8 + 9, and so on).
Similarly, because 8 is composite, so are all even numbers greater than 11: 12 = 4 + 8, 14 = 6 + 8, 16 = 8 + 8,
and so on.
21. Assume that every even integer greater than 2 is the sum of two primes, and let n be an integer greater
than 5. If n is odd, then we can write n = 3 + (n - 3), decompose n - 3 = p + q into the sum of two primes
(since rz - 3 is an even integer greater than 2), and therefore have written n = 3 + p + q as the sum of three
primes. If n is even, then we can write n = 2 + (n - 2), decompose n - 2 = p + q into the sum of two primes
(since n - 2 is an even integer greater than 2), and therefore have written n = 2 + p + q as the sum of three
primes. For the converse, assume that every integer greater than 5 is the sum of three primes, and let n be
an even integer greater than 2. By our assumption we can write n + 2 as the sum of three primes. Since
n + 2 is even, these three primes cannot all be odd, so we have n + 2 = 2 + p + q, where p and q are primes,
whence n = p + q, as desired.
23. We give a proof by contradiction. For this proof we need a fact about polynomials, namely that a nonconstant
polynomial can take on the same value only a finite number of times. (Think about its graph~a polynomial of
degree rz has at most n "wiggles'' and so a horizontal line can intersect it at most n times. Alternatively, this
statement follows from the Fundamental Theorem of Algebra, which guarantees that a polynomial of degree
n has at most n O's; if f (x) = a, then x is a zero of the polynomial f (x) - a.) Our given polynomial f can
take on the values 0 and ±1 only finitely many times, so if there is not some y such that f(y) is composite,
Supplementary Exercises
145
then there must be some x 0 such that ±f(xo) is prime, say p. Now look at f(xo + kp). When we plug
x 0 + kp in for x in the polynomial and multiply it out, every term will contain a factor of p except for the
terms that form f (x 0 ) . Therefore f (x 0 + kp) = f ( Xo) + mp = (m ± 1 )p for some integer m. As k varies,
this value can be 0 or p or -p only finitely many times; therefore it must be a different multiple of p and
therefore a composite number for some values of k, and our proof is complete.
25. gcd(10223, 33341) = gcd(10223, 2672) = gcd(2672, 2207) = gcd(2207, 465) = gcd( 465, 347) = gcd(347, 118) =
gcd(118,111) = gcd(lll, 7) =gcd(7,6) = gcd(6,1) =gcd(l,O) = 1
27. By Lemma 1 in Section 4.3, gcd(2n + 1, 3n + 2) = gcd(2n + 1, n + 1), since 2n + 1 goes once into 3n + 2
with a remainder of n + 1. Now if we divide n + 1 into 2n + 1, we get a remainder of n, so the answer must
equal gcd(n + 1, n). At this point, the remainder when dividing n into n + 1 is 1, so the answer must equal
gcd( n, 1), which is clearly 1. Thus the answer is 1.
29. It might be helpful to read the solution to Exercise 55 in Section 4.3 to see the philosophy behind this approach.
Suppose by way of contradiction that q1 , q2 , ... , qn are all the primes of the form 6k + 5. Thus q1 = 5,
q2 = 11, and so on. Let Q = 6q1q2 · · · qn - l. We note that Q is of the form 6k + 5, where k = q1q2 · · · qn - l.
Now Q has a prime factorization Q = P1P2 · ··Pt. Clearly no Pi is 2, 3, or any q1 , because the remainder
when Q is divided by 2 is 1, by 3 is 2, and by q1 is q1 - 1. All odd primes other than 3 are of the form
6k + 1 or 6k + 5, and the product of primes of the form 6k + 1 is also of this form. Therefore at least one of
the Pi 's must be of the form 6k + 5, a contradiction.
31. Let's try the strategy used in the proof of Theorem 3 in Section 4.3. Suppose that P1, P2, ... , Pn are the
only primes of the form 4k + 1. Notice that the product of primes of this form is again of this form, because
(4k1+1) (4k2 + 1) = l6k1k2 + 4k1+4k2+1 = 4( 4k1k2 + k1+k2)+1. We could try looking at 4p1p2 · · · Pn + 1,
which is again of this form. By the Fundamental Theorem of Arithmetic, it has prime factors, and clearly no
Pi is a factor. Unfortunately, we cannot be guaranteed that any of its prime factors are of the form 4k + 1,
because the product of two odd primes not of this form, namely of the form 4k + 3, is of the form 4k + 1;
indeed, (4k1 +3)(4k2 + 3) = l6k 1k2 + 12k1 + 12k2 + 9 = 4( 4k1k2 + 3k1 + 3k2 + 2) + 1. Thus the proof breaks
down at this point.
33. a) Since 2 is a factor of all three of these integers, this set is not mutually relatively prime.
b) Since 12 and 25 share no common factors, this set has greatest common divisor 1, so it is mutually
relatively prime. (It is possible for every pair of integers in a set of mutually relatively prime integers to have
a nontrivial common factor (see Exercise 34), but certainly if two of the integers in a set are relatively prime,
then the set is automatically mutually relatively prime.)
c) Since 15 and 28 share no common factors, this set has greatest common divisor 1, so it is mutually
relatively prime.
d) Since 21 and 32 share no common factors, this set has greatest common divisor 1, so it is mutually
relatively prime.
35. If n is even, then n 4 +4n is an even composite number, so we can restrict ourselves to n odd. The appearance
of 4 suggests that we might work modulo 5, so let's try that. If n is not divisible by 5, then n 4 = 1 (mod 5)
by Fermat's little theorem, and since n is odd, 4n
(-l)n = -1 (mod 5). Therefore n 4 +4n is divisible by 5.
So except for n = 1, in which case n 4 +4n = 5 is prime, the only possible values of n that can result in a prime
value for n 4 + 4n are 5, 15, 25, and so on. Maple tells us that 54 + 4 5 = 17 · 97, 15 4 + 4 15 = 36833 · 29153,
and 25 4 + 4 25 = 29 · 373 · 3121·33350257. So let us try to factor n 4 + 4n algebraically. It is not at all obvious
how to discover these factors, but if we multiply out
=
146
Chapter 4
Number Theory and Cryptography
we get n 4 + 4n. (Because n is odd, the exponent nt 1 is an integer.) It only remains to show that each of
these factors is greater than 1. The first is clearly so. The second factor takes on the values 5, 17, 65, and
n+l
305 for n = 3, 5, 7, and 9, respectively. Clearly for large n the 2n term far exceeds the -2-2-n term, and
our proof is complete.
37. The least common multiple of 6 and 15 is 30, so the set of solutions will be given modulo 30 (see Exercise 38).
Since the numbers involved here are so small, it is probably best simply to write down the solutions of
x = 4 (mod 6) and then see which, if any, of them are also solutions of x = 13 (mod 15). The solutions of
the first congruence, up to 30, are 4, 10, 16, 22, and 28. Only 28 is congruent to 13 modulo 15. Therefore
the general solution is all numbers congruent to 28 modulo 30, i.e., ... , -32, -2, 28, 58, ....
39. Maple tells us that this is true, namely that n 9 - n mod 30 = 0 for all n (we need only check n from 0 to
29). For a more human-oriented proof (conceptual rather than computational), notice that it suffices to show
that n 9 - n 0 (mod 2), n 9 - n 0 (mod 3), and n 9 - n 0 (mod 5). The first is obvious (odd minus odd,
and even minus even, are both even). The second follows from Fermat's little theorem, because n 9 = (n 3) 3
n 3 n (mod 3). The third also follows from that theorem, because n 9 = n 4 · n 5 1 · n n (mod 5).
=
=
=
=
= =
=
(mod q) and clearly qP- 1 = O (mod q). Therefore pq-l + qP- 1
1+0 = 1 (mod q). Similarly, pq-l + qP-l = 1 (mod p). It follows from the Chinese remainder theorem that
pq-l + qP-l = 1 (mod pq).
41. By Fermat's little theorem, pq-l
=1
43. Because 1 and 3 are both relatively prime to 10, if the congruence is satisfied by the correct ISBN-13, it
cannot be satisfied if one of the digits is changed. In particular, if a, is changed from x to y, then the change
in the left-hand side of the congruence is either y - x or 3(y - x), modulo 10, neither of which can be 0.
Therefore the sum can no longer be 0 modulo 10.
45. Subtract d9 from both sides of the congruence and multiply through by -1 to obtain
This is equivalent to
7(d1 + d4 + d7) + 3(d2 + ds + ds) + 9(d3 + d5)
=dg (mod 10).
Because 0:::; dg :::; 9, it follows that dg = 7(d1 + d4 + d7) + 3(d2 + d5 + ds) + 9(d3 + d5) mod 10. We calculate
with the given eight digits to conclude that d 9 = 7(1+0+0)+3(1+0+2)+9(1+0) mod 10 = 25 mod 10 = 5.
4 7. We need to find the inverse function. In other words, given (ap + b) mod 26, how does one recover p?
Working modulo 26, if we subtract b, then we will have ap. If we then multiply by an inverse of a (which
must exist since we are assuming that gcd( a, 26) = 1), we will have p back. Therefore the decryption
function is g( q) = a( q - b) mod 26, where a is an inverse of a modulo 26. In this case, a = 7 and
b = 10. Computing the inverse of 7 modulo 26 by the techniques of Section 4.4 (or by using Maple), we find
a= 15, so the decryption function is g(q) = 15(q-10) mod 26. Translating the letters into numbers, we see
that the encrypted message is 11-9-12-10-6 12-6-12-23-5 16-4-23-12-22. Applying this function,
we obtain the decrypted message 15-11-4-0-18 4-18-4-13-3 12-14-13-4-24. This translates into
PLEAS ESEND MONEY, which, after correcting the spacing, is PLEASE SEND MONEY.
49. a) The seed is 23 (X); adding this to the first character of the plaintext, 19 (T), gives 16, which is Q. Therefore
the first character of the ciphertext is Q. The next character of the keystream is the aforementioned T (19);
add this to H (7) to get 0 (A), so the next character of the ciphertext is A. We continue in this manner,
producing the encrypted message QAL HUVEM AT WVESGB.
Writing Projects
147
b) Again the seed is 23 (X); adding this to the first character of the plaintext, 19 (T), gives 16, which is Q.
Therefore the first character of the ciphertext is Q. The next character of the keystream is the aforementioned
Q (16); add this to H (7) to get 23 (X), so the next character of the ciphertext is X. We continue in this
manner, producing the encrypted message QXB EVZZL ZEVZZRFS.
WRITING PROJECTS FOR CHAPTER 4
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. As usual, the Web is an excellent resource here.
www.mersenne.org/prime.htm.
Check out the GIMPS page and then follow its links:
2. These primes are sometimes jokingly referred to as "industrial strength primes." Number theory textbooks,
such as [Ro3], would be a place to start. See also the article [Le3].
3. One can often find mathematical news reported in The New York Times and other nontechnical media. Search
an index to find a story about this topic. (While you're looking at back issues of the Times, read the January
31, 1995, article on the solution to Fermat's last theorem.)
4. Your essay should mention the RSA-129 project. See The New York Times, around the spring of 1994 (use
its index). For an expository article, try [Po].
5. There are dozens of books on computer hardware and circuit design that discuss the algorithms and circuits
used in performing these operations. If you are a computer science or computer engineering major, you
probably have taken (or will take) a course that deals with these topics. See [Ko2] and similar books.
6. A traditional history of mathematics book should be helpful here; try [Bo4] or [Ev3].
7. This topic has taken on a lot of significance recently as randomized algorithms become more and more important. The August/September 1994 issue of SIAM News (the newsletter of the Society for Industrial and
Applied Mathematics) has a provocative article on the subject. See also [Lal], or for older material in a
textbook try volume 2 of [Kn].
8. The Web has the answers. Try www. skynet. iermartin/pages/iban.html or Wikipedia.
9. The Web has the answers.
10. The author's number theory text ([Ro3]) has material on this topic. The amazing mathematician John H.
Conway (inventor of the Game of Life, among other things) has devised what he calls the Doomsday Algorithm,
and it works quite fast with practice. See [BeCo]. (Conway can determine any day of the week mentally in a
second or two.)
11. This topic gets into the news on a regular basis. Try The New York Times index. See also Writing Project 4,
above, and 12, below.
12. If I encrypt my signature with my private key, then I will produce something that will decrypt (using my
public key) as my signature. Furthermore, no one else can do this, since no one else knows my private key.
Good sources for cryptography include [Be], [MeOo], and [St2].
Chapter 4
148
Number Theory and Cryptography
13. Several excellent books have appeared in the past decade on cryptography, such as [MeOo]. Many of them,
including that one, will treat this topic.
14. Suppose that we use a prime for n. To find a private decryption key from the corresponding public encryption
key e, one would need to find a number d that is an inverse for e modulo n - 1 so that the calculation shown
before Example 9 in Section 4.6 can go through. But finding such a d is easy using the Euclidean algorithm,
because the person doing this would already know n - 1. (In particular, to find d, one can work backward
through the steps of the Euclidean algorithm to express 1 as a linear combination of e and n - 1 ; then d
is the coefficient of e in this linear combination.) The important point in the actual RSA system is that the
person trying to find this inverse will not know (p - l)(q - 1) and therefore cannot simply use the Euclidean
algorithm.
15. Start with Wikipedia.
Section 5.1
Mathematical Induction
149
CHAPTERS
Induction and Recursion
SECTION 5.1
Mathematical Induction
Understanding and constructing proofs by mathematical induction are extremely difficult tasks for most students. Do not be discouraged, and do not give up, because, without doubt, this proof technique is the most
important one there is in mathematics and computer science. Pay careful attention to the conventions to be
observed in writing down a proof by induction. As with all proofs, remember that a proof by mathematical
induction is like an essay-it must have a beginning, a middle, and an end; it must consist of complete sentences, logically and aesthetically arranged; and it must convince the reader. Be sure that your basis step
(also called the "base case") is correct (that you have verified the proposition in question for the smallest
value or values of n ), and be sure that your inductive step is correct and complete (that you have derived the
proposition for k + 1, assuming the inductive hypothesis that the proposition is true for k ).
Some, but not all, proofs by mathematical induction are like Exercises 3~17. In each of these, you are
asked to prove that a certain summation has a "closed form" representation given by a certain expression.
Here the proofs are usually straightforward algebra. For the inductive step you start with the summation for
P(k + 1), find the summation for P(k) as its first k terms, replace that much by the closed form given by the
inductive hypothesis, and do the algebra to get the resulting expression into the desired form. When doing
proofs like this, however, remember to include all the words surrounding your algebra-the algebra alone is
not the proof. Also keep in mind that P(n) is the proposition that the sum equals the closed-form expression,
not just the sum and not just the expression.
Many inequalities can be proved by mathematical induction; see Exercises 18~24, for example. The
method also extends to such things as set operations, divisibility, and a host of other applications; a sampling
of them is given in other exercises in this set. Some are quite complicated.
One final point about notation. In performing the inductive step, it really does not matter what letter we
use. We see in the text the proof of P( k) ----+ P( k + 1) ; but it would be just as valid to prove P( n) ----+ P( n + 1),
since the k in the first case and the n in the second case are just dummy variables. l-Ve will use both notations
in this Guide; in particular, we wi1l use k for the first few exercises but often use n afterwards. A student just
beginning to write proofs by mathematical induction should be even more formal than we are in this Guide
and follow the template given at the end of this section. With experience, you can relax a bit and the style
can become more informal, as long as all the steps are there.
1. We can prove this by mathematical induction. Let P( n) be the statement that the train stops at station n.
We want to prove that P(n) is true for all positive integers n. For the basis step, we are told that P(l)
is true. For the inductive step, we are told that P(k) implies P(k + 1) for each k 2 1. Therefore by the
principle of mathematical induction, P( n) is true for all positive integers 11.
3. a) Plugging in n
= 1 we have that P(l) is the statement 12 = 1 · 2 · 3/6.
b) Both sides of P(l) shown in part (a) equal 1.
c) The inductive hypothesis is the statement that
12
+ 22 + ... + k2 =
k(k
+ 1)(2k + 1).
6
Chapter 5
150
Induction and Recursion
d) For the inductive step, we want to show for each k 2'. 1 that P(k) implies P(k + 1). In other words, we
want to show that assuming the inductive hypothesis (see part ( c)) we can show
12+22+ .. ·+k2+(k+l)2= (k+l)(k+2)(2k+3).
6
e) The left-hand side of the equation in part (d) equals, by the inductive hypothesis, k(k + 1)(2k + 1)/6 +
(k + 1) 2 . We need only do a bit of algebraic manipulation to get this expression into the desired form: factor
out (k + 1)/6 and then factor the rest. In detail,
k(k + 1)(2k + 1)
+ (k + 1) 2 (by the inductive hypothesis)
(1 2 + 22 + · · · + k2) + (k + 1) 2 =
6
1
1
(2k 2 +7k + 6)
=
(k(2k+1) +
+ 1)) =
k;
6(k
k;
= k + 1(k+2)(2k + 3) = (k + l)(k + 2)(2k + 3) .
6
6
f) We have completed both the basis step and the inductive step, so by the principle of mathematical induction,
the statement is true for every positive integer n.
5. We proceed by induction. The basis step, n = 0, is true, since 12 = 1·1 · 3/3. For the inductive step assume
the inductive hypothesis that
12+32+52+ .. ·+(2k+1)2= (k+1)(2k+1)(2k+3).
3
We want to show that
12 + 32 + 52+ ... +(2k+1)2 + (2k + 3)2 = (k + 2)(2k + 3)(2k + 5)
3
(the right-hand side is the same formula with k + 1 plugged in for n ). Now the left-hand side equals, by the
inductive hypothesis, (k + 1)(2k + 1)(2k + 3)/3 + (2k + 3) 2 . We need only do a bit of algebraic manipulation
to get this expression into the desired form: factor out (2k + 3)/3 and then factor the rest. In detail,
(1 2 + 32 + 52 + ... +(2k+1) 2) + (2k + 3) 2
(k + 1)(2k + 1)(2k + 3)
.
.
.
+ (2k + 3) 2 (by the mduct1ve hypothesis)
3
2
2
3
3
= k: ((k+1)(2k+1)+3(2k+3)) = k: (2k 2 +9k+10)
=
= 2k+3((k+2)(2k+5)) = (k+2)(2k+3)(2k+5).
3
3
7. Let P(n) be the proposition 3 + 3 · 5 + 3 · 52 + · · · + 3 · 5n = 3(5n+l - 1)/4. To prove that this is true
for all nonnegative integers n, we proceed by mathematical induction. First we verify P(O), namely that
3 = 3(5 -1)/4, which is certainly true. Next we assume that P(k) is true and try to derive P(k + 1). Now
P(k
+ 1) is the formula
3(5k+ 2 - 1)
3 + 3. 5 + 3. 52 + ... + 3. 5k + 3. 5k+l =
.
4
All but the last term of the left-hand side of this equation is exactly the left-hand side of P(k), so by the
inductive hypothesis, it equals 3(5k+l - 1)/4. Thus we have
3(5k+l - 1)
3 + 3 · 5 + 3 · 52 + · · · + 3 · 5k + 3 · 5k+I =
+ 3 · 5k+l
4
= 5k+l
(~ +
4
3) -
~
4
= 5k+l . 15 - ~
4
4
= 5k+2. ~ - ~ = 3(5k+2 - 1) .
4 4
4
Section 5.1
Mathematical Induction
151
9. a) We can obtain a formula for the sum of the first n even positive integers from the formula for the sum of
the first n positive integers, since 2 + 4 + 6 + · · · + 2n = 2(1+2 + 3 + · · · + n). Therefore, using the result of
Example 1, the sum of the first n even positive integers is 2(n(n + 1)/2) = n(n + 1).
b) We want to prove the proposition P( n) : 2 + 4 + 6 + · · · + 2n = n( n + 1). The basis step, n = 1, says that
2 = 1·(1+1), which is certainly true. For the inductive step, we assume that P(k) is true, namely that
2 + 4 + 6 + ... + 2k = k(k + 1))
and try to prove from this assumption that P(k
+ 1)
is true, namely that
2 + 4 + 6 + ... + 2k + 2(k + 1) = (k + l)(k + 2).
(Note that the left-hand side consists of the sum of the first k
+ 1 even positive integers.) We have
2 + 4 + 6 + ... + 2k + 2(k + 1) = (2 + 4 + 6 + ... + 2k) + 2(k + 1)
= k(k + 1) + 2(k + 1)
(by the inductive hypothesis)
=(k+l)(k+2),
as desired, and our proof by mathematical induction is complete.
11. a) Let us compute the values of this sum for n :S 4 to see whether we can discover a pattern. For n = 1 the
sum is ~ . For n = 2 the sum is ~ + :j = ~ . For n = 3 the sum is ~ + :j + ~ = £. And for n = 4 the sum is
15/16. The pattern seems pretty clear, so we conjecture that the sum is always (2n - 1)/2n.
b) We have already verified that this is true in the base case (in fact, in four base cases). So let us assume it
for k and try to prove it for k + 1. More formally, we are letting P(n) be the statement that
1 1 1
1
2n - 1
-+-+-+···+- = - - .
2 4 8
2n
2n '
and trying to prove that P(n) is true for all n. We have already verified P(l) (as well as P(2), P(3), and
P(4) for good measure). We now assume the inductive hypothesis P(k), which is the equation displayed
above with k substituted for n, and must derive P( k + 1), which is the equation
1
1
1
2+ 4+ 8+
1
1
2k+l - 1
... + 2k + 2k+l = 2k+l
The "obvious" thing to try is to add l/2k+l to both sides of the inductive hypothesis and see whether the
algebra works out as we hope it will. We obtain
1 1 1
1)
1
2k - 1
1
2 . 2k - 2 . 1 + 1 2k+i - 1
2k+l
= 2k+l '
( 2 + 4 + S + ... + 2k + 2k+l = 2 k + 2k+l =
as desired.
13. The base case of the statement P(n): 12 -2 2 +3 2 - ··· + (-1r- 1n 2 =(-1r- 1n(n+1)/2, when n = 1, is
12 = (-1) 0 · l · 2/2, which is certainly true. Assume the inductive hypothesis P(k), and try to derive P(k+ 1):
12 - 22 + 32 - ... + ( -1 )k-1 k2 + (-1 l (k + 1)2 = (-1 )d k + 1) (k + 2)
2
.
Starting with the left-hand side of P(k
+ 1), we have
(12 - 22 + 32 - ... + (-l)k-lk2) + (-l)k(k + 1)2
= (-l)k-l k(k + l) + (-l)k(k + 1) 2 (by the inductive hypothesis)
2
= (-l)k(k + 1)((-k/2) + k + 1)
=(-l)k(k+1)(~+1)
the right-hand side of P(k
+ 1).
=(-lt(k+1)2(k+2)'
152
Chapter 5
Induction and Recursion
15. The base case of the statement P(n): 1·2+2 · 3 + · · · + n(n + 1) = n(n + l)(n + 2)/3, when n = 1, is
1 · 2 = 1 · 2 · 3/3, which is certainly true. We assume the inductive hypothesis P( k), and try to derive P( k + 1):
1. 2 + 2. 3 + ... + k(k + 1) + (k + l)(k + 2) = (k + l)(k + 2)(k + 3)
Starting with the left-hand side of P(k
3
+ 1), we have
(1. 2 + 2. 3 + ... + k(k + 1)) + (k + l)(k + 2)
2
= k(k + l)(k + ) + (k + l)(k + 2)
3
(by the inductive hypothesis)
=(k+l)(k+2)(~+1) = (k+l)(k;2)(k+3)'
the right-hand side of P(k
+ 1).
17. This proof follows the basic pattern of the solution to Exercise 3, but the algebra gets more complex. The
statement P( n) that we wish to prove is
4
4
4
1 + 2 + 3 + · .. +n 4 =
n(n+1)(2n+l)(3n 2 +3n-l)
30
'
where n is a positive integer. The basis step, n = 1, is true, since 1 · 2 · 3 · 5/30 = 1. Assume the displayed
statement as the inductive hypothesis, and proceed as follows to prove P(n + 1):
2
(14 + 24 +· .. +n4) + ( n+ l) 4 = n(n+l)(2n+1)(3n +3n-l) + (n+l )4
30
n+l
= 3() (n(2n + 1)(3n 2 + 3n - 1) + 30(n + 1) 3 )
n+l
4
3
2
= ----W-(6n + 39n + 9ln + 89n + 30)
n+l
= ----W-(n + 2)(2n + 3)(3(n + 1) 2 + 3(n + 1) -1)
The last equality is straightforward to check; it was obtained not by attempting to factor the next to last
expression from scratch but rather by knowing exactly what we expected the simplified expression to be.
19. a) P(2) is the statement that 1 + ~ < 2 - ~.
b) This is true because 5 / 4 is less than 6 / 4.
c) The inductive hypothesis is the statement that
1
1
1
1 + 4 + ... + k2 < 2 - k .
d) For the inductive step, we want to show for each k 2: 2 that P(k) implies P(k
want to show that assuming the inductive hypothesis (see part ( c)) we can show
+ 1).
1
1
1
1
1 + 4 + ... + k 2 + (k + 1) 2 < 2 - k + 1.
e) Assume the inductive hypothesis. Then we have
1
1
1
1
1
l+4+·"+ k 2 + (k+1) 2 < 2 -k+ (k+1) 2
= 2 -(~- (k~1)2)
= 2 - ( k + 2k + 1 - k)
2
k(k + 1) 2
k +k
1
=2---k(k + 1)2
k(k + 1)2
1
1
1
=2---<2--k+l
k(k+1) 2
k+l'
2
In other words, we
Section 5.1
Mathematical Induction
153
f) We have completed both the basis step and the inductive step, so by the principle of mathematical induction,
the statement is true for every positive integer n greater than 1.
21. Let P( n) be the proposition 2n > n 2 . We want to show that P( n) is true for all n > 4. The base case
is therefore n = 5, and we check that 25 = 32 > 25 = 52 . Now we assume the inductive hypothesis that
2k > k 2 and want to derive the statement that 2k+l > (k + 1) 2 . Working from the right-hand side, we have
(k + 1) 2 = k 2 +2k+1 3). Thus we have (k + 1) 2 < 2k 2 < 2 · 2k
(by the inductive hypothesis), which in turn equals 2k+l , as desired.
23. We compute the values of 2n + 3 and 2n for the first few values of n and come to the immediate conjecture
that 2n + 3 :S:: 2n for n 2: 4 but for no other nonnegative integer values of n. The negative part of this
statement is just the fact that 3 > 1, 5 > 2, 7 > 4, and 9 > 8. We must prove by mathematical induction
that 2n + 3::; 2n for all n;::: 4. The base case is n = 4, in which we see that, indeed, 11::; 16. Next assume
the inductive hypothesis that 2n + 3::; 2n, and consider 2(n + 1) + 3. This equals 2n + 3 + 2, which by the
inductive hypothesis is less than or equal to 2n + 2. But since n ;::: 1, this in turn is at most 2n + 2n = 2n+l,
precisely the statement we wished to prove.
25. We can assume that h > -1 is fixed, and prove the proposition by induction on n. Let P( n) be the proposition
1 + nh ::; (1 +hr. The base case is n = 0, in which case P(O) is simply 1 ::; 1, certainly true. Now we assume
the inductive hypothesis, that 1 + kh :S:: (1 + h)k; we want to show that 1 + (k + l)h :S:: (1 + h)k+l. Since
h > -1, it follows that 1 + h > 0, so we can multiply both sides of the inductive hypothesis by 1 + h to obtain
(l+h)(l+kh)::; (l+h)k+l. Thus to complete the proof it is enough to show that l+(k+l)h::; (l+h)(l+kh).
But the right-hand side of this inequality is the same as 1 + h + kh + kh 2 = 1 + (k + l)h + kh 2 , which is
greater than or equal to 1 + (k + l)h because kh 2 ;::: 0.
27. This exercise involves some messy algebra, but the logic is the usual logic for proofs using the principle of
mathematical induction. The basis step ( n = 1) is true, since 1 is greater than 2( V2 - 1) ~ 0.83. We assume
that
1
1 + _!_ + ... + - - > 2(v'n+l - 1)
Vn
y'2
and try to derive the corresponding statement for n + 1:
1
1
1
1 + - + .. · + - +
V2
Since by the inductive hypothesis we know that
1
1
1+-+···+-+
V2
VnTI
Vn
Vn
1
VnTI
>2(vn+2-1)
>2(vn+1-1)+
1
VnTI
,
we will be finished if we can show that the inequality
2(Vri+T-1)+
~>2(vn+2-1)
n+l
holds. By canceling the -2 from both sides and rearranging, we obtain the equivalent inequality
2(vn+2-vn+i) <
k·
n+l
which in turn is equivalent to
~
~
~
~
2(vn-r..::-vn+l)(vn+2+vn+l)
VnTI
Jn+2
n+l
< Vn+T+ Vn+T.
n+l
This last inequality simplifies to
Vn+2
2<1+ ~'
vn+ 1
which is clearly true. Therefore the original inequality is true, and our proof is complete.
154
Chapter 5
Induction and Recursion
29. Recall that Hk = 1/1+1/2 + · · · + l/k. We want to prove that H2n:::; 1 + n for all natural numbers n. We
proceed by mathematical induction, noting that the basis step n = 0 is the trivial statement H 1 = 1 :::; 1+0.
Therefore we assume that H2n :::; 1 + n; we want to show that H 2 n+1 :::; 1+(n+1). We have
H 2 n+1 =
:::;
<
=
1
1
1
+ - - - + - - - + · · · + - -1 (by definition; there are 2n fractions here)
2n + 1 2n + 2
2n+
1
1
1
(1 + n) + n + + n + + · · · + n+l (by the inductive hypothesis)
2
1 2
2
2
1
1
1
(1 + n) + - - + - - - + · · · + - - - (we made the denominators smaller)
2n + 1 2n + 1
2n + 1
2n
1 + n + - - < 1+n+1=1+(n+1).
2n + 1
H2n
31. This is easy to prove without mathematical induction, because we can observe
and either n or n + 1 is even. If we want to use the principle of mathematical
as follows. The basis step is the observation that 12 + 1 = 2 is divisible by
hypothesis, that k2 + k is divisible by 2; we must show that (k + 1) 2 + (k +
that n 2 + n = n(n + 1),
induction, we can proceed
2. Assume the inductive
1) is divisible by 2. But
(k + 1) 2 +(k+1) = k2 +2k+1+k+1 = (k 2 + k) + 2(k + 1). But now k2 + k is divisible by 2 by the inductive
hypothesis, and 2(k + 1) is divisible by 2 by definition, so this sum of two multiples of 2 must be divisible
by 2.
33. To prove that P(n): 5 l(n 5
n) holds for all nonnegative integers n, we first check that P(O) is true; indeed
5 I 0. Next assume that 5 I(n - n) , so that we can write n 5 - n = 5t for some integer t. Then we want to
prove P( n + 1), namely that 5 I((n + 1)5 - ( n + 1)) . We expand and then factor the right-hand side to obtain
-
5
(n + 1) 5
-
(n + 1) = n 5 + 5n4 + 10n3 + 10n2 +5n+1 - n - 1
= (n 5
-
n) + 5(n 4 + 2n3 + 2n 2 + n)
= 5t + 5(n 4 + 2n3 + 2n 2 + n)
4
3
(by the inductive hypothesis)
2
= 5(t + n + 2n + 2n + n).
Thus we have shown that (n + 1) 5 - ( n + 1) is also a multiple of 5, and our proof by induction is complete.
(Note that here we have used n as the dummy variable in the inductive step, rather than k. It really makes
no difference.)
We should point out that using mathematical induction is not the only way to prove this proposition;
it can also be proved by considering the five cases determined by the value of n mod 5. The reader is
encouraged to write down such a proof.
35. First let us rewrite this proposition so that it is a statement about all nonnegative integers, rather than just
the odd positive integers. An odd positive integer can be written as 2n - 1, so let us prove the proposition
P(n) that (2n - 1) 2 - 1 is divisible by 8 for all positive integers n. We first check that P(l) is true; indeed
8 I 0. Next assume that 8 l((2n - 1) 2 - 1). Then we want to prove P(n + 1), namely that 8 l((2n + 1) 2 - 1).
Let us look at the difference of these two expressions: (2n + 1) 2 -1- ((2n -1) 2 -1). A little algebra reduces
this to 8n, which is certainly a multiple of 8. But if this difference is a multiple of 8, and if, by the inductive
hypothesis, (2n - 1)2 - 1 is a multiple of 8, then (2n + 1)2 - 1 must be a multiple of 8, and our proof by
induction is complete.
37. It is not easy to stumble upon the trick needed in the inductive step in this exercise, so do not feel bad
if you did not find it. The form is straightforward. For the basis step ( n = 1 ), we simply observe that
11 1 +1 + 12 2 l-l = 121+12 = 133, which is divisible by 133. Then we assume the inductive hypothesis, that
Mathematical Induction
Section 5.1
155
11n+l+122n-i is divisible by 133, and let us look at the expression when n + 1 is plugged in for n. We want
somehow to manipulate it so that the expression for n appears. We have
11(n+l)+l+122(n+l)-1
=
11. 11n+l+144. 122n-1
=
11 · 11 n+l + (11 + 133) · 122n-i
= 11(11 n+l + 12 2n-i) + 133 · 12 2n-i.
Looking at the last line, we see that the expression in parentheses is divisible by 133 by the inductive
hypothesis, and obviously the second term is divisible by 133, so the entire quantity is divisible by 133,
as desired.
39. The basis step is trivial, as usual: Ai c;;; Bi implies that n~=i AJ c;;; n~=i BJ because the intersection of one
set is itself. Assume the inductive hypothesis that if AJ c;;; BJ for j = L 2, ... , k, then n:=i A 1 c;;; n:=i B 1 .
We want to show that if AJ c;;; B 1 for j = 1, 2, ... , k + 1, then n:~i AJ c;;; n:~i BJ. To show that one set is
a subset of another we show that an arbitrary element of the first set must be an element of the second set.
So let x E
n~~i AJ = ( n~=i AJ) n Ak+i.
Because x E
n~=i AJ,
we know by the inductive hypothesis that
x E n:=i BJ; because x E Ak+l, we know from the given fact that Ak+i
x E (n:=l BJ) n Bk+I =
<;;;;
Bk+l that x E Bk+l · Therefore
n:~i BJ.
This is really easier to do directly than by using the principle of mathematical induction. For a noninductive proof, suppose that x E n;=l AJ. Then x E AJ for each j from 1 to n. Since Aj c;;; BJ, we know
that x E B 1 . Therefore by definition, x E n;=l BJ.
41. In order to prove this statement, we need to use one of the distributive laws from set theory: (XU Y) n Z =
(X n Z) U (Y n Z) (see Section 2.2). Indeed, the proposition at hand is the generalization of this distributive
law, from two sets in the union to n sets in the union. We will also be using implicitly the associative law for
set union.
The basis step, n = 1, is the statement Ai n B
the inductive hypothesis that
= A1 n B,
which is obviously true. Therefore we assume
(Ai U A2 U · · · U An) n B = (Ai n B) U (A2 n B) U · · · U (Ann B).
We wish to prove the similar statement for n
+ 1, namely
Starting with the left-hand side, we apply the distributive law for two sets:
((Ai U A2 U .. · U An) U An+i) n B
=
((Ai U A2 U .. · U An) n B) U (An+i n B)
= ((A1 n B) U (A 2 n B) U · · · U (Ann B))
U (An+i
n B)
(by the inductive hypothesis)
= (A1 n B) u (A2 n B) U · · · u (Ann B) U (An+i n B)
43. In order to prove this statement, we need to use one of De Morgan's laws from set theory: (AU B) =An B
(see Section 2.2). Indeed, the proposition at hand is the generalization of this law, from two sets in the union
to n sets in the union. We will also be using implicitly the associative laws for set union and intersection.
The basis step, n = 1, is the statement Ai = Ai (since the union or intersection of just one set is the set
itself), and this proposition is obviously true. Therefore we assume the inductive hypothesis that
n
LJ Ak =
k=i
n
n
Ak.
k=i
Chapter 5
156
We wish to prove the similar statement for n
Induction and Recursion
+ 1, namely
LJ Ak = nAk.
n+l
n+l
k=l
k=l
Starting with the left-hand side, we group, apply De Morgan's law for two sets, and then the inductive
hypothesis:
n+l
n
k=l
k=l
LJ Ak = ( LJ Ak) U An+
i
-n - =
LJ Ak
n An+i
(by DeMorgan's law)
k=l
n
n
=n
Ak) n An+1
= (
(by the inductive hypothesis)
k=l
n+l
Ak
k=l
45. This proof will be similar to the proof in Example 10. The basis step is clear, since for n = 2, the set has
exactly one subset containing exactly two elements, and 2(2 - 1)/2 = 1. Assume the inductive hypothesis,
that a set with n elements has n(n - 1)/2 subsets with exactly two elements; we want to prove that a set
S with n + 1 elements has (n + l)n/2 subsets with exactly two elements. Fix an element a in S, and let
T be the set of elements of S other than a. There are two varieties of subsets of S containing exactly two
elements. First there are those that do not contain a. These are precisely the two-element subsets of T, and
by the inductive hypothesis, there are n( n - 1) /2 of them. Second, there are those that contain a together
with one element of T. Since T has n elements, there are exactly n subsets of this type. Therefore the total
number of subsets of S containing exactly two elements is (n(n - 1)/2) + n, which simplifies algebraically to
(n + l)n/2, as desired.
4 7. The hint gives the greedy algorithm. We reorder the locations if necessary so that x1 :::;;
x 3 :::;; · · · :::;; Xd.
The first tower is placed at position t 1 = x 1 + 1. It will serve the first building and any other buildings at
locations no greater than t 1 + 1 . Assume that we have placed tower k at tk. Then place tower k + 1 at
position tk+l = x + 1, where x is the smallest x, greater than tk + 1. Because all the buildings up to the one
at position x were already serviced by the previous towers, and the building at position x is serviced by the
new tower at tk+ 1 , all the buildings will be serviced once this process is completed.
x2 :::;;
49. The one and only flaw in this proof is in this statement, which is part of the inductive step: "the set of the
first n horses and the set of the last n horses [in the collection of n + 1 horses being considered] overlap."
The only assumption made about the number n in this argument is that n is a positive integer. When n = 1,
so that n + 1 = 2, the statement quoted is obviously nonsense: the set of the first one horse and the set of
the last one horse, in this set of two horses, are disjoint.
51. The mistake is in applying the inductive hypothesis to look at max(x - 1, y - 1), because even though x and
y are positive integers, x - 1 and y - 1 need not be (one or both could be 0). In fact, that is what happens
if we let x = 1 and y = 2 when k = 1.
53. The base case ( n
= 2) is the old "I cut and you choose'' approach. The first person cuts the cake into
two portions that she thinks are each 1/2 of the cake. The second person chooses the portion he thinks
is at least 1/2 of the cake (at least one of the pieces must satisfy that condition). For the inductive step,
157
Mathematical Induction
Section 5.1
suppose there are k + 1 people. By the inductive hypothesis, we can suppose that the first k people have
divided the cake among themselves so that each person is satisfied that he got at least a fraction 1/k of
the cake. Each of them now cuts his or her piece into k + 1 pieces of equal size. The last person gets to
choose one piece from each of the first k people's portions. After this is done, each of the first k people
is satisfied that she still has ( 1/ k) (k / ( k + 1)) = 1/ ( k + 1) of the cake. To see that the last person is
satisfied, suppose that he thought that the ith person ( 1 :S i :S k) had a portion p, of the cake, where
L~=l p, = 1. By choosing what he thinks is the largest piece from each person, he is satisfied that he has at
least
1 p,/(k+1) = (1/(k + 1))
1 p, = 1/(k + 1) of the cake.
2::7=
2::7=
55. We use the notation (i, j) to mean the square in row i and column j, where we number from the left and
from the bottom, starting at (0, 0) in the lower left-hand corner. We use induction on i + j to show that
every square can be reached by the knight. There are six base cases, for the cases when i + j :S 2. The
knight is already at (0, 0) to start, so the empty sequence of moves reaches that square. To reach (1, 0), the
knight moves successively from (0, 0) to (2, 1) to (0, 2) to (1, 0). Similarly, to reach (0, 1), the knight moves
successively from (0, 0) to (1, 2) to (2, 0) to (0, 1). Note that the knight has reached (2, 0) and (0, 2) in the
process. For the last basis step, note this path to (1, 1): (0, 0) to (1, 2) to (2, 0) to (0, 1) to (2, 2) to (0, 3)
to (1, 1). We now assume the inductive hypothesis, that the knight can reach any square (i, j) for which
i + j = k, where k is an integer greater than 1 , and we want to show how the knight can reach each square
(i, j) when i + j = k + 1. Since k + 1 2": 3, at least one of i and j is at least 2. If i 2": 2, then by the inductive
hypothesis, there is a sequence of moves ending at (i - 2, j + 1) , since i - 2 + j + 1 = i + j - 1 = k; from there
it is just one step to (i, j). Similarly, if j 2: 2, then by the inductive hypothesis, there is a sequence of moves
ending at (i + 1,j - 2), since i + 1 + j - 2 = i + j - 1 = k; from there it is again just one step to (i,j).
57. The base cases are n = 0 and n = 1, and it is a simple matter to evaluate, directly from the ''limit of difference
quotient" definition, the derivatives of x 0 = 1 and x 1 = x:
_!!__xo = lim (x
dx
h-+O
+ h)o -
d 1
l' (x
-dX=Im
X
h-+0
h
xo = lim 1 - 1 = 0 = 0.
h-+O
h
+ h)l h
Xl
= lim !!:. = 1 =
h-+O
h
1.
x-1
XO
We are told to assume that the product rule holds:
d~ (f(x) · g(x)) =
f(x) · g'(x) + g(x) · f'(x)
So we work as follows, invoking the inductive hypothesis and the base cases:
_!!__xn+l = j_(x · xn) = x · _!!__xn + xn · j_x
dx
dx
dx
dx
= x · nxn-l + xn · 1 = nxn + xn = (n + l)xn
59. We prove this by induction on k. The basis step k
= 0 is the trivial statement that
1 = 1 (mod m). Suppose
that the statement is true for k. We must show it for k + 1. So let a = b (mod m) . By the inductive
hypothesis we know that ak = bk (mod m). Then we apply Theorem 5 from Section 4.1 to conclude that
a· ak
b ·bk (mod m), which by definition says that ak+ 1
bk+ 1 (mod m), as desired.
=
=
61. Let P(n) be the proposition
[(P1-+ P2) /\ (p2-+ p3) /\ · · · /\ (Pn-1-+ Pn)]-+ [(P1 /\p2 /\ · · · APn-d-+ Pn] ·
We want to prove this proposition for all n 2: 2. The basis step, (p 1 -+ P2) -+ (p 1 -+ P2) , is clearly true (a
tautology), since every proposition implies itself. Now we assume P(n) and want to show P(n + 1), namely
[(P1 -+ P2) /\ (P2 -+ p3)/\ · · · /\ (Pn-1 -+ Pn) /\ (Pn -+ Pn+i)J -+
[(P1 /\ P2 /\ · · · /\ Pn -1 /\ Pn) -+ Pn+i] ·
Chapter 5
158
Induction and Recursion
To show this, we will assume that the hypothesis (everything in the first square brackets) is true and show
that the conclusion (the conditional statement in the second square brackets) is also true.
So we assume (P1 ---> P2) I\ (P2 ---> p3) I\··· I\ (Pn-1 ---> Pn) I\ (Pn---> Pn+1). By the associativity of A, we
can group this as ( (p1 ---> P2) I\ (P2 ---> p3) I\ · · · I\ (Pn-1 ---> Pn)) I\ (Pn ---> Pn+i). By the simplification rule,
we can conclude that the first group, (p1 ---> P2) I\ (P2 ---> p3) I\ · · · I\ (Pn-1 ---> Pn), must be true. Now the
inductive hypothesis allows us to conclude that (p 1 Ap2 A··· APn-l)---> Pn. This together with the rest of the
assumption, namely Pn ---> Pn+I, yields, by the hypothetical syllogism rule, (p1 A P2 A · · · A Pn-1) ---> Pn+I.
That is almost what we wanted to prove, but not quite. We wanted to prove that (p1A P2A· · ·l\Pn-11\Pn) --->
Pn+I · In order to prove this, let us assume its hypothesis, P1A p21\· · ·l\Pn-11\Pn. Again using the simplification
rule we obtain P1 A P2 I\··· APn-1. Now by modus ponens with the proposition (P1 I\ P2 I\··· APn-i) ---> Pn+I i
which we proved above, we obtain Pn+l · Thus we have proved (p1 A P2 A··· APn-1 Apn)---> Pn+I, as desired.
63. This exercise, as the double star indicates, is quite hard. The trick is to induct not on n itself, but rather
on log 2 n. In other words, we write n = 2k and prove the statement by induction on k. This will prove the
statement for every n that is a power of 2; a separate argument is needed to extend to the general case.
We take the basis step to be k = 1 (the case k = 0 is trivially true, as well), so that n = 2 1 = 2. In this
case the trick is to start with the true inequality ( y'a1 - yla2) 2 2 0. Expanding, we have a1 - 2fol 1,
then the inductive hypothesis tells us that we can run k - l miles, so we can run (k -1) + 2=k+1 miles.
3. a) P(8) is true, because we can form 8 cents of postage with one 3-cent stamp and one 5-cent stamp. P(9)
is true, because we can form 9 cents of postage with three 3-cent stamps. P(lO) is true, because we can form
10 cents of postage with two 5-cent stamps.
b) The inductive hypothesis is the statement that using just 3-cent and 5-cent stamps we can form j cents
postage for all j with 8 ::::; j ::::; k , where we assume that k ~ 10 .
c) In the inductive step we must show, assuming the inductive hypothesis, that we can form k
postage using just 3-cent and 5-cent stamps.
+1
cents
+ 1 cents of postage. Since k ~ 10, we know that P(k - 2) is true, that is, that we
can form k - 2 cents of postage. Put one more 3-cent stamp on the envelope, and we have formed k + 1 cents
d) We want to form k
of postage, as desired.
Chapter 5
162
Induction and Recursion
e) We have completed both the basis step and the inductive step, so by the principle of strong induction, the
statement is true for every integer n greater than or equal to 8.
5. a) We can form the following amounts of postage as indicated: 4 = 4, 8 = 4 + 4, 11 = 11, 12 = 4 + 4 + 4,
15 = 11+4, 16 = 4 + 4 + 4 + 4, 19 = 11+4 + 4, 20 = 4 + 4 + 4 + 4 + 4, 22 = 11+11, 23 = 11+4 + 4 + 4,
24 = 4 + 4 + 4 + 4 + 4 + 4' 26 = 11 + 11 + 4' 27 = 11 + 4 + 4 + 4 + 4' 28 = 4 + 4 + 4 + 4 + 4 + 4 + 4'
30 = 11+11+4 + 4, 31=11+4 + 4 + 4 + 4 + 4, 32 = 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4, 33=11+11+11. By
having considered all the combinations, we know that the gaps in this list cannot be filled. We claim that we
can form all amounts of postage greater than or equal to 30 cents using just 4-cent and 11-cent stamps.
b) Let P( n) be the statement that we can form n cents of postage using just 4-cent and 11-cent stamps. We
want to prove that P( n) is true for all n :'.'.'. 30. The basis step, n = 30, is handled above. Assume that we
can form k cents of postage (the inductive hypothesis); we will show how to form k + 1 cents of postage. If
the k cents included an 11-cent stamp, then replace it by three 4-cent stamps ( 3 · 4 = 11 + 1 ). Otherwise,
k cents was formed from just 4-cent stamps. Because k :'.'.: 30, there must be at least eight 4-cent stamps
involved. Replace eight 4-cent stamps by three 11-cent stamps, and we have formed k + 1 cents in postage
(3·11=8·4+1).
c) P( n) is the same as in part (b). To prove that P( n) is true for all n :'.'.: 30, we note for the basis step that
from part (a), P(n) is true for n = 30,31,32,33. Assume the inductive hypothesis, that P(j) is true for all
j with 30::;; j::;; k, where k is a fixed integer greater than or equal to 33. We want to show that P(k + 1) is
true. Because k - 3 :'.'.: 30, we know that P(k - 3) is true, that is, that we can form k - 3 cents of postage.
Put one more 4-cent stamp on the envelope, and we have formed k + 1 cents of postage, as desired. In this
proof our inductive hypothesis included all values between 30 and k inclusive, and that enabled us to jump
back four steps to a value for which we knew how to form the desired postage.
7. We can form the following amounts of money as indicated: 2 = 2, 4 = 2 + 2, 5 = 5, 6 = 2 + 2 + 2. By having
considered all the combinations, we know that the gaps in this list ($1 and $3) cannot be filled. We claim that
we can form all amounts of money greater than or equal to 5 dollars. Let P(n) be the statement that we can
form n dollars using just 2-dollar and 5-dollar bills. We want to prove that P( n) is true for all n :'.'.: 5. We
already observed that the basis step is true for n = 5 and 6. Assume the inductive hypothesis, that P(j) is
true for all j with 5 ::;; j ::;; k, where k is a fixed integer greater than or equal to 6. We want to show that
P( k + 1) is true. Because k - 1 :'.'.: 5, we know that P( k - 1) is true, that is, that we can form k - 1 dollars.
Add another 2-dollar bill, and we have formed k + 1 dollars, as desired.
9. Following the hint, we let P(n) be the statement that there is no positive integer b such that J2 = n/b. For
the basis step, P(l) is true because J2 > 1 :'.'.'. l/b for all positive integers b. For the inductive step, assume
that P(j) is true for all j ::;; k, where k is an arbitrary positive integer; we must prove that P(k + 1) is true.
So assume the contrary, that J2 = (k + 1)/b for some positive integer b. Squaring both sides and clearing
fractions, we have 2b 2 = (k + 1) 2 . This tells us that (k + 1) 2 is even, and so k + 1 is even as well (the square
of an odd number is odd, by Example 1 in Section 1.7). Therefore we can write k + 1 = 2t for some positive
integer t. Substituting, we have 2b 2 = 4t 2 , so b2 = 2t 2 . By the same reasoning as before, b is even, so b = 2s
for some positive integer s . Then we have J2 = (k + 1) / b = (2t) / (2s) = t / s. But t ::;; k, so this contradicts
the inductive hypothesis, and our proof of the inductive step is complete.
11. There are four base cases. If n = 1 = 4 · 0 + 1, then clearly the first player is doomed, so the second player
wins. If there are two, three, or four matches ( n = 4 · 0 + 2, n = 4 · 0 + 3, or n = 4 · 1), then the first
player can win by removing all but one match. Now assume the strong inductive hypothesis, that in games
with k or fewer matches, the first player can win if k 0, 2 or, 3 (mod 4) and the second player can win if
k
1 (mod 4). Suppose we have a game with k + 1 matches, with k :'.'.: 4. If k + 1 0 (mod 4), then the first
=
=
=
Section 5.2
163
Strong Induction and Well-Ordering
=
player can remove three matches, leaving k- 2 matches for the other player. Since k - 2 1 (mod 4), by the
inductive hypothesis, this is a game that the second player at that point (who is the first player in our game)
can win. Similarly, if k + 1 2 (mod 4), then the first player can remove one match, leaving k matches for
the other player. Since k = 1 (mod 4), by the inductive hypothesis, this is a game that the second player at
=
that point (who is the first player in our game) can win. And if k + 1 = 3 (mod 4), then the first player can
remove two matches, leaving k - 1 matches for the other player. Since k - 1 1 (mod 4), by the inductive
hypothesis, this is again a game that the second player at that point (who is the first player in our game) can
win. Finally, if k + 1 1 (mod 4) , then the first player must leave k, k - 1 , or k - 2 matches for the other
player. Since k
0 (mod 4), k - 1 3 (mod 4), and k - 2 2 (mod 4), by the inductive hypothesis, this
is a game that the first player at that point (who is the second player in our game) can win. Thus the first
player in our game is doomed, and the proof is complete.
=
=
=
=
=
13. Let P(n) be the statement that exactly n - 1 moves are required to assemble a puzzle with n pieces. Now
P(l) is trivially true. Assume that P(j) is true for all j < n, and consider a puzzle with n pieces. The final
move must be the joining of two blocks, of size k and n - k for some integer k, 1 :s; k :s; n - 1. By the
inductive hypothesis, it required k - 1 moves to construct the one block, and n - k - 1 moves to construct
the other. Therefore 1 + (k - 1) + (n - k - 1) = n - 1 moves are required in all, so P(n) is true. Notice that
for variety here we proved P( n) under the assumption that P(j) was true for j < n; so n played the role
that k + 1 plays in the statement of strong induction given in the text. It is worthwhile to understand how
all of these forms are saying the same thing and to be comfortable moving between them.
15. Let the Chomp board have n rows and n columns. We claim that the first player can win the game by making
the first move to leave just the top row and left-most column. (He does this by selecting the cookie in the
second column of the second row.) Let P(n) be the statement that if a player has presented his opponent
with a Chomp configuration consisting of just n cookies in the top row and n cookies in the left-most column
(both of these including the poisoned cookie in the upper left corner), then he can win the game. We will
prove \/nP(n) by strong induction. We know that P(l) is true, because the opponent is forced to take the
poisoned cookie at his first turn. Fix k 2 1 and assume that P(j) is true for all j :=:; k. We claim that
P(k + 1) is true. It is the opponent's turn to move. If she picks the poisoned cookie, then the game is over
and she loses. Otherwise, assume that she picks the cookie in the top row in column j, or the cookie in the
left column in row j, for some j with 2 :=:; j :=:; k + 1. The first player now picks the cookie in the left column
in row j, or the cookie in the top row in column j, respectively. This leaves the position covered by P(j - 1)
for his opponent, so by the inductive hypothesis, he can win.
17. Let P(n) be the statement that if a simple polygon with n sides is triangulated, then at least two of the
triangles in the triangulation have two sides that border the exterior of the polygon. We will prove \In 2 4 P( n).
The statement is clearly true for n = 4, because there is only one diagonal, leaving two triangles with the
desired property. Fix k 2 4 and assume that P(j) is true for all j with 4 :s; j :s; k. Consider a polygon with
k + 1 sides, and some triangulation of it. Pick one of the diagonals in this triangulation. First suppose that
this diagonal divides the polygon into one triangle and one polygon with k sides. Then the triangle has two
sides that border the exterior. Furthermore, the k-gon has, by the inductive hypothesis, two triangles that
have two sides that border the exterior of that k-gon, and only one of these triangles can fail to be a triangle
that has two sides that border the exterior of the original polygon. The only other case is that this diagonal
divides the polygon into two polygons with j sides and k + 3 - j sides for some j with 4 :s; j :s; k - 1. By
the inductive hypothesis, each of these two polygons has two triangles that have two sides that border their
exterior, and in each case only one of these triangles can fail to be a triangle that has two sides that border
the exterior of the original polygon.
164
Chapter 5
Induction and Recursion
19. Let P(n) be the statement that the area of a simple polygon with n sides and vertices all at lattice points is
given by I+ B /2 - 1, where I and B are as defined in the exercise. We will prove 'Vn;::: 3 P( n). We begin
by proving an additivity lemma. If P is a simple polygon with all vertices at the lattice points, divided into
polygons P 1 and P 2 by a diagonal, then J(P) + B(P)/2-1 = (J(P1) + B(P1)/2-1) + (J(P2) + B(P2)/2-1).
To see this, suppose there are k lattice points on the diagonal, not counting its endpoints. Then I(P) =
J(P1) + J(P2) + k and B(P) = B(P1) + B(P2) - 2k - 2; and the result follows by simple algebra. What this
says in particular is that if Pick's formula gives the correct area for P 1 and P 2 , then it must give the correct
formula for P, whose area is the sum of the areas for P1 and P2 ; and similarly if Pick's formula gives the
correct area for P and one of the P, 's, then it must give the correct formula for the other P, .
Next we prove the theorem for rectangles whose sides are parallel to the coordinate axes. Such a rectangle
necessarily has vertices at (a, b), (a, c), (d, b), and (d, c), where a, b, c, and d are integers with b < c and
a< d. Its area is clearly (c- b)(d- a). By looking at the perimeter, we see that it has B = 2(c- b + d- a),
and we see also that it has I= (c - b - l)(d - a - 1) = (c - b)(d - a) - (c - b) - (d - a)+ 1. Therefore
I+ B /2 - 1 = (c - b)( d - a) - (c - b) - (d - a) + 1 + (c - b + d - a) - 1 = (c - b)( d - a), which is the desired
area.
Next consider a right triangle whose legs are parallel to the coordinate axes. This triangle is half a
rectangle of the type just considered, for which Pick's formula holds, so by the additivity lemma, it holds for
the triangle as well. (The values of B and I are the same for each of the two triangles, so if Pick's formula
gave an answer that was either too small or too large, then it would give a correspondingly wrong answer for
the rectangle.)
For the next step, consider an arbitrary triangle with vertices at the lattice points that is not of the type
already considered. Embed it in as small a rectangle as possible. There are several possible ways this can
happen, but in any case (and adding one more edge in one case), the rectangle will have been partitioned into
the given triangle and two or three right triangles with sides parallel to the coordinate axes. See the figure for
a typical case. Again by the additivity lemma, we are guaranteed that Pick's formula gives the correct area
for the central triangle.
Note that we have now proved P(3), the basis step in our strong induction proof. For the inductive step,
given an arbitrary polygon, use Lemma 1 in the text to split it into two polygons. Then by the additivity lemma
above and the inductive hypothesis, we know that Pick's formula gives the correct area for this polygon.
Here are some good websites for more details:
planetmath.org/encyclopedia/ProofDfPicksTheorem.html
mathforum.org/library/drmath/view/65670.html
21. a) Use the left figure. Angle abp is smallest for p, but the segment bp is not an interior diagonal.
b) Use the right figure. The vertex other than b with smallest x-coordinate is d, but the segment bd is not
an interior diagonal.
Section 5.2
Strong Induction and Well-Ordering
165
c) Use the right figure. The vertex closest to b is d, but the segment bd is not an interior diagonal.
23. a) When we try to prove the inductive step and find a triangle in each subpolygon with at least two sides
bordering the exterior, it may happen in each case that the triangle we are guaranteed in fact borders the
diagonal (which is part of the boundary of that polygon). This leaves us with no triangles guaranteed to touch
the boundary of the original polygon.
b) We proved 'v'n24T(n) in Exercise 17. Since we can always find two triangles that satisfy the property,
perforce, at least one triangle does. Thus we have proved 'v'n 2 4 E( n).
25. a) The inductive step here allows us to conclude that P(3), P(5), ... are all true, but we can conclude nothing
about P(2), P(4), ....
b) We can conclude that P( n) is true for all positive integers n, using strong induction.
c) The inductive step here allows us to conclude that P(2), P(4), P(8), P(16), ... are all true, but we can
conclude nothing about P( n) when n is not a power of 2.
d) This is mathematical induction; we can conclude that P( n) is true for all positive integers n.
27. Suppose, for a proof by contradiction, that there is some positive integer n such that P(n) is not true. Let
m be the smallest positive integer greater than n for which P( m) is true; we know that such an m exists
because P( m) is true for infinitely many values of m, and therefore true for more than just 1, 2, ... , n - 1.
But we are given that P( m) ----> P( m - 1), so P( m - 1) is true. Thus m - 1 cannot be greater than n, so
m - 1 = n and P(n) is in fact true. This contradiction shows that P(n) is true for all n.
29. The error is in going from the basis step n = 0 to the next value, n = 1. We cannot write 1 as the sum of
two smaller natural numbers, so we cannot invoke the inductive hypothesis. In the notation of the "proof,"
when k = 0, we cannot write 0 + 1 = i + j where 0 :::;: i :::;: 0 and 0 :::;: j :S 0.
31. To show that strong induction is valid, let us suppose that we have a proposition 'v'nP( n) which has been
proved using it. We must show that in fact 'v'nP(n) is true (to say that a principle of proof is valid means
that it proves only true propositions). Let S be the set of counterexamples, i.e., S = { n I •P(n) }. We want
to show that S = 0. We argue by contradiction. Assume that S =I- 0. Then by the well-ordering property, S
has a smallest element. Since part of the method of strong induction is to show that P(l) is true, this smallest
counterexample must be greater than 1. Let us call it k + 1. Since k + 1 is the smallest element of S, it must
be the case that P(l) /\ P(2) /\ · · · /\ P(k) is true. But the rest of the proof using strong induction involved
showing that P(l) /\ P(2) /\ · · · /\ P(k) implied P(k + 1); therefore since the hypothesis is true, the conclusion
must be true as well, i.e., P(k
+ 1)
is true. This contradicts our assumption that k
+ 1 E S.
Therefore we
conclude that S = 0, so 'v'nP( n) is true.
33. In each case we will argue on the basis of a "smallest counterexample."
a) Suppose that there is a counterexample, that is, that there are values of n and k such that P( n, k) is not
true. Choose a counterexample with n + k as small as possible. We cannot have n = l and k = 1, because
we are given that P(l, 1) is true. Therefore either n > 1 or k > 1. In the former case, by our choice of
counterexample, we know that P(n - 1, k) is true. But the inductive step then forces P(n, k) to be true, a
contradiction. The latter case is similar. So our supposition that there is a counterexample must be wrong,
and P(n, k) is true in all cases.
b) Suppose that there is a counterexample, that is, that there are values of n and k such that P( n, k) is not
true. Choose a counterexample with n as small as possible. We cannot have n = 1, because we are given
that P(l, k) is true for all k. Therefore n
> 1. By our choice of counterexample, we know that P(n - 1, k)
166
Chapter 5
Induction and Recursion
is true. But the inductive step then forces P(n, k) to be true, a contradiction. So our supposition that there
is a counterexample must be wrong, and P( n, k) is true in all cases.
c) Suppose that there is a counterexample, that is, that there are values of n and k such that P( n, k) is not
true. Choose a counterexample with k as small as possible. We cannot have k = 1, because we are given
that P( n, 1) is true for all n. Therefore k > 1. By our choice of counterexample, we know that P( n, k - 1)
is true. But the inductive step then forces P(n, k) to be true, a contradiction. So our supposition that there
is a counterexample must be wrong, and P(n, k) is true in all cases.
35. We want to calculate the product a 1 a 2 · · · an by inserting parentheses to express the calculation as a sequence
of multiplications of two quantities. For example, we can insert parentheses into a 1a 2a 3a4a 5 to render it
(a 1 (a2a 3 ))(a4a 5 ), and then the four multiplications are a 2 · a 3 , a4 · a 5 , a 1 · (a 2 a 3 ), and finally the product of
these last two quantities. We must show that no matter how the parentheses are inserted, n -1 multiplications
will be required. If n = 1, then clearly 0 multiplications are required, so the basis step is trivial. Now assume
the strong inductive hypothesis, that for all k < n, no matter how parentheses are inserted into the product
of k numbers, k - 1 multiplications are required to compute the answer. Consider a parenthesized product
of ai through an, and look at the last multiplication. Thus we have (a 1 a 2 · · · ar) · (ar+l · · ·an), where we
do not care how the parentheses are distributed within the pairs shown here. By the inductive hypothesis, it
requires r - 1 multiplications to obtain the first product in parentheses and n - r - 1 to obtain the second
(that second product has n - r factors). Furthermore, 1 additional multiplication is needed to multiply these
two answers together. This gives a total of (r - 1) + (n - r - 1) + 1 = n - 1 multiplications for the given
problem, exactly what we needed to show.
37. Suppose that we have two such pairs, say (q, r) and (q', r'), so that a= dq + r = dq' + r', with 0 :Sr, r' < d.
We will show that the pairs are really the same, that is, that q = q' and r = r'. From dq + r = dq' + r' we
obtain d( q - q') = r' - r. Therefore d I(r' - r). But Ir' - rl < d (since both r' and r are nonnegative integers
less than d). The only multiple of d in that range is 0, so we are forced to conclude that r' = r. Then it
easily follows that q = q' as well, since q = (a - r)/d = (a - r')/d = q'.
39. This problem deals with a paradox caused by self-reference. First of all, the answer to the question is clearly
"no,'' because there are a finite number of English words, and so only a finite number of strings of fifteen words
or fewer; therefore only a finite number of positive integers can be so described, not all of them. On the other
hand, we might offer the following "proof" that every positive integer can be so expressed. Clearly 1 can be so
expressed (e.g., "one" or "the cardinality of the power set of the empty set"). By the well-ordering property,
if there is a positive integer that cannot be expressed in fifteen words or fewer, then there is a smallest such,
say s. Then the phrase "the smallest positive integer that cannot be described using no more than fifteen
English words'' is a description of s using no more than fifteen English words, a contradiction. Therefore no
such s exists, and we seem to have proved that every positive integer can be so expressed, in obvious violation
to common sense (and the argument presented above). Paradoxes like this are likely to arise whenever we try
to use language to talk about itself; the use of language in this way, while seeming to be meaningful, is in fact
nonsense.
41. We will prove this by contradiction. Suppose that the well-ordering property were false. Let S be a counterex-
ample: a nonempty set of nonnegative integers that contains no smallest element. Let P(n) be the statement
"i t/:. S for all i :S n." We will show that P( n) is true for all n (which will contradict the assertion that
S is nonempty). Now P(O) must be true, because if 0 E S then clearly S would have a smallest element,
namely 0. Suppose now that P(n) is true, so that it/:. S for i = 0, 1, ... ,n. We must show that P(n + 1)
is true, which amounts to showing that n + 1 t/:. S. If n + 1 E S, then n + 1 would be the smallest element
of S, and this would contradict our assumption. Therefore n + 1 tf:. S. Thus we have shown by the principle
Section 5.3
Recursive Definitions and Structural Induction
167
of mathematical induction that P( n) is true for all n, which means that there can be no elements of S. This
contradicts our assumption that S -1- 0, and our proof by contradiction is complete.
43. First we claim that strong induction implies the principle of mathematical induction. Suppose we have proved
P(l) and proved for all k 2 1 that P(k)-> P(k + 1) is true. This certainly implies that [P(l) /\ · · · /\ P(k)] ->
P( k + 1) is true (here we have a stronger hypothesis). Therefore by strong induction P( n) is true for all n.
By Exercise 41, the principle of mathematical induction implies the well-ordering property. Therefore by
assuming strong induction as an axiom, we can prove the well-ordering property.
SECTION 5.3
Recursive Definitions and Structural Induction
The best way to approach a recursive definition is first to compute several instances. For example, if you
are given a recursive definition of a function f, then compute f(O) through J(8) to get a feeling for what
is happening. Most of the time it is necessary to prove statements about recursively defined objects using
structural induction (or mathematical induction or strong induction), and the induction practically takes care
of itself, mimicking the recursive definition.
= 0, together with the
given fact that f(O) = 1. Then we compute f(2) by using the recursive part of the definition with n = 1,
together with the given value of f(l). We continue in this way to obtain f(3) and f(4).
1. In each case, we compute f(l) by using the recursive part of the definition with n
a) f(l)=f(0)+2=1+2=3; f(2)=f(1)+2=3+2=5; f(3)=f(2)+2=5+2=7; f(4)=f(3)+2=
7+2=9
b) f(l) = 3f(O) = 3 · 1 = 3; f(2) = 3f(l) = 3 · 3 = 9; f(3) = 3f(2) = 3 · 9 = 27; f(4) = 3f(3) = 3 · 27 = 81
c) f(l) = 2f(O) = 21 = 2; f(2) = 2J(l) = 22 = 4; f(3) = 2f( 2 ) = 24 = 16; f(4) = 2J( 3 ) = 216 = 65,536
d) f(l) = J(0) 2 + f(O)+l = 12 +1+1 = 3; J(2) = J(1) 2 + J(l)+l = 32 +3+1=13; f(3) = f(2) 2 + f(2)+1 =
13 2 + 13 + 1=183; !(4) = !(3) 2 + !(3) + 1=1832 + 183 + 1 = 33,673
3. In each case we compute the subsequent terms by plugging into the recursive formula, using the previously
given or computed values.
a) J(2) = f(l) + 3f(O) = 2 + 3(-1) = -1; f(3) = f(2) + 3f(l) = -1+3 · 2 = 5; J(4) = J(3) + 3f(2) =
5+3(-1)=2; f(5)=f(4)+3J(3)=2+3·5=17
b) f(2) = f(1) 2 f(O) = 22 · (-1) = -4; f(3) = J(2) 2 J(l) = (-4) 2 · 2 = 32; f(4) = J(3) 2 f(2) = 32 2 · (-4) =
-4096; !(5) = !(4) 2 !(3) = (-4096) 2 . 32 = 536,870,912
c) f(2) = 3f(1) 2 - 4f(0) 2 = 3 · 22 - 4 · (-1) 2 = 8; f(3) = 3f(2) 2 - 4f(1) 2 = 3 · 82 - 4 · 22 = 176;
!(4) = 3/(3) 2 -4/(2) 2 = 3·1762 -4·82 = 92,672; /(5) = 3/(4) 2 -4/(3) 2 = 3·92672 2 -4·176 2 = 25,764,174,848
d) f(2) = f(O)/J(l) = (-1)/2 = -1/2; J(3) = f(l)/J(2) = 2/(-~) = -4; f(4) = f(2)/J(3) = (-~)/(-4) =
1/8; !(5) = J(3)/J(4) = (-4)/i = -32
5. a) This is not valid, since letting n = 1 we would have f(l) = 2/(-1), but f(-1) is not defined.
b) This is valid. The basis step tells us what f(O) is, and the recursive step tells us how each subsequent value
is determined from the one before. It is not hard to look at the pattern and conjecture that f(n) = 1- n. We
prove this by induction. The basis step is f(O) = 1=1- O; and if f(k) = 1- k, then J(k + 1) = J(k) - 1 =
1 - k - 1 = 1 - (k
+ 1).
c) The basis conditions specify f (0) and f ( 1), and the recursive step gives f (n) in terms of f (n -1) for n 2: 2 ,
so this is a valid definition. If we compute the first several values, we conjecture that f(n) = 4 - n if n > 0,
but f (0) = 2. That is our "formula." To prove it correct by induction we need two basis steps: f (0) = 2,
and f(l) = 3 = 4-1. For the inductive step (with k 2: 1), J(k + 1) = f(k) -1 = (4- k) -1=4- (k + 1).
Chapter 5
168
Induction and Recursion
d) The basis conditions specify f(O) and f(l), and the recursive step gives f(n) in terms of f(n - 2) for
n ;:::: 2, so this is a valid definition. The sequence of function values is 1, 2, 2, 4, 4, 8, 8, ... , and we can
fit a formula to this if we use the floor function: f(n) = 2l(n+ll/ 2 J. For a proof, we check the base cases:
f(O) = 1 = 2l(O+l)/ 2 J and f(l) = 2 = 2l(l+ll/ 2 J. For the inductive step: f(k + 1) = 2f(k - 1) = 2 · 2lk/ 2 J =
2lk/2j+1 = 2l((k+l)+l)/2J.
e) The definition tells us explicitly what f(O) is. The recursive step specifies f(l), f(3), ... in terms of f(O),
f(2), ... ; and it also gives f(2), f(4), ... in terms of f(O), f(2), .... So the definition is valid. We compute
that f (1) = 3, f (2) = 9 , f (3) = 27, and so conjecture that f (n) = 3n . The basis step of the inductive proof
is clear. For odd n greater than 0 we have f(n) = 3f(n - 1) = 3 · 3n-l = 3n, and for even n greater than 1
we have f(n) = 9f(n - 2) = 9 · 3n- 2 = 3n. Note that we used a slightly different notation here, letting n be
the new value, rather than k + 1, but the logic is the same.
7. There are many correct answers for these sequences. We will give what we consider to be the simplest ones.
a) Clearly each term in this sequence is 6 greater than the preceding term. Thus we can define the sequence
by setting ai = 6 and declaring that an+l =an+ 6 for all n ;:::: 1.
b) This is just like part (a), in that each term is 2 more than its predecessor. Thus we have a 1 = 3 and
an+l = an + 2 for all n ;:::: 1 .
c) Each term is 10 times its predecessor. Thus we have a 1 = 10 and an+l = lOan for all n;:::: 1.
d) Just set al
=
5 and declare that an+l =an for all n ;:::: 1.
9. We need to write F(n + 1) in terms of F(n). Since F(n) is the sum of the first n positive integers (namely 1
through n ), and F(n+ 1) is the sum of the first n+ 1 positive integers (namely 1 through n+ 1 ), we can obtain
F (n + 1) from F (n) by adding n + 1. Therefore the recursive part of the definition is F (n + 1) = F (n) +n + 1.
The initial condition is a specification of the value of F(O); the sum of no positive integers is clearly 0, so we
set F(O) = 0. (Alternately, if we assume that the argument for F is intended to be strictly positive, then we
set F(l) = 1, since the sum of the first one positive integer is 1.)
11. We need to see how Pm(n + 1) relates to Pm(n). Now Pm(n + 1) = m(n + 1) = mn + m = Pm(n) + m.
Thus the recursive part of our definition is just Pm(n + 1) = Pm(n) + m. The basis step is Pm(O) = 0, since
m · 0 = 0, no matter what value m has.
13. We prove this using the principle of mathematical induction. The base case is n = 1, and in that case the
statement to be proved is just Ji = h; this is true since both values are 1. Next we assume the inductive
hypothesis, that
Ji+
h
+ · · · + hn-1
and try to prove the corresponding statement for n
Ji+
h
= hn'
+ 1 , namely,
+ · · · + hn-1 + hn+l = hn+2 ·
We have
Ji+
h
+ · · · + hn-1 + hn+l
= hn + hn+l (by the inductive hypothesis)
=
hn+2
(by the definition of the Fibonacci numbers).
15. We prove this using the principle of mathematical induction. The basis step is for n = 1, and in that case
the statement to be proved is just fofi + fih = Ji; this is true since 0 · 1+1 · 1 = 12 . Next we assume the
inductive hypothesis, that
fofi + fih + · · · + hn-lhn =fin'
Section 5.3
Recursive Definitions and Structural Induction
and try to prove the corresponding statement for n
Joli
169
+ 1 , namely,
+ lifz + · · · + fzn-ifzn + fznfzn+i + fzn+dzn+Z =
fin+Z ·
Note that two extra terms were added, since the final subscript has to be even. We have
Joli+ lifz
+ · · · + fzn-dzn + fznfzn+i + fzn+dzn+Z
=
JJn
+ fznfzn+i + fzn+dzn+Z
(by the inductive hypothesis)
= fzn(fzn
+ fzn+1) + fzn+ifzn+Z
(by factoring)
=
fznfzn+2
+ fzn+dzn+Z
(by the definition of the Fibonacci numbers)
+ fzn+1)fzn+Z
= fzn+zfzn+Z = fin+z·
= (Jzn
17. Let dn be the number of divisions used by Algorithm 1 in Section 4.3 (the Euclidean algorithm) to find
gcd(Jn+1, f n). We write the calculation in this order, since fn+i 2: fn. We begin by finding the values of dn
for the first few values of n, in order to find a pattern and make a conjecture as to what the answer is. For
n = 0 we are computing gcd(Ji, Jo) = gcd(l, 0). Without performing any divisions, we know immediately
that the answer is 1, so do= 0. For n = 1 we are computing gcd(fz,li) = gcd(l,1). One division is used
to show that gcd(l, 1) = gcd(l, 0), so di = 1. For n = 2 we are computing gcd(h, fz) = gcd(2, 1). One
division is used to show that gcd(2, 1) = gcd(l, 0), so dz = 1. For n = 3, the computation gives successively
gcd(J4 ,h) = gcd(3,2) = gcd(2,l) = gcd(l,O), for a total of 2 divisions; thus d3 = 2. For n = 4, we have
gcd(f5, f4) = gcd(5, 3) = gcd(3, 2) = gcd(2, 1) = gcd(l, 0), for a total of 3 divisions; thus d4 = 3. At this
point we see that each increase of 1 in n seems to add one more division, in order to reduce gcd(fn+l• fn)
to gcd(fn,fn-i). Perhaps, then, for n 2: 2, we have dn = n -1. Let us make that conjecture. We have
already verified the basis step when we computed that dz = 1. Now assume the inductive hypothesis, that
dn = n - 1. We must show that dn+l = n. Now dn+i is the number of divisions used in computing
gcd(fn+z, fn+i). The first step in the algorithm is to divide fn+i into fn+z. Since fn+z = f n+i + fn (this
is the key point) and fn < fn+i, we get a quotient of 1 and a remainder of fn. Thus we have, after one
division, gcd(Jn+z, fn+i) = gcd(jn+l• fn). Now by the inductive hypothesis we need exactly dn = n - 1
more divisions, since the algorithm proceeds from this point exactly as it proceeded given the inputs for the
case of n. Therefore 1 + (n - 1) = n divisions are used in all, and our proof is complete. The answer,
then, is that d 0 = 0, di = 1, and dn = n - 1 for n : :'.'. 2. (If we interpreted the problem as insisting that
we compute gcd(fn, f n+i), with that order of the arguments, then the analysis and the answer are slightly
different: d0 = 1, and dn = n for n :::'.'. 1.)
19. The determinant of the matrix A= [ ~
~] , written
IAI, is by definition ad - be; and the determinant has
the multiplicative property that IABI = IAllBI. Therefore the determinant of the matrix A=
r.
U ~]
in
On the other hand, the determinant of the matrix
Exercise 16 is 1 . 0 - 1 . 1 = -1, and IAn I = IA In = ( -1
1
[f
:: ] is by definition f n+ifn-l - f~. In Exercise 18 we showed that An is this latter matrix. The
1 1
identity in Exercise 14 follows.
'J:
21. Assume that the definitions given in Exercise 20 were as follows: the max or min of one number is itself;
max( ai, az) = a 1 if ai :::'.'. az and az if ai < az, whereas min( ai, az) = az if ai :::'.'. a2 and ai if ai < a2; and
for n :::'.'. 2,
170
Chapter 5
Induction and Recursion
and
We can then prove the three statements here by induction on n.
a) For n = 1, both sides of the equation equal -a1. For n = 2, we must show that max(-a 1, -a 2) =
- min( a 1, a 2) . There are two cases, depending on the relationship between a 1 and a 2 . If a 1 :::;; a 2 , then
-a1 2 -a2, so by our definition, max(-a1. -a2) = -a 1 . On the other hand our definition implies that
min(a1,a2) = a1 in this case. Therefore max(-a1,-a2) = -a1 = -min(a1,a2). The other case, a 1 > a 2 ,
is similar: max( -a 1, -a2) = -a2 = - min( a 1, a2). Now we are ready for the inductive step. Assume the
inductive hypothesis, that
We need to show the corresponding equality for n + 1. We have
= max(max(-a1, -a2, ... , -an), -an+1)
=max(- min(a1, a2, ... , an), -an+1)
= -min(min(a1, a2, ... , an), an+1)
= - min(a1, a2, ... , an, an+i)
(by definition)
(by the inductive hypothesis)
(by the already proved case n = 2)
(by definition).
b) For n = 1, the equation is simply the identity a 1 + b1 = a 1 + b1 . For n = 2, the situation is a little
messy. Let us consider first the case that a1 + b1 2 a2 + b2. Then max(a 1 + b1, a2 + b2) = a 1 + b1 . Also note
that ai :::;; max(a 1, b1), and b1 :::;; max(b1, b2), so that a 1 +bi :::;; max(a1, a 2) + max(b 1, b2). Therefore we have
max(a1 +bi, a2 + b2) = ai + b1 :::;; max(a1, a2) + max(b1, b2). The other case, in which ai +bi < a2 + b2, is
similar. Now for the inductive step, we first need a lemma: if u :::;; v, then max( u, w) :::;; max( v, w); this is
easy to prove by looking at the three cases determined by the size of w relative to the sizes of u and v. Now
assuming the inductive hypothesis, we have
max(a1 +bi, a2 + b2, ... , an+ bn, an+l + bn+1)
= max(max(a1 + b1, a2 + b2, ... , an+ bn), an+I + bn+1)
(by definition)
:::;; max(max(a1, a2, ... , an)+ max(b1, b2, ... , bn), an+I + bn+1)
(by the inductive hypothesis and the lemma)
:::;; max(max(a1, a2, ... , an), an+1) + max(max(b1, b2, ... , bn), bn+i)
(by the already proved case n = 2)
= max(a1, a2, ... , an, an+1)
+ max(b1, b2, ... , bn, bn+1) (by definition).
c) The proof here is exactly dual to the proof in part (b). We replace every occurrence of "max" by "min,''
and invert each inequality. The proof then reads as follows. For n = 1, the equation is simply the identity
a1 + b1 = a1 + b1 . For n = 2, the situation is a little messy. Let us consider first the case that a 1 + b1 :::;; a 2 + b2.
Then min(a1 + b1,a2 + b2) = a1 + b1. Also note that a1 2 min(a1,a 2), and b1 2 min(bi,b2), so that
a1 +b1 2 min(a1, a2)+min(b1, b2). Therefore we have min(a1 +b1, a2+b2) = a1 +b1 2 min(a1, a 2)+min(b1, b2).
The other case, in which a 1 + b1 > a 2 + b2 , is similar. Now for the inductive step, we first need a lemma: if
u 2 v, then min( u, w) 2 min( v, w); this is easy to prove by looking at the three cases determined by the size
Section 5.3
Recursive Definitions and Structural Induction
171
of w relative to the sizes of u and v. Now assuming the inductive hypothesis, we have
min(a1
+ b1, a2 + b2, ... , a + bn, an+l + bn+1)
= min(min(a1 + b1, a2 + b2, ... , a,,+ bn), an+l + bn+d (by definition)
2 min(min(a1, a2, ... , an)+ min(b1, b2 .... , b an+l + bn+I)
11
11 ),
(by the inductive hypothesis and the lemma)
2'. min(min(a1, a2, ... , an), an+d + min(min(b1, b2, ... , bn), bn+I)
(by the already proved case n = 2)
= min(a1, a2, ... , an, an+1) + min(b1, b2, ... , bn, bn+I)
(by definition).
23. We can define the set S = { x I x is a positive integer and x is a multiple of 5} by the basis step requirement
that 5 E S and the recursive requirement that if n E S, then n + 5 E S. Alternately we can mimic Example 5,
making the recursive part of the definition that x + y E S whenever :r and y are in S.
25. a) Since we can generate all the even integers by starting with 0 and repeatedly adding or subtracting 2, a
simple recursive way to define this set is as follows: 0 E S; and if x E S then x + 2 E S and x - 2 E S.
b) The smallest positive integer congruent to 2 modulo 3 is 2, so we declare 2
obtained by adding multiples of 3, so our inductive step is that if x E S, then x
E S. All the others can be
+3E
S.
c) The positive integers not divisible by 5 are the ones congruent to 1. 2, 3, or 4 modulo 5. Therefore we
can proceed just as in part (b), setting 1 E S, 2 E S, 3 E S, and 4 E S as the base cases, and then declaring
that if x E S, then x + 5 E S.
27. a) If we apply each of the recursive step rules to the only element given in the basis step, we see that (0, 1),
(1, 1), and (2, 1) are all in S. If we apply the recursive step to these we add (0, 2), (1, 2), (2, 2), (3, 2), and
(4, 2). The next round gives us (0, 3), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), and (6, 3). And a fourth set of
applications adds (0,4), (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (7,4), and (8,4).
b) Let P( n) be the statement that a ::; 2b whenever (a, b)
E S is obtained by n applications of the recursive
step. For the basis step, P(O) is true, since the only element of S obtained with no applications of the
recursive step is (0, 0), and indeed 0 ::; 2 · 0. Assume the strong inductive hypothesis that a ::; 2b whenever
(a, b) ES is obtained by k or fewer applications of the recursive step, and consider an element obtained with
k + 1 applications of the recursive step. Since the final application of the recursive step to an element (a, b)
must be applied to an element obtained with fewer applications of the recursive step, we know that a ::; 2b.
So we just need to check that this inequality implies a ::; 2(b + 1). a+ 1 ::; 2(b + 1), and a+ 2 ::; 2(b + 1). But
this is clear, since we just add 0 ::; 2. 1 ::; 2, and 2 ::; 2, respectively, to a ::; 2b to obtain these inequalities.
c) This holds for the basis step, since 0 ::::'. 0. If this holds for (a, b), then it also holds for the elements
obtained from (a, b) in the recursive step, since adding 0 ::; 2, 1 ::; 2, and 2 ::; 2, respectively, to a ::; 2b yields
a ::; 2(b + 1), a+ 1 ::; 2(b + 1), and a+ 2 ::; 2(b + 1).
29. a) Since we are working with positive integers, the smallest pair in which the sum of the coordinates is even
is (1, 1). So our basis step is (1, 1) E S. If we start with a point for which the sum of the coordinates is
even and want to maintain this parity, then we can add 2 to the first coordinate, or add 2 to the second
coordinate, or add 1 to each coordinate. Thus our recursive step is that if (a, b) E S, then (a + 2, b) E S,
(a, b + 2) E S, and (a+ 1, b + 1) E S. To prove that our definition works, we note first that (1, 1) has an
even sum of coordinates, and if (a, b) has an even sum of coordinates, then so do (a+ 2, b), (a, b + 2), and
(a + 1, b + 1) , since we added 2 to the sum of the coordinates in each case. Conversely, we must show that
if a + b is even, then (a, b) E S by our definition. We do this by induction on the sum of the coordinates. If
the sum is 2, then (a,b) = (1,1), and the basis step put (a,b) into S. Otherwise the sum is at least 4, and
172
Chapter 5
Induction and Recursion
at least one of (a - 2, b), (a, b - 2), and (a -1, b -1) must have positive integer coordinates whose sum is an
even number smaller than a + b, and therefore must be in S by our definition. Then one application of the
recursive step shows that (a, b) E S by our definition.
b) Since we are working with positive integers, the smallest pairs in which there is an odd coordinate are
(1, 1), (1, 2), and (2, 1). So our basis step is that these three points are in S. If we start with a point for
which a coordinate is odd and want to maintain this parity, then we can add 2 to that coordinate. Thus our
recursive step is that if (a, b) ES, then (a+ 2, b) ES and (a, b + 2) ES. To prove that our definition works,
we note first that (1, 1), (1, 2), and (2, 1) all have an odd coordinate, and if (a, b) has an odd coordinate,
then so do (a+ 2, b) and (a, b + 2), since adding 2 does not change the parity. Conversely (and this is the
harder part), we must show that if (a, b) has at least one odd coordinate, then (a, b) E S by our definition.
We do this by induction on the sum of the coordinates. If (a, b) = (1, 1) or (a, b) = (1, 2) or (a, b) = (2, 1),
then the basis step put (a, b) into S. Otherwise either a or b is at least 3, so at least one of (a - 2, b) and
(a, b - 2) must have positive integer coordinates whose sum is smaller than a+ b, and therefore must be in
S by our definition, since we haven't changed the parities. Then one application of the recursive step shows
that (a, b) E S by our definition.
c) We use two basis steps here, (1,6) ES and (2,3) ES. Ifwe want to maintain the parity of a+b and the
fact that b is a multiple of 3, then we can add 2 to a (leaving b alone), or we can add 6 to b (leaving a
alone). So our recursive step is that if (a, b) E S, then (a + 2, b) E S and (a, b + 6) E S. To prove that our
definition works, we note first that (1, 6) and (2, 3) satisfy the condition, and if (a, b) satisfies the condition,
then so do (a+ 2, b) and (a, b + 6), since adding 2 or 6 does not change the parity of the sum, and adding
6 maintains divisibility by 3. Conversely (and this is the harder part), we must show that if (a, b) satisfies
the condition, then (a, b) E S by our definition. We do this by induction on the sum of the coordinates. The
smallest sums of coordinates satisfying the condition are 5 and 7, and the only points are (1, 6), which the
basis step put into S, (2, 3), which the basis step put into S, and (4, 3) = (2 + 2, 3), which is in S by one
application of our recursive definition. For a sum greater than 7, either a ~ 3, or a ~ 2 and b ~ 9 (since
2 + 6 is not odd). This implies that either (a - 2, b) or (a, b - 6) must have positive integer coordinates whose
sum is smaller than a + b and satisfy the condition for being in S, and hence are in S by our definition. Then
one application of the recursive step shows that (a, b) E S by our definition.
31. The answer depends on whether we require fully parenthesized expressions. Assuming that we do not, then the
following definition is the most straightforward. Let F be the required collection of formulae. The basis step
is that all specific sets and all variables representing sets are to be in F. The recursive part of the definition
is that if a and j3 are in F, then so are a, (a) , a U (3, a n (3, and a - (3. If we insist on parentheses, then
the recursive part of the definition is that if a and (3 are in F, then so are a, (a U (3), (an (3) , and (a - (3).
33. Let D = {O, 1, 2, 3, 4, 5, 6, 7, 8, 9} be the set of decimal digits. We think of a string as either being an element
of D or else coming from a shorter string by appending an element of D, as in Definition 1. This problem is
somewhat like Example 7.
a) The basis step is for a string of length 1, i.e., an element of D. If x E D, then m(x) = x. For the recursive
step, if the string s = tx, where t E D* and x E D, then m( s) = min( m( s), x). In other words, if the last
digit in the string is smaller than the minimum digit in the rest of the string, then the last digit is the smallest
digit in the string; otherwise the smallest digit in the rest of the string is the smallest digit in the string.
b) Recall the definition of concatenation (Definition 2). The basis step does not apply, since s and t here
must be nonempty. Lett= wx, where w ED* and x ED. If w =>.,then m(st) = m(sx) = min(m(s),x) =
min(m(s), m(x)) by the recursive step and the basis step of the definition of min part (a). Otherwise, m(st) =
m((sw)x) = min(m(sw),x) by the definition of min part (a). But m(sw) = min(m(s),m(w)) by the inductive hypothesis of our structural induction, so m(st) = min(min(m(s), m(w)), x) = min(m(s), min(m(w), x))
Section 5.3
Recursive Defi.nitions and Structural Induction
173
by the meaning of min. But min( m( w), x) = m( wx) = m( t) by the recursive step of the definition of m in
part (a). Thus m(st) = min(m(s), m(t)).
35. The string of length 0, namely the empty string, is its own reversal, so we define ).. R = ).. . A string w of
length n + 1 can always be written as vy, where v is a string of length n (the first n symbols of w ), and y
is a symbol (the last symbol of w). To reverse w, we need to start with y, and then follow it by the first part
of w (namely v ), reversed. Thus we define wR = y(vR). (Note that the parentheses are for our benefit~they
are not part of the string.)
37. We set w 0 =)..(the concatenation of no copies of w should be defined to be the empty string). For i;::: 0, we
define wi+l = ww", where this notation means that we first write down w and then follow it with w".
39. The recursive part of this definition tells us that the only way to modify a string in A to obtain another string
in A is to tack a 0 onto the front and a 1 onto the end. Starting with the empty string, then, the only strings
we get are>-., 01, 0011, 000111, .... In other words, A= {on1n In 2 0}.
41. The basis step is i = 0, where we need to show that the length of w 0 is 0 times the length of w. This is true,
no matter what w is, since l(w 0 ) = l(>-.) = 0. Assume the inductive hypothesis that l(w') = i · l(w). Then
l( w"+l) = l( wwi) = l( w) + l( w"), this latter equality having been shown in Example 12. Now by the inductive
hypothesis we have l(w) + l(w") = l(w) + i · l(w) = (i + 1) · l(w), as desired.
43. This is similar to Theorem 2. For the full binary tree consisting of just the root r the result is true since
n(T) = 1 and h(T) = 0, and 1 2 2 · 0 + 1. For the inductive hypothesis we assume that n(T1) 2 2h(T1) + 1
and n(T2) 2: 2h(T2) + 1 where T1 and T2 are full binary trees. By the recursive definitions of n(T) and h(T),
we have n(T) = 1 + n(T1) + n(T2) and h(T) = 1 +max( h(T1), h(T2)). Therefore n(T) = 1 + n(T1) + n(T2) ;:::
1+2h(T1)+1+2h(T2) + 1 2 1 + 2 · max(h(T1), h(T2)) + 2 since the sum of two nonnegative numbers is at
least as large as the larger of the two. But this equals 1 + 2(max(h(T1), h(T2 )) + 1) = 1 + 2h(T), and our
proof is complete.
45. The basis step requires that we show that this formula holds when (m, n) = (0, 0). The inductive step requires
that we show that if the formula holds for all pairs smaller than (m, n) in the lexicographic ordering of N x N,
then it also holds for (m, n) . For the basis step we have a0 ,0 = 0 = 0 + 0. For the inductive step, assume
that am' ,n' = m' + n' whenever (m', n') is less than (m, n) in the lexicographic ordering of N x N. By the
recursive definition, if n = 0 then am,n = am-l,n + 1; since (m - 1, n) is smaller than (m, n), the inductive
hypothesis tells us that am-l,n = m - 1 + n, so am,n = m - 1+n+1 = m + n, as desired. Now suppose that
n > 0, so that am,n = am,n-l + 1. Again we have am,n-l = m + n - 1, so am,n = m + n - 1+1 = m + n,
and the proof is complete.
47. a) It is clear that Pm,m =Pm, since a number exceeding m can never be used in a partition of m.
b) We need to verify all five lines of this definition, show that the recursive references are to a smaller value of
m or n, and check that they take care of all the cases and are mutually compatible. Let us do the last of these
first. The first two lines take care of the case in which either m or n is equal to 1. They are consistent with
each other in case m = n = 1. The last three lines are mutually exclusive and take care of all the possibilities
for m and n if neither is equal to 1, since, given any two numbers, either they are equal or one is greater
than the other. Note finally that the third line allows m = 1; in that case the value is defined to be P 1,1 ,
which is consistent with line one, since P1 ,n = 1.
Chapter 5
174
Induction and Recursion
Next let us make sure that the logic of the definition is sound, specifically that Pm,n is being defined in
terms of P,,1 for i :::; rn and j :::; n, with at least one of the inequalities strict. There is no problem with the
first two lines, since these are not recursive. The third line is okay, since rn < n, and Pm,n is being defined
in terms of Pm,m. The fourth line is also okay, since here Pm,m is being defined in terms of Pm,m-l. Finally,
the last line is okay, since the subscripts satisfy the desired inequalities.
Finally, we need to check the content of each line. (Note that so far we have hardly even discussed what
Pm,n means!) The first line says that there is only one way to write the number 1 as the sum of positive
integers, none of which exceeds n, and that is patently true, namely as 1 = 1. The second line says that there
is only one way to write the number rn as the sum of positive integers, none of which exceeds 1, and that, too,
is obvious, namely rn = 1 + 1 + · · · + 1. The third line says that the number of ways to write rn as the sum of
integers not exceeding n is the same as the number of ways to write rn as the sum of integers not exceeding
rn as long as rn < n. This again is true, since we could never use a number from {rn + 1, rn + 2, ... , n} in
such a sum anyway. Now we begin to get to the meat. The fourth line says that the number of ways to write
rn as the sum of positive integers not exceeding rn is 1 plus the number of ways to write rn as the sum of
positive integers not exceeding m - 1. Indeed, there is exactly one way to write m as the sum of positive
integers not exceeding m that actually uses m, namely rn = m; all the rest use only numbers less than or
equal to m - 1. This verifies line four. The real heart of the matter is line five. How can we write rn as
the sum of positive integers not exceeding n? We may use an n, or we may not. There are exactly Pm,n-1
ways to form the sum without using n, since in that case each summand is less than or equal to n - 1 . If we
do use at least one n, then we have m = n + (m - n) . The number of ways this can be done, then, is the
same as the number of ways to complete the partition by writing (m - n) as the sum of positive integers not
exceeding n. Thus there are Pm-n,n ways to write rn as the sum of numbers not exceeding n, at least one
of which equals n. By the sum rule (see Chapter 6), we have Pm,n = Pm,n-1 + Pm-n,n, as desired.
c) We expand each Pm,n according to the definition. For the first problem we have (deleting the commas in
the subscripts for readability) Ps = Pss = l+Ps4 = l+P.s3+P14 = l+Ps2+P23+l = l+Ps1 +P32+P22+l =
l+l+P31 +P12 + l+P21 +1=1+1+ 1+1+ 1+ 1+1 = 7. For the second problem we have P5 = P6 5 = l+P5 5 =
l+P54+P1s = l+P53+P24+l = l+P52+P33+P22+l = l+P51 +P42+l+P32+l+P21 +1 = l+l+P41 +
P 22 +1+P31 +P12 +1+1+1 = l+l+l+l+P21 +l+l+l+l+l+l=1+1+1+1+1+1+1+1+1+1+1=11.
49. We prove this by induction on m. The basis step is m = 1, so we need to compute A(l, 2). Line four of the
definition tells us that A(l, 2) = A(O, A(l, 1)). Since A(l, 1) = 2, by line three, we see that A(l, 2) = A(O, 2).
Now line one of the definition applies, and we see that A(l, 2) = A(O, 2) = 2 · 2 = 4, as desired. For the
inductive step, assume that A(m - 1, 2) = 4, and consider A(rn, 2). Applying first line four of the definition,
then line three, and then the inductive hypothesis, we have A(m, 2) = A(m - 1, A(m, 1)) = A(m -1, 2) = 4.
51. a) We use the results of Exercises 49 and 50: A(2, 3) = A(l, A(2, 2)) = A(l, 4) = 24 = 16.
b) We have A(3,3) = A(2,A(3,2)) = A(2,4) by Exercise 49. Now one can show by induction (using the
2
result of Exercise 50) that A(2, n) is equal to 22
22
22 = 216 = 65,536.
, with n 2's in the tower.
Therefore the answer is
53. It is often the case in proofs by induction that you need to prove something stronger than the given proposition,
in order to have a stronger inductive hypothesis to work with. This is called inductive loading (see the
preamble to Exercise 74 in Section 5.1). That is the case with our proof here. We will prove the statement
"A(rn,k) > A(rn,l) if k > l for all m, k, and l," and we will use double induction, inducting first on m,
and then within the inductive step for that induction, inducting on k (using strong induction). Note that this
stronger statement implies the statement we are trying to prove-just take k = l + 1 .
Section 5.3
Recursive Definitions and Structural Induction
175
The basis step is m = 0, in which the statement at hand reduces (by line one of the definition) to the true
conditional statement that if k > l, then 2k > 2l. Next we assume the inductive hypothesis on m, namely
that A( m, x) > A( m, y) for all values of x and y with x > y. We want now to show that if k > l, then
A( m + 1, k) > A( m + 1, l). This we will do by induction on k. For the basis step, k = 0, there is nothing to
prove, since the condition k > l is vacuous. Similarly, if k = 1, then A(m + 1, k) = 2 and A(m + 1, l) = 0
(since necessarily l = 0 ), so the desired inequality holds. So assume the inductive hypothesis (using strong
induction), that A(m + l,r) > A(m + l,s) whenever k > r > s, where k;:::: 2. We need to show that
A(m + 1, k) > A(m + 1, l) if k > l. Now A(m + 1, k) = A(m, A(m + 1, k - 1)) by line four of the definition.
Since k - 1 ;:::: l, we apply the inductive hypothesis on k to yield A(m + 1, k - 1) > A(m + 1, l - 1), and
therefore by the inductive hypothesis on m, we have A( m, A( m + 1, k - 1)) > A( m, A( m + 1, l - 1)). But this
latter value equals A(m + 1, l), as long as l;:::: 2. Thus we have shown that A(m + 1, k) > A(m + 1, l) as long
as l 2: 2. On the other hand, if l = 0 or 1, then A(m + 1, l) :::; 2 (by lines two and three of the definition),
whereas A(m + 1, 2) = 4 by Exercise 49. Therefore A(m + 1, k) ;:::: A(m + 1, 2) > A(m + 1, l). This completes
the proof.
55. We repeatedly invoke the result of Exercise 54, which says that A( m + 1, j) > A( m, j). Indeed, we have
A(i,j);:::: A(i-1,j) 2: · · ·;:::: A(O,j) = 2j 2: j.
57. Let P(n) be the statement "F is well-defined at n; i.e., F(n) is a well-defined number." We need to show
that P(n) is true for all n. We do this by strong induction. First P(O) is true, since F(O) is well-defined
by the specification of F(O). Next assume that P(k) is true for all k < n. We want to show that P(n) is
also true, in other words that F(n) is well-defined. Since the definition gave F(n) in terms of F(O) through
F(n - 1), and since we are assuming that these are all well-defined (our inductive hypothesis), we conclude
that F(n) is well-defined, as desired.
59. a) This would be a proper definition if the recursive part were stated to hold for n ;:::: 2. As it stands, however,
F(l) is ambiguous.
b) This definition makes no sense as it stands; F(2) is not defined, since F(O) isn't.
c) For n = 4, the recursive part makes no sense, since we would have to know F( 4/3). Also, F(3) is
ambiguous.
d) The definition is ambiguous about n = 1, since both the second clause and the third clause seem to apply.
If the second clause is restricted to odd n 2: 3, then the sequence is well-defined and begins 1 , 2, 2, 3, 3, 3,
4, 4, 5, 4.
e) We note that F(l) is defined explicitly, but we run into problems trying to compute F(2):
F(2) = 1+F(F(l))=1 + F(2).
This not only leaves us begging the question as to what F(2) is, but is a contradiction, since
0-/= 1.
61. In each case we will apply the definition to compute log(o) , then log(l) , then logC 2 l, then logC 3 l and so on.
As soon as we get an answer no larger than 1 we stop; the last "exponent" is the answer. In other words
log* n is the number of times we need to apply the log function until we get a value less than or equal
to 1. Note that log(l) n = log n for n > 0. Similarly, log( 2 ) n = log(log n) as long as it is defined ( n > 1),
logC 3 l n = log(log(log n)) as long as it is defined ( n > 2), and so on. Normally the parentheses are understood
and omitted.
a) log(O) 2 = 2, log(l) 2 =log 2 = 1; therefore log* 2 = 1, the last "exponent."
b) log(o) 4 = 4, log(l) 4 = log 4 = 2, logC 2 l 4 = log 2 = 1; therefore log* 4 = 2, the last "exponent." We had
to take the log twice to get from 4 down to 1.
Chapter 5
176
Induction and Recursion
c) log(Ol 8 = 8, log( 1 ) 8 = log 8 = 3, log( 2 ) 8 = log 3 ~ 1.585, log( 3 ) 8 ~ log 1.585 ~ 0.664; therefore log* 8 = 3,
the last "exponent." We had to take the log three times to get from 8 down to something no bigger than 1.
d) log(o) 16 = 16, log(l) 16 = log 16 = 4, log( 2 ) 16 = log 4 = 2, log< 3 l 16 = log 2 = 1; therefore log* 16 = 3,
the last "exponent." We had to take the log three times to get from 16 down to 1.
e) log< 0 l 256 = 256, log< 1 l 256 = log 256 = 8; by part ( c), we need to take the log three more times in order
to get from 8 down to something no bigger than 1, so we have to take the log four times in all to get from
256 down to something no bigger than 1. Thus log* 256 = 4.
f) log 65536 = 16; by part ( d), we need to take the log three more times in order to get from 16 down to 1,
so we have to take the log four times in all to get from 65536 down to 1. Thus log* 65536 = 4.
g) log 22048 = 2048; taking log four more times gives us, successively, 11, approximately 3.46, approximately
1. 79, approximately 0.84. So log* 22048 = 5.
63. Each application of the function f subtracts another a from the argument. Therefore iterating this function
k times (which is what J(k) does) has the effect of subtracting ka. Therefore J(k)(n) = n - ka. Now f 0(n)
is the smallest k such that Jlkl(n) :::::; 0, i.e., n - ka:::::; 0. Solving this for k easily yields k?:: n/a. Thus
f 0(n) = In/ al (we need to take the ceiling function because k must be an integer).
65. Each application of the function f takes the square root of its argument. Therefore iterating this function
k times (which is what J(k) does) has the effect of taking the (2k)th root. Therefore J(kl(n) = n 1 12 k. Now
f2(n) is the smallest k such that J(k)(n):::::; 2, that is, n 112 •:::::; 2. Solving this for n easily yields n :S 22 k, so
k ?:: log log n, where logarithm is taken to the base 2. Thus f2 (n) = f1og log nl for n ?:: 2 (we need to take
the ceiling function because k must be an integer) and f2(1) = 0.
SECTION 5.4
Recursive Algorithms
Recursive algorithms are important theoretical entities, but they often cause a lot of grief on first encounter.
Sometimes it is helpful to "play computer" very carefully to see how a recursive algorithm works. Ironically,
however. it is good to avoid doing that after you get the idea. Instead, convince yourself that if the recursive
algorithm handles the base case correctly, and handles the recursive step correctly (gives the correct answer
assuming that the correct answer was obtained on the recursive call), then the algorithm works. Let the
computer worry about actually "recursing all the way down to the base case"!
1. First, we use the recursive step to write 5! = 5 · 4!. We then use the recursive step repeatedly to write
4! = 4 · 3!, 3! = 3 · 2!, 2! = 2 · l!, and 1! = 1 · O!. Inserting the value of O! = 1, and working back through the
steps, we see that 1! = 1·1=1, 2! = 2·1! = 2·1=2, 3! = 3 · 2! = 3 · 2 = 6, 4! = 4 · 3! = 4 · 6 = 24, and
5! = 5 . 4! = 5 . 24 = 120.
3. With this input, the algorithm uses the else clause to find that gcd(8, 13) = gcd(l3 mod 8, 8) = gcd(5, 8).
It uses this clause again to find that gcd(5, 8) = gcd(8 mod 5, 5) = gcd(3, 5), then to get gcd(3, 5) =
gcd(5 mod 3, 3) = gcd(2, 3), then gcd(2, 3) = gcd(3 mod 2, 2) = gcd(l, 2), and once more to get gcd(l, 2) =
gcd(2 mod 1, 1) = gcd(O, 1). Finally, to find gcd(O, 1) it uses the first step with a= 0 to find that gcd(O, 1) =
1. Consequently, the algorithm finds that gcd(8, 13) = 1.
Section 5.4
177
Recursive Algorithms
5. First, because n = 11 is odd, we use the else clause to see that
mpower(3, 11, 5) = (mpower(3, 5, 5) 2 mod 5 · 3 mod 5) mod 5.
We next use the else clause again to see that
mpower(3, 5, 5) = (mpower(3, 2, 5) 2 mod 5 · 3 mod 5) mod 5.
Then we use the else if clause to see that
mpower(3, 2, 5)
=
mpower(3, 1, 5) 2 mod 5.
Using the else clause again, we have
mpower(3, 1, 5)
= (mpower(3, 0, 5) 2 mod
5 · 3 mod 5) mod 5.
Finally, using the if clause, we see that mpower(3, 0, 5) = 1. Now we work backward: mpower(3, 1, 5) =
(1 2 mod 5 · 3 mod 5) mod 5 = 3, mpower(3, 2, 5) = 32 mod 5 = 4, mpower(3, 5, 5) = (4 2 mod 5 ·
3 mod 5) mod 5 = 3, and finally that mpower(3, 11, 5) = (3 2 mod 5 · 3 mod 5) mod 5 = 2. We conclude
that 3 11 mod 5 = 2.
7. The key idea for this recursive procedure is that nx = ( n - 1 )x + x. Thus we compute nx by calling the
procedure recursively, with n replaced by n - 1 , and adding x. The base case is 1 · x = x.
procedure product(n: positive integer, x: integer)
if n = 1 then return x
else return product(n - 1, x) + x
9. If we have already found the sum of the first n - 1 odd positive integers, then we can find the sum of the first
n positive integers simply by adding on the value of the nth odd positive integer. We need to realize that the
nth odd positive integer is 2n - 1, and we need to note that the base case (in which n = 1) gives us a sum
of 1 . The algorithm is then straightforward.
procedure sum of odds(n: positive integer)
if n = 1 then return 1
else return sum of odds(n - 1) + 2n - 1
11. We recurse on the size of the list. If there is only one element, then it is the smallest. Otherwise, we find
the smallest element in the list consisting of all but the last element of our original list, and compare it with
the last element of the original list. Whichever is smaller is the answer. (We assume that there is already a
function min, defined for two arguments, which returns the smaller.)
procedure smallest(a 1 , a 2 , ... , an : integers)
if n = 1 then return 1
else return min(smallest(a1, a2, ... , an-1), an)
13. We basically just take the recursive algorithm for finding n! and apply the mod operation at each step. Note
that this enables us to calculate n! mod m without using excessively large numbers, even if n! is very large.
procedure modfactorial(n, m: positive integers)
if n = 1 then return 1
else return (n · (modfactorial(n -1,m)) mod m
15. We need to worry about which of our arguments is the larger. Since we are
to make sure always to call the algorithm with the first argument less than
two stopping conditions: when a = 0 (in which case the answer is b), and
become equal (in which case the answer is their common value). Otherwise
that gcd(a, b)
given a < b as input, we need
the second. There need to be
when the two arguments have
we use the recursive condition
= gcd(a, b - a), taking care to reverse the arguments if necessary.
178
Chapter 5
Induction and Recursion
procedure gcd(a, b: nonnegative integers with a< b)
if a = 0 then return b
else if a = b - a then return a
else if a< b - a then return gcd(a, b - a)
else return gcd(b - a, a)
17. We build the recursive steps into the algorithm.
procedure multiply(x, y: nonnegative integers)
if y = 0 then return 0
else if y is even then return 2 · multiply(x, y/2)
else return 2 · multiply(x, (y - 1)/2) + x
19. We use strong induction on a, starting at a = 0. If a = 0, we know that gcd(O, b) = b for all b > 0, so
the if clause handles this basis case correctly. Now fix k > 0 and assume the inductive hypothesis-that the
algorithm works correctly for all values of its first argument less than k. Consider what happens with input
(k, b), where k < b. Since k > 0, the else clause is executed, and the answer is whatever the algorithm gives
as output for inputs (b mod k, k). Note that b mod k < k, so the input pair is valid. By our inductive
hypothesis, this output is in fact gcd(b mod k, k). By Lemma 1 in Section 4.3 (when the Euclidean algorithm
was introduced), we know that gcd(k, b) = gcd(b mod k, k), and our proof is complete.
21. For the basis step, if n = 1, then nx = x, and the algorithm correctly returns x. For the inductive step,
assume that the algorithm correctly computes kx, and consider what it does to compute (k + 1)x. The
recursive clause applies, and it recursively computes the product of k + 1 - 1 = k and x, and then adds x.
By the inductive hypothesis, it computes that product correctly, so the answer returned is kx + x = ( k + 1)x,
which is the correct answer.
23. As usual with recursive algorithms, the algorithm practically writes itself.
procedure square(n: nonnegative integer)
if n = 0 then return 0
else return square(n - 1) + 2(n - 1) + 1
The proof of correctness, by mathematical induction, practically writes itself as well. Let P( n) be the
statement that this algorithm correctly computes n 2 . Since 02 = 0, the algorithm works correctly (using the
if clause) if the input is 0. Assume that the algorithm works correctly for input k. Then for input k + 1 it
gives as output (because of the else clause) its output when the input is k, plus 2(k + 1 - 1) + 1. By the
inductive hypothesis, its output at k is k 2 , so its output at k+ 1 is k 2 +2(k+1-1) + 1 = k 2 +2k+1 = (k+ 1) 2 ,
exactly what it should be.
25. Algorithm 2 uses 2n multiplications by a, one for each factor of a in the product a 2 n The algorithm in
Exercise 24, based on squaring, uses only n multiplications (each of which is a multiplication of a number
24
by itself). For instance, to compute a = a 16 , this algorithm will compute a· a= a 2 (one multiplication),
4
2
2
then a · a = a (a second multiplication), then a 4 · a4 = as (a third), and finally as· as = a 16 (a fourth
multiplication).
27. Algorithm 2 uses n multiplications by a, one for each factor of a in the product an. The algorithm in
Exercise 26 will use O(log n) multiplications as it computes squares. Furthermore, in addition to squaring,
sometimes a multiplication by a is needed; this will add at most another O(log n) multiplications. Thus a
total of O(log n) multiplications are used altogether.
29. This is very similar to the recursive procedure for computing the Fibonacci numbers. Note that we can
combine the two base cases (stopping rules) into one.
Section 5.4
Recursive Algorithms
179
procedure sequence (n : nonnegative integer)
if n < 2 then return n + 1
else return sequence(n - 1) · sequence(n - 2)
31. The iterative version is much more efficient. The analysis is exactly the same as that for the Fibonacci
sequence given in this section. Indeed, the nth term in this sequence is actually just 2fn, as is easily shown
by induction.
33. This is essentially just Algorithm 8, with a different operation and with different (and more) initial conditions.
procedure iterative(n: nonnegative integer)
if n = 0 then return 1
else if n = 1 then return 2
else
x := 1
y := 2
:= 3
for i := 1 to n - 2
w :=x+y+z
x := y
y := z
z
z :=w
return z
35. These algorithms are very similar to the procedures for computing the Fibonacci numbers. Note that for the
recursive version, we can combine the three base cases (stopping rules) into one.
procedure recursive(n: nonnegative integer)
< 3 then return 2n + 1
else return recursive(n -1) · (recursive(n - 2)) 2 • (recursive(n - 3)) 3
if n
procedure iterative(n: nonnegative integer)
= 0 then return 1
else if n = 1 then return 3
else
x := 1
y := 3
if n
z := 5
for i := 1 to n - 2
w := z · y 2 · x 3
x := y
y := z
z :=w
return z
The recursive version is much easier to write, but the iterative version is much more efficient. In doing the
computation for the iterative version, we just need to go through the loop n - 2 times in order to compute an,
so it requires 0( n) steps. In doing the computation for the recursive version, we are constantly recalculating
previous values that we've already calculated, just as was the case with the recursive version of the algorithm
to calculate the Fibonacci numbers.
37. We use the recursive definition of the reversal of a string given in Exercise 35 of Section 5.3, namely that
(vy )R = y( vR) , where y is the last symbol in the string and v is the substring consisting of all but the last
symbol. The right-hand side of the last statement in this procedure means that we concatenate bn with the
output of the recursive call.
180
Chapter 5
Induction and Recursion
procedure reverse(b 1b2 ... bn : bit string)
if n = 0 then return .A
else return bnreverse(b1b2 ... bn-1)
39. The procedure correctly gives the reversal of .A as .A (the basis step of our proof by mathematical induction
on n ), and because the reversal of a string consists of its last character followed by the reversal of its first
n -1 characters (see Exercise 35 in Section 5.3), the algorithm behaves correctly when n > 0 by the inductive
hypothesis.
41. The algorithm merely implements the idea of Example 14 in Section 5.1. If n = 1 (the basis step here), we
simply place the one right triomino so that its armpit corresponds to the hole in the 2 x 2 board. If n > 1,
then we divide the board into four boards, each of size 2n-l x 2n-l, notice which quarter the hole occurs
in, position one right triomino at the center of the board with its armpit in the quarter where the missing
square is (see Figure 7 in Section 5.1), and invoke the algorithm recursively four times-once on each of the
2n-l x 2n-l boards, each of which has one square missing (either because it was missing to begin with, or
because it is covered by the central triomino).
43. Essentially all we do is write down the definition as a procedure.
procedure ackermann(m, n: nonnegative integers)
if m = 0 then return 2 · n
else if n = 0 then return 0
else if n = 1 then return 2
else return ackermann(m - 1, ackermann(m, n - 1))
45. We assume that sorting is to be done into alphabetical order. First the list is split into the two lists b, d, a,
f, g and h, z,p, o, k, and each of these is sorted by merge sort. Let us assume for a moment that this has been
done, so the two lists are a, b, d, f, g and h, k, o, p, z. Then these two lists are merged into one sorted list, as
follows. We compare a with h and find that a is smaller; thus a comes first in the merged list, and we pass
on to b. Comparing b with h, we find that b is smaller, so b comes next in the merged list, and we pass on
to d. We repeat this process (using Algorithm 10) until the lists are merged into one sorted list, a, b, d, f, g, h,
k, o,p, z. (It was just a coincidence that every element in the first of these two lists came before every element
in the second.)
Let us return to the question of how each of the 5-element lists was sorted. For the list b, d, a, f, g, we
divide it into the sublists b, d, a and f, g. Again we sort each piece by the same algorithm, producing a, b, d
and f, g, and we merge them into the sorted list a, b, d, f, g. Going one level deeper into the recursion, we
see that sorting b, d, a was accomplished by splitting it into b, d and a, and sorting each piece by the same
algorithm. The first of these required further splitting into b and d. One element lists are already sorted, of
course. Similarly, the other 5-element list was sorted by a similar recursive process. A tree diagram for this
problem is displayed below. The top half of the picture is a tree showing the splitting part of the algorithm.
The bottom half shows the merging part as an upside-down tree.
bdefghzpok
bdef g
hzpok
~ fg
bde
A
bd
(\d
e
/'\
f
/-.__
hzp
g
A/\
p
h
z
ok
A
o
k
Section 5.4
Recursive Algorithms
181
h~v
hkopz
ebdf ghkopz
47. All we have to do is to make sure that one of the lists is exhausted only when the other list has only one
element left in it. In this case, a comparison is needed to place every element of the merged list into place,
except for the last element. Clearly this condition is met if and only if the largest element in the combined
list is in one of the initial lists and the second largest element is in the other. One such pair of lists is {1, 2,
... , m - 1, m + n} and {m, m + 1, ... , m + n - 1}.
49. We use strong induction on n, showing that the algorithm works correctly if n = 1, and that if it works
correctly for n = 1 through n = k , then it also works correctly for n = k + 1 . If n = 1 , then the algorithm
does nothing, which is correct, since a list with one element is already sorted. If n = k + 1, then the list is
split into two lists, L 1 and L 2 . By the inductive hypothesis, mergesort correctly sorts each of these sublists,
and so it remains only to show that merge correctly merges two sorted lists into one. This is clear, from the
code given in Algorithm 10, since with each comparison, the smallest element in L 1 U L2 not yet put into L
is put there.
51. We need to compare every other element with a 1 . Thus at least n-1 comparisons are needed (we will assume
for Exercises 53 and 55 that the answer is exactly n - 1 ). The actual number of comparisons depends on the
actual encoding of the algorithm. With any reasonable encoding, it should be O(n).
53. In our analysis we assume that a 1 is considered to be put between the two sublists, not as the last element of
the first sublist (which would require an extra pass in some cases). In the worst case, the original list splits
into lists of length 3 and 0 (with a 1 between them); by Exercise 51, this requires 4 - 1 = 3 comparisons.
No comparisons are needed to sort the second of these lists (since it is empty). To sort the first, we argue in
the same way: the worst case is for a splitting into lists of length 2 and 0, requiring 3 - 1 = 2 comparisons.
Similarly, 2 - 1 = 1 comparison is needed to split the list of length 2 into lists of length 1 and 0. In all, then,
3 + 2 + 1 = 6 comparisons are needed in this worst case. (One can prove that this discussion really does deal
with the worst case by looking at what happens in the various other cases.)
55. In our analysis we assume that a 1 is considered to be put between the two sublists, not as the last element of
the first sublist (which would require an extra pass in some cases). We claim that the worst case complexity
is n( n - 1) /2 comparisons, and we prove this by induction on n. This is certainly true for n = 1, since no
comparisons are needed. Otherwise, suppose that the initial split is into lists of size k and n - k - 1. By the
inductive hypothesis, it will require (k(k -1 )/2) + ((n - k - l)(n - k - 2)/2) comparisons in the worst case to
finish the algorithm. This quadratic function of k attains its maximum value if k = 0 (or k = n - 1), namely
the value (n - l)(n - 2)/2. Also, it took n - 1 comparisons to perform the first splitting. If we add these two
quantities ((n - l)(n - 2)/2 and n - l) and do the algebra, then we obtain n(n -1)/2, as desired. Thus in
the worst case the complexity is O(n 2 ).
182
Chapter 5
SECTION 5.5
Induction and Recursion
Program Correctness
Entire books have been written on program verincation; obviously here we barely scratch the surface. In some
sense program verincation is just a very careful stepping through a program to prove that it works correctly.
Tliere should be no problem verifying anything except loops; indeed it may seem that there is nothing to
prove. Loops are harder to deal with. The trick is to find the right invariant. Once you have the invariant, it
again is really a matter of stepping through one pass of the loop to make sure that the invariant is satisned at
the end of that pass, given that it was satisfied at the beginning. Analogous to our remark in the comments
for Section 1. 7, there is a deep theorem of logic (or theoretical computer science) that says essentially that
there is no algorithm for proving correct programs correct, so the task is much more an art than a science.
Your proofs must be valid proofs, of course. You may use the rules of inference discussed in Section 1.6.
What is special about proofs of program correctness is the addition of some special rules for this setting. These
exercises (and the examples in the text) may seem overly simple, but unfortunately it is extremely hard to
prove all but the simplest programs correct.
1. We suppose that initially x
= 0. The segment causes two things to happen. First
y is assigned the value
of 1. Next the value of x + y is computed to be 0 + 1 = 1, and so z is assigned the value 1. Therefore at the
end z has the value 1, so the final assertion is satisfied.
3. We suppose that initially y = 3. The effect of the first two statements is to assign x the value 2 and z the
value 2 + 3 = 5. Next, when the if. .. then statement is encountered, since the value of y is 3, and 3 > 0
is true, the statement z := z + 1 assigns the value 5 + 1 = 6 to z (and the else clause is not executed).
Therefore at the end, z has the value 6, so the final assertion z
=6
is true.
5. We generalize the rule of inference for the if. .. then ... else statement, given before Example 3. Let p be the
initial assertion, and let q be the final assertion. If condition 1 is true, then 5 1 will get executed, so we need
(p A condition 1 ){ Si}q. Similarly, if condition 2 is true, but condition 1 is false, then 5 2 will get executed, so
we need (p A •(condition 1) A condition 2){S2 }q. This pattern continues, until the last statement we need is
(p
A •(condition 1) A •(condition 2) A · · · /\ •(condition n - 1)) {Sn }q. Given all of these, we can conclude
that pTq, where T is the entire statement. In symbols, we have
(p /\ cond1.tion 1){S1 }q
(p /\•(condition 1) /\ condition 2){S2 }q
(p
/\•(condition 1) /\•(condition 2) /\ · · · /\•(condition n - l)){Sn}q
.·. p{if condition 1 then 5 1 else if condition 2 then 5 2
..•
else Sn}q.
7. The problem is similar to Example 4. We will use the loop invariant p: '"power= x 2 - 1 and i ::=; n+ l." Now
p is true initially, since before the loop starts, i = 1 and power = 1 = x 0 = x 1 - 1 . (There is a technicality
here: we define o0 to equal 1 in order for this to be correct if a; = 0. There is no harm in this, since n > 0,
so if x = 0, then the program certainly computes the correct answer xn = 0.) We must now show that if p is
true and i ::=; n before some pass through the loop, then p remains true after that pass. The loop increments
i by one. Hence since i ::=; n before this pass through the loop, i ::=; n + 1 after this pass. Also the loop assigns
power · x to power. By the inductive hypothesis, power started with the value x 2 - 1 (the old value of i).
Therefore its new value is :r'- 1 · x = x' = x n. At
termination we have (i ::=; n + 1) /\ •(i ::=; n), so i = n + 1. Hence power= x(n+l)-l = xn, as desired.
Review Questions
183
9. We will break the problem up into the various statements that are left unproved in Example 5.
We must show that p{ Si }q, where p is the assertion that m and n are integers, and q is the proposition
p /\ (a= lnl). This follows from Example 3 and the fact that the values of m and n have not been tampered
with.
We must show that q{ S2}r, where r is the proposition q /\ (k = 0) /\ (x = 0). This is clear, since in S2,
none of m, n, or a have been altered, but k and x have been assigned the value 0.
We must show that (x = mk) /\ (k:::; a) is an invariant for the loop in S 3 . Assume that this statement
is true before a pass through the loop. In the loop, k is incremented by 1. If k < a (the condition of the
loop), then at the end of the pass, k 0; then the loop will be executed again. Within the loop x and y are
replaced by y and x mod y, respectively. According to Lemma 1 in Section 4.3, gcd(x, y) = gcd(y,x mod y).
Therefore after the execution of the loop, the value of gcd(x, y) remains what it was before. Furthermore,
since y is a remainder, it is still greater than or equal to 0. Hence p remains true-it is a loop invariant.
Furthermore, if the loop terminates, then it must be the case that y = 0. In this case we know that
gcd(x,y) = x, the desired final assertion. Therefore the program, which gives x as its output, has correctly
computed gcd(a, b). Finally (although this is not necessary to establish partial correctness), we can prove that
the loop must terminate, since each iteration causes the value of y to decrease by at least 1 (by the definition
of mod). Thus the loop can be iterated at most b times.
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 5
1. a) no
b) Sometimes yes. If the given formula is correct, then it is often possible to prove it using the principle of
mathematical induction (although it would be wishful thinking to believe that every such true formula could
be so proved). If the formula is incorrect, then induction would not work, of course; thus an incorrect formula
could not be shown to be incorrect using the principle.
c) See Exercise 9 in Section 5.1.
Chapter 5
184
Induction and Recursion
2. a) n 2 7
b) For the basis step we just check that 11 · 7 + 17 :S 27 . Fix n 2 7, and assume the inductive hypothesis,
that lln + 17 :S 2n. Then ll(n + 1) + 17=(lln+17) + 11:S2n+11 < 2n + 2n = 2n+i. The strict inequality
here follows from the fact that n ~ 4.
3. a) Carefully considering all the possibilities shows that the amounts of postage less than 32 cents that can
be achieved are 0, 5, 9, 10, 14, 15, 18, 19, 20, 23, 24, 25, 27, 28, 29, and 30. All amounts greater than
or equal to 32 cents can be achieved.
b) To prove this latter statement, we check the basis step by noting that 32 = 9 + 9 + 9 + 5. Assume that
we can achieve n cents, and consider n + 1 cents, where n ~ 32. If the stamps used for n cents included a
9-cent stamp, then replacing it by two 5-cent stamps gives us n + 1 cents, as desired. Otherwise only 5-cent
stamps were used to achieve n cents, and since n > 30, there must be at least seven such stamps. Replace
seven of the 5-cent stamps by four 9-cent stamps; this increases the amount of postage by 4 · 9 - 7 · 5 = 1 cent,
again as desired.
c) We check the base cases 32 = 3 · 9 + 5, 33 = 2 · 9 + 3 · 5, 34 = 9 + 5 · 5, 35 = 7 · 5, and 36 = 4 · 9. Fix
n 2 37 and assume that all amounts from 32 to n - 1 can be achieved. To achieve n cents postage, take
the stamps used for n - 5 cents (since n 2 37, n - 5 2 32, so the inductive hypothesis applies) and adjoin a
5-cent stamp.
d) Let n be an integer greater than or equal to 32. We want to express n as a sum of a nonnegative multiple
of 5 and a nonnegative multiple of 9. Divide n by 5 to obtain a quotient q and remainder r such that
n = 5q + r and 0::::; r :S 4. Note that since n ~ 32, q 2 6. If r = 0, then we already have n expressed in
the desired form. If r = 1, then n 2 36, so q 2 7; thus we can write n = 5q + 1 = 5(q - 7) + 4 · 9 to get
the desired decomposition. If r = 2, then we rewrite n = 5q + 2 = 5(q - 5) + 3 · 9. If r = 3, then we rewrite
n = 5q + 3 = 5(q - 3) + 2 · 9. And if r = 4, then we rewrite n = 5q + 4 = 5(q - 1) + 9. In each case we have
the desired sum.
4. See Examples 2 and 3 in Section 5.2.
5. a) Seep. 314 and Appendix 1 (Axiom 4 for the positive integers).
b) Let S be the set of positive integers that cannot be written as the product of primes. If S #- 0, then
S has a least element, c. Clearly c #- 1, since 1 is the product of no primes. Thus c is greater than 1.
Now c cannot be prime, since as such it would already be written as the product of primes (namely itself).
Therefore c is a composite number, say c =ab, where a and b are both positive integers less than c. Since c
is the smallest element of S, neither a nor b is in S. Therefore both a and b can be written as the product
of primes. But multiplying these products together patently shows that c is the product of primes. This is
a contradiction to the choice of c. Therefore our assumption that S #- 0 was wrong, and the theorem is
proved.
b) J(l) = 2, and f(n) = (n + l)f(n - 1) for all n 2 2
6. a) See Exercise 56 in Section 5.3.
7. a) See the top of p. 347.
b) See Example 4 in Section 5.3.
b) an = 3 · 2n- 3 for n ~ 3
8. a) See Exercise 57 in Section 5.3.
9. See Examples 8 and 9 in Section 5.3.
b) See Example 12 in Section 5.3.
10. a) See Example 7 in Section 5.3.
11. a) See the beginning of Section 5.4.
b) Call the sequence a 1 , a 2 ,
12. See Example 3 in Section 5.4.
... ,
an. If n = 1, then the sum(a1) = ai. Otherwise sum(a1,a2, ... ,an) =
Supplementary Exercises
185
13. a) See p. 367.
b) We split the list into the two halves: 4, 10, 1, 5, 3 and 8, 7, 2, 6, 9. We then merge sort each half by
applying this algorithm recursively and merging the results. For the first half, for example, this means splitting
4, 10, 1, 5, 3 into the two halves 4, 10, 1 and 5, 3, recursively sorting each half, and merging. For the second
half of this, for example, it means splitting into 5 and 3, recursively sorting each half, and merging. Since
these two halves are already sorted, we just merge, into the sorted list 3, 5. Similarly, we will get 1, 4, 10 for
the result of merge sort applied to 4, 10, 1. When we merge 1, 4, 10 and 3, 5, we get 1, 3, 4, 5, 10. Finally, we
merge this with the sorted second half, 2, 6, 7, 8, 9, to obtain the completely sorted list 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
c) O(nlogn); see pp. 369-370.
14. a) no
b) No-you also need to show that it halts for all inputs, and the initial and final assertions for which you
provide a proof of partial correctness need to be appropriate ones (i.e., relevant to the question of whether the
program produces the correct output).
15. See the rules displayed in Section 5.5.
16. See p. 375.
SUPPLEMENTARY EXERCISES FOR CHAPTER 5
1. Let P(n) be the statement that this equation holds. The basis step consists of verifying that P(l) is true,
which is trivial because 2/3 = 1 - (1/3 1 ). For the inductive step we assume that P(k) is true and try to
prove P(k + 1). We have
22
2
2
2
- + - + - + .. · + - + - 3
9
27
3n
3n+l
1
2
= 1 - -3n + 3n+I
=
(by the inductive hypothesis)
3
2
1 - 3n+l + 3n+l
1
= 1- 3n+l'
as desired.
3. We prove this by induction on n. If n = 1 (basis step), then the equation reads 1·2°
is the true statement 1 = 1. Assume that the statement is true for n:
1 · 2° + 2 · 21 + 3 · 22 + · · · + n · 2n-l
We must show that it is true for n
+ 1.
= (1-1) · 21 +1,
= (n - 1) · 2n + 1
Thus we have
1·2°+2 · 21 +3 · 22 + .. · + n · 2n-l + (n + 1) · 2n
= (n - 1) · 2n + 1+(n+1) · 2n
=
(2n) · 2n + 1
+1
1) - 1) · 2n+l + 1 ,
= n · 2n+I
= ( (n +
exactly as desired.
(by the inductive hypothesis)
which
Chapter 5
186
Induction and Recursion
5. We prove this by induction on n. If n = 1 (basis step), then the equation reads 1/(1·4)=1/4, which is true.
Assume that the statement is true for n:
1
1
1
n
-+-+···+----1·4 4·7
(3n - 2)(3n + 1)
3n+ 1
We must show that it is true for n + 1. Thus we have
1
1
1
1
- - + - - + ... +
+ -c------.,--,...----~
1·4 4-7
(3n-2)(3n+l)
(3(n+l)-2)(3(n+l)+l)
n
3n+l
n
--3n + 1
1
= --
+
=
+ -------
(3(n+l)-2)(3(n+l)+l)
(by the inductive hypothesis)
1
(3n + 1)(3n + 4)
= 3n ~ 1 ( n + 3n ~
4)
2
= _1_ (3n +4n+1)
3n + 1
3n+4
= _1_ ((3n + l)(n + 1))
3n + 1
3n+4
n+l
3n+4
n+l
3(n+l)+l'
exactly as desired.
7. Let P( n) be the statement 2n > n 3 . We want to prove that P( n) is true for all n > 9. The basis step is
n = 10, in which we have 210 = 1024 > 1000 = 103 . Assume P(n); we want to show P(n + 1). Then we have
(n + 1)3 = n 3 + 3n 2 + 3n + 1
~ n 3 + 3n2 + 3n 2 + 3n2
= n
3
+ 9n
(since n 2': 1)
2
3
< n + n 3 (since n > 9)
= 2n 3 < 2 · 2n (by the inductive hypothesis)
= 2n+l,
as desired.
9. This problem deals with factors in algebra. We have to be just a little clever. Let P(n) be the statement that
a - b is a factor of an - bn . We want to show that P( n) is true for all positive integers n, and of course we
will do so by induction. If n = 1, then we have the trivial statement that a - b is a factor of a - b. Next
assume the inductive hypothesis, that P( n) is true. We want to show P( n + 1), that a - b is a factor of
an+l - bn+l. The trick is to rewrite an+l - bn+l by subtracting and adding abn (and hence not changing
its value). We obtain an+l - bn+l = an+l - abn + abn - bn+l = a(an - bn) + bn(a - b). Now this expression
contains two terms. By the inductive hypothesis, a - b is a factor of the first term. Obviously a - b is a factor
of the second. Therefore a - b is a factor of the entire expression, and we are done.
11. This should be similar to Example 9 in Section 5.1. First we check the basis step: When n = 1, 5n+l +
72 n-l = 36 + 7 = 43, which is certainly divisible by 43. Assume the inductive hypothesis, that 43 divides
5n+l + 72 n-l; we must show that 43 divides 5n+ 2 + 72 n+l. We have 5n+ 2 + 72 n+l = 6 · 5n+l + 49 · 72 n-l =
6 · 5n+l + 6 · 72 n-l + 43 · 72 n-l = 6(6n+l + 72 n-l) + 43 · 72 n-l. Now the first term is divisible by 43 because the
inductive hypothesis guarantees that its second factor is divisible by 43. The second term is patently divisible
by 43. Therefore the sum is divisible by 43, and our proof by mathematical induction is complete.
187
Supplementary Exercises
13. Let P( n) be the given equation. It is certainly true for n
P(n) is true:
= 0, since it reads a = a
a+ (a+ d) +···+ (a+n d)
in that case. Assume that
(n+l)(2a+nd)
-----'-2
=-'-----'--'---
Then
a + (a + d) + · · · + (a + nd) + (a + (n + 1)d)
= (n + 1) (~a + nd) + (a + (n + 1)d)
(by the inductive hypothesis)
(n + 1)(2a + nd) + 2(a + (n + l)d)
2
(n + 1)(2a + nd) + 2a + nd + nd + 2d
2
(n + 2)(2a + nd) + (n + 2)d
2
(n + 2)(2a + (n + l)d)
2
which is exactly P( n
+ 1).
15. We use induction. If n = 1, then the left-hand side has just one term, namely 5/6, and the right-hand side is
10/12, which is the same number. Assume that the equation holds for n = k, and consider n = k+ l. (Do not
get confused by the choice of letters here! The index of summation in the problem as stated is just a dummy
variable, and since we want to use k in the inductive hypothesis, we have changed the dummy summation
index to i.) Then we have
k+l
i + 4
i + 4
k
~i(i+l)(i+2) = ~i(i+l)(i+2)
k+5
+ (k+l)(k+2)(k+3)
k(3k + 7)
k +5
(k
+
l)(k
+
)
+
(k
+
l)(k
+ 2)(k + 3)
2
2
1
= (k + l)(k + 2) .
(k(3k+7)
2
(by the inductive hypothesis)
k+5)
+ k+ 3
1
2(k + l)(k + 2)(k + 3) . (k(3k + 7)(k + 3) + 2(k + 5))
2(k + l)(k
~ 2)(k + 3) . (3k3 + 16k2 + 23k + 10)
1
2(k + l)(k + 2)(k + 3) . (3k + lO)(k + l)
2
1
2(k + 2)(k + 3) . (3k + lO)(k + 1)
(k + 1)(3(k + 1) + 7)
2((k + 1) + l)((k + 1) + 2)'
as desired.
17. When n = 1, we are looking for the derivative of g( x) = xex, which, by the product rule, is x · ex + ex =
(x + 1)e, so the statement is true for n = 1. Assume that the statement is true for n = k, that is, the kth
derivative is given by g(k) = (x + k)ex. Differentiating by the product rule gives us the (k + l)st derivative:
g(k+l) = (x + k)ex +ex= (x + (k + l))ex, as desired.
19. We look at the first few Fibonacci numbers to see if there is a pattern: Jo = 0 (even), Ji = 1 (odd),
h = 1 (odd), f3 = 2 (even), f 4 = 3 (odd), f 5 = 5 (odd), .... The pattern seems to be even-odd-odd,
repeated forever. Since the pattern has period 3, we can formulate our conjecture as follows: f n is even if
Chapter 5
188
Induction and Recursion
n = 0 (mod 3), and is odd in the other two cases. Let us prove this by mathematical induction. There are
two base cases, n = 0 and n = 1. The conjecture is certainly true in each of them, since 0 O (mod 3) and
Jo is even. and 1 -;j. 0 (mod 3) and Jo is odd. So we assume the inductive hypothesis and consider a given
=
n+ 1. There are three cases to consider, depending on the value of (n+ 1) mod 3. If n+ 1 = 0 (mod 3), then
n - 1 and n are congruent to 1 and 2 modulo 3, respectively. By the inductive hypothesis, both Jn-l and
Jn are odd. Therefore f n+I. which is the sum of these two numbers, is even, as desired. The other two cases
are similar. If n + 1 1 (mod 3), then n - 1 and n are congruent to 2 and 0 modulo 3, respectively. By the
=
inductive hypothesis, J n-l is odd and Jn is even. Therefore J n+l • which is the sum of these two numbers,
is odd, as desired. On the other hand, if n
+ 1 =2
(mod 3), then n - 1 and n are congruent to 0 and 1
modulo 3, respectively. By the inductive hypothesis, Jn-l is even and Jn is odd. Therefore Jn+l, which is
the sum of these two numbers, is odd, as desired.
21. The important point to note here is that k can be thought of as a universally quantified variable for each n.
Thus the statement we wish to prove is P(n): for every k, fkJn + Jk+iJn+i = Jn+k+l · We use mathematical
induction. If n = 0 (the first base case), then we want to prove P(O): for every k, fkJo + Jk+ifi = Jo+k+i'
which reduces to the identity fk+ 1 = fk+1, since Jo = 0 and Ji = 1. If n = 1 (the second base case),
then we want to prove P(l): for every k, fkfi + fk+ 1h = fi+k+l, which reduces to the defining recurrence
Jk + fk+i = fk+ 2, since Ji = 1 and h = 1. Now we assume the inductive hypothesis P(n) and try to prove
P(n + 1). It is a straightforward calculation, using the inductive hypothesis and the recursive definition of the
Fibonacci numbers:
fkfn+l
+ fk+1fn+2
+ fn) + fk+1Un + fn+i)
= fkJ n-l + fkJ n + fk+ifn + Jk+lfn+l
= (fkJn-l + Jk+ifn) + (fkJn + fk+ifn+l)
= Jn-l+k+l + Jn+k+l = Jn+k+2,
= fkUn-l
as desired.
23. Let P(n) be the statement z5 + Zi + · · ·+ z; = lnln+l + 2. We easily verify the two base cases, P(O) and P(l),
since 22 = 2 · 1 + 2 and 22 + 12 = 1 · 3 + 2. Next assume the inductive hypothesis and consider P( n + 1). We
have
Z5 + Zi + · · · + z;, + z;,+ 1 = lnln+i + 2 + z;+ 1
= ln+1(ln + ln+l) + 2
= ln+lln+2 + 2,
which is exactly what we wanted.
25. The identity is clearly true for n
= 1. Let us expand the right-hand side for
+ 1,
invoking the inductive
hypothesis at the appropriate point (and using the suggested trigonometric identities as well as the fact that
i 2 = -1 ):
n
+ isin(nx + x)
= cos nx cos x - sin nx sin x + i (sin nx cos x + cos nx sin x)
= cos x (cos nx + i sin nx) + sin x ( - sin nx + i cos nx)
= cos x (cos nx + i sin nx) + i sin x (i sin nx + cos nx)
= (cos nx + i sin nx) (cos x + i sin x)
cos(n + l)x + isin(n + l)x = cos(nx + x)
= (cos x + i sin x r (cos x + i sin x)
=
(cos x
+ i sin x) n+l
Supplementary Exercises
189
27. First let's rewrite the right-hand side to make it simpler to work with, namely as 2n+ 1 (n 2 - 2n + 3) - 6.
We use induction. If n = 1, then the left-hand side has just one term, namely 2, and the right-hand side is
4 · 2 - 6 = 2 as well. Assume that the equation holds for n = k, and consider n = k + 1 . Then we have
k+l
k
I:J221 = I:J221+(k+1) 2 2k+ 1
J=l
J=l
=
=
2k+l(k 2 - 2k + 3) - 6 + (k 2 + 2k + 1)2k+ 1
2k+l(2k 2 + 4) - 6
= 2k+ 2 (k 2 + 2) - 6
= 2k+ 2 ((k + 1) 2 - 2(k + 1) + 3) -
(by the inductive hypothesis)
6,
as desired.
29. One solution here is to use partial fractions and telescoping. First note that
j2
c
~ 1 = ~ ~ 1- j ~ 1) .
Therefore when summing from 1 to n, the terms being added and the terms being subtracted all cancel out
except for 1/(j -1) when j = 2 and 3, and 1/(j + 1) when j = n - 1 and n. Thus the sum is just
~(~+~-~-n~l) ·
which simplifies, with a little algebra, to the expression given on the right-hand side of the formula in the
exercise.
In the spirit of this chapter, however, we also give a proof by mathematical induction. Let P(n) be the
formula in the exercise. The basis step is for n = 2, in which case both sides reduce to 1/3. For the inductive
step assume that the equation holds for n = k, and consider n = k + 1 . Then we have
k+l
1
k
.L p _ 1 =.LP
J=l
J=l
=
1
1
-1+(k+1)2 -1
(k-1)(3k+2)
1
+ (k + l) 2
4k(k + l)
1 (by the inductive hypothesis)
(k-1)(3k+2)
1
(k-1)(3k+2)
1
= 4k(k + 1) + k 2 + 2k = 4k(k + 1) + k(k + 2)
(k - 1)(3k + 2)(k + 2) + 4(k + 1)
4k(k + l)(k + 2)
3k 3 + 5k 2
4k(k + l)(k + 2)
k(3k + 5)
4(k + l)(k + 2)
which is exactly what P(k + 1) asserts.
_
3k 2 + 5k
4(k + l)(k + 2)
((k + 1) -1)(3(k + 1) + 2)
4(k + l)(k + 2)
31. Let P(n) be the assertion that at least n+ 1 lines are needed to cover the lattice points in the given triangular
region. Clearly P(O) is true, because we need at least one line to cover the one point at (0, 0). Assume the
inductive hypothesis, that at least k + 1 lines are needed to cover the lattice points with x ~ 0, y ~ 0, and
x + y ::; k. Consider the triangle of lattice points defined by x ~ 0, y ~ 0, and x + y ::; k + 1. Because this
set includes the previous set, at least k + 1 lines are required just to cover the smaller set (by the inductive
hypothesis). By way of contradiction, assume that k + 1 lines could cover this larger set as well. Then these
lines must also cover the k + 2 points on the line x + y = k + 1, namely (0, k + 1), (1, k), (2, k - 1), ... ,
(k, 1), (k + 1, 0). But only the line x + y = k + 1 itself can cover more than one of these points, because
Chapter 5
190
Induction and Recursion
two distinct lines intersect in at most one point, and this line does nothing toward covering the lattice points
in the smaller triangle. Therefore none of the k + 1 lines that are needed to cover the lattice points in the
smaller triangle can cover more than one of the points on the line x + y = k + 1 , and this leaves at least one
point uncovered. Therefore our assumption that k + 1 line could cover the larger set is wrong, and our proof
is complete.
33. The basis step is the given statement defining B. Assume the inductive hypothesis, that Bk = MA kM- 1 . We
want to prove that Bk+l = MAk+ 1M- 1 . By definition Bk+ 1 = BBk = MAM- 1Bk = MAM- 1MAkM- 1
by the inductive hypothesis. But this simplifies, using rules for matrices, to MAIAkM- 1 = MAAkM- 1 =
MA k+l M- 1 , as desired.
35. It takes some luck to be led to the solution here. We see that we can write 3l = 3+2+ 1. We also have a recursive
definition of factorial, that (n + 1) ! = ( n + 1)n!, so we might hope to multiply each of the divisors we got at the
previous stage by n + 1 to get divisors at this stage. Thus we would have 4! = 4 · 3! = 4(3+2+1) = 12 + 8 + 4,
but that gives us only three divisors in the sum, and we want four. That last divisor, which is n + 1 can,
however, be rewritten as the sum of n and 1, so our sum for 4l is 12 + 8 + 3 + 1. Let's see if we can continue
this. We have 5l = 5 · 4l = 5(12 + 8 + 3 + 1) = 50 + 40 + 15 + 5 = 50 + 40 + 15 + 4 + 1. It seems to be
working. The basis step n = 3 is already done, so let's see if we can prove the inductive step. Assume that
we can write kl as a sum of the desired form, say kl = a 1 + a2 + · · · + ak, where each ai is a divisor of k! and
the divisors are listed in strictly decreasing order, and consider (k + 1)!. Then we have (k + l)l = (k + l)k! =
(k + l)(a1 + a2 + · · · + ak) = (k + l)a1 + (k + l)a2 + · · · + (k + l)ak = (k + l)a1 + (k + l)a2 + · · · + k · ak + ak.
Because each a, was a divisor of k!, each (k + l)ai is a divisor of (k + 1)!, but what about those last two
terms? We don't seem to have any way to know that k · ak is a factor of (k + 1) ! . Hold on, in our exploration
we always had the last divisor in our sum being 1. If so, then k · ak = k, which is a divisor of (k + l)l,
and ak = 1, so the new last summand is again 1. (Notice also that our list of summands is still in strictly
decreasing order.) So our proof by mathematical induction needs to be of the following stronger result: For
every n ~ 3, we can write n! as the sum of n of its distinct positive divisors, one of which is 1 . The argument
we have just given proves this by mathematical induction.
37. When n = 1 the statement is vacuously true. If n = 2 there must be a woman first and a man second, so
the statement is true. Assume that the statement is true for n = k, where k ~ 2, and consider k + 1 people
standing in a line, with a woman first and a man last. If the kth person is a woman, then we have that woman
standing in front of the man at the end, and we are done. If the kth person is a man, then the first k people
in line satisfy the conditions of the inductive hypothesis for the first k people in line, so again we can conclude
that there is a woman directly in front of a man somewhere in the line.
39. (It will be helpful for the reader to draw a diagram to help in following this proof.) When n = 1 there is one
circle, and we can color the inside blue and the outside red to satisfy the conditions. Assume the inductive
hypothesis that if there are k circles, then the regions can be 2-colored such that no regions with a common
boundary have the same color, and consider a situation with k+ 1 circles. Remove one of the circles, producing
a picture with k circles, and invoke the inductive hypothesis to color it in the prescribed manner. Then replace
the removed circle and change the color of every region inside this circle (from red to blue, and from blue to
red). It is clear that the resulting figure satisfies the condition, since if two regions have a common boundary,
then either that boundary was an arc of the new circle, in which case the regions on either side used to be the
same region and now the inside portion is colored differently from the outside, or else the boundary did not
involve the new circle, in which case the regions are colored differently because they were colored differently
before the new circle was restored.
191
Supplementary Exercises
41. We use induction. If n = 1 then the equation reads 1 · 1 = 1 · 2/2, which is true. Assume that the equation
is true for n and consider it for n + 1. (We use the letter n rather than k, because k is used for something
else here.) Then we have, with some messy algebra (including an application of Example 2 in Section 5.1 in
line 4),
n+l
(n+l 1)
?;(2j - 1) ~ k
n
=
?;(2j - 1)
n
.
=2:( 2J-l)
J=l
=
=
(n+l 1)
1
~ k + (2(n + 1) - 1) · n + 1
(
1
n+l
1)
n
+Lk
2n + 1
+ n+l
k=J
n
- 1 ~(2j
- 1) ) +
( n+lL.,,
J=l
(
n
~(2j
- 1)
L.,,
J=l
1
( -- . n2) + n(n + l) + 2n + 1
n+l
2
n+l
(
n -1)) + _n_
2 + 1
~
L.,,k
k=J
n+l
(by the inductive hypothesis)
2n 2 + n(n + 1) 2 + (4n + 2)
2(n + 1)
2(n + 1) 2 + n(n + 1) 2
2(n + 1)
(n+l)(n+2)
2
as desired.
43. Let T( n) be the statement that the sequence of towers of 2 is eventually constant modulo n. We will use
strong induction to prove that T(n) is true for all positive integers n. Basis step: When n = 1 (and n = 2),
the sequence of towers of 2 modulo n is the sequence of all O's. Inductive step: Suppose that k is an integer
with k ::'.': 2. Suppose that T(j) is true for 1 :::; j :::; k - 1. In the proof of the inductive step we denote the
rth term of the sequence modulo n by ar. We will split the proof of the inductive step into two cases, based
on the parity of k.
Suppose k is even. Let k = 28 q where s 2: 1 and q < k is odd. When j is large enough, a1 _2 2: s,
2
and for such j, a1 = 22a 1 _ is a multiple of 28 • It follows that for sufficiently large j, a1 = 0 (mod 28 ) .
Hence, for large enough i, 28 divides ai+ 1 - a,. By the inductive hypothesis T(q) is true, so the sequence
a 1 , a 2, a 3 , ... is eventually constant modulo q. This implies that for large enough i, q divides a,+l - a,.
Because gcd(q, 28 ) = 1 and for sufficiently large i both q and 28 divide a,+ 1 - a,, k = 2 8 q divides a,+ 1 - a,
for sufficiently large i. Hence, for sufficiently large i , ai+ 1 - a,
0 (mod k) . This means that the sequence
is eventually constant modulo k.
=
Finally, suppose k is odd. Then gcd(2, k) = 1, so by Euler's theorem (found in elementary number theory
books, such as the author's), we know that 2 1 or x < 0, then the factors have different signs, so the inequality does not hold. Thus the only
solutions are x = 0 and x = 1. The corresponding solutions for y are therefore also 0 and 1. Thus the only
time we have both x 2: y 2 and y 2: x 2 is when x = y; this means that the relation is antisymmetric. It is
transitive. Suppose x 2: y 2 and y 2: z 2 . Again the second inequality implies that both sides are nonnegative,
so we can square both sides to obtain y 2 2: z4 . Combining these inequalities gives x 2: z 4 . Now we claim that
it is always the case that z 4 2: z 2 ; if so, then we combine this fact with the last inequality to obtain x 2: z 2 ,
so x is related to z. To verify the claim, note that since we are working with integers, it is always the case
that z 2 2: lzl (equality for z = 0 and z = 1, strict inequality for other z). Squaring both sides gives the
desired inequality.
9. Each of the properties is a universally quantified statement. Because the domain is empty, each of them is
vacuously true.
11. The relations in parts (a), (b), and (e) all have at least one pair of the form (x,x) in them, so they are not
irreflexive. The relations in parts (c), (d), and (f) do not, so they are irreflexive.
13. According to the preamble to Exercise 11, an irreflexive relation is one for which a is never related to itself;
i.e., \fa((a,a) ~ R).
a) Since we saw in Exercise 5a that \fa((a, a) ER), clearly R is not irreflexive.
b) Since there are probably pages a with no links at all, and for such pages it is true that there are no common
links found on both page a and page a, this relation is probably not irreflexive.
c) This relation is not irreflexive, because for any page a that has links on it, (a, a) ER.
d) This relation is not irreflexive, because for any page a that has links on it that are ever cited, (a, a) E R.
15. The relation in Exercise 3a is neither reflexive nor irreflexive. It contains some of the pairs (a, a) but not all
of them.
17. Of course many answers are possible. The empty relation is always irreflexive ( x is never related to y ). A less
trivial example would be (a, b) E R if and only if a is taller than b. Since nobody is taller than him/herself,
we always have (a, a) ~ R.
19. The relation in part (a) is asymmetric, since if a is taller than b, then certainly b cannot be taller than a.
The relation in part (b) is not asymmetric, since there are many instances of a and b born on the same day
(both cases in which a = b and cases in which a -/=- b ), and in all such cases, it is also the case that b and a
were born on the same day. The relations in part ( c) and part ( d) are just like that in part (b), so they, too,
are not asymmetric.
21. According to the preamble to Exercise 18, an asymmetric relation is one for which (a, b) E R and (b, a) E R
can never hold simultaneously, even if a = b. Thus R is asymmetric if and only if R is antisymmetric and
also irreflexive.
a) not asymmetric since ( -1, 1) E R and (1, -1) E R
b) not asymmetric since ( -1, 1) E R and (1, -1) E R
c) not asymmetric since (-1, 1) ER and (1, -1) ER
d) not asymmetric since (0, 0)
E
R
Chapter 9
316
Relations
e) not asymmetric since (2, 1) ER and (1, 2) ER
f) not asymmetric since (0, 1) ER and (1, 0) ER
g) not asymmetric since (1, 1) ER
h) not asymmetric since (2, 1) ER and (1, 2) ER
23. According to the preamble to Exercise 18, an asymmetric relation is one for which (a, b) ER and (b, a) E R
can never hold simultaneously. In symbols, this is simply Va Vb..,( (a, b) E R /\ (b, a) E R). Alternatively,
VaVb ((a, b) ER---> (b, a) tj. R).
25. There are mn elements of the set A x B, if A is a set with m elements and B is a set with n elements.
A relation from A to B is a subset of A x B. Thus the question asks for the number of subsets of the set
A x B, which has mn elements. By the product rule, it is 2mn.
27. a) By definition the answer is { (b, a) I a divides b}, which, by changing the names of the dummy variables,
can also be written { (a, b) I b divides a } . (The universal set is still the set of positive integers.)
b) By definition the answer is { (a, b)
J
a does not divide b } . (The universal set is still the set of positive
integers.)
29. The inverse relation is just the graph of the inverse function. Somewhat more formally, we have R- 1 =
{ (f(a), a) I a EA} = { (b, 1- 1 (b)) I b EB}, since we can index this collection just as easily by elements of
B as by elements of A (using the correspondence b = f(a) ).
31. This exercise is just a matter of the definitions of the set operations.
a) the set of pairs (a, b) where a is required to read b in a course or has read b
b) the set of pairs (a, b) where a is required to read b in a course and has read b
c) the set of pairs (a, b) where a is required to read b in a course or has read b, but not both; equivalently,
the set of pairs (a, b) where a is required to read b in a course but has not done so, or has read b although
not required to do so in a course
d) the set of pairs (a, b) where a is required to read b in a course but has not done so
e) the set of pairs (a, b) where a has read b although not required to do so in a course
33. To find So R we want to find the set of pairs (a, c) such that for some person b, a is a parent of b, and b
is a sibling of c. Since brothers and sisters have the same parents, this means that a is also the parent of c.
Thus S o R is contained in the relation R. More specifically, (a, c) E S o R if and only if a is the parent of c,
and c has a sibling (who is necessarily also a child of a). To find Ro S we want to find the set of pairs (a, c)
such that for some person b, a is a sibling of b, and b is a parent of c. This is the same as the condition that
a is the aunt or uncle of c (by blood, not marriage).
35. a) The union of two relations is the union of these sets. Thus R 2 U R 4 holds between two real numbers if R 2
holds or R 4 holds (or both, it goes without saying). Since it is always true that a ::;: b or b ::;: a, R2 U R4 is
all of R 2 , i.e., the relation that always holds.
b) For (a, b) to be in R3 U R 6 , we must have a < b or a =/= b. Since this happens precisely when a =/= b, we
see that the answer is R 6 .
c) The intersection of two relations is the intersection of these sets. Thus R 3 n R 6 holds between two real
numbers if R 3 holds and R6 holds as well. Thus for (a, b) to be in R 3 n R6, we must have a < b and a =/= b.
Since this happens precisely when a
< b, we see that the answer is
R3 .
d) For (a, b) to be in R4 n R 6 , we must have a::;: b and a=/= b. Since this happens precisely when a< b, we
see that the answer is R 3 .
Section 9.1
Relations and Their Properties
317
e) Recall that R 3 - R 6 = R 3 n R 6 . But R 6 = R 5 , so we are asked for R 3 n R 5 . It is impossible for a < b
and a= b to hold at the same time, so the answer is 0, i.e., the relation that never holds.
f) Reasoning as in part ( e), we want R6 n R3 = R6 n R2, which is clearly Ri (since a =f. b and a :::: b precisely
when a> b).
g) Recall that
R6 n R 3 = R 3 .
h) Recall that
Rs n R2 = Rs .
R2 EB
Thus
R3 EB
Thus
R6 = (R2 n R5) u (R6
our answer is R 5 U R3
Rs = (R3 n Rs) U (Rs
our answer is R3 U R5
n R2). We see that R2 n R6 = R2 n R5 = R5, and R6 n R2 =
= R4 .
n R3). We see that R 3 n Rs = R3 n R 6 = R 3 , and Rs n R 3 =
= R4 .
37. Recall that the composition of two relations all defined on a common set is defined as follows: (a, c) E S o R
if and only if there is some element b such that (a, b) E R and (b, c) E S. We have to apply this in each case.
a) For (a, c) to be in R2 o Ri, we must find an element b such that (a, b) E R 1 and (b, c) E R 2 . This means
that a> b and b ~ c. Clearly this can be done if and only if a> c to begin with. But that is precisely the
statement that (a, c) E R 1 . Therefore we have R 2 o R 1 = R 1 .
b) For (a, c) to be in R2 o R 2 , we must find an element b such that (a, b) E R 2 and (b, c) E R 2 . This means
that a :::: b and b :::: c. Clearly this can be done if and only if a :::: c to begin with. But that is precisely the
statement that (a, c) E R 2 . Therefore we have R 2 o R 2 = R 2 . In particular, this shows that R 2 is transitive.
c) For (a, c) to be in R3 o R 5 , we must find an element b such that (a, b) E R 5 and (b, c) E R 3 . This means
that a = b and b < c. Clearly this can be done if and only if a < c to begin with (choose b = a). But that is
precisely the statement that (a, c) E R 3 . Therefore we have R 3 o R 5 = R 3 . One way to look at this is to say
that Rs, the equality relation, acts as an identity for the composition operation (on the right-although it is
also an identity on the left as well).
d) For (a, c) to be in R4 o R 1 , we must find an element b such that (a, b) E R 1 and (b, c) E R 4 . This means
that a > b and b ::::; c. Clearly this can always be done simply by choosing b to be small enough. Therefore
we have R 4 o R 1
= R2 ,
the relation that always holds.
e) For (a, c) to be in R 5 o R 3 , we must find an element b such that (a, b) E R 3 and (b, c) E R 5 . This means
that a < b and b = c. Clearly this can be done if and only if a < c to begin with (choose b = c). But that
is precisely the statement that (a, c) E R 3 . Therefore we have Rs o R 3 = R 3 . One way to look at this is to
say that Rs , the equality relation, acts as an identity for the composition operation (on the left-although it
is also an identity on the right as well).
f) For (a, c) to be in R3 o R6, we must find an element b such that (a, b) E R 6 and (b, c) E R 3 . This means
that a =f. b and b < c. Clearly this can always be done simply by choosing b to be small enough. Therefore
we have R 3 o R6 = R 2 , the relation that always holds.
g) For (a, c) to be in R4 o R 6 , we must find an element b such that (a, b) E R 6 and (b, c) E R 4 . This means
that a =f. b and b ::::; c. Clearly this can always be done simply by choosing b to be small enough. Therefore
we have R 4 o R 6 = R 2 , the relation that always holds.
h) For (a, c) to be in R6 o R6, we must find an element b such that (a, b) E R6 and (b, c) E R 6 . This means
that a =f. b and b =f. c. Clearly this can always be done simply by choosing b to be something other than a
or c. Therefore we have R 6 o R 6 = R 2 , the relation that always holds. Note that since the answer is not R 6
itself, we know that R 6 is not transitive.
39. One earns a doctorate by, among other things, writing a thesis under an advisor, so this relation makes sense.
(We ignore anomalies like someone having two advisors or someone being awarded a doctorate without having
an advisor.) For (a, b) to be in R 2 , we must find a c such that (a, c) ER and (c, b) ER. In our context, this
says that b got his/her doctorate under someone who got his/her doctorate under a. Colloquially, a is the
academic grandparent of b, or b is the academic grandchild of a. Generalizing, (a, b) E Rn precisely when there
is a sequence of n+ 1 people, starting with a and ending with b, such that each is the advisor of the next person
Chapter 9
318
Relations
in the sequence. People with doctorates like to look at these sequences (and trace their ancestry) back as far as
they can. There is an excellent website for doing so in mathematics (www .genealogy .math.ndsu.nodak. edu).
41. a) The union of two relations is the union of these sets. Thus R 1 U R 2 holds between two integers if R 1
holds or R2 holds (or both, it goes without saying). Thus (a, b) E R 1 U R 2 if and only if a= b (mod 3) or
a= b (mod 4). There is not a good easier way to state this, other than perhaps to say that a - b is a multiple
of either 3 or 4, or to work modulo 12 and write a - b 0, 3, 4, 6, 8, or 9 (mod 12).
b) The intersection of two relations is the intersection of these sets. Thus R 1 n R 2 holds between two integers
if R 1 holds and R 2 holds. Thus (a, b) E R 1 n R 2 if and only if a= b (mod 3) and a= b (mod 4). Since this
means that a - b is a multiple of both 3 and 4, and that happens if and only if a - b is a multiple of 12, we
=
can state this more simply as a= b (mod 12).
c) By definition R 1 - R 2 = R 1 n R 2 . Thus this relation holds between two integers if R 1 holds and R 2
does not hold. We can write this in symbols by saying that (a, b) E R 1 - R 2 if and only if a
b (mod 3)
and a-:;/:. b (mod 4). We could, if we wished, state this working modulo 12: (a,b) E R 1 - R 2 if and only if
a - b = 3, 6, or 9 (mod 12).
=
d) By definition R2 - R 1 = R2 n R1 . Thus this relation holds between two integers if R 2 holds and R 1
does not hold. We can write this in symbols by saying that (a, b) E R 2 - R 1 if and only if a = b (mod 4)
and a -:;/:. b (mod 3). We could, if we wished, state this working modulo 12: (a, b) E R 2 - R 1 if and only if
a - b 4 or 8 (mod 12) .
=
e) We know that R 1 tBR2 = (R1 - R 2) U (R2 -R1 ), so we look at our solutions to part (c) and part (d). Thus
this relation holds between two integers if R 1 holds and R2 does not hold, or vice versa. We can write this in
symbols by saying that (a, b) E R 1 ffi R 2 if and only if (a= b (mod 3) and a-:;/:. b (mod 4)) or (a= b (mod 4)
and a-:;/:. b (mod 3) ). We could, if we wished, state this working modulo 12: (a, b) E R1 ffi R 2 if and only if
a - b = 3, 4, 6, 8 or 9 (mod 12). We could also say that a - b is a multiple of 3 or 4 but not both.
43. A relation is just a subset. A subset can either contain a specified element or not; half of them do and half
of them do not. Therefore 8 of the 16 relations on {O, 1} contain the pair (0, 1). Alternatively, a relation on
{O, 1} containing the pair (0, 1) is just a set of the form {(O, 1)} U X, where XS: {(O, 0), (1, 0), (1, 1)}. Since
this latter set has 3 elements, it has 23 = 8 subsets.
45. This is similar to Example 16 in this section.
a) A relation on a set S with n elements is a subset of S x S. Since S x S has n 2 elements, we are asking
2
for the number of subsets of a set with n 2 elements, which is 2n • In our case n = 4, so the answer is
216 = 65,536.
b) In solving part (a), we had 16 binary choices to make-whether to include a pair (x,y) in the relation or
not as x and y ranged over the set {a, b, c, d}. In this part, one of those choices has been made for us: we
must include (a.a). We are free to make the other 15 choices. So the answer is 215
= 32,768.
See Exercise 47
for more problems of this type.
47. These are combinatorics problems, some harder than others. Let A be the set with n elements on which the
relations are defined.
a) To specify a symmetric relation, we need to decide, for each unordered pair {a, b} of distinct elements
of A, whether to include the pairs (a, b) and (b, a) or leave them out; this can be done in 2 ways for each
such unordered pair. Also, for each element a E A, we need to decide whether to include (a, a) or not, again
2 possibilities. We can think of these two parts as one by considering an element to be an unordered pair with
repetition allowed. Thus we need to make this 2-fold choice C(n + 1, 2) times, since there are C(n + 2 - 1, 2)
ways to choose an unordered pair with repetition allowed. Therefore the answer is 2°Cn+l, 2 ) = 2n(n+l)/ 2 .
Section 9.1
Relations and Their Properties
319
b) This is somewhat similar to part (a). For each unordered pair {a, b} of distinct elements of A, we have
a 3-way choice-either include (a, b) only, include (b, a) only, or include neither. For each element of A we
have a 2-way choice. Therefore the answer is 3C(n, 2 )2n = 3n(n-l)/ 2 2".
c) As in part (b) we have a 3-way choice for a -/=- b. There is no choice about including (a, a) in the
relation-the definition prohibits it. Therefore the answer is 3C(n, 2 ) = 3n(n-l)/ 2 .
d) For each ordered pair (a, b), with a-/=- b (and there are P(n, 2) such pairs), we can choose to include (a, b)
or to leave it out. There is no choice for pairs (a, a). Therefore the answer is 2P(n, 2 ) = 2n(n-l).
e) This is just like part (a), except that there is no choice about including (a, a). For each unordered pair of
distinct elements of A, we can choose to include neither or both of the corresponding ordered pairs. Therefore
the answer is 2C(n, 2 ) = 2n(n-lJ/ 2 .
f) We have complete freedom with the ordered pairs (a, b) with a -/=- b, so that part of the choice gives us
2P(n, 2 ) possibilities, just as in part (d). For the decision as to whether to include (a,a), two of the 2"
possibilities are prohibited: we cannot include all such pairs, and we cannot leave them all out. Therefore the
2
2
2
answer is 2P(n, 2 l(2" - 2) = 2" -"(2" - 2) = 2" - 2n -n+1.
49. The second sentence of the proof asks us to "take an element b E A such that (a, b) E R .'' There is no
guarantee that such an element exists for the taking. This is the only mistake in the proof. If one could
be guaranteed that each element in A is related to at least one element, then symmetry and transitivity
would indeed imply reflexivity. Without this assumption, however, the proof and the proposition are wrong.
As a simple example, take the relation 0 on any nonempty set. This relation is vacuously symmetric and
transitive, but not reflexive. Here is another counterexample: the relation {(1, 1), (1, 2), (2, 1), (2. 2)} on the
set {1,2,3}.
51. We need to show two things. First, we need to show that if a relation R is symmetric, then R = R- 1 , which
means we must show that R <:;;; R- 1 and R- 1 <:;;; R. To do this, let (a, b) E R. Since R is symmetric, this
implies that (b,a) ER. But since R- 1 consists of all pairs (a,b) such that (b,a) ER, this means that
(a, b) E R- 1 . Thus we have shown that R <:;;; R- 1 . Next let (a, b) E R- 1 . By definition this means that
(b,a) ER. Since R is symmetric, this implies that (a,b) ER as well. Thus we have shown that R- 1 <:;;; R.
Second we need to show that R = R- 1 implies that R is symmetric. To this end we let (a, b) E R
and try to show that (b, a) is also necessarily an element of R. Since (a, b) E R, the definition tells us that
(b, a) E R- 1 . But since we are under the hypothesis that R = R- 1 , this tells us that (b, a) E R, exactly as
desired.
53. Suppose that R is reflexive. We must show that R- 1 is reflexive, i.e., that (a,a) E R- 1 for each a EA. Now
since R is reflexive, we know that (a, a) ER for each a ER. By definition, this tells us that (a, a) E R- 1 , as
desired. (Interchanging the two a's in the pair (a, a) leaves it as it was.) Conversely, if R- 1 is reflexive, then
(a,a) E R- 1 for each a EA. By definition this means that (a,a) ER (again we interchanged the two a's).
55. We prove this by induction on n. The case n = 1 is trivial, since it is the statement R = R. Assume the
inductive hypothesis that R" = R. We must show that R"+l = R. By definition R"+ 1 = R" o R. Thus our
task is to show that R" o R <:;;; R and R <:;;; R" o R. The first uses the transitivity of R, as follows. Suppose
that (a, c) E R" o R. This means that there is an element b such that (a, b) E R and (b, c) E R". By the
inductive hypothesis, the latter statement implies that (b, c) E R. Thus by the transitivity of R, we know
that (a, c) ER, as desired.
Next assume that (a, b) E R. We must show that (a, b) E R" o R. By the inductive hypothesis, Rn = R,
and therefore R" is reflexive by assumption. Thus (b, b) ER". Since we have (a, b) ER and (b, b) ER", we
have by definition that (a, b) is an element of R" o R, exactly as desired. (The first half of this proof was not
really necessary, since Theorem 1 in this section already told us that R" <:;;; R for all n.)
Chapter 9
320
57. We use induction on n, the result being trivially true for n
=
Relations
1. Assume that Rn is reflexive; we must show
that Rn+I is reflexive. Let a EA, where A is the set on which R is defined. By definition Rn+I
= RnoR.
By
the inductive hypothesis, Rn is reflexive, so (a, a) E Rn. Also, since R is reflexive by assumption, (a, a) E R.
Therefore by the definition of composition, (a, a) E Rn o R, as desired.
59. It is not necessarily true that R 2 is irreflexive when R is. We might have pairs (a, b) and (b, a) both in R, with
a# b; then it would follow that (a, a) E R 2 , preventing R 2 from being irreflexive. As the simplest example,
let A= {1,2} and let R = {(1,2),(2,1)}. Then R is clearly irreflexive. In this case R 2 = {(1,1),(2,2)},
which is not irreflexive.
SECTION 9.2
n-ary Relations and Their Applications
This section is a brief introduction to relational models for data bases. The exercises are straightforward and
similar to the examples. Projections are formed by omitting certain columns, and then eliminating duplicate
rows. Joins are analogous to compositions of relations.
1. We simply need to find solutions of the inequality, which we can do by common sense. The set is { (1, 2, 3),
(1,2,4),(1,3,4), (2,3,4)}.
3. The 5-tuples are just the lines of the table. Thus the relation is
{(Nadir, 122, 34, Detroit. 08: 10), (Acme, 221, 22, Denver, 08: 17), (Acme, 122, 33, Anchorage, 08: 22),
(Acme, 323, 34, Honolulu, 08: 30), (Nadir, 199, 13, Detroit, 08: 47), (Acme, 222, 22, Denver, 09: 10),
(Nadir, 322, 34, Detroit, 09: 44)}.
5. We need to find a field that, when used along with the Airline field uniquely specifies a row of the table.
Certainly FlighLnumber is one such field, since there is only one line of the table for each pair (Airline,
FlighLnumber); no airline has the same flight number for two different flights. Gate and Destination do not
qualify, however, since Nadir has two flights leaving from Gate 34 going to Detroit. Finally, Departure_time is
a key by itself (no two flights leave at the same time), so it and Airline form a composite key as well.
7. a) A school would be giving itself a lot of headaches if it didn't make the student ID number different for each
student, so student ID is likely to be a primary key.
b) Name could very easily not be a primary key. Two people named Jennifer Johnson might easily both be
students.
c) Phone number is unlikely to be a primary key. Two roommates, or two siblings living at home, will likely
have the same phone number, and they might both be students at that school.
9. a) Everyone has a different Social Security number, so that field will be a primary key.
b) It is unlikely that (name, street address) will be a composite key. Somewhere in the United States there
could easily be two people named Jennifer Johnson both living at 123 Washington Street. In order for this to
work, there must never be two people with the same name who happen to have the same street address.
c) For this to work, we must never have two people with the same name living together. Given the size of the
country, one would doubt that this would work. For example, a husband and wife each named Morgan White
would make this not a composite key, as would a mother and daughter living at home with the same name.
11. The selection operator picks out all the tuples that match the criteria. The 5-tuples in Table 8 that have
Detroit as their destination are (Nadir, 122, 34, Detroit, 08: 10), (Nadir, 199, 13, Detroit, 08: 47), and (Nadir,
322, 34, Detroit, 09: 44).
Section 9.2
n-ary Relations and Their Applications
321
13. The selection operator picks out all the tuples that match the criteria. The 5-tuples in Table 8 that have
Nadir as their airline are (Nadir, 122, 34, Detroit, 08: 10), (Nadir, 199, 13, Detroit, 08: 47), and (Nadir, 322, 34,
Detroit, 09: 44). The 5-tuples in Table 8 that have Denver as their destination are (Acme, 221, 22, Denver,
08: 17) and (Acme, 222, 22, Denver, 09: 10). We need the union of these two lists: (Nadir, 122, 34, Detroit,
08: 10), (Nadir, 199, 13, Detroit, 08: 47), (Nadir, 322, 34, Detroit, 09: 44), (Acme, 221, 22, Denver, 08: 17), and
(Acme, 222, 22, Denver, 09: 10).
15. The subscripts on the projection mapping notation indicate which columns are to be retained. Thus if we
want to delete columns 1, 2, and 4 from a 6-tuple, we need to use the projection P 3 , 5 ,6 .
17. The table uses columns 1 and 4 of Table 8. We start by deleting columns 2, 3, and 5 from Table 8. At this
point, rows 5, 6 and 7 are duplicates of earlier rows, so they are omitted (rather than being listed twice).
Therefore the answer is as follows.
Airline
Destination
Nadir
Detroit
Acme
Denver
Acme
Anchorage
Acme
Honolulu
19. We need to find rows of Table 9 the last two entries of which are identical to the first two entries of rows of
Table 10. We combine each such pair of rows into one row of our new table. For instance, the last two entries
in the first row of Table 9 are 1092 and 1 . The first two entries in the second row of Table 10 are also 1092
and 1. Therefore we combine them into the row 23, 1092, 1, 2, 2 of our new table, whose columns represent
Supplier, Part_number, Project, Quantity, and Color_code. The new table consists of all pairs found in this
way.
Supplier
23
23
23
31
31
32
32
33
ParLnumber
1092
1101
9048
4975
3477
6984
9191
1001
Project
1
3
4
3
2
4
2
1
Quantity
2
1
12
6
25
10
80
14
Color_code
2
1
2
2
2
1
4
8
21. Both sides of this equation pick out the subset of R consisting of those n-tuples satisfying both conditions
C1 and C 2 . This follows immediately from the definition of the selection operator.
23. Both sides of this equation pick out the set of n-tuples that satisfy three conditions: they are in R, they
are in S, and they satisfy condition C. This follows immediately from the definitions of intersection and the
selection operator.
25. Both sides of this equation pick out the m-tuples consisting of ilh, i~h, ... , i~ components of n-tuples in
either R or S (or, of course, both). This follows immediately from the definitions of union and the projection
operator.
27. Note that we lose information when we delete columns. Therefore we might be taking something away when
we form the second set of m-tuples that might not have been taken away if all the original information is
Chapter 9
322
Relations
there (forming the first set of m-tuples). A simple example would be to let R = {(a,b)} and S = {(a,c)},
n = 2, m = 1, and i 1 = 1. Then R - S = R, so P 1 (R - S) = P 1 (R) = {(a)}. On the other hand,
P1(R) = P1(S) ={(a)}, so P1(R) - P1(S) = 0.
29. This is similar to Example 13.
a) Since two databases are listed in the "FROM" field, the first operation is to form the join of these two
databases, specifically the join J 2 of these two databases. We then apply the selection operator with the
condition "Quantity :".'.: 10 ." This join will have eight 5-tuples in it. Finally we want just the Supplier and
Project, so we are forming the projection Pl,3.
b) Four of the 5-tuples in the joined database have a quantity of no more than 10. The output, then, is the
set of the four 2-tuples corresponding to these fields: (23, 1), (23, 3), (31, 3), (32, 4).
31. A primary key is a domain whose value determines the values of all the other domains. For this relation, this
does not happen. The third domain (the modulus) is not a primary key, because, for example, 1
and 2
= 12
=11 (mod 10)
(mod 10), so the triples (1, 11, 10) and (2, 12, 10) are both in the relation. Knowing that the
third component of a triple is 10 does not tell us what the other two components are. Similarly, the triples
(1, 11. 10) and (1, 21. 10) are both in the relation, so the first domain is not a key; and the triples (1, 11, 10)
and ( 11, 11, 10) are both in the relation, so the second domain is not a key.
SECTION 9.3
Representing Relations
l\Jatrices and directed graphs provide useful ways for computers and humans to represent relations and manipulate them. Become familiar with working with these representations and the operations on them (especially
the matrix operation for forming composition) by working these exercises. Some of these exercises explore
how properties of a relation can be found from these representations.
1. In each case we use a 3 x 3 matrix, putting a 1 in position (i, j) if the pair (i, j) is in the relation and a 0 in
position (i,j) if the pair (i,j) is not in the relation. For instance, in part (a) there are l's in the first row,
since each of the pairs (1.1), (1, 2), and (1, 3) are in the relation, and there are O's elsewhere.
a)
b)
c)
d)
3. a) Since the (1,l)th entry is a 1, (1,1) is in the relation. Since (1,2)th entry is a 0, (1,2) is not in the
relation. Continuing in this manner, we see that the relation contains (1, 1), (1, 3), (2, 2), (3, 1), and (3, 3).
b) (1,2). (2,2), and (3,2)
c) (1,1), (1.2), (1,3), (2,1), (2,3), (3,1), (3,2), and (3,3)
5. An irreflexive relation (see the preamble to Exercise 11 in Section 9.1) is one in which no element is related
to itself. In the matrix, this means that there are no 1 's on the main diagonal (position mii for some i).
Equivalently, the relation is irreflexive if and only if every entry on the main diagonal of the matrix is 0.
7. For reflexivity we want all 1's on the main diagonal; for irreflexivity we want all O's on the main diagonal; for
symmetry, we want the matrix to be symmetric about the main diagonal (equivalently, the matrix equals its
transpose); for antisymmetry we want there never to be two l's symmetrically placed about the main diagonal
(equivalently, the meet of the matrix and its transpose has no l's off the main diagonal); and for transitivity
we want the Boolean square of the matrix (the Boolean product of the matrix and itself) to be "less than or
equal to" the original matrix in the sense that there is a 1 in the original matrix at every location where there
is a 1 in the Boolean square.
Section 9.3
Representing Relations
323
a) Since there are all l's on the main diagonal, this relation is reflexive and not irreflexive. Since the matrix is
symmetric, the relation is symmetric. The relation is not antisymmetric~look at positions (1, 3) and (3, 1).
Finally, the Boolean square of this matrix is itself, so the relation is transitive.
b) Since there are both O's and l's on the main diagonal, this relation is neither reflexive nor irreflexive. Since
the matrix is not symmetric, the relation is not symmetric (look at positions (1, 2) and (2, 1), for example).
The relation is antisymmetric since there are never two l's symmetrically placed with respect to the main
diagonal. Finally, the Boolean square of this matrix is itself, so the relation is transitive.
c) Since there are both O's and l's on the main diagonal, this relation is neither reflexive nor irreflexive. Since
the matrix is symmetric, the relation is symmetric. The relation is not antisymmetric~look at positions (1, 3)
and (3, 1), for example. Finally, the Boolean square of this matrix is the matrix with all l's, so the relation
is not transitive ( 1 is related to 2, and 2 is related to 1, but 2 is not related to 2).
9. Note that the total number of entries in the matrix is 100 2 = 10,000.
a) There is a 1 in the matrix for each pair of distinct positive integers not exceeding 100, namely in position
(a, b) where a > b. Thus the answer is the number of subsets of size 2 from a set of 100 elements, i.e.,
C(lOO, 2) = 4950.
b) There is a 1 in the matrix at each position except the 100 positions on the main diagonal. Therefore the
answer is 100 2 - 100 = 9900.
c) There is a 1 in the matrix at each entry just below the main diagonal (i.e., in positions (2, 1), (3, 2), ... ,
(100, 99). Therefore the answer is 99.
d) The entire first row of this matrix corresponds to a= 1. Therefore the matrix has 100 nonzero entries.
e) This relation has only the one element (1, 1) in it, so the matrix has just one nonzero entry.
11. Since the relation R is the relation that contains the pair (a, b) (where a and b are elements of the appropriate
sets) if and only if R does not contain that pair, we can form the matrix for R simply by changing all the l's
to O's and O's to l's in the matrix for R.
13. Exercise 12 tells us how to do part (a) (we take the transpose of the given matrix MR, which in this case
happens to be the matrix itself). Exercise 11 tells us how to do part (b) (we change l's to O's and O's to l's
in MR)· For part (c) we take the Boolean product of MR with itself.
a)
b)
c)
[: : :l
15. We compute the Boolean powers of MR; thus MR2
MR4 = M~l = MR8M~l.
a)
[! : ~l
b)
[~ ; :J
c)
= M~J =
MR 8 MR, MRa
= M~l =
MR 8 M~J, and
[: ;:l
17. The matrix for the complement has a 1 wherever the matrix for the relation has a 0, and vice versa. Therefore
the number of nonzero entries in MR is n 2 - k, since these matrices have n rows and n columns.
19. In each case we need a vertex for each of the elements, and we put in a directed edge from x to y if there
is a 1 in position (x, y) of the matrix. For simplicity we have indicated pairs of edges between the same two
vertices in opposite directions by using a double arrowhead, rather than drawing two separate lines.
Chapter 9
324
1~4
2~3
(a)
1~4
2
0
3
(b)
0
Relations
1V14
2~3 2~3
(c)
(d)
21. In each case we need a vertex for each of the elements, and we put in a directed edge from x to y if there
is a 1 in position (x, y) of the matrix. For simplicity we have indicated pairs of edges between the same two
vertices in opposite directions by using a double arrowhead, rather than drawing two separate lines.
1v2
3~4
(a)
(c)
23. We list all the pairs (x, y) for which there is an edge from x to y in the directed graph:
{(a, b), (a, c), (b, c), (c, b)}.
25. We list all the pairs (x, y) for which there is an edge from x to y in the directed graph:
{(a,c),(b,a),(c,d), (d,b)}.
27. We list all the pairs (x, y) for which there is an edge from x to y in the directed graph:
{(a,a), (a,b), (a,c), (b,a), (b,b), (b,c),(c,a), (c,b),(d,d)}.
29. An asymmetric relation is one for which it never happens that a is related to b and simultaneously b is related
to a, even when a = b. In terms of the directed graph, this means that we must see no loops and no closed
paths of length 2 (i.e., no pairs of edges between two vertices going in opposite directions).
31. Recall that the relation is reflexive if there is a loop at each vertex; irreflexive if there are no loops at all;
symmetric if edges appear only in anti parallel pairs (edges from one vertex to a second vertex and from the
second back to the first); antisymmetric if there is no pair of antiparallel edges; and transitive if all paths
of length 2 (a pair of edges (x, y) and (y, z)) are accompanied by the corresponding path of length 1 (the
edge (x, z) ). The relation drawn in Exercise 23 is not reflexive but is irreflexive since there are no loops. It is
not symmetric, since, for instance, the edge (a, b) is present but not the edge (b, a). It is not antisymmetric,
since both edges (b, c) and (c, b) are present. It is not transitive, since the path (b, c), (c, b) from b to b is
not accompanied by the edge (b, b). The relation drawn in Exercise 24 is reflexive and not irreflexive since
there is a loop at each vertex. It is not symmetric, since, for instance, the edge (b, a) is present but not the
edge (a, b). It is antisymmetric, since there are no pairs of anti parallel edges. It is transitive, since the only
nontrivial path of length 2 is bac, and the edge (b, c) is present. The relation drawn in Exercise 25 is not
reflexive but is irreflexive since there are no loops. It is not symmetric, since, for instance, the edge (b, a) is
present but not the edge (a, b). It is antisymmetric, since there are no pairs of anti parallel edges. It is not
transitive, since the edges (a, c) and (c, d) are present, but not (a, d).
33. Since the inverse relation consists of all pairs (b, a) for which (a, b) is in the original relation, we just have to
take the digraph for R and reverse the direction on every edge.
Section 9.4
Closures of Relations
325
35. We prove this statement by induction on n. The basis step n = 1 is tautologically true, since M~l =MR.
Assume the inductive hypothesis that M~l is the matrix representing Rn. Now M~+l] =MR 8 M~l. By
the inductive hypothesis and the assertion made before Example 5, that MsoR = MR 8 Ms, the right-hand
side is the matrix representing Rn o R. But Rn o R = Rn+l , so our proof is complete.
SECTION 9.4
Closures of Relations
This section is harder than the previous ones in this chapter. Warshall's algorithm, in particular, is fairly
tricky, and Exercise 27 should be worked carefully, following Example 8. It is easy to forget to include the
loops (a, a) when forming transitive closures "by hand."
1. a) The reflexive closure of R is R together with all the pairs (a, a). Thus in this case the closure of R is
{(0,0),(0,1),(1, 1),(1,2),(2,0),(2,2),(3,0),(3,3)}.
b) The symmetric closure of R is R together with all the pairs (b, a) for which (a, b) is in R. For example,
since (1, 2) is in R, we need to add (2, 1). Thus the closure of R is { (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2),
(2,0),(2,1), (2,2), (3,0)}.
3. To form the symmetric closure we need to add all the pairs (b, a) such that (a, b) is in R. In this case, that
means that we need to include pairs (b, a) such that a divides b, which is equivalent to saying that we need
to include all the pairs (a, b) such that b divides a. Thus the closure is { (a, b) I a divides b or b divides a } .
5. We form the reflexive closure by taking the given directed graph and appending loops at all vertices at which
there are not already loops.
7. We form the reflexive closure by taking the given directed graph and appending loops at all vertices at which
there are not already loops.
9. We form the symmetric closure by taking the given directed graph and appending an edge pointing in the
opposite direction for every edge already in the directed graph (unless it is already there); in other words, we
append the edge (b,a) whenever we see the edge (a,b). We have labeled the figures below (a), (b), and (c),
corresponding to Exercises 5, 6, and 7, respectively.
c
(a)
d
c
(b)
(c)
326
Chapter 9
Relations
11. We are asked for the symmetric and reflexive closure of the given relation. We form it by taking the given
directed graph and appending (1) a loop at each vertex at which there is not already a loop and (2) an edge
pointing in the opposite direction for every edge already in the directed graph (unless it is already there). We
have labeled the figures below (a), (b), and (c), corresponding to Exercises 5, 6, and 7, respectively.
(a)
(b)
(c)
13. The symmetric closure of R is RuR- 1 . The matrix for R- 1 is
Mk, as we saw in Exercise 12 in Section 9.3.
The matrix for the union of two relations is the join of the matrices for the two relations, as we saw in
Section 9.3. Therefore the matrix representing the symmetric closure of R is indeed MR V
Mk.
15. If R is already irreflexive, then it is clearly its own irreflexive closure. On the other hand if R is not irreflexive,
then there is no relation containing R that is irreflexive, since the loop or loops in R prevent any such relation
from being irreflexive. Thus in this case R has no irreflexive closure. This exercise shows essentially that the
concept of "irreflexive closure" is rather useless, since no relation has one unless it is already irreflexive (in
which case it is its own ''irreflexive closure").
17. A circuit of length 3 can be written as a sequence of 4 vertices, each joined to the next by an edge of the
given directed graph, ending at the same vertex at which it began. There are several such circuits here, and
we just have to be careful and systematically list them all. There are the circuits formed entirely by the loops:
aaaa, cccc, and eeee. The triangles abea and adea also qualify. Two circuits start at b: bccb and beab.
There are two more circuits starting at c, namely ccbc and cbcc. Similarly there are the circuits deed, eede
and edee, as well as the other trips around the triangle: eabe, dead, and eade.
19. The way to form these powers is first to form the matrix representing R, namely
MR=
r~ ~ ~ ~ ~] ,
1 0 0
0 1 0
and then take successive Boolean powers of it to get the matrices representing R 2 , R 3 , and so on. Finally, for
part ( f) we take the join of the matrices representing R, R 2 , ... , R 5 . Since the matrix is a perfectly good
way to express the relation, we will not list the ordered pairs.
a) The matrix for R 2 is the Boolean product of the matrix displayed above with itself, namely
0
0
1
1
MR'~ M~I ~ rl ~ ~ ~ !1
b) The matrix for R 3 is the Boolean product of the first matrix displayed above with the answer to part (a),
namely
Mn,~ M~I ~ [i
1 1
0 0
1 1
1 1 1
0 1 0
1
0
0
~]
Section 9.4
Closures of Relations
327
c) The matrix for R 4 is the Boolean product of the first matrix displayed above with the answer to part (b),
namely
1 0 1
1 1 1
MR4 = M~l = 1 1 1 1 1
[1 0 1 1 1
1 1 1 1 1
d) The matrix for R 5 is the Boolean product of the first matrix displayed above with the answer to part ( c),
namely
~ ~1
MR
5
=
M~l [i ~ i i ill ·
=
1
1
1
1
6
e) The matrix for R is the Boolean product of the first matrix displayed above with the answer to part ( d),
namely
1
1
1
1
!l
1
f) The matrix for R* is the join of the first matrix displayed above and the answers to parts (a) through ( d),
namely
1 1 1
1 1 1
1 1 1
MR•
1 1 1
1 1 1
~MRVM~l VM~l VM~l VM~] ~ [j
j]
21. a) The pair (a, b) is in R 2 if there is a person c other than a or b who is in a class with a and a class with b.
Note that it is almost certain that (a, a) is in R 2 , since as long as a is taking a class that has at least one
other person in it, that person serves as the "c."
b) The pair (a, b) is in R 3 if there are persons c (different from a) and d (different from b and c) such that
c is in a class with a, c is in a class with d, and d is in a class with b.
c) The pair (a, b) is in R* if there is a sequence of persons, c0 , c 1 , c2 , ... , Cn, with n 2: 1, such that
Cn = b, and for each i from 1 ton, Ci-l f. Ci and Ci-l is in at least one class with Ci·
co =
a,
23. Suppose that (a, b) E R*; then there is a path from a to b in (the digraph for) R. Given such a path, if R
is symmetric, then the reverse of every edge in the path is also in R; therefore there is a path from b to a
in R (following the given path backwards). This means that (b, a) is in R* whenever (a, b) is, exactly what
we needed to prove.
25. Algorithm 1 finds the transitive closure by computing the successive powers and taking their join. We exhibit
our answers in matrix form as MR V M~J V ... V M~J = MR· .
a)
[~
b)
[l
1
0
0
0
0
0
0
0
0
1
0
0
0
1
0
1
!l [i
v
~] v [l
0
1
0
1
0
0
0
0
1
0
0
0
0
0
1
0
~] v [~
~] v [l
1
0
1
0
0
0
0
0
0
1
0
1
0
1
0
1
~] [~
~] [l
v
v
0
1
0
1
0
0
0
0
1
0
1
0
0
0
1
0
~] [l
~] [l
1 1
1 1
1 1
1 1
0 0
0 1
0 1
0
1
l
Il
328
Chapter 9
c)
[O0 0I 1I l1 l v [O0 00 0I
0
0
0
0
0
0
1
0
0
0
0
0
0
0
Relations
l I]
l
[o
o
o
l
[o
o
o
ol
[o
ovoooovoooo
1
1
0 0 0 01
0000
=
0
0000
0000
0
0
0
0
0
0
[j
1
1
1
0
0
1
1
0
Note that the relation was already transitive, so its transitive closure is itself.
d)
[l
0
0
0
1
1 0
1 0
]
v
[j
1 0
1 0
1 1
0 1
1
1
1
1
}[j j]v [j
1
1
1
0
1
1
1
1
1
1
1
1
il
1
1
1 1
1 1
!l
27. In Warshall's algorithm (Algorithm 2 in this section), we compute a sequence of matrices W 0 (the matrix
representing R), W 1 , W 2 , ... , W n, the last of which represents the transitive closure of R. Each matrix
Wk comes from the matrix Wk-1 in the following way. The (i,j)th entry of Wk is the "V" of the (i,j)th
entryofWk-l with the"/\'' ofthe (i,k)th entry and the (k,j)th entryofWk-l· Wewillexhibitoursolution
by listing the matrices W 0 , W 1 , W 2 , W 3 , W 4 , in that order; W 4 represents the answer. In each case
W o is the matrix of the given relation. To compute the next matrix in the solution, we need to compute it
one entry at a time, using the equation just discussed (the "V" of the corresponding entry in the previous
matrix with the "/\ '' of two entries in the old matrix), i.e., as i and j each go from 1 to 4, we need to write
down the (i, j)th entry using this formula. Note that in computing Wk the kth row and the kth column are
unchanged, but some of the entries in other rows and columns may change.
a)
b)
c)
1
1
0 1
1
0
0 0
0 0
0 0
0 0
1
1
~] [~
Note that the relation was already transitive, so each matrix in the sequence was the same.
d)
0 0
1
1
1
1
1
1
ll
[j
1
1
1
1
1
1
1
1
ll
29. a) We need to include at least the transitive closure, which we can compute by Algorithm 1 or Algorithm 2 to
be (in matrix form)
[H H] .
All we need in addition is the pair (2, 2) in order to make the relation
1 1 0 1
reflexive. Note that the result is still transitive (the addition of a pair (a, a) cannot make a transitive relation
no longer transitive), so our answer is
[~
~ ~ ~i ·
0 0 1 0
1 1 0 1
Section 9.5
Equivalence Relations
329
1
b) The symmetric closure of the original relation is represented by
0011
[
tnu.,itive closure of this 'elation, namely [
~ ~ ~ ~]
0
0
0
~ ~]
· We need at le"-'t the
· Since it is aL'K> eymmctdc, we file done. Note
that it would not have been correct to find first the transitive closure of the original matrix and then make
it symmetric, since the pair (2, 2) would be missing. What is going on here is that the transitive closure
of a symmetric relation is still symmetric, but the symmetric closure of a transitive relation might not be
transitive.
c) Since the answer to part (b) was already reflexive, it must be the answer to this part as well.
31. Algorithm 1 has a loop executed O(n) times in which the primary operation is the Boolean product compu-
tation (the join operation is fast by comparison). If we can do the product in O(n 2 ·8 ) bit operations, then
the number of bit operations in the entire algorithm is O(n · n 2 · 8 ) = O(n 3 ·8 ). Since Algorithm 2 does not use
the Boolean product, a fast Boolean product algorithm is irrelevant, so Algorithm 2 still requires O(n 3 ) bit
operations.
33. There are two ways to go. One approach is to take the output of Algorithm 1 as it stands and then make sure
that all the pairs (a, a) are included by forming the join with the identity matrix (specifically set B := B VIn).
See the discussion in Exercise 29a for the justification. The other approach is to insure the reflexivity at the
beginning by initializing A := Mr V In; if we do this, then only paths of length strictly less than n need to
be looked at, so we can change the n in the loop to n - 1.
35. a) No relation that contains R is not reflexive, since R already contains all the pairs (0, 0), (1, 1), and (2, 2).
Therefore there is no "nonreflexive" closure of R.
b) Suppose S were the closure of R with respect to this property. Since R does not have an odd number
of elements, S =/. R, so S must be a proper superset of R. Clearly S cannot have more than 5 elements,
for if it did, then any subset of S consisting of R and one element of S - R would be a proper subset of
S with the property; this would violate the requirement that S be a subset of every superset of R with the
property. Thus S must have exactly 5 elements. Let T be another superset of R with 5 elements (there
are 9 - 4 = 5 such sets in all). Thus T has the property, but S is not a subset of T. This contradicts the
definition. Therefore our original assumption was faulty, and the closure does not exist.
SECTION 9.5 Equivalence Relations
This section is extremely important. If you do nothing else, do Exercise 9 and understand it, for it deals with
the most common instances of equivalence relations. (See the comments in our solution below for some added
insight.) Exercise 16 is interesting-it hints at what fractions really are (if understood properly) and perhaps
helps to explain why children (and adults) usually have so much trouble with fractions: they really involve
equivalence relations. Spend some time thinking about fractions in this context. (See also Writing Project 4
for this chapter.)
It is usually easier to understand equivalence relations in terms of the associated partition-it's a more
concrete visual image. Thus make sure you understand exactly what Theorem 2 says. Look at Exercise 67 for
the relationship between equivalence relations and closures.
Chapter 9
330
Relations
1. In each case we need to check for reflexivity, symmetry, and transitivity.
a) This is an equivalence relation; it is easily seen to have all three properties. The equivalence classes all
have just one element.
b) This relation is not reflexive since the pair (1, 1) is missing. It is also not transitive, since the pairs (0, 2)
and (2, 3) are there, but not (0, 3).
c) This is an equivalence relation. The elements 1 and 2 are in the same equivalence class; 0 and 3 are each
in their own equivalence class.
d) This relation is reflexive and symmetric, but it is not transitive. The pairs (1, 3) and (3, 2) are present,
but not (1, 2).
e) This relation would be an equivalence relation were the pair (2, 1) present. As it is, its absence makes the
relation neither symmetric nor transitive.
3. As in Exercise 1, we need to check for reflexivity, symmetry, and transitivity.
a) This is an equivalence relation, one of the general form that two things are considered equivalent if they
have the same "something" (see Exercise 9 for a formalization of this idea). In this case the "something" is
the value at 1.
b) This is not an equivalence relation because it is not transitive. Let f(x) = 0, g(x) = x, and h(x) = 1
for all x E Z. Then f is related tog since f(O) = g(O), and g is related to h since g(l) = h(l), but f is
not related to h since they have no values in common. By inspection we see that this relation is reflexive and
symmetric.
= 0 =f. 1. It is not
symmetric, since if j(x)- g(x) = 1, then g(x)- f(x) = -1=f.1. It is not transitive, since if f(x)- g(x) = 1
c) This relation has none of the three properties. It is not reflexive, since f(x) - f(x)
and g(x) - h(x) = 1, then f(x) - h(x) = 2 =f. 1.
d) This is an equivalence relation. Two functions are related here if they differ by a constant. It is clearly
reflexive (the constant is 0). It is symmetric, since if f (x) - g( x) = C, then g( x) - f (x) = -C. It is transitive,
since if f(x) - g(x) = C 1 and g(x) - h(x) = C 2 , then f(x) - h(x) = C 3 , where C3 = C1 + C 2 (add the first
two equations).
e) This relation is not reflexive, since there are lots of functions f (for instance, f(x) = x) that do not have
the property that f(O) = f(l). It is symmetric by inspection (the roles of f and g are the same). It is not
transitive. For instance, let f(O) = g(l) = h(O) = 7, and let f(l) = g(O) = h(l) = 3; fill in the remaining
values arbitrarily. Then f and g are related, as are g and h, but f is not related to h since 7 =f. 3.
5. Obviously there are many possible answers here. We can say that two buildings are equivalent if they were
opened during the same year; an equivalence class consists of the set of buildings opened in a given year (as
long as there was at least one building opened that year). For another example, we can define two buildings to
be equivalent if they have the same number of stories; the equivalence classes are the set of 1-story buildings,
the set of 2-story buildings, and so on (one class for each n for which there is at least one n-story building).
In our third example, partition the set of all buildings into two classes~those in which you do have a class this
semester and those in which you don't. (We assume that each of these is nonempty.) Every building in which
you have a class is equivalent to every building in which you have a class (including itself), and every building
in which you don't have a class is equivalent to every building in which you don't have a class (including
itself).
7. Two propositions are equivalent if their truth tables are identical. This relation is reflexive, since the truth
table of a proposition is identical to itself. It is symmetric, since if p and q have the same truth table, then
q and p have the same truth table. There is one technical point about transitivity that should be noted. We
need to assume that the truth tables, as we consider them for three propositions p, q, and r, have the same
Section 9.5
331
Equivalence Relations
atomic variables in them. If we make this assumption (and it cannot hurt to do so, since adding information
about extra variables that do not appear in a pair of propositions does not change the truth value of the
propositions), then we argue in the usual way: if p and q have identical truth tables, and if q and r have
identical truth tables, then p and r have that same common truth table. The proposition T is always true;
therefore the equivalence class for this proposition consists of all propositions that are always true, no matter
what truth values the atomic variables have. Recall that we call such a proposition a tautology. Therefore
the equivalence class of T is the set of all tautologies. Similarly. the equivalence class of F is the set of all
contradictions.
9. This is an important exercise, since very many equivalence relations are of this form. (In fact, all of them
are-see Exercise 10. A relation defined by a condition of the form '' x and y are equivalent if and only if
they have the same ... '' is an equivalence relation. The function f here tells what about x and y are "the
same.")
a) This relation is reflexive, since obviously f(x) = f(x) for all x EA. It is symmetric, since if f(x) = f(y),
then f(y) = J(x) (this is one of the fundamental properties of equality). It is transitive, since if f(x) = J(y)
and J(y)
= f(z),
then J(x)
= f(z)
(this is another fundamental property of equality).
b) The equivalence class of xis the set of ally EA such that f(y) = f(x). This is by definition just the
inverse image of f(x). Thus the equivalence classes are precisely the sets 1- 1 (b) for every bin the range
off.
11. This follows from Exercise 9, where
bits.
f is the function that takes a bit string of length 3 or more to its first 3
f is the function that takes a bit string of length 3 or more to the ordered
pair (b 1 , b3 ), where b1 is the first bit of the string and b3 is the third bit of the string. Two bit strings agree on
their first and third bits if and only if the corresponding ordered pairs for these two strings are equal ordered
pairs.
13. This follows from Exercise 9, where
15. By algebra, the given condition is the same as the condition that f ( (a, b)) = f ( (c, d)), where f ( (x, y)) = x -y.
Therefore by Exercise 9 this is an equivalence relation. If we want a more explicit proof, we can argue as
follows. For reflexivity, ( (a, b), (a, b)) E R because a + b = b + a. For symmetry, ( (a, b), (c, d)) E R if and
only if a+ d = b + c, which is equivalent to c + b = d +a, which is true if and only if ((c, d), (a, b)) ER. For
transitivity, suppose ( (a, b), (c, d)) E R and ( (c, d), (e, f)) E R. Thus we have a+ d = b + c and c + e = d + f.
Adding, we obtain a + d + c + e = b + c + d + f. Simplifying, we have a + e = b + f, which tells us that
((a, b), (e, f)) ER.
17. a) This follows from Exercise 9, where the function f from the set of differentiable functions (from R to R)
to the set of functions (from R to R) is the differentiation operator-i.e., f of a function g is the function
g' . The best way to think about this is that any relation defined by a statement of the form '"a and b are
equivalent if they have the same whatever" is an equivalence relation. Here "whatever" is ''derivative"; in the
general situation of Exercise 9, "whatever" is "function value under f ."
b) We are asking for all functions that have the same derivative that the function f(x) = x 2 has, i.e., all
functions of x whose derivative is 2x. In other words, we are asking for the general antiderivative of 2x, and
we know that J 2x = x 2 + C, where C is any constant. Therefore the functions in the same equivalence class
as J(x) = x 2 are all the functions of the form g(x) = x 2 + C for some constant C. Indefinite integrals in
calculus, then, give equivalence classes of functions as answers, not just functions.
19. This follows from Exercise 9, where the function f from the set of all URLs to the set of all Web pages is the
function that assigns to each URL the Web page for that URL.
Chapter 9
332
Relations
21. We need to observe whether the relation is reflexive (there is a loop at each vertex), symmetric (every edge
that appears is accompanied by its antiparallel mate-an edge involving the same two vertices but pointing
in the opposite direction), and transitive (paths of length 2 are accompanied by the path of length 1-i.e.,
edge-between the same two vertices in the same direction). We see that this relation is not transitive, since
the edges (c, d) and (d, c) are missing.
23. As in Exercise 21, this relation is not transitive, since several required edges are missing (such as (a, c) ).
25. This follows from Exercise 9, with
f
being the function from bit strings to nonnegative integers given by
f(s) =the number of l's ins.
27. Only parts (a) and (b) are relevant here, since the others are not equivalence relations.
a) An equivalence class is the set of all people who are the same age. (To really identify the equivalence class
and the equivalence relation itself, one would need to specify exactly what one meant by "the same age." For
example, we could define two people to be the same age if their official dates of birth were identical. In that
case, everybody born on April 25, 1948, for example, would constitute one equivalence class.)
b) For each pair (m, f) of a man and a woman, the set of offspring of their union, if nonempty, is an equivalence
class. In many cases, then, an equivalence class consists of all the children in a nuclear family with children.
(In real life, of course, this is complicated by such things as divorce and remarriage.)
29. The equivalence class of 011 is the set of all bit strings that are related to 011, namely the set of all bit strings
that have the same number of l's as 011. In other words, it is the (infinite) set of all bit strings with exactly
2 l's: {11, 110, 101, 011, 1100, 1010, 1001, ... }.
31. Since two strings are related if they agree beyond their first 3 bits, the equivalence class of a bit string xyzt,
where x, y, and z are bits, and t is a bit string, is the set of all bit strings of the form x' y' z't, where x', y',
and z' are any bits.
a) the set of all bit strings of length 3 (take t = ,\ in the formulation given above)
b) the set of all bit strings of length 4 that end with a 1
c) the set of all bit strings of length 5 that end 11
d) the set of all bit strings of length 8 that end 10101
33. This is like Example 15. Each bit string of length less than 4 is in an equivalence class by itself ( [A]R4 =
{A}, [O]R 4 = {0}, [l]R4 = {l}, [00]R4 = {00}, [Ol]R4 = {01}, ... , [lll]R4 = {111}). This accounts for
1 + 2 + 4 + 8 = 15 equivalence classes. The remaining 16 equivalence classes are determined by the bit strings
of length 4:
[0000] R 4 = {0000, 00000, 00001, 000000, 000001, 000010, 000011, 0000000, ... }
[0001]R 4
= {0001, 00010, 00011, 000100, 000101, 000110, 000111, 0001000, ... }
[0010]R 4 = {0010, 00100, 00101, 001000, 001001, 001010, 001011, 0010000, ... }
[llll]R, = {llll,11110,lllll,111100,111101,111110,llllll,1111000, ... }
=
35. We have by definition that [n]s = { i I i
n (mod 5) } .
a) [2]s = { i I i = 2 (mod 5)} = {... , -8, -3, 2, 7, 12, ... }
b) [3]s
= {i Ii= 3
c) [6]s = {
(mod 5)} = { .. .,-7,-2,3,8, 13, ... }
i Ii= 6 (mod 5)} = { ... , -9, -4, 1, 6, 11, ... }
Ii= -3 (mod 5)} = {... , -8, -3, 2, 7, 12, ... }
d) [-3]s = { i
(the same as [2]s)
Section 9.5
Equivalence Relations
333
37. This is very similar to Example 14. There are 6 equivalence classes, namely
[0]6 ={ ... ' -12, -6, 0, 6, 12, ... } '
[1]6 ={ ... '-11, -5, 1, 7, 13, ... } '
[2]6 ={ ... ' -10, -4, 2, 8, 14, ... } '
[3]6 ={ ... '-9, -3, 3, 9, 15, ... } '
[4]6 ={ ... ' -8, -2,4, 10, 16, ... } '
[5]6 ={ ... '-7, -1, 5, 11, 17, ... } .
Another way to describe this collection is to say that it is the collection of sets { 6n
k = 0,1,2,3,4,5.
+k
I n E Z } for
39. a) We observed in the solution to Exercise 15 that (a, b) is equivalent to (c, d) if a - b = c - d. Thus because
1-2 = -1, we have [(1,2)] = {(a,b) I a-b= -1} = {(1,2),(3,4),(4,5),(5,6), ... }.
b) Since the equivalence class of (a, b) is entirely determined by the integer a - b, which can be negative,
positive, or zero, we can interpret the equivalences classes as being the integers. This is a standard way to
define the integers once we have defined the whole numbers.
41. The sets in a partition must be nonempty, pairwise disjoint, and have as their union all of the underlying set.
a) This is not a partition, since the sets are not pairwise disjoint (the elements 2 and 4 each appear in two
of the sets) .
b) This is a partition.
c) This is a partition.
d) This is not a partition, since none of the sets includes the element 3.
43. In each case, we need to see that the collection of subsets satisfy three conditions: they are nonempty, they
are pairwise disjoint, and their union is the entire set of 256 bit strings of length 8.
a) This is a partition, since strings must begin either 1 or 0, and those that begin 0 must continue with either
0 or 1 in their second position. It is clear that the three subsets satisfy the conditions.
b) This is not a partition, since these subsets are not pairwise disjoint. The string 00000001, for example,
contains both 00 and 01.
c) This is clearly a partition. Each of these four subsets contains 64 bit strings, and no two of them overlap.
d) This is not a partition, because the union of these subsets is not the entire set. For example, the string
00000010 is in none of the subsets.
e) This is a partition. Each bit string contains some number of l's. This number can be identified in exactly
one way as of the form 3k, the form 3k + 1, or the form 3k + 2, where k is a nonnegative integer; it really is
just looking at the equivalence classes of the number of l's modulo 3.
45. In each case, we need to see that the collection of subsets satisfy three conditions: they are nonempty, they
are pairwise disjoint, and their union is the entire set Z x Z.
a) This is not a partition, since the subsets are not pairwise disjoint. The pair (2, 3), for example, is in both
of the first two subsets listed.
b) This is a partition. Every pair satisfies exactly one of the conditions listed about the parity of x and y,
and clearly these subsets are nonempty.
c) This is not a partition, since the subsets are not pairwise disjoint. The pair (2, 3), for example, is in both
of the first two subsets listed. Also, (0, 0) is in none of the subsets.
d) This is a partition. Every pair satisfies exactly one of the conditions listed about the divisibility of x and
y by 3, and clearly these subsets are nonempty.
334
Chapter 9
Relations
e) This is a partition. Every pair satisfies exactly one of the conditions listed about the positiveness of x and
y, and clearly these subsets are nonempty.
t) This is not a partition, because the union of these subsets is not all of Z x Z. In particular, (0, 0) is in
none of the parts.
47. In each case, we need to list all the pairs we can where both coordinates are chosen from the same subset. We
should proceed in an organized fashion, listing all the pairs corresponding to each part of the partition.
a) {(0,0), (1, 1),(1,2),(2. 1), (2,2), (3,3), (3,4), (3,5), (4,3),(4,4),(4,5), (5,3),(5,4), (5,5)}
{(0,0), (0, 1), (1,0), (1, 1), (2,2),(2,3),(3,2), (3,3),(4,4),(4,5), (5,4),(5,5)}
c) {(0,0), (0, 1), (0,2), (1,0), (1, 1), (1,2), (2,0), (2, 1), (2,2), (3,3), (3,4), (3,5), (4,3), (4,4), (4,5), (5,3),
b)
(5, 4), (5, 5)}
d)
{(0,0), (1, 1), (2,2),(3,3),(4,4),(5,5)}
49. We need to show that every equivalence class modulo 6 is contained in an equivalence class modulo 3. We claim
that in fact, for each n E Z, [n]6 (,;:; [nh. To see this suppose that m E [n]6. This means that m = n (mod 6),
i.e., that m - n is a multiple of 6. Then perforce m - n is a multiple of 3, so m = n (mod 3), which means
that m E [nh.
51. By the definition given in the preamble to Exercise 49, we need to show that every set in the first partition is
a subset of some set in the second partition. Let A be a set in the first partition. So A is the set of all bit
strings of length 16 that agree on their last eight bits. Pick a particular element x of A, and suppose that the
last four bits of x are abed. Then the set of all bit strings of length 16 whose last four bits are abed is one
of the sets in the second partition, and clearly every string in A is in that set, since every string in A agrees
with x on the last eight bits, and therefore perforce agrees on the last four bits.
53. We are asked to show that every equivalence class for R 31 is a subset of some equivalence class for Rs. Let
[x]R 31 be an arbitrary equivalence class for R31. We claim that [x]R31 <:;; [x]R8 ; proving this claim finishes the
proof. To show that one set is a subset of another set, we choose an arbitrary element y in the first set and
show that it is also an element of the second set. In this case since y E [x]R 3 , , we know that y is equivalent
to x under R 31 , that is, that either y = x or y and x are each at least 31 characters long and agree on their
first 31 characters. Because strings that are at least 31 characters long and agree on their first 31 characters
perforce are at least 8 characters long and on their first 8 characters, we know that either y = x or y and x
are each at least 8 characters long and agree on their first 8 characters. This means that y is equivalent to x
under Rs, that is, that y E [x]R8 •
55. We need first to make the relation symmetric, so we add the pairs (b, a), (e, a), and (e, d). Then we need to
make it transitive, so we add the pairs (b,e), (e,b), (a,a), (b,b), (e,e), (d,d), and (e,e). (In other words,
we formed the transitive closure of the symmetric closure of the original relation.) It happens that we have
already achieved reflexivity, so we are done; if there had been some pairs (x, x) missing at this point, we would
have added them as well. Thus the desired equivalence relation is the one consisting of the original 3 pairs
and the 10 we have added. There are two equivalence classes, {a, b, e} and {d, e}.
57. a) The equivalence class of 1 is the set of all real numbers that differ from 1 by an integer. Obviously this is
the set of all integers.
b) The equivalence class of 1/2 is the set of all real numbers that differ from 1/2 by an integer, namely
1/2, 3/2, 5/2, etc., and -1/2, -3/2, etc. These are often called
{ (2n + 1)/2 In E Z}, among other ways.
half-integers. We could write this set as
Equivalence Relations
Section 9.5
335
59. This problem actually deals with a branch of mathematics called group theory; the object being studied here
is related to a certain dihedral group. If this fascinates you, you might want to take a course with a title like
Abstract Algebra or Modern Algebra, in which such things are studied in depth.
In order to have a way to talk about specific colorings, let us agree that a sequence of length four, each
element of which is either r or b, represents a coloring of the 2 x 2 checkerboard, where the first letter denotes
the color of the upper left square, the second letter denotes the color of the upper right square, the third letter
denotes the color of the lower left square, and the fourth letter denotes the color of the lower right square. For
example, the board in which every square is red except the upper right would be represented by rbrr. There
are really only four different rotations, since after the rotation we need to end up with another checkerboard
(and we can assume that the edges of the board are horizontal and vertical). If we rotate our sample coloring
goo clockwise, then we obtain the coloring rrrb; if we rotate it 180°, then we obtain the coloring rrbr; if we
rotate it 270° clockwise (or goo counterclockwise), then we obtain the coloring brrr; and if we rotate it 360°
clockwise (or 0°-i.e., not at all), then we obtain the coloring rbrr itself back. Note also that some colorings
are invariant (i.e., unchanged) under rotations in addition to the 360° one; for example, bbbb is invariant
under all rotations, and brrb is invariant under a 180° rotation. Similarly there are four reflections: around
the center vertical axis of the board, around the center horizontal axis, around the lower-left-to-upper-right
diagonal, and around the lower-right-to-upper-left diagonal. For example, applying the vertical axis reflection
to rrbb yields itself, while applying the lower-left-to-upper-right diagonal reflection results in brbr.
The definition of equivalence for this problem makes the proof rather messy, since both rotations and
reflections are involved, and it is required that we reduce everything to just one or two operations. In fact,
we claim that there are only eight possible motions of this square: clockwise rotations of 0°, goo, 180°, or
270°, and reflections through the vertical, horizontal, lower-left-to-upper-right, and lower-right-to-upper-left
diagonals. To verify this, we must show that the composition of every two of these operations is again an
operation in our list. Below is the "group table" that shows this, where we use the symbols rO, rgo, rl80,
r270, Jv, Jh, Jp, and Jn for these operations, respectively. (The mnemonic is that r stands for "rotation,"
J stands for "flip," and v, h, p, and n stand for "vertical," "horizontal", "positive-sloping,'' and ''negativesloping,'' respectively.) It is read just like a multiplication table, with the operation o meaning ''followed by."
For example, if we first perform rgo and then perform J h, then we get the same result as if we had just
performed Jp (try it!).
0
rO
rgo
rl80
r270
Jv
Jh
Jp
Jn
rO
rgo
rl80
rO
rgo
rl80
rgo
rl80
r270
r270
r270
Jv
Jh
Jv
Jh
Jv
Jn
Jh
Jp
rO
fp
fp
rl80
r270
Jh
Jp
Jv
Jn
rl80
rO
rgO
Jp
Jv
Jn
Jh
rgo
r270
rO
Jn
Jn
rl80
r270
rO
rgo
Jh
fv
Jn
Jp
rgo
r270
rl80
Jn
Jh
Jp
Jv
r270
rgo
rl80
rO
r270
rO
Jp
Jn
Jh
Jv
rO
rgo
rl80
Jn
fp
Jv
Jh
So the result of this computation is that we can consider only these eight moves, and not have to worry about
combinations of them-every combination of moves equals just one of these eight.
a) To show reflexivity, we note that every coloring can be obtained from itself via a 0° rotation. In technical
terms, the 0° rotation is the identity element of our group. To show symmetry, we need to observe that
rotations and reflections have inverses: If C 1 comes from C 2 via a rotation of n° clockwise, then C 2 comes
from C 1 via a rotation of n° counterclockwise (or equivalently, via a rotation of (360 - n) 0 clockwise);
and every reflection applied twice brings us back to the position (and therefore coloring) we began with.
336
Chapter 9
Relations
And transitivity follows from the fact that the composition of two of these operations is again one of these
operations.
b) The equivalence classes are represented by colorings that are truly distinct, in the sense of not being
obtainable from each other via these operations. Let us list them. Clearly there is just one coloring using four
red squares, and so just one equivalence class, [rrrr]. Similarly there is only one using four blues, [bbbb]. There
is also just one equivalence class of colorings using three reds and one blue, since no matter which corner the
single blue occupies in such a coloring, we can rotate to put the blue in any other corner. Thus our third and
fourth equivalence classes are [rrrb] and [bbbr]. Note that each of them contains four colorings. (For example,
[rrrb] = {rrrb, rrbr, rbrr, brrr} .) This leaves only the colorings with two reds and two blues to consider. In
every such coloring, either the red squares are adjacent (i.e., share a common edge), such as in bbrr, or they
are not (e.g., brrb). Clearly the red squares are adjacent if and only if the blue ones are, since the only pairs of
nonadjacent squares are (lower-left,upper-right) and (upper-left,lower-right). It is equally clear that there are
only two colorings in which the red squares are not adjacent, namely rbbr and brrb, and they are equivalent
via a 90° rotation (among other transformations). So our fifth equivalence class is [rbbr] = {rbbr,brrb}.
Finally, there is only one more equivalence class, and it contains the remaining four colorings (in which the
two red squares are adjacent and the two blue squares are adjacent), namely {rrbb, brbr, bbrr, rbrb}, since each
of these can be obtained from each of the others by a rotation. In summary we have partitioned the set of
24 = 16 colorings (i.e., r-b strings of length four) into six equivalence classes, two of which have cardinality
one, three of which have cardinality four, and one of which has cardinality two.
One final comment. We saw in the solution to part (b) that only rotations are needed to show the
equivalence of every pair of equivalent colorings using just red and blue. This means that we are actually
dealing with just part of the dihedral group here. If more colors had been used, then we would have needed
to use the reflections as well. A complete discussion would get us into Polya's theory of enumeration, which
is studied in advanced combinatorics classes.
61. It is easier to write down a partition than it is to list the pairs in an equivalence relation, so we will answer
the question using this notation. Let the set be {1, 2, 3}. We want to write down all possible partitions of this
set. One partition is just { {1, 2, 3}}, i.e., having just one set (this corresponds to the equivalence relation in
which every pair of elements are related). At the other extreme, there is the partition {{l }, {2}, {3}}, which
corresponds to the equality relation (each x is related only to itself). The only other way to split up the
elements of this set is into a set with two elements and a set with one element, and there are clearly three ways
to do this, depending on which element we decide to put in the set by itself. Thus we get the partitions (pay
attention to the punctuation!) { {l, 2}, {3}}, {{1, 3}, {2}}, and { {2, 3}, {1}}. If we wished to list the ordered
pairs, we could; for example, the relation corresponding to {{2,3},{l}} is {(2,2),(2,3),(3,2),(3,3),(1,1)}.
We found five partitions, so the answer to the question is 5.
63. We do get an equivalence relation. The issue is whether the relation formed in this way is reflexive, transitive
and symmetric. It is clearly reflexive, since we included all the pairs (a, a) at the outset. It is clearly transitive,
since the last thing we did was to form the transitive closure. It is symmetric by Exercise 23 in Section 9.4.
65. We end up with the relation R that we started with. Two elements are related if they are in the same set
of the partition, but the partition is made up of the equivalence classes of R, so two elements are related
precisely if they are related in R.
67. We make use of Exercise 63. Given the relation R, we first form the reflexive closure R' of R by adding
to R each pair (a, a) that is not already there. Next we form the symmetric closure R" of R', by adding,
for each pair (a, b) E R' the pair (b, a) if it is not already there. Finally we apply Warshall's algorithm (or
Section 9.6
Partial Orderings
337
Algorithm 1) from Section 9.4 to form the transitive closure of R". This is the smallest equivalence relation
containing R.
69. The exercise asks us to compute p( n) for n = 0, 1, 2, ... , 10. In doing this we will use the recurrence relation,
building on what we have already computed (namely p( n - j - 1), noting that n - j - 1 < n), as well as using
(n - 1)!
the binomial coefficients C( n - 1, j) = . (
. . We organize our computation in the obvious way. using
J! n - l - J)!
the formula in Exercise 68.
p(O) = 1
(the initial condition)
p(l) = C(O, O)p(O) = 1 · 1 = 1
p(2) = C(l, O)p(l) + C(l, l)p(O) = 1·1+1·1 = 2
p(3) = C(2, O)p(2) + C(2, l)p(l) + C(2, 2)p(O) = 1·2+2·1+1·1 = 5
p(4) = C(3, O)p(3) + C(3, l)p(2) + C(3, 2)p(l) + C(3, 3)p(O) = 1·5 + 3 · 2 + 3 · 1+1·1=15
p(5) = C( 4, O)p( 4) + C( 4, l)p(3) + C( 4, 2)p(2) + C( 4. 3)p(l) + C( 4, 4)p(O)
= 1 . 15 + 4. 5 + 6 . 2 + 4 . 1 + 1 . 1 = 52
p(6) = C(5, O)p(5) + C(5, l)p(4) + C(5, 2)p(3) + C(5, 3)p(2) + C(5, 4)p(l) + C(5, 5)p(O)
= 1 . 52 + 5 . 15 + 10 . 5 + 10 . 2 + 5 . 1 + 1 . 1 = 203
p(7) = C(6, O)p(6) + C(6, l)p(5) + C(6, 2)p( 4) + C(6, 3)p(3) + C(6, 4)p(2) + C(6, 5)p(l) + C(6, 6)p(O)
= 1 . 203 + 6 . 52 + 15 . 15 + 20 . 5 + 15 . 2 + 6 . 1 + 1 . 1 = 877
p(8) = C(7, O)p(7) + C(7, l)p(6) + C(7, 2)p(5) + C(7, 3)p( 4) + C(7, 4)p(3) + C(7, 5)p(2)
+ C(7, 6)p(l) + C(7, 7)p(O)
= 1 . 877 + 7. 203 + 21 . 52 + 35 . 15 + 35 . 5 + 21 . 2 + 7. 1 + 1 . 1 = 4140
p(9) = C(8, O)p(8) + C(8, l)p(7) + C(8, 2)p(6) + C(8, 3)p(5) + C(8, 4)p( 4) + C(8, 5)p(3)
+ C(8, 6)p(2) + C(8, 7)p(l) + C(8, 8)p(O)
= 1 . 4140 + 8. 877 + 28. 203 + 56. 52 + 70. 15 + 56. 5 + 28. 2 + 8. 1+1 . 1 = 21147
p(lO) = C(9, O)p(9) + C(9, l)p(8) + C(9, 2)p(7) + C(9, 3)p(6) + C(9, 4)p(5) + C(9, 5)p( 4)
+ C(9, 6)p(3) + C(9, 7)p(2) + C(9, 8)p(l) + C(9, 9)p(O)
= 1 . 21147 + 9. 4140 + 36. 877 + 84. 203 + 126. 52
+ 126 . 15 + 84 . 5 + 36 . 2 + 9 . 1 + 1 . 1 = 115975
SECTION 9.6
Partial Orderings
Partial orderings (or "partial orders"-the two phrases are used interchangeably) rival equivalence relations in
importance in mathematics and computer science. Again, try to concentrate on the visual image-in this case
the Hasse diagram. Play around with different posets to become familiar with the different possibilities; not
all posets have to look like the less than or equal relation on the integers. Exercises 32 and 33 are important,
and they are not difHcult if you pay careful attention to the definitions.
338
Chapter 9
Relations
1. The question in each case is whether the relation is reflexive, antisymmetric, and transitive. Suppose the
relation is called R.
a) Clearly this relation is reflexive because each of 0, 1, 2, and 3 is related to itself. The relation is also
antisymmetric, because the only way for a to be related to b is for a to equal b. Similarly, the relation is
transitive, because if a is related to b, and b is related to c, then necessarily a= b = c so a is related to c
(because the relation is reflexive). This is just the equality relation on {O, 1, 2, 3}; more generally, the equality
relation on any set satisfies all three conditions and is therefore a partial ordering. (It is the smallest partial
ordering; reflexivity insures that every partial ordering contains at least all the pairs (a, a) .)
b) This is not a partial ordering, because although the relation is reflexive, it is not antisymmetric (we have
2 R 3 and 3 R 2, but 2 -=F 3 ), and not transitive ( 3 R 2 and 2 R 0, but 3 is not related to 0 ).
c) This is a partial ordering, because it is clearly reflexive; is antisymmetric (we just need to note that (1, 2)
is the only pair in the relation with unequal components); and is transitive (for the same reason).
d) This is a partial ordering because it is the "less than or equal to" relation on {1, 2, 3} together with the
isolated point 0.
e) This is not a partial ordering. The relation is clearly reflexive, but it is not antisymmetric ( 0 R 1 and 1 R 0,
but 0 -=F 1) and not transitive ( 2 R 0 and 0R1, but 2 is not related to 1).
3. The question in each case is whether the relation is reflexive, antisymmetric, and transitive.
a) Since nobody is taller than himself, this relation is not reflexive so (S, R) cannot be a poset.
b) To be not taller means to be exactly the same height or shorter. Two different people x and
the same height, in which case x Ry and y Rx but x
-=F
y could have
y, so R is not antisymmetric and this is not a poset.
c) This is a poset. The equality clause in the definition of R guarantees that R is reflexive. To check
antisymmetry and transitivity it suffices to consider unequal elements (these rules hold for equal elements
trivially). If a is an ancestor of b, then b cannot be an ancestor of a (for one thing, an ancestor needs to be
born before any descendant), so the relation is vacuously antisymmetric. If a is an ancestor of b, and b is an
ancestor of c, then by the way ''ancestor" is defined, we know that a is an ancestor of b; thus R is transitive.
d) This relation is not antisymmetric. Let
b Ra, but a
-=F
a and b be any two distinct friends of yours. Then a Rb and
b.
5. The question in each case is whether the relation is reflexive, antisymmetric, and transitive.
a) The equality relation on any set satisfies all three conditions and is therefore a partial partial ordering.
(It is the smallest partial partial ordering; reflexivity insures that every partial order contains at least all the
pairs (a, a).)
b) This is not a poset, since the relation is not reflexive, not antisymmetric, and not transitive (the absence
of one of these properties would have been enough to give a negative answer).
c) This is a poset, as explained in Example 1.
d) This is not a poset. The relation is not reflexive, since it is not true, for instance, that 2 ,.\'2. (It also is not
antisymmetric and not transitive.)
7. a) This relation is {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 3)}. It is not antisymmetric because (1, 2) and (2, 1)
are both in the relation, but 1 -=F 2. We can see this visually by the pair of 1's symmetrically placed around
the main diagonal at positions (1, 2) and (2, 1). Therefore this matrix does not represent a partial order.
b) This matrix represents a partial order. Reflexivity is clear. The only other pairs in the relation are (1, 2)
and (1. 3), and clearly neither can be part of a counterexample to antisymmetry or transitivity.
c) A little trial and error shows that this relation is not transitive ( (4, 1) and (1, 3) are present, but not (4, 3))
and therefore not a partial order.
Section 9.6
Partial Orderings
339
9. This relation is not transitive (there are arrows from a to b and from b to d, but there is no arrow from a
to d) , so it is not a partial order.
11. This relation is a partial order, since it has all three properties-it is reflexive (there is an arrow at each
point), antisymmetric (there are no pairs of arrows going in opposite directions between two different points),
and transitive (there is no missing arrow from some x to some z when there were arrows from x to y and y
to z ).
13. The dual of a poset is the poset with the same underlying set and with the relation defined by declaring a
related to b if and only if b :Sa in the given poset.
a) The dual relation to :S is 2':, so the dual poset is ( {O, 1, 2}, :::::) . Explicitly it is the set { (0, 0), (1, 0), (1, 1),
(2,0),(2,1),(2,2)}.
b) The dual relation to 2: is ::; , so the dual poset is (Z, ::;) .
c) The dual relation to ~ is <;;;;, so the dual poset is (P(Z), <;;;;).
d) There is no symbol generally used for the "is a multiple of" relation, which is the dual to the "divides''
relation in this part of the exercise. If we let R be the relation such that aRb if and only if b Ia, then the
answer can be written (z+, R).
15. We need to find elements such that the relation holds in neither direction between them. The answers we give
are not the only ones possible.
a) One such pair is { 1} and {2} . These are both subsets of {0, 1, 2} , so they are in the poset, but neither is
a subset of the other.
b) Neither 6 nor 8 divides the other, so they are incomparable.
17. We find the first coordinate (from left to right) at which the tuples differ and place first the tuple with the
smaller value in that coordinate.
a) Since 1=1 in the first coordinate, but 1 < 2 in the second coordinate, (1, 1, 2) < (1, 2, 1).
b) The first two coordinates agree, but 2 < 3 in the third, so (0, 1, 2, 3) < (0, 1, 3, 2).
c) Since 0 < 1 in the first coordinate, (0, 1, 1, 1, 0) < (1, 0, 1, 0, 1).
19. All the strings that begin with 0 precede all those that begin with 1. The 0 comes first. Next comes 0001,
which begins with three O's, then 001, which begins with two O's. Among the strings that begin 01, the order
is 01 < 010 < 0101 < 011. Putting this all together, we have 0 < 0001 < 001 < 01 < 010 < 0101 < 011 < 11.
21. This is a totally ordered set, so the Hasse diagram is linear.
15
11
10
5
2
0
23. We put x above y if y divides x. We draw a line between x and y, where y divides x, if there is no number
z in our set, other than x or y, such that y I z /\ z Ix. Note that in part (b) the numbers other than 1 are
all (relatively) prime, so the Hasse diagram is short and wide, whereas in part ( d) the numbers all divide one
another, so the Hasse diagram is tall and narrow.
Chapter 9
340
Relations
8
4
7
2
64
48
32
36
16
8
4
2
3
(c)
2
(d)
25. We need to include every pair (x,y) for which we can find a path going upward in the diagram from x toy.
We also need to include all the reflexive pairs (x, x). Therefore the relation is the following set of pairs:
{(a,a), (a,b),(a,c), (a,d). (b,b),(b,c),(b,d),(c,c),(d,d)}.
27. The procedure is the same as in Exercise 25: {(a,a),(a,d),(a,e),(a,f),(a,g),(b,b),(b,d),(b,e),(b,f),(b,g),
(c,c),(c,d),(c,e), (c,f),(c,g),(d,d),(e,e),(f,f),(g,d),(g,e), (g,f),(g,g)}.
29. In this problem X ~ Y when X <:;;; Y. For (X, Y) to be in the covering relation, we need X to be a proper
subset of Y but we also must have no subset strictly between X and Y. For example, ({a}, {a, b, c}) is
not in the covering relation, since {a} C {a, b} and {a, b} C {a, b, c}. With this understanding it is easy to
list the pairs in the covering relation: (0,{a}), (0,{b}), (0,{c}), ({a},{a,b}), ({a},{a,c}), ({b},{a,b}),
({b}, {b, c}), ({c}, {a, c}), ({c}, {b, c}), ({a, b}, {a, b, c}), ({a, c}, {a, b, c}), and ( {b, c}, {a, b, c}).
31. Let (S, j) be a finite poset. We claim that this poset is just the reflexive transitive closure of its covering
relation. Suppose that (a, b) is in the reflexive transitive closure of the covering relation. Then either a = b
or a -< b (in which cases certainly a j b) or else there is a sequence a -< a 1 -< a 2 -< · · · -< an -< b, in which
case again a ~ b, by the transitivity of ~. Conversely, suppose that a ~ b. If a = b, then (a, b) is certainly
in the reflexive transitive closure of the covering relation. If a -< b and there is no z such that a -< z -< b,
then (a, b, ) is in the covering relation and again therefore in its reflexive transitive closure. Otherwise, let
a -< a 1 -< a 2 -< · · · -< an -< b be a longest possible sequence of this form; since the poset is finite, there must
be such a longest sequence. Then no intermediate elements can be inserted into this sequence (to do so would
lengthen it), so each pair (a,a 1 ), (a 1 ,a 2 ), ... , (an,b) is in the covering relation, so again (a,b) is in its
reflexive transitive closure. This completes the proof. Note how the finiteness of the poset was crucial here. If
we let S be the set of all subsets of N (the set of natural numbers) under the subset relation, then we cannot
recover S from its covering relation, since nothing in the covering relation allows us to relate a finite set to
an infinite one; thus for example we could not recover the relationship {1, 2} C N.
33. It is helpful in this exercise to draw the Hasse diagram.
a) Maximal elements are those that do not divide any other elements of the set. In this case 24 and 45 are
the only numbers that meet that requirement.
b) Minimal elements are those that are not divisible by any other elements of the set. In this case 3 and 5
are the only numbers that meet that requirement.
Section 9.6
341
Partial Orderings
c) A greatest element would be one that all the other elements divide. The only two candidates (maximal
elements) are 24 and 45, and since neither divides the other, we conclude that there is no greatest element.
d) A least element would be one that divides all the other elements. The only two candidates (minimal
elements) are 3 and 5, and since neither divides the other, we conclude that there is no least element.
e) We want to find all elements that both 3 and 5 divide. Clearly only 15 and 45 meet this requirement.
f) The least upper bound is 15 since it divides 45 (see part ( e)).
g) We want to find all elements that divide both 15 and 45. Clearly only 3, 5, and 15 meet this requirement.
h) The number 15 is the greatest lower bound, since both 3 and 5 divide it (see part (g) ).
35. To help us answer the questions, we will draw the Hasse diagram, with the commas and braces eliminated in
the labels, for readability.
234
12
34
4
a) The maximal elements are the ones without any elements lying above them in the Hasse diagram, namely
{1,2}, {1,3,4}, and {2,3,4}.
b) The minimal elements are the ones without any elements lying below them in the Hasse diagram, namely
{ 1} , {2} , and {4} .
c) There is no greatest element, since there is more than one maximal element, none of which is greater than
the others.
d) There is no least element, since there is more than one minimal element, none of which is less than the
others.
e) The upper bounds are the sets containing both { 2} and {4} as subsets, i.e., the sets containing both 2
and 4 as elements. Pictorially, these are the elements lying above both {2} and {4} (in the sense of there
being a path in the diagram), namely {2,4} and {2,3,4}.
f) The least upper bound is an upper bound that is less than every other upper bound. We found the upper
bounds in part (e), and since {2,4} is less than (i.e., a subset of) {2,3,4}, we conclude that {2,4} is the
least upper bound.
g) To be a lower bound of both {l, 3, 4} and {2, 3, 4}, a set must be a subset of each, and so must be a subset
of their intersection, {3, 4}. There are only two such subsets in our poset, namely {3, 4} and {4}. In the
diagram, these are the points which lie below (in the path sense) both {1,3,4} and {2,3,4}.
h) The greatest lower bound is a lower bound that is greater than every other lower bound. We found the
lower bounds in part (g), and since {3,4} is greater than (i.e., a superset of) {4}, we conclude that {3,4} is
the greatest lower bound.
37. First we need to show that lexicographic order is reflexive, i.e., that (a, b) ::5 (a, b); this is true by fiat, since we
defined ::5 by adding equality to -<. Next we need to show antisymmetry: if (a, b) ::5 (c, d) and (a, b) =/=- (c, d),
then (c, d) f._ (a, b). By definition (a, b) -< (c, d) if and only if either a -< c, or a = c and b -< d. In the first
case, by the antisymmetry of the underlying relation, we know that c /, a, and similarly in the second case
we know that d /, d. Thus there is no way that we could have (c, d) -< (a, b) . Finally, for transitivity, let
(a, b) ::5 (c, d) ::5 (e, J). We want to show that (a, b) ::5 (e, J). If one of the given inequalities is an equality,
then there is nothing to prove, so we may assume that (a, b) -< (c, d) -< (e, !) . If a -< c, then by the transitivity
of the underlying relation, we know that a -< e and so (a, b) -< (e, f). Similarly, if c -< e, then again a -< e
Chapter 9
342
Relations
and so (a, b) --< (e, f). The only other way for the given inequalities to hold is if a= c = e and b--< d--<
this case the latter string of inequalities implies that b --< f and so again by definition (a, b) --< (e, f).
f. In
39. First we must show that :::5 is reflexive. Since s :::5 1 s and t :::5 2 t by the reflexivity of these underlying
partial orders, ( s, t) :::5 (s, t) by definition. For antisymmetry, assume that ( s, t) :::5 (u, v) and (u, v) :::5 ( s, t) .
Then by definition s :::5 1 u and t :::5 2 v, and u :::5 1 s and v :::5 2 t. By the antisymmetry of the underlying
relations, we conclude that s = u and t = v, whence ( s, t) = (u, v) . Finally, for transitivity, suppose that
(s,t) :::S (u,v) :::5 (w,x). This means thats :::51 u :::51 wand t :::52 v :::52 x. The transitivity of the underlying
partial orders tells us thats :::5 1 wand t :::52 x, whence by definition (s,t) :::5 (w,x).
41. a) We argue essentially by contradiction. Suppose that m 1 and m 2 are two maximal elements in a poset that
has a greatest element g; we will show that m 1 = m 2 . Now since g is greatest, we know that m 1 :::S g, and
similarly for m 2 . But since each m, is maximal, it cannot be that m, --< g; hence m 1 = g = m2.
b) The proof is exactly dual to the proof in part (a), so we just copy over that proof, making the appropriate
changes in wording. To wit: we argue essentially by contradiction. Suppose that m 1 and m 2 are two minimal
elements in a poset that has a least element l; we will show that m 1 = m 2 . Now since l is least, we know that
l :::5 m 1 , and similarly for m 2 . But since each m, is minimal, it cannot be that l --< m,; hence m1 = l = m 2 .
43. In each case, we need to check whether every pair of elements has both a least upper bound and a greatest
lower bound.
a) This is a lattice. If we want to find the l.u.b. or g.l.b. of two elements in the same vertical column of the
Hasse diagram, then we simply take the higher or lower (respectively) element. If the elements are in different
columns, then to find the g.l.b. we follow the diagonal line upward from the element on the left, and then
continue upward on the right, if necessary to reach the element on the right. For example, the l.u.b. of d and
c is f; and the l.u.b. of a and e is e. Finding greatest lower bounds in this poset is similar.
b) This is not a lattice. Elements b and c have f, g, and h as upper bounds, but none of them is a l.u.b.
c) This is a lattice. By considering all the pairs of elements, we can verify that every pair of them has a l.u.b.
and a g.l.b. For example, b and e have g and a filling these roles, respectively.
45. As usual when trying to extend a theorem from two items to an arbitrary finite number, we will use mathematical induction. The statement we wish to prove is that if S is a subset consisting of n elements from a
lattice, where n is a positive integer, then S has a least upper bound and a greatest lower bound. The two
proofs are duals of each other, so we will just give the proof for least upper bound here. The basis is n = 1,
in which case there is really nothing to prove. If S = {x}, then clearly x is the least upper bound of S. The
case n = 2 could be singled out for special mention also, since the l.u.b. in that case is guaranteed by the
definition of lattice. But there is no need to do so. Instead, we simply assume the inductive hypothesis, that
every subset containing n elements has a 1.u.b., and prove that every subset S containing n + 1 elements also
has a l.u.b. Pick an arbitrary element x ES, and let S' = S - {x}. Since S' has only n elements, it has a
l.u.b. y, by the inductive hypothesis. Since we are in a lattice, there is an element z that is the l.u.b. of x
and y. We will show that in fact z is the least upper bound of S. To do this, we need to show two things:
that z is an upper bound, and that every upper bound is greater than or equal to z. For the first statement,
let w be an arbitrary element of S; we must show that w :::5 z. There are two cases. If w = x, then w :::5 z
since z is the l.u.b. of :r and y. Otherwise, w E S', and so w :::S y because y is the l.u.b. of S'. But since z
is the l.u.b. of x and y, we also have y :::5 z. By transitivity, then, w :::S z. For the second statement, suppose
that u is any other upper bound of S; we must show that z :::5 u. Since u is an upper bound of S, it is also
an upper bound of x and y. But since z is the least upper bound of x and y, we know that z :::5 u.
47. The needed definitions are in Example 25.
Section 9.6
Partial Orderings
343
a) No. The authority level of the first pair (1) is less than or equal to (less than, in this case) that of the
second (2); but the subset of the first pair is not a subset of that of the second.
b) Yes. The authority level of the first pair (2) is less than or equal to (less than, in this case) that of the
second (3); and the subset of the first pair is a subset of that of the second.
c) The classes into which information can flow are those classes whose authority level is at least as high as
Proprietary, and whose subset is a superset of {Cheetah, Puma}. We can list these classes: (Proprietary,
{Cheetah, Puma}), (Restricted, {Cheetah, Puma}), (Registered, {Cheetah, Puma}), (Proprietary, {Cheetah,
Puma, Impala}), (Restricted, {Cheetah, Puma, Impala}), and (Registered, {Cheetah, Puma, Impala}).
d) The classes from which information can flow are those classes whose authority level is at least as low
as Restricted, and whose subset is a subset of {Impala, Puma}, namely (Nonproprietary,{Impala,Puma}),
(Proprietary, {Impala,Puma}), (Restricted, { Impala,Puma}), ( Nonproprietary, {Impala}), (Proprietary,
{Impala}) , (Restricted, {Impala}) , (Non proprietary, {Puma}) , (Proprietary, {Puma}) , (Restricted, {Puma}) ,
(Nonproprietary,0), (Proprietary,0), and (Restricted,0).
49. Let II be the set of all partitions of a set S, with a relation :::S defined on II according to the referenced
preamble: a partition P 1 is a refinement of P2 if every set in P 1 is a subset of one of the sets in P2 . We
need to verify all the properties of a lattice. First we need to show that (II, :::S) is a poset, that is, that :::S is
reflexive, antisymmetric, and transitive. For reflexivity, we need to show that P :::S P for every partition P.
This means that every set in P is a subset of one of the sets in P, and this is trivially true, since every set is
a subset of itself. For antisymmetry, suppose that P 1 :::S P 2 and P2 :::S Pi . We must show that P 1 = P 2 . By
the equivalent roles played here by P 1 and P2 , it is enough to show that every T E P 1 (where T <;;;: S) is also
an element of P2. Suppose we have such a T. Then since P 1 :::S P2 , there is a set T' E P2 such that T <;;;: T'.
But then since P 2 :::S P 1 , there is a set T" E P 1 such that T' <;;;: T". Putting these together, we have T <;;;: T".
But P1 is a partition, and so the elements of P 1 are nonempty and pairwise disjoint. The only way for this to
happen if one is a subset of the other is for the two subsets T and T" to be the same. But this implies that
T' (which is caught in the middle) is also equal to T. Thus TE P2 , which is what we were trying to show.
Finally, for transitivity, suppose that P 1 :::S P2 and P 2 :::S P 3 . We must show that P 1 :::S P3 . To this end, we
take an arbitrary element T E P 1 . Then there is a set T' E P2 such that T <;;;: T'. But then since P2 :::S P 3 ,
there is a set T" E P 3 such that T' <;;;: T". Putting these together, we have T <;;;: T". This demonstrates that
P1 :::S P3.
Next we have to show that every two partitions Pi and P 2 have a least upper bound and a greatest
lower bound in II. We will show that their greatest lower bound is their "coarsest common refinement",
namely the partition P whose subsets are all the nonempty sets of the form T 1 n T 2 , where T 1 E P 1 and
T2 E P2. As an example, if P1 = {{1, 2, 3}, {4}, {5}} and P2 = {{1, 2}, {3, 4}, {5}}, then the coarsest common
refinement is P = {{1, 2}, {3}, {4}, {5}}. First, we need to check that this is a partition. It certainly is a set
of nonempty subsets of S. It is pairwise disjoint, because the only way an element could be in T1 n T 2 n T{ n T~
if T 1 n T 2 -=/:. T{ n T~ is for that element to be in both T 1 n T{ and T 2 n T~ , which means that T 1 = T{ and
T 2 = T~ , a contradiction. And it covers all of S, because if x E S, then x E T 1 for some T 1 E P 1 , and x E T 2
for some T2 E P2, and so x E T1 n T2 E P. Second, we need to check that P is a refinement of both P1
and P2 . So suppose T E P. Then T = T1 n T2 , for some T 1 E P 1 and T 2 E P2 . It follows that T <;;;: T1
and T <;;;: T 2 . But then T 1 and T 2 satisfy the requirements in the definition of refinement. Third, we need to
check that if P' is any other common refinement of both P 1 and P2 , then P' is also a refinement of P. To
this end, suppose that T E P'. Then by definition of refinement, there are subsets T 1 E P 1 and T 2 E P 2 such
that T <;;;: T1 and T <;;;: T2. Therefore T <;;;: T1 n T 2 . But T 1 n T 2 E P, and our proof for greatest lower bounds
is complete.
It's a little harder to state the definition of the least upper bound (which again we'll call P) of two given
partitions P 1 and P2 . Essentially it is just the set of all minimal nonempty subsets of S that do not "split
344
Chapter 9
Relations
apart'' any element of either P 1 or P 2 . (In the example above, it is {{1, 2, 3, 4}, {5}} .) It will be a little
easier if we define it in terms of an equivalence relation rather than a partition. Note that from this point of
view, one equivalence relation is a refinement of a second equivalence relation if whenever two elements are
related by the first relation, then they are related by the second. The equivalence relation determining P is
the relation in which x E S is related to y E S if there is a "path" (a sequence) x = x 0 , x 1 , x 2 , ... , Xn = y,
for some n ?: 0, such that for each i from 1 to n, Xi-1 and Xi are in the same element of partition P 1 or of
partition P 2 (in other words, x,_ 1 and x, are related either by the equivalence relation corresponding to P 1
or by that corresponding to P2 ). It is clear that this is an equivalence relation: it is reflexive by taking n = 0;
it is symmetric by following the path backwards; and it is transitive by composing paths. It is also clear that
P1 (and P 2 similarly) is a refinement of this partition, since if two elements of S are in the same equivalence
class in P1 , then we can take n = 1 in our path definition to see that they are in the same equivalence class
in P. Thus P is an upper bound of both P 1 and P 2 . Finally, we must show that P is the least upper bound,
that is, a refinement of every other upper bound. This is clear from our construction: we only forced two
elements of S to be related (i.e., in the same class of the partition) when they had to be related in order to
enable P 1 and P 2 to be refinements. Therefore if two elements are related by P, then they have to be related
by every equivalence relation (partition) Q of which both P 1 and P 2 are refinements; so P is a refinement
of Q.
51. This follows immediately from Exercise 45. To be more specific, according to Exercise 45, there is a least
upper bound (respectively, a greatest lower bound) for the entire finite lattice. This element is by definition
a greatest element (respectively, a least element).
53. We need to show that every nonempty subset of z+ x z+ has a least element under lexicographic order.
Given such a subset S, look at the set S 1 of positive integers that occur as first coordinates in elements of S.
Let m 1 be the least element of S 1 , which exists since z+ is well-ordered under :::; . Let S' be the subset of
S consisting of those pairs that have m 1 as their first coordinate. Thus S' is clearly nonempty, and by the
definition of lexicographic order, every element of S' is less than every element in S - S'. Now let 82 be the
set of positive integers that occur as second coordinates in elements of S', and let m 2 be the least element of
S2 . Then clearly the element (m 1 , m 2 ) is the least element of S' and hence is the least element of S.
55. If x is an integer in a decreasing sequence of elements of this poset, then at most \x\ elements can follow x in
the sequence, namely integers whose absolute values are \xi - 1, \xi - 2, ... , 1, 0. Therefore there can be no
infinite decreasing sequence. This is not a totally ordered set, since 5 and -5, for example, are incomparable;
from the definition given here, it is neither true that 5 -< -5 nor that -5 -< 5, because neither one of \5\ or
\ - 51 is less than the other (they are equal).
57. We know from elementary arithmetic that Q is totally ordered by <, and so perforce it is a partially ordered
set. To be precise, to find which of two rational numbers is larger, write them with a positive common
denominator and compare numerators. To show that this set is dense, suppose x < y are two rational
numbers. Let z be their average, i.e., (x + y) /2. Since the set of rational numbers is closed under addition
and division, z is also a rational number, and it is easy to show that x < z < y.
59. Let (S, ::S) be a partially ordered set. From the definitions of well-ordered, totally ordered, and well-founded,
it is clear that what we have to show is that every nonempty subset of S contains a least element if and only
if there is no infinite decreasing sequence of elements a1, a2, a 3 , ... in S (i.e., where a,+l - 0, then we now
know that there are paths of arbitrary length from v, to vJ , since we can loop around Vk as long as we please;
in this case we set w!~l to oo. If wlkk-l] = 0, then we do not yet have such looping, so we set w!~l to the
larger of w!~-l] and w!~-l]
+ wl~-l].
(Initially we set W
0
equal to the matrix representing the relation.)
21. There are 52 partitions in all, but that is not the question. If there are to be three equivalence classes, then
the classes must have sizes 3, 1, 1 or 2, 2, 1. There are C(5, 3) = 10 partitions into one set with 3 elements
and the other two sets of 1 element each, since the only choice involved is choosing the 3-set. There are
C(5, 2)C(3, 2)/2 = 15 ways to partition our set into sets of size 2, 2, and 1; we need to choose the 2 elements
for the first set of size 2 , then we need to choose the 2 elements from the 3 remaining for the second set of
size 2 , except that we have overcounted by a factor of 2 , since we could choose these two 2-sets in either
order. Therefore there are 10 + 15 = 25 partitions into three classes.
23. There is no question that the collection defined here is a refinement of each of the given partitions, since each
set A, n BJ is a subset of A, and of BJ. We must show that it is actually a partition. By construction, each
of the sets in this collection is nonempty. To see that their union is all of S, let s E S. Since P1 and P2 are
partitions of S, there are sets A, and BJ such that s E A, and s E BJ. Therefore s E A, n B 1 , which shows
that s is in one of the sets in our collection. Finally, to see that these sets are pairwise disjoint, simply note
that unless i = i' and j = j', then (A, n BJ) n (A,, n BJ') = (A, n A,,) n (BJ n BJ') is empty, since either
(A, n A,,) or (BJ n BJ') is empty.
25. The subset relation is a partial order on every collection of sets, since it is reflexive, antisymmetric, and
transitive. Here the collection of sets happens to be R(S).
27. We need to find a total order compatible with this partial order. We work from the bottom up, writing down
a task (vertex in the diagram) and removing it from the diagram, so that at each stage we choose a vertex
with no vertices below it. One such order is: Find recipe -< Buy seafood -< Buy groceries -< Wash shellfish
-< Cut ginger and garlic -< Clean fish -< Steam rice -< Cut fish -< Wash vegetables -< Chop water chestnuts
-< Make garnishes -< Cook in wok -< Arrange on platter -< Serve.
29. Since every subset of an antichain is clearly an antichain, we will list only the maximal antichains; the actual
answers will be everything we list together with all the subsets of them.
349
Supplementary Exercises
a) Here every two elements are comparable except c and d. Thus the maximal antichains are {c, d}, {a},
and {b}. (There are three more anti chains which are subsets of these: {c}, {d}, and 0.)
b) Here the maximal antichains are {a}, {b, c}, {c, e}, and {d, e}.
c) In this case there are only three maximal antichains: {a, b, c}, {d, e, J}, and {g}.
31. Let C be a maximal chain. We must show that C contains a minimal element of S. Since C can itself be
viewed as a finite poset (being a subset of a poset), it contains a minimal element m. We need to show that m
is also a minimal element of S. If it were not, then there would be another element a E S such that a --< m.
Now we claim that CU {a} is a chain, which will contradict the maximality of C. We need to show that a is
comparable to every element of C. We already know that a is comparable to m. Let x be any other element
of C. Since m is minimal in C, it cannot be that x--< m; thus since x and m have to be comparable (they
are both in C), it must be that m--< x. Now by transitivity we have a--< x, and we are done.
33. Consider the relation R on the set of mn
+1
people given by (a, b) E R if and only if a is a descendant
of or equal to b. This makes the collection into a poset. In the terminology of Exercise 32, if there is not a
subset of n + 1 people none of whom is a descendant of any other, then k ::::; n, since such a subset is certainly
an antichain. Therefore the poset can be partitioned into k ::::; n chains. Now by the generalized pigeonhole
principle, at least one of these chains must contain at least m + 1 elements, and this is the desired list of
descendants.
35. Recall the definition of well-founded from the preamble to Exercise 55 in Section 9.6-that there is no infinite
decreasing sequence. We must show that under this hypothesis, and if \:.Ix( (\:./y(y --< x --+ P(y))) --+ P( x)),
then P( x) is true for all x E S. We give a proof by contradiction. If it does not hold that P( x) is true for
all x E S, let x 1 be an element of S such that P(x 1 ) is not true. Then by the conditional statement given
above, it must be the case that \:./y(y --< x 1 --+ P(y)) is not true. This means that there is some y with y --< x 1
such that P(y) is not true. Rename this y as x 2 • So we know that P(x 2 ) is not true. Again invoking the
conditional statement, we get an x 3 --< x 2 such that P(x 3 ) is not true. And so on forever. This contradicts
the well-foundedness of our poset. Therefore P(x) is true for all x ES.
37. We assume that R is reflexive and transitive on A, and we must show that Rn R- 1 is reflexive, symmetric,
and transitive. Reflexivity is easy: if a E A, then we know that (a, a) E R, so by the definition of R- 1 as
the reverses of the pairs in R, we know that (a, a) E R- 1 as well, whence it follows that (a, a) E Rn R- 1 .
Every relation of the form Rn R- 1 is symmetric, no matter what R is, since if (a, b) ER, then (b, a) E R- 1
and vice versa. For transitivity, suppose that (a, b) E Rn R- 1 and (b, c) E Rn R- 1 . We must show that
(a, c) E Rn R- 1 . Since (a, b) E R and (b, c) E R, and since R is transitive, (a, c) E R. Similarly, since
(a, b) E R- 1 and (b, c) E R- 1 , (b, a) ER and (c, b) ER. Again, since R is transitive, (c, a) E R, and hence
(a, c) E R- 1 . Putting these two parts together, we conclude that (a, c) ER n R- 1 , as desired.
39. There is not much to show in this exercise, since the definitions of greatest lower bound and least upper bound
exhibit these properties by their very form.
a) The g.l.b. of x and y was defined to be the greatest element that is a lower bound of both x and y. The
roles of x and y in this statement are symmetric, so it follows immediately that x /\ y = y /\ x. Similarly for
least upper bound.
b) By definition, (x /\ y) /\ z is a lower bound of x, y, and z that is greater than every other common lower
bound (this is how we proceeded in Exercise 45 of Section 9.6). Since x, y, and z play interchangeable roles
in this statement, grouping does not matter, so x /\ (y /\ z) is the same element. Similarly for l.u.b.
c) The two statements are duals, so we will prove just the first one; the proof of the second can be obtained
formally simply by exchanging each symbol and word for its dual. To show that x /\ (x V y) = x, we must
Chapter 9
350
Relations
show that x is the greatest lower bound of x and x V y. Clearly x is a lower bound for x, and since x Vy is
by definition greater than or equal to x, x is a lower bound for it as well. Therefore x is a lower bound. But
every other lower bound for x has to be less than x , so x is the greatest lower bound.
d) Obviously x is a lower (upper) bound for itself and itself, and the greatest (least) such.
41. There is nothing very deep going on here-it's just a matter of applying the definitions.
a) Since 1 is the only element greater than or equal to 1, it is the only upper bound for 1 and therefore the
only possible value of the least upper bound of x and 1.
b) Clearly x is a lower bound for both x and 1 (since x ::S 1), and clearly no other lower bound can be
greater than x , so x /\ 1 = x .
c) This is the dual to part (b). We formed the following proof on the word processor used to produce this
solutions manual by copying the words in the solution to part (b) and replacing each word and symbol by its
dual: Clearly x is an upper bound for both x and 0 (since 0 ::S x ), and clearly no other upper bound can be
smaller than x, so x V 0 = x.
d) This is the dual of part (a): Since 0 is the only element less than or equal to 0, it is the only lower bound
for 0 and therefore the only possible value of the greatest lower bound of x and 0.
43. One way to solve this problem is to play around with some small examples. Here is one counter-example
that the author obtained in this way. The lattice has as its elements 0, {1}, {2}, {3}, {1, 2}, {2, 3}, and
{1, 2, 3}, with, as usual, the relation <;;;. (Draw its Hasse diagram!) It is easy to check that every two elements
have both a least upper bound and a greatest lower bound (note that 0 is a lower bound for the whole lattice,
and {l,2,3} is an upper bound for the whole lattice). Take x = {1}, y = {2}, and z = {3}, and compute
both sides of the equation x V (y /\ z) = (x Vy)/\ (x V z). Note that since we do not have the full subset
lattice, least upper bounds are not just unions. The left-hand side is x, since y /\ z = 0. The right-hand side
is the greatest lower bound of {1, 2} and {1, 2, 3}, which is {1, 2}. Since these are different, the lattice is not
distributive.
45. Yes. First, recall from Example 22 in Section 9.6 that x /\ y is the greatest common divisor (gcd) of x
and y, while x Vy is their least common multiple (lcm). We can analyze this problem by looking at prime
factorizations. The power to which a prime p appears in the gcd of two numbers is the minimum of the powers
to which it appears in the two numbers. Similarly, the power to which p appears in the lcm is the maximum
of the powers to which it appears in the two numbers. Thus if we let a, b, and c represent the powers to
which p appears in x, y, and z, respectively, the first identity we need to prove is
max( a, min(b, c)) =min( max( a, b), max( a, c)).
We consider the several cases. If a is the largest of the three numbers, then both sides equal a. If a is the
smallest, then both sides equal the smaller of b and c. Otherwise, we can suppose without loss of generality
(since the roles of b and c are symmetric) that b ~ a ~ c, in which case we easily compute that both sides
equal a. The proof of the other statement is dual to this proof. The result now follows from the fundamental
theorem of arithmetic, since numbers are determined by their prime factorizations.
47. As might be expected from the name, the complement of a subset X <;;;: S is its complement S - X. To prove
this, we need to prove that XV (S - X) = 1 and X /\ (S - X) = 0, which translated into our particular
setting reads: XU (S - X) = S and X n (S - X) = 0. But these are trivially true.
49. Think of the rectangular grid as representing elements in a matrix. Thus we number from top to bottom and
within that from left to right. For example, (2, 4) is the element in row 2, column 4. The partial order is
that (a, b) ::S (c, d) if a ~ c and b ~ d. Note that (1, 1) is the least element under this relation. The rules
Writing Projects
351
for Chomp as explained in Chapter 1 coincide with the rules stated in the preamble here. But now we can
identify the point (a, b) with the natural number pa-lqb-l for all a and b with 1 :Sa :Sm and 1 :Sb :Sn.
This identifies the points in the rectangular grid with the set S in this exercise, and the partial order ~ just
described is the same as the divides relation, because pa-lqb-l I pc-lqd-l if and only if the exponent on p
on the left does not exceed the exponent of p on the right, and similarly for q.
WRITING PROJECTS FOR CHAPTER 9
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. See the same references as suggested for fuzzy logic in Writing Project 2 of Chapter 1, as well as [Zi].
2. There are numerous textbooks on databases. Thy to consult one that is fairly recent, because in many areas
of computer science, progress is so fast that books soon become out-dated. You will find them in the QA 76.9
area of the library's shelves. Two recommended ones are are [Dal] and [Mal].
3. Thy author or key-word search in an appropriate database (e.g., one provided by Mathematical Reviews, which
is available on the Web as MathSciNet). Consult the oldest reference you can find that talks about these
topics, and it will probably lead you to the original sources. Simultaneous discovery occurs in many branches
of intellectual pursuit, not just mathematics and computer science. See Writing Project 17 in Chapter 11 for
something along the same lines.
4. The abstraction and difficulty here is part of what makes fractions hard for many children (and some adults)
to handle. Be careful to avoid 0 in the denominator! You should be able to figure this out without consulting
other sources, and it is a good project to work on with other people.
5. See the hints for Writing Project 3.
6. Entire books have been written on security issues in computer systems ([Pf], for one), and it should not be
hard to find a chapter or two on the subject in many more general books (try [Del]).
7. Textbooks on project scheduling should be a good source of information. See [Mo], for example. Scheduling
is a topic in a branch of mathematics known as Operations Research. It has its own journals, conferences,
subspecialties, software, etc.
8. See the suggestions for Writing Project 7.
9. We have hinted at duality in many of the exercise solutions in this Guide. A book on lattice theory (such as
[Grl]) will make the concept more precise.
10. As mentioned in the previous suggestion, you can find entire books on lattice theory. In fact, Mathematical
Reviews (which is available on the Web as MathSciNet) devotes a whole category (numbered 06) to lattices
and other kinds of ordered sets and ordered algebraic structures.
352
Chapter 10
Graphs
CHAPTERlO
Graphs
SECTION 10.1
Graphs and Graph Models
The examples and exercises give a good picture of the ways in which graphs can model various real world
applications. In constructing graph models you need to determine what the vertices will represent, what the
edges will represent, whether the edges will be directed or undirected, whether loops should be allowed, and
whether a simple graph or multigraph is more appropriate.
1. In part (a) we have a simple graph, with undirected edges, no loops or multiple edges. In part (b) we have a
multigraph, since there are multiple edges (making the figure somewhat less than ideal visually). In part ( c) we
have the same picture as in part (b) except that there is now a loop at one vertex; thus this is a pseudograph.
Boston
Boston
Boston
Miami
(a)
(c)
(b)
In part ( d) we have a directed graph, the directions of the edges telling the directions of the flights; note that
the anti parallel edges (pairs of the form (u, v) and (v, u)) are not parallel. In part ( e) we have a directed
multigraph, since there are parallel edges.
Boston
Miami
Miami
(d)
(e)
3. This is a simple graph; the edges are undirected, and there are no parallel edges or loops.
5. This is a pseudograph; the edges are undirected, but there are loops and parallel edges.
7. This is a directed graph; the edges are directed, but there are no parallel edges. (Loops and antiparallel
edges-see the solution to Exercise ld for a definition-are allowed in a directed graph.)
9. This is a directed multigraph; the edges are directed, and there is a set of parallel edges.
353
Graphs and Graph Models
Section 10.1
11. In a simple graph, edges are undirected. To show that R is symmetric we must show that if uRv, then vRu.
If uRv, then there is an edge associated with {u, v}. But {u, v} = {v, u}, so this edge is associated with
{v, u} and therefore vRu. A simple graph does not allow loops; that is if there is an edge associated with
{u, v}, then u f= v. Thus uRu never holds, and so by definition R is irreflexive.
13. In each case we draw a picture of the graph in question. All are simple graphs. An edge is drawn between two
vertices if the sets for the two vertices have at least one element in common. For example, in part (a) there
is an edge between vertices A 1 and A 2 because there is at least one element common to A1 and A2 (in fact
there are three such elements). There is no edge between Ai and A3 since Ai n A3 = 0.
~
A4~A1
A5
(b)
A5
(c)
15. We draw a picture of the graph in question, which is a simple graph. Two vertices are joined by an edge if
we are told that the species compete (such as robin and mockingbird) but there is no edge between pairs of
species that are not given as competitors (such as robin and blue jay).
robin~lue Jf:lY
hermit thrush
r
"'\mockingbird
nuthf:ltch .___... hf:ll ry woodpecker
17. Here are the persons to be included, listed in order of birth year: Aristotle (384-322 B.C.E.), Euclid (325-265
B.C.E. ), Eratosthenes (276-194 B.C.E. ), al-Khowarizmi (780-850), Fibonacci (1170-1250), Maurolico (14941575), Mersenne (1588-1648), Descartes (1596-1650), Fermat (1601-1665), Goldbach (1690-1764), Stirling
(1692-1770), Bezout (1730-1783), Gauss (1777-1855), Lame (1795-1870), De Morgan (1806-1871), Lovelace
(1815-1852), Boole (1815-1864), and Dodgson (1832-1898). We draw the graph by connecting two people if
their date ranges overlap. Note that there is a complete subgraph (see Section 10.2) consisting of the last six
people listed. A few of the vertices are isolated (again see Section 10.2). In all our graph has 17 vertices and
22 edges. A graph like this is called an interval graph, since each vertex can be associated with an interval
of real numbers; it is a special case of an intersection graph, where two vertices are adjacent if the sets
associated with those vertices have a nonempty intersection (see Exercise 13).
Fermat~escartes
Lame
Goldbach
Bezout
Mersenne
De Morgan
Aristotle
•
Euclid
Fibonacci
•
•
Gauss
Eratosthenes
Maurolico
al-Khowarizmi
Dodgson
Boole
354
Chapter 10
Graphs
19. We draw a picture of the graph in question, which is a directed graph. We draw an edge from u to v if we
are told that u can influence v. For instance the Chief Financial Officer is an isolated vertex since she is
influenced by no one and influences no one.
Chief Fin. Off
•
D.c Opers.~Pros.
Dir Mrkt.
Dir R&D
21. We draw a picture of the graph in question, which is a directed graph. We draw an edge from u to v if we
are told that u beat v.
Tigers,~lue Jays
Orioles~Card1nals
23. We could compile a list of phone numbers (the labels on the vertices) in the February call graph that were
not present in January, and a list of the January numbers missing in February. For each number in each list,
we could make a list of the numbers they called or were called by, using the edges in the call graphs. Then we
could look for February lists that were very similar to January lists. If we found a new February number that
had almost the same calling pattern as a defunct January number, then we might suspect that these numbers
belonged to the same person, who had recently changed his or her number.
25. For each e-mail address (the labels on the vertices), we could make a list of the other addresses they sent
messages to or received messages from. If we see two addresses that had almost the same communication
pattern, then we might suspect that these addresses belonged to the same person, who had recently changed
his or her e-mail address.
27. The vertices represent the people at the party. Because it is possible that a knows b's name but not vice
versa, we need a directed graph. We will include an edge associated with (u, v) if and only if u knows v's
name. There is no need for multiple edges (either a knows b's name or he doesn't). One could argue that
we should not clutter the model with loops, because obviously everyone knows her own name. On the other
hand, it certainly would not be wrong to include loops, especially if we took the instructions literally.
29. We should use a directed graph, with the vertices being the courses and the edges showing the prerequisite
relationship. Specifically, an edge from u to v means that course u is a prerequisite for course v. Courses
that do not have any prerequisites are the courses with in-degree 0, and courses that are not the prerequisite
for any other courses have out-degree 0. An interesting question would be how to model courses that are
co-requisites (in two different senses-either courses u and v must be taken at the same time, or course u
must be taken before course v or in the same semester as course v).
31. For this to be interesting, we want the graph to model all marriages, not just ones that are currently active.
(In the latter case, for the Western world, there would be at most one edge incident to each vertex.) So we
let the set of vertices be a set of people (for example, all the people in North America who lived at any point
in the 20th century), and two vertices are joined by an edge if the two people were ever married. Since laws
in the 20th century allowed only marriages between persons of the opposite sex, and ignoring complications
caused by sex-change operations, we note that this graph has the property that there are two types of vertices
(men and women), and every edge joins vertices of opposite types. In the next section we learn that the word
used to describe a graph like this is bipartite.
Section 10.2
Graph Terminology and Special Types of Graphs
355
33. We draw a picture of the directed graph in question. There is an edge from u to v if the assignment made
in u can possibly influence the assignment made in v. For example, there is an edge from S 3 to S 6 , since
the assignment in S 3 changes the value of y, which then influences the value of z (in S4) and hence has a
bearing on 8 6 . We assume that the statements are to be executed in the given order, so, for example, we do
not draw an edge from 85 to 82.
35. The vertices in the directed graph represent people in the group. We put a directed edge into our directed
graph from every vertex A to every vertex B =/=- A (we do not need loops), and furthermore we label that edge
with one of the three labels L, D, or N . Let us see how to incorporate this into the mathematical definition.
Let us call such a thing a directed graph with labeled edges. It is defined to be a triple (V, E, f), where (V, E)
is a directed graph (i.e., V is a set of vertices and E is a set of ordered pairs of elements of V) and f is a
function from E to the set { L, D, N}. Here we are simply thinking of f (e) as the attitude of the person at
the tail (initial vertex-see Section 10.2) of e toward the person at the head (terminal vertex) of e.
SECTION 10.2
Graph Terminology and Special Types of Graphs
Graph theory is sometimes jokingly called the "theory of definitions," because so many terms can be--and
have been-defined for graphs. A few of the most important concepts are given in this section; others appear
in the rest of this chapter and the next, in the exposition and in the exercises. As usual with definitions,
it is important to understand exactly what they are saying. You should construct some examples for each
definition you encounter-examples both of the thing being defined and of its absence. Some students find it
useful to build a dictionary as they read, including their examples along with the formal definitions.
The handshaking theorem (that the sum of the degrees of the vertices in a graph equals twice the number
of edges), although trivial to prove, is quite handy, as Exercise 55, for example, i1lustrates. Be sure to look
at Exercise 43, which deals with the problem of when a sequence of numbers can possibly be the degrees of
the vertices of a simple graph. Some interesting subtleties arise there, as you will discover when you try to
draw the graphs. Many arguments in graph theory tend to be rather ad hoc, really getting down to the nitty
gritty, and Exercise 43c is a good example. Exercise 51 is really a combinatorial problem; such problems
abound in graph theory, and entire books have been written on counting graphs of various types. The notion
of complementary graph, introduced in Exercise 59, will appear again later in this chapter, so it would be
wise to look at the exercises dealing with it.
1. There are 6 vertices here, and 6 edges. The degree of each vertex is the number of edges incident to it.
Thus deg(a) = 2, deg(b) = 4, deg(c) = 1 (and hence c is pendant), deg(d) = 0 (and hence d is isolated),
deg(e) = 2, and deg(!)= 3. Note that the sum of the degrees is 2 + 4 + 1+0 + 2 + 3 = 12, which is twice
the number of edges.
3. There are 9 vertices here, and 12 edges. The degree of each vertex is the number of edges incident to it.
Thus deg(a) = 3, deg(b) = 2, deg(c) = 4, deg(d) = 0 (and hence dis isolated), deg(e) = 6, deg(!) = 0
(and hence f is isolated), deg(g) = 4, deg(h) = 2, and deg(i) = 3. Note that the sum of the degrees is
3 + 2 + 4 + 0 + 6 + 0 + 4 + 2 + 3 = 24, which is twice the number of edges.
Chapter 10
356
Graphs
5. By Theorem 2 the number of vertices of odd degree must be even. Hence there cannot be a graph with 15
vertices of odd degree 5. (We assume that the problem was meant to imply that the graph contained only
these 15 vertices.)
7. This directed graph has 4 vertices and 7 edges. The in-degree of vertex a is deg- (a) = 3 since there are
3 edges with a as their terminal vertex; its out-degree is deg+ (a) = 1 since only the loop has a as its
initial vertex. Similarly we have deg-(b) = 1, deg+(b) = 2, deg-(c) = 2, deg+(c) = 1, deg-(d) = 1, and
deg+ (d) = 3 . As a check we see that the sum of the in-degrees and the sum of the out-degrees are equal (both
are equal to 7).
9. This directed multigraph has 5 vertices and 13 edges. The in-degree of vertex a is deg- (a) = 6 since there
are 6 edges with a as their terminal vertex; its out-degree is deg+(a) = 1. Similarly we have deg-(b) = 1,
deg+(b) = 5, deg-(c) = 2, deg+(c) = 5, deg-(d) = 4, deg+(d) = 2, deg-(e) = 0, and deg+(e) = 0 (vertex e
is isolated). As a check we see that the sum of the in-degrees and the sum of the out-degrees are both equal
to the number of edges ( 13 ).
11. To form the underlying undirected graph we simply take all the arrows off the edges. Thus, for example, the
edges from e to d and from d to e become a pair of parallel edges between e and d.
•f
13. Since a person
collaborators v
by that vertex.
single-authored
person.
is joined by an edge to each of his or her collaborators, the degree of v is the number of
has. Similarly, the neighborhood of a vertex is the set of coauthors of the person represented
An isolated vertex represents a person who has no coauthors (he or she has published only
papers), and a pendant vertex represents a person who has published with just one other
15. Since there is a directed edge from u to v for each call made by u to v, the in-degree of v is the number
of calls v received, and the out-degree of u is the number of calls u made. The degree of a vertex in the
undirected version is just the sum of these, which is therefore the number of calls the vertex was involved in.
17. Since there is a directed edge from u to v to represent the event that u beat v when they played, the in-degree
of v must be the number of teams that beat v, and the out-degree of u must be the number of teams that u
beat. In other words, the pair (deg+(v),deg-(v)) is the win-loss record of v.
19. Model the friendship relation with a simple undirected graph in which the vertices are people in the group,
and two vertices are adjacent if those two people are friends. The degree of a vertex is the number of friends
in the group that person has. By Exercise 18, there are two vertices with the same degree, which means that
there are two people in the group with the same number of friends in the group.
21. To show that this graph is bipartite we can exhibit the parts and note that indeed every edge joins vertices
in different parts. Take { e} to be one part and {a, b, c, d} to be the other (in fact there is no choice in the
matter). Each edge joins a vertex in one part to a vertex in the other. This graph is the complete bipartite
graph K 1 , 4 .
Section 10.2
Graph Terminology and Special Types of Graphs
357
23. To show that a graph is not bipartite we must give a proof that there is no possible way to specify the parts.
(There is another good way to characterize nonbipartite graphs, but it takes some notions not introduced until
Section 10.4.) We can show that this graph is not bipartite by the pigeonhole principle. Consider the vertices
b, c, and f. They form a triangle~each is joined by an edge to the other two. By the pigeonhole principle,
at least two of them must be in the same part of any proposed bipartition. Therefore there would be an edge
joining two vertices in the same part, a contradiction to the definition of a bipartite graph. Thus this graph
is not bipartite.
An alternative way to look at this is given by Theorem 4. Because of the triangle, it is impossible to color
the vertices to satisfy the condition given there.
25. As in Exercise 23, we can show that this graph is not bipartite by looking at a triangle, in this case the triangle
formed by vertices b, d, and e. Each of these vertices is joined by an edge to the other two. By the pigeonhole
principle, at least two of them must be in the same part of any proposed bipartition. Therefore there would
be an edge joining two vertices in the same part, a contradiction to the definition of a bipartite graph. Thus
this graph is not bipartite.
27. a) The bipartite graph has vertices h, s, n, and w representing the support areas and P, Q, R, and S
representing the employees. The qualifications are modeled by the bipartite graph with edges Ph, Pn, Pw,
Qs, Qn, Rn, Rw, Sh, and Ss.
b) Since every vertex representing an area has degree at least 2, the condition in Hall's theorem is satisfied
for sets of size less than 3. We can easily check that the number of employees qualified for each of the four
subsets of size 3 is at least 3, and clearly the number of employees qualified for each of the subsets of size 4
has size 4.
c) The answer is not unique; one complete matching is {Pn, Qs, Rw, Sh}, which is easily found by inspection.
29. The partite sets are the set of women ({Tina, Uma, Vandana, Xia, Zelda}) and the set of men ( { Anil, Barry,
Emilio, Sandeep, Teja} ). We will use first letters for convenience (but J for Teja). The given information tells
us that we have edges AV, AZ, BT, BX, BU, ET, EZ, JT, JZ, ST, and SV in our graph. We do not
put an edge between a man and a woman he is not willing to marry. By inspection we find that the condition
in Hall's theorem is violated by {U, X}, because these two vertices are adjacent only to B. In other words,
only Barry is willing to marry Uma and Xia, so there can be no matching.
31. We model this with an undirected bipartite graph, with the men and the women represented by the vertices in
the two parts and an edge between two vertices if they are willing to marry each other. By Hall's theorem, it
is enough to show that for every set S of women, the set N(S) of men willing to marry them has cardinality
at least ISi. A clever way to prove this is by counting edges. Let m be the number of edges between S and
N(S). Since every vertex in S has degree k, it follows that m = kl SI. Because these edges are incident to
N(S), it follows that m ::; klN(S)I. Combining these two facts gives klSI ::; klN(S)I, so IN(S)I ?: ISi, as
desired.
33. a) By definition, the vertices are a, b, c, and f, and the edges are all the edges of the given graph joining
vertices in this list, namely ab, af, be, and bf.
b) Contracting edge bf merges the vertices b and f into a new vertex; call it x. Edges ab and af are
replaced by edge ax; edges eb and ef are replaced by edge ex; and edge cb is replaced by edge ex. Vertex
d continues to be an isolated vertex in the contracted graph.
35. a) Obviously Kn has n vertices. It has C(n, 2) = n(n - 1)/2 edges, since each unordered pair of distinct
vertices is an edge.
358
Chapter 10
Graphs
b) Obviously Cn has n vertices. Just as obviously it has n edges.
c) The wheel Wn is the same as Cn with an extra vertex and n extra edges incident to that vertex. Therefore
it has n + 1 vertices and n + n = 2n edges.
d) By definition Km,n has m + n vertices. Since it has one edge for each choice of a vertex in the one part
and a vertex in the other part, it has mn edges.
e) Since the vertices of Qn are the bit strings of length n, there are 2n vertices. Each vertex has degree n,
since there are n strings that differ from any given string in exactly one bit (any one of the n different bits
can be changed). Thus the sum of the degrees is n2n. Since this must equal twice the number of edges (by
the handshaking theorem), we know that there are n2n /2 = n2n-i edges.
37. In each case we just record the degrees of the vertices in a list, from largest to smallest.
a) Each of the four vertices is adjacent to each of the other three vertices, so the degree sequence is 3, 3, 3, 3.
b) Each of the four vertices is adjacent to its two neighbors in the cycle, so the degree sequence is 2, 2, 2, 2.
c) Each of the four vertices on the rim of the wheel is adjacent to each of its two neighbors on the rim, as well
as to the middle vertex. The middle vertex is adjacent to the four rim vertices. Therefore the degree sequence
is 4, 3, 3, 3, 3.
d) Each of the vertices in the part of size two is adjacent to each of the three vertices in the part of size three,
and vice versa, so the degree sequence is 3, 3, 2, 2, 2.
e) Each of the eight vertices in the cube is adjacent to three others (for example, 000 is adjacent to 001, 010,
and 100. Therefore the degree sequence is 3, 3, 3, 3, 3, 3, 3, 3.
39. Each of the n vertices is adjacent to each of the other n - 1 vertices, so the degree sequence is simply
n - 1, n - 1, ... , n - 1, with n terms in the sequence.
41. The number of edges is half the sum of the degrees (Theorem 1). Therefore this graph has (5
2 + 1)/2 = 7 edges. A picture of this graph is shown here (it is essentially unique).
+2+2+2+
43. There is no such graph in part (b), since the sum of the degrees is odd (and also because a simple graph with
5 vertices cannot have any degrees greater than 4). Similarly, the odd degree sum prohibits the existence of
graphs with the degree sequences given in part (d) and part (f). There is no such graph in part (c), since
the existence of two vertices of degree 4 implies that there are two vertices each joined by an edge to every
other vertex. This means that the degree of each vertex has to be at least 2, and there can be no vertex of
degree 1. The graphs for part (a) and part ( e) are shown below; one can draw them after just a little trial
and error.
(a)
(e)
45. We need to prove two conditional statements. First, suppose that di , d 2 , ... , dn is graphic. We must show
that the sequence whose terms are d2 - 1, d 3 - 1, ... , dd 1 +1 - 1, dd 1 + 2 , dd 1 +3, ... , dn is graphic once it
is put into nonincreasing order. Apparently what we want to do is to remove the vertex of highest degree
(di) from a graph with the original degree sequence and reduce by 1 the degrees of the vertices to which it is
Section 10.2
Graph Terminology and Special Types of Graphs
359
adjacent, but we also want to make sure that those vertices are the ones with the highest degrees among the
remaining vertices. In Exercise 44 it is proved that if the original sequence is graphic, then in fact there is a
graph having this degree sequence in which the vertex of degree di is adjacent to the vertices of degrees d 2 ,
d3, ... , dd, +1 . Thus our plan works, and we have a graph whose degree sequence is as desired.
Conversely, suppose that di , d 2 , ... , dn is a nonincreasing sequence such that the sequence d 2 - 1,
d3 - 1, ... , dd, + i - 1, dd, +2 , dd, +3 , ... , dn is graphic once it is put into nonincreasing order. Take a graph
with this latter degree sequence, where vertex Vi has degree di - 1 for 2 ~ i ~ di + 1 and vertex vi has degree
di for di + 2 ~ i ~ n. Adjoin one new vertex (call it vi), and put in an edge from vi to each of the vertices
V2, V3, ... , Vd, +i. Then clearly the resulting graph has degree sequence di, d2, ... , dn.
47. Let di, d2, ... , dn be a nonincreasing sequence of nonnegative integers with an even sum. We want to
construct a pseudograph with this as its degree sequence. Even degrees can be achieved using only loops, each
of which contributes 2 to the count of its endpoint; vertices of odd degrees will need a non-loop edge, but
one will suffice (the rest of the count at that vertex will be made up by loops). Following the hint, we take
vertices vi, v2, ... , Vn and put ld,/2 J loops at vertex Vi, for i = 1, 2, ... , n. For each i, vertex Vi now has
degree either d, (if di is even) or d, - 1 (if di is odd). Because the original sum was even, the number of
vertices falling into the latter category is even. If there are 2k such vertices, pair them up arbitrarily, and put
in k more edges, one joining the vertices in each pair. The resulting graph will have degree sequence di, d 2 ,
· · ·, dn.
49. We will count the subgraphs in terms of the number of vertices they contain. There are clearly just 3 subgraphs
consisting of just one vertex. If a subgraph is to have two vertices, then there are C(3, 2) = 3 ways to choose
the vertices, and then 2 ways in each case to decide whether or not to include the edge joining them. This
gives us 3 · 2 = 6 subgraphs with two vertices. If a subgraph is to have all three vertices, then there are 23 = 8
ways to decide whether or not to include each of the edges. Thus our answer is 3 + 6 + 8 = 17.
51. This graph has a lot of subgraphs. First of all, any nonempty subset of the vertex set can be the vertex
set for a subgraph, and there are 15 such subsets. If the set of vertices of the subgraph does not contain
vertex a, then the subgraph can of course have no edges. If it does contain vertex a, then it can contain or
fail to contain each edge from a to whichever other vertices are included. A careful enumeration of all the
possibilities gives the 34 graphs shown below.
DODOO
LJD[JDDJ
DDDDLJ
[J [J CJ LJ CJ
360
Chapter 10
Graphs
Dl D [] lSl lfSl
~lf:lD:l
DD. LJLLJ
d
[] [SJ ~
cd
l5J
53. a) The complete graph Kn is regular for all values of n :::: 1, since the degree of each vertex is n - 1.
b) The degree of each vertex of Cn is 2 for all n for which Cn is defined, namely n :::: 3, so Cn is regular for
all these values of n.
c) The degree of the middle vertex of the wheel W n is n, and the degree of the vertices on the "rim" is 3.
Therefore l-Vn is regular if and only if n = 3. Of course W3 is the same as K 4 .
d) The cube Qn is regular for all values of
n:::: 0, since the degree of each vertex in Qn is n. (Note that Qo
is the graph with 1 vertex.)
55. If a graph is regular of degree 4 and has n vertices, then by the handshaking theorem it has 4n/2 = 2n edges.
Since we are told that there are 10 edges, we just need to solve 2n = 10. Thus the graph has 5 vertices. The
complete graph K 5 is one such graph (and the only simple one).
57. \Ve draw the answer by superimposing the graphs (keeping the positions of the vertices the same).
59. a) The complement of a complete graph is a graph with no edges.
b) Since all the edges between the parts are present in Km,n, but none of the edges between vertices in the
same part are, the complement must consist precisely of the disjoint union of a Km and a Kn, i.e., the graph
containing all the edges joining two vertices in the same part and no edges joining vertices in different parts.
c) There is really no better way to describe this graph than simply by saying it is the complement of Cn.
One representation would be to take as vertex set the integers from 1 to n, inclusive, with an edge between
distinct vertices i and j as long as i and j do not differ by ±1, modulo n.
d) Again, there is really no better way to describe this graph than simply by saying it is the complement of
Qn. One representation would be to take as vertex set the bit strings of length n, with two vertices joined by
an edge if the bit strings differ in more than one bit.
= v(v - 1)/2 edges, and since G has all the edges of Kv that G is missing, it is clear
that G has [v(v - 1)/2] - e edges.
61. Since Kv has C(v, 2)
63. If G has n vertices, then the degree of vertex v in G is n - 1 minus the degree of v in G (there will be
an edge in G from v to each of the n - 1 other vertices that v is not adjacent to in G). The order of the
sequence will reverse, of course, because if d, :::: d1 , then n - l - d, :S: n - l - d1 . Therefore the degree sequence
of G will be n - 1 - dn, n - 1 - dn-l, ... , n - 1 - d2, n - 1 - d1.
Section 10.3
Representing Graphs and Graph Isomorphism
361
65. Consider the graph G U G. Its vertex set is clearly the vertex set of G; therefore it has n vertices. If u and
v are any two distinct vertices of GU G, then either the edge between u and v is in G, or else by definition
it is in G. Therefore by definition of union, it is in G U G. Thus by definition G U G is the complete graph
Kn·
67. These pictures are identical to the figures in those exercises, with one change, namely that all the arrowheads
are turned around. For example, rather than there being a directed edge from a to b in #7, there is an edge
from b to a. Note that the loops are unaffected by changing the direction of the arrowhead-a loop from a
vertex to itself is the same, whether the drawing of it shows the direction to be clockwise or counterclockwise.
69. It is clear from the definition of converse that a directed graph G = (V, E) is its own converse if and only
if it satisfies the condition that (u, v) E E if and only if (u, v) E E. But this is precisely the definition of
symmetry for the associated relation.
71. Our picture is just like Figure 13, but with only three vertices on each side.
P(0,0)
P(l,O)
P(2,0)
P(0,1)
P(l,1)
P(2,1)
P(0,2)
P(l,2)
P(2,2)
73. Suppose P(i,j) and P(k, l) need to communicate. Clearly by using Ii- kl hops we can move from P(i,j) to
P(k,j). Then using IJ - ll hops we can move from P(k,j) to P(k,l). In all we used Ii - kl+ lj- ll hops.
But each of these absolute values is certainly less than m, since all the indices are less than m. Therefore the
sum is less than 2m, so it is 0( m).
SECTION 10.3
Representing Graphs and Graph Isomorphism
Human beings can get a good feeling for a small graph by looking at a picture of it drawn with points in
the plane and lines or curves joining pairs of these points. If a graph is at all large (say with more than a
dozen vertices or so), then the picture soon becomes too crowded to be useful. A computer has little use
for nice pictures, no matter how small the vertex set. Thus people and machines need more precise-more
discrete-representations of graphs. In this section we learned about some useful representations. They are
for the most part exactly what any intelligent person would come up with, given the assignment to do so.
The only tricky idea in this section is the concept of graph isomorphism. It is a special case of a more
general notion of isomorphism, or sameness, of mathematical objects in various settings. Isomorphism tries
to capture the idea that all that really matters in a graph is the adjacency structure. If we can find a way
to superimpose the graphs so that the adjacency structures match, then the graphs are, for all purposes that
matter, the same. In trying to show that two graphs are isomorphic, try moving the vertices around in your
mind to see whether you can make the graphs look the same. Of course there are often lots of things to help.
For example, in every isomorphism, vertices that correspond must have the same degree.
A good general strategy for determining whether two graphs are isomorphic might go something like this.
First check the degrees of the vertices to make sure there are the same number of each degree. See whether
vertices of corresponding degrees follow the same adjacency pattern (e.g., if there is a vertex of degree 1
adjacent to a vertex of degree 4 in one of the graphs, then there must be the same pattern in the other, if the
Chapter 10
362
Graphs
graphs are isomorphic). Then look for triangles in the graphs, and see whether they correspond. Sometimes,
if the graphs have lots of edges, it is easier to see whether the complements are isomorphic (see Exercise 46).
If you cannot find a good reason for the graphs not to be isomorphic (an invariant on which they differ),
then try to write down a one-to-one and onto function that shows them to be isomorphic (there may be more
than one such function); such a function has to have vertices of like degrees correspond, so often the function
practically writes itself. Then check each edge of the first graph to make sure that it corresponds to an edge
of the second graph under this correspondence.
Unfortunately, no one has yet discovered a really good algorithm for determining graph isomorphism that
works on all pairs of graphs. Research in this subject has been quite active in recent years. See Writing
Project 10.
1. Adjacency lists are lists of lists. The adjacency list of an undirected graph is simply a list of the vertices of the
given graph, together with a list of the vertices adjacent to each. The list for this graph is as follows. Since,
for instance, b is adjacent to a and d, we list a and d in the row for b.
Vertex
Adjacent vertices
a
b,c,d
a,d
a,d
a,b,c
b
c
d
3. To form the adjacency list of a directed graph, we list, for each vertex in the graph, the terminal vertex of
each edge that has the given vertex as its initial vertex. The list for this directed graph is as follows. For
example, since there are edges from d to each of b, c, and d, we put those vertices in the row for d.
Initial vertex
a
b
Terminal vertices
a,b,c,d
d
c
d
b,c,d
a,b
5. For Exercises 5-8 we assume that the vertices are listed in alphabetical order. The matrix contains a 1 as
entry (i, j) if there is an edge from vertex i to vertex j; otherwise that entry is 0.
[; ~ ~ il
7. This is similar to Exercise 5. Note that edges have direction here, so that, for example, the (1, 2) entry is a 1
since there is an edge from a to b, but the (2, 1) entry is a 0 since there is no edge from b to a. Also, the
(1, 1) entry is a 1 since there is a loop at a, but the (2, 2) entry is a 0 since there is no loop at b.
[! ~ ~ ~]
9. We can solve these problems by first drawing the graph, then labeling the vertices, and finally constructing
the matrix by putting a 1 in position (i, j) whenever vertices i and j are joined by an edge. It helps to
choose a nice order, since then the matrix will have nice patterns in it.
Section 10.3
Representing Graphs and Graph Isomorphism
363
a) The order of the vertices does not matter, since they all play the same role. The matrix has O's on the
diagonal, since there are no loops in the complete graph.
b) We put the vertex in the part by itself first.
[l
~ ~ ~ ~1
0 0 0 0
c) We put the vertices in the part of size 2 first. Notice the block structure.
o0 o0
11
1 1 0
[1 1 0
1 1 0
1
1
11
1
0
0
0
0
0
0
d) We put the vertices in the same order in the matrix as they are around the cycle.
[~ ~ ~ !l
e) We put the center vertex first. Note that the last four columns of the last four rows represent a C4 .
o1
01 11 01 111
1 1 0 1 0
[1 0 1 0 1
1
1
0
1
0
f) We can label the vertices by the binary numbers from 0 to 7. Thus the first row (also the first column) of
this matrix corresponds to the string 000, the second
this is an 8 x 8 matrix.
0 1 1 0
1 0 0 1
1 0 0 1
0 1 1 0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
to the string 001, and so on. Since Q 3 has 8 vertices,
1
0
0
0
0
1
1
0
0
1
0
0
1
0
0
1
0
0
1
0
1
0
0
0
0
0
1
0
1
1
1 0
11. This graph has four vertices and is directed, since the matrix is not symmetric. We draw the four vertices as
points in the plane, then draw a directed edge from vertex i to vertex j whenever there is a 1 in position
(i, j) in the given matrix.
~b
a~
Chapter 10
364
Graphs
13. We use alphabetical order of the vertices for Exercises 13-15. If there are k parallel edges between vertices i
and j, then we put the number k into the (i,j)th entry of the matrix. In this exercise, there is only one pair
of parallel edges.
15. This is similar to Exercise 13. In this graph there are loops, which are represented by entries on the diagonal.
For example, the loop at c is shown by the 1 as the (3, 3)th entry.
[
~ ~
2
1
~ 0~i
1
1 2 0 1
17. Because of the numbers larger than 1, we need multiple edges in this graph.
6
19. We use alphabetical order of the vertices. We put a 1 in position (i, j) if there is a directed edge from vertex
i to vertex j; otherwise we make that entry a 0. Note that loops are represented by l's on the diagonal.
o
1
0 1
~ ~i
[0 1 1 1
1
0 0 0
21. This is similar to Exercise 19, except that there are parallel directed edges. If there are k parallel edges from
vertex i to vertex j, then we put the number k into the (i, j) th entry of the matrix. For example, since there
are 2 edges from a to c, the (1, 3)th entry of the adjacency matrix is 2; the loop at c is shown by the 1 as
the (3, 3)th entry.
23. Since the matrix is not symmetric, we need directed edges; furthermore, it must be a directed multigraph
because of the entries larger than 1. For example, the 2 in position (3, 2) means that there are two parallel
edges from vertex c to vertex b .
Section 10.3
Representing Graphs and Graph Isomorphism
365
25. Since the matrix is symmetric, it has to be square, so it represents a graph of some sort. In fact, such a
matrix does represent a simple graph. The fact that it is a zero-one matrix means that there are no parallel
edges. The fact that there are O's on the diagonal means that there are no loops. The fact that the matrix is
symmetric means that the edges can be assumed to be undirected. Note that such a matrix also represents a
directed graph in which all the edges happen to appear in antiparallel pairs (see the solution to Exercise ld
in Section 10.l for a definition), but that is irrelevant to this question; the answer to the question asked is
"yes."
27. In an incidence matrix we have one column for each edge. We use alphabetical order of the vertices. Loops
are represented by columns with one 1; other edges are represented by columns with two l's. The order in
which the columns are listed is immaterial.
Exercise 13
[~
0 0 0
1 1 1
1 0 0
0 1 1
Exercise 15
[~
1
0
1
0
~]
1 1 0 0
0 0 1 1
1 0 0 1
0 1 0 0
Exercise 14
[~
1
1
0
0
1
1
0
0
1 0 0 0
0 1 0 0
0 1 1 1
1 0 1 1
~]
~]
0 0 0
1 1 0
0 0 1
1 1 0
29. In an undirected graph, each edge incident to a vertex j contributes 1 in the jth column; thus the sum of
the entries in that column is just the number of edges incident to j. Another way to state the answer is that
the sum of the entries is the degree of j minus the number of loops at j, since each loop contributes 2 to the
degree count.
In a directed graph, each edge whose terminal vertex is j contributes 1 in the lh column; thus the sum
of the entries in that column is just the number of edges that have j as their terminal vertex. Another way
to state the answer is that the sum of the entries is the in-degree of j .
31. Since each column represents an edge, the sum of the entries in the column is either 2, if the edge has 2
incident vertices (i.e., is not a loop), or 1 if it has only 1 incident vertex (i.e., is a loop).
33. a) The incidence matrix for Kn has n rows and C( n, 2) columns. For each
there is a column with l's in rows i and j and O's elsewhere.
and j with 1 :::; i < j :::; n,
b) The matrix looks like this, with n rows and n columns.
1 0 0
1 1 0
0 1
0 0
0 1 1
0 0
0
0
0
1
0 0 0
0 0 0
0
1 0
1 1
c) The matrix looks like the matrix for Cn, except with an extra row of O's (which we have put at the end),
since the vertex "in the middle" is not involved in the edges "around the outside,'' and n more columns for
the "spokes." We show some extra space between the rim edge columns and the spoke columns; this is for
Chapter 10
366
Graphs
human convenience only and does not have any bearing on the matrix itself.
1 0 0
1 1 0
0 1 1
0 0 1
0
0
0
0
0
0
0
1 0
1 1
0 0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0 0
1 0
0 1
0 0
0
0
0
0
0 0 0
0 0 0
1 1 1
0
1
1
d) This matrix has m + n rows and mn columns, one column for each pair (i,j) with 1
1 j S n. We have put in some extra spacing for readability of the pattern.
s
1 1
0 0
1
0
0
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1 0
0 1
0
0
1 0
0 1
0
0
1 0
0 1
0
0
0
1
0
1
0
1
0
0
0
0
s
i S m and
35. These graphs are isomorphic, since each is the 5-cycle. One isomorphism is f(u1) = v1, f(u2) = v3, f(u3) =
V5, l(u4) =v2, and l(us) =V4.
37. These graphs are isomorphic, since each is the 7-cycle (this is just like Exercise 35).
39. These two graphs are isomorphic. One can see this visually-just imagine "moving" vertices u 1 and u 4
into the inside of the rectangle, thereby obtaining the picture on the right. Formally, one isomorphism is
l(u1) = V5, l(u2) = v2, l(u3) = v3, l(u4) = V5, l(us) = V4, and l(u5) = v1.
41. These graphs are not isomorphic. In the first graph the vertices of degree 3 are adjacent to a common vertex.
This is not true of the second graph.
43. These are isomorphic. One isomorphism is l (u1) = V1, l( u2) = Vg, l (u3) = V4, l (u4) = V3, l( us) = V2,
f(u5) =Vs, f(u1) = v1, f(us) = 115, f(ug) = 1110, and l(u10) = 115.
45. We must show that being isomorphic is reflexive, symmetric, and transitive. It is reflexive since the identity
function from a graph to itself provides the isomorphism (the one-to-one correspondence)-certainly the identity function preserves adjacency and nonadjacency. It is symmetric, since if f is a one-to-one correspondence
that makes G1 isomorphic to G2 , then 1- 1 is a one-to-one correspondence that makes G2 isomorphic to G 1 ;
that is, 1- 1 is a one-to-one and onto function from Vi to Vi such that c and d are adjacent in G 2 if and
only if l- 1 (c) and l- 1 (d) are adjacent in G 1 . It is transitive, since if l is a one-to-one correspondence that
makes G 1 isomorphic to G2 , and g is a one-to-one correspondence that makes G2 isomorphic to G3, then
go l is a one-to-one correspondence that makes G 1 isomorphic to G 3 .
47. If a vertex is isolated, then it has no adjacent vertices. Therefore in the adjacency matrix the row and column
for that vertex must contain all O's.
Section 10.3
Representing Graphs and Graph Isomorphism
367
49. Let Vi and Vi be the two parts, say of sizes m and n, respectively. We can number the vertices so that
all the vertices in Vi
columns. Since there
all be O's. Similarly,
rows must all be O's.
come before all the vertices in Vi. The adjacency matrix has m + n rows and m + n
are no edges between two vertices in Vi, the first m columns of the first m rows must
since there are no edges between two vertices in Vi, the last n columns of the last n
This is what we were asked to prove.
51. There are two such graphs, which can be found by trial and error. (We need only look for graphs with 5
vertices and 5 edges, since a self-complementary graph with 5 vertices must have C(5, 2)/2 = 5 edges. If
nothing else, we can draw them all and find the complement of each. See the pictures for the solution of
Exercise 47d in Section 10.4.) One such graph is C 5 . The other consists of a triangle, together with an edge
from one vertex of the triangle to the fourth vertex, and an edge from another vertex of the triangle to the
fifth vertex.
53. If Cn is to be self-complementary, then Cn must have the same number of edges as its complement. We know
that Cn has n edges. Its complement has the number of edges in Kn minus the number of edges in Cn,
namely C(n, 2)-n = [n(n-1)/2]-n. If we set these two quantities equal we obtain [n(n-1)/2]-n = n, which
has n = 5 as its only solution. Thus C 5 is the only Cn that might be self-complementary-our argument
just shows that it has the same number of edges as its complement, not that it is indeed isomorphic to its
complement. However, it we draw C 5 and then draw its complement, then we see that the complement is
again a copy of C5 . Thus n = 5 is the answer to the problem.
55. We need to enumerate these graphs carefully to make sure of getting them all-leaving none out and not
duplicating any. Let us organize our catalog by the degrees of the vertices. Since there are only 3 edges, the
largest the degree could be is 3, and the only graph with 5 vertices, 3 edges, and a vertex of degree 3 is a
Ki, 3 together with an isolated vertex. If all the vertices that are not isolated have degree 2, then the graph
must consist of a C 3 and 2 isolated vertices. The only way for there to be two vertices of degree 2 (and
therefore also 2 of degree 1) is for the graph to be three edges strung end to end, together with an isolated
vertex. The only other possibility is for 2 of the edges to be adjacent and the third to be not adjacent to
either of the others. All in all, then, we have the 4 possibilities shown below.
See [ReWi] for more information about graph enumeration problems of this sort (such as Exercises 54,
56, and 68 in this section, Exercise 47 in Section 10.4, and supplementary exercises 2, 31, 32, and 40) .
•
57. a) Both graphs consist of 2 sides of a triangle; they are clearly isomorphic.
b) The graphs are not isomorphic, since the first has 4 edges and the second has 5 edges.
c) The graphs are not isomorphic, since the first has 4 edges and the second has 3 edges.
59. There are at least two approaches we could take here. One approach is to have a correspondence not only
of the vertices but also of the edges, with incidence (and nonincidence) preserved. In detail, we say that two
pseudographs G 1 = (Vi, E 1 ) and G 2 = (Vi, E 2 ) are isomorphic if there are one-to-one and onto functions
f : Vi -> Vi and g : E1 -> E2 such that for each vertex v E Vi and edge e E E1 , v is incident to e if and
only if f(v) is incident to g(e).
Another approach is simply to count the number of edges between pairs of vertices. Thus we can define
G1 = (Vi, E1) to be isomorphic to G2 = (Vi, E2) if there is a one-to-one and onto function f: Vi ->Vi such
that for every pair of (not necessarily distinct) vertices u and v in Vi , there are exactly the same number of
Chapter 10
368
Graphs
edges in E 1 with {u, v} as their set of endpoints as there are edges in E 2 with {! (u), f (v)} as their set of
endpoints.
61. We can tell by looking at the loop, the parallel edges, and the degrees of the vertices that if these directed graphs
are to be isomorphic, then the isomorphism has to be f (ui) = V3, f (u 2) = V4, f (u 3) = v 2 , and f (u4) = v1 .
We then need to check that each directed edge (uiiu 1 ) corresponds to a directed edge (f(ui),f(u 1 )). We
check that indeed it does for each of the 7 edges (and there are only 7 edges in the second graph). Therefore
the two graphs are isomorphic.
63. If there is to be an isomorphism, the vertices with the same in-degree would have to correspond, and the edge
between them would have to point in the same direction, so we would need u 1 to correspond to v3 , and u 2 to
correspond to 111. Similarly we would need u3 to correspond to v4 , and u 4 to correspond to v 2 . If we check
all 6 edges under this correspondence, then we see that adjacencies are preserved (in the same direction), so
the graphs are isomorphic.
65. If
f is an isomorphism from a directed graph C to a directed graph H, then f is also an isomorphism from
cc
to He. This is clear, because (u.v) is an edge of cc if and only if (v,u) is an edge of C if and only if
(f(v),f(u)) is an edge of H if and only if (f(u),f(v)) is an edge of He.
67. A graph with a triangle will not be bipartite, but cycles of even length are bipartite. So we could let one graph
be C6 and the other be the union of two disjoint copies of C 3 .
69. Suppose that the graph has v vertices and e edges. Then the incidence matrix is a v x e matrix, so its
transpose is an e x v matrix. Therefore the product is a v x v matrix. Suppose that we denote the typical
entry of this product by ai1 . Let t,k be the typical entry of the incidence matrix; it is either a 0 or a 1. By
definition
e
a,) =
2..:
tikt1k.
k=l
We can now read off the answer from this equation. If i-=/:- j, then ai1 is just a count of the number of edges
incident to both i and j -in other words, the number of edges between i and j . On the other hand aii is
equal to the number of edges incident to i .
71. Perhaps the simplest example would be to have the graphs have all degrees equaling 2. One way for this to
happen is for the graph to be a cycle. But it will also happen if the graph is a disjoint union of cycles. The
smallest example occurs when there are six vertices. If G 1 is the 6-cycle and G 2 is the union of two triangles,
then the degree sequences are (2, 2, 2, 2, 2, 2) for both, but obviously the graphs are not isomorphic. If we
want a connected example, then look at Exercise 41, where the degree sequence is (3, 3, 2, 2, 1, 1, 1, 1) for each
graph.
SECTION 10.4
Connectivity
Some of the most important uses of graphs deal with the notion of path, as the examples and exercises in
this and subsequent sections show. It is important to understand the definitions, of course. Many of the
exercises here are straightforward. The reader who wants to get a better feeling for what the arguments in
more advanced graph theory are like should tackle problems like Exercises 35-38.
Section 10.4
Connectivity
369
1. a) This is a path of length 4, but it is not simple, since edge {b, c} is used twice. It is not a circuit, since it
ends at a different vertex from the one at which it began.
b) This is not a path, since there is no edge from
c to a.
c) This is not a path, since there is no edge from b to a.
d) This is a path of length 5 (it has 5 edges in it). It is simple, since no edge is repeated. It is a circuit since
it ends at the same vertex at which it began.
3. This graph is not connected-it has three components.
5. This graph is not connected. There is no path from the vertices in one of the triangles to the vertices in the
other.
7. A connected component of an acquaintanceship graph represent a maximal set of people with the property
that for any two of them, we can find a string of acquaintances that takes us from one to the other. The word
"maximal" here implies that nobody else can be added to this set of people without destroying this property.
9. If a person has Erdos number n, then there is a path of length n from that person to Erdos in the collaboration
graph. By definition, that means that that person is in the same component as Erdos. Conversely, if a person
is in the same component as Erdos, then there is a path from that person to Erdos, and the length of a shortest
such path is that person's Erdos number.
11. a) Notice that there is no path from a to any other vertex, because both edges involving a are directed
toward a. Therefore the graph is not strongly connected. However, the underlying undirected graph is clearly
connected, so this graph is weakly connected.
b) Notice that there is no path from c to any other vertex, because both edges involving c are directed
toward c. Therefore the graph is not strongly connected. However, the underlying undirected graph is clearly
connected, so this graph is weakly connected.
c) The underlying undirected graph is clearly not connected (one component has vertices b, f, and e), so
this graph is neither strongly nor weakly connected.
13. The strongly connected components are the maximal sets of phone numbers for which it is possible to find
directed paths between every two different numbers in the set, where the existence of a directed path from
phone number x to another phone number y means that x called some number, which called another number,
... , which called y. (The number of intermediary phone numbers in this path can be any natural number.)
15. In each case we want to look for large sets of vertices all which of which have paths to all the others. For these
graphs, this can be done by inspection. These will be the strongly connected components.
a) Clearly {a, b, J} is a set of vertices with paths between all the vertices in the set. The same can be said of
{c, d, e} . Every edge between a vertex in the first set and a vertex in the second set is directed from the first,
to the second. Hence there are no paths from c, d, or e to a, b, or f, and therefore these vertices are not in
the same strongly connected component. Therefore these two sets are the strongly connected component.
b) The circuits a, e, d, c, b, a and a, e, d, h, a show that these six vertices are all in the same component. There
is no path from f to any of these vertices, and no path from g to any other vertex. Therefore f and g are
not in the same strong component as any other vertex. Therefore the strongly connected components are
{a, b, c, d, e, h}, {!}, and {g}.
c) It is clear that a and i are in the same strongly connected component. If we look hard, we can also
find the circuit b, h, f, g, d, e, d, b, so these vertices are in the same strongly connected component. Because of
edges ig and hi, we can get from either of these collections to the other. Thus {a, b, d, e, f, g, h, i} is a strong
component. We cannot travel from c to any other vertex, so c is in a component by itself.
Chapter 10
370
Graphs
17. The hardest part of this exercise is figuring out what we need to prove. It is enough to prove that if the strong
components of u and v are not disjoint then they are the same. So suppose that w is a vertex that is in
both the strong component of u and the strong component of v. (It is enough to consider the vertices in
these components, because the edges in a strong component are just all the edges joining the vertices in that
component.) This means that there are directed paths (in each direction) between u and w and between v
and w. It follows that there are directed paths from u to v and from v to u, via w. Suppose x is a vertex
in the strong component of u. Then x is also in the strong component of v, because there is a path from x
to v (namely the path from x to u followed by the path from u to v) and vice versa.
19. One approach here is simply to invoke Theorem 2 and take successive powers of the adjacency matrix
A~ ~ ~
[;
l
The answers are the off-diagonal elements of these powers. An alternative approach is to argue combinatorially
as follows. Without loss of generality, we assume that the vertices are called 1, 2, 3, 4, and the path is to run
from 1 to 2 . A path of length n is determined by choosing the n - 1 intermediate vertices. Each vertex in
the path must differ from the one immediately preceding it.
a) A path of length 2 requires the choice of 1 intermediate vertex, which must be different from both of the
ends. Vertices 3 and 4 are the only ones available. Therefore the answer is 2.
b) Let the path be denoted 1, x, y, 2 . If x = 2 , then there are 3 choices for y. If x = 3, then there are 2
choices for y; similarly if x = 4. Therefore there are 3 + 2 + 2 = 7 possibilities in all.
c) Let the path be denoted 1, x, y, z, 2. If x = 3, then by part (b) there are 7 choices for y and z. Similarly
if x = 4. If x = 2, then y and z can be any two distinct members of {1,3,4}, and there are P(3,2) = 6
ways to choose them. Therefore there are 7 + 7 + 6 = 20 possibilities in all.
d) Let the path be denoted 1,w,x,y,z,2. If w = 3, then by part (c) there are 20 choices for x, y, and z.
Similarly if w = 4. If w = 2, then x must be different from 2, and there are 3 choices for x. For each of
these there are by part (b) 7 choices for y and z. This gives a total of 21 possibilities in this case. Therefore
the answer is 20 + 20 + 21 = 61.
21. Graph G has a triangle ( u 1 , u 2 , u 3 ). Graph H does not (in fact, it is bipartite). Therefore G and H are not
isomorphic.
23. The drawing of G clearly shows it to be the cube Q 3 . Can we see H as a cube as well? Yes-we can view the
outer ring as the top face, and the inner ring as the bottom face. We can imagine walking around the top face
of G clockwise (as viewed from above), then dropping down to the bottom face and walking around it counterclockwise, finally returning to the starting point on the top face. This is the path u1, u2, u7, u6, u5, u4, u3, us, u1.
The corresponding path in H is v 1 , v2, V3, V4, vs, vs, V7, V6, v1. We can verify that the edges not in the path
do connect corresponding vertices. Therefore G ~ H.
25. As explained in the solution to Exercise 19, we could take powers of the adjacency matrix
A=
0
0
0
1
1
1
0 0
0 0
0 0
1 1
1 1
1 1
1 1 1
1 1 1
1 1 1
0 0 0
0 0 0
0 0 0
The answers are found in location (1, 2), for instance. Using the alternative approach is much easier than in
Exercise 19. First of all, two nonadjacent vertices must lie in the same part, so only paths of even length can
Section 10.4
Connectivity
371
join them. Also, there are clearly 3 choices for each intermediate vertex in a path. Therefore we have the
following answers:
a) 31 = 3
c) 33
b) 0
= 27
d) 0
27. There are two approaches here. We could use matrix multiplication on the adjacency matrix of this directed
graph (by Theorem 2), which is
1
0
1
0 0 0
1 0 0
0 0 0
0
2
1
1
3
Thus we can compute A for part (a), A for part (b ), and so on, and look at the (1, 5)th entry to determine
the number of paths from a to e. Alternately, we can argue in an ad hoc manner, as we do below.
a) There is just 1 path of length 2, namely a, b, e.
b) There are no paths of length 3, since after 3 steps, a path starting at a must be at b, c, or d.
c) For a path of length 4 to end at e, it must be at b after 3 steps. There are only 2 such paths, a, b, a, b, e
and a, d, a, b, e.
d) The only way for a path of length 5 to end at e is for the path to go around the triangle bee. Therefore
only the path a, b, e, c, b, e is possible.
e) There are several possibilities for a path of length 6. Since the only way to get to e is from b, we are asking
for the number of paths of length 5 from a to b. We can go around the square (a, b, e, d, a, b ), or else we can
jog over to either b or d and back twice-there being 4 ways to choose where to do the jogging. Therefore
there are 5 paths in all.
f) As in part ( d), it is clear that we have to use the triangle. We can either have a, b, a, b, e, c, b, e or
a, d, a, b, e, c, b, e or a, b, e, c, b, a, b, e. Thus there are 3 paths.
29. The definition given here makes it clear that u and v are related if and only if they are in the same componentin other words f(u) = f(v) where f(x) is the component in which x lies. Therefore by Exercise 9 in Section 9.5
this is an equivalence relation.
31. A cut vertex is one whose removal splits the graph into more components than it originally had (which is 1
in this case). Only vertex c is a cut vertex here. If it is removed, then the resulting graph will have two
components. If any other vertex is removed, then the graph remains connected.
33. There are several cut vertices here: b, c, e, and i. Removing any of these vertices creates a graph with more
than one component. The removal of any of the other vertices leaves a graph with just one component.
35. Without loss of generality, we can restrict our attention to the component in which the cut edge lies; other
components of the graph are irrelevant to this proposition. To fix notation, let the cut edge be uv. When the
cut edge is removed, the graph has two components, one of which contains v and the other of which contains u.
If v is pendant, then it is clear that the removal of v results in exactly the component containing u-a
connected graph. Therefore v is not a cut vertex in this case. On the other hand, if v is not pendant, then
there are other vertices in the component containing v-at least one other vertex w adjacent to v. (We are
assuming that this proposition refers to a simple graph, so that there is no loop at v.) Therefore when v 1s
removed, there are at least two components, one containing u and another containing w.
37. If every component of G is a single vertex, then clearly no vertex is a cut vertex (the removal of any of them
actually decreases the number of components rather than increasing it). Therefore we may as well assume
372
Chapter 10
Graphs
that some component of G has at least two vertices, and we can restrict our attention to that component; in
other words, we can assume that G is connected. One clever way to do this problem is as follows. Define the
distance between two vertices u and v, denoted d( u, v), to be the length of a shortest path joining u and v.
Now choose u and v so that d( u, v) is as large as possible. We claim that neither u nor v is a cut vertex.
Suppose otherwise, say that u is a cut vertex. Then v is in one component that results after u is removed,
and some vertex w is in another. Since there is no path from w to v in the graph with u removed, every
path from w to v must have passed through u. Therefore the distance between w and v must have been
strictly greater than the distance between u and v . This is a contradiction to the choice of u and v, and our
proof by contradiction is complete.
39. This problem is simply asking for the cut edges of these graphs.
a) The link joining Denver and Chicago and the link joining Boston and New York are the cut edges.
b) The following links are the cut edges: Seattle-Portland, Portland-San Francisco, Salt Lake City-Denver,
New York-Boston, Boston-Bangor, Boston-Burlington.
41. A vertex basis will be a set of people who collectively can influence everyone, at least indirectly, but none of
whom influences another member of that set (otherwise the set would not be minimal). The set consisting
of Deborah is a vertex basis, since she can influence everyone except Yvonne directly, and she can influence
Yvonne indirectly through Brian.
43. Since there can be no edges between vertices in different components, G will have the most edges when each
of the components is a complete graph. Since Kn, has C(ni, 2) edges, the maximum number of edges is the
sum given in the exercise.
45. Before we give a correct proof here, let us look at an incorrect proof that students often give for this exercise.
It goes something like this. ''Suppose that the graph is not connected. Then no vertex can be adjacent to
every other vertex, only to n - 2 other vertices. One vertex joined to n - 2 other vertices creates a component
with n - 1 vertices in it. To get the most edges possible, we must use all the edges in this component. The
number of edges in this component is thus C (n - 1, 2) = (n - 1) (n - 2) / 2, and the other component (with only
one vertex) has no edges. Thus we have shown that a disconnected graph has at most (n - l)(n - 2)/2 edges,
so every graph with more edges than that has to be connected." The fallacy here is in assuming-without
justification-that the maximum number of edges is achieved when one component has n - 1 vertices. What
if, say, there were two components of roughly equal size? Might they not together contain more edges? We will
see that the answer is "no," but it is important to realize that this requires proof-it is not obvious without
some calculations.
Here is a correct proof, then. Suppose that the graph is not connected. Then it has a component with k
vertices in it, for some k between 1 and n-1, inclusive. The remaining n- k vertices are in one or more other
components. The maximum number of edges this graph could have is then C(k, 2) + C(n - k, 2), which, after
a bit of algebra, simplifies to k 2 - nk + (n 2 - n) /2. This is a quadratic function of k. It is minimized when
k = n/2 (the k coordinate of the vertex of the parabola that is the graph of this function) and maximized at
the endpoints of the domain, namely k = 1 and k = n - 1. In the latter cases its value is (n - l)(n - 2)/2.
Therefore the largest number of edges that a disconnected graph can have is (n - l)(n - 2)/2, so every graph
with more edges than this must be connected.
47. We have to enumerate carefully all the possibilities.
a) There is obviously only 1, namely K 2 , the graph consisting of two vertices and the edge between them.
b) There are clearly 2 connected graphs with 3 vertices, namely K 3 and K 3 with one edge deleted, as
shown.
Section 10.4
373
Connectivity
c) There are several connected graphs with n = 4. If the graph has no circuits, then it must either be a path
of length 3 or the "star" K 1 , 3 . If it contains a triangle but no copy of C4, then the other vertex must be
pendant-only 1 possibility. If it contains a copy of C4 , then neither, one, or both of the other two edges
may be present-3 possibilities. Therefore the answer is 2 + 1 + 3 = 6. The graphs are shown below.
d) We need to enumerate the possibilities in some systematic way, such as by the largest cycle contained in
the graph. There are 21 such graphs, as can be seen by such an enumeration, shown below. First we show
those graphs with no circuits, then those with a triangle but no C4 or C 5 , then those with a C4 but no C5 ,
and finally those with a C 5 . In doing this problem we have to be careful not only not to leave out any graphs,
but also not to list any twice.
+
LS1 LS:-. 75:
LJ rsr- w
• • • • •
• •
I
•
~
~r=rw
OO®W
@~@@
49. In each case we just need to verify that the removal of an edge will not disconnect the graph.
a) Removing an edge from a cycle leaves a path, which is still connected.
b) Removing an edge from the cycle portion of the wheel leaves that portion still connected as in part (a),
and the central vertex is clearly still connected to it as well. Removing a spoke leaves the cycle intact and the
central vertex still connected to it as well.
c) Let u, v, a, b be any four vertices of Km,n with u and v in one part and a and bin the other. They are
connected by the 4-cycle uavb. Removing one edge will not disconnect this 4-cycle, so these vertices are still
connected, and the entire graph is therefore still connected. Note that we needed m, n ?: 2 for this to work
(and for the statement to be true).
d) Think of Qn as two copies of Qn-l with corresponding vertices joined by an edge. Without loss of
generality we can assume that the removed edge is one of the edges joining corresponding vertices. Since each
Qn-l is connected and at least one edge remains joining the two copies, the resulting graph is connected.
374
Chapter 10
Graphs
51. If G is complete, then removing vertices one by one leaves a complete graph at each step, so we never get a
disconnected graph. Conversely, if G is not complete, say with edge uv missing, then removing all the vertices
except u and v creates the disconnected graph consisting of just those two vertices.
53. Without loss of generality, assume m ::::; n. We can disconnect Km,n by removing the m vertices in the
smaller part. To see that removing fewer than m vertices will not disconnect the graph, note that given any
two vertices ·u and v, there are m paths that pairwise share nothing except their endpoints; these paths are
of length 2 if u and v are in the same part and of length 1 or 3 if they are in different parts. Removing fewer
than m vertices can cut at most m - 1 of these paths, so the resulting graph is still connected. Therefore
K(Km,n) = min(m,n). It is also clear that >.(Km,n)::::; min(m,n), because we can disconnect the graph by
removing all the edges incident to a vertex in the larger part. By the inequality stated after Example 9,
>.(Km,n) :'.'.: K(Km,n). Therefore >.(Km,n) = min(m, n) as well.
55. Let G be a graph with n vertices. Note that K( G) ::::; n - 1. Suppose a smallest edge cut C (i.e., one with
ICI = >.( G)) leaves a nonempty proper subset S of the vertices of G disconnected from the complementary
set S' = V - S. If xy is an edge of G for every x E S and y E S', then the size of C is ISllS'I, which is
at least n - 1 (it is this small only if [S[ = 1 or n - 1), so K(G) :::; .A(G) in this case. Otherwise, let x ES
and y E S' be nonadjacent vertices. Let T consist of all neighbors of x in S' together with all vertices of
S - {x} with neighbors in S' . Then T is a vertex cut, because it separates x and y . Now look at the edges
from x to T n S' and one edge from each vertex of T n S to S'; this gives us ITI distinct edges that lie in
c, so .A(G) =[Cl 2 ITI :'.'.: K(G).
57. We need to look at successive powers of the adjacency matrix until we find one in which the (1, 6)th entry is
not 0. Since the matrix is
A=
0
1
0
1
1
0
1
0
1
0
1
1
0
1
0
1
0
1
1
0
1
0
1
0
1 0
1 1
0 1
1 0
0 1
1 0
we see that the (1, 6)th entry of A 2 is 2. Thus there is a path of length 2 from a to f (in fact 2 of them).
On the other hand there is no path of length 1 from a to f (i.e., no edge), so the length of a shortest path
is 2.
59. Let the simple paths P1 and P2 be u = xo, X1, ... , Xn = v and u = yo, y 1, ... , Ym = v, respectively. The
paths thus start out at the same vertex. Since the paths do not contain the same set of edges, they must
diverge eventually. If they diverge only after one of them has ended, then the rest of the other path is a simple
circuit from v to v. Otherwise we can suppose that x 0 = Yo, x 1 = y 1 , ... , Xi = Yi, but Xi+i -=/=- Yi+l . To form
our simple circuit, we follow the path Yi, Yi+l, Yi+ 2 , and so on, until it once again first encounters a vertex
on P1 (possibly as early as Yi+l' no later than Ym ). Once we are back on P 1 , we follow it along-forwards or
backwards, as necessary-to return to Xi. Since Xi = Yi, this certainly forms a circuit. It must be a simple
circuit, since no edge among the Xk's or the yz's can be repeated (P1 and P 2 are simple by hypothesis) and
no edge among the Xk's can equal one of the edges Yz that we used, since we abandoned P2 for Pi as soon
as we hit P 1 .
61. Let A be the adjacency matrix of a given graph G. Theorem 2 tells us that Ar counts the number of paths of
length r between vertices. If an entry in Ar is greater than 0, then there is a path between the corresponding
vertices in G. Suppose that we look at A+ A 2 +A 3 + · · ·+An- l , where n is the number of vertices in G. If
there is a path between a pair of distinct vertices in G, then there is a path of length at most n - 1 , so this sum
Section 10.5
Euler and Hamilton Paths
375
will have a positive integer in the corresponding entry. Conversely, if there is no path, then the corresponding
entry in every summand will be 0, and hence the entry in the sum will be 0. Therefore the graph is connected
(i.e., there is a path between every pair of distinct vertices in G) if and only if every off-diagonal entry in this
sum is strictly positive. To determine whether G is connected, therefore, we just compute this sum and check
to see whether this condition holds.
63. We have to prove a statement and its converse here. One direction is fairly easy. If the graph is bipartite, say
with parts A and B, then the vertices in every path must alternately lie in A and B. Therefore a path that
starts in A, say, will end in B after an odd number of steps and in A after an even number of steps. Since
a circuit ends at the same vertex where it starts, the length must be even. The converse is a little harder.
We suppose that all circuits have even length and want to show that the graph is bipartite. We can assume
that the graph is connected, because if it is not, then we can just work on one component at a time. Let v
be a vertex of the graph, and let A be the set of all vertices to which there is a path of odd length starting
at v, and let B be the set of all vertices to which there is a path of even length starting at v. Since the
component is connected, every vertex lies in A or B. No vertex can lie in both A and B, since then following
the odd-length path from v to that vertex and then back along the even-length path from that vertex to v
would produce an odd circuit, contrary to the hypothesis. Thus the set of vertices has been partitioned into
two sets. Now we just need to show that every edge has endpoints in different parts. If xy is an edge where
x E A, then the odd-length path from v to x followed by xy produces an even-length path from v to y, so
y EB (and similarly if x EB).
65. Suppose the couples are Bob and Carol Sanders, and Ted and Alice Henderson (these were characters in a
movie from 1969). We represent the initial position by (BCT A•, 0), indicating that all four people are on
the left shore along with the boat (the dot). We want to reach the position (0, BCT A•). Positions will be
the vertices of our graph, and legal transitions will be the edges. If Bob and Carol take the boat over, then
we reach the position (TA, BC•). The only useful transition at that point is for someone to row back. Let's
try Bob; so we have (BT A•, C). If Bob and Ted now row to the right shore, we reach (A, BCT•). Ted can
take the boat back to fetch his wife, giving us (TA•, BC) and then (0, BCT A•). Notice that this path never
violates the jealousy conditions imposed in this problem. The entire graph model would have many more
positions, but we just need one path.
SECTION 10.5 Euler and Hamilton Paths
An Euler circuit or Euler path uses every edge exactly once. A Hamilton circuit or Hamilton path uses every
vertex exactly once (not counting the circuit's return to its starting vertex). Euler and Hamilton circuits and
paths have an important place in the history of graph theory, and as we see in this section they have some
interesting applications. They provide a nice contrast-there are good algorithms for finding Euler paths (see
also Exercises 50-53), but computer scientists believe that there is no good (efficient) algorithm for finding
Hamilton paths.
Most of these exercises are straightforward. The reader should at least look at Exercises 16 and 17 to see
how the concept of Euler path applies to directed graphs-these exercises are not hard if you understood the
proof of Theorem 1 (given in the text before the statement of the theorem).
Chapter 10
376
1. Since there are four vertices of odd degree (a, b, c, and e) and 4
nor an Euler path.
Graphs
> 2, this graph has neither an Euler circuit
3. Since there are two vertices of odd degree (a and d), this graph has no Euler circuit, but it does have an
Euler path starting at a and ending at d. We can find such a path by inspection, or by using the splicing
idea explained in this section. One such path is a, e, c, e, b, e, d, b, a, c, d.
5. All the vertex degrees are even, so there is an Euler circuit. We can find such a circuit by inspection, or by
using the splicing idea explained in this section. One such circuit is a, b, c, d, c, e, d, b, e, a, e, a.
7. All the vertex degrees are even, so there is an Euler circuit. We can find such a circuit by inspection, or by
using the splicing idea explained in this section. One such circuit is a, b, c, d, e, f, g, h, i, a, h, b, i, c, e, h, d, g,
c,a.
9. No, an Euler circuit does not exist in the graph modeling this hypothetical city either. Vertices A and B
have odd degree.
11. Assuming we have just one truck to do the painting, the truck must follow an Euler path through the streets
in order to do the job without traveling a street twice. Therefore this can be done precisely when there is
an Euler path or circuit in the graph, which means that either zero or two vertices (intersections) have odd
degree (number of streets meeting there). We are assuming, of course, that the city is connected.
13. In order for the picture to be drawn under the conditions of Exercises 13-15, the graph formed by the picture
must have an Euler path or Euler circuit. Note that all of these graphs are connected. The graph in the
current exercise has all vertices of even degree; therefore it has an Euler circuit and can be so traced.
15. See the comments in the solution to Exercise 13. This graph has 4 vertices of odd degree; therefore it has no
Euler path or circuit and cannot be so traced.
17. If there is an Euler path, then as we follow it through the graph, each vertex except the starting and ending
vertex must have equal in-degree and out-degree, since whenever we come to the vertex along some edge, we
leave it along some edge. The starting vertex must have out-degree 1 greater than its in-degree, since after
we have started, using one edge leading out of this vertex, the same argument applies. Similarly, the ending
vertex must have in-degree 1 greater than its out-degree, since until we end, using one edge leading into this
vertex, the same argument applies. Note that the Euler path itself guarantees weak connectivity; given any
two vertices, there is a path from the one that occurs first along the Euler path to the other, via the Euler
path.
Conversely, suppose that the graph meets the degree conditions stated here. By Exercise 16 it cannot
have an Euler circuit. If we add one more edge from the vertex of deficient out-degree to the vertex of deficient
in-degree, then the graph now has every vertex with its in-degree equal to its out-degree. Certainly the graph
is still weakly connected. By Exercise 16 there is an Euler circuit in this new graph. If we delete the added
edge, then what is left of the circuit is an Euler path from the vertex of deficient in-degree to the vertex of
deficient out-degree.
19. For Exercises 18-23 we use the results of Exercises 16 and 17. By Exercise 16, we cannot hope to find an
Euler circuit since vertex b has different out-degree and in-degree. By Exercise 17, we cannot hope to find an
Euler path since vertex b has out-degree and in-degree differing by 2.
Section 10.5
Euler and Hamilton Paths
377
21. This directed graph satisfies the condition of Exercise 17 but not that of Exercise 16. Therefore there is no
Euler circuit. The Euler path must go from a to e. One such path is a, d, e, d, b, a, e, c, e, b, c, b. e.
23. There are more than two vertices whose in-degree and out-degree differ by 1, so by Exercises 16 and 17, there
is no Euler path or Euler circuit.
25. The algorithm is very similar to Algorithm 1. The input is a weakly connected directed multigraph in which
either each vertex has in-degree equal to its out-degree, or else all vertices except two satisfy this condition
and the remaining vertices have in-degree differing from out-degree by 1 (necessarily once in each direction).
We begin by forming a path starting at the vertex whose out-degree exceeds its in-degree by 1 (in the second
case) or at any vertex (in the first case). We traverse the edges (never more than once each), forming a path,
until we cannot go on. Necessarily we end up either at the vertex whose in-degree exceeds its out-degree (in
the first case) or at the starting vertex (in the second case). From then on we do exactly as in Algorithm 1,
finding a simple circuit among the edges not yet used, starting at any vertex on the path we already have;
such a vertex exists by the weak connectivity assumption. We splice this circuit into the path, and repeat the
process until all edges have been used.
27. a) Clearly K 2 has an Euler path but no Euler circuit. For odd n > 2 there is an Euler circuit (since the
degrees of all the vertices are n -1, which is even), whereas for even n > 2 there are at least 4 vertices of odd
degree and hence no Euler path. Thus for no n other than 2 is there an Euler path but not an Euler circuit.
b) Since Cn has an Euler circuit for all n, there are no values of n meeting these conditions.
c) A wheel has at least 3 vertices of degree 3 (around the rim), so there can be no Euler path.
d) The same argument applies here as applied in part (a). In more detail, Q 1 (which is the same as K 2 ) is
the only cube with an Euler path but no Euler circuit, since for odd n > 1 there are too many vertices of odd
degree, and for even n > 1 there is an Euler circuit.
29. Just as a graph with 2 vertices of odd degree can be drawn with one continuous motion, a graph with 2m
vertices of odd degree can be drawn with m continuous motions. The graph in Exercise 1 has 4 vertices of
odd degree, so it takes 2 continuous motions; in other words, the pencil must be lifted once. We could do
this, for example, by first tracing a, c, d, e, a, b and then tracing c, b, e. The graphs in Exercises 2-7 all have
Euler paths, so no lifting is necessary.
31. It is clear that a, b, c, d, e, a is a Hamilton circuit.
33. There is no Hamilton circuit because of the cut edges ( {c, e}, for instance). Once a purported circuit had
reached vertex e, there would be nowhere for it to go.
35. There is no Hamiltonian circuit in this graph. If there were one, then it would have to include all the edges
of the graph, because it would have to enter and exit vertex a, enter and exit vertex d, and enter and exit
vertex e. But then vertex c would have been visited more than once, a contradiction.
37. This graph has the Hamilton path a, b, c, f, d, e. This simple path hits each vertex once.
39. This graph has the Hamilton path
f, e, d, a, b, c.
41. There are eight vertices of degree 2 in this graph. Only two of them can be the end vertices of a Hamilton
path, so for each of the other six their two incident edges must be present in the path. Now if either all four
of the "outside" vertices of degree 2 (a, c, g, and e) or all four of the ''inside'' vertices of degree 2 ( i , k,
Chapter 10
378
Graphs
l, and n) are not end vertices, then a circuit will be completed that does not include all the vertices-either
the outside square or the middle square. Therefore if there is to be a Hamilton path then exactly one of the
inside corner vertices must be an end vertex, and each of the other inside corner vertices must have its two
incident edges in the path. Without loss of generality we can assume that vertex i is an end, and that the
path begins i, o, n, m, l, q, k, j. At this point, either the path must visit vertex p, in which case it gets stuck,
or else it must visit b, in which case it will never be able to reach p. Either case gives a contradiction, so
there is no Hamilton path.
43. It is easy to write down a Hamiltonian path here; for example, a, d, g, h, i,
f,
c, e, b.
45. A Hamilton circuit in a bipartite graph must visit the vertices in the parts alternately, returning to the part
in which it began. Therefore a necessary condition is certainly m = n. Furthermore K 1 , 1 does not have a
Hamilton circuit, so we need n ::'.". 2 as well. On the other hand, since the complete bipartite graph has all
the edges we need, these conditions are sufficient. Explicitly, if the vertices are a 1 , a 2 , ... , an in one part and
b1, b2, ... , bn in the other, with n 2: 2, then one Hamilton circuit is a1, b1, a 2, b2, ... , an, bn, a 1 .
47. For Dirac's theorem to be applicable, we need every vertex to have degree at least n/2, where n is the number
of vertices in the graph. For Ore's theorem, we need deg( x) + deg(y) ::'.". n whenever x and y are not adjacent.
a) In this graph n = 5. Dirac's theorem does not apply, since there is a vertex of degree 2, and 2 is smaller
than n/2. Ore's theorem also does not apply, since there are two nonadjacent vertices of degree 2, so the
sum of their degrees is less than n. However, the graph does have a Hamilton circuit-just go around the
pentagon. This illustrates that neither of the sufficient conditions for the existence of a Hamilton circuit given
in these theorems is necessary.
b) Everything said in the solution to part (a) is valid here as well.
c) In this graph n = 5, and all the vertex degrees are either 3 or 4, both of which are at least n/2. Therefore
Dirac's theorem guarantees the existence of a Hamilton circuit. Ore's theorem must apply as well, since
(n/2) + (n/2) = n; in this case, the sum of the degrees of any pair of nonadjacent vertices (there are only two
such pairs) is 6, which is greater than or equal to 5.
d) In this graph n = 6, and all the vertex degrees are 3, which is (at least) n/2. Therefore Dirac's theorem
guarantees the existence of a Hamilton circuit. Ore's theorem must apply as well, since (n/2) + (n/2) = n; in
this case, the sum of the degrees of any pair of nonadjacent vertices is 6.
Although not illustrated in any of the examples in this exercise, there are graphs for which Ore's theorem
applies, even though Dirac's does not. Here is one: Take K 4 , and then tack on a path of length 2 between two
of the vertices, say a, b, c. In all, this graph has five vertices, two with degree 3, two with degree 4, and one
with degree 2. Since there is a vertex with degree less than 5/2, Dirac's theorem does not apply. However,
the sum of the degrees of any two (nonadjacent) vertices is at least 2 + 3 = 5, so Ore's theorem does apply
and guarantees that there is a Hamilton circuit.
49. The trick is to use a Gray code for n to build one for n
+ 1.
We take the Gray code for n and put a 0 in
front of each term to get half of the Gray code for n + 1; we put a 1 in front to get the second half. Then
we reverse the second half so that the junction at which the two halves meet differ in only the first bit. For
a formal proof we use induction on n. For n = 1 the code is 0, 1 (which is not really a Hamilton circuit in
Q1). Assume the inductive hypothesis that C1, C2, ... , C2n is a Gray code for n. Then Oc1, Oc2, ... , Oc2n, lc2n,
... , lc2 , lc 1 is a Gray code for n + 1.
51. Turning this verbal description into pseudocode is straightforward, especially if we allow ourselves lots of words
in the pseudocode. We build our circuit (which we think of simply as an ordered list of edges) one edge at a
time, keeping track of the vertex v we are at; the subgraph containing the edges we have not yet used we will
Section 10.5
Euler and Hamilton Paths
379
call H. We assume that the vertices of G are listed in some order, so that when we are asked to choose an
edge from v meeting certain conditions, we can choose the edge to the vertex that comes first in this order
among all those edges meeting the conditions. (This avoids ambiguity, which an algorithm is not supposed to
have.)
procedure fieury(G: connected multigraph with all degrees even)
v := first vertex of G
circuit :=the empty circuit
H:=G
while H has edges
Let e be an edge in H with v as one of its endpoints,
such that e is not a cut edge of H, if such an edge
exists; otherwise let e be any edge in H with v as
one of its endpoints.
v :=other endpoint of e
Add e to the end of circuit
Remove e from H
return circuit { circuit is an Euler circuit}
53. If every vertex has even degree, then we can simply use Fleury's algorithm to find an Euler circuit, which is
by definition also an Euler path. If there are two vertices with odd degree (and the rest with even degree),
then we can add an edge between these two vertices and apply Fleury's algorithm (using this edge as the first
edge to make it easier to find later), then delete the added edge.
55. A Hamilton circuit in a bipartite graph would have to look like a 1 ,b 1 ,a 2 ,b2 , ... ,akibk,a 1 , where each ai is
in one part and each bi is in the other part, since the only edges in the graph join vertices in opposite parts.
In the Hamilton circuit, no vertex is listed twice (except for the final a 1 ), and every vertex is listed, so the
total number of vertices in the graph must be 2k, which is not an odd number. Therefore a bipartite graph
with an odd number of vertices cannot have a Hamilton circuit.
57. We draw one vertex for each of the 12 squares on the board. We then draw an edge from a vertex to each
vertex that can be reached by moving 2 units horizontally and 1 unit vertically or vice versa. The result is as
shown.
59. First let us try to find a reentrant knight's tour. Looking at the graph in the solution to Exercise 57 we see
that every vertex on the left and right edge has degree 2. Therefore the 12 edges incident to these vertices
would have to be in a Hamilton circuit if there were one. If we draw these 12 edges, however, we see that they
form two circuits, each with six edges. Therefore there is no re-entrant knight's tour. However, we can splice
these two circuits together by using an edge from a middle vertex in the top row to a middle vertex in the
bottom row (and omitting two edges adjacent to this edge). The result is the knight's tour shown here.
3
6
9
12
8
11
4
1
5
2
7
10
380
Chapter 10
Graphs
61. We give an ad hoc argument by contradiction, using the notation shown in the following diagram. We think
of the board as a graph and need to decide which edges need to be in a purported Hamilton path.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
There are only two moves from each of the four corner squares. If we put in all of the edges 1-10, 1-7, 16-10,
and 16-7, then a circuit is complete too soon, so at least one of these edges must be missing. Without loss of
generality, then, we may assume that the endpoints of the path are 1 and either 4 or 13, and that the path
contains all of the edges 1-10, 10-16, and 16-7. Now vertex (square) 3 has edges only to squares 5, 10, and
12; and square 10 already has its two incident edges. Therefore 3-5 and 3-12 must be in the Hamilton path.
Similarly, edges 8-2 and 8-15 must be in the path. Now square 9 has edges only to squares 2, 7, and 15. If
there were to be edges to both 2 and 15, then a circuit would be completed too soon (2-9-15-8-2). Therefore
the edge 9-7 must be in the path, thereby giving square 7 its full complement of edges. But now square 14 is
forced to be joined in the path to squares 5 and 12, and this completes a circuit too soon (5-14-12-3-5). Since
we have reached a contradiction, we conclude that there is no Hamilton path.
63. An m x n board contains mn squares. If both m and n are odd, then it contains an odd number of squares.
By Exercise 62, the corresponding graph is bipartite. Exercise 55 told us that the graph does not contain a
Hamilton circuit. Therefore there is no re-entrant knight's tour (see Exercise 58b).
65. This is a proof by contradiction. We assume that G satisfies Ore's inequality that deg(x) + deg(y) ~ n
whenever x and y are nonadjacent vertices in G, but G does not have a Hamilton circuit. We will end up
with a contradiction, and therefore conclude that under these conditions, G must have a Hamilton circuit.
a) Since G does not have a Hamilton circuit, we can add missing edges one at a time in such a way that
we do not obtain a graph with a Hamilton circuit. We continue this process as long as possible. Clearly it
cannot go on forever, because once we've formed the complete graph by adding all missing edges, there is a
Hamilton circuit (recall that n :'.:': 3). Whenever the process stops, we have obtained a graph H with the
desired property. (Note that H might equal G itself-in other words, we add no edges. However, H cannot
be complete, as just noted.)
b) Add one more edge to H. By the construction in part (a), we now have a Hamilton circuit, and clearly
this circuit must use the edge we just added. The path consisting of this circuit with the added edge omitted
is clearly a Hamilton path in H.
c) Clearly v1 and Vn are not adjacent in H, since H has no Hamilton circuit. Therefore they are not adjacent
in G. But the hypothesis was that the sum of the degrees of vertices not adjacent in G was at least n. This
inequality can be rewritten as n - deg( Vn) :::; deg( v1). But n - deg( Vn) is just the number of vertices not
adjacent to Vn .
d) Let's make sure we understand what this means. If, say, v7 is adjacent to v1 , then v 6 is in S. Note that
v1 E S, since v2 is adjacent to v1. Also, Vn is not in S, since there is no vertex following Vn in the Hamilton
path. Now each one of the deg( v1) vertices adjacent to v1 gives rise to an element of S, so S contains deg( v1)
vertices.
e) By part (c) there are at most deg(vi) - 1 vertices other than Vn not adjacent to Vn, and by part (d)
there are deg( v1) vertices in S, none of which is Vn. So S has more vertices other than Vn than there are
vertices not adjacent to Vn; in other words, at least one vertex of S is adjacent to Vn. By definition, if Vk is
Section 10.6
Shortest-Path Problems
this vertex, then H contains edges
381
VkVn
and v 1vk+l. Note that 1
4. By Exercise 32, the thickness of Kn is
at least
C(n,2) = n(n-1)/2 = n(n- l) = ~ (n+ 1 + _2_)
3n-6
3n-6
6(n-2)
6
n-2
rounded up. Since this quantity is never an integer, it equals one more than itself rounded down, namely
1 (
6
2 )
n+7
2
n + 1 + n - 2 + 1 = -6- + 6(n - 2)
rounded down. The last term can be ignored: it is always less than 1/6 and therefore will not influence the
rounding process (since the first term has denominator 6). Thus we have proved that the thickness of Kn is
at least l(n + 7)/6J.
35. This follows immediately from Exercise 34, since Km,n has mn edges and m+n vertices and, being bipartite,
has no triangles.
37. We can represent the surface of a torus with a rectangle, thinking of the right-hand edge as being equal to
the left-hand edge, and the top edge as being equal to the bottom edge. For example, if we travel out of the
rectangle across the right-hand edge about a third of the way from the top, then we immediately reenter the
rectangle across the left-hand edge about a third of the way from the top. The picture below shows K 3 ,3
drawn on this surface. Note that the edges that seem to leave the rectangle really reenter it from the opposite
side.
Section 10.8
Graph Coloring
SECTION 10.8
389
Graph Coloring
Like the problem of finding Hamilton paths, the problem of finding colorings with the fewest possible colors
probably has no good algorithm for its solution. In working these exercises, for the most part you should
proceed by trial and error, using whatever insight you can gain by staring at the graph (for instance, finding
large complete subgraphs). There are also some interesting exercises here on coloring the edges of graphs-see
Exercises 21-26. Exercises 29-31 are worth looking at, as well: they deal with a fast algorithm for coloring a
graph that is not guaranteed to produce an optimal coloring.
1. We construct the dual graph by putting a vertex inside each region (but not in the unbounded region), and
drawing an edge between two vertices if the regions share a common border. The easiest way to do this is
illustrated in our answer. First we draw the map, then we put a vertex inside each region and make the
connections. The dual graph, then, is the graph with heavy lines.
The number of colors needed to color this map is the same as the number of colors needed to color the dual
graph. Since A, B, C, and D are mutually adjacent, at least four colors are needed. We can color each of
the vertices (i.e., regions) A, B, C, and D a different color, and we can give Ethe same color as we give C.
3. We construct the dual as in Exercise 1.
As in Exercise 1, the number of colors needed to color this map is the same as the number of colors needed
to color the dual graph. Three colors are clearly necessary, because of the triangle ABC, for instance.
Furthermore three colors suffice, since we can color vertex (region) A red, vertices B, D, and F blue, and
vertices C and E green.
5. For Exercises 5-11, in order to prove that the chromatic number is k, we need to find a k-coloring and to
show that (at least) k colors are needed. Here, since there is a triangle, at least 3 colors are needed. Clearly
3 colors suffice, since we can color a and d the same color.
7. Since there is a triangle, at least 3 colors are needed. Clearly 3 colors suffice, since we can color a and c the
same color.
9. Since there is an edge, at least 2 colors are needed. The coloring in which b, d, and e are red and a and c
blue shows that 2 colors suffice.
11. Since there is a triangle, at least 3 colors are needed. It is not hard to construct a 3-coloring. We can let a,
f, h, j, and n be blue; let b, d, g, k, and m be green; and let c, e, i, l, and o be yellow.
13. If a graph has an edge (not a loop, since we are assuming that the graphs in this section are simple), then
its chromatic number is at least 2. Conversely, if there are no edges, then the coloring in which every vertex
receives the same color is proper. Therefore a graph has chromatic number 1 if and only if it has no edges.
Chapter 10
390
Graphs
15. In Example 4 we saw that the chromatic number of Cn is 2 if n is even and 3 if n is odd. Since the wheel
vVn is just Cn with one more vertex, adjacent to all the vertices of the Cn along the rim of the wheel, Wn
clearly needs exactly one more color than Cn (for that middle vertex). Therefore the chromatic number of
vVn is 3 if n is even and 4 if n is odd.
17. Consider the graph representing this problem. The vertices are the 8 courses, and two courses are joined by
an edge if there are students taking both of them. Thus there are edges between every pair of vertices except
the 7 pairs listed. It is much easier to draw the complement than to draw this graph itself; it is shown below.
473
185
213•
102.--~~~---:•195
101
We want to find the chromatic number of the graph whose complement we have drawn; the colors will be the
time periods for the exams. First note that since Math 185 and the four CS courses form a Ks (in other
words, there are no edges between any two of these in our picture), the chromatic number is at least 5. To
show that it equals 5, we just need to color the other three vertices. A little trial and error shows that we
can make Math 195 the same color as (i.e., have its final exam at the same time as) CS 101; and we can make
Math 115 and 116 the same color as CS 473. Therefore five time slots (colors) are sufficient.
19. We model the problem with the intersection graph of these sets. Note that every pair of these intersect except
for C4 and Cs. Thus the graph is K 6 with that one edge deleted. Clearly its chromatic number is 5, since
we need to color all the vertices different colors, except that C4 and Cs may have the same color. In other
words, 5 meeting times are needed, since only committees C4 and Cs can meet simultaneously.
21. Note that the number of colors needed to color the edges is at least as large as the largest degree of a vertex,
since the edges at each vertex must all be colored differently. Hence if we can find an edge coloring with that
many colors, then we know we have found the answer. In Exercise 5 there is a vertex of degree 3, so the edge
chromatic number is at least 3. On the other hand, we can color {a, c} and {b, d} the same color, so 3 colors
suffice. In Exercise 6 the 6 edges incident to g must all get different colors. On the other hand, it is not
hard to complete a proper edge coloring with only these colors (for example, color edge {a, f} with the same
color as used on {d, g} ) , so the answer is 6. In Exercise 7 the answer must be at least 3; it is 3 since edges
that appear as parallel line segments in the picture can have the same color. In Exercise 8 clearly 4 colors
are required, since the vertices have degree 4. In fact 4 colors are sufficient. Here is one proper 4-coloring
(we denote edges in the obvious shorthand notation): color 1 for ac, be, and df; color 2 for ae, bd, and cf;
color 3 for ab, cd, and e f; and color 4 for ad, bf, and ce. In Exercise 9 the answer must be at least 3;
it is easy to construct a 3-coloring of the edges by inspection: {a, b} and {c, e} have the same color, {a, d}
and { b, c} have the same color, and {a, e} and {c, d} have the same color. In Exercise 10 the largest degree
is 6 (vertex i has degree 6 ); therefore at least 6 colors are required. By trial and error we come up with this
coloring using 6 colors (we use the obvious shorthand notation for edges); there are many others, of course.
Assign color 1 to ag, cd, and hi; color 2 to ab, cf, dg, and ei; color 3 to bh, cg, di, and ef; color 4 to
ah, ci, and de; color 5 to bi, ch, and f g; and color 6 to ai, be, and gh. Finally, in Exercise 11 it is easy
to construct an edge-coloring with 4 colors; again the edge chromatic number is the maximum degree of a
vertex.
Despite the appearances of these examples, it is not the case that the edge chromatic number of a graph
is always equal to the maximum degree of the vertices in the graph. The simplest example in which this is not
Section 10.8
Graph Coloring
391
true is K 3 . Clearly its edge chromatic number is 3 (since all three edges are adjacent to each other), but its
maximum degree is 2 . There is a theorem, however, stating that the edge chromatic number is always equal
to either the maximum degree or one more than the maximum degree.
23. a) The n-cycle's edges are just like the n-cycle's vertices (each adjacent to the next as we go around the
cycle), so the edge chromatic number is the same, namely 2 if n is even and 3 if n is odd, as in Example 4.
b) The edge chromatic number is at least n, because the radial edges are all pairwise adjacent and therefore
must all have distinct colors. Suppose we call these colors 1 through n proceeding clockwise. We need no
additional colors for the edges of the cycle, because we can color the edge adjacent to the spokes colored 1
and 2 with color 3 and proceed clockwise with colors 4, 5, ... , n - 1, n, 1, and 2. Therefore x' (Wn) = n.
25. Two edges that have the same color share no endpoints. Therefore if more than n/2 edges were colored the
same, the graph would have more than 2(n/2) = n vertices.
27. This problem can be modeled with the intersection graph of the sets of steps during which the variables must
be stored. This graph has 7 vertices, t through z; there is an edge between two vertices if the two variables
they represent must be stored during some common step. The answer to the problem is the chromatic number
of this graph. Rather than considering this graph, we look at its complement (it has a lot fewer edges). Here
two vertices are adjacent if the sets (of steps) do not intersect. The only edges are {u,w}, {u,x}, {u,y},
{u,z}, {v,x}, {x,z}. Note that there are no edges in the complement joining any two of {t,v,w,y,z}, so
that these vertices form a K 5 in the original graph. Thus the chromatic number of the original graph is at
least 5. To see that it is 5, note that vertex u can have the same color as w, and x can have the same color
as z (these pairs appear as edges in the complement). Since the chromatic number is 5, we need 5 registers,
with variables u and w sharing a register, and vertices x and z sharing one.
29. First we need to list the vertices in decreasing order of degree. This ordering is not unique, of course; we will
pick e,a,b,c,f,h,i,d,g,j. Next we assign color 1 toe, and then to f and d, in that order. Now we assign
color 2 to a, c, i, and g, in that order. Finally, we assign color 3 to b, h and j , in that order. Thus the
algorithm gives a 3-coloring. Since the graph contains triangles, we know that this is the best possible, so the
algorithm "worked" here (but it need not always work-see Exercise 27).
31. A simple example in which the algorithm may fail to provide a coloring with the minimum number of colors
is C6 , which of course has chromatic number 2. Since all the vertices are of degree 2, we may order them v1 ,
v4, v2, v3, v5, V6, where the edges are { V1, v2}, {v2, v3}, {V3, v4}, {V4, v5}, { V5, V6}, and {v1, v6}. Then V1
gets color 1, as does V4. Next V2 and v 5 get color 2; and then V3 and v6 must get color 3.
33. We need to show that the wheel Wn when n is an odd integer greater than 1 can be colored with four colors,
but that any graph obtained from it by removing one edge can be colored with three colors. Four colors
are needed to color this graph, because three colors are needed for the rim (see Example 4), and the center
vertex, being adjacent to all the rim vertices, will require a fourth color. To complete the proof that Wn is
chromatically 4-critical, we must show that the graph obtained from Wn by deleting one edge can be colored
with three colors. There are two cases. If we remove a rim edge, then we can color the rim with two colors,
by starting at an endpoint of the removed edge and using the colors alternately around the portion of the rim
that remains. The third color is then assigned to the center vertex. On the other hand, if we remove a spoke
edge, then we can color the rim by assigning color #1 to the rim endpoint of the removed edge and colors #2
and #3 alternately to the remaining vertices on the rim, and then assign color #1 to the center.
392
Chapter 10
Graphs
35. We give a proof by contradiction. Suppose that G is chromatically k-critical but has a vertex v of degree
k - 2 or less. Remove from G one of the edges incident to v. By definition of "k-critical," the resulting
graph can be colored with k - 1 colors. Now restore the missing edge and use this coloring for all vertices
except v. Because we had a proper coloring of the smaller graph, no two adjacent vertices have the same
color. Furthermore, v has at most k - 2 neighbors, so we can color v with an unused color to obtain a proper
(k - 1)-coloring of G. This contradicts the fact that G has chromatic number k. Therefore our assumption
was wrong, and every vertex of G must have degree at least k - 1.
37. a) Note that vertices d, e, and f are mutually adjacent. Therefore six different colors are needed in a 2-tuple
coloring, since each of these three vertices needs a disjoint set of two colors. In fact it is easy to give a coloring
with just six colors: Color a, d, and g with {1,2}; color c and e with {3,4}; and color band f with {5,6}.
Thus x2(G) = 6.
b) This one is trickier than part (a). There is no coloring with just six colors, since if there were, we would
be forced (without loss of generality) to color d with {l, 2}; e with {3, 4}; f with {5, 6}; then g with {l, 2},
b with {5,6}, and c with {3,4}. This gives no free colors for vertex a. Now this may make it appear that
eight colors are required, but a little trial and error shows us that seven suffice: Color a with {2, 4}; color
b and f with {5, 6}; color d with {l, 2}; color c with {3, 7}; color e with {3, 4}; and color g with {l, 7}.
Thus x2(H) = 7.
c) This is similar to part (a). Here nine colors are necessary and sufficient, since a, d, and g can get one
set of three colors; b and f can get a second set; and c and e can get a third set. Clearly nine colors are
necessary to color the triangles.
d) First we construct a coloring with 11 colors: Color a with {3, 6, 11}; color b and f with {7, 8, 9}; color
d with {1, 2, 3}; color c with {4, 5, 10}; color e with {4, 6, 11}; and color g with {l, 2, 5}. To prove that
x3 (H) = 11, we must show that it is impossible to give a 3-tuple coloring with only ten colors. If such a
coloring were possible, without loss of generality we could color d with {l, 2, 3}, e with {4, 5, 6}, f with
{7, 8. 9}, and g with {l, 2, 10}. Now nine colors are needed for the three vertices a, b, and c, since they form
a triangle; but colors 1 and 2 are already used in vertices adjacent to all three of them. Therefore at least
9 + 2 = 11 colors are necessary.
39. The frequencies are the colors, the zones are the vertices, and two zones that are so close that interference
would be a problem are joined by an edge in the graph. Then it is clear that a k-tuple coloring is exactly an
assignment of frequencies that avoids possible interference.
41. We use induction on the number of vertices of the graph. Every graph with five or fewer vertices can be colored
with five or fewer colors, since each vertex can get a different color. That takes care of the basis case(s). So
we assume that all graphs with k vertices can be 5-colored and consider a graph G with k + 1 vertices. By
Corollary 2 in Section 10.7, G has a vertex v with degree at most 5. Remove v to form the graph G'. Since
G' has only k vertices, we 5-color it by the inductive hypothesis. If the neighbors of v do not use all five
colors, then we can 5-color G by assigning to v a color not used by any of its neighbors. The difficulty arises
if v has five neighbors, and each has a different color in the 5-coloring of G'. Suppose that the neighbors of
v, when considered in clockwise order around v, are a, b, c, m, and p. (This order is determined by the
clockwise order of the curves representing the edges incident to v .) Suppose that the colors of the neighbors
are azure, blue, chartreuse, magenta, and purple, respectively. Consider the azure-chartreuse subgraph (i.e.,
the vertices in G colored azure or chartreuse and all the edges between them). If a and c are not in the same
component of this graph, then in the component containing a we can interchange these two colors (make the
azure vertices chartreuse and vice versa), and G' will still be properly colored. That makes a chartreuse, so
we can now color v azure, and G has been properly colored. If a and c are in the same component, then
there is a path of vertices alternately colored azure and chartreuse joining a and c. This path together with
Review Questions
393
edges av and vc divides the plane into two regions, with b in one of them and m in the other. If we now
interchange blue and magenta on all the vertices in the same region as b, we will still have a proper coloring
of G', but now blue is available for v. In this case, too, we have found a proper coloring of G. This completes
the inductive step, and the theorem is proved.
43. We follow the hint. Because the measures of the interior angles of a pentagon total 540°, there cannot be as
many as three interior angles of measure more than 180° (reflex angles). If there are no reflex angles, then
the pentagon is convex, and a guard placed at any vertex can see all points. If there is one reflex angle, then
the pentagon must look essentially like figure (a) below, and a guard at vertex v can see all points. If there
are two reflex angles, then they can be adjacent or nonadjacent (figures (b) and (c)); in either case, a guard
at vertex v can see all points. (In figure (c), choose the reflex vertex closer to the bottom side.) Thus for all
pentagons, one guard suffices, so g(5) = 1.
v
(a)
(c)
45. The figure suggested in the hint (generalized to have k prongs for any k ~ 1) has 3k vertices. Consider
the set of points from which a guard can see the tip of the first prong, the set of points from which a guard
can see the tip of the second prong, and so on. These are disjoint triangles (together with their interiors).
Therefore a separate guard is needed for each of the k prongs, so at least k guards are needed. This shows
that g(3k) ~ k = l3k/3J. To handle values of n that are not multiples of 3, let n = 3k + i, where i = 1
or 2. Then obviously g(n) ~ g(3k) ~ k = Ln/3J.
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 10
1. a) See pp. 641 ~642 and Table 1 in Section 10.
b) See Exercise 1 in Section 10 .1.
2. See all the examples Section 10.l.
3. See Theorem 1 in Section 10.2.
4. See Theorem 2 in Section 10.2.
5. See Theorem 3 in Section 10.2.
6. a) See Example 5 in Section 10.2.
c) See Example 6 in Section 10.2.
e) See Example 8 in Section 10.2.
7. a) n, C(n, 2)
b) m+n, mn
b) See Example 13 in Section 10.2.
d) See Example 7 in Section 10.2.
c) n, n
d) n + 1, 2n
8. a) See p. 656.
c) (See also Example 12 and Exercise 66 in Section 10.2.) The following algorithm is an efficient way to
determine whether a connected graph can be 2-colored (which is the same thing as saying that it is bipartite);
apply it to each component of the given graph. First color any vertex red. Then color all vertices adjacent
to this vertex blue. Then look at all vertices adjacent to these just-colored blue vertices. If any of them are
already colored blue, then stop and declare the graph not to be bipartite; otherwise color all the uncolored
ones red. Next look at all vertices adjacent to all the vertices just colored red. If any of them are already
Chapter 10
394
Graphs
colored red, then stop and declare the graph not to be bipartite; otherwise color all the uncolored ones blue.
Continue in this way until no more vertices can be colored. If we get this far, then (this component of) the
graph is bipartite. (If uncolored vertices remain, they are in a different component, so we can repeat the entire
process starting with any uncolored vertex.)
9. a) adjacency lists, adjacency matrices, incidence matrices
b) Look at Figure 1 in Section 10.3, and let G be the graph consisting of the vertices and edges shown there,
together with edges {b, c} and {b, e}. Its adjacency lists are shown on p. 668 (Table 1), once we add c and e
to the list of adjacent vertices of b, and add b to the list for c and for e. Its adjacency matrix and incidence
matrix are as follows (using alphabetical order):
10. a) See p. 672.
b) Seep. 672; number of vertices, number of edges, degrees of vertices, existence of triangle ( C3 as subgraph),
and existence of Hamilton circuit are all invariants.
c) See Figure 10 in Section 10.3.
d) no
11. a) See p. 681.
b) See p. 682.
12. a) See p. 669.
b) See p. 673.
c) See Theorem 2 in Section 10.4.
b) See pp. 693-694.
c) All edges are in the same component and there are at most two vertices of odd degree (see Theorems 1 and
2 in Section 10.5).
d) All edges are in the same component and there are no vertices of odd degree (see Theorem 1 in Section 10.5).
13. a) See p. 694.
14. a) See p. 698.
b) the existence of a cut edge or a cut vertex
15. finding the shortest highway route between two cities; finding the cheapest way to lease telephone lines to join
two communications centers, using intermediate switching centers
16. a) See pp. 709-712.
17. a) Seep. 719.
b) See Exercise 5, third part, in Section 10.6.
b) K 6
18. a) If the planar graph has v vertices, e edges, and c components, and is embedded in the plane so as to form
r regions, then v - e + r = 1 + c. (The theorem as stated in the text-Theorem 1 in Section 10.7-assumes
that c = 1, but this is the more general statement.)
b) By Corollary 1 in Section 10. 7, for every planar graph with at least three vertices we know that e ::::; 3v - 6
(connectivity need not be assumed). Thus we can show that a graph is nonplanar by showing that it has
too many edges, namely more than 3v - 6 edges. For example, K 6 is nonplanar, since it has 15 edges, and
15~3·6-6.
19. See Theorem 2 in Section 10.7. A graph is planar if and only if it does not contain a subgraph homeomorphic
to K5 or
K3,3.
20. a) See p. 728.
b) n
c) 2 if n is even, 3 if n is odd
d) 2
21. Every planar graph can be 4-colored (i.e., has chromatic number at most 4), but K 6 , for instance, requires
six colors.
Supplementary Exercises
395
22. See Examples 5-7 in Section 10.8.
SUPPLEMENTARY EXERCISES FOR CHAPTER 10
1. Every vertex has degree 50. Thus the sum of the degrees must be 50 · 100
=
5000. By the handshaking
theorem, the graph therefore has 5000/2 = 2500 edges.
3. Both graphs have a lot of symmetry to them, and the degrees of the vertices are the same, so we might hope
that they are isomorphic. Let us try to form the correspondence f. First note that there is a 4-cycle u 1 , u 5 ,
u 2 , u6, u1 in the first graph. Suppose that we try letting it correspond to the 4-cycle v1, v 2 , v 3, v4, v 1 in the
second graph. Thus we let f (u1) = V1 , f (u5) = v2 , f (u2) = V3 , and f (u6) = V4 . The rest of the assignments
are forced: since U7 is the other vertex adjacent to u 1 , we must let f (u 7 ) = v 6 , since v6 is the other vertex
adjacent to v1 (which is f(u1)). Similarly, f(u3) = v7, f(us) =Vs, and f(u4) = v5. Now we just have to
check that the vertices corresponding to the vertices in the 4-cycle v 5, v6, v 7, vs, v 5 in the second graph form
a 4-cycle in that order. Since these vertices form the 4-cycle u 4, u 7, u 3, u 8, u 4 , our correspondence works.
5. These graphs are isomorphic, although the isomorphism is hard to find. One approach that can lead to the
isomorphism is to draw the complement of each graph. The complements are a little simpler than the original
graphs, since they have fewer edges. When we do this, it is easy to give a planar representation of each. Then
by looking at the sizes of the regions we can find an isomorphism. One such correspondence is u 1 .__., v 5 ,
u2 .__., V4, u3 .__., V7, U4 .__., V3, us .__., vs, U6 .__., v2, U7 .__., v6, and Us .__., v1 . We just need to check that all the
edges are preserved by this correspondence.
7. It follows immediately from the definition that the complete m-partite graph with parts n 1 , n 2 , ... , nm has
n 1 + n2 + · · · +nm = 2.:::': 1 n, vertices. We will organize the count of the edges by looking at which parts the
edges join. Fix 1 ::::; i < j ::::; m, and consider the edges between the ith part and the jth part. It is easy to
see from the product rule that there are n,n1 edges. Therefore to get all the edges, we have to add all these
products, for all possible pairs (i, j). Thus the number of edges is I:i::;z 1 , and let F be a formula of length n. Then
F arises from part (ii) of the definition, so F consists of * X Y, for some operator * and some formulae X
and Y. By the inductive hypothesis, the number of symbols in X exceeds the number of operators there
by 1, and the same holds for Y . If we add and note that there is one more operator in F than in X and Y
combined, then we see that the number of symbols in F exceeds the number of operators in F by 1 , as well.
that the statement is true for formulae of length less than n
33. Any string of length n, using these six characters, is a well-formed formula as long as two conditions are
met: if we read the string from left to right, the number of symbols is always at least 1 greater than the
number of operators; and in all there is one more symbol than operator. We are asked to write down six such
strings, with n
2 7. One such set is xxxx+++, xxxxx++++, xx+xx++, xxxx+xx++++,
xx x +xx++ +, and xx+ xx x + + +.
SECTION 11.4 Spanning Trees
The spanning tree algorithms given here provide systematic methods for searching through graphs, and they
are the foundation of many other, more complicated, algorithms. The concept of a spanning tree is quite
simple and natural, of course: the problem comes with finding spanning trees efficiently. The reader should
pay attention to the exercises on backtracking (Exercises 26-30) to get a feel both for the ideas behind it and
for its inefficiency for large problems.
1. The graph has m edges. The spanning tree has n-1 edges. Therefore we need to remove m-(n-1) edges.
3. We have to remove edges, one at a time. We can remove any edge that is part of a simple circuit. The answer
is by no means unique. For example, we can start by removing edge {a, d}, since it is in the simple circuit
adcba. Then we might choose to remove edge {a, g}. We can continue in this way and remove all of these
edges: {b,e}, {b,J}, {b,g}, {d,e}, {d,g}, and {e,g}. At this point there are no more simple circuits, so we
have a spanning tree.
5. ThisissimilartoExercise3. Hereisonepossiblesetofremovals: {a,b}, {a,d}, {a,!}, {b,c}, {b,d}, {c,d},
{d,e}, {d,g}, {d,j}, {e,g}, {e,j}, {f,g}, {h,j}, {h,k}, and {j,l}. As a check, note that there are 12
vertices and 26 edges. A spanning tree must have 11 edges, so we need to remove 15 of them, as we did.
7. In each case we show the original graph, with a spanning tree in heavier lines. These were obtained by trial
and error. In each case except part ( c), our spanning tree is a simple path (but other answers are possible).
In part ( c), of course, the graph is its own spanning tree.
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422
Chapter 11
Trees
9. We can remove any one of the four edges in the square on the left, together with any one of the four edges in
the square on the right. Therefore there are 4 · 4 = 16 different spanning trees, shown here.
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11. We approach this problem in a rather ad hoc way.
a) Every pair of edges in K 3 forms a spanning tree, so there are C(3, 2) = 3 such trees.
b) There are 16 spanning trees; careful counting is required to see this. First, let us note that the trees can
take only two shapes: the star K 1 ,3 and the simple path of length 3. There are 4 different spanning trees of
the former shape, since any of the four vertices can be chosen as the vertex of degree 3. There are P( 4, 4) = 24
orders in which the vertices can be listed in a simple path of length 3, but since the path can be traversed
in either of two directions to yield the same tree, there are only 12 trees of this shape. Therefore there are
4 + 12 = 16 spanning trees of K 4 altogether.
c) Note that K 2 , 2 = C 4 . A tree is determined simply by deciding which of the four edges to remove. Therefore
there are 4 spanning trees.
d) By the same reasoning as in part (c), there are 5 spanning trees.
13. If we start at vertex a and use alphabetical order, then the depth-first search spanning tree is unique. We
start at vertex a and form the path shown in heavy lines to vertex i before needing to backtrack. There are
no unreached vertices from vertex h at this point, but there is an unreached vertex (j) adjacent to vertex g.
Thus the tree is as shown in heavy lines.
6~1
b~J
15. The procedure is the same as in Exercise 13. The spanning tree is shown in heavy lines.
Section 11.4
Spanning '!Tees
423
17. a) We start at the vertex in the middle of the wheel and visit a neighbor--one of the vertices on the rim.
From there we move to an adjacent vertex on the rim, and so on all the way around until we have reached
every vertex. Thus the resulting spanning tree is a path of length 6.
b) We start at any vertex, visit a neighbor, then a new neighbor, and so on until we have reached every vertex.
Thus the resulting spanning tree is a path of length 5.
c) We start at a vertex in the part with four vertices, move to a vertex in the part with three vertices, then
back to a vertex in the larger part, and so on. The resulting spanning tree is a path of length 6.
d) Depending on what order we choose to visit the vertices, we may or may not need to backtrack in doing
this depth-first search. In most cases, the resulting tree is a path of length 7, but we could, for example, come
up with the following tree.
19. With breadth-first search, the initial vertex is the middle vertex, and the n spokes are added to the tree as
this vertex is processed. Thus the resulting tree is K 1 ,n. With depth-first search, we start at the vertex in the
middle of the wheel and visit a neighbor--one of the vertices on the rim. From there we move to an adjacent
vertex on the rim, and so on all the way around until we have reached every vertex. Thus the resulting
spanning tree is a path of length n.
21. With breadth-first search, we fan out from a vertex of degree m to all the vertices of degree n as the first
step. Next a vertex of degree n is processed, and the edges from it to all the remaining vertices of degree m
are added. The result is what is called a "double star": a K 1 ,n-l and a K 1 ,m-l with their centers joined by
an edge. With depth-first search, we travel back and forth from one partite set to the other until we can go
no further. If m = n or m = n - 1, then we get a path of length m + n - 1. Otherwise, the path ends while
some vertices in the larger partite set have not been visited, so we back up one link in the path to a vertex v
and then successively visit the remaining vertices in that set from v. The result is what is called a ''broom'':
a path with extra pendant edges coming out of one end of the path (our vertex v).
23. The question is simply asking for a spanning tree of the graph shown. There are of course many such spanning
trees. One that the airline would probably not like to choose is the tree that consists of the path Bangor,
Boston, New York, Detroit, Chicago, Washington, Atlanta, St. Louis, Dallas, Denver, San Diego, Los Angeles,
San Francisco, Seattle. All the other 18 flights, then, would be discontinued. This set of flights would not be
useful to a New York to Washington flyer, for example, nor to one who wants to fly from Chicago to Seattle.
A more practical approach would be to build the tree up from nothing, using key short flights, such as the one
between Detroit and Chicago. After 13 such edges had been chosen, without creating any simple circuits, we
would have the desired spanning tree, and the other flights would be discontinued.
25. We prove this statement by induction on the length of a shortest path from v to u. If this length is 0, then
v = u, and indeed, v is at the root of the tree. Assume that the statement is true for all vertices w for which
a shortest path to v has length n, and let u be a vertex for which a shortest path has length n + 1. Clearly
u cannot be at a level less than n + 1, because then a shorter path would be evident in the tree itself. On the
other hand, if we let w be the penultimate vertex in a shortest path from v to u of length n + 1, then by
the inductive hypothesis, we know that w is at level n of the tree. Now when vertex w was being processed
by the breadth-first search algorithm, either u was already in the tree (and therefore adjacent to a vertex at
level at most n) or u was one of the vertices put into the tree adjacent to w. In either case, u is adjacent to
a vertex at level at most n and therefore is at level at most n + 1 .
Chapter 11
424
Trees
27. Label the squares of the n x n chessboard with coordinates (i, j) , where i and j are integers from 1 to n,
inclusive.
a) For the 3 x 3 board, we start our search by placing a queen in square (1, 1). The only possibility for
a queen in the second column is square (3, 2). Now there is no place to put a queen in the third column.
Therefore we backtrack and try placing the first queen in square (2, 1). This time there is no place to put a
queen in the second column. By symmetry, we need not consider the initial choice of a queen in square (3, 1)
(it will be just like the situation for the queen in square (1, 1), turned upside down). Therefore we have shown
that there is no solution.
b) We start by placing a queen in square (1, 1). The first place a queen might then reside in the second
column is square (3, 2), so we place a queen there. Now the only free spot in the third column is (5, 3), the
only free spot in the fourth column is (2, 4), and the only free spot in the fifth column is (4, 5). This gives us
a solution. Note that we were lucky and did not need to backtrack at all to find this solution.
c) The portion of the decision tree corresponding to placing the first queen in square (1, 1) is quite large here,
and it leads to no solution. For example, the second queen can be in any of the squares (3, 2), (4, 2), (5, 2),
or (6, 2). If the second queen is in square (3, 2), then the third can be in squares (5, 3) or (6, 3). After several
backtracks we find that there is no solution with one queen in square (1, 1). Next we try square (2, 1) for
the first queen. After a few more backtracks, we are led to the solution in which the remaining queens are in
squares (4,2), (6,3), (1,4), (3,5) and (5,6).
29. Assume that the graph has verticet>
V1 , v2 , ... , Vn . In looking for a Hamilton circuit we may as well start
building a path at u1 . The general step it> as follows. We extend the path if we can, to a new vertex (or to
u1 if this will complete the Hamilton circuit) adjacent to the vertex we are at. If we cannot extend the path
any further, then we backtrack to the last previous vertex in the path and try other untried extensions from
that vertex. The procedure for Hamilton paths is the same, except that we have to try all possible starting
vertices, and we do not allow a return to the starting vertex, stopping instead when we have a path of the
right length.
31. We know that every component of the graph has a spanning tree. The union of these spanning trees is clearly
a spanning forest for the graph, since it contains every vertex and two vertices in the same component are
joined by a path in the spanning tree for that component.
33. First we claim that the spanning forest will use n - c edges in all. To see this, let n, be the number of vertices
in the ith component, for i = 1, 2, ... , c. The spanning forest uses n, -1 edges in that component. Therefore
the spanning forest uses 2:(n, -1) = (I:n,) - c = n - c edges in all. Thus we need to remove m - (n - c)
edges to form a spanning forest.
35. We execute the breadth-first search algorithm, starting with the vertex v1 from which we wish to find the
shortest path. Each vertex v is assigned a value, L( v), which will be the length of a shortest path from v1 to v.
We initialize L( v1 ) = 0. and as each vertex w is added to the tree in Algorithm 2, we set L( w) = 1 + L( v) .
37. We carry out the breadth-first search algorithm, marking each vertex as we encounter it. The set of vertices
encountered (and all the edges joining those vertices) form one component. Once BFS has concluded we look
for an unmarked vertex and repeat the process starting from an unmarked vertex; this gives us a second
component. When there are no more unmarked vertices, we are done. This works just as well for DFS.
39. A connected simple graph has only one spanning tree if the graph is itself a tree (clearly its only spanning
tree is itself in this case). On the other hand, if a connected simple graph is not a tree, then it has a simple
circuit containing k ~ 3 edges, and one can find a spanning tree of the graph containing any k - 1 of these
edges but not the other (see Exercise 23 in Section 11.5).
Section 11.4
Spanning Trees
425
41. In effect we use the depth-first search algorithm on each component. In more detail, once that procedure
wants to stop, have it search through the list of vertices in the graph to find one that is not yet in the forest.
If there is such a vertex, then repeat the process starting from that vertex. Continue this until all the vertices
have been included in the forest.
43. If an edge uv is not followed while we are processing vertex u during the depth-first search process, then it
can only be the case that the vertex v had already been visited. There are two cases. If vertex v was visited
after we started processing u, then, since we are not finished processing u yet, v must appear in the subtree
rooted at u (and hence must be a descendant of u). On the other hand, if the processing of v had already
begun before we started processing u, then why wasn't this edge followed at that time? It must be that we
had not finished processing v, in other words, that we are still forming the subtree rooted at v, so u is a
descendant of v, and hence v is an ancestor of u.
45. Certainly these two procedures produce the identical spanning trees if the graph we are working with is a tree
itself, since in this case there is only one spanning tree (the whole graph). This is the only case in which that
happens, however. If the original graph has any other edges, then by Exercise 43 they must be back edges
and hence join a vertex to an ancestor or descendant, whereas by Exercise 34, they must connect vertices at
the same level or at levels that differ by 1. Clearly these two possibilities are mutually exclusive. Therefore
there can be no edges other than tree edges if the two spanning trees are to be the same.
47. Since the edges not in the spanning tree are not followed in the process, we can ignore them. Thus we can
assume that the graph was a rooted tree to begin with. The basis step is trivial (there is only one vertex),
so we assume the inductive hypothesis that breadth-first search applied to trees with n vertices have their
vertices visited in order of their level in the tree and consider a tree T with n + 1 vertices. The last vertex to
be visited during breadth-first search of this tree, say v, is the one that was added last to the list of vertices
waiting to be processed. It was added when its parent, say u, was being processed. We must show that v
is at the lowest (bottom-most, i.e., numerically greatest) level of the tree. Suppose not; say vertex x, whose
parent is vertex w, is at a lower level. Then w is at a lower level than u. Clearly v must be a leaf, since any
child of v could not have been seen before v is seen. Consider the tree T' obtained from T by deleting v.
By the inductive hypothesis, the vertices in T' must be processed in order of their level in T' (which is the
same as their level in T, and the absence of v in T' has no effect on the rest of the algorithm). Therefore
u must have been processed before w, and therefore v would have joined the waiting list before x did, a
contradiction. Therefore v is at the bottom-most level of the tree, and the proof is complete.
49. We modify the pseudocode given in Algorithm 2 by initializing m to be 0 at the beginning of the algorithm,
and adding the statements "m := m + l" and "assign m to vertex v" after the statement that removes vertex
v from L.
51. This is similar to Exercise 43. If a directed edge uv is not followed while we are processing its tail u during
the depth-first search process, then it can only be the case that its head v had already been visited. There are
three cases. If vertex v was visited after we started processing u, then, since we are not finished processing u
yet, v must appear in the subtree rooted at u (and hence must be a descendant of u), so we have a forward
edge. Otherwise, the processing of v must have already begun before we started processing u. If it had not
yet finished (i.e., we are still forming the subtree rooted at v ), then u is a descendant of v, and hence v is an
ancestor of u (we have a back edge). Finally, if the processing of v had already finished, then by definition
we have a cross edge.
53. There are five trees here, so there are C(5, 2) = 10 questions. Let T be the tree in Figure 3c, and let T1
through T 4 be the trees in Figure 4, reading from left to right. We will discuss one of the pairs at length and
426
Chapter 11
Trees
simply report the other answers. Let d(T, T1) denote the distance between trees T and T 1 . Note that tree
T1 has edges {a, e}, {c, g}, and {e, J} that tree T does not. Since the trees have the same number of edges,
there must also be 3 edges in T that are not in T1 (we do not need to list them). Therefore d(T, T1 ) = 6.
Similarly we have d(T, T2) = 4, d(T, T3) = 4, d(T, T4) = 2, d(T1, T2) = 4, d(T1, T3) = 4, d(T1, T4) = 6,
d(T2,T3) = 4, d(T2,T4) = 2, and d(T3,T4) = 4.
55. Let ei = { u, v}. The graph T2 U {ei} contains a (unique) simple circuit C containing edge e 1 . Now T 1 - { e 1}
has two components, one of which contains u and the other of which contains v. We travel along the circuit C,
starting at u and not using edge ei first, until we first reach a vertex in the component of T 1 - { ei} that
contains v; obviously we must reach such a vertex eventually, since we eventually reach v itself. The edge we
last traversed is e 2 . Clearly T2 U {ei} - { e 2} is a tree, since e 2 is on C. On the other hand, T 1 - { ei} U {e 2}
is also a tree, since e2 reunites the two components of T1 - { e 1}.
Onb Onb
57. Rooted spanning trees are easy to find in all six figures, as these pictures show. There are of course many
other possible correct answers. In the first five cases the tree is a path. In the last case, the root is d.
c
d
c
d
( 1 B)
~
] ~.
~
(20)
( 19)
~
•c
~
] ~.
~
•c
~
(21)
(22)
59. By Exercise 16 in Section 10.5, we know that such a directed graph has an Euler circuit. Now we traverse
the Euler circuit, starting at some vertex v (which will be our root), and delete from the circuit every edge
that has as its terminal vertex a vertex we have already visited on this traversal. The graph that remains is a
rooted spanning tree; there is a path from the root to every other vertex, and there can be no simple circuits.
61. According to Exercise 60, a directed graph contains a circuit if and only if there are any back edges. We can
detect back edges as follows. Add a marker on each vertex v to indicate what its status is: not yet seen (the
initial situation), seen (i.e., put into T) but not yet finished (i.e., visit(v) has not yet terminated), or finished
(i.e., visit(v) has terminated). A few extra lines in Algorithm 1 will accomplish this bookkeeping. Then to
determine whether a directed graph has a circuit, we just have to check when looking at edge uv whether the
status of v is "seen." If that ever happens, then we know there is a circuit; if not, then there is no circuit.
Section 11.5
Minimum Spanning Trees
SECTION 11.5
427
Minimum Spanning Trees
The algorithms presented here are not hard, once you understand them. The two algorithms are almost
identical, the only real difference being in the set of edges available for inclusion in the tree at each step. In
Prim 's algorithm, only those edges that are adjacent to edges already in the, tree (and not completing simple
circuits) may be added (so that, as a result, all the intermediate stages are trees). In Kruskal's algorithm, any
edge that does not complete a simple circuit may be added (so that the intermediate stages may be forests and
not trees). The reader might try to discover other methods for finding minimum spanning trees, in addition
to the ones in this section.
1. We want a minimum spanning tree in this graph. We apply Kruskal's algorithm and pave the following edges:
Oasis to Deep Springs, Lida to Gold Point, Lida to Goldfield, Silver Peak to Goldfield, Oasis to Dyer, Oasis
to Silver Peak, Manhattan to Tonopah, Goldfield to Tonopah, Gold Point to Beatty, and Tonopah to Warm
Springs. At each stage we chose the minimum weight edge whose addition did not create a simple circuit
(Kruskal's algorithm).
3. We start with the minimum weight edge {e, !} . The least weight edges incident to the tree constructed so far
are edges {c, !} and { e, h}, each with weight 3, so we add one of them to the tree (we will break ties using
alphabetical order, so we add {c, !} ). Next we add edge {e, h}, and then edge {h, i}, which has a smaller
weight but has just become eligible for addition. The edges continue to be added in the following order (note
that ties are broken using alphabetical order): {b,c}, {b,d}, {a,d}, and {g,h}. The total weight of the
minimum spanning tree is 22.
5. Kruskal's algorithm will have us include first the links from Atlanta to Chicago, then Atlanta to New York,
then Denver to San Francisco (the cheapest links). The next cheapest link, from Chicago to New York,
cannot be included, since it would form a simple circuit. Therefore we next add the link from Chicago to San
Francisco, and our network is complete.
7. The edges are added in the following order (with Kruskal's algorithm, we add at each step the shortest edge
that will not complete a simple circuit): {e,J}, {a,d}, {h,i}, {b,d}, {c,J}, {e,h}, {b,c}, and {g,h}. The
total weight of the minimum spanning tree is 22.
9. A graph with one edge obviously cannot be the solution, and a graph with two edges cannot either, since a
simple connected graph with two edges must be a tree. On the other hand, if we take a triangle ( K 3 ) and
weight all the edges equally, then clearly there are three different minimum spanning trees.
11. Ifwe simply replace each of the occurrences of the word "minimum" with the word ''maximum" in Algorithm 1,
then the resulting algorithm will find a maximum spanning tree.
13. We use an analog of Kruskal's algorithm, adding at each step an edge of greatest weight that does not create
a simple circuit. The answer here is unique. It uses edges {a, c}, {b, d}, { b, e}, and {c, e}.
15. There are numerous possible answers. One uses edges {a, d}, {b, !} , { c, g}, {d, p}, {e, !} , {!, j}, {g, k},
{h,l}, {i,j}, {i,m}, {j,k}, {j,n}, {k,l}, {k,o}, and {o,p}, obtained by choosing at each step an edge of
greatest weight that does not create a simple circuit.
17. If we want a second "shortest" spanning tree (which may, of course, have the same weight as the "shortest"
tree), then we need to use at least one edge not in some minimum spanning tree T that we have found. One
way to force this is, for each edge e of T, to apply a minimum spanning tree algorithm to the graph with e
deleted, and then take a tree of minimum weight among all of these. It cannot equal T, and so it must be a
second "shortest" spanning tree.
428
Chapter 11
Trees
19. The proof that Prim's algorithm works shows how to take any minimum spanning tree T and, if T is not
identical to the tree constructed by Prim's algorithm, to find another minimum spanning tree with even more
edges in common with the Prim tree than T has. The core of the proof is in the last paragraph, where we
add an edge ek+I to T and delete an edge e. Now if all the edges have different weights, then the result of
this process is not another minimum spanning tree but an outright contradiction. We conclude that there are
no minimum spanning trees with any edges not in common with the Prim tree, i.e., that the only minimum
spanning tree is the Prim tree.
21. We simply apply Kruskal's algorithm, starting not from the empty tree but from the tree containing these two
edges. We can add the following edges to form the desired tree: {c, d}, { k, l}, {b, f}, { c, g}, {a, b}, {f, j},
{a,e}, {g,h}, and {b,c}.
23. The algorithm is identical to Kruskal's algorithm (Algorithm 2), except that we replace the statement "T :=
empty graph" by the assignment to T initially of the specified set of edges, and, instead of iterating from 1
to n - 1, we iterate from 1 to n - 1 - s, where s is the number of edges in the specified set. It is assumed
that the specified set of edges forms no simple circuits.
25. a) First we need to find the least expensive edges incident to each vertex. These are the links from New York
to Atlanta, Atlanta to Chicago, and Denver to San Francisco. The algorithm tells us to choose all of these
edges. At the end of this first pass, then, we have a forest of two trees, one containing the three eastern cities,
the other containing the two western cities. Next we find the least expensive edge joining these two trees,
namely the link from Chicago to San Francisco, and add it to our growing forest. We now have a spanning
tree, and the algorithm has finished. Note, incidentally, that this is the same spanning tree that we obtained
in Example 1; by the result of Exercise 19, since the weights in this graph are all different, there was only one
minimum spanning tree.
b) On the first pass, we choose all the edges that are the minimum weight edges at each vertex. This set
consists of {a,b}, {b,f}, {c,d}, {a,e}, {c,g}, {g,h}, {i,j}, {f,j}, and {k,l}. At this point the forest has
three components. Next we add the lowest weight edges connecting these three components, namely {h, l}
and {b, c}, to complete our tree.
27. Let e 1 , e 2 , ... , en-I be the edges of the tree S chosen by Sollin's algorithm in the order chosen (arbitrarily
order the edges chosen at the same stage). Let T be a minimum spanning tree that contains all the edges
e 1 , e 2 , ... , ek for as large a k as possible. Thus 0 ::; k ::; n - 1. If k = n - 1, then S = T and we have
shown that S is a minimum spanning tree. Otherwise we will construct another minimum spanning tree T'
which contains edges e 1 , e 2 , ... , ek, ek+l , contradicting the choice of T and completing the proof.
Let S' be the forest constructed by Sollin's algorithm at the stage before ek+I is added to S. Let u be
the endpoint of ek+l that is in a component C of S' responsible for the addition of ek+l (i.e., so that ek+I
is the minimum weight edge incident to C). Let v be the other endpoint of ek+l. We let P be the unique
simple path from u to v in T. We follow P until we come to the first edge e' not in S'. Thus e' is also
incident to C. Since the algorithm chose to add ek+l on behalf of C, we know that w(ek+I) ::; w(e'), and
that e' was added on behalf of the component of its other endpoint. Now if e' is not in {e 1 , ... , ek}, then we
go on to the next paragraph. Otherwise, we continue following P until we come to the first edge e" not in
S'. Thus e" is also incident to C', but was added on behalf of another component C", and w(e') ::; w(e").
We continue in this way until we come to an edge e(r) not in {e 1 , ... , ek}.
Finally, let T' = TU {ek+ 1 } - { e(r)}. Then by stringing together the inequalities we have obtained along
the way, we know that w(ek+I) ::; w(e(r)). It follows that w(T') ::; w(T), so T' is again a minimum spanning
tree. Furthermore, T' contains e 1 , e 2 , ... , ek, ek+I, and we have our desired contradiction.
Review Questions
429
29. Suppose that there are r trees in the forest at some intermediate stage of Sollin's algorithm. Each new tree
formed during this stage will then contain at least two of the old trees, so there are at most r /2 new trees.
In other words, we have to reduce the number of trees by at least r - (r/2) = r/2. Since each edge added at
this stage reduces the number of trees by exactly one, we must add at least r /2 edges. Finally, the number of
edges added is of course an integer, so it is at least r /21 .
r
31. This follows easily from Exercises 29 and 30. After k stages of Sollin's algorithm, since the number of trees
begins at n and is at least halved at each stage, there are at most n/2k trees. Thus if n :::; 2k, then this
quantity is less than or equal to 1, so the algorithm has terminated. In other words, the algorithm terminates
after at most k stages if k ~ log n, which is what we wanted to prove.
33. Suppose by way of contradiction that a minimum spanning tree T contains edge e = uv that is the maximum
weight edge in simple circuit C. Delete e from T. This creates a forest with two components, one containing
u and the other containing v . Follow the edges of the path C - { e} , starting at u. At some point this path
must jump from the component of T - {e} containing u to the component of T - {e} containing v, say
using edge f. This edge cannot be in T, because e can be the only edge of T joining the two components
(otherwise there would be a simple circuit in T). Because e is the edge of greatest weight in C, the weight
of f is smaller. The tree formed by replacing e by f in T therefore has smaller weight, a contradiction.
35. The reverse-delete algorithm must terminate and produce a spanning tree, because the algorithm never disconnects the graph and upon termination there can be no more simple circuits. The edge deleted at each stage
of the algorithm must have been the edge of maximum weight in whatever circuits it was a part of. Therefore
by Exercise 33 it cannot be in any minimum spanning tree. Since only edges that could not have been in
any minimum spanning tree have been deleted, the result must be a minimum spanning tree. Actually the
assumption that the edge weights all be distinct can be avoided; see [K1Ta05] for details.
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 11
1. a) Seep. 746.
b) Seep. 746.
2. No, for each ordered pair of vertices u and v, there is a unique simple path from u to v.
3. See Examples 5-8 in Section 11.1.
4. a) Seep. 747.
b) Seep. 747.
c) Seep. 748.
d) Figure 8a in Section 11.1 is such a tree. Its root is a. The parent of each vertex is the vertex immediately
above it; thus the parent of b is a, the parent of c is also a, the parent of d is b, and so on. The children
of a vertex are the vertices immediately below it; thus the children of a are b and c, the children of b are d
and e, the only child of h is j, e has no children, and so on. The internal vertices are the ones with children:
a, b, c, d, h, i, and l. The leaves are the vertices without children: e, f, g, j, k, and m.
5. a) n - 1
b) If c is the number of components, then e
=n-
c.
6. a) Seep. 748.
b) mi+l; (m-l)i+l; seeTheorem4inSection 11.1.
7. a) See p. 753.
b) See p. 753.
c) between 1 and mh, inclusive
Chapter 11
430
8. a) See pp. 757-758.
Trees
b) Repeatedly apply Algorithm 1 in Section 11.2 to insert items one by one.
c) If we insert the items in the order given, we obtain the following tree.
grosbeek
nuthetch
kingfisher
9. a) See p. 762.
b) See pp. 762-763.
10. a) See p. 773, 775 and 776.
b) See Examples 2-4 in Section 11.3.
11. a) Build the expression tree. Its preorder traversal gives prefix form; its postorder traversal gives postfix form;
and its inorder traversal gives infix form (if we assume that there are only binary operations, and if we put a
pair of parentheses around the expression for every subtree that is not a leaf).
b) Here is the expression tree.
+
/~+
-
/ \3
x
I
/""
/ \4
x
1'
/ \3
-
/ \y
x
c) prefix:
+-
x 3 + / x 4 i - x y 3;
postfix: x 3 - x 4 / x y - 3 i
++
12. See Theorem 1 in Section 11.2 and the discussion preceding it.
13. a) See pp. 763-764.
b) The answer is not unique, because there is a choice of whether to make the subtree containing A and B
or the leaf C the left subtree at the second step. Assuming we choose the former, the code for A is 000; that
for B is 001; that for C is 01; and the code for D is 1.
14. The tree is too large to draw here.
(It can be completed by referring to the solution to Exercise 33 in
Section 11.2.) The children of the root are the positions 4, 31, 21, 11, and 1. Each of these has value -1,
except for the position with just one pile with one stone. That is a winning position for the first player, since
it is a leaf and the second player has no move. So the first player wins the game by removing the pile of four
stones at her first move. Thus the value of the root of the tree is 1.
15. a) See p. 785.
b) all connected ones
c) road-plowing, communication network reinforcement
16. a) See pp. 787-789 (depth-first search) and pp. 789-791 (breadth-first search).
b) See Exercises 15 and 16 in Section 11.4.
17. a) See Example 6 in Section 11.4.
18. a) See p. 798.
b) Apply this algorithm to W5 .
b) minimum cost communications network, shortest connecting road configuration
19. a) See Algorithms 1 and 2 in Section 11.5.
b) See Examples 2 and 3 in Section 11.5.
Supplementary Exercises
431
SUPPLEMENTARY EXERCISES FOR CHAPTER 11
1. There are of course two things to prove here. First let us assume that G is a tree. We must show that G
contains no simple circuits (which is immediate by definition) and that the addition of an edge connecting
two nonadjacent vertices produces a graph that has exactly one simple circuit. Clearly the addition of such
an edge e = {u, v} produces a graph with a simple circuit, namely u, e, v, P, u, where P is the unique simple
path joining v to u in G. Since P is unique, moreover, this is the only simple circuit that can be formed.
To prove the converse, suppose that G satisfies the given conditions; we want to prove that G is a tree,
in other words, that G is connected (since one of the conditions is already that G has no simple circuits). If
G is not connected, then let u and v lie in separate components of G. Then edge {u, v} can be added to
G without the formation of any simple circuits, in contradiction to the assumed condition. Therefore G is
indeed a tree.
3. Let P be a longest simple path in a given tree T. This path has length at least 1 as long as T has at least
one edge, and such a longest simple path exists since T is finite. Now the vertices at the ends of P must both
have degree 1 (i.e., be pendant vertices), since otherwise the simple path P could be extended to a longer
simple path.
5. Since the sum of the degrees of the vertices is twice the number of edges, and since a tree with n vertices has
n - 1 edges, the answer is 2n - 2.
7. One way to prove this is simply to note that the conventional way of drawing rooted trees provides a planar
embedding. Another simple proof is to observe that a tree cannot contain a subgraph homeomorphic to K 5
or K 3 ,3 , since these must contain simple circuits. A third proof, of the fact that a tree can be embedded in
the plane with straight lines for the edges, is by induction on the number of vertices. The basis step (a tree
with one vertex) is trivial (there are no edges). If tree T contains n + 1 vertices, then delete one vertex of
degree 1 (which exists by Exercise 3), embed the remainder (by the inductive hypothesis), and then draw the
deleted vertex and reattach it with a short straight line to the proper vertex.
9. Since the colors in one component have no effect on the colors in another component, it is enough to prove
this for connected graphs, i.e., trees. In order to color a tree with 2 colors, we simply view the tree as rooted
and color all the vertices at even-numbered levels with one color and all the vertices at odd-numbered levels
with another color. Since every edge connects vertices at adjacent levels, this coloring is proper.
11. A B-tree of degree k and height h has the most leaves when every vertex not at level h has as many children
as possible, namely k children. In this case the tree is simply the complete k-ary tree of height h, so it has
kh leaves. Thus the best upper bound for the number of leaves of a B-tree of degree k and height h is kh.
To obtain a lower bound, we want to have as few leaves as possible. This is accomplished when each vertex
has as few children as possible. The root must have at least 2 children. Each other vertex not at level h must
have at least lk/21 children. Thus there are 2lk/2lh-l leaves, so this is our best lower bound. (Of course if
h = 0, then the tree has exactly 1 leaf.)
13. We follow the instructions, constructing each tree from the previous trees as indicated.
•
I
Chapter 11
432
Trees
15. Since Bk+l is formed from two copies of Bk, one shifted down one level, the height increases by 1 as k
increases by 1. Since Bo had height 0, it follows by induction that Bk has height k.
17. Since the root of Bk+l is the root of Bk with one additional child (namely the root of the other Bk), the
degree of the root increases by 1 as k increases by 1. Since Bo had a root with degree 0, it follows by
induction that Bk has a root with degree k.
19. We follow the recursive definition in drawing these trees. For example, we obtain 8 4 by taking a copy of 8 3
(on the left), adding one more child of the root (the right-most one), and putting a copy of 8 3 rooted at this
new child.
s 1-tree
21. We prove this by induction on k. If k = 0 or 1, then the result is trivial. Assume the inductive hypothesis
that the Sk_ 1 -tree can be formed in the manner indicated, and let T be an Sk-tree. Then T consists of a
copy of an Sk-l tree Tk-l with root rk-l, together with another copy of an Sk-l tree whose root is made
a child of rk-l. Now by the inductive hypothesis, this latter Sk_ 1 -tree can be formed from a handle v and
disjoint trees T 0 , T1 , ... , Tk_ 2 by connecting v to r 0 and r, to r,+1 for i = 0, 1, ... , k - 3. Since our tree
T is formed by then joining rk- 2 to rk-l, our tree is formed precisely in the manner desired, and the proof
by induction is complete.
23. Essentially we just want to do a breadth-first search of the tree, starting at the root, and considering the
children of a vertex in the order from left to right. Thus the algorithm is to perform breadth-first search,
starting at the root of the tree. Whenever we encounter a vertex, we print it out.
25. First we determine the universal addresses of all the vertices in the tree. The root has address 0. For every
leaf address, we include an address for each prefix of that address. For example, if there is a leaf with address
4.3.7.3, then we include vertices with addresses 4, 4.3, and 4.3.7. The tree structure is constructed by making
all the vertices with positive integers as their addresses the children of the root, in order, and all the vertices
with addresses of the form A.i, where A is an address and i is a positive integer, the children of the vertex
whose address is A, in order by i.
27. For convenience, let us define a "very simple circuit" to be a set of edges that form a circuit all of whose
vertices are distinct except that (necessarily) the last vertex is the same as the first. Thus a cactus is a graph
in which every edge is in no more than one very simple circuit.
a) This is a cactus. The edge at the top is in no simple circuit. The three edges at the bottom are only in
the triangle they form.
b) This is not a cactus, since the edge in the upper right-hand corner, for instance, is in more than one very
simple circuit: a triangle and a pentagon.
c) This is a cactus. The edges in each of the three triangles are each in exactly one very simple circuit.
433
Supplementary Exercises
29. Adding a very simple circuit (see the solution to Exercise 27 for the definition) does not give the graph any
more very simple circuits, except for the one being added. Thus the new edges are each in exactly one very
simple circuit, and the old edges are still in no more than one.
31. There is clearly a spanning tree here which is a simple path (a, b, c, f, e, d, for instance); since each degree is
1 or 2, this spanning tree meets the condition imposed.
33. There is clearly a spanning tree here which is a simple path (a, b, c, f, e, d, i, h, g, for instance); since each
degree is 1 or 2, this spanning tree meets the condition imposed.
35. We need to label these trees so that they satisfy the condition. We work by trial and error, using some common
sense. For example, the labels 1 and n need to be adjacent in order to obtain the difference n - 1.
3
1
1
4
2
3
• • • •
(a}
5
3
4
• I2 ••
(b}
(c}
1
7
2
6
T
(d)
37. We count the caterpillars by drawing them all, using the length of the longest path to organize our work. In
fact every tree with six vertices is a caterpillar. They are the five trees shown with heavy lines in our solution
to Exercise 33c in Section 11.1, together with the star K 1 , 5 . Thus the answer is 6.
39. a) The frequencies of the bits strings are 0.81 for 00, 0.09 for 01 and for 10, and 0.01 for 11. The resulting
Huffman code uses 0 for 00, 11 for 01, 100 for 10, and 101 for 11. (The exact coding depends on how
ties were broken, but all versions are equivalent.) Thus in a string of length n, the average number of bits
required to send two bits of the message is 1·0.81+2 · 0.09 + 3 · 0.09 + 3 · 0.01 = 1.29, so the average number
of bits required to encode the string is 1.29 · (n/2) = 0.645n.
b) The frequencies of the bits strings are 0.729 for 000, 0.081 for 001, 010, and 100, 0.009 for 011, 101,
and 110, and 0.001 for 111. The resulting Huffman code uses 0 for 000, 100 for 001, 101 for 010, 110 for
100, 11100 for 011, 11101 for 101, 11110 for 110, and 11111 for 111. (The exact coding depends on how
ties were broken, but all versions are equivalent.) Thus in a string of length n, the average number of bits
required to send three bits of the message is 1 · 0. 729 + 3 · 3 · 0.081 + 5 · 3 · 0.009 + 5 · 0.001 = 1.598, so the
average number of bits required to encode the string is 1.598 · (n/3) ~ 0.533n.
41. Let T be a minimum spanning tree. If T contains e, then we are done. If not, then adding e to T creates
a simple circuit, and then deleting another edge e' of this simple circuit gives us another spanning tree T'.
Since w ( e) ::;: w (e') , the weight of T' is no larger than the weight of T. Therefore T' is a minimum spanning
tree containing e.
43. The proof uses the same idea as in the solution to Exercise 18 in Section 11.5. Suppose that edge e is the
edge of least weight incident to vertex v, and suppose that T is a spanning tree that does not include e. Add
e to T, and delete from the simple circuit formed thereby the other edge of the circuit that contains v. The
result will be a spanning tree of strictly smaller weight (since the deleted edge has weight greater than the
weight of e ). This is a contradiction, so T must include e.
45. Because paths in trees are unique, an arborescence T of a directed graph G is just a subgraph of G that is
a tree rooted at r, containing all the vertices of G, with all the edges directed away from the root. Thus the
in-degree of each vertex other than r is 1. To show the converse, it is enough to show that for each v E V
there is a unique directed path from r to v. Because the in-degree of each vertex other than r is 1 , we can
434
Chapter 11
Trees
follow the edges of T backwards from v. This path can never return to a previously visited vertex, because
that would create a simple circuit. Therefore the path must eventually stop, and it can stop only at r, the
vertex whose in-degree is not necessarily 1 (in fact r has to have in-degree 0). Following this path forward
gives the path from r to v required by the definition of arborescence.
47. a) We just run the breadth-first search algorithm, starting from v and respecting the directions of the edges,
marking each vertex that we encounter as reachable.
b) Clearly running breadth-first search on
again starting at v, respecting the directions of the edges,
and marking each vertex that we encounter, will identify all the vertices from which v is reachable.
cconv,
c) By definition, the strong component of G containing v consists of all vertices w such that w is reachable
from v and v is reachable from w. Therefore the set of vertices marked by both of the algorithms above
constitute the vertices in the strong component of G containing v. (The edges in this strong component, of
course, are just all the edges joining vertices in this strong component.) So choose a vertex v1 and find the
strong component containing this vertex (it might be as small as {vi}). Then choose another vertex v 2 not
yet in a strong component and find the strong component of v 2 . Repeat until all vertices have been included.
Recall that by Exercise 17 in Section 10.4, strong components are pairwise disjoint.
WRITING PROJECTS FOR CHAPTER 11
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. You can probably find something useful in [BiLl]. Also, take a look at [WhCl].
2. Look up ''phylogenetic tree" on the Web.
3. This Wikipedia site might be a place to start:
en.wikipedia.org/wiki/Hierarchical_clustering_of_networks
4. Consult books on data structures or algorithms, such as [Kr] or [Ma2].
5. There is a relevant article in [De2], pp. 290-295.
6. See hints for Writing Project 4.
7. A Web search will turn up some good expositions.
8. An elementary exposition can be found in [Gr2]. See also algorithm or data structures texts, such as [Sm].
9. A Web search will turn up some good sites, including one by Dr. A. N. Walker.
10. See books on parallel processing, such as [Le2], to learn what meshes of trees are and how they are applied.
11. Books are beginning to appear in this new field; see [Ma5], for instance.
12. Good books on algorithms will contain material for this. See [CoLe], for example.
13. Good books on algorithms will contain material for this. See [CoLe], for example.
Writing Projects
435
14. Each search company wants to keep the details of its techniques secret, but you should be able to find some
general information on the Web. See also [Sz].
15. This problem is much more difficult than the plain minimum spanning tree problem. You may have to go to
the research literature here (search the Mathematical Reviews, database available on the Web as MathSciNet,
for keywords "spanning tree" and "constrained" or "degree").
16. Any good data structures or algorithms book would have loads of material on this topic, including those
mentioned in Writing Project 4 (or others given in the reference list). See also volume 3 of the classic [Kn].
17. See the article [Gr He].
18. See books on random graph theory ([Bo2] or [Pal]), or the article [Ti].
436
Chapter 12
Boolean Algebra
CHAPTER12
Boolean Algebra
SECTION 12.1
Boolean Functions
The first 28 of these exercises are extremely straightforward and should pose no difficulty. The next four relate
to the interconnection between duality and De Morgan's laws; they are a bit subtle. Boolean functions can be
proved equal by tables of values (as we illustrate in Exercise 13), by using known identities (as we illustrate
in Exercise 11), or by taking duals of equal Boolean functions (Exercise 28 justifies this). To show that two
Boolean functions are not equal, we need to find a counterexample, i.e., values of the variables that give the
two functions different values. Exercises 35-43 deal with abstract Boolean algebras, and the proofs are of the
formal, "symbol-pushing" variety.
1. a) 1 · 0 = 1 · 1
=1
c)O·O=l·O=O
b) l+l=l+O=l
d) (1
+ o) =I= o
3. a) We compute (1·1) + (0 · 1+0) = 1+(0+0) = 1+(1+0) = 1+1=1.
b) Following the instructions, we have (T AT) V (•(FA T) VF)
=T.
5. In each case, we compute the various components of the final expression and put them together as indicated.
For part (a) we have
x
y
z
x
xy
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
0
0
0
0
1
1
1
1
0
0
0
0
1
1
0
0
For part (b) we have
For part ( c) we have
x
y
z
yz
x+yz
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
0
0
1
0
0
0
1
1
1
1
1
0
0
0
Section 12.1
437
Boolean Functions
x
y
z
'fl
x'fj
xyz
xyz
xy+ xyz
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
0
0
1
1
0
0
1
1
0
0
1
1
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
0
1
1
1
1
1
1
1
For part ( d) we have
x
y
z
'fl
z
yz
'flz
yz+yz
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
0
0
0
1
0
0
0
0
0
0
1
0
0
0
1
1
0
0
1
1
0
0
1
x(yz
+ yz)
1
0
0
1
0
0
0
0
7. In each case, we note from our solution to Exercise 5 which vertices need to be blackened in the cube, as in
Figure 1.
(8)
111
100
(b)
111
100
010
101
010
011
000
001
(c)
000
111
100
001
(d)
111
100
101
010
011
000
001
000
001
9. By looking at the definitions, we see that this equation is satisfied if and only if x
x=y=l.
= y, i.e., x = y = 0 or
11. First we ''factor" out an x by using the identity and distributive laws: x + xy = x · 1 + x · y = x · (1 + y).
Then we use the commutative law, the domination law, and finally the identity law again to write this as
x . (y + 1) = x . 1 = x.
13. Probably the simplest way to do this is by use of a table, as in Example 8. We list all the ppssible values for
the triple (x, y, z) (there being eight such), and for each compute both sides of this equation. For example,
for x = y = z = 0 we have x'fj + yz + xz = 0 · 1 + 0 · 1 + 1 · 0 = 0 + 0 + 0 = 0, and similarly xy
We do this for all eight lines of the table to conclude that the two functions are equal.
t yz + xz = 0.
438
Chapter 12
x
y
z
xy+yz+xz
xy+yz+xz
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
0
1
1
1
1
1
1
0
0
1
1
1
1
1
1
0
Boolean Algebra
= x and x + x = x. There are only four things to check: 0 · 0 = 0,
15. The idempotent laws state that x · x
0 + 0 = 0, 1 · 1 = 1, and 1 + 1 = 1, all of which are part of the definitions. The relevant tables, exhibiting
these calculations, have only two rows.
17. The domination laws state that x + 1 = 1 and x · 0 = 0. There are only four things to check: 0 + 1 = 1,
0 · 0 = 0, 1 + 1 = 1, and 1 · 0 = 0, all of which are part of the definitions. The relevant tables, exhibiting
these calculations, have only two rows.
19. We can verify each associative law by constructing the relevant table, which will have eight rows, since there
are eight combinations of values for x, y, and z in the equations x + (y + z) = (x + y) + z and x(yz) = (xy)z.
Rather than write down these tables, let us observe that in the first case, both sides are equal to 1 unless
x = y = z = 0 (in which case both sides equal 0), and, dually, in the second case, both sides are equal to 0
unless x = y = z = 1 (in which case both sides equal 1 ).
21. We construct the relevant tables (as in Exercise 13) and compute the quantities shown. Since the fourth and
seventh columns are equal, we conclude that (xy) = x
conclude that (x + y) = xy.
+ y;
since the ninth and tenth columns are equal, we
x
y
xy
(xy)
x
y
x+y
x+y
1
1
0
0
1
0
1
0
1
0
0
0
0
1
1
1
0
0
1
1
0
1
0
1
0
1
1
1
1
1
1
0
---
(x
+ y)
0
0
0
1
xy
0
0
0
1
23. The zero property states that x · x = 0. There are only two things to check: 0 ·0 = 0 · 1 = 0 and l
The relevant table, exhibiting this calculation, has only two rows.
·I =
l ·0
= 0.
25. We could prove these by constructing tables, as in Exercise 21. Instead we will argue directly.
a) The left-hand side is equal to 1 if x i- y. In this case the right-hand side is necessarily 1 · 0 = 1, as well.
On the other hand if x = y = 1, then the left-hand side is by definition equal to 0, and the right-hand side
equals 1·0 = O; similarly if x = y = 0, then the left-hand side is by definition equal to 0, and the right-hand
side equals 0 · 1 = 0.
b) The left-hand side is equal to 1 if x i- y. In this case the right-hand side is necessarily 1 + 0 = 1 or
0 + 1 = 1, as well. On the other hand if x = y, then the left-hand side is by definition equal to 0, and the
right-hand side equals 0 + 0 = 0.
27. a) We can prove this by constructing the appropriate table, as in Exercise 21. What we will find is that each
side equals 1 if and only if an odd number of the variables are equal to 1 . Thus the two functions are equal.
b) This is not an identity. If we let x
side is 1 EB 1 = 0.
= y = z = 1, then the left-hand side is 1 + 0 = 1, while the right-hand
c) This is not an identity. If we let x
right-hand side is 0 + 1 = 1.
=
y
= 1 and
z
= 0, then the left-hand side is 1 EB 1 = 0, while the
Section 12.1
439
Boolean Functions
29. Let B be a Boolean expression representing F, and let D be the dual of B. We want to show that for every
set of values assigned to the variables x 1 , x 2 , ... , Xn, the value D equals the opposite of the value of B
with the opposites of these values assigned to x 1 , x2, ... , Xn. The trick is to look at B. Then by repeatedly
applying De Morgan's laws from the outside in, we see that B is the same as the expression obtained by
replacing each occurrence of Xi in D by Xi. Thus for any values of x1, x 2 , ... , Xn, the value of D is the
same as the value of B for the corresponding value of x1 , x2 , ... , Xn. This tells us that B represents the
function whose values are exactly those of the function represented by D when the opposites of each of the
values of the variables Xi are used, and that is exactly what we wanted to prove.
31. Because of the stated condition, we are free to specify F(l, y, z) for all pairs (y, z), but then all the values of
F(O, y, z) are thereby determined. There are 4 such pairs (y, z) (each one can be either 0 or 1 ), and for each
such pair we have 2 choices as to the value of F (l, y, z) . Therefore the answer is 24 = 16.
33. We need to replace each 0 by F, 1 by T, + by V, · (or Boolean product implied by juxtaposition) by /\,
and by -, . We also replace x by p and y by q so that the variables look like they represent propositions,
and we replace the equals sign by the logical equivalence symbol. Thus for the first De Morgan law in Table 5,
xy = x + y becomes -,(p /\ q) = -,p V -,q, which is the first De Morgan law in Table 6 of Section 1.3. Dually,
x + y = xy becomes -,(p V q) =: -,p /\ -,q for the other De Morgan law.
35. We need to play around with the symbols until the desired results fall out. To prove that x V x
compute x V (x /\ x) in two ways. On the one hand,
= x, let us
xv (x /\ x) =xv O = x
by the complement law and the identity law. On the other hand, by using the distributive law followed by the
complement and identity laws we have
xv (x /\ x) =(xv x) /\(xv x)
= (x V x) /\ 1 = x V x.
By transitivity of equality, x = x V x. The other property is the dual of this one, and its proof can be obtained
by formally replacing every V with /\ and replacing every 0 with 1 , and vice versa. Thus our proof, shortened
to one line, becomes
x
= x /\ 1 = x /\ (x v x) = (x /\ x) v (x /\ x) = (x /\ x) v o = x v x.
37. By Exercise 36 we know that the complement of an element is that unique element that obeys the complement
laws. Therefore to show that 0 = 1 we just need to prove that 0 V 1 = 1 and 0 /\ 1 = 0. But these follow
immediately from the identity laws (and, for the first, the commutative law). The other half follows in the
same manner.
39. Before we prove this, we will find the following lemma useful: x /\ 0 = 0 and x V 1 = 1 for all x. To prove the
first half of the lemma, we invoke the results of Exercises 35 and 36 and compute as follows:
x /\ o = x /\ (x /\ x)
=(x/\x)/\x
= x /\ x = 0.
The second half is similar, using the duality replacements mentioned in the solution to Exercise 35.
440
Chapter 12
Boolean Algebra
Now for the exercise at hand, by Exercise 36 it is enough to show that the claimed complements behave
correctly. That is, we must show that (x Vy) V (x A Y) = 1 and (x Vy) A (x A y) = 0. For the second of
these we compute as follows, using the lemma at the end and using the various defining properties freely (in
particular, we use distributivity, associativity and commutativity a lot).
(xv y) A (x A y) = 'fJ A
[x A (xv y)]
= 'fJ A [(x Ax) v (x A y)]
= y /\ [O V (x /\ y)]
= 'jj /\ x /\ y = x /\ (y /\ y)
=xAO=O
The first statement is proved in a similar way:
(xv y) v (x /\ y)
= yv
[xv (x /\ y)]
= y v [(x v x) /\ (x v Y)]
= y V [1 A (x V 'fJ)]
=yv
x v 'iJ = (y v Y) v x
=lVx=l
And of course the other half of this problem is proved in a manner completely dual to this.
41. Using the hypothesis, we compute as follows. x = x V 0 = x V (x Vy) = (x V x) Vy= x Vy= 0. Similarly
for y, by the commutative law. Note that we used the result of Exercise 35. The other statement is proved in
the dual manner: x = x /\ 1 = x /\ (x /\ y) = (x /\ x) /\ y = x /\ y = 1 and similarly for y.
43. Most of this work was done in the Supplementary Exercises for Chapter 9 (or else is immediate from the
definition). We need to verify the five laws (each one consisting of a dual pair, of course). The identity laws
are Exercise 41. The complement laws are part of the definition of "complemented". The associative and
commutative laws are Exercise 39. Finally, the distributive laws are again part of the definition.
SECTION 12.2
Representing Boolean Functions
The first six exercises are straightforward practice dealing with sum-of-products expansions. These are obtained by writing down one product term for each combination of values of the variables that makes the
function have the value 1, and taking the sum of these terms. The dual to the sum-of-products expansion is
discussed in Exercises 7-11, and these should be looked at. Since the concept of complete sets of operators
plays an important role in the logical circuit design in sections to follow, Exercises 12-20 are also important.
1. a) We want x, 'jj, and z all to have the value 1; therefore we take the product x'jj z. The other parts are
similar, so we present only the answers.
b) xyz
c) xy z
d) xyz
3. a) We want the function to have the value 1 whenever at least one of the variables has the value 1. There are
seven minterms that achieve this, so the sum has seven summands: xy z+xyz+xy z+xy z+xyz+xyz+xy z.
b) Here is another way to think about this problem (rather than just making a table and reading off the
minterms that make the value equal to 1 ). If we expand the expression by the distributive law (and use the
commutative law), we get xy+yz. Now invoking the identity laws, the law that s+s = 1, and the distributive
and commutative laws again, we write this as xyl + lyz = xy(z + z) + (x + x)yz = xyz + xyz + xyz + xyz.
Finally, we use the idempotent law to collapse the first and third term, to obtain our answer: xyz+xyz+xyz.
Section 12.2
Representing Boolean Functions
441
c) We can use either the straightforward approach or the idea used in part (b). The answer is x y z + x y z
+
xyz+xyz.
d) The method discussed in part (b) works well here, to obtain the answer x y z + x yz.
5. We need to list all minterms that have an odd number of the variables without bars (and hence an odd number
with bars). There are C(4, 1) +C(4, 3) = 8 terms. The answer is wxyz+wxy z+wxy z+wxy z+wxyz+
wxyz+wxyz+wxyz.
7. This exercise is dual to Exercise 1.
x
will have the value 0 if and only if x = 1. Similarly, y will have the value 0 if and only if
y = 1. Therefore the expression x + y + z will have the value 0 precisely in the desired case. The remaining
parts are similar, so we list only the answers.
a) Note that
b) x+y+z
c) x+y+z
9. By the definition of "+ ," the sum Y1 + · · · + Yn has the value 0 if and only if each Yi = 0. This happens
precisely when Xi = 0 for those cases in which Yi = Xi and Xi = 1 in those cases in which Yi = Xi.
11. a) This function is already written in its product-of-sums form (with one factor).
y = 0 or both x and z equal 0. Therefore we need maxterms
x + y + z, x + y + z, x + y + z , and x + y + z (to take care of y = 0), and also x + y + z . Therefore the
b) This function has the value 0 in case
answer is the product of these five maxterms.
c) This function has the value 0 in case x = 0. Therefore we need four maxterms, and the answer is
(x + y + z)(x + y + z)(x + y + z)(x + y + z).
d) Let us indicate another way to solve problems like this. In Exercise 3d we found the sum-of-products
expansion of this function. Suppose that we take take sum-of-products expansion of the function that is the
opposite of this one. It will have all the minterms other than the ones in the answer to Exercise 3d, so it will
be xy z+x yz+xy z+xyz+xy z+xyz. If we now take the complement of this (put a big bar over it), then
we will have an expression for the function we want. Then we push the complementations inside, using De
Morgan's laws and the fact that = s. This will give us the desired product-of-sums expansion. Formally, all
we do is put parentheses around the minterms, erase all the plus signs, put plus signs between all the variables
(where there used to be implied products), and change every complemented variable to its uncomplemented
s
version and vice versa. The answer is thus
(x + y + z)(x + y + z)(x + y + z)(x + y + z)(x + y + z)(x + y + z).
13. To do this exercise we need to use De Morgan's law to replace st by (s + t). Thus we just do this formally
in the expressions in Exercise 12, and we obtain the answers. It is also good to simplify double complements,
of course.
a) This is already in the desired form, having no products.
b) x+y(x+z) =x+ (y+ (x+z)) =x+ (y+ (x+z))
c) This is already in the desired form, having no products.
d) x(x+y+z) = (x+(x+y+z)) = (x+(x+y+z))
15. a) We use the definition of l · If
the corresponding values of x.
x
= 1, then
x l x = 0; and if x = 0, then x l x = 1. These are precisely
b) We can construct a table to look at all four cases, as follows. Since the fifth and sixth columns are equal,
the expressions are equivalent.
Chapter 12
442
x y xlx
1
1
1
0
0
0
0
0
1
0
1
1
YlY
0
1
0
1
(x l x) l (y l y)
Boolean Algebra
xy
1
1
0
0
0
0
0
0
c) We can construct a table to look at all four cases, as follows. Since the fourth and fifth columns are equal,
the expressions are equivalent.
x y xly
(x 1 y) 1 (x 1 y)
x+y
1
1
1 1
0
1
1
1 0
0
1
1
0 1
0
1
0
0
0 0
17. a) Since x + y + z = (x + y) + z, we first write this as ((x + y) I (x + y)) I (z I z), using the identity in
Exercise 14c. Then we rewrite x + y as (x Ix) I (y I y), using the same identity. This gives us
(((x
Ix) I (y I y)) I ((x Ix) I (y I y))) I (z I z).
b) First we write this as ((x + z) I y) I ((x + z) I y), using the identity in Exercise 14b. Then we rewrite x + z
as (x Ix) I (z I z), using the identity in Exercise 14c. This gives us
(((x
Ix) I (z I z)) I y) I (((x Ix) I (z I z)) I y).
c) There are no operators, so nothing needs to be done; the expression as given is the answer.
d) First we write this as (x I fl) I (x I fl), using the identity in Exercise 14b. Then we rewrite
the identity in Exercise 14a. This gives us (x I (y I y)) I (x I (y I y)).
y as y I y, using
19. We claim that it is impossible to write a Boolean expression for x involving x, +, and ·. The reason is that
if x = 1, then any combination of these two operators applied to x (and the results of previous calculations)
can yield only the value 1. But I = 0. (This problem is somewhat harder if we allow the use of the constants
0 and 1 . The argument given here is then no longer valid, since x · 0 = 0. Nevertheless it is possible to
show that even with these constants allowed, the set {+, ·} is not functionally complete---i.e., it is impossible
to write down a Boolean expression using only the operators + and · (together with x , 0, and 1) that is
equivalent to x. What one can do is to prove by induction on the length of the expression that any such
Boolean expression that has the value 1 when x = 0 must also have the value 1 when x = 1.)
SECTION 12.3 Logic Gates
This section is a brief introduction to circuit design using AND and OR gates and inverters. In real life,
circuits will have thousands of these components, but you will get some of the flavor in these exercises. Notice
particularly Exercise 9, which shows how circuits already constructed can be further combined to give more
complex, useful circuits. If we want to get by with fewer types of gates (but more gates), then we can use
NOR or NAND gates, as illustrated in Exercises 15~18. Exercise 20 introduces the notion of the depth of a
circuit.
Section 12.3
443
Logic Gates
1. The output of the OR gate at the top is x
the output of the circuit is (x + y )y.
+ y.
This and y are the inputs to the final AND gate. Therefore
3. The idea is the same as in the previous two exercises. The final output is an OR with two inputs. The first
of these inputs is the result of inverting xy, and the second is z
+ x.
Therefore the answer is (xy) + (z + x) .
+ y + z, x + y + z, and x + y + z. Therefore the output from the
(x + y + z) + (x + y + z) + (x + y + z). It is not hard to see that this is always 1, since if
5. The outputs from the three OR gates are x
final OR gate is
x = 1 then the output of the top initial OR gate is 1, whereas if x
OR gate is 1.
=0
then the output of the middle initial
7. Let v, w, x, y, and z be the votes of the five individuals, with a 1 representing a yes vote and a 0 representing
a no vote. Then the majority will be a yes vote (represented by an output of 1) if and only if there are at
least three yes votes. Thus we make an AND gate for each of the C(5, 3) = 10 triples of voters, and combine
the outputs from these 10 gates with an OR. We turn the picture on its side for convenience.
9. The circuit is identical to Figure 10, expanded by two more units to accommodate the two additional bits. To
get the computation started, x 0 and y 0 are the inputs to the half adder. Thereafter, the carry bit from each
column is input, together with the next pair (xio y,) to a full adder to find the output and carry for the next
column. The final carry bit ( c4 ) is the final answer bit ( s 5 ).
XO
Yo
x,
Y1
X2
Y2
X3
Y3
X4
Y4
so
s,
s2
S3
S4
S5
11. We will construct the full subtractor directly. Suppose the input bits are x. y, and b, where we are computing
x - y with borrow b. The output is a bit, z, and a borrow from the next column, b'. Then by looking
at the eight possibilities for the bits x, y, and b, we see that z = x y b + x y b + x y b + xy b; and that
b' = xyb+xyb+xyb+xyb. Therefore a full subtractor can be formed by using AND gates, OR gates, and
inverters to represent these expressions. We obtain the circuits shown below.
444
Chapter 12
Boolean Algebra
13. The first number is larger than the second if X1 > Y1 (which means x 1 = 1 and y 1 = 0), or if x 1 = y 1 (which
means either that x1 = 1 and Y1 = 1, or that x1 = 0 and Y1 = 0) and also Xo > Yo (which means xo = 1
and Yo = 0 ). We translate this sentence into a circuit in the obvious manner, obtaining the picture shown.
15. Note that in this exercise the usual operation symbol
a) By Exercise 14a in Section 12.2,
x = x I x.
I is used for the
Therefore the gate for
NAND operation.
x is as shown below.
b) By Exercise 14c in Section 12.2, x + y = (x Ix) I (y I y). Therefore the gate for x + y is as shown below.
c) By Exercise 14b in Section 12.2, xy
= (x I y) I (x I y). Therefore the gate for xy is as shown below.
Section 12.4
Minimization of Circuits
445
x
y
x
xy
y
d) First we note that x EB y = xy + xy
= ((xy) (xy)) = (xY) I (xy) = (x I y) I (x I y).
We constructed the
gate for inverting in part (a) . Therefore the gate for xy is as shown below.
x@y
17. We know that the sum bit in the half adder is s = xy + x y. The answer to Exercise 15d shows precisely this
gate constructed from NAND gates, so it gives us this part of the answer. Also, the carry bit in the half adder
is c = xy. The answer to Exercise 15c shows precisely this gate constructed from NAND gates, so it gives us
this part of the answer.
19. We can set this up so that the value of Xi "gets through" to a final OR gate if and only if (c 1 c0 )2 = i. For
example, we want x2 to get through if and only if c 1 = 1 and c0 = 0, since the Base 2 numeral for 2 is 10.
We can do this by combining the x, input with either c0 or its inversion and either c 1 or its inversion, using
an AND gate with three inputs. Thus the output of each of these gates is either 0 (if (c 1 c0 )2 -=/:: i) or is the
value of Xi (if (c1 c0 )2 = i). So if we combine the result of these four outputs, using an OR gate, we will get
the desired result (since at most one of the four outputs can possibly be nonzero). Here is the circuit.
SECTION 12.4
Minimization of Circuits
The two methods presented here for minimizing circuits are really the same, just looked at from two different
points of view-one geometric (K-maps) and one algebraic (the Quine-McCluskey method). In each case the
idea is to get larger blocks, which represent simpler terms, to cover several minterms. The calculations can
get very messy, but if you follow the examples and organize your work carefully, you should not have trouble
with them. The hard part in these algorithms from a theoretical point of view is finding the set of products
that cover all the minterms. In these small examples, this tends not to be a problem.
446
Chapter 12
Boolean Algebra
1. a) The K-map we draw here has the variable x down the side and variable y across the top.
:rn
y
g
b) The upper left-hand corner cell (whose minterm is x y) and the lower right-hand corner cell (whose minterm
is xy) are adjacent to this cell.
3. The 2 x 2 square is used in each case. We put a 1 in those cells whose minterms are listed.
y
g
:EB
(a)
y
g
xGIJ
x[]ij
(b)
y
g
:rn
x~
(c)
5. a) We can draw a K-map for three variables in the manner shown here, with the x variable down the side and
the y and z variables across the top, in the order shown. We have placed a 1 in the requested position.
b) There are three cells adjacent to every cell (since there are three variables). The minterms of the adjacent
cells can be read off the picture:
x y z, x y z, and x y z.
7. The 2 x 4 square is used in each case. We put a 1 in those cells whose minterms are listed.
yz yz gz gz
(a)
:EEEE
(b)
xG-f.m
x~
(c)
9. In the figure below we have drawn the K-map. For example, since one of the terms was xz, we put a 1 in
each cell whose address contained x and z. Note that this meant two cells, one for y and one for y. Each cell
with a 1 in it is an implicant, as are the pairs of cells that form blocks, namely xy, xz, and yz. Since each
cell by itself is contained in a block with two cells, none of them is prime. Each of the mentioned blocks with
two cells is prime, since none is contained in a larger block. Furthermore, each of these blocks is essential,
since each contains a cell that no other prime implicant contains: xy contains xyz, xz contains x yz, and
yz contains x y z.
yz yz gz gz
x~
xrn=IJ
11. The figure below shows the 4-cube Q4 , labeled as requested. Compare with Figure 1 in Section 12.1. A
complemented Boolean variable corresponds to 0, and an uncomplemented Boolean variable corresponds to 1.
For example, the lower right corner of the sub-3-cube on the left corresponds to 0001.
Section 12.4
Minimization of Circuits
wxyz
447
wxyz
The 3-cube on the right corresponds to w, since all of its vertices are labeled w. Similarly, the 3-cube given
by the top surface of the whole figure represents x; the 3-cube given by the back surface of the whole figure
represents y; and the 3-cube given by the right surfaces of both the left and the right 3-cube represents z. We
show the last of these with heavy lines below. The "opposite 3-face" in each case represents the complemented
literal.
wxyz
wxyz
The 2-cube (i.e., square) that represents wz, in the same way, is the set of vertices that have w and z as
part of their labels, rather than w and/ or z. This is the right face of the 3-cube on the right. Similarly, the
2-cube that represents xy is bottom rear, and the 2-cube that represents yz is front left.
13. a) We can draw a K-map for four variables in the manner shown here, with the w and x variables down
the side and the y and z variables across the top, in the order shown. We have placed a 1 in the requested
position.
yz yz gz gz
wx
wx
Wx
Wx
1
b) There are four cells adjacent to every cell (since there are four variables). The minterms of the adjacent
cells can be read off the picture:
bottom row) wxyz.
wxyz, wxyz, wxyz, and (recalling that the top row is adjacent to the
15. A K-map for five variables needs 32 cells in all. We arrange them as shown below, following the discussion in
Example 6. Reread that example to understand what rows and columns have to be considered adjacent. (We
use rectangles rather than squares to save vertical space.)
a) We want to put a 1 in all cells the correspond to x 1 , x 2 , x 3 , and x 4 all being uncomplemented. Looking
at the diagram, we see that we need l's in precisely the two cells shown.
448
Chapter 12
x1Xz
x,x2
x1x2
x,x2
~~~
~~~
1
1
~~~
~~~
~~~
~~~
~~~
Boolean Algebra
~~~
1--~--1~~--1-~~--1-~~-+~~-+-~~-+-~~-+-~~~
1--~--1~~-+~~-+~~-t-~~-t-~~-+-~~-+-~~~
'--~----'~~---'-~~-'-~~--'-~~-'-~~_,_~~-'-~~~
b) We want to put a 1 in all cells the correspond to
x1 being complemented and X3 and xs both being
uncomplemented. Looking at the diagram, we see that we need l's in precisely the four cells shown. Note that
these are all adjacent, even though they don't look adjacent, since the first and fourth columns are considered
adjacent.
x1x2
x,x2
x1x2
1
1
x,x2
1
1
c) We want to put a 1 in all cells the correspond to x 2 and x 4 both being uncomplemented. Looking at the
diagram, we see that we need l's in precisely the eight cells shown.
x1x2
~~~
1
~~~
1
1
1
~~~
~~~
~~~
~~~
~~~
1
~~~
1
1
1
x,x2
x1x2
x,x2
d) We want to put a 1 in all cells the correspond to x 3 and x 4 both being complemented. Looking at the
diagram, we see that we need l's in precisely the eight cells shown.
~~~
~~~
~~~
~~~
x1x2
x,x2
x,xz
x,x2
~~~
1
~~~
1
1
1
1
1
1
1
~~~
~~~
e) We want to put a 1 in all cells the correspond to x 3 being uncomplemented. Looking at the diagram, we
see that we need l's in precisely the 16 cells shown.
~~~
~~~
~~~
~~~
x1x2
1
1
1
1
x,x2
1
1
1
1
x1x2
1
1
1
1
x,xz
1
1
1
1
~~~
~~~
~~~
~~~
f) We want to put a 1 in all cells the correspond to x 5 being complemented. Looking at the diagram, we see
that we need l's in precisely the 16 cells shown.
Section 12.4
Minimization of Circuits
449
X1X2
1
1
1
1
X1X2
1
1
1
1
x1x2
1
1
1
1
X1X2
1
1
1
1
17. a) There are clearly 2n cells in a K-map for n variables, since in specifying a minterm each variable can
appear either complemented or uncomplemented. Thus the answer is 26 = 64.
b} There are n cells adjacent to each cell in the K-map for n variables, because each of the variables can be
changed (from complemented to uncomplemented or vice versa) to produce an adjacent cell. Thus the answer
is 6.
19. Clearly we need to consider the first and fourth rows adjacent. The columns are more complicated, since each
column needs to be adjacent to three others (those that differ from it in one bit). Thus we must declare the
following columns adjacent (using the obvious notation): 1-4, 1-12, 1-16, 2-11, 2-15, 3-6, 3-10, 4-9, 5-8,
5-16, 6-15, 7-10, 7-14, 8-13, 9-12, 11-14, and 13-16. The complexity of this makes it virtually impossible
for a human to use this visual aid.
21. We could ignore the stated condition, but then our circuit would be more complex than need be. If we use the
condition, then there are really only three inputs; let us call them x, y, and z, where z represents Marcus's
vote and x and y represent the votes of the unnamed people. Since Smith and Jones always vote against
Marcus, a majority will occur if either Marcus assents with both of the other two (i.e., the minterm x y z ), or
else if Marcus votes no but at least one of the other two vote yes (which we can represent by (x + y)z). Thus
we design our circuit to be the OR of these two expressions.
23. We organize our work as in the text.
a)
Term
1
xyz
2
x'[jz
String
011
001
Step 1
Term
String
(1,2)xz
0-1
In this case we have one product that covers both of the minterms, so this product (x z) is our answer.
b}
Term
1
2
3
4
xyz
xyz
xyz
xyz
String
111
110
011
010
Step 1
Term
String
11(1,2)xy
-11
(1, 3) y z
(2, 4) yz
-10
01(3,4)xy
Step 2
Term
(1,2,3,4)y
Again we have one product that covers all the minterms, so that is our answer ( y).
String
-1-
Chapter 12
450
c)
Term
xyz
xyz
xyz
xyz
xyz
1
2
3
4
5
Boolean Algebra
Step 1
Term
String
(1,4)xz
1-0
(2, 4) xy
10-01
(2,5)yz
0-1
(3, 5) x z
String
110
101
011
100
001
(Note that we reordered the minterms so that the number of l's decreased as we went down the list.) No
further combinations are possible at this point, so there are four products that must be used to cover our five
minterms, each product covering two minterms. Clearly, then, two of these are not enough, but it is easy to
find three whose sum covers all the minterms. One possible answer is to choose the first, third, and fourth of
the products, namely x z + y z + x z.
d)
Term
xyz
xyz
xyz
xyz
xyz
xyz
1
2
3
4
5
6
Step 1
Term
String
1-1
(1,2)xz
-11
(l,3)yz
(2,4)xy
1001(3,5)xy
(4,6)yz
-00
0-0
(5, 6) xz
String
111
101
011
100
010
000
(Note that we reordered the minterms so that the number of l's decreased as we went down the list.) No
further combinations are possible at this point, so there are six products that must be used to cover our six
minterms, each product covering two minterms. Clearly, then, two of these are not enough, but it is not hard
to find three whose sum covers all the minterms. One possible answer is to choose the first, fourth, and fifth
of the products, namely x z + x y + yz.
25. We follow the procedure and notation given in the text.
a)
Term
wxyz
wxyz
wxyz
wxyz
wxyz
1
2
3
4
5
Step 1
Term
(1, 2) w x z
(2,3)wxy
(2,5)wyz
String
1111
1101
1100
1010
1001
String
11-1
1101-01
The three products in the last column as well as minterm #4 are possible products in the desired expansion,
since they are not contained in any other product. We make a table of which products cover which of the
original minterms.
wxz
wxy
wyz
wxyz
1
2
3
x
x
x
x
x
4
5
x
x
Since only the first of our products covers minterm #1, it must be included. Similarly, the other three products
must be included since they are the only ones that cover minterms #3, #4, and #5. If we do include them
all, then of course all the minterms are covered. Therefore our answer is w x z + w x y + w y z + w x y z.
Section 12.4
Minimization of Circuits
451
b)
Term
wxyz
wx'i}z
wxyz
wxyz
wxyz
wxyz
1
2
3
4
5
6
Step 1
Tenn
String
(2,4) x'i}z -101
(4,6)wyz 0-01
String
1110
1101
1011
0101
0010
0001
Since minterms #1, #3, and #5 are not contained in any others, these, along with the two products in the
last column, are the products that we look at to cover the original minterms. It is not hard to see that all five
are needed to cover all the original minterms. Therefore the answer is x'i}z+wyz+wxyz+wxyz+wxyz.
c)
1
2
3
4
5
6
7
8
Term
String
wxyz
wxyz
wx'i}z
wxyz
wxyz
wxyz
wxyz
wxyz
1111
Step 1
Term
String
(1, 2) wxy 111- {=
(1,3)wxz
(3,4)wyz
(3,5)xyz
(4, 6) wxy
(4,8)xyz
(5,8)wyz
1110
1101
1001
0101
1000
0010
0001
11-1
1-01
-101
100-
Step 2
Term
(3, 4, 5, 8) y z
String
--01
{=
{=
-001
0-01
The product in the last column, as well as the products in Step 1 that are marked with an arrow, as well
as minterm #7, are possible products in the desired expansion, since they are not contained in any other
product. We make a table of which products cover which of the original minterms.
'i}z
wxy
wxz
wxy
wxyz
1
2
x
x
x
3
4
5
x
x
x
6
8
7
x
x
x
x
x
In order to cover minterms #5, #2, #6, and #7, we need the first, second, fourth, and fifth of the products
in this table. If we include these four, then all the minterms are covered, and we do not need the third one.
Therefore our answer is 'iJ z + w x y + w x 'iJ + w x y z.
d)
Term
1
2
3
4
5
6
7
8
9
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
String
1111
1110
1101
1011
0111
1010
0011
0010
0001
Step 1
Term
String
(1, 2) wxy 111(1,3)wxz 11-1 {=
(1, 4) w y z 1-11
(1,5)xyz -111
(2,6)wyz
(4, 6) wxy
(4, 7)xyz
(5,7)wyz
(6,8)xyz
(7,8)wxy
(7, 9) wx z
1-10
101-011
0-11
-010
00100-1
{=
Step 2
Term
(1,2,4,6)wy
(1,4,5,7)yz
String
1-1--11
(4,6, 7,8)xy
-01-
452
Chapter 12
Boolean Algebra
The products in the last column, as well as the products in Step 1 that are marked with an arrow, are possible
products in the desired expansion, since they are not contained in any other product. We make a table of
which products cover which of the original minterms.
wy
yz
xy
wxz
wxz
1
2
x
x
x
3
5
4
x
x
6
x
8
9
x
x
x
x
7
x
x
x
x
x
x
In order to cover minterms #2, #5, #8, #3, and #9, we need the first, second, third, fourth, and fifth of the
products in this table, respectively-i.e., all of them. Therefore our answer is w y + y z + x y + w x z + wx z.
27. Using the method of Exercise 26, we draw the following picture, putting a 0 in each cell that represents a
maxterm in our product-of-sums expansion.
We then combine them into larger blocks, as shown, obtaining two large blocks: the entire first row, which
represents the factor x, and the entire first column, which represents the factor (y + z). Since neither of these
blocks alone covers all the maxterms, we need to use both. Therefore the simplified product is x(y + z).
29. We need to review Example 8 and note that the various positions in the 4 x 4 square correspond to the
various decimal digits. For example, the digit 6 corresponds to the box labeled w x y z. There are "don't
care" positions as well, for 4-digit binary numerals that exceed 9. (We do not care what output results for
such inputs.) Using the techniques of this section, we obtain the following maximal blocks.
yz yz gz gz
wx
d '-d
wx 'd
Lo.:;.;
wx
wx
d (d
d
l1
CD
\.1
1
Each one corresponds to a minterm, and our minimal sum-of-products expansion is therefore w z + x y z +
xyz + wx'[jz. Since we were asked for a circuit, we turn the products into AND gates and the sum into one
big OR gate.
Review Questions
453
31. We can cover all the l's with two large blocks here, one consisting of the middle four cells, and the other
consisting of the four corners (it is a block because of the wrap-around nature of the figure). In doing so, we
happened to have covered all the d's as well, but that is irrelevant. The point is that we obtained the largest
possible blocks in this manner, and if we had chosen not to cover the d's, then the blocks would have been
smaller. It is clear from this covering that the minimal sum-of-products expansion is just
xz + x z.
33. A formal proof here proceeds by induction on n. If n = 1, then we are looking at the 1-cube, which is a line
segment, labeled 0 at one end and 1 at the other end. The only possible value of k is also 1, and if the literal
is x 1 , then the subcube we have is the 0-dimensional subcube consisting of the endpoint labeled 1, and if
the literal is x1 , then the subcube we have is the 0-dimensional subcube consisting of the endpoint labeled 0.
Now assume that the statement is true for n; we must show that it is true for n + 1. If the literal Xn+i (or
its complement) is not part of the product, then by the inductive hypothesis, the product when viewed in
the setting of n variables corresponds to an (n - k )-dimensional subcube of the n-dimensional cube, and the
Cartesian product of that subcube with the line segment [O, 1] gives us a subcube one dimension higher in
our given (n + 1)-dimensional cube, namely having dimension (n + 1) - k, as desired. On the other hand, if
the literal Xn+l (or its complement) is part of the product, then the product of the remaining k - 1 literals
corresponds to a subcube of dimension n - (k - 1) = (n + 1) - k in the n-dimensional cube, and that slice,
at either the 1-end or the 0-end in the last variable, is the desired subcube.
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 12
1. Seep. 812.
2. 16 (see Table 3 in Section 12.1)
3. Seep. 813.
4. a) See p. 816.
b) See p. 816.
5. See pp. 819-820.
6. a) See p. 821.
b) This set is not functionally complete; see Exercise 19 in Section 12.2.
c) yes, such as {!} (see Exercise 16 in Section 12.2)
7. See Example 3 in Section 12.3.
8. See Figure 8 in Section 12.3.
9. yes-NAND gates or NOR gates
10. a) See pp. 830-833.
b) The following figure shows that the simplification is
x
y + y z + y z.
yz yz gz gz
:~
xlli!Uill_J
11. a) See pp. 833-834.
b) The following figure shows that the simplification is w x + y z + w z + w x
yz yz gz gz
wx
T
wx -1
Wx J_
Wx
-!.
1
1 (1
l!_
LD
y.
454
Chapter 12
Boolean Algebra
12. a) See p. 836.
b) The following figure shows that the simplification is w
+ x y.
The resulting circuit needs just one AND
gate, one OR gate, and no inverters.
yz yz gz gz
wx
~
wx
d
~ d
di
d
1)
1
Wx
Wx
r--Ii'
13. a) See pp. 837-841.
b)
1
2
3
4
5
Term
String
xyz
xyz
xyz
xyz
xyz
110
010
100
001
000
Step 1
Term
String
(1,2)yz
-10
(1, 3) xz
1-0
0-0
(2, 5) xz
(3, 5) yz
-00
00(4,5)xy
Step 2
Term
(1, 2, 3, 5) z
String
--0
The product z in the last column covers all the minterms except #4, and the fifth product in Step 1 (xy)
covers it. Thus the answer is z + xy.
SUPPLEMENTARY EXERCISES FOR CHAPTER 12
1. a) Suppose that the equation holds. If x = 1, then the left-hand side is 1; therefore the right-hand side
must be 1, as well, and that forces y and z to be 1. Similarly, if x = 0, then equality holds if and only if
y=z=O. Thereforetheonlysolutionsare (1,1,1) and (0,0,0).
b) If x = 1, then the right-hand side equals 1, so in order for there to be equality we need y + z = 1, whence
y = 1 or z = 1. Thus (1, 1, 0), (1, 0, 1), and (1, 1, 1) are all solutions. If x = 0, then the left-hand side is 0,
so we need yz = 0 in order for equality to hold, whence y = 0 or z = 0. Thus (0,0,0), (0,0, 1), and (0, 1,0)
are also solutions.
c) There are no solutions here. If any of the variables equals 1, then the left-hand side is 0 and the right-hand
side is 1; if all the variables are 0, then the left-hand side is 1 and the right-hand side is 0.
3. In each case we form F(x 1 , . .. , Xn) and simplify. If we get back to what we originally started with, then
the function is self-dual; if what we obtain is not equivalent to what we began with, then the function is not
self-dual. The simplification is done using the identities for Boolean algebra, especially De Morgan's laws.
a)
x=
x, so the function is self-dual.
b) (xy+xy) = (xy+xy), which is the complement of what we originally had. Thus this is as far from
being self-dual as it can possibly be.
c) (x + y) =
xy =
x y, which is certainly not equivalent to x
+ y.
Therefore this is not self-dual.
d) We first simplify the expression, using the distributive law and the fact that x
function as F(x,y) = y. Now, as in part (a), we see that it is indeed self-dual.
+x =
1 to rewrite our
5. The reasoning here is essentially the same as in Exercise 31 in Section 12.1. To specify all the values of a
self-dual function, we are free to specify the values of F(l, x 2 , x 3 , ... , Xn), and we can do this in 22 n - i ways,
since there are 2n-l different elements at which we can choose to make the function value either 0 or 1.
Once we have specified these values, the values of F(O, x 2 , x 3 , ... , Xn) are all determined by the definition of
self-duality, so no further choices are possible. Therefore the arniwer is 22 n - i .
Supplementary Exercises
455
7. a) At every point in the domain, it is certainly the case that if F(x 1, ... , Xn) = 1, then (F+G)(x 1 , ... , Xn) =
F(xi, ... , Xn) + G(x1, ... , Xn) = 1 + G(x1, ... , Xn) = 1, no matter what value G has at that point. Thus by
definition F ::; F + G.
b) This is dual to the first part. At every point in the domain, it is certainly the case that if F(x 1 , ... , Xn) = 0,
then (FG)(xi, .. . , Xn) = F(x1, . .. , Xn)G(x1, ... , Xn) = 0 · G(x1, ... , Xn) = 0, no matter what value G has
at that point. The contrapositive of this statement is that if (FG)(x 1, ... ,xn) = 1, then F(x 1, ... ,xn) = 1.
Thus by definition FG ::; F.
9. We need to show that this relation is reflexive, antisymmetric, and transitive. That F ::; F (reflexivity) is
simply the tautology "if F(x1, ... , Xn) = 1, then F(x 1, ... , Xn) = l." For antisymmetry, suppose that F::; G
and G::; F. Then the definition of the relation says that F(x 1, ... , xn) = 1 if and only if G(x 1 , ... , Xn) = 1,
which is the definition of equality between functions, so F = G. Finally, for transitivity, suppose that F ::; G
and G::; H. We want to show that F::; H. So suppose that F(x 1 , ... , Xn) = 1. Then by the first inequality
G(x1, ... , Xn) = 1, whence by the second inequality H(x 1, ... , Xn) = 1, as desired.
= 1, y = z = 0 works for all three.
a) We have 1 I (0 I 0) = 1 I 1 = 0, whereas (11 0) I 0 = 1 I 0 = 1.
b) We have 1 l (0 l 0) = 1 l l = 0, whereas (1 l 0) l (1 l 0) = 0 l 0 = 1.
c) We have 1 l (0 I 0) = 1 l 1 = 0, whereas (1 l 0) I (1 l 0) = 0 I 0 = 1.
11. None of these are identities. The counterexample x
13. This is clear from the definitions. The given operation applied to x and y is defined to be 1 if and only if
x = y, while XOR applied to x and y is defined (preamble to Exercise 24 in Section 12.1) to be 1 if and only
if x # y.
15. We show this to be true with a table. Since the fifth and seventh columns are equal, the equation is an
identity.
x
y
z
x8y
(x8y)8z
y8z
x 8 (y 8 z)
1
1
1
1
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
1
1
0
0
1
0
1
1
0
1
0
1
0
0
0
1
1
0
0
1
1
0
1
1
0
0
0
0
0
0
1
1
17. In each case we can actually list all the functions.
a) The only function values we can get are x, y, 0, 1,
not give us anything new. Therefore the answer is 6.
b) Since s · s = s , s · 1 = s , and s · 0
Therefore the answer is 5 .
x, and y, since applying complementation twice does
= 0 for all s , the only functions we can get are x ,
y , 0 , 1 , and x y .
c) By duality the answer here has to be the same as the answer to part (b), namely 5.
d) We can get the 6 distinct functions x, y, 0, 1 , x y and x + y. Any further applications of these operations,
however, returns us to one of these functions. For example, x y + x = x.
19. The sum bit is the exclusive OR of the inputs, and the carry bit is their product. Therefore we need only two
gates to form the half adder if we allow an XOR gate and an AND gate.
Chapter 12
456
Boolean Algebra
21. We need to figure out which combinations of values for x 1 , x 2 , and x 3 cause the inequality -x 1 +x 2 +2x 3 2: 1/2
to be satisfied. Clearly this will be true if x 3 = l. If x 3 = 0, then it will be true if and only if x 2 = 1 and
x 1 = 0. Thus a Boolean expression for this function is x 3 + x 1 x 2 .
23. We prove this by contradiction. Suppose that a, b, and T are such that ax+ by 2: T if and only if x EB y = 1,
i.e., if and only if either x = 1 and y = 0, or else x = 0 and y = 1. Thus for the first case we need a 2: T,
and for the second we need b 2: T. Since we need ax + by < T for x = y = 0, we know that T
particular b is positive. Therefore we have a+ b > a 2: T, which contradicts the fact that 1EB1
a+ b:::; T).
> 0. Hence in
= 0 (requiring
WRITING PROJECTS FOR CHAPTER 12
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. Martin Gardner is probably the best writer of mathematical material for the general public. For this project,
look at [Gal].
2. Try a good book on circuits, such as [Ch].
3. Try a good book on circuits, such as [Ch].
4. Try a good book on circuits, such as [Ch].
5. Try a good book on circuits, such as [Ch].
6. Try a good book on circuits, such as [Ch].
7. See Section 9.5 of [HiPel].
8. Try a good book on switching theory, such as [Kol].
9. Try a good book on switching theory, such as [Kol]. Also see [HiPel].
10. Try a good book on switching theory, such as [Kol].
11. The classic paper on this algorithm is [RuSa], but good books on circuits should mention it (the second author
of that paper has written some-check his Web page).
12. Try a good book on switching theory, such as [Kol].
Section 13.1
Languages and Grammars
457
CHAPTER 13
Modeling Computation
SECTION 13.1
Languages and Grammars
There is no magical way to come up with the grammars to generate a language described in English. In
particular, Exercises 15 and 16 are challenging and very worthwhile. Exercise 21 shows how grammars can
be combined. In constructing grammars, we observe the rule that every production must contain at least one
nonterminal symbol on the left. This allows us to know when a derivation is completed-namely, when the
string we have generated contains no nonterminal symbols.
1. The following sequences of lines show that each is a valid sentence.
a) sentence
noun phrase intransitive verb phrase
article adjective noun intransitive verb phrase
article adjective noun intransitive verb
the adjective noun intransitive verb
the happy noun intransitive verb
the happy hare intransitive verb
the happy hare runs
b) sentence
noun phrase intransitive verb phrase
article adjective noun intransitive verb phrase
article adjective noun intransitive verb adverb
the adjective noun intransitive verb adverb
the sleepy noun intransitive verb adverb
the sleepy tortoise intransitive verb adverb
the sleepy tortoise runs adverb
the sleepy tortoise runs quickly
c) sentence
noun phrase transitive verb phrase noun phrase
article noun transitive verb phrase noun phrase
article noun transitive verb noun phrase
article noun transitive verb article noun
the noun transitive verb article noun
the tortoise transitive verb article noun
the tortoise passes article noun
the tortoise passes the noun
the tortoise passes the hare
d) sentence
noun phrase transitive verb phrase noun phrase
article adjective noun transitive verb phrase noun phrase
article adjective noun transitive verb noun phrase
458
Chapter 13
Modeling Computation
article adjective noun transitive verb article adjective noun
the adjective noun transitive verb article adjective noun
the sleepy noun transitive verb article adjective noun
the sleepy hare transitive verb article adjective noun
the sleepy hare passes article adjective noun
the sleepy hare passes the adjective noun
the sleepy hare passes the
happy noun
the sleepy hare passes the
happy tortoise
3. Since runs is only an intransitive verb, it can only occur in a sentence of the form noun phrase intransitive verb phrase. Such a sentence cannot have anything except an adverb after the intransitive verb,
and the sleepy tortoise cannot be an adverb.
5. a) It suffices to give a derivation of this string. We write the derivation in the obvious way. S ==> IA ==>
lOB ==> lOlA ==> lOlOB ==> 10101.
b) This follows from our solution to part ( c), because 10110 has two 1's in a row and is not of the form
discussed there.
c) Notice that the only production with A on the left is A---+ OB. Furthermore, the only productions with
B on the left are B ---+ IA and B ---+ 1 . Combining these, we see that we can eliminate B and replace
these three rules by A ---+ OIA and A ---+ 01. This tells us that every string in the language generated by G
must end with some number of repetitions of 01 (at least one). Furthermore, because of the rules S---+ OA
and S ---+ IA, the string must start with either a 0 or a 1 preceding the repetitions of 01. Therefore the
strings in this language consist of a 0 or a 1 followed by one or more repetitions of 01. We can write this as
{ O(Ol)n In 2': 0} U { l(Ol)n In 2': 0}
7. We write the derivation in the obvious way. S ==> OSI ==> OOSll ==> OOOSlll ==> 000111. We used the rule
S ---+ OSI in the first three steps and S ---+ ,\ in the last step.
9. a) Using G 1 , we can add O's on the left or l's on the right of S. Thus we have S ==>OS==> OOS ==> OOSl ==>
OOSll ==> OOSlll ==> OOSllll ==> 001111.
b) In this grammar we must add all the O's first to S, then change to an A and add the 1's, again on the
left. Thus we have S ==>OS==> OOS ==> OOIA ==> 0011A ==> 00111A ==> 001111.
11. First we apply the first rule twice and the rule S---+ ,\to get OOABAB. We can then apply the rule BA---+ AB,
to get OOAABB. Now we can apply the rules OA ---+ 01 and IA ---+ 11 to get 0011BB; and then the rules
IB ---+ 12 and 2B ---+ 22 to end up with 001122, as desired.
13. In each case we will list only the productions, because V and T will be obvious from the context, and S
speaks for itself.
a) For this finite set of strings, we can simply have S
---+
0, S
---+
1, and S
---+
11 .
b) We assume that "only l's" includes the case of no l's. Thus we can take simply S---+ IS and S---+ >..
c) The middle can be anything we like, and we will let A represent the middle. Then our productions are
S---+ OAI, A-+ IA, A-+ OA, and A-+>..
d) We will let A represent the pairs of l's. Then our productions are S---+ OA, A---+ llA, and A---+>..
15. a) We need to add the O's two at a time. Thus we can take the rules S
---+
BOO and S
---+ ,\.
b) We can use the same first rule as in part (a), namely S---+ BOO, to increase the number of O's. Since the
string must begin 10, we simply adjoin to this the rule S ---+ 10.
Section 13.1
Languages and Grammars
459
c) We need to add O's and l's two at a time. Furthermore, we need to allow for O's and l's to change their
order. Since we cannot have a rule 01 __, 10 (there being no nonterminal symbol on the left), we make up
nonterminal analogs of 0 and 1, calling them A and B, respectively. Thus our rules are as follows: S __, AAS,
S __, BBS, AB __, BA, BA __, AB, S __, ,\, A __, 0, and B __, 1. (There are also totally different ways to
approach this problem, which are just as effective.)
d) This one is fairly simple: S __, OOOOOOOOOOA, A __, OA, A __, .A. This assures at least 10 O's and allows
for any number of additional O's.
e) We need to invoke the trick used in part (c) to allow O's and l's to change their order. Furthermore, since
we need at least one extra 0, we use S __, A as our vanishing condition, rather than S __, ,\. Our solution,
then, is S_,AS, S_,ABS, S_,A, AB_,BA, BA_,AB, A_,O,and B_,l.
f) This is identical to part ( e), except that the vanishing condition is S __, ,\, rather than S __, A, and there
is no rule S __, AS.
g) We just put together two copies of a solution to part (e), one in which there are more O's than l's, and
one in which there are more l's than O's. The rules are as follows: S __,ABS, S __, T, S __, U, T __,AT,
T __, A, U __, BU, U __, B, AB __, BA, BA __, AB, A __, 0, and B __, 1.
17. In each case we will list only the productions, because V and T will be obvious from the context, and S
speaks for itself.
a) It suffices to have S __,OS and S __,.A.
b) We let A represent the string of l's. Thus we take S __,AO, A__, IA, and A__, .A. Notice that A__, Al
works just as well A__, IA here, so either one is fine.
c) It suffices to have S __, OOOS and S __, ,\.
19. a) This is a type 2 grammar, because the left-hand side of each production has a single nonterminal symbol.
It is not a type 3 grammar, because the right-hand side of the productions are not of the required type.
b) This meets the definition of a type 3 grammar.
c) This is only a type 0 grammar; it does not fit the definition of type 1 because the right side of the second
production does not maintain the context set by the left side.
d) This is a type 2 grammar, because the left-hand side of each production has a single nonterminal symbol.
It is not a type 3 grammar, because the right-hand side of the productions are not of the required type.
e) This meets the definition of a type 2 grammar. It is not of type 3, because of the production A__, B.
f) This is only a type 0 grammar; it does not fit the definition of type 1 because the right side of the second
production does not maintain the context set by the left side.
g) This meets the definition of a type 3 grammar. Note, however, that it does not meet the definition of a
type 1 grammar because of S __, ,\.
h) This is only a type 0 grammar; it does not fit the definition of type 1 because the right side of the third
production does not maintain the context set by the left side.
i) This is a type 2 grammar because each left-hand side is a single nonterminal. It is not type 3 because of
the production B __, ,\.
j) This is a type 2 grammar because each left-hand side is a single nonterminal. It is not type 3; each of the
productions violates the conditions imposed for a type 3 grammar.
21. Let us assume that the nonterminal symbols of G 1 and G 2 are disjoint. (If they are not, we can give those
in G 2 , say, new names so that they will be; obviously this does not change the language that G 2 generates.)
Call the start symbols S1 and S2. In each case we will define G by taking all the symbols and rules for G1
and G2, a new symbol S, which will be the start symbol for G, and the rules listed below.
a) Since we want strings that either G 1 or G 2 generate, we add the rules S __, S 1 and S __, S 2 .
Chapter 13
460
Modeling Computation
b) Since we want strings that consist of a string that G 1 generates followed by a string that G 2 generates,
we add the rule S----; S1S2.
c) This time we add the rules S ----; S1S and S ----; >.. This clearly gives us all strings that consist of the
concatenation of any number of strings that G 1 generates.
23. We simply translate the derivations we gave in the solution to Exercise 1 to tree form, obtaining the following
pictures.
--------------------sentence
noun phrase
intransitive verb phrase
/I----article adjective noun
I
I
the
I
intrunsitive verb
I
happy
I
hare
runs
(a)
sentence
noun phrase
intransitive verb phrase
~~
adverb
/I-----
article adjective noun
I
I
the
sleepy
intransitive verb
I
I
tortoise
I
runs
quickly
(b)
-------1-------sentence
noun phruse
/~
truns. verb. phruse
I
noun phrase
/~
art.
noun
trans. verb
art.
I
I
I
I
the
tortoise
passes
(c)
noun
I
the
hare
-------1-------sentence
noun phrase
/I~
art. adj. noun
I
I
I
the sleepy hare
trans. verb. phrase
noun phrase
/I~
I
trans. verb
art. adj. noun
I
I
passes
I
I
the happy tortoise
(d)
25. We can assume that the derivation starts S :::;. AB :::;. CaB :::;. cbaB, or S :::;. AB :::;. CaB :::;. baB. This shows
that neither the string in part (b) nor the string in part ( d) is in the language, since they do not begin cba
or ba. In order to derive the string in part (a), we need to turn B into ba, and this is easy, using the rule
B ----; Ba and then the rule B ----; b. Finally, for part ( c), we again simply apply these two rules to change B
into ba.
27. This is straightforward. The - is the sign and the 109 is an integer, so the tree starts as shown. Then we
decompose the integer 109 into the digit 1 and the integer 09, then in turn to the digit 0 and the integer
(digit) 9.
~~
I
---------
I~~
1
I
I
0
I
9
Section 13.l
Languages and Grammars
461
29. a) Note that a string such as "34." is not allowed by this definition, but a string such as -02. 780 is. This
is pretty straightforward using the following rules. As can be seen, we are using (integer) to stand for a
nonnegative integer.
S--+ (sign)(integer)
S --+ (sign) (integer). (positive integer)
(sign) --+ +
(sign) --+ (integer) --+ (integer) (digit)
(integer) --+ (digit)
(positive integer) --+ (integer) (nonzero digit) (integer)
(positive integer) --+ (integer) (nonzero digit)
(positive integer) --+ (nonzero digit) (integer)
(positive integer) --+ (nonzero digit)
(digit) --+ (nonzero digit)
(digit) --+ 0
(nonzero digit) --+ 1
(nonzero digit) --+ 2
(nonzero digit) --+ 3
(nonzero digit) --+ 4
(nonzero digit) --+ 5
(nonzero digit) --+ 6
(nonzero digit) --+ 7
(nonzero digit) --+ 8
(nonzero digit) --+ 9
b) We combine rows of the previous answer with the same left-hand side, and we change the notation to
produce the answer to this part.
(signed decimal number)::= (sign)(integer) I (sign)(integer).(positive integer)
(sign) ::= + I (integer) ::= (integer)(digit) (digit)
(positive integer) ::= (integer) (nonzero digit) (integer) (integer) (nonzero digit)
(nonzero digit) (integer) (nonzero digit)
(digit)::= (nonzero digit) I 0
(nonzero digit)::= 1 I 2 3 I 4 5 I 6 I 7 I 8 9
c) We easily produce the following tree.
J
J
J
J
J
J
J
I
/~
I
I
I
I
I
4
I
3
31. a) We can think of appending letters to the end at each stage:
(identifier) ::= ( lcletter) I (identifier) ( lcletter)
(lcletter) ::=a I b c I · · · I z
b) We need to be more explicit here than in part (a) about how many letters are used:
(identifier) ::= ( lcletter) ( lcletter) (lcletter) I (lcletter) ( lcletter) (lcletter) (lcletter) I
( lcletter) ( lcletter) (lcletter) ( lcletter) ( lcletter) I
J
Chapter 13
462
Modeling Computation
( lcletter) ( lcletter) ( lcletter) ( lcletter) ( lcletter) ( lcletter)
(lcletter) ::= a I b I c I · · · I z
c) This is similar to the part (b), allowing for two types of letters:
(identifier) ::= ( ucletter) I (ucletter) (letter) I (ucletter) (letter) (letter)
( ucletter) (letter) (letter) (letter)
I
I (ucletter) (letter) (letter) (letter) (letter) I
( ucletter) (letter) (letter) (letter) (letter) (letter)
(letter) ::= ( lcletter) I ( ucletter)
(lcletter) ::= a I b I c I · · · I z
(ucletter) ::=A
I BI CI··· I Z
d) This is again similar to previous parts. We need to invent a name for "digit or underscore."
(identifier) ::= ( lcletter) ( dzgitorus) (alphanumeric) (alphanumeric) (alphanumeric) I
( lcletter) ( digitorus) (alphanumeric) (alphanumeric) (alphanumeric) (alphanumeric)
(digitorus) ::=(digit) I (alphanumeric) ::= (letter) I (digit)
(letter) ::= (lcletter) I ( ucletter)
(lcletter) ::=a I b I c I · · · I z
(ucletter) ::=A I BI CI · · · I Z
(digit) ::= o I i I 2 I · · · I 9
33. We create a name for ''letter or underscore" and then define an identifier to consist of one of those, followed
by any number of other allowed symbols. Note that an underscore by itself is a valid identifier, and there is
no prohibition on consecutive underscores.
(identifier)::= (letterorus) I (identifier)(symbol)
(symbol) ::= (letterorus) I (digit)
(letterorus) ::= (letter) I _
(letter) ::= ( lcletter) I (ucletter)
( lcletter) ::= a I b I c I · · · I z
(ucletter) ::=A I BI CI ···I Z
(digit) ::= o 11 I 2 I · · · I 9
35. We assume that leading O's are not allowed in the whole number part, since the problem explicitly mentioned
them only for the decimal part. Our rules have to allow the optional sign using the question mark, the integer
part consisting of one or more digits, not beginning with a 0 unless 0 is the entire whole number part, and
then either the decimal part or not. Note that the decimal part has a decimal point followed by zero or more
digits.
numeral ::= sign? nonzerodigit digit* decimal? I sign? 0 decimal?
decimal::= .digit*
digit ::= 0 I nonzerodigit
sign::=+ I nonzerodigit ::= 1 2 I · · · I 9
J
37. We can simplify the answer given in Exercise 33 using the asterisk for repeating optional elements.
identifier ::= letterorus symbol*
symbol ::= letterorus I digit
letterorus ::= letter I _
letter ::= lcletter I ucletter
lcletter ::= a b c I · · · I z
ucletter ::=A I B IC I · · · I Z
digit ::= o 11 I 2 I · · · I 9
J
J
Section 13.1
Languages and Grammars
463
39. a) This string is generated by the grammar. The substring be* is a term, since it consists of the factor b
followed by the factor c followed by the mulOperator *. Thus the entire expression consists of two terms
followed by an addOperator. We can show the steps in the following sequence:
(expression)
(term) (term) (add Operator)
(!actor) (factor) (factor) ( mulOperator) (add Operator)
(identifier) (identifier) (identifier) ( mulOperator) (add Operator)
abc*+
b) This string is not generated by the grammar. The second plus sign needs two terms preceding it, and xy+
can only be deconstructed to be one term.
c) This string is generated by the grammar. The substring xy- is a factor, since it is an expression, namely
the term x followed by the term y followed by the add Operator - . Thus the entire expression consists of
two factors followed by a mulOperator. We can show the steps in the following sequence:
(expression)
(term)
(factor) (factor) ( mulOperator)
(expression) (!actor) ( m ulOperator)
(term) (term) (add Operator) (!actor) ( mulOperator)
(!act or) (!actor) (add Operator) (!actor) ( m ulOperator)
(identifier) (identifier) (add Operator) (identifier) ( mulOperator)
XY-Z*
d) This is similar to part (c). The entire expression consists of two factors followed by a mulOperator; the
first of these factors is just w, and the second is the term x y z - *. That term, in turn, deconstructs as in
previous parts. We can show the steps in the following sequence:
(expression)
(term)
(!actor) (!actor) ( m ulOperator)
(factor) (expression) ( mulOperator)
(!actor) (term) ( mulOperator)
(factor) (factor) (factor) ( mulOperator) ( mulOperator)
(factor) (factor) (expression) ( mulOperator) ( mulOperator)
(factor) (factor) (term) (term) (add Operator) ( mulOperator) ( mulOperator)
(factor) (factor) (factor) (factor) (add Operator) ( mulOperator) ( mulOperator)
(identifier) (identifier) (identifier) (identifier) (add Operator) ( mulOperator) ( mulOperator)
wxyz-*/
e) This string is generated as follows (similar to previous parts of this exercise):
(expression)
(term)
(factor) (factor) ( mulOperator)
(!actor) (expression) ( m ulOperator)
(factor) (term) (term) ( addOperator) ( mulOperator)
(factor) (factor) (factor) ( addOperator) ( mulOperator)
(identifier) (identifier) (identifier) (add Operator) ( mulOperator)
ade-*
41. The answers will depend on the grammar given as the solution to Exercise 40. We assume here that the answer
to that exercise is very similar to the preamble to Exercise 39. The only difference is that the operators are
Chapter 13
464
Modeling Computation
placed between their operands, rather than behind them, and parentheses are required in expressions used as
factors.
a) This string is not generated by the grammar, because the addition operator can only be applied to two
terms, and terms that are themselves expressions must be surrounded by parentheses.
b) This string is generated by the grammar. The substrings a/band c/d are terms, so they can be combined
to form the expression. We show the steps in the following sequence:
(expression)
(term) (add Operator) (term)
(factor) ( mulOperator) (!actor) (add Operator) (!actor) ( mulOperator) (factor)
(identifier) ( m ulOperator) (identifier) (add Operator) (identifier) ( m ulOperator) (identifier)
a/b + c/d
c) This string is generated by the grammar. The substring (n + p) is a factor, since it is an expression
surrounded by parentheses. We show the steps in the following sequence:
(expression)
(term)
(factor) ( mulOperator) (factor)
(factor) ( mulOperator) ( (expression))
(!actor) ( m ulOperator) ( (term) (add Operator) (term))
(!actor) ( mulOperator) ((factor) (add Operator) (!actor))
(identifier) ( mulOperator) ( (identifier) ( addOperator) (identifier))
m*(n+p)
d) There are several reasons that this string is not generated, among them the fact that it is impossible for
an expression to start with an operator in this grammar.
e) This is very similar to part ( c):
(expression)
(term)
(factor) ( mulOperator) (factor)
( (expression)) ( mulOperator) ( (expression))
((term) (addOperator) (term) )(mulOperator)( (term) (addOperator) (term))
((factor) (add Operator) (!actor)) ( mulOperator) ((factor) (add Operator) (!actor))
( (identifier) (add Operator) (identifier)) ( mulOperator) ( (identifier) (add Operator) (identifier))
(m + n)
* (p -
SECTION 13.2
q)
Finite-State Machines with Output
Finding finite-state machines to do specific tasks is in essence computer programming. There is no set method
for doing this. You have to think about the problem for awhile, ask yourself what it might be useful for
the states to represent, and then very carefully proceed to construct the machine. Expect to have several
false starts. "Bugs" in your machines are also very common. There are of course many machines that will
accomplish the same task. The reader should look at Exercises 20~25 to see that it is also possible to build
finite-state machines with the output associated with the states, rather than the transitions.
Section 13.2
Finite-State Machines with Output
465
1. We draw the state diagrams by making a node for each state and a labeled arrow for each transition. In
part (a), for example, since under input 1 from state s2 we are told that we move to state s 1 and output
a 0, we draw an arrow from s2 to s1 and label it 1, 0. It is assumed that s 0 is always the start state.
01,1
~@
0,0
0,0
@
1,1
0,0
1,0
(1.1)
0°)
~
(;;\
start
(;;\
1,1
~---1~,1--@
'·'l ~I'·
''
6)
>ts;\
~
~
l,O
1,1
(c)
3. a) The machine starts in state s 0 . Since the first input symbol is 0, the machine moves to state s 1 and gives
0 as output. The next input symbol is 1 , so the machine moves to state s 2 and gives 1 as output. The next
input is 1, so the machine moves to state s 1 and gives 0 as output. The fourth input is 1, so the machine
moves to state s 2 and gives 1 as output. The fifth input is 0, so the machine moves to state s 1 and gives 0
as output. Thus the output is 01010.
b) The machine starts in state so. Since the first input symbol is 0, the machine moves to state s 1 and gives
0 as output. The next input symbol is 1, so the machine moves to state so and gives 1 as output. The next
input is 1, so the machine stays in state s 0 and gives 0 as output. The fourth input is 1, so the machine
again stays in state so and gives 0 as output. The fifth input is 0, so the machine moves to state s 1 and
gives 0 as output. Thus the output is 01000.
c) The machine starts in state so. Since the first input symbol is 0, the machine stays in state s0 and gives
1 as output. The next input symbol is 1, so the machine moves to state s 4 and gives 1 as output. The next
input is 1 , so the machine moves to state s 0 and gives 0 as output. The fourth input is 1, so the machine
moves to state s 4 and gives 1 as output. The fifth input is 0, so the machine moves to state s 1 and gives 1
as output. Thus the output is llOll.
5. a) The machine starts in state so. Since the first input symbol is 0, the machine moves to state s 1 and gives
1 as output. (This is what the arrow from so to s 1 with label 0, 1 means.) The next input symbol is 1.
Because of the edge from s 1 to s 0 , the machine moves to state so and gives 1 as output. The next input
is 1 . Because of the loop at s 0 , the machine stays in state s 0 and gives output 0. The same thing happens
on the fourth input symbol. Therefore the output is 1100 (and the machine ends up in state s 0 ).
b) This is similar to part (a). The first two symbols of input cause the machine to output two O's and remain
in state s 0 . The third symbol causes an output of 1 as the machine moves into state s 1 . The fourth input
466
Chapter 13
Modeling Computation
takes us back to state so with output 1. The next four symbols of input cause the machine to give output
0110 as it goes to states s0 , s 1 , s 0 , and s0 , respectively. Therefore the output is 00110110.
c) This is similar to the other parts. The machine alternates between states s 0 and s 1 , outputting 1 for each
input. Thus the output is 11111111111.
7. We model this machine as follows. There are four possible inputs, which we denote by 5, 10, 25, and b,
standing for a nickel, a dime, a quarter, and a button labeled by a kind of soda pop, respectively. (Actually
the model is a bit more complicated, since there are three kinds of pop, but we will ignore that; to incorporate
the kind of pop into the model, we would simply have three inputs in place of just b.) The output can either
be an amount of money in cents~O, 5, 10, 15, 20, or 25--or can be a can of soda pop, which we denote c.
There will be eight states. Intuitively, state s, will represent the state in which the machine is indebted to the
customer by 5i cents. Thus s0 , the start state, will represent that the machine owes the customer nothing;
state s1 will represent that the machine has accepted 5 cents from the customer, and so on. State s 7 will
mean that the machine owes the customer 35 cents, which will be paid with a can of soda pop, at which time
the machine will return to state s 0 , owing nothing. The following picture is the state diagram of this machine,
simplified even further in that we have eliminated quarters entirely for sake of readability. For example, the
transition from state s 6 ( 30 cents credit) on input of a dime is to state s 7 ( 35 cents credit) with the return
of 5 cents in change. We have also used a to stand for any monetary input: if you deposit any amount when
the machine already has your 35 cents, then you get that same amount back. Thus the transition a, a really
stands for three transitions: 5, 5 and 10, 10 and 25, 25.
9. We draw the diagram for this machine. Intuitively, we need four states, corresponding to the four possibilities
for what the last two bits have been. In our picture, state s 1 corresponds to the last two bits having been
00; state s 2 corresponds to the last two bits having been 01; state s 3 corresponds to the last two bits having
been 10; state s 4 corresponds to the last two bits having been 11. We also need a state so to get started, to
account for the delay. Let us see why some of the transitions are what they are. If you are in state s3, then
the last two bits have been 10. If you now receive an input 0, then the last two bits will be 00, so we need
to move to state s 1 . Furthermore, since the bit received two pulses ago was a 1 (we know this from the fact
that we are in state s 3 ), we need to output a 1. Also, since we are told to output 00 at the beginning, it is
right to have transitions from so as shown.
If we look at this machine, we observe that states s 0 and s 1 are equivalent, i.e., they cause exactly the same
transitions and outputs. Therefore a simpler answer would be a machine like this one, but without state s 0 ,
where state s1 is the start state.
Section 13.2
467
Finite-State Machines with Output
11. This machine is really only part of a machine; we are not told what happens after a successful log-on. Also,
the machine is really much more complicated than we are indicating here, because we really need a separate
state for each user. We assume that there is only one user. We also assume that an invalid user ID is rejected
immediately, without a request for a password. (The alternate assumption is also reasonable, that the machine
requests a password whether or not the ID is valid. In that case we obtain a different machine, of course.)
We need only two states. The initial state waits for the valid user ID. We let i be the valid user ID, and we
let j be any other input. If the input is valid, then we enter state s 1 , outputting the message e: "enter your
password." If the input is not valid, then we remain in state so, outputting the message t: "invalid ID; try
again." From state s 1 there are only two relevant inputs: the valid password p and any other input q. If the
input is valid, then we output the message w: "welcome" and proceed. If the input is invalid, then we output
the message a: "invalid password; enter user ID again" and return to state s 0 to await another attempt at
logging-on.
13. This exercise is similar to Exercise 7. We let state s, for i = 0, 1, 2, 3, 4 represent the fact that 5i cents has
been deposited. When at least 25 cents has been deposited, we return to state s 0 and open the gate. Nickels
(input 5), dimes (input 10) and quarters (input 25) are available. We let o and c be the outputs: the gate
is opened (for a limited time, of course), or remains closed. After the gate is opened, we return to state s 0 .
(We assume that the gate closes after the car has passed.)
5,o
1O,o
25,o
15. The picture for this machine would be too complex to draw. Instead, we will describe the machine verbally,
and even then we won't give every last gory detail. We assume that possible inputs are the ten digits. We will
let s 0 be the start state and let s 1 be the state representing a successful call (so we will not list any outputs
from s 1 ). From s 0 , inputs of 2, 3, 4, 5, 6, 7, or 8 send the machine back to s 0 with output of an error message
for the user. From so an input of 0 sends the machine to state s 1 , with the output being that the 0 is sent
to the network. From so an input of 9 sends the machine to state s 2 with no output; from there an input
of 1 sends the machine to state s3 with no output; from there an input of 1 sends the machine to state s 1
with the output being that the 911 is sent to the network. All other inputs while in states s 2 or s 3 send the
machine back to s 0 with output of an error message for the user. From s 0 an input of 1 sends the machine
to state s 4 with no output; from s 4 an input of 2 sends the machine to state s 5 with no output; and this
path continues in a similar manner to the 911 path, looking next for 1, then 2, then any seven digits, at which
point the machine goes to state s 1 with the output being that the ten-digit input is sent to the network. Any
"incorrect" input while in states s 5 or s 6 (that is, anything except a 1 while in s 5 or a 2 while in s 6 ) sends
the machine back to s 0 with output of an error message for the user. Similarly, from s 4 an input of 8 followed
by appropriate successors drives us eventually to s 1 , but inappropriate outputs drive us back to s 0 with an
error message. Also, inputs while in state s 4 other than 2 or 8 send the machine back so with output of an
error message for the user.
17. We interpret this problem as asking that a 1 be output if the conditions are met, and a 0 be output otherwise.
For this machine, we need to keep track of what the last two inputs have been, and we need four states to
468
Chapter 13
Modeling Computation
"store" this information. Let the states s 3 , s 4 , s 5 , and S6 be the states corresponding to the last two inputs
having been 00, 10, 01, and 11, respectively. We also need some states to get started-to get us into one
of these four states. There are only two cases in which the output is 1: if we are in states s 3 or s 5 (so that
the last two inputs have been 00 or 01) and we receive a 1 as input. The transitions in our machine are the
obvious ones. For example, if we are in state s 5 , having just read 01, and receive a 0 as input, then the last
two symbols read are now 10, so we move to state s4.
~
~:
1,0
@~(9~(9
''qt0?1'··
(g~@~(g~
~1,0
As in Exercise 9, we can actually get by with a smaller machine. Note that here states s 1 and s4 are equivalent,
as are states s 2 and s 6 • Thus we can merge each of these pairs into one state, producing a machine with
only five states. At that point, furthermore, state s 0 is equivalent to the merged s 2 and s 6 , so we can omit
state so and make this other state the start state. The reader is urged to draw the diagram for this simpler
machine.
19. We need some notation to make our picture readable. The alphabet has 26 symbols in it. If a is a letter, then
by a we mean any letter other than a. Thus an arrow labeled a really stands for 25 arrows. The output
is to be 1 when we have just finished reading the word computer. Thus we need eight states, to stand for
the various stages of having read part of that word. The picture below gives the details, except that we have
omitted all the outputs except on inputs r and r; all the omitted ones are intended to be 0. The reader might
contemplate why this problem would have been harder if the word in question were something like baboon.
r, 1
f O
21. We construct the state table by having one row for each state. The arrows tell us what the values of the
transition function are. For example, since there is an arrow from s 0 to s 1 labeled 0, the transition from s 0
on input 0 is to s 1 . Similarly, the transition from s 0 on input 1 is to s 2 . The output function values are
shown next to each state. Thus the output for state s 0 is 1, the output for state s 1 is 1, and the output for
state s 2 is 0. The table is therefore as shown here.
State
Input
1
0
so
s1
Sz
S1
S1
so
Sz
s1
Sz
Output
1
1
0
23. a) The input drives the machine successively to states s 1 , s 0 , s 1 , and s 0 . The output is the output of the
start state, followed by the outputs of these four states, namely 11111.
b) The input drives the machine to state s 2 , where it remains because of the loop. The output is the output
of the start state, followed by the output at state s 2 six times, namely 1000000.
c) The states visited after the start state are, in order, s 2 , s 2 , s 2 , s 1 , s 0 , s 2 , s 2 , s 1 , s 0 , s 2 , and s2.
Therefore the output is 100011001100.
Section 13.3
Finite-State Machines with No Output
469
25. We can use a machine with just two states, one to indicate that there is an even number of l's in the input
string, the other to indicate that there is an odd number of l's in the string. Since the empty string has an
even number of l's, we make s 0 (the start state) the state for an even number of l's. The output for this
state will be 1, as directed. The output from state s 1 will be 0 to indicate an odd number of l's. The input 1
will drive the machine from one state to the other, while the input 0 will keep the machine in its current state.
The diagram below gives the desired machine.
(:>
start
>G) ____.@
1
SECTION 13.3
(:>
0
Finite-State Machines with No Output
As in the previous section, many of these exercises are really exercises in programming. There is no magical
way to become a good programmer, but experience helps. The converse problem is also hard-finding a good
verbal description of the set recognized by a given finite-state automaton.
1. a) This is the set of all strings ab, where a E A and b E B. Thus it contains precisely 000, 001 , 1100, and
1101.
b) This is the set of all strings ba, where a E A and b E B. Thus it contains precisely 000, 0011, 010, and
0111.
c) This is the set of all strings a 1 a 2 , where a 1 E A and a 2 E A. Thus it contains precisely 00, 011, 110, and
1111.
d) This is the set of all strings b1b2 b3 , where each b, EB. Thus it contains precisely 000000, 000001, 000100,
000101, 010000, 010001, 010100 and 010101.
3. Two possibilities are of course to let A be this entire set and let B = { >.}, and to let B be this entire set and
let A= {>.}. Let us find more. With a little experimentation we see that A={>., 10} and B = {10, 111, 1000}
also works, and it can be argued that there are no other solutions in which >. appears in either set. Finally,
there is the solution A = {l, 101} and B = {O, 11, 000}. It can be argued that there are no more. (Here is
how the first of these arguments goes. If >. E A, then necessarily >. tf:_ B. Hence the shortest string in B has
length at least 2, from which it follows that 10 E B. Now since the only other string in AB that ends with
10 is 1010, the only possible other string in A is 10. This leads to the third solution mentioned above. On
the other hand, if A E B, then A ~ A, so it must be that the shortest string in A is 10. This forces 111 to
be in A , and now there can be no other strings in B. The second argument is similar.)
5. a) One way to write this answer is { (10r I n
the string 10.
=
0, 1, 2, ... } . It is the concatenation of zero or more copies of
b) This is like part (a). This set consists of all copies of zero or more concatenations of the string 111 .
In other words, it is the set of all strings of l's of length a multiple of 3. In symbols, it is { (lll)n I n =
0, 1, 2, ... } = { 13 n I n = 0, 1, 2, ... }.
c) A little thought will show that this consists of all bit strings in which every 1 is immediately preceded by
a 0. No other restrictions are imposed, since 0 E A.
d) Because the 0 appears only in 101, the strings formed here have the property that there are at least two
l's between every pair of O's in the string, and the string begins and ends with a 1. All strings satisfying this
property are in A* .
Chapter 13
470
Modeling Computation
7. This follows directly from the definition. Every string w in A* consists of the concatenation of one or more
strings from A. Since A <:;;: B, all of these strings are also in B, so w is the concatenation of one or more
strings from B, i.e., is in B*.
9. a) This set contains all bit strings, so of course the answer is yes.
b) This set contains all strings consisting of any number of l's, followed by any number of O's, followed by
any number of l's. Since 11101 is such a string, the answer is yes.
c) Any string belonging to this set must start 110, and 11101 does not, so the answer is no.
d) All the strings in this set must in particular have even length. The given string has odd length, so the
answer is no.
e) The answer is yes. Just take one copy of each of the strings 111 and 0, together with the required string 1.
f) The answer is yes again. Just take 11 from the first set and 101 from the second.
11. In each case we will list the states in the order that they are visited, starting with the initial state. All we
need to do then is to note whether the place we end up is a final state ( s 0 or s 3 ) or a nonfinal state. (It is
interesting to note that there are no transitions to s 3 , so this state can never be reached.)
a) We encounter s 0 s 1s 2s 0 , so this string is accepted.
b) We encounter
sososos1s2,
c) We encounter
d) We encounter
sos 1s 0s 1s 0 s 1s 2 s 0 ,
so this string is not accepted.
so this string is accepted.
sosos1s2sos1s2sos1s2, so this string is not accepted.
13. a) The set in question is the set of all strings of zero or more O's. Since the machine in Figure 1 has s 0 as a
final state, and since there is a transition from so to itself on input 0, every string of zero or more O's will
leave the machine in state s 0 and will therefore be accepted. Therefore the answer is yes.
b) Since this set is a subset of the set in part (a), the answer must be yes.
c) One string in this set is the string 1 . Since an input of 1 drives the machine to the nonfinal state s 1 , not
every string in this set is accepted. Therefore the answer is no.
d) One string in this set is the string 01. Since an input of 01 drives the machine to the nonfinal state s 1 ,
not every string in this set is accepted. Therefore the answer is no.
e) The answer here is no for exactly the same reason as in part (d).
f) The answer here is no for exactly the same reason as in part ( c).
15. We use structural induction on the input string y. The basis step is y = ,\, and for the inductive step
we write y = wa, where w E I* and a E I. For the basis step, we have xy = x, so we must show that
f(s,x) = f(f(s,x),A). But part (i) of the definition of the extended transition function says that this is
true. We then assume the inductive hypothesis that the equation holds for shorter strings and try to prove
that f(s,xwa) = f(f(s,x),wa). By part (ii) of the definition, the left-hand side of this equation equals
f(f(s,xw),a). By the inductive hypothesis (because w is shorter than y), f(s,xw) = f(f(s,x),w), so
f(f(s, xw), a)= J(f(f(s, x), w), a). On the other hand, the right-hand side of our desired equality is, by part
(ii) of the definition, equal to f(f(f(s,x),w),a). We have shown that the two sides are equal, and our proof
is complete.
17. The only final state is s 2 , so we need to determine which strings drive the machine to state s 2 . Clearly the
strings 0, 10, and 11 do so, as well as any of these strings followed by anything else. Thus we can write the
answer as {O, 10, 11}{0, l}*.
19. A string is accepted if and only if it drives this machine to state s 1 . Thus the string must consist of zero or
more O's, followed by a 1, followed by zero or more l's. In short, the answer is {om 1n I m 2: 0 A n 2: 1 } .
Section 13.3
Finite-State Machines with No Output
471
21. Because so is final, the empty string is accepted.
The strings that drive the machine to final state s3
are precisely {O}{l}*{O}. There are three ways to get to final state s 4 , and once we get there, we stay
there. The path through s2 tells us that strings in { 10, 11}{0, 1} * are accepted. The path sos1 s3s4
tells us that strings in {0}{1}*{01}{0,1}* are accepted. And the path sos1s3s5s4 tells us that strings in
{0}{1}*{00}{0}*{1}{0, 1}* are accepted. Thus the language recognized by this machine is {A}U{O}{l}*{O}U
{10, 11}{0, 1}* u {0}{1}*{01}{0, 1}* u {0}{1}*{00}{0}*{1}{0, 1}*.
23. We want to accept only the strings that begin 01 . Let s 2 be the only final state, and put transitions from s2
to itself on either input. We want to reach s2 after encountering 01, so put a transition from the start state
s 0 to s 1 on input 0, and a transition from s 1 to s2 on input 1. Finally make a "graveyard" state s3, and
have the other transitions from so and s 1 (as well as both transitions from s3) lead to s 3 .
25. We can have a sequence of three states to record the appearance of 101. State s1 will signify that we have
just seen a 1; state s2 will signify that we have just seen a 1 followed by a 0; state S3 will be the only final
state and will signify that we have seen the string 101. Put transitions from s 3 to itself on either input (it
doesn't matter what follows the appearance of 101 ). Put a transition from the start state so to itself on
input 0, because we are still waiting for a 1. Put a transition from s 0 to s 1 on input 1 (because we have
just seen a 1 ). From s 1 on input 0 we want to go to state s 2 , but on input 1 we stay at s 1 because we have
still just seen a 1. Finally, from s 2 , put a transition on input 1 to the final state s 3 (success!), but on input
0 we have to start over looking for 101, so this transition must be back to s 0 .
27. We can let state Si, for i = 0, 1, 2, 3 represent that exactly i O's have been seen, and state s4 will represent
that four or more O's have been seen. Only s 3 will be final. For i = 0, 1, 2, 3, we transition from s, to itself
on input 1 and to s,+l on input 0. Both transitions from s 4 are to itself.
29. We can let state s,, for i = 0, 1, 2, 3 represent that i consecutive 1's have been seen. Only s3 will be final.
For i = 0, 1, 2, we transition from s, to Si+i on input 1 but back to s0 on input 0. Both transitions from s3
are to itself.
31. This is a little tricky. We want states at the start that prevent us from accepting a string if it does not start
with 11. Once we have seen the first two 1's, we can accept the string if we do not encounter a 0 (after all,
the strings 11 and 111 do satisfy the condition). We can also accept the string if it has anything whatsoever
in the middle, as long as it ends 11. The machine shown below accomplishes all this. Note that S3 is a
graveyard state, and state s4 is where we "start over" looking for the final 11.
33. We need just two states, s 0 to represent having seen an even number of O's (this will be the start state,
because to begin we have seen no O's), and s 1 to represent having seen an odd number of O's (this will be the
only final state). The transitions are from each state to itself on input 1, and from each state to the other on
input O.
472
Chapter 13
Modeling Computation
35. This is similar to Exercise 33, except that we need to look for the initial 0. Note that s 3 is the graveyard.
37. We prove this by contradiction. Suppose that such a machine exists, with start state so and other state
s1 . Because the empty string is not in the language but some strings are accepted, we must have s 1 as the
only final state, with at least one transition from so to s1 . Because the string 0 is not in the language, any
transition from so on input 0 must be to itself, so there must be a transition from s0 to s 1 on input 1. But
this cannot happen, because the string 1 is not in the language. Having obtained a contradiction, we conclude
that no such finite-state automaton exists.
39. We want the new machine to accept exactly those strings that the original machine rejects, and vice versa. So
we simply change each final state to a nonfinal state and change each nonfinal state to a final state.
41. We use exactly the same machine as in Exercise 25, but make so, s 1 , and s 2 the final states and make s 3
nonfinal.
43. First some general comments on Exercises 43-49: In general it is quite hard to describe succinctly languages
recognized by machines. An ad hoc approach is usually best. In this exercise there is only one final state, s2 ,
and only three ways to get there, namely on input 0, 01, or 11. Therefore the language recognized by this
machine is {O, 01, 11}.
45. Clearly the empty string is accepted. There are essentially two ways to get to the final state s2 • We can
go through state s 1 , and every string of the form on1 m, where n and m are positive integers, will take us
through state s1 on to s2 . We can also bypass state s 1 , and every string of the form 01 m for m 2 0 will
take us directly to Sz. Thus our answer is {A} u { on1 m I n, m 2 1} u { 01 m I m 2 0}. Note that this can
also be written as { .x, o} u { on 1m \ n, m 2 1 } .
47. First it is easy to see that all strings of the form ion for n 2 0 can drive the machine to the final state s 1 .
Next we see that all strings of the form ioniom for n, m 2 0 can drive the machine to state s 3 . No other
strings can drive the machine to a final state. Therefore the answer is {ion \ n 2 0} U { 1oniom \ n, m 2 0}.
49. We notice first that state s 2 is a final state, that once we get there, we can stay there, and that any string
that starts with a 0 can lead us there. Therefore all strings that start with a 0 are in the language. If the
string starts with a 1, then we must go first to state s 1 . If we ever leave state s 1 , then the string will not be
accepted, because there are no paths out of s 1 that lead to a final state. Therefore the only other strings that
are in the language are the empty string (because s0 is final) and those strings that can drive the machine to
state s1, namely strings consisting of all l's (we've already included those of the form 01*). Therefore the
language accepted by this machine is the union of the set of all strings that start with a 0 and the set of all
strings that have no O's.
51. One way to do Exercises 50-54 is to construct a machine following the proof of Theorem 1. Rather than do
that, we construct the machines in an ad hoc way, using the answers obtained in Exercises 43-47. Since A, 0,
and 1 are accepted by the nondeterministic automaton in Exercise 44, we make states s 0 , s 1 , and s 2 in the
Section 13.3
Finite-State Machines with No Output
473
following diagram final. States 8 3 and 8 4 provide for the acceptance of strings of the form 1no for all n 2:: 1 .
State 8 5 , the graveyard state, assures that no other strings are accepted.
53. This machine is practically deterministic already, since there are no cases of ambiguous transitions (a given
input allowing transition to more than one state). All that keeps this machine from being deterministic is that
there are no transitions from certain states on certain inputs. Therefore to make this machine deterministic,
we just need to add a "graveyard" state, 8 3 , with transitions from s0 on input 0 and from 8 1 on input 1 to
this graveyard state, and transitions from s 3 to itself on input 0 or 1. The graveyard state is not final, of
course.
55. a) We want to accept only the string 0. Let s 1 be the only final state, where we reach s 1 on input 0 from
the start state so. Make a "graveyard" state s 2 , and have all other transitions (there are five of them in all)
lead there.
b) This uses the same idea as in part (a), but we need a few more states. The graveyard state is s4 . See the
picture for details.
start>G)-...:.o--4@
©
o
}~
1
©
01 @Jo,1
c) In the picture of our machine, we show a transition to the graveyard state whenever we encounter a 0.
The only final state is s 2 , which we reach after 11 and remain at as long as the input consists just of 1's.
start)@
1
@
1
©)
~'l/
6)]0,1
57. Intuitively, the reason that a finite-state automaton cannot recognize the set of bit strings containing an equal
number of O's and l's is that there is not enough "memory" in the machine to keep track of how many extra
O's or l's the machine has read so far. Of course, this intuition does not constitute a proof-maybe we are
just not being clever enough to see how a machine could do this with a finite number of states. Instead, we
must give a proof of this assertion. See Exercises 22-25 of Section 13.4 for a development of what are called
"pumping lemmas" to handle various problems like this. (See also Example 6 in Section 13.4.)
The natural way to prove a negative statement such as this is by contradiction. So let us suppose that we
do have a finite-state automaton M that accepts precisely the set of bit strings containing an equal number of
O's and l's. We will derive a contradiction by showing that the machine must accept some illegal strings. The
474
Chapter 13
Modeling Computation
idea behind the proof is that since there are only finitely many states, the machine must repeat some states as
it computes. In this way, it can get into arbitrarily long loops, and this will lead us to a contradiction. To be
specific, suppose that AJ has n states. Consider the string on+ 1 1n+i. As the machine processes this string,
it must encounter the same state more than once as it reads the first n + 1 O's (by the pigeonhole principle).
Say that it hits state s twice. Then some positive number, say k, of O's in the input drives M from state
s back to state s. But then the machine will end up at exactly the same place after reading on+i+k1 n+I as
it will after reading on+ 1 1n+l, since those extra k O's simply drove it in a loop. Therefore since M accepts
on+i1 n+i, it also accepts on+I+k1n+i. But this is a contradiction, since this latter string does not have the
same number of 0 's as 1 's.
59. We know from Exercise 58d that the equivalence classes of Rk are a refinement of the equivalence classes
of Rk-I for each positive integer k. The equivalence classes are finite sets, and finite sets cannot be refined
indefinitely (the most refined they can be is for each equivalence class to contain just one state). Therefore
this sequence of refinements must stabilize and remain unchanged from some point onward. It remains to
show that as soon as we have Rn = Rn+ I, then Rn = Rm for all m > n, from which it follows that Rn = R*,
and so the equivalence classes for these two relations will be the same. By induction, it suffices to show that
if Rn = Rn+l , then Rn+l = Rn+2. By way of contradiction, suppose that Rn+l i- Rn+2. This means that
there are states s and t that are (n + 1)-equivalent but not (n + 2)-equivalent. Thus there is a string x of
length n + 2 such that, say, f(s, x) is final but f(t, x) is nonfinal. Write x =aw, where a EI. Then f(s, a)
and f (t, a) are not (n + 1)-equivalent, because w drives the first to a final state and the second to a nonfinal
state. But f(s, a) and f(t, a) are n-equivalent, because s and t are (n + 1)-equivalent. This contradicts the
fact that Rn = Rn+l , and our proof is complete.
61. a) By the way the machine Af was constructed, a string will drive M from the start state to a final state if
and only if that string drives M from the start state to a final state.
b) For a proof of this theorem, see a source such as Introduction to Automata Theory, Languages, and
Computation (2nd Edition) by John E. Hopcroft, Rajeev Motwani, and Jeffrey D. Ullman (Addison Wesley,
2000).
SECTION 13.4
Language Recognition
Finding good verbal descriptions of the set of strings generated by a regular expression is not easy; neither
is finding a good regular expression for a given verbal description. What Kleene's theorem says is that these
problems of "programming" in regular expressions are really the same as the programming problems for
machines discussed in the previous section. The pumping lemma, discussed in Exercise 22 and the three
exercises that follow it, is an important technique for proving that certain sets are not regular.
1. a) This regular expression generates all strings consisting of zero or more l's, followed by a lone 0.
b) This regular expression generates all strings consisting of zero or more l's, followed by one or more O's.
c) This set has only two elements, 111 and 001.
d) This set contains all strings in which the O's come in pairs.
e) This set consists of all strings in which every 1 is preceded by at least one 0, with the proviso that the
string ends in a 1 if it is not the empty string.
f) This gives us all strings of length at least 3 that end 00.
Section 13.4
Language Recognition
475
3. In each case we try to view 0101 as fitting the regular expression description.
a) The strings described by this regular expression have at most three "blocks" of different digits-a 0, then
some l's, then some O's. Thus we cannot get the string 0101, which has four blocks.
b) The l's that might come between the first and second 0 in any string described by this regular expression
must come in pairs (because of the (11)* ). Therefore we cannot get 0101. Alternatively, note that every
string described by this regular expression must have odd length.
c) We can get this string as 0(10) 1 11 .
d) We can get this string as 01 10(1), where the final 1 is one of the choices in (OU 1).
e) We can get this string as (01) 2 (11) 0 .
f} We cannot get this string, because every string with any l's at all described by this regular expression must
end with 10 or 11.
g) We cannot get this string, because every string described by this regular expression must end with 11.
h) We can get this string as 01(01)1°, where the second 01 is one of the choices in (01U0).
5. a) We just need to take a union: 0 U 11 U 010.
b} More simply put, this is the set of strings of five or more O's, so the regular expression is 000000* .
c) We can use (0 U 1) to represent any symbol and (00 U 01 U 10 U 11) to represent any string of even length.
We need one symbol followed by any string of even length, so we can take (0 U 1)(00 U 01U10 U 11)*.
d) The one 1 can be preceded and/or followed by any number of O's, so we have 0*10*.
e) This one is a little harder. In order to prevent 000 from appearing, we must have every group of one or
two O's followed by a 1 (if we note that the entire string ends with a 1 as well). Thus we can break our string
down into groups of 1, 01, or 001, and we get (1 U 01 U 001 )* as our regular expression.
7. a) We can translate "one or more O's" into 00*. Therefore the answer is 00*1.
b) We can translate "two or more symbols" into (0U1)(0U1)(0U1)*. Therefore the answer is (0 U 1)(0 U
1)(0 u 1)*0000*.
c} A little thought tells us that we want all strings in which all the O's come before all the l's or all the l's
come before all the O's. Thus the answer is O* 1 * U 1 *O*.
d) The string of l's can be represented by 11(111)*; the string of O's, by (00)*. Thus the answer is
11(111 )* (00)*.
9. a) The simplest solution here is to have just the start state s 0 , nonfinal, with no transitions.
b) The simplest solution here is to have just the start state s 0 , final, with no transitions.
c) The simplest solution here is to have just two states-the nonfinal start state s 0 (since we do not want to
accept the empty string) and a final state s 1 -and just the one transition from s 0 to s 1 on input a.
11. We can prove this by induction on the length of a regular expression for A. If this expression has length 1, then
it is either 0 or >. or x (where x is some symbol in the alphabet). In each case A is its own reversal, so there is
nothing to prove. There are three inductive steps. If the regular expression for A is BC, then A = BC, where
B is the set generated by B and C is the set generated by C. By the inductive hypothesis, we know that there
are regular expressions B' and C' that generate BR and CR, respectively. Now AR= (BC)R = (CR)(BR).
Therefore a regular expression for AR is C'B'. The case of union is handled similarly. Let the regular
expression for A be BU C, with B, C, B', and C' as before. Then a regular expression for AR is B' UC',
since clearly (BU C)R = (BR) U (CR). Finally, if the regular expression for A is B*, then, with the same
notation as before, it is easy to see that (B')* is a regular expression for AR.
13. a) We can build machines to recognize O* and 1 * as shown in the second row of Figure 3. Next we need
to put these together to make a machine that recognizes O* 1 * . We place the first machine on the left and
Chapter 13
476
Modeling Computation
the second machine on the right. We make each final state in the first machine nonfinal (except for the start
state, since >. E 0*1 * ), but leave the final states in the second machine final. Next we copy each transition
to a state that was formerly final in the first machine into a transition (on the same input) to the start state
of the second machine. Lastly, since ,\ E O* , we add the transition from the start state to the state to which
there is a transition from the start state of the machine for 1 * . The result is as shown. (In all parts of this
exercise we have not put names on the states in our state diagrams.)
b) This machine is quite messy. The upper portion is for 0, and the lower portion is for 11. They are
combined to give a machine for 0 U 11. Finally, to incorporate the Kleene star, we added a new start state
(on the far left), and adjusted the transitions according to the procedure shown in Figure 2.
c) This is similar to the other parts. We grouped the expression as 01 * U (00*)1. The answer is as shown.
~~
Oo-O~©
01-0
~
o
U1
~
0
15. We choose as the nonterminal symbols corresponding to states s 0 , s 1 , and s 2 the symbols S, A, and B,
respectively. Thus S is our start symbol. The terminal symbols are of course 0 and 1. We construct the rules
for our grammars by following the procedure described in the proof of the second half of Theorem 2: putting
in rules of the form X ____, aY for each transition from the state corresponding to X to the state corresponding
to Y, on input a, and putting in a rule of the form X ____, a for a transition from the state corresponding to
X to the final state, on input a. Specifically, since there is a transition from s 0 to s 1 on input 0, we include
the rule S ____, OA. Similarly, the other transitions give us the rules S ____, IB, A ____, OB, A ____, IB, B ____, OB,
and B ____, lB. Also, the transition to the final state from S on input 0 gives rise to the rule S ____, 0. Thus
our grammar contains these seven rules.
17. This is similar to Exercise 15-see the discussion there for the approach. We let C correspond to state s 3 .
The set of rules contains S ____, OC, S ____, IA, A ____, IA, A ____, OC, B ____, OB, B ____, IB, C ____, OC, and C ____, IB
(for the transitions from a state to another state on a given input), as well as S ____, 1, A ____, 1, B ____, 0, B ____, 1,
and C ____, 1 (for the transitions to the final states that can end the computation).
Section 13.4
Language Recognition
477
19. This is clear, since the operation of the machine is exactly mimicked by the grammar. If the current string in
the derivation in the grammar is v1 V2 ... VkAs, then the machine has seen input v 1v2 ... Vk and is currently
in state s. If the current string in the derivation in the grammar is V1 v2 ... Vk, then the machine has seen
input v1 v2 ... Vk and is currently in some final state. Hence the machine accepts precisely those strings that
the grammar generates. (The empty string does not fit this discussion, but it is handled separately-and
correctly-since we take S --+ >. as a production if and only if we are supposed to.)
21. First suppose that the language recognized by M is infinite. Then the length of the words recognized by M
must be unbounded, since there are only a finite number of symbols. Thus l(x) is greater than the finite
number ISi for some word x E L(M).
Conversely, let x be such a word, and let so , Si 1 , Si 2 , ••• , s,n be the sequence of states that the machine
goes through on input x, where n = l(x) and s,n is a final state. By the pigeonhole principle, some state
occurs twice in this sequence, i.e., there is a loop from this state back to itself during the computation. Let y
be the substring of x that causes the loop, so that x = uyv. Then for every nonnegative integer k, the string
uykv is accepted by the machine M (i.e., is in L(M) ), since the computation is the same as the computation
on input x, except that the loop is traversed k times. Thus L(M) is infinite.
23. We apply the pumping lemma in a proof by contradiction. Suppose that this set were regular. Clearly it
contains arbitrarily long strings. Thus the pumping lemma tells us that for some strings u, v -I- >., and w,
the string uviw is in our set for every i. Now if v contains both O's and l's, then uv 2 w cannot be in the set,
since it would have a 0 following a 1, which no string in our set has. On the other hand, if v contains only
O's (or only l's), then for large enough i, it is clear that uv'w has more than (or less than) twice as many
O's as l's, again contradicting the definition of our set. Thus the set cannot be regular.
25. We will give a proof by contradiction, using the pumping lemma. Following the hint, let x be the palindrome
0N10N, for some fixed N > ISi, where S is the set of states in a machine that recognizes palindromes. By
the lemma, we can write x = uvw, with l(uv)::::; ISi and l(v) ::'.:: 1, so that for all i, uv'w is a palindrome.
Now since 0N10N = uvw and l(uv)::::; ISi < N, it must be the case that vis a string consisting solely of O's,
with the 1 lying in w. Then uv 2w cannot be a palindrome, since it has more O's before its sole 1 than it has
O's following the 1.
27. It helps to think of L/x in words-it is the set of "ends" of strings in L that start with the string x; in other
words, it is the set of strings obtained from strings in L by stripping away an initial piece x. To show that
11 and 10 are distinguishable, we need to find a string z such that llz E L and lOz tt L or vice versa. A
little thought and trial and error shows us that z = 1 works: 111 tt L but 101 E L. To see that 1 and 11
are indistinguishable, note that the only way for lz to be in L is for z to end with 01, and that is also the
only way for l lz to be in L.
29. By Exercise 28, if two strings are distinguishable, then they drive the machine from the start state to different
states. Therefore, if x1, x 2 , ... , Xn are all distinguishable, the states f(so, x1), f(so, x2), ... , f(so, xn) are
all different, so the the machine has at least n states.
31. We claim that any two distinct strings of the same length are distinguishable with respect to the language P
of all palindromes. Indeed, if x and y are distinct strings of length n, let z = xR (the reverse of string x ).
Then xz E P but yz tt P. Note that there are 2n different strings of length n. By Exercise 29, this tells
us that any deterministic finite-state automaton for recognizing palindromes must have at least 2n states.
Because n is arbitrary (we want our machine to recognize all palindromes), this tells us that no finite-state
machine can recognize P.
Chapter 13
478
SECTION 13.5
Modeling Computation
Turing Machines
In this final section of the textbook, we have studied a machine that has all the computing capabilities possible
(if one believes the Church-Turing thesis). Most of these exercises are really programming assignments, and
the programming language you are stuck with is not a nice, high-level, structured language like Java or C, nor
even a nice assembly language, but something much messier and less efficient. One point of the exercises is to
convince you that even in this horrible setting you can, with enough time and patience, instruct the computerthe Turing machine-to do whatever you wish computationally. Keep in mind that in many senses, a Turing
machine is just as powerful as any computer running programs written in any language. One reason for talking
about Turing machines at all, rather than just using high-level languages, is that their simplicity makes it
feasible to prove some very interesting things about them (and therefore about computers in general). For
example, one can prove that computers cannot solve the halting problem (see also Section 3.1), and one can
prove that a large class of problems have efficient algorithmic solutions if and only if certain very specific
problems, such as a decision version of the traveling salesman problem, do (the NP-complete problems-see
also Section 3.3). This is part of what makes Turing machines so important in theoretical computer science,
and time spent becoming acquainted with them will not go unrewarded as you progress in this field.
1. We will indicate the configuration of the Turing machine using a notation such as O[s 2]1Bl. This string of
symbols means that the tape is blank except for a portion which reads OlBl from left to right; that the
machine is currently in state s 2 ; and that the tape head is reading the left 1 (the currently scanned symbol
will always be the one following the bracketed state information).
a) The initial configuration is [s 0 ]0011. Because of the five-tuple (s 0 , 0, s 1, 1, R) and the fact that the machine
is in state s 0 and the tape head is looking at a 0, the machine changes the 0 to a 1 (i.e., writes a 1 in that
square), moves to the right, and enters state s 1 . Therefore the configuration at the end of one step of the
computation is l[s 1]011. Next the transition given by the five-tuple (s 1,0,s 2, 1,L) occurs, and we reach the
configuration [s 2]1111. There are no five-tuples starting with s 2 , so the machine halts at this point. The
nonblank portion of the tape contains 1111 .
b) The initial configuration is [s 0 ]101. Because of the five-tuple (s 0 , l,s1,0,R) and the fact that the machine
is in state s 0 and the tape head is looking at a 1, the machine changes the 1 to a 0, moves to the right, and
enters state s 1 . Therefore the configuration at the end of one step of the computation is O[s 1]01. At this time
transition (s 1,0,s 2,1,L) kicks in, resulting in configuration [s 2 ]011, and the machine halts, with 011 on its
tape.
c) We seem to have the idea from the first two parts, so let us just list the configurations here, using the notation
"----+'' to show the progression from one to the next. [so]llBOl ----+ O[s 1]1B01 ----+ OO[s1]B01 ----+ O[s2]0001.
Therefore the final output is 00001.
d) [s 0 ]B----+ O[s 1]B----+ [s2]00. So the final tape reads 00.
3. Note that all motion is from left to right.
a) The machine starts in state s 0 and sees the first 1. Therefore using the second five-tuple, it replaces the
1 by a 0, moves to the right, and enters state s 1 . Now it sees the second 1, so, using the fifth five-tuple, it
replaces the 1 by a 1 (i.e., leaves it unchanged), moves to the right, and enters state s 0 . The third five-tuple
now tells it to leave the blank it sees alone, move to the right, and enter state s 2 , which is a final (accepting)
state (because it is not the first state in any five-tuple). Since there are no five-tuples telling the machine what
to do in state s 2 , it halts. Note that 01 is on the tape, and the input was accepted.
b) When in state s 0 the machine skips over O's, ignoring them, until it comes to a 1. When (and if) this
happens, the machine changes this 1 to a 0 and enters state s 1 . Note also that if the machine hits a blank
( B) while in state s0 or s 1 , then it enters the final (accepting) state s 2 . Next note that s1 plays a role
similar to that played by so, causing the machine to skip over O's, but causing it to go back into state s 0 if
and when it encounters a 1. In state s 1 , however, the machine does not change the 1 it sees to a 0. Thus
Section 13.5
Turing Machines
479
the machine will alternate between states s 0 and s 1 as it encounters l's in the input string, changing half
of these l's to O's. To summarize, if the machine is given a bit string as input, it scans it from left to right,
changing every other occurrence of a 1, if any, starting with the first, to a 0, and otherwise leaving the string
unchanged; it halts (and accepts) when it comes to the end of the string.
5. a) The machine starts in state s 0 and sees the first 1. Therefore using the first five-tuple, it replaces the 1
by a 0, moves to the right, and enters state s 1 . Now it sees the second 1, so, using the second five-tuple, it
replaces the 1 by a 1 (i.e., leaves it unchanged), moves to the right, and stays in state s 1 . Since there are no
five-tuples telling the machine what to do in state s1 when reading a blank, it halts. Note that 01 is on the
tape, and the input was not accepted, because s 1 is not a final state; in fact, there are no final states (states
that begin no 5-tuples).
b) This is essentially the same as part (a). The first 1 (if any) is changed to a 0 and the others are left alone.
The input is not accepted.
7. The machine needs to search for the first 0 and when (and if) it finds it, replace it with a 1. So let's have the
machine stay in its initial state (so) as long as it reads l's, constantly moving to the right. If it ever reads a
0 it will enter state s 1 while changing the 0 to a 1. No further action is required. Thus we can get by with
just the following two five-tuples: (s 0 , 0, s1, 1, R) and (so, 1, s 0 , 1, R). Note that if the input string consists of
just l's, then the machine eventually sees the terminating blank and halts.
9. The machine should scan the tape, leaving it alone until it has encountered the first 1. At that point, it needs
to enter a phase in which it changes all the l's to O's, until it reaches the end of the input. So we'll have
tuples (so, 0, so, 0, R) and (so, 1, si, 1, R) to complete the first phase, and then have tuples (s 1, 0, s 1, 0, R) and
(s1, 1, s1, 0, R) to complete the second phase. When the machine encounters the end of the input (a blank on
the tape) it halts, since there are no transitions given with a blank as the scanned symbol.
11. We can have the machine scan the input tape until it reaches the first blank, "remembering" what the last
symbol was that it read. Let us use state s0 to represent that last symbol's being a 1, and s 1 to represent its
being a 0. It doesn't matter what gets written, so we'll just leave the tape unchanged as we move from left to
right. Thus our first few five-tuples are (s 0 , 0, s 1, 0, R), (s 0 , 1, s 0 , 1, R), (s 1, 0, s 1, 0, R), (s 1, 1. s 0 , 1, R). Now
suppose the machine encounters the end of the input, namely the blank at the end of the input string. If it is
in state so, then the last symbol read was not a 0, so we want to not accept the string. If it is in state s 1 ,
then the last symbol read was a 0, so we want to accept the string. Recall that the convention presented in
this section was that acceptance is indicated by halting in a final state, i.e., one with no transitions out of it.
So let's add the five-tuple (s 1, B, s 2 , B, R) for accepting when we should. To make sure we don't accept when
we shouldn't, we need do nothing else, because the machine will halt in the nonfinal state s 0 in this case.
An alternative approach to this problem is to have the machine scan to the right until it reaches the end
of the tape, then back up, "look" at the last symbol, and take the appropriate action.
13. This is very similar to Exercise 11. We want the machine to "remember" whether it has seen an even number
of l's or not. We'll let s 0 be the state representing that an even number of l's have been seen (which is
of course true at the start of the computation), and let s 1 be the state representing that an odd number of
l's have been seen. So we put in the following tuples: (s 0 ,0,s0 ,0,R), (s 0 ,l,s 1,l,R), (s 1,0,s 1,0,R), and
(s1, 1, so, 1, R). When the machine encounters the terminating blank, we want it to accept if it is in state s0 ,
so we add the tuple (s 0 , B, s 2 , B, R). Thus the machine will halt in final state s 2 if the input string has an
even number of O's, and it will halt in nonfinal state s 1 otherwise.
480
Chapter 13
Modeling Computation
15. You need to play with this machine to get a feel for what is going on. After doing so, you will understand
that it operates as follows. If the input string is blank or starts with a 1, then the machine halts in state
s0 , which is not final, and therefore every such string is not accepted (which is a good thing, since it is not
in the set to be recognized). Otherwise the initial 0 is changed to an Al, and the machine skips past all the
intervening O's and l's until it either comes to the end of the input string or else comes to an M (which, as
we will see, has been written over the right-most remaining bit). At this point it backs up (moves left) one
square and is in state s 2 . Since the acceptable strings must have a 1 at the right for each 0 at the left, there
had better be a 1 here if the string is acceptable. Therefore the only transition out of state s 2 occurs when
this square contains a 1. If it does, then the machine replaces it with an !vl, and makes its way back to the
left. (If this square does not contain a 1, then the machine halts in the nonfinal state s 2 , as appropriate.)
On its way back, it stays in state s 3 as long as it sees l's, then stays in s 4 as long as it sees O's. Eventually
either it encounters a 1 while in state s4 , at which point it (appropriately) halts without accepting (since
the string had a 0 to the right of a 1 ); or else it reaches the right-most M that had been written over a 0
near the beginning of the string. If it is in state s 3 when this happens, then there are no more O's in the
string, so it had better be the case (if we want to accept this string) that there are no more l's either; this
is accomplished by the transitions (s 3 , l\1, s 5 , l\1, R) and (s 5 , l\1, s6 , M, R), and s 6 is a final state. Otherwise,
the machine halts in nonfinal state s.5 . If it is in state s4 when this M is encountered, then we need to start
all over again, except that now the string will have had its left-most remaining 0 and its right-most remaining
1 replaced by ]\f's. So the machine moves (staying in state S4) to the left-most remaining 0 and goes back
into state s 0 to repeat the process.
17. This will be similar to the machine in Example 3, in that we will change the digits one at a time to a new
symbol M. We can't work from the outside in as we did there, however, so we'll replace all three digits from
left to right. Furthermore, we'll put a new symbol, E, at the left end of the input in order to tell more easily
when we have arrived back at the starting point. Here is our plan for the states and the transitions that will
accomplish our goal. State s 9 is our (accepting) final state. States s 0 and s 1 will write an E to the left of the
initial input and return to the first input square, entering state s 2 . (If, however, the tape is blank, then the
machine will accept immediately, and if the first symbol is not a 0, then it will reject immediately.) The fivetuples are (s 0 , B, sg, B, L), (s 0 , 0, s1, 0, L), and (s 1, B, s2, E, R). State s2 will skip past any M's until it finds
the first 0, change it to an l\l, and enter state s 3 . The transitions are (s 2 , J\.1, s 2 , M, R) and (s 2 , 0, s 3 , M, R) .
Similarly, state s 3 will skip past any remaining O's and any M's until it finds the first 1, change it to
an l\f, and enter state s 4 . The transitions are (s3, 0, s3, 0, R), (s3, A1, s3, M, R) and (s3, 1, s4, M, R). State
s 4 will do the same for the first 2 (skipping past remaining l's and M's, and ending in state S5 ), with
transitions (s 4 , 1, s 4 , 1, R), (s 4 , l\l, s 4 , !vl, R) and (s 4 , 2, s 5 , M, R). State s 5 then will skip over any remaining
2's and (if there is any chance of accepting this string) encounter the terminating blank. The transitions are
(s 5 ,2,s 5 ,2,R) and (s 5 ,B,s 6 ,B,L). Note that once this blank has been seen, we back up to the last symbol
before it and enter state s 6 . There are now two possibilities. If the scanned square is an M, then we should
accept if and only if the entire string consists of ]\;f's at this point. We will enter state s 8 to check this,
with the transition (s 6 , l\l, s 8 , l\l, L) . Otherwise, there will be a 2 here, and we want to go back to the start
of the string to begin the cycle all over; we'll use state s 7 to accomplish this, so we put in the five-tuple
(s 6 , 2, s7 , 2, L). In this latter case, the machine should skip over everything until it sees the marker E that we
put at the left end of the input, then move back to the initial input square, and start over in state s 2 . The
transitions (s 7,0,s7,0,L), (s 7,l,s7,l,L), (s7,2,s 7,2,L), (s7,Jl.;f,s 7,M,L), and (s7,E,s2,E,R) accomplish
this. But if we entered state s 8 , then we need to make sure that there is nothing but M's all the way back
to the starting point; we add the five-tuples (s 8 ,Jl.;1,s 8 ,l\1,L) and (s 8 ,E,s 9 ,E,L), and we're finished.
19. Recall that functions are computed in a funny way using unary notation. The string representing n is a string
of n + 1 l's. Thus we want our machine to erase three of these l's (or all but one of them, if there are
Section 13.5
Turing Machines
481
fewer than four), and then halt. One way to accomplish this is as follows. If n :::: 3, then the five-tuples
(s 0 ,l,s 1,B,R), (s 1,l,s 2,B,R), (s2,l,s 3,B,R), and (s3,l,s4,l,R) will do the trick (s4 is just a halting
state). To account for the possibilities that n < 3, we add transitions (s 1 ,B,s4,l,R), (s2,B,s4,l,R), and
(s 3, B, s 4, 1, R). In each of these three cases, we needed to restore one 1 before halting (since the "answer"
was to be 0).
2 5. If it finds that n 2 5, then it needs to leave exactly
four l's on the tape (according to our rules for representing numbers in unary); otherwise it needs to leave
exactly one 1. We'll use states s 0 through s 6 for this task, with the following five-tuples, which erase the
21. The machine here first needs to "decide" whether n
tape as they move from left to right through the input: (s 0 ,l,s 1 ,B,R), (s 1,l,s 2,B,R), (s 1 ,B,s6 ,B,R),
(s 2, 1, s3, B, R), (s 2, B, s 6 , B, R), (s3, 1, s4, B, R), (s 3, B, s5, B, R), (s4, 1, s 5, B, R), (s4, B, s5, B, R). At this
point, the machine is either in state s 5 (and n 2 5), or in state s 6 with a blank tape (and n < 5). To
finish in the latter case, we just write a 1 and halt: (s 6 , B, s 10 , 1, R). For the former case, we erase the rest
of the tape, write four l's, and halt: (s 5, 1, s 5, B, R), (ss, B, s7, 1, R), (s7, B, ss, 1, R), (ss, B, sg, 1, R), and
(sg, B, s10, 1, R).
23. We start with a string of n + 1 l's, and we want to end up with a string of 3n + 1 l's. Our idea will be to
replace the last 1 with a 0, then for each 1 to the left of the 0, write a pair of new l's to the right of the 0.
To keep track of which l's we have processed so far, we will change each left-side 1 to a 0 as we process it.
At the end, we will change all the O's back to l's. Basically our states will mean the following (''first'' means
"first encountered"): s 0 , scan right for last 1 ; s 1 , change the last 1 to 0; s2 , scan left to first 1; s 3 , scan
right for end of input (having replaced the 1 where we started with a 0); s 3 and s 4 , write the two more l's;
s 5 , scan left to first O; s 6 , replace the remaining O's with l's; s7, halt.
The needed five-tuples are as follows: (s 0 ,l,s 0 ,l,R), (so,B,s 1 ,B,L), (s 1,l,s2,0,L), (s2,0.s2,0,L),
(s2,l,s3,0,R), (s2,B,s5,B,R), (s3,0,s 3,0,R), (s3,l,s3,l,R), (s3,B,s4,l,R), (s4,B,s5,l,L),
(ss,l,ss,l,L), (ss,O,s2,0,L), (s5,0,s5,l,R), (s5,l,s7,l,R), (s5,B,s7,B,R).
25. The idea here is to match off the l's in the two inputs (changing the l's to O's from the left, say, to keep
track), until one of them is exhausted. At that point, we need to erase the larger input entirely (as well as
the asterisk) and change the O's back to l's. Here is how we'll do it. In state s 0 we skip over any O's until
we come to either a 1 or the *. If it's the *, then we know that the second input ( n 2 ) is at least as large as
the first ( n 1 ), so we enter a clean-up state s 5 , which erases the asterisk and all the O's and l's to its right.
The five-tuples for this much are (so, 0, so, 0, R), (so,*, s5, B, R), (ss, 1, s5, B, R), and (s5, 0, s5, B, R). Once
this erasing is finished, we need to go over to the part of the tape where the first input was and change all the
O's back to l's; the following transitions accomplish this: (s 5 ,B,s6 ,B,L), (s5,B,s5,B,L), (s5,0,s7,l,L),
and (s7,0,s 7,l,L). Eventually the machine halts in state s7 when the blank preceding the original input is
encountered.
The other possibility is that the machine encounters a 1 while in state so. We want to change this 1 to
a 0, skip over any remaining l's as well as the asterisk, skip over any O's to the right of the asterisk (these
represent parts of n 2 that have already been matched off against equal parts of n 1 ), and then either find a 1
in n 2 (which we change to a 0) or else come to the blank at the end of the input. Here are the transitions:
(so, 1, s1, 0, R), (s 1, 1, s1, 1, R), (s 1, *, s2, *, R), (s2, 0, s2,0,R), (s2, 1, s3, 0, L), (s2, B, s4,B,L). At this point
we are either in state s 3 , ready to go back for the next iteration, or in state s 4 ready for some cleanup. In
the former case, we want to skip back over the nonblank symbols until we reach the start of the string, so
we add five-tuples (s 3, *, s 3, *, L), (s 3, 0, s 3, 0, L), (s 3, 1, s 3, 1, L), and (s 3, B, so, B, R). In the latter case, we
know that the first string is longer than the second. Therefore we want to change the O's in the second input
string back to l's and then erase the asterisk and remnants of the first input string. Here are the transitions:
(s4,0,s4,l,L), (s4,*,Ss,B,L), (ss,O,ss,B,L), (ss,l,ss,B,L).
Chapter 13
482
Modeling Computation
27. The discussion in the preamble tells how to take the machines from Exercises 22 and 18 and create a new
machine. The only catch is that the tape head needs to be back at the leftmost 1. Suppose that Sm, where m
is the largest index, is the state in which the Turing machine for Exercise 22 halts after completing its work,
and suppose that we have designed that machine so that when the machine halts the tape head is reading the
leftmost 1 of the answer. Then we renumber each state in the machine for Exercise 18 by adding m to each
subscript, and take the union of the two sets of five-tuples.
29. If the answer is yes/no, then the problem is a decision problem.
a) No, the answer here is a number, not yes or no.
b) Yes, the answer is either yes or no.
c) Yes, the answer is either yes or no.
d) Yes, the answer is either yes or no.
31. This is a fairly hard problem, which can be solved by patiently trying various combinations. The following
five-tuples will do the trick: (s 0 , B, s 1, 1, L), (so, 1, s1, 1, R), (s 1, B, so, 1, R).
GUIDE TO REVIEW QUESTIONS FOR CHAPTER 13
1. a) See p. 849.
b) Seep. 849.
2. a) See p. 850.
b) { 0 3 n 1 I n 2:
o}
c) The vocabulary is { S, 0, 1} ; the terminals are T
= { 0, 1} ; the start symbol is S; and the productions are
S-+Sl and S-+O.
c) Seep. 851.
b) a grammar that contains a production like AB --+ C
d) a grammar that contains a production like Sa--+ Sbc
e) Seep. 851.
f) a grammar that contains a production like S--+ SS
3. a) See p. 851.
4. a) See p. 851.
b) See p. 851.
c) See Example 8 in Section 13. l.
5. a) See p. 854.
b) See Example 14 in Section 13.l.
6. a) Seep. 851 (machines with output) and p. 867 (machines without output, called finite-state automata). See
also p. 863 for comments on other types of finite-state machines.
b) Have three states and only one input symbol, Q (quarter). The start state s 0 has a transition to state s 1
on input Q and outputs nothing; state s 1 has a transition to state s 2 on input Q and outputs nothing; state
s 2 has a transition back to state s1 on input Q and outputs a drink.
7. 1*u1 *00
8. Have four states, with only s 2 final. From the start state s 0 , go to a graveyard state s1 on input 0, and go
to state s 2 on input 1 . From both states s2 and S3, go to s2 on input 1 and to s3 on input 0.
9. a) See p. 866.
b) the set of all strings in which all the maximal blocks of consecutive l's (if any) have an even number of l's
10. a) See p. 867.
b) See p. 868.
11. a) See p. 873.
b) See Theorem 1 in Section 13.3.
12. a) See p. 879.
b) Seep. 879.
Supplementary Exercises
483
13. See Theorem 1 in Section 13.4.
14. See the proof of Theorem 2 in Section 13.4.
15. See Example 6 in Section 13.4.
16. See p. 889.
17. See p. 891.
18. See p. 892.
19. See p. 895. The halting problem is unsolvable; see p. 895.
SUPPLEMENTARY EXERCISES FOR CHAPTER 13
1. a) We simply need to add two O's on the left and three l's on the right at the same time. Thus the rules can
be S _, 008111 and S _, >..
b) We need to add two O's for every 1 and also allow the symbols to change places at will. Following the
trick in our solution to Exercise 15c in Section 13.1, we let A and B be nonterminal symbols representing 0
and 1 , respectively. Our rules are S _, AAB S, AB _, BA, BA _, AB , A _, 0, B _, 1, and S _, >..
c) Our trick here is first to generate a string that looks like Ew( wR), with A in the place of 0, and B in the
place of 1 , in the second half. The rules S _, ET, T _, OT A, T _, IT B, and T _, >. will accomplish this
much. Then we force the A's and B's to march to the left, across all the O's and l's, until they bump into
the left-hand wall (E), at which point they turn into their terminal counterparts. Finally, the wall disappears.
The rules for doing this are OA _, AO, IA _, Al, OB _, BO, IB _, Bl, EA _, EO, EB _, El, and E _, >..
3. For part (a) note that (()) can come from (B), which in turn can come from (A), which can come from B,
and we can start S :::;, A :::;, B. Thus the tree can be as shown in the first picture. For part (b) we need to
use the rule A _, AB early in the derivation, with the A turning into (), and the B turning into ( ()). The
ideas in part ( c) are similar.
5
5
I
I
A
A
I
B
/1""(
A
)
I
~~
A
I
B
/I""<
A
)
I
B
(/".)
(a)
(c)
(b)
5. The idea is that the rules enable us to add O's to either the right or the left. Thus we can get three O's in
many ways, depending on which side we add the O's on.
/""'
/"'5
0
5
5
I
0
0
484
Chapter 13
Modeling Computation
7. It is not true that IABI is always equal to IAI · IBI, since a string in AB may be formed in more than
one way. After a little experimentation, we might come up with the following example to show that IABI
need not equal IBAI and that IABI need not equal IAI · JBI. Let A = {O, 00}, and let B = {01, 1}. Then
AB = {01, 001, 0001} (there are only 3 elements, not 2 · 2 = 4, since 001 can be formed in two ways), whereas
BA= {010, 0100, 10, 100} has 4 elements.
9. This is clearly not necessarily true. For example, we could take A = V* and B = V. Then A* is again V* ,
so it is true that A* i:;;; B*, but of course A<£_ B (for one thing, A is infinite and B is finite).
11. In each case we apply the definition to rewrite h(E) in terms of h applied to the subexpressions of E.
a) h(O*l) = max(h(O*), h(l)) = max(h(O) + 1, 0) = max(O + 1, 0) = 1
b) h(O*l *) = max(h(O*), h(l *)) = max(h(O) + 1, h(l) + 1) = max(O + 1, 0 + 1)
c) h((O*Ol)*) = h(O*Ol) + 1=1+1=2
=1
d) This is similar to part ( c); the answer is 3 .
e) There are three "factors,'' and by the definition we need to find the maximum value that h takes on them.
It is easy to compute that these values are 1, 2, and 2, respectively, so the answer is 2.
f) A calculation similar to that in part ( c) shows that the answer is 4.
13. We need to have states to represent the number of l's read in so far. Thus Bi, for i = 0, 1, 2, 3, will "mean"
that we have seen exactly i l's so far, and s 4 will signify that we have seen at least four l's. We draw only
the finite-state automaton; the machine with output is exactly the same, except that instead of a state being
designated final, there is an output for each transition; all the outputs are 0 except for the outputs to our
final state, and all of the outputs to this final state are 1.
15. This is similar to Exercise 13, except that we need to return to the starting state whenever we encounter a 0,
rather than merely remaining in the same state. As in Exercise 13, we draw only the automaton, since the
machine with output is practically the same.
17. a) To specify a machine, we need to pick a start state (this can be done in n ways), and for each pair
(state, input) (and there are nk such pairs), we need to choose a state and an output. By the product rule,
therefore, the answer is n · n nk · m nk. (We are answering the question as it was asked. A much harder question
is to determine how many "really" different machines there are, since two machines that really do the same
thing and just have different names on the states should perhaps be considered the same. We will not pursue
this question.)
b) This is just like part (a), except that we do not need to choose an output for each transition, only an
output for each state. Thus the term mnk needs to be replaced by mn, and the answer is n · nnk · mn.
19. This machine has no final states. Therefore no strings are accepted. Any deterministic machine with no final
states will be equivalent to this one. We show one such machine below.
Supplementary Exercises
485
21. The answers are not unique, of course.
There are two ways to approach this exercise. We could simply
apply the algorithm inherent in the proof of Kleene's theorem, but that would lead to machines much more
complicated than they need to be. Alternately, we just try to be clever and make the machines "do what the
expressions say." This is essentially computer programming, and it takes experience to be able to do it well.
In part (a), for example, we want to accept every string of O's, so we make the start state a final state, with
returns to this state on input 0; and we want to accept every string that has this beginning and then consists
of any number of copies of 10-which is precisely what the rest of our machine does. These pictures can
either be viewed as nondeterministic machines, or else for all the missing transitions we assume a transition
to a new state (the graveyard), which is not final and which has transitions to itself on both inputs. Also, as
usual, having two labels on an edge is an abbreviation for two edges, one with each label.
a) This one is pretty simple. State so represents the condition that only O's have been read so far; it is final.
After we have read in as many O's as desired, we still want to accept the string if we read in any number of
copies of 10. This is accomplished with the other two states.
b) In this machine, we keep returning to s 0 as long as we are reading 01 or 111, corresponding to the first
factor in our regular expression. Then we move to s4 for the term 10* , and finally to s 5 for the factor
(0 u 1).
c) Note that the inner star in this regular expression is irrelevant; we get the same set whether it is there or
not. Our machine returns us to s 0 after we have read either 001 or 11, so we accept every string consisting
of any number of copies of these strings.
23. We invoke the power of Kleene's theorem here. If A is a regular set, then there is a deterministic finite
automaton that accepts A. If we take the same machine but make all the final states nonfinal and all the
nonfinal states final, then the result will accept precisely A. Therefore A is regular.
25. See the comments for Exercise 21. Here the problem is even harder, since we are given just verbal descriptions
of the sets. Thus there is no general algorithm we can invoke. We just have to be clever programmers. See
the comments on the solution to Exercise 21 for how to interpret missing transitions.
a) The top part of our machine (as drawn) takes us to a graveyard if there are more than three consecutive
O's at the beginning. The rest assures that there are at least two consecutive 1 's.
486
Chapter 13
Modeling Computation
b) This one is rather complicated. The states represent what has been seen recently in the input. For example,
states s 2 and s 6 represent the condition in which the last two symbols have been 10. Thus if we encounter
a 1 from either of these states, we move to a graveyard. (Note that we could have combined states s 3 and
into one, or, under our conventions, we could have omitted them altogether; the answers to these exercises
are by no means unique.) States s 0 , s 2 and s 5 all represent conditions in which an even number of symbols
have been read in, whereas s1 , S4 and s 6 represent conditions in which an odd number of symbols have been
read.
s7
c) This one is really not as bad as it looks. The first row in our machine (as drawn) represents conditions
before any O's have been read; the second row after one 0, and the third row after two or more O's. The
horizontal direction takes care of looking for the blocks of l's.
27. Suppose that {IP I p is prime} is regular. Then by the pumping lemma, we can find a prime p such that
lP = uvw, with l( v) 2 1, so that uv'w is a string of a prime number of l's for all i. If we let a be the
number of l's in uw and b > 0 the number of l's in v, then this means that a + bi is prime for all i. In
other words, the gap between two consecutive primes (once we are looking at numbers greater than a) is at
most b. This contradicts reality, however, since for every n, all the numbers from n! + 2 through n! + n are
not prime-in other words the gaps between primes can be arbitrarily large.
29. The idea here is to match off the l's in the two inputs (changing the l's to O's from the left, say, to keep
track), until one of them is exhausted. At that point, we need to erase the smaller input entirely (as well as
the asterisk) and change the O's back to l's. Here is how we'll do it. In state s 0 we skip over any O's until we
come to either a 1 or the *. If it's the *, then we know that the second input ( n 2 ) is at least as large as the
first ( n 1 ), so we enter a clean-up state s 5 , which erases the asterisk and all the O's and l's to its left. The
five-tuples for this much are (s 0 , 0, s0 , 0, R), (s 0 , *, s 5 , B, L), and (s 5 , 0, s 5 , B, L). Once this erasing is finished,
we need to go over to the part of the tape where the second input was and change all the O's back to l's;
the following transitions accomplish this: (s 5 , B, s 6 , B, R), (s 6 , B, s 6 , B, R), (s 6 , 0, s 7 , 1, R), (s 7 , 0, s 7 , 1, R),
Writing Projects
and (s 7 , 1, s 7 , 1, R). Eventually the machine halts in state
encountered.
487
s7
when the blank following the original input is
The other possibility is that the machine encounters a 1 while in state so. We want to change this 1 to
a 0, skip over any remaining l's as well as the asterisk, skip over any O's to the right of the asterisk (these
represent parts of n 2 that have already been matched off against equal parts of n 1 ), and then either find a 1
in n 2 (which we change to a 0) or else come to the blank at the end of the input. Here are the transitions:
(s 0 ,l,8 1 ,0,R), (8 1 ,1,8 1 ,1,R), (8 1 ,*,8 2 ,*,R), (82,0,82,0,R), (82,l,83,0,L), (82,B,84,B,L). At this point
we are either in state 8 3 , ready to go back for the next iteration, or in state 8 4 ready for some cleanup.
In the former case, we want to skip back over the nonblank symbols until we reach the start of the string,
so we add five-tuples (s3, *, s 3, *, L), (s3, 0, s3, 0, L), (s3, 1, s3, 1, L), and (s3, B, so, B, R). In the latter case,
we know that the first string is longer than the second. Therefore we want to erase remnants of the second
input string and the asterisk, and change the O's in the first input string back to l's. Here are the transitions:
(8 4 , 0, 84 , B, L), (8 4 , *, 88 , B, L), (8 8 , 0, 88 , 1, L), (s 8 , 1, s 8, 1, L). The machine halts in state ss when the blank
preceding the original input is encountered.
WRITING PROJECTS FOR CHAPTER 13
Books and articles indicated by bracketed symbols below are listed near the end of this manual. You should
also read the general comments and advice you will find there about researching and writing these essays.
1. See the chapter on generative grammars in [De2]. Lindenmeyer systems are a special kind of generative
grammar.
2. Most textbooks on programming languages should discuss this, as well as books on the specific languages
mentioned. The call number for programming languages is QA 76.7. As usual, you can find websites with this
information using a search engine; for example, search for the three keywords backus, naur, and java.
3. See [BeKa], for example.
4. One book on network protocols that discusses finite state machines is [Ho3].
5. Mehryar Mohri of the Department of Computer Science at the Courant Institute of Mathematical Sciences is
an expert in this area. See his Web page.
6. The Wikipedia article on finite-state machines is a good place to start.
7. There are several textbooks on automata theory and finite-state machines of all kinds that cover topics
such as this. Try [Brl], [Co], [DeDe], [HoUl], or [LePa], for example. These are also good sources for
supplementing the material in this chapter. This subject in general (including Turing machines, computability,
and computational complexity), is usually called "the theory of computation," and, again, there are numerous
books with essentially this title; see [Si] for a fairly recent and readable one that takes you from the beginning
to a fairly advanced level.
8. Again, there are entire books on this subject (see [Pr Du], for instance). The Game of Life was invented by the
British mathematician John H. Conway, and was the subject of several articles in Martin Gardner's Scientific
American column during the 1970s. Three of them are collected in [Ga2], which also mentions other books
and articles. See also [BeCo], which covers lots of solitaire and two-person games, as well as Life.
9. See standard references on automata theory, such as those mentioned in Writing Project 7.
Chapter 13
488
Modeling Computation
10. See standard references on automata theory, such as those mentioned in Writing Project 7.
11. Turing's first article is [Tu2], and it is actually quite readable. Keep in mind as you read it that real computers
had not yet been invented.
12. See standard references on automata theory, such as those mentioned in Writing Project 7.
13. This actually opens up the door to most of the important modern-day research in theoretical computer science.
For an elementary, nontechnical account, see [Gr2]. See the references given in Writing Project 4 for more
detail. The big question is whether deterministic Turing machines can compute functions as efficiently as
nondeterministic ones. You should definitely consult [GaJo], the classic work in this area.
14. See the references given for Writing Project 13.
15. See standard references on automata theory, such as those mentioned in Writing Project 7.
16. Searching under "lambda calculus" or ''recursive function theory" in your library should turn up a place to
start.
17. See standard references on automata theory, such as those mentioned in Writing Project 7.
18. See standard references on automata theory, such as those mentioned in Writing Project 7.
489
Appendixes
APPENDIXES
APPENDIX 1
Axioms for the Real Numbers and the Positive Integers
These facts are all things you know to be true. The point here is to prove them rigorously using the axioms
and the theorems stated or proved in this section. Throughout this section we will omit parentheses that
clarify order of operations (writing a· b + c · d rather than (a· b) + (c · d), for instance).
1. This proof is similar to the proof of Theorem 1, that the additive identity for real numbers is unique. In fact,
we can just mimic that proof, changing addition to multiplication and 0 to 1 throughout. So suppose that
1' is also a multiplicative identity for the real numbers. This means that 1' · x = x · 1' = x whenever x is a
real number. In particular, letting x = 1, we have 1 · 1' = 1. By the multiplicative identity law, we also have
1 · 1' = 1'. It follows that 1' = 1, because both equal 1 · 1'. This shows that 1 is the unique multiplicative
identity for the real numbers.
3. To show that a number equals -( x · y), the additive inverse of x · y, it suffices to show that this number plus
x · y equals 0, because Theorem 2 guarantees that additive inverses are unique. For the first part, then, we
have
(-x) · y + x · y = (-x + x) · y (by the distributive law)
= 0·y
(by the inverse law)
= y · 0 (by the commutative law)
= 0 (by Theorem 5).
The second part is essentially identical.
5. This is similar to Exercise 3. We will show that (-x) · (-y) has the property that when we add it to -(x · y)
we get 0. This will guarantee that it is the unique additive inverse of -( x · y); but because x · y is the additive
inverse of -(x · y), this means that it must equal x · y, which is what we are asked to show.
(-x) · (-y)
+ (-(x · y)) = (-x) · (-y) + (-x) · y (by Exercise 3)
= (-x) · ((-y) + y) (by the distributive law)
= (-x) · 0
= 0
(by the inverse law)
(by Theorem 5)
7. By definition, -(-x) is the additive inverse of -x. But-xis the additive inverse of x, so xis the additive
inverse of -x. By Theorem 2 additive inverses are unique, so we must have -( -x) = x.
9. This is similar to Exercise 3. We will show that -x - y has the property that when we add it to x
+y
we
get 0. This will guarantee that it is the unique additive inverse of x + y and so must equal -(x + y), which
is what we are asked to show.
(-x - y) + (x + y) = ((-x) + (-y)) + (x + y) (by definition of subtraction)
= ((-y) + (-x)) + (x+y)
(by the commutative law)
= (-y)+((-x)+(x+y))
(by the associative law)
=(-y)+((-x+x)+y)
(by the associative law)
= (-y) + (0 + y) (by the inverse law)
= (-y) + y (by the identity law)
= 0 (by the inverse law)
Appendixes
490
11. By definition of division and uniqueness of multiplicative inverses (Theorem 4) it suffices to prove that ( (w / x) +
(y / z)) · (x · z) = w · z + x · y. But this follows after several steps, using the definition that division is the same
as multiplication by the inverse, as well as the distributive law and the associative, commutative, identity, and
inverse laws for multiplication: ((w · (1/x)) + (y · (1/z))) · (x · z) = (w · (1/x)) · (x · z) + (y · (1/z)) · (x · z) =
(w·(l/x))·(x·z)+(y·(l/z))·(z·x) = w·((l/x)·(x·z))+y·((l/z)·(z·x)) = w·(((l/x)·x)·z)+y·(((l/z)·z)·x) =
w · (1 · z) + y · (1 · x) = w · z + y · x = w · z + x · y.
> 0 and y > 0, then x · y > 0. By the multiplicative compatibility law, we have
x · y > 0 · y, but by the commutative law and Theorem 5, 0 · y = 0.
13. We must show that if x
15. First let's prove a lemma, that if z < 0, then -z > 0. This follows immediately by adding -z to both sides of
the hypothesis. Now given x > y and -z > 0, we have x · (-z) > y · (-z) by the multiplicative compatibility
law. But by Exercise 3 this is equivalent to -(x · z) > -(y · z). By the additive compatibility law we can
add x · z and y · z to both sides and obtain another valid inequality. Then we apply the various laws in the
obvious ways to simplify and obtain x · z < y · z.
17. The additive compatibility law tells us that w + y < x + y and (together with the commutative law) that
x + y < x + z. By the transitivity law, this gives the desired conclusion.
19. A simple proof using Theorem 8 is given in the answer key in the textbook. For a proof from first principles,
we can use the completeness property and a proof by contradiction. Suppose that n · x ::;: 1 for all positive
integers n. Then the set S = { n · x I n is a positive integer} is bounded above and therefore has a least
upper bound; call it b. Now for all positive integers n we have (n + 1) · x::;: b, so n · x + x::;: b. This tells us
that n · x ::;: b - x, which says that b - x is also an upper bound for S, contradicting the definition of least
upper bound. Therefore it must be true that n · x > 1 for some positive integer n.
21. The proof practically writes itself if we just use the definitions. We want to show that if (w, x)
rv (
w', x')
and
(y, z) rv (y', z')' then (w+y, x+z) rv (w'+y', x'+z') and that (w·y+x·z, x·y+w·z) rv (w'·y'+x'·z', x'·y'+w'·z').
Thus we are given that w + x' = x + w' and that y + z' = z + y' , and we want to show that w + y + x' + z' =
x + z + w' + y' and that w · y + x · z + x' · y' + w' · z' = x · y + w · z + w' · y' + x' · z' (note that grouping does not
matter, because of the associative law). For the first of the desired conclusions, add the two given equations.
For the second, rewrite the given equations as w - x = w' - x' and y - z = y' - z', multiply them, and do
the algebra.
Appendixes
491
APPENDIX 2
Exponential and Logarithmic Functions
This material should all be familiar from high school algebra. When working with exponents it is important to
remember the rules bx by = bx+y and (bx)Y = bxy. When working with logarithms it is important to remember
= logb x + logb y and logb xY = y logb x .
a) We use the facts that 21 = 2 and 2x+y = 2x · 2Y. Thus we have 2 · 22 = 21 . 22 = 21+ 2 = 23 .
that logb (xy)
1.
b) We use the second part of Theorem 1 to write (2 2 ) 3
= 2C 2 ·3 ) = 26 .
c) There is no way to use a rule of exponents to simplify this. However, since 22
=
4, we can write this as 24 .
= (log4 x)/(log 4 2) = (log4 x)/(1/2) = 2log4 x = 2y.
b) This is just like part (a). We have log8 x = (log 4 x)/(log4 8) = (log4 x)/(3/2) = (2 log4 x)/3 = 2y/3.
c) This is just like part (a). We have log 16 x = (log 4 x)/(log4 16) = (log 4 x)/2 = y/2.
3. a) We use Theorem 3. Thus log 2 x
5. We can draw these graphs by plotting points, following the general shape shown in Figure 1. For part (a) the
graph is similar to the graph of f(x) = 2x. For part (b) the graph is similar to the graph of J(x) = (1/2)x.
Finally for part ( c) the function is constant, since ix = 1 for all x.
8
6
4
2
-2 -1
0
(a)
APPENDIX 3
2
-2 -1
0
(b)
2
-2 -1
0
2
(c)
Pseudocode
Obviously the text cannot give an entire course in programming in just a few pages. One nice thing about
pseudocode is that it is readable even when the reader has no specialized knowledge of the conventions being
followed.
1. In the first case, the value of a is first set to the value of b, and then the value of b is changed. Thus the final
value of a is the same as the original value of b. In the second case, the value of b is changed first (to the
original value of c), then this new value is assigned to a. Thus the final value of a is the same as the original
value of c.
3. According to the definition, the way this loop is executed is that first i is assigned the initial value . Then
as long as i is less than or equal to final value, the statement in the loop is executed, and the value of i is
incremented by 1. Thus the code is equivalent to the following code.
i := initial value
while i ::::; final value
statement
i := i + 1
A Guide to Proof- Writing
492
A Guide to Proof-Writing
by Ron Morash, University of Michigan-Dearborn
At the end of Section 1.8, the text states, "We have not given a procedure that can be used for proving theorems
in mathematics. It is a deep theorem of mathematical logic that there is no such procedure." This is true, but does
not mean that proof-writing is purely an art, so that only those with exceptional talent and insight can possibly
write proofs. Most proofs that students are asked to write in elementary courses fall into one of several categories,
each calling for a systematic approach that can be demonstrated, imitated, and eventually mastered. We present
some of these categories and techniques for working within them, organized as follows. This material supplements
that found in the text and is intended to help get you started creating your own proofs. Also, studying the material
in this Guide will help you understand better the proofs you read.
1. Deducing conclusions having the form "For every x, if P(x), then Q(x)."
1.1. Direct proof
1.1.1. Propositions having no hypothesis
1.1.2. Propositions having one or more hypotheses
1.1.3. Disproving false propositions having conclusions of the form "\ix[P(x)-+ Q(x)]"
1.1.4. The tactic of division into cases
1.1.5. Proving equality of sets
1.2. Indirect proof
1.2.1. Proof by contraposition
1.2.2. Proof by contradiction
1.2.3. Deriving conclusions of the form "q or r"
2. Remarks on additional methods of proof
2.1. Deducing conclusions having the form "For every x, there exists y such that P(x,y).''
2.2. Proof by mathematical induction
1.
Deducing conclusions having the form "For every x, if P(x), then Q(x)."
Many defining properties in mathematics have the form Vx[P(x) -+ Q(x)], representing the idea "All P's are
Q's." (Cf. Examples 23, 26, and 27 in Section 1.3 of the text.) Some definitions involving this form are:
(i) A set A is a subset of a set B: In symbols, A~ B if and only if Vx[(x EA) -+ (x EB)]. This is read in
words, "A is a subset of B if and only if, for every x, if x E A, then x E B." Less formally, A is a subset of B if
and only if every element of A is also an element of B. (Cf. Definition 3 in Section 2.1 of the text.)
(ii) A function f is one-to-one: f is one-to-one if and only if, for every x 1 and x 2 in the domain of
f(xi) = f(x 2 ), then x 1 = x2. (Cf. Definition 5 in Section 2.3 of the text.)
(iii) A relation Ron a set A is symmetric: R is symmetric if and only if, for every x, y EA,
f, if
if (x, y) ER, then
(y, x) E R. (Cf. Definition 4 in Section 9.1 of the text.)
Many mathematical propositions that students are asked to prove have as their conclusion a statement involving
a definition of the form just described. Some examples are:
(a) Prove that for all sets A and B, A~ AU B.
(b) Prove that for all sets X, Y, and Z, if X
~
Y, then X
nZ
~
Y
n Z.
(c) Prove that for every function f whose domain and codomain are subsets of the set of real numbers, if f is
strictly increasing (cf. Definition 6 in Section 2.3), then f is one-to-one.
(d) Prove that for all relations R 1 and R 2 on a set A, if R 1 and R 2 are symmetric, then the relation R 1
is symmetric.
n R2
Note that the desired conclusion in each of the propositions (a)-(d) is a statement involving one of the definitions
(i)-(iii). Furthermore, propositions (b), (c), and (d) each have a hypothesis, a statement we are allowed to assume
true and whose assumed truth, presumably, will play a role in deriving the conclusion. We begin our study of
proof-writing methods by considering the very broad category known as direct proof.
A Guide to Proof- Writing
493
Direct proof
1.1.
An argument in which we prove a proposition in its originally-stated form is called a direct proof. Some forms
of direct proof are discussed in Sections 1.6 and 1. 7 of the text. In the sections of this Guide that follow, we present
various techniques for creating direct proofs. Attempting to write a direct proof of a proposition is usually our first
line of attack. Direct proof contrasts with indirect proof, in which we prove a proposition by proving a different,
but logically equivalent, form of the original proposition. We will introduce indirect proofs later, but will focus, in
Examples 1-11, on various approaches to direct proofs.
1.1.1.
Propositions having no hypothesis
Direct proofs of propositions like (a), having no hypothesis, tend to be simpler in their structure than the
proofs that are required for propositions (b)-(d). Examples 1 and 2 demonstrate proofs for this simpler case.
Example 1. Prove Proposition (a): For all sets A and B, A~ AU B.
Solution. The proof proceeds as follows: Let A and B be arbitrary sets. To prove A ~ AU B, let x be
an arbitrarily chosen element of A. [Note: We are assuming that x E A.] We must prove that x E AU B. By
the definition of "union," this means we must prove that either x E A or x E B. Since we know x E A, by our
assumption, the desired conclusion x E A or x E B follows immediately. I
Let's dissect the proof in Example 1 and analyze what we did. Our starting point "
assume x E A
"
is an application of one of the most widely-used approaches to proof-writing, known as the choose method. The
basic approach to deriving a conclusion of the form \fx[P(x)----> Q(x)J, is to begin by choosing an arbitrary object
(giving it a specific name such as "x") for which it is assumed that P(x) is true. Our goal is to deduce that Q(x)
must consequently also be true. In Example 1, P(x) is the assumption "x EA" and Q(x) is the desired conclusion
"x E AU B." We make the following additional observations:
•The object xis a fixed, but arbitrarily chosen, element of the universe of discourse of P(x) and Q(x). We
do not assign any specific value to x; rather we give the name "x" to a generic object [that is assumed to satisfy
the propositional function P(x)] and use that name to keep track of the object as we proceed through the steps of
the proof. The power of this approach is that any conclusion we draw about "this x" applies to every object a for
which the assumption P(a) is true. This is valid by the rule of universal generalization; see Table 2 of Section 1.6.
•In the first part of a proof of a conclusion of the form \fx[P(x)----> Q(x)], called the "setting-up" of the proof,
we choose x, assume P(x) is true, and then write out what it would mean for Q(x) to be true (in our example, "...
to prove x EAU B, we must prove that either x EA or x EB"). Learning the process of setting up a proof in this
category provides a fairly standardized, predictable, and almost mechanical beginning of a prospective direct proof.
Furthermore, once we have written out these details, the remainder of the proof-the path from the assumption
P(x) to the desired conclusion Q(x)-is sometimes obvious.
•In the proof in Example 1, the path from what we assumed (i.e., x EA) to the conclusion (i.e., x EAU B)
was obvious. In our proof, we stated that the conclusion "follows immediately." But was there something more
than just "common sense" to justify that conclusion? Yes! The rule of inference p----> p V q ("Law of Addition") is
the underlying logical tool that justifies this step. This rule and other rules of inference are stated in the text, in
Table 1 of Section 1.6. In the sample proofs that follow, we will make explicit reference to the rules of inference
used (in an increasingly less obvious way as the proofs become more complex), even though it is common practice
to apply these rules only implicitly, that is, without specific mention. To become proficient at writing proofs, you
need to know how to use these rules of inference and when to use them.
Example 2. Prove that for all sets X and Y, X n (YU X)
Discussion. Let X and
of X n (YU X). [We could
setting-up of the proof. Now
begin by analyzing what our
~
Y.
Y be arbitrary sets. To prove X n (YU X) ~ Y, let a be an arbitrarily chosen element
also say "Assume a E X n (YU X)."] We must prove a E Y. [This concludes the
we must figure out how to get from the assumption to the conclusion. To do that, we
assumption means.]
Since a E X n (YU X), we know that a E X and a E YU X. The latter, in turn, tells us that either a E Y
or a EX, that is, either a E Y or a t:J. X. [Note that the preceding sentence makes the first mention of the desired
494
A Guide to Proof- Writing
conclusion a E Y.] Now, can we infer the conclusion a E Y from the known "either a E Y or a tJ_ X"? This would
require a rule of inference "(p V q) ___. p" (the converse of the Law of Addition). This is not a valid inference since
(p V q) ___.pis not a tautology, so this approach does not work. Note, however, that not only do we know "either
a E Y or a t/:- X" from our assumption, but we also know that a E X. Thus, what we know from our assumption
has the form (p V q) /\ •q. [Note: p is "a E Y" and q is "a tf:_ X," so •q is "a E X."] Does Table 1 in Section 1.6
give us a conclusion that follows from this premise? It does! The Law of Disjunctive Syllogism, [(p V q) /\ •q] ___. p,
enables us to draw the conclusion p, that is, a E Y, the desired conclusion. [Note: As stated in Table 1, the roles
of p and q are reversed from what we have here, but that is of no consequence.] I
Before moving on, we rewrite the preceding proof, leaving out explanatory comments. What remains provides
a representative view of what a typical proof looks like:
"Let X and Y be arbitrary sets. To prove X n (YU X) c:;; Y, assume a EX n (YU X). We must prove a E Y.
By our assumption, we know a EX and a E YU X; therefore a EX and either a E Y or a EX. Thus we know
that either a E Y or at/:- X; but we also know that a EX, so at/:- Xis false. Hence we conclude a E Y, as desired."
At this point, you may wish to try some relevant exercises in the text, such as Exercises 16(a,c) and 18(a,b,c)
in Section 2.2. You should find the principles from Examples 1 and 2 helpful in attempting these exercises.
1.1.2.
Propositions having one or more hypotheses
As we work through the steps of a prospective proof, the tools at our disposal in moving toward a desired
conclusion are
• the assumption(s) we are entitled to make at the outset in setting up the proof,
• assumed axioms and previously-proved theorems (if any), and
•rules of inference from logic, such asp - t (pV q), used in Example 1, and [(pV q) /\ •q]
(See Table 1 in Section 1.6 for additional rules of this type.)
-t
p, used in Example 2.
In addition to these, most propositions we are asked to prove contain
• one or more hypotheses, statements whose truth is to be assumed in the proof and which, we expect, will be
used as part of the argument leading to the conclusion.
Example 3 provides our first instance of a proposition in which a hypothesis is to be assumed in deriving a
conclusion of the form \fx[P(x) ___. Q(x)].
Example 3. Prove Proposition (b): For all sets X, Y, and Z,
if X c:;; Y, then X n Z c:;; Y n z.
Proof. Let X, Y, and Z be sets such that X c:;; Y. To prove X n Z <:;;; Y n Z, assume b EX n Z. To prove
b E Y n Z, we must prove b E Y and b E Z. [This marks the end of "setting up the proof." Now we must return
to our assumption and the hypothesis, and begin to analyze what they mean and what information we can draw
from them.] By our assumption, we know that b E X and b E Z, so, in particular, b E Z, one of our two desired
conclusions. Furthermore, since b E X (part of our assumption) and since X c:;; Y [here we are, for the first time,
bringing in the hypothesis], we may conclude that b E Y, our other desired conclusion. I
The following feature of the proof in Example 3 is very important. In setting up the argument at the outset,
we applied the choose method to the desired conclusion, not the hypothesis. Thus our initial statement was "...
assume b E X n Z." A common mistake by beginning students is to begin with " ... assume b E X ... ," erroneously
focusing at the start of the proof on the hypothesis rather than on the desired conclusion. Note that we did not
employ the hypothesis until the very end of the proof!
In the last sentence of the proof in Example 3, we concluded b E Y from knowing b E X and X <:;;; Y. Let us
consider why this conclusion is justified. The truth of X <:;;; Y means that the proposition \fx[(x EX) ___. (x E Y)] is
true. Thus, in particular, the proposition (b E X) ___. (b E Y) is true, where b is the specific object we are working
with in the proof. Since b EX is true and the "if... then" statement (b E X) ___. (b E Y) is true, the truth of b E Y
follows from the rule of inference modus ponens (cf. Table 1 in Section 1.6 of the text). Note, once again, that a
rule of inference has played an important, though implicit, role in a proof!
A Guide to Proof- Writing
495
The principles discussed thus far apply to every proof of a proposition whose conclusion has the logical form
\fx[P(x) ---+ Q(x)], and not just to proofs that one set is a subset of another. Examples 4 and 5 illustrate this.
Example 4. Prove that every nonconstant linear function f(x)
= li1x +
B, li1 =/= 0, is one-to-one.
Proof. Let Jl.,f be a nonzero real number. Let x 1 and x 2 be real numbers and assume that f(x1) = f(x 2). We
must prove that x1 = x2. Since f(xi) = Mx1 +Band f(x2) = li1x2 + B, we have J\,fx1 + B = lvlx2 + B. By a rule
of elementary algebra, if Mx 1 + B = li1x 2 + B, then li1x 1 = Mx 2. Since li1x 1 = lifx 2 and M =/= 0, by hypothesis,
we conclude by another rule of elementary algebra that x 1 = x 2 , as desired. I
Example 5. Prove Proposition (d): For all relations R 1 and R 2 on a set A, if R 1 and R 2 are symmetric, then
the relation Ri n R2 is symmetric.
Proof. Let A be an arbitrary set and let R 1 and R 2 be symmetric relations on A. To prove that the relation
Ri n R2 is symmetric, let x and y be arbitrary elements of A and assume that (x, y) E R 1 n R 2. We must prove
(y,x) E Ri nR2, that is, (y,x) E Ri and (y,x) E R 2. [End of set-up!] Now since (x,y) E R 1 nR 2 (by assumption),
we know that (x,y) E R 1 and (x,y) E R 2. Since (x,y) E R 1 and R 1 is symmetric (by hypothesis), (y,x) E R 1. This
is one of our desired conclusions. Since (x, y) E R 2 and R 2 is symmetric (by hypothesis), (y, x) E R 2, the second of
our two desired conclusions. With this, the proposition is proved. I
At this point, you may want to practice applying the principles from Examples 3-5 in the following exercises:
•Prove that for all sets A and B, if An B =A, then A<:;;; B.
•Prove that for all sets A, B, and C, if A<:;;; Band B
the text.)
<:;;;
C, then A<:;;; C. (This is Exercise 17 in Section 2.1 of
•Prove that the function g(x) = x 3 is one-to-one.
1.1.3.
Disproving false propositions having conclusions of the form \fx[P(x)---+ Q(x)]
Sometimes we are faced with a proposition that we must either "prove or disprove." We are not told in advance
whether the proposition is true. If it is false, then it will of course be impossible to write a correct proof of the
proposition. Time spent trying to do so may provide insight, but cannot ultimately lead to a valid proof. Example 6
illustrates how we should approach this type of problem.
Example 6. Prove or disprove the converse of Proposition (b): For all sets X, Y, and Z,
then X <:;;; Y. (Cf. Example 3.)
li X n Z
<:;;;
Y n Z,
Discussion. Suppose we try first to approach this proposition in the manner of previous examples. Our setting
up of a "proof" would read as follows: "Let X, Y, and Z be arbitrary sets such that X n Z <:;;; Y n z. To prove
X <:;;; Y, let w E X; we must prove w E Y." At this point, we must return to the hypothesis X n Z <:;;; Y n Z and
ask whether, in combination with the assumption w E X, it leads to the conclusion w E Y. If we could get w to
lie in X n Z, then we could invoke the hypothesis to conclude w E Y n Z, which would imply w E Y, the desired
conclusion. However, we know from our assumption only that w E X; in order to conclude w EX n Z, we would
need to know that w E Z, which we do not!
With this, our attempt to write a direct proof breaks down, leaving us with two possibilities. Either there is
another route to a proof, or perhaps, the proposition we are trying to prove is false. If we do not know whether a
general proposition is true or false, and our initial attempts at a proof fail, we should do some experimenting to
see whether we can find a counterexample, i.e., a specific example that contradicts the truth of the proposition.
Before we can do that, we must formulate precisely the negation of the proposition. Logically the negation of
a proposition "for every x, if P(x), then Q(x)" is "there exists x such that P(x) but not Q(x)." In symbols,
-Nx[P(x) ---+ Q(x)] is equivalent to 3x[P(x) /\ -iQ(x)]. In this example, the negation is "there exist sets X, Y,
and Z such that X n Z <:;;; Y n Z, but X is not a subset of Y.'' Can we find specific sets X, Y, and Z satisfying
this statement? Consider the sets X = {4, 7, 8, 11 }, Y = {2, 7, 8, 9, 11 }, and Z = {l, 7, 8, 9, 10, 11, 12}. Note that
X n Z = {7, 8, 11} and Y n Z = {7, 8, 9, 11}, so X n Z <:;;; Y n Z. However Xis clearly not a subset of Y. Hence we
have a counterexample; the proposition in question is false! I
Note that a single counterexample to a general proposition is sufficient to prove that proposition false. This
is a far cry from what is required to prove a general proposition true, when the domain (universe of discourse) is
A Guide to Proof- Writing
496
infinite. Since we can never exhaust all the possible examples, no number of specific cases that affirm a proposition
are sufficient to establish its truth in general. We must write a general proof in order to do that-the process of
writing such proofs is the major topic you are now studying, and one to which we will return shortly.
In the last sentence of the Discussion of Example 6, we stated that "X is clearly not a subset of Y ." How do
we justify this statement formally? Recall that "X ~ Y" is defined by "'v'w[(w E X) --+ (w E Y)]." Hence the
proposition ":::Jw[(w EX)/\ (w r:J_ Y)]" corresponds to "Xis not a subset of Y." For the sets X = {4, 7, 8, 11} and
Y = {2, 7,8,9, 11}, we note that, choosing w = 4, we have 4 EX, but 4 r:J_ Y. This particular example is all that
is needed to conclude that X is not a subset of Y. (Note incidentally that the choice of w used to prove that X is
not a subset of Y, w = 4, has the property that w r:J_ Z. This is not surprising since, in our initial attempt to prove
the false proposition in Example 6, the obstacle we could not overcome was that our arbitrarily chosen w did not
need to lie in Z.)
The following exercises are germane to the issues raised by Example 6 and the paragraphs following it.
• Prove or disprove: For all sets X, Y, and Z, XU (Y n Z)
~
(XU Y) n z.
•Prove or disprove: For all sets X, Y, and Z, if X ~ Z, then XU (Y n Z) ~ (XU Y) n Z.
1.1.4.
The tactic of division into cases
As propositions we are asked to prove become more complex, we must expand our arsenal of tools that are
effective for proceeding toward the desired conclusion of a proposition, once we have finished setting up the argument.
The applications of the choose method that occur in Examples 7 and 8, which follow, demonstrate a new such tool,
known as division into cases, which is useful in some proofs.
Example 7. Prove that for all sets A and B, (An B) U (An B)
~
A.
Proof. Let A and B be arbitrary sets. To prove (An B) U (An B) ~A, assume x E (An B) U (An B). We
must prove x E A. By our assumption, we know that either x E An B or x E An B, that is, either x E A and
x EB, or else x EA and x E B. [Note: We don't know which of these two is the case, but we do know that at least
one of them must be true.] Hence, at this point, we divide the argument into two exhaustive cases:
Case I Suppose that x E A and x E B. Then, in particular, x E A [by the rule of inference (p /\ q)
the desired conclusion obtains in this case.
Case II
--+
p], so
Suppose that x E A and x E B. Then, again, x E A, so the desired conclusion is again verified.
Under either of the only two possible cases, we have x E A, the desired conclusion. I
Example 8. Prove that for all sets A and B, A~ (An B) U (An B).
Proof. Let A and B be arbitrary sets. To prove A ~ (An B) U (An B), assume x E A. We must prove
x E (An B) U (An B). To do this, we must prove that either x E An B or x E An B, that is, either x E A and
x E B, or else x E A and x E B. [Recall that our assumption is that x E A. This assumption involves only the
set A. The problem we must solve is how to bring the relationship between x and the set B into the discussion.]
We note that, necessarily, either x E B or x r:J_ B, by the tautology p V •p. Having noted this, we consider two cases:
Case I Suppose that x E B. Then, since x E A [by our assumption], we have x E A and x E B, one of the
two alternatives in our desired conclusion.
Case II Suppose that x r:J_ B. Equivalently, x EB. Then, since x EA, we have x EA and x EB, the other of
the two alternatives in our desired conclusion. I
Compare the proofs in Examples 7 and 8, both involving division into cases. The second proof illustrates
somewhat more creativity than does the first. It requires a slightly more active role on our part, involving the
tautology p V •p. The truth of this tautology is "common sense." Once the idea of bringing this division into cases
into the argument is suggested, virtually anyone would agree that it is correct and, furthermore, is an effective step
at this stage. The difficult part for you, as a beginning student just learning to write proofs, is thinking of this idea
on your own.
A Guide to Proof- Writing
497
Many complex proofs require that some creative idea be brought in from outside the basic structure (i.e., the
setting up) of the argument. This is the aspect of proof-writing that is not mechanical. It is learned from experience
and by an active interest in the "why" in mathematics. It is fostered by developing the habit of having firmly in
mind all the statements, pertaining to the problem at hand, that we know to be true, and by being willing to try
to apply these statements until we find one that works.
Here are some exercises that involve the choose method and division into cases:
•Prove that for all sets X, Y, and Z, if X <:;;Zand Y <:;; Z, then XU Y <:;; Z.
•Prove that for all sets A, B, and C, if A<:;;; B, then AU C <:;;;BU C.
•Prove that for all sets X, Y, and Z, if X n Z ~ Y n Zand X n
1.1.5.
Z
~Y n
Z,
then X ~ Y.
Proving equality of sets
Three approaches to proving equality of sets are discussed in Examples 14-18 in Section 2.2 of the text. Of
these, the first, known as mutual inclusion (introduced in Example 10), is the most generally applicable. In addition,
it expands naturally on the earlier material in this Guide, so we give that approach additional emphasis here.
A formal version of the definition of equality of sets given in the text (cf. Definition 2 in Section 2.1) is
A= B
if and only if \ix[(x EA)+---+ (x EB)].
The latter proposition is equivalent to
\fx{[(x EA)
--->
(x E B)] /\ [(x E B)
--->
(x E A)]},
which, in turn, is equivalent to
{\ix[(x E A)
--->
(x E B)]} /\ {\ix[(x E B)
--->
(x E A)]},
which is the definition of
A<:;;; B
and
B
~A.
We may prove that two sets are equal by proving that each is a subset of the other. We illustrate this approach
in Examples 9 and 10.
Example 9. Prove that for all sets A and B, (An B) U (An B) =A.
Proof. We may prove the desired equality by proving mutual inclusion, i.e., that each of the two sets is a
subset of the other. The inclusion (AnB)U(AnB) <:;;;A, however, is precisely what we proved already in Example 7.
The other inclusion A ~ (An B) U (An B) was proved in Example 8. Having written these two proofs, we have
established the desired equality. I
In Example 10, we encounter a conclusion of equality that is preceded by a hypothesis.
Example 10. Prove that for all sets A and B, if A~ B, then AU B = B.
Proof. Let A and B be arbitrary sets such that A<:;;; B. We may prove AU B
AUB<;;;B.
=B
by proving B <:;;;AU Band
1. B <:; ; AU B is essentially Proposition (a), proved earlier in Example 1. [Note that the hypothesis of the
theorem is not required to establish this inclusion. Like Example 6 in Section 1. 7 of the text, the proof in this
direction is trivial.]
2. To prove AU B <:; ; B, given the hypothesis A<:;;; B, we proceed by the choose method. Assume x EAU B.
We must prove thatx EB. By our assumption, we know that either x EA or x EB. Since we do not know which
of these two statements is true, we divide the argument into cases:
Case I Suppose that x EA. Then, since A<:;;; B, by hypothesis, we have x EB, as desired. [Recall the middle
paragraph of the discussion between Examples 3 and 4, on the role of modus ponens.]
Case II
Suppose that x E B. Since this is the desired conclusion, the result is trivially true in this case.
A Guide to Proof- Writing
498
We conclude x EB, as desired. I
In some circumstances, it is possible to prove set equality using a single chain of valid equations, thus avoiding
the sometimes cumbersome mutual-inclusion approach. One such circumstance is in proving any of the set identities
in Table 1 in Section 2.2 of the text. Each of these results is a set-theory version of a corresponding equivalence of
propositions in logic, covered in Section 1.3. Each can be proved using the approach of Example 12 in Section 2.2
of the text.
Another approach to proving an equality of sets is to use other equalities proved previously. Suppose the
following identities of set theory have already been proved:
[1]
ForallsetsA, B, andC, An(BuC)=(AnB)u(AnC).
For every set B, BUB= U
(U represents the universal set.)
[2]
[3]
For every set C, C n U = C.
On these bases, we can give a proof of the proposition proved in Example 9 that does not use mutual inclusion.
Example 11. Prove that for all sets A and B, (An B) U (An B) =A.
Proof. Let A and B be arbitrary sets. Then we have
(An B) u (An B) = An (Bu B)
=A nu
=A,
(by [1])
(by [2])
(by [3])
as desired. I
Note that, in applying the result [1], we used the special case C = B of the identity [1] (which is the property
of distributivity of intersection over union-see Table 1 in Section 2.2 of the text). The technique of using a special
case of a known result is called specialization. Whenever you find yourself saying "in particular," in making an
inference from a known general fact in the course of an argument, you are using the specialization tactic. Like
division into cases, specialization is a sometimes-useful technique for proceeding beyond the initial setting up of a
proof toward the desired conclusion. Formally, it is justified by the rule of universal instantiation, shown in Table 2
of Section 1.6.
Indirect proof
1.2.
Sometimes it is convenient, or even necessary, to prove a form of a proposition that is different from the original,
but logically equivalent to it. Whenever we write a proof in such a form, we are writing an indirect proof. Three
common forms of indirect proof are based on three logical equivalences of pairs of propositions:
• (•q) __, (•p) is equivalent to p __, q
[4]
• ( •p) __, (q /\ •q) is equivalent to p
[5]
• (p /\ •q) __, r is equivalent to p __, (q V r).
[6]
The equivalence [4] is the basis for the form of an indirect proof known as proof by contraposition. The
equivalence [5] justifies the form of an indirect proof called proof by contradiction. The equivalence [6] underlies a
standard approach to deriving a conclusion involving alternatives, i.e., having the form "either q or r."
1.2.1.
Proof by contraposition
Sometimes it is difficult to see how to prove a proposition of the form 'v'x[P(x) __, Q(x)] by starting with the
assumption that P(x) is true. (Recall Section 1.1.2 of this Guide.) What can you do if you cannot see how to
deduce the conclusion Q(x) from the assumption P(x) and any additional given hypotheses (if there are any)?
Sometimes, in such cases, assuming the negation •Q(x) of the conclusion provides a better match with known facts
or the other given hypotheses, and the two together lead readily to the negation •P(x) of the original assumption.
An argument in this form is an instance of proof by contraposition. See Example 3 in Section l. 7 of the text for an
example of such a proof. Our Example 12 provides another illustration of the method.
A Guide to Proof- Writing
499
Example 12. Prove Proposition (c): For every function f whose domain and codomain are subsets of the set
of real numbers, if f is strictly increasing, then f is one-to-one.
Discussion. Let f be any function that is strictly increasing. To show that f is one-to-one using the original
form of the definition, we would let x 1 and x 2 be real numbers in the domain off and assume f(xi) = f(x 2). We
would then have to prove x1 = x2. We have completed the setting up of a direct proof, but have no way of using the
hypothesis that f is strictly increasing [i.e., " ... if x 1 < x2, then f(x 1) < J(x 2)"]. The assumption f(xi) = f(x 2)
simply does not "match up" with the "if part" of the hypothesis in a way that permits us to proceed anywhere
from that hypothesis.
However, suppose we decide instead to derive the contrapositive of the definition of one-to-one. Under this
approach, we will begin by assuming that x1-:/=- x 2 . Our goal will then be to prove that f(x 1)-:/=- f(x 2). We proceed
from here as follows. Since x 1 -:/=- x2 (by assumption), then it must be that either x1 < x2 or x2 < x1. We consider
two cases:
Case I If X1 < x2, then since f is strictly increasing, we may conclude f(xi)
f(x1) -:/=- f(x2), as desired.
< f(x 2), so, in particular,
Case II If x 2 < x 1 , then since f is strictly increasing, we may conclude f(x 2) < f(x 1), so f(x 1) -:/=- f(x 2),
again as desired. I
Another situation in which a proof by contraposition is sometimes called for is one in which the conclusion
of the proposition has a much simpler logical form than does the hypothesis. We illustrate this in the following
example.
Example 13. Suppose that a is a real number satisfying the property 'VM>O (/al < M). Then a= 0.
Discussion. Note the simple form of the conclusion. Rather than involving a definition having an "if. .. then"
form, as in most of our earlier examples, it is simply the flat statement that a = 0. If we try to begin a direct
proof by focusing on the desired conclusion, then there is really no place to begin, no basis for making the kind of
assumption that is needed to get the proof "off the ground."
So instead we proceed by contraposition. Our approach will be to assume a-:/=- 0 and try to deduce the negation
of the hypothesis. This negation may be formulated (cf. Table 2 in Section 1.4 of the text)
[7]
:JM>O (/a/ ::::: M).
We need only produce a positive M whose value does not exceed that of the absolute value of the nonzero a. We
take M = iai, noting that this value of M clearly satisfies the condition [7]. I
Here is a third circumstance in which a proof by contraposition is appropriate. Suppose a proposition of the
form "if p and q, then r" is known to be true, and we are asked to prove that p and the negation of r together
imply the negation of q. We may always proceed, using contraposition, by assuming that the negation of q is false,
that is, that q is true. Then since p is true, by hypothesis, we have that p and q are both true, so, by the known
proposition, we may conclude that r is true, contradicting the fact that -ir is one of the hypotheses. You will have
an opportunity to apply this approach in the third and fourth of the exercises that follow.
•Prove that for all sets A and B, if A<;;; B, then B <;;;A.
• Prove that if a linear function f (x)
= M x +B
is one-to-one, then M -:/=- 0.
• Suppose it is known that "every sum or difference of two integers is an integer." Use this result to prove that
for all real numbers x and y, if x is an integer and x + y is an integer, then y is an integer. Prove also that the sum
of an integer and a noninteger must be a noninteger.
•Prove that for all sets A and B and for every object x, if x EA and x
1.2.2.
tJ_
An B, then x
tJ_
B.
Proof by contradiction
The idea behind the equivalence [5] is that we may prove a conclusion p by showing that the denial of p leads
to a contradiction. Actually, proof by contraposition is a form of proof by contradiction. For if, in a proof that p
implies q, we assume the truth of p (as we are entitled to do) and then use the negation of q to derive -ip, then
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A Guide to Proof- Writing
we have obtained the contradiction p /\ -.p. Another circumstance in which proof by contradiction is the standard
approach is any proof of a theorem in set theory in which the conclusion asserts that some set equals the empty set.
Example 14. Prove that for all sets A and B, if
A~
B, then An B = 0.
Discussion. A direct approach would be to establish the equality A n B = 0 using mutual inclusion. Indeed
the containment in one direction, 0 ~ A n B, is true automatically, based on the principle that the empty set is
a subset of every set (this result is vacuously true-recall Example 5 in Section 1.7 of the text). However, for the
containment in the other direction, An B ~ 0, the approach "assume x E An B ... we must prove x E 0" is
doomed to failure, since the conclusion "x E 0" can never be reached.
Since we are unable to write a direct proof, we proceed by contradiction. Let A and B be arbitrary sets such
that A ~ B. Assume that An B # 0. Then there exists some object that lies in An B; let us call it c. Since
c EA n B, we know that c EA and c EB. Since c EA and A~ B, by hypothesis, we have c EB. Thus we have
c EB and c EB, soc EB and c tf_ B. This is a contradiction of the form p /\ -.p, so our proof is complete . I
A classic example of a proof by contradiction is provided in the text in Example 10 of Section 1.7, which shows
that y'2 is irrational. Here are some exercises:
•Prove that for every set A, An
A= 0.
• Prove that for all sets A and B, if (B n A) U (B n A)
= B, then A= 0.
• Prove that for all sets A and B, (AU B) n (AU B) n (AU B) n (AU B)
1.2.3.
= 0.
Deriving conclusions of the form "q or r"
The equivalence [6] becomes relevant to the writing of proofs when we must derive a conclusion involving
alternatives, q or r. For a proposition of this type, there may be no circumstance under which we can be sure
which of the alternatives is true, only that at least one of them must be true under every circumstance in which
the hypothesis is true. Because of this, we are unable to determine whether to set up a direct proof based on the
conclusion q or on the conclusion r. (Indeed, usually no such proof is possible.) Fortunately there is an indirect
approach, based on the equivalence [6], that enables us to get around this difficulty. Rather than attempting to
prove directly that q or r follows from p, we may replace this problem by the problem of showing that r follows
from p and -.q '(or else that q follows from p and -.r-either approach will do the job).
A classic example of a proof in this category is the following theorem from elementary algebra: "For all real
numbers x and y, if xy = 0, then x = 0 or y = 0." Clearly we should set up this proof by letting x and y be real
numbers such that xy = 0. But at this point there is no evident way of proceeding toward the conclusion that one
or the other of x and y (we know not which) equals 0. The escape is to make the additional assumption that x # 0,
with the goal of proving that therefore y must equal 0. Since x # 0, its reciprocal 1/x must exist, and we may
write the chain of equations y = 1 · y = [(1/x)(x)](y) = (1/x)(xy) = (1/x)(O) = 0, soy= 0, as desired. [Note that
this chain of equations also uses the facts that multiplication of real numbers is associative and that the product of
every real number with 0 equals 0.]
A problem in set theory in which this approach is sometimes useful is proving that one set is a subset of the
union of two other sets. This is demonstrated in Example 15.
Example 15. Prove that for all sets A and B, A~ BU (An B).
Proof. Let A and B be arbitrary sets. To prove A~ BU (An B), assume that x E A. We must prove that
x E BU (An B), that is, either x E B or x E An B. Since our desired conclusion is now seen to have the form
"either q or r," we take the approach suggested by the equivalence [6], and assume that x tf_ B. Our goal now
becomes to prove that, on the basis of this additional assumption, it must be true that x E An B, that is, x E A
and x EB. We already know x EA, by our initial assumption in the proof. As for x EB, that follows immediately
from our additional assumption x ¢:. B. •
If a desired conclusion has more than two alternatives, the strategy suggested by [6] is generalized as follows:
Assume the negation of all but one of the alternative conclusions and, on that basis, try to prove that the remaining
one must be true. We illustrate this in Example 16.
A Guide to Proof- Writing
501
Example 16. Prove that for all sets A and B, if Ax B =Bx A, then either A= 0 or B = 0 or A= B.
Sketch of Proof. [Note first that the notation Ax B refers to the cartesian product of sets A and B, defined
as the set of all ordered pairs (a, b), where a EA and b EB. This definition is the basis of the content of Chapter 6
of the text, on relations.] Let A and B be any sets such that Ax B = B x A. Assume further that A -1- 0 and
B -1- 0. With these two additional assumptions, our goal then becomes to prove that the third alternative A = B
must be true. The remainder of this proof is left as an exercise. I
The following exercises provide the opportunity to use the strategy suggested by the equivalence [6].
• Complete the proof in Example 16.
•Prove that if A, B, and Care any sets such that Ax B =Ax C, then either A= 0 or B
•Prove that if A and B are any sets such that Ax B
•Prove that for all sets X, Y, and Z, (XU Y) n Z
2.
= 0,
~XU
then either A= 0 or B
(Y
= C.
= 0.
n Z).
Remarks on additional methods of proof
Not all propositions we may wish to prove have conclusions involving the form \lx[P(x) ---. Q(x)]. Nonetheless,
beginning students who are able to write correctly the proofs called for in the exercises in Section 1 of this Guide
are well prepared to deal with the new issues that arise in writing other types of proofs. One reason for this is
that many of the tactics (e.g., division into cases, specialization) and strategies (e.g., the choose method, indirect
proof), highlighted in Section 1, have application beyond proving propositions whose conclusion is of the form
\lx[P(x)---. Q(x)J. In this section, we discuss briefly two additional types of propositions.
2.1.
Deducing conclusions having the form "For every x, there exists y such that P(x, y)."
Many defining properties in mathematics have one of the forms 3x P(x) or \Ix 3y P(x, y). Elementary definitions
of these types include:
(i) Let m and n be integers. We say that m divides n, denoted m In, if and only if there exists an integer p
such that n =mp. (Cf. Definition 1 in Section 4.1 of the text.)
(ii) A function f is onto: f is onto if and only if, for every yin the codomain off, there exists x in the domain
off such that J(x) = y. (Cf. Definition 7 in Section 2.3 of the text.)
(iii) A real number x is said to be rational if and only if there exist integers p and q, with q -1- 0, such that
x = p/q.
Many important mathematical propositions whose proofs should be within the capabilities of students working
through this Guide have as their conclusion a statement involving one of the preceding definitions. Some examples
are:
(a) Prove that if m, n, and pare integers such that m divides n and m divides p, then m divides n
+ p.
(b) Prove that if functions f and g, having the real numbers as their domain and codomain, are both onto,
then their composition fog is also onto.
(c) Prove that if x and y are rational, then xy and x + y are rational.
The new issue involved in proving propositions like (a)-( c) is existence. At a key point of each of these proofs,
we must "produce,'' or define, an appropriate object of the type whose existence is asserted in the desired conclusion.
In doing this, it is important to realize that, for a conclusion of the form \Ix 3y P(x, y), they whose existence is to
be proved usually depends on the given x; we should expect it to be defined in terms of x or else in terms of some
other object that is defined in terms of x. This principle is demonstrated in Examples 17 and 18, which follow.
Example 17. Prove the first part of Proposition (c): If x and y are rational, then xy is rational.
Discussion. Assume that real numbers x and y are rational. To prove that their product xy is rational, we
must show that xy = p/ q, where p and q are integers with q -1- 0. Our job in this proof is to produce, literally to
A Guide to Proof- Writing
502
build, the integers p and q whose quotient p/q equals xy. As in most proofs, once the argument is set up, we must
next assess what we have available to work with. In the case of a proof of existence, this includes asking whether
what we have to work with provides any "building blocks." We have at our disposal only the hypotheses that x
and y are rational. This means we can state that there exist integers P1 and q1, with q1 =I- 0, such that x = p 1/ q1;
and there exist integers P2 and q2, with q2 =I- 0, such that y = P2/q2. We note that, therefore, xy = (pi/q1)(p2/q2),
which, by rules of algebra, equals (p 1p2)/(q1q2). Noting that P1P2 and q1q2 are necessarily integers and that q1q2 =I- 0
(Why?), we declare that p = p 1p 2 and q = q1 q2 are the required integers. I
Example 18. Prove Proposition (b): If functions f and g, having the real numbers as their domain and
codomain, are both onto, then their composition f o g is also onto.
and g are onto. To prove that their composition f o g is onto, let z be
an arbitrary real number. We must prove that there exists x E R such that (! o g)(x) = z. Now since f is onto,
we know that, corresponding to the given z, there exists a real number y such that z = f(y). Next, since g is
onto, then corresponding to this y, there must exist a real number w such that y = g(w). Note therefore that
z = f(y) = f(g(w)) = (f o g)(w). Hence our choice of the desired real number x becomes evident, namely choose
x=w. I
Proof. Assume that the functions
f
Theorem 1 in Section 4.1 of the text contains several propositions related to Definition (i), including a proof
of Proposition (a). You should study that proof, noting its similarities to the proofs in Examples 17 and 18, and
then attempt the following exercises.
• Prove parts (ii) and (iii) of Theorem 1 in Section 4.1 of the text.
• Prove that for every integer n, there exists an integer m such that m \ n.
• Prove that for every positive integer m, there exists a positive integer n such that m \ n.
•Prove the second part of Proposition (c): If x and y are rational, then x
+y
is rational.
• Prove that if f and g are functions having the real numbers as their domain and codomain, and if
onto, then f is onto.
2.2.
f
o g is
Proof by mathematical induction
We use mathematical induction to prove a proposition whose conclusion has the form \fnP(n), where n
is a positive integer (or, sometimes, a nonnegative integer). Thus proof by induction is an appropriate approach
when the universe of discourse for a predicate quantified by "for every" is the set of all positive integers. If you
review earlier sections of this Guide, you will note that this has not usually been the case in most of the examples
and exercises covered, so induction would not have been an appropriate approach at those earlier stages.
Note that the inductive step in every proof by mathematical induction involves a proposition of the form
\fn[P(n) --> Q(n)], where Q(n) is P(n + 1). Thus the basic approach to be taken in the second part of a proof
by induction is the same approach that was emphasized throughout Section 1 of this Guide, namely the choose
method. We start by letting n be an arbitrary positive integer for which it is assumed that P(n) is true. We must
prove, on the basis of that assumption and whatever else is available (e.g., hypotheses), that P(n + 1) is also true.
For more on mathematical induction, see Sections 5.1 and 5.2 of the text.
Writing Projects
503
General Advice on the Writing Projects
If your instructor assigns one or more of the writing projects, you are fortunate. Written communication skills
are of utmost importance in today's world of information. In doing the research for such essays, you will become
familiar with the literature in many areas of mathematics and computer science, and you will hone your library
and information gathering skills. In this section we offer some helpful advice and provide a list of information
resources-including books, articles, and Internet resources-to get you started. At the end of the solutions section
of each chapter in this Guide, we give specific suggestions of where you might look when working on the various
writing projects. We do not guarantee that you will find exactly what you are looking for in the references we
suggest, but at least our pointers will start you in the right direction. Tracking down the information is half the
challenge!
Here are several ideas and points to bear in mind as you do the research for the writing projects:
• The first place to search for material on any of the writing projects is probably the World Wide Web. In
fact, the existence of the World Wide Web and search engines makes it embarrassingly easy to find sources of
information on any topic one desires. To use a search engine, type in one or more key words or phrases (such
as "graph theory" inside quotation marks), and the search engine looks over the billions of websites around the
world to find those that mention these words or phrases; then you can visit those sites. The whole process takes
only a few seconds. One particularly good search engine is Google (www. google. com). It shows you the "best"
hits (those sites that contain your words or phrases most prominently, and/or are sites that many people link
to).
• You should definitely check the website for this textbook: www.m.hhe.com/rosen. There you will find many useful links that can get you started on researching the writing projects. Notice the various Web icons throughout
the textbook, which indicates relevant material on this website.
• Most libraries have on-line search facilities that allow you to look for key words in titles of books in their
collection. For example, to find resources on fuzzy sets or fuzzy logic, you could search on the word fuzzy. You
can also search for authors or titles, of course. Ask a librarian for assistance if necessary. Also, catalogs to
many university libraries (and the Library of Congress) are available on the Internet; see www. libdex. com.
• The following library research technique should come in handy. If the source you are looking at does not deal
in enough detail with the topic you are investigating, then consult the references given in that source. Continue
this process backwards as deeply and broadly as necessary. Of course this is particularly easy to do on the
Web.
• There is a comprehensive set of brief reviews of essentially every mathematical research paper and book written
since 1940 (and being kept up-to-date at the rate of over 60,000 items per year), in a journal called Mathematical
Reviews, published out of Ann Arbor, Michigan, by the American Mathematical Society. It comes in electronic
form on the Web (called MathSciNet) and in printed volumes (which few institutions subscribe to any longer).
Ask your librarian whether your institution is a subscriber, and if so you can usually use computers on campus
to gain access. MathSciNet lets you search by title, author, keywords, and subject area, and you can quickly
follow leads from one review to another.
• We shouldn't need to mention obvious things, like using the index and table of contents of any book you
consult. When looking up items in an index, don't forget to try possible variations of what you are looking
for (e.g., you may find one of the entries "induction, mathematical" or "mathematical induction" but not the
other).
• An excellent source for many of these writing projects is [MiRo] (see the bibliography that follows), which
is published as a supplement to this textbook and is also available on the text's companion website. It has
articles covering various parts of pure and applied discrete mathematics, at levels varying from elementary to
intermediate. It is worth browsing through this book, even if you do not find anything in it relevant to a project
you are working on. It will give you a feeling for the breadth of the subject you are studying.
• Popular accounts of mathematical topics often make their way into The New York Times. This premier of
American newspapers has a detailed index, which is available in most libraries that carry the newspaper; it
504
Writing Projects
can also be accessed from the Web, at www.nytimes.com/ref/membercenter/nytarchive .html. One prolific
writer of mathematical articles is Gina Kolata. The Times also produces on a regular basis a special edition
of mathematics-related articles; ask your library or mathematics department whether they have a recent issue.
The Times and most other major newspapers are also available on the Web, usually for free.
• Many of the essays assigned in this textbook deal with the history of mathematical topics. Most books on the
general history of mathematics are filed under the call letters QA 21. See [Bo4] and [Ev3] for two good sources.
There are also wonderful extensive collections of essays about mathematics, both historical and expository.
A classic is the four-volume treatise [Ne]. A more recent one of high quality is [DaHe]. Perhaps the best
resource for the history of mathematics is the MacTutor History of Mathematics archive on the Web; its URL
is www-history. mes. st-and. ac. uk. It has biographies of hundreds of mathematicians, as well as references,
articles, links, and an unbelievable amount of information.
• The Mathematical Association of America (MAA) has a website with lots of interesting articles (updated
monthly) and special sections for students. Its URL iswww.maa.org.
• Some of these projects go into depth on various topics in discrete mathematics. There are several good, more
advanced textbooks on combinatorics and graph theory, such as [Bol], [BoMu], [Br2], [ChLe], [Rol], and [Tul].
The library classifications here are QA 164 and QA 166, where you will also find specialized books, research
monographs, and conference proceedings. In addition, there are dozens of other discrete mathematics textbooks
at a level comparable with or slightly more or less advanced than your textbook. An excellent one is [Gr2]. It
has comprehensive discussions of most discrete mathematical topics and a wide variety of interesting problems,
including some challenging and open-ended ones. It also has a bibliography of 335 books and articles, and
a detailed index that will lead you to the right source for further reading. Another, slightly different, more
advanced book to take a look at (if nothing else, for its style!) is [GrKn].
• There is an intimate relationship between discrete mathematics and computer science. Computer science
books of all sorts, whether dealing with hardware and circuit issues, programming, data structures, algorithms,
complexity, theoretical foundations, operating systems, compilers, artificial intelligence, or other topics, may
well be relevant to many of these projects. QA 76 is where many such books are housed in the library, although
specialized topics will have their own call numbers (e.g., Q 335 for artificial intelligence or the high TK 7800's
for circuit design). Our list that follows includes several textbooks on data structures and algorithms. Another
lively source is [De2], a collection of essays on various aspects of computer science and related mathematics,
each with references for further reading. You will find those essays relevant to a large number of the writing
projects, and you should definitely try to have a look at this collection.
Here are several points to bear in mind about writing essays (whether in mathematics or in other subjects):
• All the rules and advice you have learned over the years about good writing apply to technical writing as well
as to other forms of prose. It is often more difficult to express mathematical ideas clearly and precisely, so do
not expect these writing projects to be easy.
• Know your reader! Keep in mind for whom you are writing, and pitch the level to that audience. Think about
how much you will assume your reader knows and how much you will need to fill in. (When in doubt, do not
assume the reader knows much.)
• Organize, organize, organize! Essays need to have an introduction, a body, and a conclusion. If the work is
going to be long, it probably makes sense to have labeled sections covering the different points. Make an outline
of what you plan to say, and think a lot about how to order it, both before you start writing and throughout
the process.
• Use a word processor. This makes it much easier to revise and edit your work numerous times, until it is
just the way you want it. Make sure to take advantage of special features like spelling, grammar and usage
checkers. Pay some attention to the format (fonts, spacing, layout, etc.); most word processors let you design a
very pleasing document. Print your essay on a laser-quality printer if you can. If your essay will contain much
mathematical symbolism, try to use a mathematical word processor or typesetter. The best of these is TEX,
Writing Projects
505
which it would definitely be worth your while to learn to use (although it is not easy). To give you an idea of
how nice 'IEX can look, note that this solutions manual was produced using 'IEX• Finally, be careful to give credit to the sources you use. Plagiarism has become a major problem, and if you
copy material from the Web or other sources and present it as your own, you are stealing another person's
property. The consequences can include suspension from your school. When in doubt, ask you instructor about
proper procedures for citations.
List of References for the Writing Projects
[AlSp] Noga Alon and Joel H. Spencer, The Probabilistic Method, with an Appendix on Open Problems by
Paul Erdos (Wiley, 1982; second edition, 2000)
[ApHa] K. Appel and W. Haken, "The solution of the four-color-map problem,'' Scientific American 237,4
(1977) 108-121
(Bal] Roland C. Backhouse, Program Construction and Verification (Prentice-Hall, 1986)
[Ba2] Paul Bachmann, Analytische Zahlentheorie (Leipzig, 1900)
[Ba3] Albert Laszlo Barabcisi, Linked: The New Science of Networks (Perseus, 2002)
[BaGo]
Hans Bandemer and Siegfried Gottwald, Fuzzy Sets, Fuzzy Logic, Fuzzy Methods with Applications
(Wiley, 1995)
[BeKa] Kenneth R. Beesley and Lauri Karttunen, Finite State Morphology (Center for the Study of Language
and Information, 2003)
(BeCo] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for Your Mathematical
Plays, in two volumes (Academic Press, 1982)
[Bel) Richard Bellman, Dynamic Programming, (Princeton University Press, 1957)
(Be] Albrecht Beutelspacher, Cryptology (Mathematical Association of America, 1994)
(BiLl]
Norman L. Biggs, E. Keith Lloyd, and Robin J. Wilson, Graph Theory 1736-1936 (Clarendon Press,
1976)
[Bol] Kenneth P. Bogart, Introductory Combinatorics, second edition (Harcourt Brace, 1990)
[BoDo) Kenneth P. Bogart and Peter G. Doyle, "Nonsexist solution of the menage problem," The American
Mathematical Monthly 93 (1986) 514-519
(Bo2] Bela Bollobcis, Random Graphs (Academic Press, 1985)
[Bo3] Bela Bollobas, "Random graphs,'' in Bela Bollobas, ed., Probabilistic Combinatorics and Its Applications,
Proceedings of Symposia in Applied Mathematics 44 (American Mathematical Society, 1991) 1-20
[BoMu)
John A. Bondy and U. S. R. Murty, Graph Theory with Application (American Elsevier, 1976)
(Bo4] Carl B. Boyer, A History of Mathematics, second edition (Wiley, 1991)
[BrBr) Gilles Brassard and Paul Bratley, Algorithmics: Theory and Practice (Prentice-Hall, 1988)
[Brl] J. Glenn Brookshear, Theory of Computation: Formal Languages, Automata, and Complexity (PrenticeHall, 1988)
(Br2] Richard A. Brualdi, Introductory Combinatorics, second edition (North Holland, 1992)
506
Writing Projects
[Cal] Lewis Carroll, Lewis Carroll's Symbolic Logic (Crown, 1977)
[Ca2] Lewis Carroll, Mathematical Recreations of Lewis Carroll (Dover, 1958)
[Ca3] Lewis Carroll, Symbolic Logic and the Game of Logic (Dover, 1958)
[ChLe] Gary Chartrand and Linda Lesniak, Graphs f3 Digraphs, fourth edition (Chapman & Hall/CRC, 2005)
[ChOe] Gary Chartrand and Ortrud R. Oellermann, Applied and Algorithmic Graph Theory (McGraw-Hill,
1993)
[Ch] Paul M. Chirlian, Analysis and Design of Integrated Electronic Circuits, second edition (Harper & Row,
1987)
[Co]
[CoLe]
Daniel I. A. Cohen, Introduction to Computer Theory (Wiley, 1986)
Thomas H. Cormen, Charles E. Leiserson, and Ronald L Rivest, Introduction to Algorithms, second
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[Dal] C. J. Date, An Introduction to Database Systems (Addison-Wesley, 1990)
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[DaHe] Philip
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[De2] A. K. Dewdney, The New Turing Omnibus: 66 Excursions in Computer Science (Freeman, 1993)
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[Du] David A. Duffy, Principles of Automated Theorem Proving (Wiley, 1991)
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[En] Herbert Enderton, A Mathematical Introduction to Logic, second edition (Harcourt/ Academic Press,
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[Evl] Shimon Even, Algorithmic Combinatorics (Macmillan, 1973)
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[Fe] Guillaume Fertin et al., Combinatorics of Genome Rearrangements (MIT Press, 2009)
[FiWi] S. Fiorini and Robin James Wilson, Edge-colourings of graphs (Pitman, 1977)
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[Gal] Martin Gardner, Logic Machines and Diagrams (McGraw-Hill, 1958)
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(Gol]
507
Michael R. Garey and David S. Johnson, Computers and Intractability: A Guide to the Theory of
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Alan M. Gibbons, Efficient Parallel Algorithms (Cambridge University Press, 1988)
Solomon W. Golomb, Polyominoes: Puzzles, Patterns, Problems, and Packings, second edition (Princeton University Press, 1994)
[Go2] Ronald J. Gould, "Updating the Hamiltonian problem-a survey," Journal of Graph Theory 15 (1991)
121-151
[GrHe] Ronald L. Graham and Pavol Hell, "On the history of the minimum spanning tree problem," Annals of
the History of Computing 7 (1985) 43-57
[GrKn]
Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for
Computer Science, second edition (Addison-Wesley, 1994)
[GrRo] Ronald L. Graham, Bruce L. Rothschild, and Joel H. Spencer, Ramsey Theory, second edition (Wiley,
1990)
[Grl]
George Gratzer, Lattice Theory: First Concepts and Distributive Lattices (Freeman, 1971)
[Gr2] Jerrold W. Grossman, Discrete Mathematics: An Introduction to Concepts, Methods, and Applications
(Macmillan, 1990)
[GrZe]
Jerrold W. Grossman and R. Suzanne Zeitman, "An inherently iterative computation of Ackermann's
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[GrSh]
Branko Griinbaum and G. C. Shephard, Tilings and Patterns (Freeman, 1987)
[Gu] Richard K. Guy, Unsolved Problems in Number Theory, second edition (Springer-Verlag, 1994)
[HaRi]
H. Halberstam and H.-E. Richert, Sieve Methods (Academic Press, 1974)
[Hal] Paul R. Halmos, Naive Set Theory (Springer-Verlag, 1974)
[HaMa] Frank Harary and John S. Maybee, eds., Graphs and Applications: Proceedings of the First Colorado
Symposium on Graph Theory (Wiley, 1985)
[Ha2] David Harel, Algorithmics: The Spirit of Computing (Addison-Wesley, 1987)
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[HiPel]
Frederick J. Hill and Gerald R. Peterson, Computer Aided Logical Design with Emphasis on VLSI,
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[HiPe2]
Peter Hilton and Jean Pedersen, "Catalan numbers, their generalization, and their uses," The Mathematical Intelligencer 13,2 (1991) 64-75
(Hol]
C. A. R. Hoare, "An axiomatic basis for computer programming," Communications of the Association
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[Ho2]
John Hogger, Essentials of Logic Programming (Oxford University Press, 1990)
[Ho3]
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(HoUI] John E. Hopcroft and Jeffrey D. Ullman, Introduction to Automata Theory, Languages, and Computation
(Addison-Wesley, 1979)
[Ka] Abraham Kandel, Fuzzy Mathematical Techniques with Applications (Addison-Wesley, 1986)
Writing Projects
508
[Kn] Donald E. Knuth, The Art of Computer Programming, in three volumes, some in second edition
(Addison-Wesley, 1968-73)
[KoSc] Johannes Kobler, Uwe Schoning, and Jacobo Toran, The Graph Isomorphism Problem: Its Structural
Complexity (Birkhiiuser, 1993)
[Kol] Zvi Kohavi, Switching and Finite Automata Theory, second edition (McGraw-Hill, 1978)
[Ko2] Israel Koren, Computer Arithmetic Algorithms (Prentice-Hall, 1993)
[Ko3] Bart Kosko, Fuzzy Thinking: The New Science of Fuzzy Logic (Hyperion, 1993)
[Kr] Robert L. Kruse, Data Structures and Program Design, second edition (Prentice-Hall, 1987)
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[La2] Clement W. H. Lam, "How reliable is a computer-based proof?" Mathematical Intelligencer 12,l (1990)
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[Le3] Arjen K. Lenstra, "Primality testing,'' in Carl Pomerance, ed., Cryptology and Computational Number
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[LePa] Harry R. Lewis and Christos H. Papadimitriou, Elements of the Theory of Computation (Prentice-Hall,
1981)
[Li] Mario Livio, The Golden Ratio: The Story of Phi, the World's Most Astonishing Number (Broadway,
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[Mal] David Maier, The Theory of Relational Databases (Computer Science Press, 1983)
[Ma2] Udi Manber, Introduction to Algorithms (Addison-Wesley, 1989)
[Ma3] Eli Maor, e: The Story of a Number (Princeton University Press, 1994)
[Ma4] Eli Maor, To Infinity and Beyond (Birkhiiuser, 1987)
[Ma5] Thomas Maufer, Deploying IP Multicast
in
the Enterprise (Prentice-Hall, 1998)
[Mc] James A. McHugh, Algorithmic Graph Theory (Prentice-Hall, 1990)
[McCh] L. E. McMahon, L. L. Cherry, and R. Morris, "Statistical text processing," Bell System Technical Journal
57 (1978) 2137-2154
[McFr] Daniel McNeill and Paul Freiberger, Fuzzy Logic (Simon and Schuster, 1993)
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(CRC Press, 1996)
[Mel] Susan Merritt, "An Inverted Taxonomy of Sorting Algorithms," Communications of the ACM 20 (1985)
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509
[Me2] W. Meyer, "Huffman codes and data compression," UMAP Journal 5 (1984) 278-296
(MiRo] John G. Michaels and Kenneth H. Rosen, Applications of Discrete Mathematics (McGraw-Hill, 1991)
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[Mo] Joseph J. Moder, Project Management with CPM, PERT, and Precedence Diagramming (Van Nostrand
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[Ne]
James R. Newman, The World of Mathematics, in four volumes (Simon and Schuster, 1956)
[Ni] Ivan Niven, Irrational Numbers (Wiley, 1956)
[O'R] Joseph O'Rourke, Computational Geometry in C, second edition (Cambridge University Press, 1998)
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(Pa2] Cheng Dong Pan and Cheng Biao Pan, Goldbach Conjecture (Science Press, 1992)
[PeWi] Marko Petkovsek, Herbert S. Wilf, and Doron Zeilberger, A= B (A. K. Peters, 1996)
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Proceedings of Symposia in Applied Mathematics 42 (American Mathematical Society, 1990) 27-47
[PrDu]
Kendall Preston, Jr., and Michael J. B. Duff, Modern Cellular Automata: Theory and Applications
(Plenum Press, 1984)
[Ral] Anthony Ralston, "De Bruijn sequences~a model example of the intersection of discrete mathematics
and computer science," Mathematics Magazine 55 (1982) 131-143
[Ra2] K. Ramachandra, "Many famous conjectures on primes; meagre but precious progress of a deep nature,"
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[Ri2] John Riordan, Combinatorial Identities (Wiley, 1968)
[Rol] Fred S. Roberts, Applied Combinatorics (Prentice-Hall, 1984)
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510
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[Sa2]
Andras Sarkozy, ''Unsolved problems in number theory," Periodica Mathematica Hungarica 42 (2001)
17-35
[Sc]
E. D. Schell, "Samuel Pepys, Isaac Newton, and Probability,'' The American Statistician 14 (1960) 27-30
[Si]
Michael Sipser, An Introduction to the Theory of Computation (PWS, 1997)
[Sk]
Steven Skiena, Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica (Addison-Wesley, 1990)
[Sl]
N. J. A. Sloane and Simon Plouffe, Encyclopedia of Integer Sequences (Academic Press, 1995)
[Sm] Jeffrey D. Smith, Design and Analysis of Algorithms (PWS-Kent, 1989)
[St]
[St2]
Thomas A. Standish, Data Structure Techniques (Addison-Wesley, 1980)
Douglas R. Stinson, Cryptography: Theory and Practice (CRC Press, 1995)
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[Sz]
Bohdan 0. Szuprowicz, Search Engine Technologies for the World Wide Web and Intranets (Computer
Technology Research Corp., 1997)
[Ti]
G. Tinhofer, "Generating graphs uniformly at random," in G. Tinhofer et al., eds., Computational Graph
Theory, Computing Supplementum (Springer-Verlag, 1990) 235-255
[Tul] Alan Tucker, Applied Combinatorics, third edition (Wiley, 1995)
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[Ty2]
Thomas Tymoczko, "The four-color problem and its philosophical significance,'' Journal of Philosophy
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[WhCl]
J. R. C. White and M. J. Clugston, ''The enumeration of isomers-with special reference to the stereoisomers of Decane," Journal of Chemical Education 70 (1993) 874-876
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[Wi2] Herbert Wilf, generatingfunctionology (Academic Press, 1990)
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Beineke and Robin J. Wilson, eds., Selected Topics in Graph Theory (Academic Press, 1978) 83-101
[Zi] H.-J. Zimmermann, Fuzzy Set Theory and Its Applications (Kluwer, 1991)
511
Sample Tests
Sample Tests
This section of the Student's Solutions Guide contains sample tests based on the text. A test is
included for each of the 13 chapters. Solutions are provided, of course.
You can use these tests to prepare for a midterm or final examination in your course, perhaps
simulating an actual test environment. Also, you can use them for general review or to see
whether you already know certain material from previous study.
It is understood that general directions for a test in a mathematics course such as this
include showing your work and justifying your answers. Remember that mathematics involves
communication as well as computation.
Each sample test has been printed on a separate page.
512
Sample Tests
Sample Test for Chapter 1
1. Consider the proposition "Alice will win the game only if she plays by the rules."
(a) Restate this proposition in English in three different equivalent ways.
(b) State the converse of this proposition.
( c) State the contrapositive of this proposition.
( d) Suppose that Alice plays by the rules but loses. Determine with justification whether the original proposition is true or false.
2. Determine with justification whether p-+ (q V r) and (p /\ -iq)-+ r are logically equivalent.
3. If the propositional function E(p, t) is "Person p does task t correctly," write a proposition in symbols using
quantifiers that expresses the idea ''Nobody's perfect."
4. Consider the propositions 'v'x3yP(x,y) and 3y'v'xP(x,y).
(a) Write out the first of these completely in English.
(b) Repeat part (a) for the second of the propositions.
( c) Give an example that shows that the two propositions need not be logically equivalent; explain how your
example shows this.
5. Prove or disprove each statement.
(a) If x is an irrational number, then 3x
(b) If x is an irrational number, then 3x
+2
2
is an irrational number.
+2
is an irrational number.
6. Prove or disprove each of the following propositions.
(a) If n is a multiple of 4 and k is a multiple of 3, then nk is a multiple of 12.
(b) If n is a multiple of 4 and k is a multiple of 3, then n
+ k is a multiple of 7.
7. Use a proof by contraposition or a proof by contradiction to show that if 3n + 5 is a multiple of 7, then n is
not a multiple of 7.
Sample Tests
513
Solutions for Chapter 1 Sample Test
1. (a) If Alice wins the game, she plays by the rules. Playing by the rules is necessary for Alice to win the game.
Winning the game is a sufficient condition for having played by the rules.
(b) If Alice plays by the rules, then she will win the game.
( c) If Alice doesn't play by the rules, then she won't win the game.
(d) It is true, since the hypothesis is false and the conclusion is true.
2. The only case in which the first proposition is false is the case in which the hypothesis is true and the conclusion
is false. This means that p is true and q and r are both false. The only case in which the second proposition
is false is again the case in which the hypothesis is true and the conclusion is false. Here this means that p and
•q must both be true and r must be false-which is the same as saying that p is true and q and r are both
false. Since these two propositions are true in exactly the same cases, they are logically equivalent.
3. We want to express the idea that everybody makes mistakes-in other words, that everybody does some task
incorrectly. More precisely, for every person, there is some task that that person does not do correctly. Therefore
the answer is Vp3hE(p, t). Another way to express this is that it is not the case that there exists a perfect
person, i.e., a person who does every task correctly. In symbols, -dpVtE(p, t).
4. (a) For every x there exists a y such that P(x, y).
(b) There exists a y such that for every x, P(x,y).
( c) Let P( x, y) be y > x, where the context is real numbers. Then for every x there is certainly a y larger
than x, for example, y = x + 1. The y depends on x. Thus the first proposition is true. However, there
does not exist a constant y that is larger than every x, so the second proposition is false.
5. (a) We will use a proof by contraposition. If 3x+2 were a rational number r, then we would have x = (r-2)/3.
Since the rational numbers are closed under the operations of subtraction and division (by a nonzero
number), this means that x is rational. Thus we have proved the contrapositive of the desired statement,
and our proof by contraposition is complete.
(b) This is not true. If x =
the statement.
v12, then
3x 2
+ 2 = 8, which is a rational number. This counterexample disproves
6. (a) This is true. The hypotheses tell us that n = 4s and k = 3t for some integers s and t. Therefore
nk = 12st, so nk is a multiple of 12.
(b) This statement is false. For a counterexample, let n = 4 and k = 6. Then n is divisible by 4 and k is
divisible by 3, but n + k = 10 is not divisible by 7.
7. Assume the given hypothesis that 3n + 5 is a multiple of 7, and suppose to the contrary of what we wish to
show that n is a multiple of 7. Then n = 7k for some integer k. This means that 3n+5 = 2lk+5 = 7(3k)+5,
which tells us that 3n + 5 leaves a remainder of 5 when divided by 7 and hence is not divisible by 7. This
contradicts the hypothesis that 3n+5 is a multiple of 7. Therefore our supposition was wrong, and we conclude
that n is not a multiple of 7.
Sample Tests
514
Sample Test for Chapter 2
1. Prove or disprove that
A- (BU C) =(A - B) n (A- C), where A, B, and C are sets.
2. Suppose that A= P({a,b,c,d,e}) and B = P({c,d,e,f,g}). (Recall that P(X) is the power set of X.)
Compute IA U Bl. [Hint: Determine first what An B is. Then use the fact that IP(X)I = 2n if IXI = n.]
3. Let A= {1,2,3,4,5,6} and B = {a,b,c}.
(a) Give an example of a function from A to B that is neither one-to-one nor onto, and explain why it meets
these conditions.
(b) State the domain, codomain, and range of the function you gave in part (a).
6
4. Compute
L l%J.
n=O
5. Let S be the set of countries of the world, and let C be the set of cities of the world.
(a) Give an example of a function
f from C to S.
(b) Give an example of a function
g
from S to C.
(c) One off o g(Venice) and go !(Venice) makes sense and one does not. Compute the value of the one that
does, and explain why the other one does not.
6. (a) Is the set of all finite bit strings countable or uncountable? Prove that your answer is correct.
(b) Is the set of all infinite bit strings countable or uncountable? Prove that your answer is correct.
7. Give an example of two 2 x 2 matrices A and B such that AB =J BA. (Make sure to justify your answer.)
Sample Tests
515
Solutions for Chapter 2 Sample Test
c
c
1. A - (Bu C) = An Bu =An (B n C) = An An B n
= (An B) n (An C) = (A - B) n (A - C)' using,
respectively, the definition of difference, De Morgan's law, idempotence, the commutativity and associativity
of intersection, and again the definition of difference. Both sides of this equality consist of those elements in A
that fail to be in either B or C.
2. Following the hint, we note that An B = P( { c, d, e}), since a set is a subset of {a, b, c, d, e} and also a subset
of {c, d, e, f, g} if and only if its elements are chosen from among {c, d, e}. Therefore IA n Bl = 23 . On the
other hand, IAI = 25 and IBI = 25 . Therefore IA U Bl = IAI + IBI - IA n Bl = 25 + 25 - 23 = 32 + 32 - 8 = 56.
3. (a) If we send every element of A to c, for example (i.e., define the function by f(x) = c for all x EA), then
this function is not one-to-one since f(l) = f(2), and not onto since for no x EA does f(x) = b.
(b) The domain of this function is A; its codomain is B, and its range is { c}.
4. We need to plug in the values of n from 0 to 6 and add: l0/2J + ll/2J + l2/2J + l3/2J + l4/2J + l5/2J + l6/2J =
0 + 0 + 1+1+2 + 2 + 3 = 9.
5. (a) The obvious choice is to let f(x) = the country in which city x lies. Thus we would have !(Perth) =
Australia, and f (Vienna) = Austria, for example. One alternative (not very interesting) correct answer is
to let f (x) = France for all cities x. There are a huge number of correct answers.
(b) One obvious choice is to let g(x) =the capital of country x. Thus we would have g(Argentina) =Buenos
Aires, and g(Russia) = Moscow, for example. One alternative (not very interesting) correct answer is to
let g(x) = Detroit for all countries x. Again, there are a huge number of correct answers.
( c) The expression f o g(Venice) makes no sense because Venice is not a country and so not an element of S,
which is the domain of g. However g o f (Venice) = g(f (Venice)) = g(Italy) = Rome.
6. (a) We can list all finite bit strings in order of increasing length, with only a finite number of each length: >.,
0, 1 , 00, 01, 10, 11 , 000, 001, . . . . This provides a one-to-one correspondence between the set of positive
integers and the set of all finite bit strings, which by definition means that the set of all finite bit strings
is countable: 1 +-+ >., 2 +-+ 0, 3 +-+ 1, 4 +-+ 00, 5 +-+ 01, 6 +-+ 10, 7 +-+ 11, 8 +-+ 000, 9 +-+ 001, ....
(b) This set is uncountable. The proof is by contradiction, using Cantor's diagonal argument. Assume that the
set of all infinite bit strings is countable, that is, that there exists a one-to-one correspondence f from the
set of positive integers to the set of all infinite bit strings-in other words, a list of all infinite bit strings.
For example, it might happen that f(l) = 010101 ... , f(2) = 110110110 ... , and so on. Form a new bit
string w whose ith bit is a 0 if the ith bit of f(i) is 1, and whose ith bit is a 1 if the ith bit of f(i)
is 0. In our example, w starts out 10 .... Then w differs from every element in the list, since it differs
from f(i) in the ith bit. In other words, our list is incomplete, and f is not a one-to-one correspondence
because it is not onto. Having obtained a contradiction, we conclude that the set of all infinite bit strings
is uncountable.
7. If we choose two matrices more or less at random, then we are very likely to find the desired pair, since it
almost never happens that the product is the same when the order of multiplication is reversed. Suppose
A = [
Then
AB = [
which are clearly not equal.
~ ~]
and
B =
~
and
BA = [
;]
U ~] .
~ ~]
,
516
Sample Tests
Sample Test for Chapter 3
1. (a) Write an algorithm to find the sum aia1
· · · + anal + ana2 + ana3 + · · · + anan.
+ aia2 + aia3 + · · · + aian + a2a1 + a2a2 + a2a3 + · · · + a2an +
(b) Analyze the time complexity of the algorithm you developed in part (a).
2. Explain how the binary search algorithm would look for 7 in the list (1, 2, 3, 4, 8, 10).
3. Suppose that A, B, and C are 3 x 7, 7 x 2, and 2 x 5 matrices of numbers, respectively. Is it more efficient
to compute the product ABC as (AB)C or as A(BC)? Justify your answer by computing the number of
multiplications of numbers needed each way.
4. Let f(n) = 10n 2 +Sn and g(n) = n 3
+ 2.
(a) State the definition of f(n) being O(g(n)).
(b) Determine with justification whether f(n) is O(g(n)) and also whether g(n) is O(f(n)).
5. (a) Explain in a sentence or two what a greedy algorithm is.
(b) Give an example of an optimization problem for which a greedy algorithm gives the correct solution.
( c) Give an example of an optimization problem for which a greedy algorithm does not give the correct solution,
and show how it fails.
Sample Tests
517
Solutions for Chapter 3 Sample Test
1. (a) We can write the algorithm in pseudocode as follows. It is straightforward: we just have a loop within a
loop to compute each product a,a3 , and accumulate the sum in the variable sum.
procedure SumOJProducts(a 1 , a 2 , ... , an : integers)
sum:= 0
for i := 1 ton
for j := 1 ton
sum := sum + a,a3
return sum
(b} The outer loop is executed n times, and for each pass through the outer loop the inner loop is executed n
times. Therefore the assignment statement is executed n 2 times. Since essentially no other work is done
by this algorithm, the number of steps is O(n 2 ). [It is interesting to note that the sum can be obtained in
O(n) steps by adding all the ai's together and then squaring the sum.]
2. The active portion of the list starts as the entire list, (1, 2, 3, 4, 8, 10). The sought-after item, 7, is compared
to the middle term in the list (actually the term just below the middle, since there are an even number of
items), namely 3. Since 7 > 3, we restrict the search to the portion of the list above 3, namely (4, 8, 10), and
repeat the process. This time we compare 7 with 8, and since 7 f 8, we restrict the list to (4, 8). The next
comparison shows that 7 > 4, so the active portion of the list becomes (8). At this point there is only one
number left, so we check to see whether this number equals 7. Since it does not, we conclude that 7 is not in
the original list.
3. To multiply A by B, we will need 3 · 7 · 2 = 42 multiplications. The result is a 3 x 2 matrix. To multiply it
by C will require 3 · 2 · 5 = 30 multiplications. This gives a total of 42 + 30 = 72 steps. On the other hand, if
we multiply B by C first, we use 7 · 2 · 5 = 70 multiplications just for that, and then another 3 · 7 · 5 = 105 to
multiply A by the 7 x 5 matrix BC. This method uses a total of 70 + 105 = 175 steps. Therefore the first
method is faster (more than twice as fast).
+ 5nl::; Cln 3 + 21 for all n > k.
we observe that l10n 2 + 5nl = 10n 2 + 5n::; l0n 3 + 5n 3 = l5n 3 ::;
4. (a) There exist constants C and k such l10n 2
15ln3 + 21
for all n ::'.:: 1. Thus we have witnesses C = 15 and k = 1. On the other hand,
+ 2 is not 0(10n 2 + 5n).
2
3
To see this, note that the ratio of these two functions, (n + 2)/(10n + 5n), is greater than n 3 /(15n 2 ) for
all n ::'.:: 1. This latter quantity equals n/15, and clearly it cannot be bounded above by a constant. Hence
there is no constant C such that ln 3 + 21 ::; Cl10n 2 + 5nl for all sufficiently large n.
(b} To see that f(n) is O(g(n)),
n3
5. (a) A greedy algorithm selects at each step what seems to be the "best choice"-the choice that optimizes at
that step----with the hope that this sequence of choices provides the global optimum.
(b) To schedule as many talks as possible in a lecture hall, given the starting and finishing times of the talks,
sort the talks according to increasing finishing time. At each step, select the talk with the earliest finishing
time among the talks that do not conflict with those already selected.
( c) To make change using as few coins as possible if quarters, dimes, and pennies are available, choose the
largest denomination possible, reduce the amount to be given by that value, and repeat until the desired
amount has been reached. This algorithm will use six coins to make change for 30 cents (a quarter and
five pennies), whereas three dimes would have sufficed.
518
Sample Tests
Sample Test for Chapter 4
1. Compute the following.
(a) gcd(742, 1908)
{b) the prime factorization of 3080
( c) an inverse of 4 modulo 13
(d) -50 mod 7
2. Show that if a= b (mod m) and c
=d (mod m) then a+ c =b + d (mod m).
3. (a) Express 425 in base 6.
(b) Express the binary numeral (101001000)2 in base 10.
4. Suppose that a message is encrypted using the shift cipher function f(p) = (p+ 10) mod 26, where the letters
A through Z are represented by the integers 0 through 25. The message "ZKCC DRSC DOCD" is received.
Determine what the original message was.
5. Members of the American Contract Bridge League are assigned a 7-digit player number a 1 a 2 a 3 a 4 a 5 a 6 a 7 • The
seventh digit is a check digit, and the following equivalence must hold: 7a 1 +6a2 + 5a 3 + 4a4 + 3a 5 + 2a6 + a1
0 (mod 11). What is a player's full player number if the first six digits are 285319?
=
Sample Tests
519
Solutions for Chapter 4 Sample Test
1. (a) We use the Euclidean algorithm:
gcd(106, 0) = 106.
gcd(742, 1908)
= gcd(7 42, 424) = gcd( 424, 318) = gcd(318, 106) =
(b) This number is clearly even, so we divide by 2. The quotient is even again, so we divide by 2 again, and
repeat this process once more, obtaining 3080 = 23 · 385. Now clearly 385 is divisible by 5, so we have
3080 = 23 · 5 · 77 = 23 · 5 · 7 · 11, and this last product contains only primes.
(c) In order to find this, we need to perform the Euclidean algorithm to find gcd(13, 4) and keep track of the
results, in order to write 1 in the form 4s + 13t. Only one step is needed, since we see that 13 = 3 · 4 + 1,
so we have 1 = 13 + (-3) · 4. From this we immediately see that -3 is an inverse of 4 modulo 13, since
the last equation shows that 1 - ( -3) · 4 is divisible by 13, i.e., that 1 = (-3) · 4 (mod 13). If we want a
positive answer, we can add 13, obtaining 10. Note that 4 · 10 = 40 = 1 (mod 13).
( d) We just divide -50 by 7, obtaining -8 (note that we need to have a positive remainder, so that -7 is not
enough), with a remainder of 6. In other words, -50 = (-8) · 7 + 6. Then -50 mod 7 is the remainder,
namely 6.
2. To say that a = b (mod m) is to say that a - b = sm for some integer s (i.e., a - b is a multiple of m).
Similarly, to say that c d (mod m) is to say that c - d = tm for some integer t (i.e., c - d is also a multiple
of m ). Adding these two equations and doing some trivial algebra, we have (a+ c) - (b+ d) = (s + t)m, which
tells us immediately from the definition that a+ c = b + d (mod m), as desired.
=
3. (a) We note that the relevant powers of 6 are 6° = 1, 6 = 6, 62 = 36, and 63 = 216. Thus we see that 425
contains one 63 , leaving 425 - 216 = 209, that 209 contains five 62 's, leaving 209 - 5 · 36 = 29. Similarly,
29 = 4 · 6 + 5, so this number contains four 6's and 5 left over. Therefore 425 = (1545)5. An alternative
way to do this is to divide repeatedly by 6 and note the remainders. These are the digits of the base 6
expansion of 425, from right to left.
(b) We see by eye-straining counting that the leftmost digit in this numeral represents 28 . Proceeding from
there, we see that the number represented by this expansion is then 28
+ 26 + 23 = 256 + 64 + 8 = 328.
4. To decrypt, we need the inverse function for p, which is clearly g(q) = (q - 10) mod 26. The encrypted
message, translated into numbers, is 25-10-2-2 3-17-18-2 3-14-2-3 (remembering that Af--t 0, Bf--t 1, and so
on). Subtracting 10 (and adding 26 if the difference is negative, in order to compute the answer modulo 26 ),
we obtain the numerical version of the plaintext: 15-0-18-18 19-7-8-18 19-4-18-19. This is quickly seen to read
PASS THIS TEST.
5. The following equivalence must hold:
7 · 2 + 6 · 8 + 5 · 5 + 4 · 3 + 3 · 1 + 2 · 9 + a7
This simplifies to 120 + a 7
number is 2853191.
= 0 (mod 11)
or 10 + a 7
= 0 (mod 11).
=0 (mod 11)
Therefore the check digit is 1, and the full
520
Sample Tests
Sample Test for Chapter 5
1. (a) State the principle of mathematical induction completely.
(b) Suppose that a sequence is defined recursively by A 0 = 2 and An = (A,,_ 1 +1)/2 for all n
that this sequence is decreasing (i.e., An < An-l for all n ).
~
1. Show
2. (a) State the principle of strong induction completely.
(b) For each integer n greater than 1 , let s (n) be the sum of the primes in the prime factorization of n (with
repetitions). So, for example, s(12) = 7, because the prime factorization of 12 is 2·2·3, and 2+2+3 = 7.
Prove that n ~ s (n) for all n ~ 2 . [Hint: First show that if x and y are integers greater than 1 , then
x · y ~ x + y; do this by starting with the inequality (x - l)(y -1) ~ 1. Second, how is s(x · y) related to
s(x) + s(y)?]
3. Give a recursive definition of the set of all postages that can be obtained using only 25-cent stamps and 44-cent
stamps.
4. A certain function of two nonnegative integer variables, C(a, b), defined whenever a~ b, has the property that
C(a, b) = C(a - 1, b) + C(a - 1, b - 1) for all nonnegative integers a and b, with a> b > 0, with the further
conditions that C(a, a) = C(a, 0) = 1 for all a.
(a) Write in pseudocode a recursive algorithm for computing this function.
(b) Compute C(4,2) using this definition.
5. Verify that the program segment
if n is odd then y := (x
else x := (x + y)/2
+ y)/2
is correct with respect to the initial assertion "n is an integer and 0 :::::; y - x :::::;
is an integer and 0 :Sy - x :S E/2."
E ,"
and the final assertion "n
Sample Tests
521
Solutions for Chapter 5 Sample Test
1. (a) If S (n) is a predicate defined for all positive integers n, and if S ( 1) is true, and if for every positive
integer k, S(k) implies S(k + 1), then S(n) is true for every n.
(b) Let S(n) be the statement that An < An-I. We will show by mathematical induction that S(n) holds
for all n. By the definition of this sequence, we see that A 0 = 2 and A 1 =(Ao+ 1)/2 = 1.5, so S(l) is
true; this takes care of the basis step. For the inductive step, assume the inductive hypothesis that S(k)
is true, i.e., Ak < Ak-l. We want to show that S(k + 1) is true, i.e., Ak+ 1 < Ak. Adding 1 to both sides
of the given inequality and dividing by 2, we obtain another valid inequality, (Ak + 1)/2 < (Ak-1 + 1)/2.
But by the definition of this sequence, this says precisely that Ak+ 1 < Ak, as desired.
2. (a) If S(n) is a predicate defined for all positive integers n, and if S(l) is true, and if for every positive
integer k, Vj5_kS(j) implies S(k+l),then S(n) is true for every n.
(b) Following the hint, suppose that x and y are integers greater than 1. Then x - 1 ;:::: 1 and y - 1 ;:::: 1,
so 1 = 1·1 5_ (x - l)(y - 1) = x · y - x - y + 1, from which it follows that x · y ;:::: x + y. We will prove
that n;:::: s(n) for all n;:::: 2 by strong induction. The basis step is trivial, because s(2) = 2. Suppose that
s(j) ;:::: j whenever 2 5_ j 5_ k. We must show that s(k + 1) ;:::: k + 1. If k + 1 is prime, then it is its own
prime factorization, so s(k + 1) = k + 1. Otherwise, k + 1 can be factored, say as x · y, where 2 5_ x 5_ k
and 2 5_ y 5_ k. The prime factorization of k + 1 is the product of the prime factorizations of x and y, so
clearly s(k + 1) = s(x) + s(y). Both x and y are in the range of the inductive hypothesis, so we know that
x;:::: s(x) and y;:::: s(y). Putting this all together, we have k + 1 = x · y;:::: x + y;:::: s(x) + s(y) = s(k + 1),
and our proof by strong induction is complete.
3. By using no stamps, we can get 0 cents postage. This our basis case. The recursive part of the definition is
that if we can obtain p cents postage, then we can also obtain p + 25 cents postage and p + 44 cents postage.
4. (a) We just incorporate the definition into the pseudocode.
procedure C(a, b: nonnegative integers with a;:::: b)
if b = 0 or b = a then return 1
else return C(a - 1, b) + C(a - 1, b - 1)
(b) We apply the recursive definition to compute that C(4, 2) = C(3, 2) + C(3, 1) = [C(2, 2) + C(2, 1)] +
[C(2, 1) + C(2, O)] = 1+2C(2,1) + 1=2+2[C(l, 1) + C(l, O)] = 2 + 2(1+1) = 6.
5. Since n is not changed, the final assertion that n is an integer is clearly true if the initial assertion that n
is an integer is true. Suppose that y - x had some value before the program segment was executed. If n is
odd, then after the segment is executed the new value of y - x is [(x + y)/2] - x which equals (y - x)/2. If
n is not odd, then after the segment is executed the new value of y - x is y - [(x + y)/2] which again equals
(y - x) /2. Thus in either case the value of y - x was cut in half. Therefore if y - x was between 0 and E
before the program segment was executed (our initial assertion), then y - x is between 0 and E/2 after the
program segment is executed (our final assertion); this statement is what we were asked to prove.
522
Sample Tests
Sample Test for Chapter 6
1. Determine the number of bit strings of length 10 that either start with six consecutive O's or end with six
consecutive O's. (Recall that "or'' is used in the inclusive sense in mathematics, unless specifically stated
otherwise.)
2. There are more than 311,000,000 people in the United States. Prove that at least six of them were born during
the exact same minute of the same hour of the same day of the same year.
3. Determine the number of 5-card hands from a standard deck of cards that contain two cards from each of two
suits and an ace from a third suit.
4. How many ways are there to distribute 25 identical Christmas tree ornaments among 5 (distinct) children, if
each child must get at least 3 ornaments and no child may get 13 or more?
5. How many ways are there to arrange 4 oranges, 5 apples, 6 pears, and 7 peaches in a row?
6. Give a combinatorial proof that I.:~=O 2kC(n, k) = 3n by computing in two different ways the number of ways
to donate some (possibly all or none) of your collection of n distinct baseball cards to 2 different museums.
[Hint: each card can go to one of 3 places; on the other hand, you can donate k cards in all.]
7. Find the coefficient of x 10 in (3x
+ 2) 15 .
523
Sample Tests
Solutions for Chapter 6 Sample Test
1. If a 10-bit string is to start 000000, then we are free to choose the last four bits, so there are 24
= 16 such
strings. Similarly, there are 16 strings that end with six consecutive O's. Furthermore, there is one string
( 0000000000) that meets both of these conditions, so it is counted twice. By the inclusion-exclusion principle,
the answer is therefore 16 + 16 - 1 = 31.
2. Certainly at least 300,000,000 of the 311,000,000 people are less than 100 years old. Thus there are at most
100 · 365.25 · 24 · 60 = 52,596,000 possible minutes for these people to have been born during. By the generalized
pigeonhole principle, there are therefore at least l300000000/52596000l = 15.70 .. ·l = 6 people with the same
minute of birth.
3. We need to make several choices in order to describe such a hand. First, there are C( 4, 2) = 6 ways to choose
the two suits. There are C(13, 2) = 78 ways to choose the two cards from each of these suits, so there are 78 2
ways to specify these four cards in all, once the suits have been selected. Finally, there are 2 ways to decide
which of the other two suits the ace is to come from. Therefore by the multiplication principle the answer is
6. 78 2 • 2 = 73,008.
4. After the required 15 ornaments are distributed ( 3 to each of the 5 children), 10 ornaments remain. There
are C(5 + 10 - 1, 10) = C(14, 10) = C(14, 4) = 1001 ways to distribute these 10 ornaments among the 5
children. However, 5 of these distributions are forbidden-giving all 10 to one child-since this would give
that child a total of 13 ornaments, in violation of the condition stated in the problem. Therefore the answer is
1001 - 5 = 996.
5. This problem asks for the number of permutations of 22 objects, consisting of 4 objects of one type, 5 of
a second type, 6 of a third type, and 7 of a fourth type. By Theorem 3 in Section 6.5, the answer is
22!/(4!5!6!7!) ~ 1.1 x 10 11 .
6. For the right-hand side we just note that each card can end up in one of 3 places-Museum #1, Museum #2,
or still with me. Therefore there are 3n ways to distribute the cards. On the other hand, for each k from
0 to n, I can decide to donate k cards. There are C( n, k) ways to choose the cards to donate, and then for
each card chosen there are 2 ways to decide which museum it will go to. Therefore there are C(n, k)2k ways
to donate k cards. Summing over all possible values of k (certainly different values of k lead to mutually
exclusive outcomes), we obtain the left-hand side of the identity. Therefore the two sides are equal.
7. By the binomial theorem, the term containing x 10 is the term
(\,5)3 10 25 = 3003. 59049. 32 = 5,674,372,704.
(155 ) ( 3x) 10 25 .
Therefore the coefficient of x 10 is
524
Sample Tests
Sample Test for Chapter 7
1. What is the probability that a positive integer less than 100 picked at random has at least five prime factors?
(The factors need not be distinct; for example, 24 has four prime factors, since 24
= 2 · 2 · 2 · 3.)
2. Suppose that a fair die is rolled and a card is drawn from a standard deck of cards. What is the probability
that the number shown on the die is the same as the number shown on the card (if any-note that only the
numbers 2 through 10 are shown on cards, since aces, kings, queens, and jacks are represented by letters).
3. What is the probability of getting exactly four heads when a fair coin in flipped eight times?
4. A fair red die and a fair blue die are rolled. What is the expected value of the sum of the number on the red
die plus twice the number on the blue die?
5. Suppose that the probability that a thumbtack lands point up is 0.4. What is the expected number of times
we must toss the tack until it lands point up?
6. In Sun and Fun Retirement Community, 603 of the residents play golf, 453 of the golfers smoke, and 303 of
the non-golfers smoke. If a random non-smoker is selected, what is the probability that he or she plays golf?
Report the answer to the nearest whole percent.
Sample Tests
525
Solutions for Chapter 7 Sample Test
1. It is easy to see by considering all possible cases that the only numbers meeting this condition are 26
5
4
4
3
2
2 = 32, 2 · 3 = 48, 2 · 5 = 80, and 2 · 3
in all, the desired probability is 5/99.
= 72.
= 64,
Since there are 5 outcomes in this event, out of 99 outcomes
2. For this event to occur, the die must show something other than a 1, and the probability of this happening is
5/6. Once a 2 through 6 is rolled, the probability that a card of that rank is drawn is 4/52 = 1/13, since there
are four cards of each rank. (Note that the draw of the card is independent of the roll of the die.) Therefore
the answer is (5/6)(1/13) = 5/78 ~ 6.4%.
3. The number of heads follows a binomial distribution. Therefore, by Theorem 2 in Section 7.2, the answer is
4
4
C(8,4) (~) (~) = 70/256 ~ 27%.
4. Let Xr and Xb be the random variables for the numbers shown on the dice. We are asked for E(Xr+Xb+Xb).
Since these dice are fair, we know that E(Xr) = E(Xb) = (1+2+3 + 4 + 5 + 6)/6 = 3.5. By linearity of
expectation (note that independence is not required), we have E(Xr + Xb + Xb) = 3.5 + 3.5 + 3.5 = 10.5.
5. The number of tosses required has a geometric distribution with p
tion 7.4, the expected value is 1/0.4 = 2.5. medskip
=
0.4. Therefore, by Theorem 4 in Sec-
6. Let G be the event that a randomly chosen resident plays golf. We know that p( G) = 0.60 and therefore
p(G) = 0.40. Let S be the event that a randomly chosen resident smokes. We are told that p(S I G) = 0.45
and p(S I G) = 0.30. From these last facts, we know that p(S I G) = 0.55 and p(S I G) = 0.70. We are asked
for p( G I S). We use Bayes' theorem:
p(S I G)p(G)
( I S)_
- p(S I G)p(G) + p(S I G)p(G)
p G
_
(0.55)(0.60)
- (0.55)(0.60) + (0.70)(0.40)
~
~ 54%
Sample Tests
526
Sample Test for Chapter 8
Sn be the number of ways to write the positive integer n as the sum of an ordered list of 2's and 3's. For
example, S8 = 4, since 8 = 2 + 2 + 2 + 2 = 2 + 3 + 3 = 3 + 2 + 3 = 3 + 3 + 2.
(a) Write down a recurrence relation for {sn}·
1. Let
(b) Write down initial conditions that, together with the recurrence relation, determine the entire sequence
{sn}·
( c) Compute
s11 .
2. Solve the recurrence relations an
= 4an-l + 12an-2 with initial conditions
a0
= 3 and a 1 =
l.
3. How many positive integers less than or equal to 10,000 are divisible by 2, 5, or 7?
4. Let Dn denote the number of derangements of n objects.
(a) Define what a derangement is.
(b) State an explicit formula for calculating Dn, and use it to compute D 5 .
( c) Five married couples are having a party. The names of the men are written on slips of paper and put into
a hat, and each woman draws a name at random (the slips are not replaced), so that each woman gets one
name. Each woman dances the first dance with the person whose name she drew. How likely is it that no
one dances with his or her spouse?
5. Suppose that f satisfies the divide-and-conquer relation f(n) = 3f(n/2)+2n and the initial condition f(l) = 0.
(a) Compute f(32).
(b) What can be said about the asymptotic behavior off if we know that
f
is an increasing function? Give
a good estimate of the form "J(n) is O(g(n)) ."
6. How many ways are there to distribute seven distinct Christmas tree ornaments to four distinct children if each
child must get at least one ornament?
7. What is the generating function for the number of ways to make change using pennies, nickels, and dimes, if
no more than two nickels and no more than five pennies may be used? Give your answer in closed form.
Sample Tests
527
Solutions for Chapter 8 Sample Test
1. (a) A sequence that sums to n can begin with a 2 and continue with a sequence that sums to n - 2, or it can
begin with a 3 and continue with a sequence that sums to n - 3. Therefore Sn = Sn-2 + Sn-3. This is
valid if n ?: 4 .
(b) We need initial conditions for s 1 , s 2 , and s 3 . It is clear that s 1 = 0, since there is no way to write 1 as
the sum of 2's and 3's. It is equally clear that s 2 = 1 and s 3 = 1.
(c) We start with the initial conditions and successively compute using the recurrence relation: s 4 = s 2 + s 1 =
1 + 0 = 1 , S5 = S3 + S2 = 1 + 1 = 2, S5 = S4 + S3 = 1 + 1 = 2 , S7 = S5 + S4 = 2 + 1 = 3,
Sg = S6 + S5 = 2 + 2 = 4, Sg = S7 + S5 = 3 + 2 = 5, S11 = Sg + Sg = 5 + 4 = 9.
2. We write down the characteristic equation r 2 - 4r -12 = 0 and find by factoring that the roots are 6 and -2.
Therefore the general solution is an = A· 5n + B · (-2)n. The initial conditions tell us that 3 = A+ B and
1 = 6A-2B, whence we find A= 7/8 and B = 17/8. Therefore the solution is an = (7 /8) · 5n + (17 /8) · (-2r.
As a check we can compute a2=4·1+12·3 = 40 from the recurrence relation and a 2 = (7 /8)·36+(17 /8)·4 = 40
using the formula.
3. Using the principle of inclusion-exclusion, we need to compute as follows:
l
lOOOOJ+llOOOOJ+llOOOOJ-llOOOOJ-llOOOOJ-llOOOOJ+llOOOOJ
2
5
7
2·5
2·7
5.7
2·5·7
= 5000 + 2000 + 1428 - 1000 - 714 - 285 + 142 = 6571
4. (a) A derangement of the objects i 1 , i 2 , ... , in is a permutation of these objects, so that for no k is ik the
kth object. In other words, the first object does not end up first, the second object does not end up second,
... , and the nth object does not end up last.
(b) By the formula developed in this chapter, we have
D
n
= nl
·
[1 - !__1! + !__2! - · · · + (- l)n !__]
n! ·
For n = 5 we compute
D5 = 120 [ 1 -
l1 + 21 - 61 +
1
1]
24 - 120
= 44.
(c) No one dances with his or her spouse if and only if the permutations of the husbands determined by the
wives' selection is a derangement. Therefore the desired probability is D 5 /5! = 44/120 ~ 0.367.
5. (a) We compute successively f(2) = 3f(l) + 2 · 2 = 3 · 0 + 4 = 4, f(4) = 3f(2) + 2 · 4 = 3 · 4 + 8 = 20,
f(8) = 3f( 4) + 2. 8 = 3. 20 + 16 = 76' f (16) = 3f (8) + 2 · 16 = 3. 76 + 32 = 260' !(32) = 3f (16) + 2. 32 =
3 . 260 + 64 = 844.
(b) The master theorem in Section 8.3 applies, with a = 3, b = 2, c = 2, and d = 1. Since a > bd ( 3 > 21 ) ,
we have f(n) is O(n10gb a) = O(n10g 3 ) ~ O(n 1 6 ).
6. We are asked for the number of onto functions from a 7-set to a 4-set. By Theorem 1 in Section 8.6, the answer
is 47 - C(4, 1)37 + C(4, 2)2 7 - C(4, 3)1 7 = 8400.
7. The generating function for choosing pennies is 1 + x + x 2 + · · · + x 5 , since anywhere from 0 to 5 pennies may
be used. The generating function for choosing nickels is 1 + x 5 + x 10 , since anywhere from 0 to 2 nickels may
be used and a nickel is worth 5 cents. The generating function for choosing dimes is 1 + x 10 + x 20 + x 30 + . · . ,
since any number of dimes may be used. A choice consists of making each of these choices, so the generating
function we seek is the product of these functions, which we can put into closed form by using the identities
given in Table 1 of Section 8.4:
1 - x 6 1 - x 15
1
(1 + x + x 2 + · · · + x 5 )(1 + x 5 + x 10 )(1 + x 10 + x 20 + x 30 + · · ·) = - - . - - 1- x
1 - x 5 1 - xlO ·
The coefficient of xn in the power series for this function gives the number of ways to make change of n cents.
(A computer algebra package will report that this series starts 1 +x+x 2 +x 3 +x 4 +2x 5 +x 6 +x7 +x 8 +x 9 +3x 10 ,
so, for example, there are 3 ways to give change of 10 cents under the rules laid out here.)
Sample Tests
528
Sample Test for Chapter 9
1. Define the relation R on the set of real numbers by xRy if and only if x 3 = -y 3 .
(a) Determine whether R is reflexive, symmetric, antisymmetric, and/or transitive.
(b) State a simple rule for when two numbers are related by R o R.
2. Explain how to tell whether a relation is symmetric under each of the following representations.
(a) matrix
(b) digraph
3. Find the following closures of the relation
(a) reflexive
(b) symmetric
<
on the integers, stating each as succinctly as possible.
( c) transitive symmetric
4. Define a relation on the set of real numbers by setting xRy if and only if x 2
-
5x
+6=
y2
-
5y
+ 6.
(a) Show that R is an equivalence relation.
(b) Write down the equivalence class [2] .
(c) Write down a "formula" (explicit description) for [r] that holds for an arbitrary real number r.
5. (a) Write down the definition of what it means for a collection
(b) Explain how a partition
of R).
7r
7r
to be a partition of a set A.
on a set A determines an equivalence relation R on A (give an explicit definition
(c) The set 7r = {{1,3,5},{2,4},{6}} is a partition of {1,2,3,4,5,6}. Write down the equivalence relation
determined by this partition, and then find the partition determined by this equivalence relation.
6. Consider the partial order on the set of all nonempty subsets of {a, b, c}.
(a) Identify all the maximal elements, greatest elements, minimal elements, and least elements of this poset.
(b) Find a total order compatible with this partial order.
( c) Give a simple way to determine the least upper bound of two elements of this poset (which will work for
every poset consisting of a collection of sets with the <;;; relation).
7. (a) State what additional properties a poset must have in order for it to be a lattice.
(b) Give an example of a poset which is not a lattice but in which every pair of elements has an upper bound
and a lower bound.
Sample Tests
529
Solutions for Chapter 9 Sample Test
1. (a) Since 13
3
-/:- -1 , R is not reflexive. Suppose that x 3 = -y 3 . Then multiplying both sides by -1 gives
-x3 ; this says that xRy --+ yRx, so R is symmetric. The relation is not antisymmetric, since
lR(-1) and (-l)Rl, but 1-/:- -1. Finally, R is not transitive, since lR(-1) and (-l)Rl, but 1 is not
related to 1 .
us y3
=
(b) By taking the cube root of both sides of x 3 = -y3 , we obtain the equivalent equation x = -y (since
cubing is a one-to-one function). We claim that Ro R is the identity relation. Indeed, every number is
related to itself by Ro R, since xR(-x) and (-x)Rx. Conversely, if x(R o R)z, then for some y we have
x = -y and y = -z; it follows that x = -(-z) = z.
2. (a) Look at each entry not on the main diagonal, and compare it to the entry in the corresponding location
after reflecting across the main diagonal. In order for the relation to be symmetric, these entries must be
the same (both 0 or both 1 ) in each case.
(b) All nonloop edges must appear in anti parallel pairs; in other words, whenever there is an edge from one
vertex to another vertex, an edge in the opposite direction (from the second vertex back to the first) must
also be present.
3. (a) We need to arrange for each integer to be related to itself as well as to the integers to which it is already
related. Therefore the answer is that x is related to y if and only if x :S y. In brief, the reflexive closure
of < is ::; .
(b) We need to arrange for x to be related to y when y is related to x, i.e., y < x (in addition to those cases
in which x is already related to y). Thus x is related to y in the symmetric closure if and only if x < y
or y <
x, or, more succinctly, x-/:-
y.
(c) The transitive symmetric closure must contain at least the pairs in the symmetric closure, which we just
found to be { (x, y) I x -/:- y } . In addition, for every x, we have x -/:- x + 1 and x + 1 -/:- x, so x will have
to be related to x in the transitive closure. But this means that every pair of numbers are related, so the
transitive symmetric closure is all of Z x Z, i.e., the relation that always holds.
4. (a) By Exercise 9 in Section 9.5, every relation defined by "xRy if and only if f(x) = f(y)" is an equivalence
relation. That is the situation here, with f(x) = x 2 - 5x + 6. Therefore this is an equivalence relation.
(b) We need to find the set of all numbers x such that x 2 - 5x + 6 = 22
equation for x gives x = 2 and x = 3. Therefore [2] = {2,3}.
-
5·2+ 6
= 0. Solving this quadratic
(c) This is like the previous part. We need to solve x 2 -5x+6 = r 2 -5r+6, which reduces to x 2 -r 2 -5x+5r = 0,
or, after factoring, (x + r)(x - r) - 5(x - r) = 0, or more simply (x - r)(x + r - 5) = 0. This equation
clearly has two solutions, namely x = r and x = 5 - r. Therefore [r] = {r, 5 - r}. For each r other than
2 ~ , this equivalence class has two elements.
5. (a)
7r
is a set of nonempty, pairwise disjoint subsets of A whose union is A.
(b) For x, y E A, xRy if and only if x and y are in the same set in the partition, i.e., 3B E 7r(x E B /\ y E B).
(c) The equivalence relation determined by
7r is R = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5),
(2, 2), (2, 4), (4, 2), (4, 4), (6, 6)}, and the partition determined by R (its set of equivalence classes) is then 7r.
6. (a) The only maximal element, which is also a greatest element, is the set itself, {a, b, c}. There is no least
element, but there are three minimal elements, namely the sets {a}, { b}, and {c}.
(b) The answer is not unique; here is one such order: {b}-< {a}-< {a,b}-< {c}-< {a,c}-< {b,c}-< {a,b,c}.
(c) The least upper bound of two sets is their union.
7. (a) Every pair of elements must have a least upper bound and a greatest lower bound.
(b) We need to find a situation where upper bounds and lower bounds exist, but in which least upper bounds
or greatest lower bounds do not. One example is to take the following sets, under the ~ relation: 0, {l},
{2}, {1,2,3}, {1,2,4}, {1,2,3,4}. Then 0 is a lower bound for every two of these, and {1,2,3,4} is
an upper bound. However, {1} and {2} have no least upper bound, since {1,2,3} and {1,2,4} are both
upper bounds, but neither is less than the other.
Sample Tests
530
Sample Test for Chapter 10
1. The intersection graph of a collection of sets is the graph whose vertices are the sets in the collection, in which an
edge joins two sets if the two sets are not disjoint. Consider the intersection graph of the set of all two-element
subsets of {1, 2, 3, 4}. You will find that this graph has 6 vertices and 12 edges.
(a) Draw a picture of this graph.
(b) Determine the degree of each vertex, and verify that the handshaking theorem is satisfied.
(c) Write down the adjacency matrix of this graph.
( d) Determine whether this graph is bipartite.
( e) What is wrong with the following "proof" that this graph is planar? Since v = 6 and e = 12, the inequality
for planar graphs e ~ 3v - 6 is satisfied; therefore the graph is planar. Determine whether this is in fact
a planar graph.
(f) Find the chromatic number of this graph.
(g) Determine whether this graph has a Hamilton circuit and whether it has an Euler circuit.
2. There are two notions of connectivity in directed graphs.
(a) Define what it means for a directed graph to be strongly connected, and define what it means for a directed
graph to be weakly connected.
(b) Give an example of a directed graph that is weakly connected but not strongly connected, and give an
example of a directed graph that is not weakly connected.
3. Find all nonisomorphic graphs with 6 vertices and 4 edges.
4. Show how to apply Dijkstra's algorithm to find the shortest path from a to z in the following weighted graph.
d
e
Sample Tests
531
Solutions for Chapter 10 Sample Test
1. (a) There is more than one way to draw this picture. We have labeled each vertex with the two-element subset
of {1, 2, 3, 4} that it represents (omitting the comma and braces for simplicity).
(b) We observe that each of the 6 vertices has degree 4, so that there should be 6 · 4/2 = 12 edges, as there
are.
( c) The adjacency matrix is as follows, where we are using the following order of the vertices: 12, 13, 14, 23,
24, 34.
0
1
1
1 1 1
0 1 1
1 0 0
1
1
1
0
1
0
1
1
0 0 1 1
1 0 1 1 0
1
1
0
0
1
1
1
(d) A bipartite graph can contain no triangle (K3 ). This graph contains many triangles, such as 12-13-23.
Therefore it is not bipartite.
( e) The theorem being applied here gives a necessary condition for a graph to be planar, not a sufficient
condition. There are many graphs that satisfy this condition and yet are not planar ( K 3 ,3 , for instance).
In fact, though, the graph is planar, as we see in our drawing above.
(f) Clearly at least 3 colors are required to properly color this graph, since it has triangles. On the other
hand, if we put 12 and 34 into one color class, 13 and 24 into a second, and 14 and 23 into a third, then
we have colored the graph with 3 colors. Therefore the chromatic number is 3.
(g) This graph must have an Euler circuit because it is connected and the degree of every vertex is even. We
can also easily find a Hamilton circuit in this graph by trial and error: 12-13-34-23-24-14-12.
2. (a) A directed graph is strongly connected if given any two vertices u and v there is a path from u to v. (Note
that this really means that there are paths both ways, since we can interchange the roles of u and v.)
A directed graph is weakly connected if the underlying undirected graph is connected-in other words,
given any two vertices u and v, there is a path from u to v in the underlying directed graph (i.e., if the
directions on the edges are ignored).
(b) The digraph consisting of just two vertices u and v and the directed edge from u to v (i.e., edge (u, v))
is weakly connected but not strongly connected (because there is no path from v to u). The digraph
consisting of these two vertices and no edges is not even weakly connected, because there is no path from
u to v in the underlying undirected graph (since it has no edges).
Sample Tests
532
3. There are 9 nonisomorphic graphs with 6 vertices and 4 edges, as shown here.
•
•
•
•
•
4. The following table shows the operation of the algorithm. Each line shows the labels on the vertices as they
change from iteration to iteration. The number is the distance from vertex a (the source) and the letter is the
vertex from which the currently known best path reaches this vertex. An asterisk indicates that the vertex was
selected during that iteration (the vertex not yet selected having the smallest label).
a:
o*
b:
00
2a*
c:
00
d:
00
5a
3a
e:
00
f:
00
z:
00
00
5a
3a*
4d*
00
00
00
6b
6d
6b
17b
17b
6d
6d*
5c*
17b
llf
We now read off the path in reverse, from z back to e, back to d, back to a. Its length is given by the label
on z, namely 8.
533
Sample Tests
Sample Test for Chapter 11
1. (a) State the definition of a tree.
(b) State the definition of a rooted tree.
( c) Give an example of two trees that are isomorphic as trees but not isomorphic as rooted trees.
2. Suppose that a full 4-ary tree has 27 internal vertices.
(a) How many leaves does it have?
(b) What is the smallest height it could possibly have?
(c) What is the largest height it could possibly have?
3. Insert the words of the following sentence into a binary search tree in the order they are encountered, using
alphabetical order to define the order relation in the tree. The only thing we have to fear is fear itself. (Note
that the second occurrence of the word fear is merely found by the insertion algorithm and thereby skipped.)
4. Consider the expression 4 * 5 - 4 * (4 + 1)
full parenthesization.
* 2.
Recall the usual rules for precedence of operations in absence of
(a) Draw the expression tree for this expression.
(b) Write this expression in prefix form, and show how to evaluate it directly in that form.
( c) Write this expression in postfix form.
(d) Write this as a fully parenthesized infix expression.
5. Consider the complete bipartite graph whose vertex sets are {2, 3, 5, 8} and {1, 4, 7, 9, 10}. In the following
parts, use vertex number to break ties in selecting vertices at each point in an algorithm.
(a) Find the breadth-first search spanning tree of this graph, starting at vertex 1.
(b) Find the depth-first search spanning tree of this graph, starting at vertex 1.
(c) Consider this a weighted graph by assigning the weight lu -vi to edge uv. (For example, edge {5, 9} has
weight 4, but edge {5,8} is not present.) Find a minimum spanning tree of this weighted graph, using
Kruskal's algorithm.
Sample Tests
534
Solutions for Chapter 11 Sample Test
1. (a) A tree is a connected undirected graph with no simple circuits.
(b) A rooted tree is a tree in which one vertex has been specified as the root. It can be thought of as a directed
graph, in which all edges are considered to be directed away from the root.
( c) Let T 1 and T2 both consist of vertices a, b, and c, with edges ab and be. As trees they are clearly
isomorphic (in fact they are equal). Make T1 a rooted tree by specifying a as the root, and make T2 a
rooted tree by specifying b as the root. As rooted trees these are not isomorphic, since in T1 the root has
degree 1, whereas in T2 the root has degree 2.
2. (a) Since each internal vertex has 4 children, the total number of children must be 4 · 27 = 108. Therefore
there are a total of 109 vertices, since only the root is not a child. Thus it has 109 - 27 = 82 leaves.
(b) Recall that the height of an m-ary tree is at least pogm
height of this tree is at least pog 4 821 = 4.
Zl , where
l is the number of leaves. Therefore the
(c) The tree's maximum height would be achieved if only one vertex at each level had children. Therefore the
27 internal vertices would occur at levels 0, 1 , ... , 26, and the height would be 27.
3. The tree will end up looking like this:
the
/~
only
thing
~
/
heve
~~
f eer
we
/
is
to
""'
itself
4. (a) The outermost operation is the subtraction, and within the second term, the first multiplication is done
before the second. We are led to the following tree.
*
~~ *
~~
/\
4
/""*'""-
5
4
2
+
/\
4
1
(b) We traverse the tree in preorder, obtaining - * 4 5 * * 4 + 4 1 2. To evaluate this, we can look for the first
occurrence of an operator immediately followed by two operands (numbers) and replace these three items
by the result of the operation applied to the operands (from left to right). If we do this on the leftmost occurrence at each stage, then we have successively - 20 * * 4 + 4 1 2 = - 20 * * 4 5 2 = - 20 * 20 2 =
- 2040 = -20.
( c) We traverse the tree in postorder, obtaining 4 5 * 4 41+*2 * - .
(d) ((4 * 5) - ((4 * (4 + 1)) * 2))
Sample Tests
535
5. (a) We include the following edges on the first pass: 1-2, 1-3, 1-5, and 1-8. Then we "fan out" from
vertex 2, adding the edges 2-4, 2-7, 2-9 and 2-10, completing the tree.
(b) The search starts by adding edge 1-2 and proceeding from vertex 2. It then moves to the first unused
vertex adjacent to vertex 2, namely vertex 4, adding edge 2-4 to the tree and continuing from vertex 4.
Continuing in this way, we add the following edges: 4-3, 3-7, 7-5, 5-9, 9-8, and 8-10.
(c) Several edges of weight 1 are added: 1-2, 3-4, 4-5, 7-8, and 8-9. Then some edges of weight
2 that do not form a simple circuit are added: 1-3, 5- 7, and 8-10. Since we have formed a tree
(having added 9 - 1 = 8 edges to the 9 vertices already present), we are done. The total weight is
1+1+1+1+1+2+2+2= 11.
536
Sample Tests
Sample Test for Chapter 12
1. Use Boolean identities to simplify x
+ x + y.
2. Find a sum-of-products expansion for the Boolean expression in problem 1.
3. Draw a combinatorial circuit for the Boolean expression obtained in problem 2.
4. Consider the Boolean expression x y z + x y z + x y z
+ x y z + x y z + x y z.
(a) Draw a K-map for this expression, circle the blocks, and find a minimal expansion by finding a minimal
set of blocks covering all the 1's in your diagram.
(b) Use the Quine-McCluskey method to find a minimal expansion.
5. One of the following equations always holds in a Boolean algebra, but the other need not always hold. Prove
the valid identity, and give an example of a Boolean algebra in which the invalid one fails.
(a) x/\(yVz)=(x/\y)Vz
(b) (x/\y) =xVy
Sample Tests
537
Solutions for Chapter 12 Sample Test
+ (xY). Then we can apply the distributive law (addition
over multiplication) to expand this into (x + x)(x + Y). Next we use the identity that x + x = 1 to rewrite this
as 1 · (x + y), and finally use the identity law to obtain x + y.
1. First we apply De Morgan's law to write this as x
2. We just saw that this expression equals x + y. In order to rewrite this in sum-of-products form, we can multiply
the first term by 1, written as y + y, and we can multiply the second term by 1, written as x + x. This gives
us x(y + Y) + (x + x)y. Then expanding by the distributive law we have xy + xy + xy + xy, which upon
deleting the duplication by the idempotent law gives us the desired answer: x y + x y + xy.
3. The appropriate inputs (some complemented) are fed to the three AND gates, and the outputs from these three
gates provide the input to the OR gate.
x
xy
y
x
y
x
>[>o
>[>o
y
>[>o
)D
)
~ xy+xy+xy
xy
xy
4. (a) The picture below shows that two blocks of four squares-the top row and the 2 x 2 square consisting of the
first and last columns-cover all the terms. Therefore the solution is the sum of the Boolean expressions
for these two blocks, namely x + z.
yz yz gz gz
x (1
~
1
1)
1
~
(b) Here is the Quine-McCluskey calculation. We see that the two terms in the last column cover all the
minterms, so the answer is x + z.
1
2
3
4
5
6
Term
xyz
xyz
xyz
xyz
xyz
xyz
String
111
110
101
011
100
001
Step 1
Term
String
11(1, 2) x y
1-1
(l,3)xz
(l,4)yz
-11
(2, 5) xz
1-0
(3, 5) xy
10-01
(3, 6) y z
0-1
(4, 6) x z
Step 2
Term
String
(1,2,3,5)x
(1, 3, 4, 6) z
1---1
5. (a) This is invalid. As an example, take x = 0 and z = 1 (y can be anything). Then the left-hand side is
0 /\ (y V z) = 0, while the right-hand side is (x /\ y) V 1=1.
(b) This follows from De Morgan's law and the double complement law: ( x /\ Y)
=
x Vy
=
x V y.
Sample Tests
538
Sample Test for Chapter 13
1. Consider the following grammar. The terminal symbols are a and b; the nonterminal symbols are S, A, B,
and C; the start symbol is S; and the productions are S __..,ABC; A__.., Aa; A__.., a; B __..,BB; B __.., b;
C __.., >.; C __.., aaC.
(a) Describe the language generated by this grammar.
(b) Which of the terms type 0, type 1, type 2, and type 3 apply to this grammar?
( c) Find a deterministic finite-state machine that recognizes the language generated by this grammar.
(d) Describe this language using a regular expression.
2. Write down the rules for a grammar, using Backus-Naur form, that will generate all positive integer multiples
of 5. Leading O's are not permitted. Assume that the rule
(nonzero digit) ::= 1 I 2 I 3 I 4 I 5 I 6 I 7 I 8 I 9
has already been included.
3. Describe in two or three sentences how nondeterministic finite-state automata are really no more powerful than
deterministic ones. Be specific about the construction involved.
4. State precisely the relationship among regular grammars, regular sets, and finite-state machines.
5. State whether the following statement is true or false, and explain in a short paragraph, being as specific as
possible. Every set of strings can be recognized by some finite-state automaton.
6. Construct a Turing machine that computes the function J(n 1 ,n2 ) = n 2 -n 1 whenever nz ::'.'.: n 1 ::'.'.: 0, using the
notational convention for representing nonnegative integers and inputs to functions adopted in this chapter. It
does not matter how the machine behaves if n 1 > n 2 .
Sample Tests
539
Solutions for Chapter 13 Sample Test
1. (a) From the first production we see that the strings generated by this language seem to have three blocks to
them, spawned by A, B, and C, respectively. It is clear that A leads to any string of one or more a 's.
Similarly, B leads to any string of one or more b's. Finally, C is seen to lead to a string consisting of an
even number (possibly 0) of a's. Thus we can describe this language as { ambna 2 k I m > 0, n > 0, k 2: 0}.
(b) This is certainly a type 0 grammar, since type 0 grammars have no restrictions. It is also a type 1 grammar,
since no production shortens unless it has >. as its right-hand side. In fact, it is a context-free (type 2)
grammar, since the left-hand sides are all single nonterminal symbols. It is not a type 3 (regular) grammar,
since, for instance, the first production is not of the required form.
( c) The following nondeterministic machine does the trick, since it leads to an accepting state, s 2 or s 4 ,
precisely when there have been some a's followed by some b's, followed by an even number of a's. To
make this machine deterministic, simply add a graveyard state, s 5 , with transitions to it on all inputs not
shown here (for example, from state s 3 on input b ).
stert
(d) The following regular expression is easily seen to work: aa*bb*(aa)*.
2. The key is to realize that a positive integer multiple of 5 is either 5 itself or else consists of a positive integer
followed by either a 0 or a 5. We can define a digit by (digit) ::= 0 I (nonzero digit), positive integer
recursively by (positive integer) ::= (nonzero digit) I (positive integer)(digit), and finally multiple of five by
(multiple of five)::= 5 I (positive integer)5 I (positive integer)O.
3. A nondeterministic finite-state automaton can be converted into a deterministic finite-state automaton that
recognizes exactly the same language. This latter machine has as its states all the sets of states of the former
machine, with a transition from A to B when B is the set of all states that can be reached from states in A on
the given input, and has as its final states all the sets of states that include final states of the nondeterministic
machine.
4. Kleene's theorem tells us that regular sets (i.e., sets that can be formed using concatenations, unions, and Kleene
closures starting from singletons and the empty set) are precisely the sets recognized by finite-state machines.
Another theorem of this chapter tells us that these are precisely the sets of strings that can be generated by
regular grammars. Explicit constructions allow one to go back and forth among a regular expression describing
the language, a regular grammar generating the language, and a finite state machine accepting the language.
5. The statement is false. For one thing, the orders of infinity of these two collections are different: there are
an uncountable number of sets of strings and only a countable number of finite-state machines. More to the
point, though, the pumping lemma and special cases of it allow one to show that infinite languages accepted
by finite-state machines have to have a certain kind of periodicity to them, since the computations in these
machines will need to cycle around and around in the same loop for arbitrarily many repetitions. In particular,
the set of strings of the form on 1n does not have this property, because there is no way for the cycling to
preserve the same number of O's as l's and at the same time have all the O's precede all the l's.
6. The idea is that we need to erase n 1 + 1 l's from both inputs to perform this subtraction, and then change the
* to a 1 in order to take into account the fact that n is represented by n + 1 1's. For example, if the input is
111*1111111, then we want 6 - 2 = 4, so we need the output to be 11111. As in Example 3 in Section 13.5, we
will wander between the ends of the string, erasing l's from each end, until the first input ( n 1 ) is exhausted.
Here are some five-tuples that will do the job: (so, 1, s 1 , B, R), (s 0 , *, s 4 , 1, R), (s 1 , 1, s 1 , 1, R), (s 1 , *, s1, *, R),
(s1,B,s2,B,L), (s2,l,s3,B,L), (s3,l,s3,l,L), (s3,*,s3,*,L), (s3,B,so,B,R).
540
Common Mistakes in Discrete Mathematics
Common Mistakes in Discrete Mathematics
In this section of the Guide we list many common mistakes that people studying discrete mathematics sometimes
make. The list is organized chapter by chapter, based on when they first occur, but sometimes mistakes made early
in the course perpetuate in later chapters. Also, some of these mistakes are remnants of misconceptions from high
school mathematics (such as the impulse to assume that every operation distributes over every other operation).
In most cases we describe the mistake, give a concrete example, and then offer advice about how to avoid
it. Note that additional advice about common mistakes in given, implicitly or explicitly, in the solutions to the
odd-numbered exercises, which constitute the bulk of this Guide.
Solving Problems in Discrete Mathematics
Before getting to the common mistakes, we offer some general advice about solving problems in mathematics, which
are particularly relevant to working in discrete mathematics. The problem-solving process should consist of the
following steps. (This four-step approach is usually attributed to the mathematician George P6lya (1887-1985).)
1. Read and understand the problem at hand. Play around with it to get a feeling for what is going on and what
is being asked.
2. Apply one or more problem-solving strategies to attack the problem. Do not give up when one particular tactic
doesn't work. This phase of the process can take a long time.
3. Carefully write up the solution when you have solved the problem. Make sure to communicate your ideas
clearly.
4. Look back on what you have done. Make sure that your answer is correct (think of creative ways to test it).
Also consider other ways of solving the problem, and think about how you might generalize your results.
Your bag of problem-solving strategies may include drawing a picture or diagram; looking at special cases or
simpler instances of your problem; looking at related problems; searching for patterns; making tables of what you
know (and in general, being organized); giving names to what you don't know and writing equations about it (or,
more generally, applying the mathematical tools you have learned in algebra and other courses); working backwards
and setting subgoals ("I could solve this problem ifl could do such-and-such"); using trial and error (usually called
"guess and check" in educational circles); using indirect reasoning and looking for counterexamples ("if this weren't
true, then what would have to happen?"); jumping out of the system and trying something totally different; or just
going away from the problem and coming back to it later. You will find it useful to read over this description of
the problem-solving process repeatedly during your study of discrete mathematics.
Here is an example of applying the problem-solving process to a problem in discrete mathematics. Suppose
we want to find the number of squares (of all sizes) on a checkerboard. We understand this to mean not only the
obvious 64 little squares, but also the 2 x 2 squares (of which there will be many), the 3 x 3 squares, and so on,
up to the entire 8 x 8 board itself. We should draw a picture to see what's going on here. Let's begin by playing
with a smaller version of this problem, using a board of size 2 x 2. Then obviously there are 4 small squares and
the entire board, for a total of 4 + 1 = 5. Let's try a 3 x 3 board. Here there are 9 little squares, and it is easy
to see that there are 4 squares of size 2 x 2 (nuzzled in the upper left, the upper right, the lower left, or the lower
right), as well as the entire board; so the answer here is 9 + 4 + 1 = 14. A pattern seems to be emerging-we
seem to be adding perfect squares to get the answer. Maybe for the 4 x 4 board there will be 16 + 9 + 4 + 1 = 30
squares in all. We draw the 4 x 4 picture and verify that this is correct. In fact, we can see exactly what is
going on, the way the upper left corner of a k x k square can be in any of the first 5 - k rows and first 5 - k
columns of the board, for k = 1, 2, 3, 4, and so there are (5 - k) 2 squares of size k in the 4 x 4 board, exactly
as our sum indicated. We have now solved the problem and can write up a solution explaining why the answer is
82 + 72 + 62 + 52 + 4 2 + 32 + 22 + 12 = 204. In the looking back stage (step 4) we would certainly want to notice that
for an n x n board, there are L~=l i 2 squares. To continue our investigation, we might want to explore such further
questions as allowing rectangles rather than squares, looking at rectangular checkerboards rather than square ones
(or counting triangles in boards made up of triangles), or moving to 3-dimensional space and counting cubes in a
large cube. Notice how we followed the process outlined above and used many of the strategies listed.
Common Mistakes in Discrete Mathematics
541
List of Common Mistakes
If students or instructors have items to add to the lists below, please let the author know (visit the companion
website for Discrete Mathematics and Its Applications at www.mhhe.com/rosen).
Chapter 1
• Incorrectly translating English statements into symbolic form. There are many errors of this type. For example,
there are difficulties with the use of the word "or" in English; be sure to differentiate between inclusive and
exclusive versions (see page 5 of the text). A conditional statement is quite different from a conjunction, but
some speakers fail to distinguish them; to say that B will happen if A happens is quite different from saying
that A and/or B will happen. Perhaps the most common mistake is confusing p----> q with q----> p. To say, for
example, that I will go to the movie if I finish my homework means something quite different from asserting
that I will go to the movie only if I finish my homework.
• Incorrectly negating compound statements without using De Morgan's laws-in effect saying, for example, that
•(p V q) is logically equivalent to •P V •q, or that •(p A q) is logically equivalent to •PA •q. For example,
if it is not true that John is over 18 years old or lives away from home, then it is true that he is not over 18
years old and (not or) he does not live away from home. The correct statements are that -, (p V q) is logically
equivalent to •p A •q, and that •(p A q) is logically equivalent to •p V •q. This mistake is a general instance
of assuming that every operation distributes over every other operation, here that negation distributes over
disjunction (or conjunction).
• Misinterpreting the meaning of the word "any" in a mathematical statement. This word is ambiguous in many
situations, and so should usually be avoided in mathematical writing. If you are not sure whether the writer
meant "every" or "some" when the word "any" was used, get the statement clarified. As a corollary, of course,
you should avoid using this word yourself. Here is an example: What would one mean if she defined a purple
set of integers to be one "in which any integer in the set has at least three distinct prime divisors"? Does "any''
mean "every" here (in which case the set {30, 40} is not purple), or does "any" mean "some" here (in which
case the set {30,40} is purple)?
• Incorrectly writing the symbolic form of an existential statement as 3x(A(x) ----> B(x)) instead of 3x(A(x) A
B(x)). For example, the symbolic form of "There exists an even number that is prime" is 3x(E(x) A P(x)),
not 3x(E(x)----> P(x)), where we are letting E(x) mean "xis even" and P(x) mean ''xis prime." As a rule
of thumb, existential quantifiers are usually followed by conjunctions.
• Incorrectly writing the symbolic form of a universal statement as \lx(A(x)AB(x)) instead of \lx(A(x)----> B(x)).
For example, the symbolic form of "Every odd number is prime" is \fx(O(x)----> P(x)), not \lx(O(x) AP(x)),
where we are letting O(x) mean "x is odd" and P(x) mean "x is prime." As a rule of thumb, universal
quantifiers are usually followed by conditional statements.
• Incorrectly putting predicates inside predicates, such as P(O(x)). For example, if P(x) means "x is prime,"
and O(x) means "xis odd,'' then it would never make sense to write P(O(x)) in trying to express a statement
such as "xis an odd prime" or to write \Ix P(O(x)) to say "all odd numbers are prime." The notation P(O(x))
would mean that the assertion that x is odd is a prime number, and clearly an assertion isn't any kind of
number at all. Functional notation has a wonderful internal beauty and consistency to it-the thing inside the
parentheses has to be what the thing outside the parentheses applies to.
• Failure to change the quantifier when negating a quantified proposition, especially in English. For example, the
negation of the statement that some cats like liver is not the statement that some cats do not like liver; it is
that no cats like liver, or that all cats dislike liver.
• Overusing the term "by definition" in justifying statements in a proof. For example, Franklin Roosevelt was
not the President of the United States at the start of the country's entry into World War II in December, 1941,
"by definition"; he was the President because he had been inaugurated as such early in 1941 and had not died
or left office.
• Not going back to carefully check the definitions in justifying statements in a proof. For example, if one is trying
to prove something about odd integers, then it is important to correctly use the meaning of that notion (that
an odd integer is one that can be written as 2k + 1 for some integer k) at one or more places in the proof.
• Incorrectly starting a proof by assuming what is to be proved. A common occurrence of this in an earlier course
is trying to prove trigonometric identities by starting with the identity and using algebra to reach A = A; this
Common Mistakes in Discrete Mathematics
542
is not valid. Similarly, if we are trying to prove a set identity in Chapter 2, such as A <;;; AU B, it would be
invalid to start with the statement A <;;; A U B.
• Invalidly assuming that a few (or even a large number of) examples of a universally quantified proposition
imply that the proposition is true. There is an example from number theory of an intriguing proposition about
a positive integer n that is true for every n ::; 4,000,000 with the sole exception of n = 1969. A proof of the
universally quantified proposition 'v'x P(x) consists of showing that the property P(x) holds no matter what
x is chosen from the domain (universe of discourse).
• Incorrectly assuming that an arbitrary object has a particular property when all you know is that there exists
an object with the property. For example, suppose we are trying to prove the assertion that x 2 always leaves
a remainder of 1 when divided by 8. It would be invalid to start our proof by assuming that x = 2n + 1 for
some integer n; even though some integers have this property of being odd, it is not true that all of them do,
so we would be proving the assertion to be true only some of the time, rather than always.
Chapter 2
• Incorrectly forming complements of sets without using De Morgan's laws-in effect saying, for example, that
A n B = A n B . The correct statements are A n B = A U B and A U B = A n B. This is another general
instance of assuming that every operation distributes over every other operation, here that complementation
distributes over intersection (or union). Students sometimes make similar errors in algebra, such as asserting,
incorrectly, that a 2 + b2 = a + b or sin( a + {3) = sin a + sin f3.
v
• Using incorrect notation regarding elements and subsets of power sets and confusing the notions of "element"
and ''subset" when dealing with a power set. If A is a subset of S, then A is an element of the power set
of S. For example, if S = {p, q, r}, then {p, r} <;;; S, so {p, r} E P(S). On the other hand, {p, r} 3v - 6 ), then it cannot be planar. What we cannot conclude
is the converse-it is not a theorem that if e ::::; 3v - 6, then the graph has to be planar.
• Using the nonexistent word "vertice" instead of the correct word "vertex" to talk about yust one of the dots in
a graph. Similarly, there is no such thing as a "matrice."
• Mistakenly thinking that a graph is nonplanar just because it is drawn one way with two edges crossing. If it
is possible to redraw the graph without edges crossing, then the graph is planar. For example, K 4 is planar,
even though drawing it as the vertices of a square with straight line segments representing the edges causes
a crossing in the middle of the picture. (Redraw it as the vertices of a triangle with one more vertex in the
interior.)
• Invalidly concluding that once one has found a coloring of a graph with n colors, its chromatic number has to
be n. In fact, all we know in that case is that its chromatic number is at most n. It may be possible to find
another coloring with fewer than n colors. For example, one could color C4 with four colors (a different color
for each vertex), but its chromatic number is in fact 2.
• Mistakenly believing that greedy algorithms always produce the optimal solution to a problem. It often happens
that the simple-minded greedy approach does find the best solution (e.g., in looking for minimum spanning
trees in Section 11.5), but often the greedy approach fails (e.g., in finding a coloring of a graph using as few
colors as possible).
• Failing to recognize that writing down a procedure doesn't guarantee that it does what you want it to do. For
example, one cannot write down a greedy algorithm to color a graph and then claim without justification that
this procedure finds the coloring with the fewest possible colors. It is not hard to find a counterexample.
Common Mistakes in Discrete Mathematics
549
Chapter 11
• Incorrectly setting up a decision tree for a problem such as identifying counterfeit coins by weighing them, and
thereby drawing the wrong conclusions. Each possible situation must correspond to a path from the root of the
tree to a leaf.
• Not realizing what type of tree is needed for a particular mathematical model. Issues to consider are whether
there should be a root (a starting point for some process), whether the children of a vertex are ordered, and
whether each child should be classified as a right child or a left child.
• Incorrectly omitting parentheses in expressions written in infix notation. In absence of a default order of
operations, an expression such as AnBUC is ambiguous, since it might mean either (AnB)UC or An(BUC),
and these are not the same sets. With prefix or postfix notation, no such ambiguities arise.
• Forgetting that when doing an inorder traversal of an ordered rooted tree that is not binary, the root of each
subtree comes after the first subtree but before all the other subtrees. For a binary tree, inorder traversal is
rather obvious-left, root, right. When the tree isn't binary, we can still define inorder traversals, but the
definition isn't as natural.
• When applying Prim's algorithm for finding minimum spanning trees, forgetting that edges become eligible for
inclusion in the tree gradually (as opposed to Kruskal's algorithm, in which they are all eligible from the start).
If there is a low-cost edge that does not currently have any endpoint in the tree constructed so far, then it
cannot yet be added to the tree. When one of its endpoints finally becomes part of the tree, it suddenly
becomes eligible and can then be added to the tree if it is the lowest cost edge currently eligible. It is easy to
overlook such edges when performing the algorithm.
Chapter 12
• Being off by one level of abstraction when thinking about Boolean Junctions. A Boolean function with n
variables can be represented by a table with 2n rows; therefore there are 2 2 different Boolean functions with
n variables.
n
• Putting inverters in the wrong place when building combinational circuits. If we want to invert the value of the
output of a gate, the inverter needs to go after the gate.
• Forgetting to apply De Morgan's laws correctly when evaluating the output of a combinational circuit. The
output of a circuit is a certain Boolean expression of the input variables. When simplifying this expression, it
is important to remember than xy = x + y and x + y = xy.
• Not finding the largest possible blocks when looking for minimum Boolean expressions using K-maps. Since
there is no known efficient algorithm for solving this problem in general (with more than just a few variables),
it should not be surprising that this procedure seems to involve some ugly "guessing" to it.
• Not finding the best cover when looking for minimum Boolean expressions using the Quine-McCluskey method.
Since there is no known efficient algorithm for solving this problem in general (with more than just a few
variables), it should not be surprising that this procedure seems to involve some ugly "guessing" to it. It
might be very hard to make sure that a covering we have found with, say, five minterms is really the best
possible-that there isn't another covering with four minterms.
Chapter 13
• Incorrectly constructing grammars to generate a desired language. There is no algorithm for doing this (this
statement is a theorem in the theory of computation, similar to Turing's theorem on the unsolvability of the
halting problem). Constructing grammars is like writing computer programs, and all the advice given in a
programming course (such as thinking from the top down in a structured way) applies.
• Incorrectly constructing finite-state machines (including Turing machines) to perform a desired task. There
is no algorithm for doing this (this statement is a theorem in the theory of computation, similar to Turing's
theorem on the unsolvability of the halting problem). Constructing machines is like writing computer programs,
and all the advice given in a programming course (such as thinking from the top down in a structured way)
applies.
550
Common Mistakes in Discrete Mathematics
• Not including all the strings that are accepted by a given finite-state automaton. Sometimes students will follow
some paths that the machine can take to reach an accepting state and forget to consider others. This will lead
to a claim that the language recognized by this automaton is a proper subset of what it really is. Make sure
to "play computer" and follow all the branches.
• Forgetting to have one arrow leaving each state for each input symbol when constructing deterministic fl nit estate automata. You usually want to have a "graveyard" state to which the machine goes when it is clear
that the input is not acceptable. There needs to be a loop from the graveyard state to itself for each alphabet
symbol.
• Failing to realize that a nondeterministic finite-state automaton can accept a string even when some computation
paths on a certain input drive the machine to a nonaccepting state. As long as at least one path leads to an
accepting state, the input string is accepted.
• Failing to keep track of all the possible states in which a nondeterministic finite-state automaton can enter at
each step, when constructing the corresponding deterministic automaton. Make good use of all your fingers in
analyzing what can happen!
• Failing to check that a machine or a grammar or a regular expression presented as the solution of some problem
actually works. This is similar to debugging a computer program. Many test cases should be tried, so that you
can be confident that your machine or grammar or expression really works.
• When constructing Turing machines, forgetting to include all the cases. If the machine can ever reach a certain
state and be viewing a particular input symbol, then a transition is needed to handle that case. Using top-down
programming methodology is advisable to make sure your machines do what you want them to do.
• Getting so bogged down in the details of constructing Turing machines that you lose sight of the main points
of the theory. The main point is given in the Church-Turing thesis: that every conceivable computation
can be performed by any reasonable computational model, be it a Turing machine, your favorite high-level
programming language, or yourself working with pencil and paper. And on that note of keeping the "big
picture" in mind, we'll bring this list of common mistakes to a close.
Appendixes
• Incorrectly thinking that certain operations distribute over certain other operations. For example, it is not true
that Va+/)
= Va+ Vb
or
Va 2 + b2 = a + b or
ax
+ aY
= ax+y or log( a + b)
= log a + log b.
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Logical and: p /\ q is true when both p and q are true, is false when at least one of p and q is false.
Logical or (inclusive): p V q is true when at least one of p and q is true, is false when both p and q are false.
Exclusive or: p EB q is true when exactly one of p and q is true, is false otherwise.
Conditional statement (implication): p-> q = "If p, then q" = "p only if q" = "pis a sufficient condition for q"
= "q is a necessary condition for p." p-> q is false when pis true and q is false, is true otherwise. •(p-> q) = p/\( •q).
p-> q is equivalent to its contrapositive •q-> •p, but not to its converse q-> p or its inverse •p-> •q.
Biconditional statement: p +-+ q, means (p-> q) /\ (q-> p), usually read "if and only if" and sometimes written
"iff" in English.
De Morgan's laws: •(p V q) = (•p) /\ (•q); •(p /\ q) = (•p) V (•q).
The basic logical operations can be represented by gates:
x~~x,J,y
x~x x-7~xy x~~x+y x~~xly Y~v'-',
~
Y-7LJ7
inverter
AND
Y~v ,
Y~L/,
OR
NANO
NOR
They can be combined to make combinational circuits to represent any logical expression.
Quantifiers: Vx(P(x)-> Q(x)) ="for all x, if P(x) then Q(x)"; 3x(P(x) /\ Q(x)) ="there exists an x such that
P(x) and Q(x)." Here P(x) and Q(x) are propositional functions, and there is always a domain or universe
of discourse, either implicit or explicitly stated, over which the variable ranges.
Negations of quantified propositions: •\:/xP(x) = 3x•P(x); --,:JxP(x) = \:/x•P(x).
Theorem: a proposition that can be proved; lemma: a simple theorem used to prove other theorems; proof:
a demonstration that a proposition is true; corollary: a proposition that can be proved as a consequence of a
theorem that has just been proved.
A valid argument-one using correct rules of inference based on tautologies-will always give correct conclusions if
the hypotheses used are correct. Invalid arguments, relying on fallacies, such as affirming the conclusion, denying
the hypothesis, begging the question, or circular reasoning, can lead to false conclusions.
Some rules of inference: [p/\ (p-> q)] -> q (modus ponens); [•q /\ (p-> q)] -> •p (modus tollens); [(p-> q) /\
(q-> r)]-> (p-> r) (hypothetical syllogism); [(pV q) /\ (•p)]-> q (disjunctive syllogism); {P(a) /\ Vx [P(x)->
Q(x)]}-> Q(a) (universal modus ponens); {•Q(a) /\\:Ix [P(x) -> Q(x)]}-> •P(a) (universal modus tollens);
(\:Ix P(x)) -> P(c) (universal instantiation); (P(c) for an arbitrary c) ->\:Ix P(x) (universal generalization);
(3x P(x))-> (P(c) for some element c) (existential instantiation); (P(c) for some element c)-> 3x P(x) (existential generalization).
Trivial proof: a proof of p -> q that just shows that q is true without using the hypothesis p.
Vacuous proof: a proof of p -> q that just shows that the hypothesis p is false.
Direct proof: a proof of p-> q that shows that the assumption of the hypothesis p implies the conclusion q.
Proof by contraposition: a proof of p -> q that shows that the assumption of the negation of the conclusion q
implies the negation of the hypothesis p (i.e., proof of contrapositive).
Proof by contradiction: a proof of p that shows that the assumption of the negation of p leads to a contradiction.
Proof by cases: a proof of (P1 V p 2 V · · · V Pn) -> q that shows that each conditional statement Pi -> q is true.
Statements of the form p +-+ q require that both p -> q and q -> p be proved. It is sometimes necessary to give two
separate proofs (usually a direct proof or a proof by contraposition); other times a string of equivalences can be
constructed starting with p and ending with q: p +-+ P1 +-+ P2 +-+ · · · +-+ Pn +-+ q.
To give a constructive proof of 3xP(x) is to show how to find an element x that makes P(x) true. Nonconstructive existence proofs are also possible, often using proof by contradiction.
One can disprove a universally quantified proposition \fxP(x) simply by giving a counterexample, i.e., an object
x such that P(x) is false. One cannot prove it with an example, however.
Fermat's last theorem: There are no positive integer solutions of xn
+ yn = zn
if n
> 2.
An integer is even if it can be written as 2k for some integer k; an integer is odd if it can be written as 2k + 1 for
some integer k; every number is even or odd but not both. A number is rational if it can be written as p/q, with
p an integer and q a nonzero integer.
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Empty set: the set with no elements, { }, is denoted 0; this is not the same as {0}, which has one element.
Subset: A t;;; B = Vx(x E A --+ x E B); proper subset: A c B = (A t;;; B) /\(A -=I- B) (i.e., B has at least one
element not in A).
Equality of sets: A= B = (A t;;; B /\ B t;;; A) = Vx(x E A +--> x E B).
Power set: P(A) = { B I B ~A} (the set of all subsets of A). A set with n elements has 2n subsets.
Cardinality: ISi =number of elements in S.
Some specific sets: R is the set of real numbers, all of which can be represented by finite or infinite decimals;
N = {O, 1, 2, 3.... } (natural numbers); Z = { ... , -2, -1, 0, 1, 2, ... } (integers); z+ = {1, 2, ... } (positive integers);
Q = { p/ q I p, q E Z /\ q -=I- 0 } (rational numbers); q+ = { p/ q I p, q E z+ } (positive rational numbers).
Set operations: Ax B = {(a, b) I a EA/\ b EB} (Cartesian product); A= the set of elements in the universe
not in A (complement); An B = { x Ix EA/\ x EB} (intersection); AU B = { x Ix EA V x EB} (union);
A - B =An B (difference); A EBB= (A- B) U (B - A) (symmetric difference).
Inclusion-exclusion (simple case): IA U Bl= IAI
+ IBI - IA n Bl.
De Morgan's laws for sets: An B =Au B; Au B =An B.
A function I from A (the domain) to B (the codomain) is an assignment of a unique element of B to each
element of A. Write I : A --+ B. Write l(a) = b if bis assigned to a. Range of I is { l(a) I a E A}; I is onto
(surjective) =range(!) = B; I is one-to-one (injective) = Va 1 Va2[l(ai) = l(a 2 ) --+ a 1 = a 2].
If I is one-to-one and onto (bijective), then the inverse function 1- 1 : B--+ A is defined by 1- 1 (y) = x = J(x) = y.
If f: B--+ C and g: A--+ B, then the composition fog is the function from A to C defined by f og(x) = J(g(x)).
Rounding functions: lxJ =the largest integer less than or equal to x (floor function); lxl =the smallest
integer greater than or equal to x (ceiling function).
n
Summation notation: La, = a 1
+ a 2 + · · · +an.
i=l
. .
.
~·
n(n+l)
S um o ffi rst n positive mtegers: L...J = 1+2 + · · · + n =
.
2
J=l
Sum of squares of first n positive integers:
tj
2 = 12 + 22 + ... + n 2 = n(n + l~( 2 n + l).
J=l
n
Sum of geometric progression: """"'ar1 = a + ar + ar 2 + · · · + arn =
L..
J=O
arn+I - a
r-1
if r -=I- 1.
Two sets have the same cardinality if there is a bijection between them. We say that IAI ~ /Bl if there is a
one-to-one function from A to B.
A set is countable if it is finite or there is a bijection from the positive integers to the set~in other words, if
the elements of the set can be listed a 1 , a 2 , . . . . Sets of the latter type are called countably infinite, and their
cardinality is denoted N0 . The empty set, the integers, and the rational numbers are countable; the set of real
numbers and the power set of the set of natural numbers are uncountable. The union of a countable
number of countable sets is countable.
The Schroder-Bernstein theorem states that if IAI ~ IBI and IBI ~ IAI, then IAI = IBI. In other words, ifthere
is a one-to-one function from A to Band there is a one-to-one function from B to A, then there is a one-to-one and
onto function from A to B.
Matrix multiplication: The (i,j)th entry of AB is I:;=l a,tbt1 for 1 ~ i ~ m and 1 ~ j ~ n, where A is an
m x k matrix and B is a k x n matrix. Identity matrix In with 1's on main diagonal and O's elsewhere is the
multiplicative identity.
Cardinality arguments can be used to show that some functions are uncomputable.
Matrix addition (+), Boolean meet (/\) and join (V) are done entry-wise; Boolean matrix product (8) is like matrix
multiplication using Boolean operations.
Transpose: At is the matrix whose (i,j)th entry is a1 , (the (j, i)th entry of A); A is symmetric if At= A.
Crib Sheets
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An algorithm is a finite sequence of precise instructions for performing a computation or solving a problem
Algorithms can be expressed in pseudocode.
Most algorithms have the following properties: having input, having output, definiteness, correctness, finiteness,
effectiveness, generality.
Algorithms that make what seems to be the "best" choice at each step are called greedy algorithms. Sometimes
they work; sometimes they don't. For example, the greedy change-making algorithm works for American coins, but
does not work for some other combinations of denominations.
There are important algorithmic paradigms besides greedy, including brute force (examine all possible solutions
in order to determine the best solution) and some that will be studied in later chapters (dynamic programming,
probabilistic algorithms, and divide-and-conquer).
The halting problem is unsolvable: There is no algorithm to test whether a given computer program with a
given input will ever halt.
Big-0 notation: "f(x) is O(g(x))" means :JC:Jk\ix(x > k---> lf(x)I S: Clg(x)I). Big-0 of a sum is largest (fastest
growing) of the functions in the sum; big-0 of a product is the product of the big-O's of the factors. If f is O(g),
then g is O(f) ("big-Omega"). If f is both big-0 and big-Omega of g, then f is 8(g) ("big-Theta").
Little-0 notation: This was introduced in the exercise set. We say that f(x) is o(g(x)) iflimx_, 00 f(x)/g(x) = 0.
Powers grow faster than logs: (logn)c is O(xd) but not the other way around, where c and d are positive
numbers.
If fi(x) is O(g1(x)) and h(x) is O(g2(x)), then (!1 + h)(x) is O(max(g1(x),g2(x))) and (f1h)(x) is O(g1(x)g2(x)).
logn! is O(nlogn).
Binary search has time complexity O(logn), whereas linear search has (worst case and average case) time
complexity O(n); both have space complexity 0(1) (not counting input). Bubble sort and insertion sort have
O(n 2 ) worst case time complexity.
Matrix multiplication by the standard algorithm has time complexity O(m 1m 2 m 3) if the matrices have dimensions
m1 x m 2 and m 2 x m3. More efficient algorithms can reduce the complexity of multiplying two n x n matrices from
O(n 3 ) to O(nv'7).
Important complexity classes include polynomial (nb), exponential (bn for b > 1), and factorial (n!).
A problem that can be solved by an algorithm with polynomial worst-case time-complexity is called tractable;
otherwise they are called intractable.
The class P is the class of tractable problems. The class NP consists of problems for which it is possible to check
solutions (as opposed to finding solutions) in polynomial time. Clearly P ~ NP. The P versus NP problem
asks whether in fact P =NP; no one knows the answer.
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Divisibility: a I b means a -1- 0 /\ :3c(ac = b) (a is a divisor or factor of b; bis a multiple of a).
Base b representations: (an-lan-2 ... a2a1ao)b = an-1bn-l + · · · + a2b 2 + aib + ao. To convert from base 10 to
base b, continually divide by band record remainders as a 0 , a 1 , a 2, ... (b = 8 is octal; b = 16 is hexadecimal,
using A through F for digits 10 through 15). Convert from binary to octal by grouping bits by threes, from the
right, to hexadecimal by grouping by fours.
Addition of two binary numerals each of n bits ((an_ 1an_ 2 ... a 2a 1a 0 )2) requires O(n) bit operations. Multiplication requires O(n 2) bit operations if done naively, O(n 1 585 ) steps by more sophisticated algorithms.
Division "algorithm": \fa \Id> 0 :Jq :Jr(a = dq + r /\ 0 :::; r < d); q is the quotient and r is the remainder; we write
a mod d for the remainder. Example: -18 = 5 · ( -4) + 2, so -18 mod 5 = 2.
Congruent modulo m: a = b (mod m) iff m Ia - b iff a mod m = b mod m. One can do arithmetic in
Zm = {0, 1, ... , m - 1} by working modulo m. There are fast algorithms for computing bn mod m, based on
successive squaring.
Integer n > 1 is prime iff its only factors are 1 and itself (2, 3, 5, 7, ... ); otherwise it is composite (4, 6, 8, 9, ... ).
There are infinitely many primes, but it is not known whether there are infinitely many twin primes (primes that
differ by2) or whether every even positive integer greater than 2 is the sum of two primes (Goldbach's conjecture)
or whether there are infinitely many Mersenne primes (primes of the form 2P - 1).
Naive test for primeness (and method for finding prime factorization): To find prime factorization of n,
successively divide it by all primes less than y7i (2, 3, 5, ... ); if none is found, then n is prime. If a prime factor pis
found, then continue the process to find the prime factorization of the remaining factor, namely n/p; this time the
trial divisions can start with p. Continue until a prime factor remains. The prime number theorem states that
there are approximately n/ ln n primes less than or equal ton.
Fundamental theorem of arithmetic: Every integer greater than 1 can be written as a product of one or more
primes, and the product is unique except for the order of the factors. (Proof based on fact that if a prime divides
a product of integers, then it divides at least one of those integers.)
Euclidean algorithm for greatest common divisor: gcd(x, y) = gcd(y, x mod y) if y -1- O; gcd(x, 0) = x. Using
extended Euclidean algorithm or working backwards, one can find Bezout coefficients and write gcd(a, b) = sa+tb.
Two integers are relatively prime if their greatest common divisor (gcd) is l. The integers ai, a 2, ... , an are
pairwise relatively prime iff gcd( ai, a1 ) = 1 whenever 1 :::; i < j :::; n.
Chinese remainder theorem: If m 1 , m 2, ... , mn are pairwise relatively prime, then the system \fi(x =
(mod mi)) has unique solution modulo m 1 m 2 · · ·mn. Application: handling very large integers on a computer.
ai
Fermat's little theorem: av- 1 = 1 (mod p) if p is prime and does not divide a. The converse is not true; for
example 2340 = 1 (mod 341), so 341 (=11·31) is a pseudoprime.
If a and bare positive integers, then there exist integers sand t such that as+bt = gcd(a, b) (linear combination).
=
This theorem allows one to compute the multiplicative inverse a of a modulo b (i.e., aa 1 (mod b)) as long as
a and bare relatively prime, which enables one to solve linear congruences ax= c (mod b).
A primitive root modulo a prime pis an integer r in Zp such that every nonzero element of Zp is a power of r.
Discrete logarithms: logr a = e modulo p if re mod p = a and 1 :::; e :::; p - l.
A common hashing function: h(k) = k mod m, where k is the key.
Check digits, for error-correcting codes like UPCs, involve modular arithmetic.
Pseudorandom numbers can be generated by the linear congruential method: Xn+l = (axn + c) mod m,
where x 0 is arbitrarily chosen seed. Then {xn/m} will be rather randomly distributed numbers between 0 and l.
Shift cipher: f(p) = (p + k) mod 26 [A
f(p) = (ap + b) mod 26 with gcd(a, 26) = l.
f--7
0, B
f--7
1, ... ]. Julius Caesar used k = 3. Affine cipher uses
RSA public key encryption system: An integer M representing the plaintext is translated into an integer C
representing the ciphertext using the function C = l'vfe mod b, where n is a public number that is the product of
two large (maybe 100-digit or so) primes, and e is a public number relatively prime to (p-l)(q-1); the primes p and
q are kept secret. Decryption is accomplished via M = Cd mod n, where d is an inverse of e modulo (p - 1) (q - 1).
It is infeasible to compute d without knowing p and q, which are infeasible to compute from n. Similar methods
can be used for key exchange protocols and digital signatures.
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The well-ordering property: Every nonempty set of nonnegative integers has a least element.
Principle of mathematical induction: Let P(n) be a propositional function in which the domain (universe of
discourse) is the set of positive integers. Then if one can show that P(l) is true (basis step or base case) and
that for every positive integer k the conditional statement P( k) --+ P( k + 1) is true (inductive step), then one has
proved VnP(n). The hypothesis P(k) in a proof of the inductive step is called the inductive hypothesis. More
generally, the induction can start at any integer, and there can be several base cases.
Strong induction: Let P(n) be a propositional function in which the domain (universe of discourse) is the set of
positive integers. Then if one can show that P(l) is true (basis step or base case) and that for every positive
integer k the conditional statement [P(l) A P(2) A··· A P(k)] --+ P(k + 1) is true (inductive step), then one
has proved VnP(n). The hypothesis Vj"5_k P(j) in a proof of the inductive step is called the (strong) inductive
hypothesis. Again, the induction can start at any integer, and there can be several base cases.
Sometimes inductive loading is needed, where we must prove by mathematical induction or strong induction
something stronger than the desired statement so as to have a powerful enough inductive hypothesis (this concept
was introduced in the exercises).
Inductive or recursive definition of a function J with the set of nonnegative integers as its domain: specification
of J(O), together with, for each n > 0, a rule for finding J(n) from values of J(k) fork< n. Example: O! = 1 and
(n + 1)! = (n + 1) · n! (factorial function).
Inductive or recursive definition of a set S: a rule specifying one or more particular elements of S, together with
a rule for obtaining more elements of S from those already in it. It is understood that S consists precisely of those
elements that can be obtained by applying these two rules.
Structural induction can be used to prove facts about recursively defined objects.
Fibonacci numbers: Jo,
Ji, h, ... : Jo= 0, Ji = 1, and Jn =Jn-I+ Jn-2 for all n 2 2.
Lame's theorem: The number of divisions used by the Euclidean algorithm to find gcd(a, b) is O(log b).
An algorithm is recursive if it solves a problem by reducing it to an instance of the same problem with smaller
input. It is iterative if it is based on the repeated use of operations in a loop.
There is an efficient recursive algorithm for computing modular powers (bn mod m), based on computing
bln/ 2 J mod m.
Merge sort is an efficient recursive algorithm for sorting a list: break the list into two parts, recursively sort each
half, and merge them together in order. It has 0( n log n) time complexity in all cases.
A program segment S is partially correct with respect to initial assertion p and final assertion q, written
p{ S}q, if whenever pis true for the input values of S and S terminates, q is true for the output values of S.
A loop invariant for while condition S is an assertion p that remains true each time S is executed in the loop;
i.e., (p A condition){S}p. If pis true before the program segment is executed, then p and -,condition are true after
it terminates (if it does). In symbols, p{ while condition S}(-,condition A p).
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Sum rule: Given t mutually exclusive tasks, if task i can be done in n, ways, then the number of ways to do
exactly one of the tasks is ni + n2 + · · · + nt.
IA1 U A2 U · · · U Anl = IA1I + IA2I + · · · + IAnl·
of inclusion-exclusion: IA U Bl= IAI + IBI - IA n BJ.
Size of union of disjoint sets:
Two-set case
Product rule: If a task consists of successively performing t tasks, and if task i can be done in n, ways (after
previous tasks have been completed), then the number of ways to do the task is n 1 · n 2 · · · nt.
A set with n elements has 2n subsets (equivalently, there are 2n bit strings of length n).
Tree diagrams can be used to organize counting problems.
Pigeonhole principle: If more than k objects are placed in k boxes, then some box will have more than 1 object.
Generalized version: If N objects are placed ink boxes, then some box will have at least IN/kl objects.
Ramsey number R( m, n) is the smallest number of people there must be at a party so that there exist either m
mutual friends or n mutual enemies (assuming each pair of people are either friends or enemies). R(3, 3) = 6.
r-permutation of set with n objects: ordered arrangement of r of the objects from the set (no repetitions allowed);
there are P(n, r) = nf/(n - r)! such permutations.
r-combination of set with n objects: unordered selection (i.e., subset) of r of the objects from the set (no
repetitions allowed); there are C(n,r) = n!/[r!(n - r)!] such combinations. Alternative notation is G), called
binomial coefficient.
Pascal's identity: C(n, k - 1) + C(n, k) = C(n + 1, k) if n ~ k ~ 1; allows construction of Pascal's triangle of
binomial coefficients, using C(n, 0) = C(n, n) = 1 along the sides.
Combinatorial identities often have combinatorial proofs: C(n, r) = C(n, n-r); (a+ b)n = "L,~=O C(n, k)an-kbk
(binomial theorem), with corollary "L,~=o C(n, k) = 2n; C(m+n, r) = "L,~=O C(m, r-k)C(n, k) (Vandermonde's
identity).
Number of r-permutations of an n-set with repetitions allowed is nr; number of r-combinations of an n-set
with repetitions allowed is C(n + r - 1, r). This latter value is the same as the number of solutions in
nonnegative integers to x 1 + x 2 + · · · + Xn = r.
Permutations with indistinguishable objects: Number of n-permutations of an n-set with n, indistinguishable
objects of type t for 1::::; t::::; k is n!/(n 1 !n2! · · · nk!). This also gives the number of ways to distribute n distinguishable
objects into k distinguishable boxes so that box t gets nt objects.
For distributing distinguishable object into distinguishable boxes, use product rule (or the formula n!/(n 1 !n2! · · · nk!)
if the number in each box is specified). For distributing indistinguishable object into distinguishable boxes, use
formula for the number of combinations with repetitions allowed. For distributing distinguishable object into
indistinguishable boxes, there is no good closed formula; Stirling numbers of the second kind are involved.
Distributing indistinguishable object into indistinguishable boxes involves partitions of positive integers, and
there is no good closed formula.
There are good algorithms for finding the lexicographically "next" permutation or combination and thereby for
generating all permutations or combinations.
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If all outcomes are equally likely in a sample space S with n outcomes, then the probability of an event E is
= IEl/n; more generally, if p(si) is probability of ith outcome si, then p(E) = Ls,EEp(si)·
p(E)
Probability distributions satisfy these conditions: 0 :::; p( s) :::; 1 for each s
E
S, and Ls ES p( s) = 1.
For complementary event, p(E) = 1 - p(E); for union of two events (either one or both happen), p(E U F) =
p(E) + p(F) - p(E n F); for independent events, p(E n F) = p(E)p(F).
The conditional probability of E given F (probability that E will happen after it is known that F happened) is
p(EIF) = p(E n F)/p(F).
Bernoulli trials: If only two outcomes are success and failure, withp(success) = p andp(failure) = q = 1-p, then
the binomial distribution applies, with probability of exactly k success inn trials being b(k; n, p) = C( n, k )pkqn-k.
Bayes' theorem: If E and Fare events such that p(E) :j:. 0 and p(F) :j:. 0, then
p
p(E I F)p(F)
- p(E I F)p(F) + p(E I F)p(F)
(FIE) -
A random variable assigns a number to each outcome. Expected value (expectation) of random variable Xis
E(X) = L~= 1 p(si)X(s,); alternatively, E(X) = LrEX(S)P(X = r)r Expected number of successes for n Bernoulli
trials is pn.
Variance of random variable Xis V(X) = L~= 1 p(si)(X(s,)-E(X)) 2 ; variance can also be computed as V(X) =
E(X 2 ) - E(X)2; square root of variance is standard deviation a(X); variance of number of successes for n
Bernoulli trials is npq.
X1 and X2 are independent if p(X1
= r1 and X2 = r2) = p(X1 = r1) · p(X2 = r2).
In this case E(XY)
=
E(X)E(Y).
Expectation is linear even when the variables are not independent. This means that the expectation of a sum is
sum of expectations, and E(aX + b) = aE(X) + b.
Variance of a sum is the sum of the variances (V(X1 +X2)
= V(X1) + V(X2)) when the variables are independent.
The random variable X that gives the number of flips needed before a coin lands tails, when the probability of tails
is p and the flips are independent, has the geometric distribution: p(X = k) = (1 - p)k-lp for k = 1, 2, ... ;
E(X) = l/p.
Chebyshev's inequality: p(IX(s) - E(X)I ~ r):::; V(X)/r 2.
A probabilistic algorithm is an algorithm that might give the incorrect answer but only with small probability.
For example, there are good probabilistic tests as to whether or not a natural number is prime.
The probabilistic method is a proof technique that shows the existence of an object with a given property by
showing that there is a nonzero probability of choosing such an object if choices are made at random. For example,
there is a probabilistic proof that the Ramsey number R( k, k) is at least 2k/ 2.
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A recurrence relation for a sequence a 0 , a 1 , a2, ... , is a formula expressing each an in terms of previous terms
(for all n > n 0 ); initial conditions specify ao through an 0 ; a solution to such a system is an explicit formula for
an in terms of n that satisfies the recurrence relation and initial conditions.
The Fibonacci numbers are defined recursively by Jo = 0, Ji = 1, and f n = fn-l + f n-2 for n ;::: 2; continues
h = 1, h = 2, /4 = 3, fs = 5, f6 = 8, h = 13, /8 = 21, ... ; explicit formula is that f n equals nearest integer to
((1 + y'5)/2)n /y'S.
A recurrence relation is linear of degree k if it is of the form an = c1an-l + c2an-2 + · · · + ckan-k + f(n); if all
the c, 's are constants, then it has constant coefficients; if f(n) is identically 0, then it is homogeneous. Such a
recurrence relation and k initial conditions completely determine the sequence.
Recurrence relations of degree 1 can often be solved by iteration. Given an expressed in terms of an-1, rewrite
an-l in this equation using the same recurrence relation with n - 1 in place of n. This expresses an in terms of
an_ 2. Then rewrite an_ 2 in terms of an_ 3 , again using the recurrence relation. Continue in this way, noting the
pattern that evolves, until finally you have an written explicitly in terms of a 1 (or a 0 ), probably as a series. This
gives the explicit solution (preferably with the series summed in closed form).
To solve linear homogeneous recurrence relation with constant coefficients: (1) write down the characteristic
equation rk - c1rk-l - c2rk- 2 - · · · - Ck = 0 and find all its roots, with multiplicities; (2) each distinct root
(characteristic root) r gives rise to a solution an = rn; if a root is repeated, occurring s times, then there are
solutions an = n'rn for i = 0, 1, ... , s - 1; (3) take arbitrary linear combination of all solutions so obtained, with
coefficients a 1 , a 2 , ... , ak; ( 4) plug in the k initial conditions to solve for the a's.
To solve linear nonhomogeneous recurrence relation with constant coefficients: (1) solve the associated homogeneous recurrence relation (with the f(n) term omitted) to obtain a general solution a~h) with some yet-tobe-calculated constants a,; (2) obtain a particular solution a~) of the nonhomogeneous recurrence relation using
the method of undetermined coefficients (the form to use depends on f(n) and on the solution of the associated
homogeneous recurrence relation); (3) write down the general solution: an = a~h) +a~{>); (4) plug in the k initial
conditions to solve for the a 's.
Divide-and-conquer relation: f(n) = af(n/b) + g(n). If f is an increasing function satisfying this relation
whenever n is a power of b, where g(n) =end, then f(n) = O(nd) if a< bd, f(n) = O(ndlogn) if a= bd, and
f(n) = O(n 10gba) if a> bd (master theorem).
Divide-and-conquer algorithms work by dividing a problem into simpler non-overlapping subproblems; dynamic programming algorithms work by dividing a problem into simpler overlapping subproblems. Matrix
multiplication can be done in O(n10g 7 ) ~ O(n 2 · 8 ) steps, rather than the naive O(n 3 ), using a divide-and-conquer
algorithm. Talks in a lecture hall can be scheduled using dynamic programming to maximize total attendance.
Generating functions are expressions of the form f (x) = 2::%°=o akxk, associated with an infinite sequence {ak}.
They can be used to solve recurrence relations, prove combinatorial identities, and solve counting problems. To
model a combinatorial situation (such as counting how many ways there are to distribute cookies), let ak be the
quantity of interest (the answer when there are k cookies); choices are modeled by adding (corresponding to "or"
situations-the person can get 1, 2, or 3 cookies) or multiplying (corresponding to "and" situations-Tom, Dick, and
Jane must each receive cookies) polynomials in x. The combinatorics is then replaced by the algebra of multiplying
out the polynomials or obtaining a closed form expression. The most important generating function in applications
is 1/(1 - ax)n = 2::%°=o C(n + k - 1, k)akxk. Partial fractions must sometimes be used when solving recurrence
relations using generating functions.
Inclusion-exclusion principle: (for n = 3) IAUBUCI = IAI + IBI + ICl- IAnBl-IAnCl-IBnCI + IAnBnCI;
(general case) IA1 u A2 u ... u Anl = L, IA,I - Li 0 and
u = v, is simple if no edge occurs more than once. A graph is connected if every pair of vertices is joined by a path;
digraph is strongly connected if every pair of vertices is joined by a path in each direction, weakly connected if
underlying undirected graph is connected. Components are maximal connected subgraphs. Removal of cut edge
(bridge) or cut vertex (articulation point) creates more components.
Vertex connectivity of a graph G: /\',( G) = size of a smallest vertex cut (set of edges whose removal disconnects G); G is k-connected if /\',(G) ~ k; edge connectivity: >.(G) =size of a smallest edge cut.
(i,j)th entry of Ar, where A is adjacency matrix, counts numbers of paths of length r from i to j.
An Euler circuit [path] is simple circuit [path] containing all edges. Connected graph has an Euler circuit [path]
if and only if the vertex degrees are all even [the vertex degrees are all even except for at most two vertices]. Splicing
algorithm or Fleury's algorithm finds them efficiently.
A Hamilton path is path containing all vertices exactly once; Hamilton path together with edge back to starting
vertex is Hamilton circuit. No good necessary and sufficient conditions for existence of these, or algorithms for
finding them, are known. Qn has Hamilton circuit for all n ~ 2 (Gray code).
Weighted graphs have lengths assigned to edges; one can find shortest path from u to v (minimum sum of
weights of edges in the path) using Dijkstra's algorithm.
A planar graph is a graph having a planar representation (drawing in plane without edges crossing); Kuratowski's theorem: graph is planar if and only if it has no subgraph homeomorphic to (formed by performing
elementary subdivisions on edges of) Ks or K3, 3.
Euler's formula: Given planar representation with v vertices, e edges, c components, r regions, v - e + r = c + 1;
corollary: in planar graph with at least 3 vertices, e ::=; 3v - 6.
A graph is colored by assigning colors to vertices with adjacent vertices getting distinct colors; minimum number
of colors required is chromatic number. Every planar graph can be colored with four colors. There are also
applications of edge colorings (adjacent edges must get different colors).
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A tree is a connected undirected graph with no simple circuits; characterized by having unique simple path between
every pair of vertices, and by being connected and satisfying e = v - 1. A forest is an undirected graph with no
simple circuits-each component is a tree, and e = v - number of components. Every tree has at least two vertices
of degree 1.
A rooted tree is a tree with one vertex specified as root; can be viewed as directed graph away from root. If uv is
a directed edge, then u is parent and v is child; ancestor, descendant, sibling defined genealogically. Vertices
without children are leaves; others are internal. Draw trees with root at the top, so that vertices occur at levels,
with root at level O; height is maximum level number. A tree is balanced if all leaves occur only at bottom or
next-to-bottom level, complete if only at bottom level. The subtree rooted at a is the tree involving a and all
its descendants.
An m-ary tree is a rooted tree in which every vertex has at most m children (binary tree when m = 2); a full
m-ary tree has exactly m children at each internal vertex. Full m-ary tree with i internal vertices and l leaves has
n = i + l = mi+ 1 vertices. An m-ary tree with height h satisfies l ::::; mh, so h ::;::: Jlogm ll (equality in latter
inequality if tree is balanced).
An ordered rooted tree has an order among the children of each vertex, drawn left-to-right; in ordered binary
tree, each child is a right child or left child, and subtree rooted at right [left] child is called right [left] subtree.
A binary search tree {BST) is binary tree with a key at each vertex so that at each vertex, all keys in left subtree
are less and all keys in right subtree are greater than key at the vertex; O(logn) algorithm for insertion and search
in BST.
Decision trees provide lower bounds on number of questions an algorithm needs to ask to accomplish its task for
all inputs (e.g., coin-weighing, searching).
Binary trees can be used to encode prefix codes, binary codes for symbols so that no code word is the beginning
of another code word. Huffman codes are efficient prefix codes for data compression.
Game trees can be used to find optimal strategies for two-person games, using the minmax principle. Value of
a leaf is payoff to first player. Value of an internal vertex at an even level (square) is maximum of values of its
children; value of an internal vertex at an odd level (circle) is minimum of values of its children.
Universal address system: root is labeled O; children of root are labeled 1, 2, ... ; children of vertex labeled x
are labeled x.1, x.2, .... Addresses are ordered using preorder traversal.
Preorder visits root, then subtrees (recursively) in preorder; postorder visits subtrees (recursively) in postorder,
then root; inorder visits first subtree (recursively) in inorder, then root, then remaining subtrees (recursively) in
inorder.
The expression tree for a calculation has constants at leaves, operations at internal vertices (evaluated by applying
operation to values of its children).
Prefix (Polish) notation corresponds to preorder traversal of expression tree (operator precedes operands); postfix (reverse Polish) notation corresponds to postorder traversal of expression tree (operands precede operator);
both of these allow unambiguous expressions without using parentheses. Infix form is normal notation but requires
full parenthesization (corresponds to inorder traversal of expression tree).
Naive sorting routines like bubble sort require 0( n 2 ) steps in worst case; the best that can be done with comparisonbased sorting is O(nlogn) (a huge improvement), using something like merge sort.
A spanning tree is a tree containing all vertices of a connected given graph; can be found by depth-first search
(recursively search the unvisited neighbors) or breadth-first search (fan out). Edges not in depth-first search
spanning tree (back edges) join vertices to ancestors or descendants. Edges not in breadth-first search spanning
tree (cross edges) join vertices at same level or level differing by one. Depth-first search can be modified to
implement backtracking algorithms for exhaustive consideration of all cases of a problem (like graph coloring).
Minimum spanning trees can be found in weighted graphs using greedy Prim or Kruskal algorithms (choose
least costly edge at each stage that doesn't get you into trouble).
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Boolean operations: sum, product, and complementation defined on {O, 1} by 0 + 0 = 0, 0 + 1 = 1 + 0 =
1+1 = 1, 1·1=1, 0 · 1 = 1·0=0·0 = 0 (also write product using concatenation, without the dot), 0 = 1, I= 0;
also XOR defined by lEBl = OE:l:JO = 0 and lEBO = OEBl = 1, NAND defined by l I l = 0 and l I 0 = 0 I 1=0 I 0 = 1,
NOR defined by 0 l 0 = 1 and 1 l 0 = 0 l 1 = 1 l 1 = 0. A Boolean algebra is an abstraction of this, with
operations V, /\, and-, which also applies to other situations (e.g., sets).
Boolean operations obey same identities (commutative, associative, idempotent, distributive, De Morgan, etc.) with
U replaced by +, n replaced by ., U replaced by 1, and 0 replaced by 0.
Boolean variables are variables taking on only values 0 and 1; Boolean expression is expression made up from
Boolean constants (0 and 1), variables and operations combined in usual ways with parentheses where desired to
override the natural precedence that products are evaluated before sums.
Dual of Boolean expression: interchange 0 and 1, + and · ; dual of an identity is an identity.
Boolean functions are functions from n-tuples of variables to {0, 1}. They can be represented using Boolean
expressions, and in particular, in disjunctive normal form as sums of products or in conjunctive normal
form as products of sums. In sum of products form, each product is a minterm y1y2 · · · Yn, where each y, is a
literal, either x, or x,. Two expressions are called equivalent if they compute the same function.
Set of operators is functionally complete if every Boolean function can be represented using them. Examples are
{+,-. -}, {+, -}, {·, -}, {I}, and {l}.
The standard Boolean operations can be represented by gates:
x~x x~f\
'"XY
~
y~LJ"
inverter
AND
x~~x+y x~f\~x1y x~~x-!,y
y~~
y~~
OR
NAND
y~~
NOR
They can be combined to make combinational circuits to represent any Boolean function, such as (full) adders
(take two bits and a carry and produce a sum bit and a carry) or half-adders (same, without the carry as input).
Minimization of circuits: given Boolean function in sum of products form, find an expression as simple as
possible to represent it (i.e., use as few literals and operations as possible, meaning using few gates to produce the
circuit). Geometric method (Karnaugh maps or K-maps) and tabular method (Quine-McCluskey procedure)
organize this task efficiently for small n. Typical K-maps with blocks circled:
yz yz gz gz
x
1
1
1
1
1)
1 (1 j
lL
J.
Typical Quine-McCluskey calculation:
1
2
3
4
5
6
7
8
yz
wxy
wxz
wxy
wxyz
Term
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
2
1
String
1111
1110
1101
1001
0101
1000
0010 {=
0001
4
3
x
x
x
x
x
x
x
Step 1
Term
(1,2)wxy
(1, 3) w x z
(3,4)wyz
(3,5)xyz
(4,6)wxy
(4, 8) xy z
(5, 8) wy z
5
x
6
x
Step 2
String
111- {=
11-1 {=
1-01
-101
100- {=
-001
0-01
7
Term
(3, 4, 5, 8) y z
8
x
x
y z + w x y + w x y + w x y z covers all minterms
String
--01
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A vocabulary or alphabet V is just a finite nonempty set of symbols; strings of symbols from V, including the
empty string ,\, are words; the set of all words is denoted V*; a language is any subset of V*.
If L 1 and L 2 are languages, then so are the union L 1 U L 2, intersection Lin L2, concatenation LiL2 = { uv I
u E L 1 /\ v E L 2 } (L 2 =LL, etc.), Kleene closure L* ={A} UL U L 2 U L 3 ···,complement L = V* - L.
Phrase-structure grammar: G = (V, T, S, P), where Vis a vocabulary, T <;;;: Vis the set of terminal symbols
(the nonterminal symbols are N = V - T), SE Vis the start symbol, Pis a set of productions, which are
rules of the form w1 -+ w2 , where w1 , w2 E V*. (The convention is to use capital letters for the nonterminals.)
Derivations: If a string u can be transformed to a string v by applying some production (i.e., replacing a substring
w1 in u by w2 where w1 -+ w2 is a production), then write u =} v and say v is directly derivable from u; if
u1 =} u 2 =} · · · =} Un, then write u1 =*Un and say Un is derivable from u1.
The language generated by G is the set of strings of terminal symbols derivable from S (the start symbol).
Types of grammars: type 0-no restrictions; type 1 (context-sensitive )-productions are all of the form
lAr -+ lwr, where A is a nonterminal symbol, l and r are strings of zero or more terminal or nonterminal symbols,
and w is a nonempty string of terminal or nonterminal symbols, or of the form S -+ >. as long as S does not appear
on the right-hand side of any other production; type 2 ( context-free)-left side of each production must be single
nonterminal symbol; type 3 (regular)-only allowed productions are A-+ bB, A-+ b, and S-+ >.,where A, B,
and S are nonterminals, b is terminal, and S is start symbol; each type is included in previous type. Productions in
type 2 grammars can also be represented in Backus-Naur form; nonterminals have angled brackets surrounding
them; a vertical bar means "or"; arrows are replaced by::=; e.g., (A) ::= (A)c(A)(B) I 3 (B)(C).
J
Derivation (or parse) tree shows transformation of start symbol into string of terminals (the leaves, read from
left to right), invoking a production at each internal vertex.
Palindrome: string that reads the same forward as backward, i.e., w = wR (string whose first half is the reverse of
its last half, either of the form uuR or uxuR for x a symbol). The set of palindromes is context-free but not regular.
Finite state machine with output on the transitions (Mealy machines): M = (S, I, 0, f, g, s 0 ), where Sis
set of states, I and 0 are input and output alphabets, so E S is start state, f : S x I -+ S and g : S x I -+ 0
are transition function and output function; can be represented in state table, or by state diagram with
i, o labeling edge (s, t) if f(s, i) = t and g(s, i) = o. Machine "moves" from state to state as it reads an input string,
producing an output string of the same length; can be used for language recognition (output symbol 1 if input
string read so far is in language, 0 if not).
Finite state machine with output on the states (Moore machines): same as Mealy machine except g :
S-+ 0 assigns an output to every state rather than to every transition; output is one symbol longer than input.
Finite state machine with no output (deterministic automaton): Jvf = (SJ, f, s 0 , F)-same as Mealy
machine except that there is no output function but rather a set of final states F <;;;: S. A string w is recognized
or accepted by M if Mends up in a final state on input w; the language recognized or accepted by Jvl, written
L(M), is the set of all strings accepted by M. (Plural of "automaton" is ''automata.")
Nondeterministic automaton: same as deterministic one except that the transition function f sends a state
and input symbol to a set of states; think of the machine as choosing which state to go into next from among
the possibilities provided by f. A string is accepted if some sequence of choices leads to a final state at the end
of the input; L( M) defined as before. For automata of either type, we say that M 1 and M 2 are equivalent if
L(M1 ) = L(M2 ). Theorem: Given a nondeterministic finite automaton, there is an equivalent deterministic one.
Regular expressions over a set I are built up from symbols for the elements of I, a symbol for the empty set,
and a symbol for the empty string by the operations of concatenation, union, and Kleene closure. The expressions
represent the corresponding sets of strings. Regular sets (languages) are sets represented by regular expressions.
Regular languages are closed under intersection, union, concatenation, Kleene closure, complement; context-free
languages are closed under union, concatenation, Kleene closure.
Theorem: A set is regular if and only if it is generated by some regular grammar if and only if it is accepted by
some finite automaton.
Pumping lemma: If z is a string in L(M) of length longer than the number of states in M, then we can write
z = uvw, with v =I- >.so that uv'w E L(M) for all i. This allows us to prove, for example, that { on1 n I n = 1, 2, 3, ... }
is not regular.
A Turing machine is specified by a set of 5-tuples (s, x, s', x', d): if in states scanning symbol x on tape, it writes
x' on tape, enters state s', and moves tape head R or L according to d. TM's can recognize all computable (Type 0)
languages, can compute all computable functions (Church-Turing thesis).
Notes
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