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CS145 Midterm II Review Notes V3.0

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==========================================================================
==========================================================================
;;
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;;
2017 Fall CS 145 Midterm II Review Notes V3.0
|
;;
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;;
Asst. 6~8
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;;
Lecture/Tutorial Notes
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;;
Prof: Gordon Cormack
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;;
ISA: Ashish Mahto
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;;
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;;
By David Duan
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;;
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;;
Outline
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;;
Part A: Asymptotic Analysis
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;;
Part B: Abstract data type
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;;
Part C: Higher order functions
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;;
Part D: Other minor topics
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;;
Part E: Appendix
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;;
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==========================================================================
;; If you find any errors (I'm sure there are a lot of them), please
|
;;
email j32duan@edu.uwaterloo.ca, David Duan. Thanks in advance.
|
==========================================================================
;; Release Note
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;;
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;; V0.X
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;;
Notes from lecture
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;;
Notes from tutorial
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;;
Ashish's messages (which are, indeed, full of wisdom xd)
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;;
My own research
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;;
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;; V1.X
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;;
Raw note
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;;
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;; V2.0 Update: (Approx 1k lines)
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;;
Finished part A, B, and C.
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;;
Still need to finish foldr and foldl.
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;;
I'll do that tomorrow after asking Ashish questions.
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;;
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;; V2.1 Update: (Approx 1.5k lines)
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;;
Added foldr, foldl, and foldr vs. foldl.
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;;
Added or/list, disjoin/list, and f^n
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;;
Changed formatting
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;;
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;; V2.2 Update: (Approx 2k lines)
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;;
Added Part D, including D1, D2, D3
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;;
Added Part E, including E1, practice questions from TUT Oct.25
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;;
Need to add: assignment 6~8
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;;
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;; V2.3 Update: (Approx 2k lines)
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;;
Modified A7
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;;
Modifed Part E, but haven't finished yet.
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;;
Modified Part C, added filter and map
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;;
Fixed typo
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;;
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;; V3.0 Release: 2155 lines
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;;
Modified Part E
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;;
Fixed typo
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;;
|
==========================================================================
;; Big thanks to Teresa Kang and Steven Wong for pointing out countless
;; typos and logic errors.
==========================================================================
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Part A: Asymptotic Analysis
Table of Contents
A1. Introduction
A2. Examples
A3. /subset and /strict-subset
A4. Order
A5. Examples
A6. Examples for using definitions to prove statements involving
Big-O notation.
A7. Additional remarks
-------------------------------------------------------------------------A1. Introduction.
Defn: (Informal)
f(x) /in O(g(x)) iff O(f(x)) <= O(g(x)).
Defn: (Formal)
There exists some constant x0 and c such that f(x) < c*g(x)
for every x > x0.
Defn: (About Big-O notation)
O(g(n)) is the set of functions f(n) such that there exists
constant c and n0 such that for all n >= n0, f(n) <= c*g(n).
Remark:
1. c*g(x) is an upper bound for f(x) when x > x0.
2. O(g(x)) serves as a placeholder for all f(x) /in O(g(x)).
3. If O(f(x)) <= O(g(x)), then for all h(x) /in O(f(x)),
h(x) /in O(g(x)).
-------------------------------------------------------------------------A2. Examples.
Ex.
x
x^2
f(x)

/in
/in
/in

O(x)
O(x^2)
O(f(x))

Ex.
O(f(x)) + k = O(f(x))
O(f(x)) * k = O(f(x))
Ex.
O(k) = O(1)
O(log_n(x)) = O(log(x))
O(x) + O(x^2) + ... + O(x^n) = O(x^n)
-------------------------------------------------------------------------A3. /subset and /strict-subset
Thm:
O(x) /subset O(x).
O(x) /subset O(x^m) for any m >= 1.
Defn:
O(f(x)) /strict-subset O(g(x)) if
a. O(f(x)) /subset O(g(x)), and
b. O(g(x)) /not /subset O(f(x))
Thm:
O(a^x) /strict-subset O((a+e)^x)
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O(x^n) /strict-subset O(a^x) given that a > 1
-------------------------------------------------------------------------A4. Order
O(1) <
<
<
<
<
<
<
<

O(log n)
O(n^k)
O(n)
O(n log n)
O(n^p), p>1
O(a^n), a>1
O(b^n), b>a
O(n!)

-------------------------------------------------------------------------A5. Examples
Ex.
|%Racket_file
|
| (define (dedupe lst)
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(cond
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[(empty? lst) empty]
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[(member (first lst) (rest lst))
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(dedupe (rest lst))]
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[else
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(cons (first lst) (dedupe (rest lst)))]))
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|%END
Note: The function member takes O(n), and we apply it to every
element in the list, so overall the running time is O(n^2).
Ex.
|%Racket_file
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| (define (append l1 l2)
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(cond
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[(empty? l1) l2]
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[else (cons (first l1) (append (rest l1) l2))]))
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|%END
Note: The function cons takes O(1), and we apply it to every
element in list1, so overall the running time is O(n)
where n is the size of list1.
Ex.
|%Racket_file
|
| (define (reverse l)
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(cond
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[(empty? l) empty]
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[else (append (reverse rest l)
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(list first l))]))
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|%END
Note: This is a bad list recursion. We apply append and
reverse which both takes O(n) time on each element,
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so overall it will be O(n^2).
-------------------------------------------------------------------------A6. Examples for using definitions to prove statements involving
Big-O notation.
Ex. Prove "3n^2 + 6n is O(n^2)" using definition 1.
We need to start with "there exists c and n0", but we don't
know what value would make the rest of the statement true,
so we leave them as symbolic constants and accumulate
information about it, as long as we make sure to specify
its value by the end of the proof.
Remark:
This part is very similar to our proof in math 147 where
we assume epsilon and delta/N exists, then do calculations
until we find an appropriate pair.
The next step is doing algebra. Assume our c works, we want
to find an appropriate n0.
Remark:
This is similar to assuming our N/delta works and work
towards epsilon.
|
|
|
|
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|
At
is
By
is
n0

3n^2 + 6n
6n
6
6 / (c-3)

<=
<=
<=
<=

c * n^2
(c-3) * n^2
(c-3) * n
n

this point, we can let c be an arbitrary number that
greater than 3, so the left side would be positive.
the Archimedean principle, the set of natural numbers
not bounded, so we can always find some natural number
such that n >= n0 implies 6 / (c-3) <= n.

Since the last inequality satisfies the requirement of
"3n^2 + 6n <= c*n^2 for all n>n0 for some c in real number
and n in natural number", we can conclude that 3n^2 + 6n is
indeed O(n^2).
Remark:
Our proof works only because everything we did was
reversible (we performed the same operations on both sides).
This is not always the case with inequalities. For example,
if we use the face that we can increase the larger side of an
inequality, we would not be able to work backward.
Ex. Prove that "3n^2-6n is not O(n)."
Express this using definition, we are saying "not" there exists
c and n0 such that for all n>n0, 3n^2-6n <= cn. This is equivalent
of saying, there exists some n > n0 that does not make our
statement true.
Let c in R. We want to prove "there exists n such that for any c,
3n^2-6n is greater than cn."
|
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|
|
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|

3n^2 - 6n >
3n^2 >
n >

cn
(c+6) * n
(c+6) / 3

There are two restrictions for n. First we have n > n0, and secondly
we have n > (c+6)/3, so we can let n > max{n0, (c+6)/3}. This way, we
have found an appropriate counter example of n that supports our
statement.
!IMPORTANT! EXAM PREPARATION
"""
At least one of the problem will be in the following form:
For each of a number of pairs of definitions for f(x) and g(x),
you must submit a module to do the following:
->

Provide a function (findx c x0) that consumes positive values c
and x0 and produces x >= x0 such that f(x) > c * g(x). If no such
value exists, produce 'impossible.

->

Provide values c and x0 for which (findx c x0) produces 'impossible.
If there is no such pair of values, define c and x0 both to have
the value 'none.

If your implementation of findx is correct, you will be able to
find suitable values of c and x0 iff f(x) is O(g(x)). In either case,
you should justify your answer using embedded comments.
"""
Note:
To prove that f(x) is O(g(x)), assume c exists, find appropriate n,
and rewrite the proof in the normal order. This is similar to limits.
To prove that f(x) is not O(g(x)), try to prove the negation of
it. That is, find an example of n such that for any c, f(x) > O(g(x)).
-------------------------------------------------------------------------A7. Additional remarks
Remark:
We sometimes use the equal sign to express the relationship,
as in "3n^2 + 6n = O(n^2)". But O(n^2) is not a function nor
an algebraic object, so this equal sign does not share any
properties such as reflexivity as the normal mathematical
equality sign.
Remark:
Running time of common Racket functions
Note:
O(1): cons, first, rest, list, make-foo, foo-x, foo?, eq?
Note:
O(n), where n = (size x):
(append x y) ;; independent of size of y
(length x)
(member e x)
(reverse x)
Note:
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O(n*T(n)), where n = (size x) and f has O(T(n)) time
(foldr f e x)
(foldl f e x) ;; requires O(n) space
(map f x)
Note:
sort: O(n log n)
quicksort: on average O(n log n), worst O(n^2)
(equal? x y): O(n), where n = min((size x), (size y))
==========================================================================
==========================================================================
Part B: Abstract Data Type
Defn:
ADT: A set of values and a finite set of operations
(functions), defined entirely by the behavior of the
operators.
Remark:
Constrast to: Concrete Data Types
- Defined by its representations.
- Has infinite number of operations.
- For example, list of number '(10, 20, 30)
-------------------------------------------------------------------------Ex. Abstract implementation of set.
Remark: Big thanks to Teresa for pointing out a crucial mistake.
Note:
We represent the set as a function.
|%Racket_file
|
| (provide make-empty-set insert-set member-set)
|
| (define (make-empty-set) (lambda (e) false))
|
| (define (member-set e s) (s e))
|
| (define (insert-set e s)
| (lambda (x)
|
(if (= e x) true (s x))))
|
| ;; You need to define what equality means.
|
|%END
Note:
1. This implementation is essentially a member function,
it tests if the argument passed in is in the set or not.
2. (make-empty-set) creates a lambda function which returns
false no matter what argument is passed in.
3. (member-set e s) returns the result of function application
s applied onto e, ie. returns (s e).
4. (insert-set e s) returns a lambda function which takes in
one argument; its output depends on if the argument x is equal
to the argument e we passed in in the first place. The variable
e is what's stored in the lambda function. The variable x is
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newly-consumed and will be checked if it's in the set already.
5. If it is still confusing, let's walk through some examples.
-------------------------------------------------------------------------Ex. (member-set 5 (make-empty-set))
This function checks if 5 is in the empty set.
|%STEPPER
|
|
(member-set 5 (make-empty-set))
|
((make-empty-set) 5)
|
((lambda (e) false) 5)
|
false
|
|%END
-------------------------------------------------------------------------Ex. (insert-set 5 (make-empty-set))
This function inserts 5 into an empty set.
|%STEPPER
|
| (insert-set 5 (make-empty-set))
|
| ;; local_var: e = 5, s = (make-empty-set)
|
((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
5 (make-empty-set))
|
| ;; (make-empty-set) = (lambda (e) false)
| (lambda (x) (if (= 5 x) true ((lambda (e) false) x)))
|
|%END
Note: Instead of returning a value, we got a lambda function.
This function consumes an additional variable x, then
compares if it equals 5. If yes, the function returns true.
Otherwise our variable x gets passed into the next layer
of lambda function, which in this case is the empty
set function that always returns false.
-------------------------------------------------------------------------Ex. (insert-set 3 (insert-set 5 (make-empty-set)))
This function inserts 3 into a set containing 5.
|%STEPPER
| (insert-set 3 (insert-set 5 (make-empty-set)))
|
|
;; local_var: e = 3, s = (insert-set 5 (make-empty-set))
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
3 (insert-set 5 (make-empty-set)))
|
| ;; substitution for (insert-set 5 (make-empty-set))
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
3
|
(lambda (e s)
|
(lambda (x)
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|
(if (= e x) true (s x))))
|
5 (make-empty-set))
|
| ;; substitution for (make-empty-set)
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
3
|
(lambda (e s)
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(lambda (x)
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(if (= e x) true (s x))))
|
5
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(lambda (e) false))
|
| ;; Now 5 gets consumed to produce a new function
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
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3
|
(lambda (x)
|
(if (= 5 x) true ((lambda (e) false) x))))
|
| ;; Now 3 gets consumes to produce a new function
| (lambda (x)
|
(if (= 3 x)
|
true
|
(lambda (x)
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(if (= 5 x) true ((lambda (e) false) x))) x))
|
|%END
Note:
Take a look at what our output function does.
This lambda function (call it lambda1) takes in 1 argument
x and compares x with 3. If they are equal,
we return true. Otherwise we call the next lambda function
(call it lambda2), which takes in the same x and compare
it with 5. If x = 5 then return true, otherwise call
the next lambda function, in our case it's the empty
set/function so it always returns false.
Note that every time the if statement fails, we are calling
(s x). The s is the lambda function, and we are feeding it with
an argument x, which is the variable right after the close
bracket of lambda function.
-------------------------------------------------------------------------Ex. (insert-set 5 (insert-set 3 (insert-set 5 (make-empty-set))))
This function (tries to) insert the number 5 into a set containing
number 3 and 5.
|%STEPPER
|
|
(insert-set 5 (insert-set 3 (insert-set 5 (make-empty-set))))
|
| ;; local_var: e = 5, s = (insert-set 3 (insert-set 5
| ;;
(make-empty-set))))
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
5 (insert-set 3 (insert-set 5 (make-empty-set))))
|
| ;; substitution for (insert-set 3 (insert-set 5 (make-empty-set)))
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| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
5
|
((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
3
|
(insert-set 5 (make-empty-set))))
|
| ;; substition for both insert-set 5 and the empty set.
|
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
5
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((lambda (e s)
|
(lambda (x)
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(if (= e x) true (s x))))
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3
|
((lambda (e s)
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(lambda (x)
|
(if (= e x) true (s x))))
|
5
|
(lambda (e) false))))
|
| ;; now consumes 5 to produce a new function
|
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
5
|
((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
3
|
(lambda (x)
|
(if (= 5 x) true
|
((lambda (e) false)) x))))
|
| ;; consumes 3
|
| ((lambda (e s)
|
(lambda (x)
|
(if (= e x) true (s x))))
|
5
|
(lambda (x)
|
(if (= 3 x) true
|
((lambda (x)
|
(if (= 5 x) true
|
((lambda (e) false) x))) x))))
|
| ;; consumes 5
|
|
(lambda (x)
|
(if (= 5 x) true
|
((lambda (x)
|
(if (= 3 x) true
|
((lambda (x)
|
(if (= 5 x) true
|
((lambda (e) false) x))) x))) x)))
|
|%END
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Note:
This lambda has similar logic as the last one, except
it actually contains two functions checking for 5.
Nevertheless it doesn't affect the set operations since
if the argument is 5, we will directly return true.
Note: To further see that this implementation is like a
member function, try ((insert 5 (make-empty-set)) 5) and see
what would happen.
==========================================================================
==========================================================================
Part C: Higher-order function
Table of Contents
C1. compose
C2. disjoin
C3. foldr
C4. foldl
C5. foldr vs. foldl
C6. or/list
C7. disjoin/list
C8: f^n
C9. currying
C10. map
C11. filter
-------------------------------------------------------------------------C1. compose
Note:
Consider the contract (f : (b : B) -> C).
This is the type contract for a generic one-argument function.
It takes an argument b of type B and returns an output of type C.
Now suppose we have another function g with type contract
(g : (a : A) -> B). We can define a new function called
compose, which represents the function composition of
function g and f:
Implementation:
;; Type contract:
;; (compose : (f : B -> C) -> (g : A -> B) => (A -> C))
|%Racket_file
|
| (define compose
|
(lambda (f g)
|
(lambda (x)
|
(f (g x)))))
|
|%END
Remark:
- Arguments:
f : B -> C
g : A -> B
- Return:
A lambda function
- Arguments:
x : A
10

CS145 Midterm II Review Notes V3.0

631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693

- Return:
(f (g x)) : C
Ex.
|%Racket_file
|
| > (define neg-sqrt (compose - sqrt))
| > neg-sqrt
| #
|
| > (neg-sqrt 3)
| -1.732.. ;; first (sqrt 3), then (- (sqrt 3))
|
| > ((compose sqrt -) 3)
| 0+1.732..i ;; first (- 3), then (sqrt (- 3))
|
|%END
Note:
Compare and contrast the following function with the compose
defined above.
Implementation:
;; Type contract:
;; (compose-then-compute : (f : B -> C) -> (g : A -> B) -> (a : A) => C)
|%Racket_file
|
| (define compose-then-compute
|
(lambda (f g a)
|
(f (g a))))
|
|%END
Remark:
- Arguments:
f : B -> C
g : A -> B
a : A
- Return:
(f (g a)) : C
Remark:
The difference is that compose returns a function, where
compose-then-compute returns a value.
-------------------------------------------------------------------------C2. disjoin
Note:
The function disjoin consumes two functions that returns
booleans and returns a function that returns the disjunction
(or-value).
Implementation:
|%Racket_file
|
| (define disjoin
|
(lambda (f g)
|
(lambda (x)
11

CS145 Midterm II Review Notes V3.0

694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756

|
|
|%END

(or (f x) (g x)))))

Remark:
- Arguments:
f : A -> Bool
g : A -> Bool
- Return:
A function
- Argument:
x : A
- Return:
(or (f x) (g x)) : Bool
Ex.
Check if the list has length less than 2?
First thought:
|
|

(or (empty? lst)
(empty? (rest lst)))

Note that the second argument inside the or
function is applying two 1-arg functions onto
the same argument, thus we can rewrite it using
compose:
|
|
|

(empty? (rest lst)
;; is equivalent to
((compose empty? rest) lst)

Also, since we are applying two functions onto
the same argument ((compose empty? rest) produces
a function!), we can rewrite the whole thing
using compose:
|
|
|
|

(or (empty? lst)
(compose empty? rest) lst)
;; is equivalent to
((disjoin empty? (compose empty? rest)) lst)

This way, we can define our new function:
Implementation:
|%Racket_file
|
| (define len<2?
|
(disjoin empty? (compose empty? rest)))
|
| (len<2? '(1 2)) => False
|
|%END
-------------------------------------------------------------------------C3: foldr
Thm:
Main characteristic of foldr: PURE RECURSION
Implementation:
12

CS145 Midterm II Review Notes V3.0

757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819

|%Racket_file
|
| (define (foldr f z ls)
|
(cond
|
[(empty? ls) z]
|
[else (f (first ls) (foldr f z (rest ls)))]))
|
|%END
Ex.
|%Racket_file
|
| (foldr cons '() '(1 2 3))
|
| (foldr + 0 '(1 2 3))
|
|%END
|%STEPPER
|
| (foldr cons '() '(1 2 3))
| (cons 1 (foldr cons '() '(2 3)))
| (cons 1 (cons 2 (foldr cons '() '(3))))
| (cons 1 (cons 2 (cons 3 (foldr cons '() '()))))
| (cons 1 (cons 2 (cons 3 '())))
| (cons 1 (cons 2 '(3)))
| (cons 1 '(2 3))
| '(1 2 3)
|
| (foldr + 0 '(1 2 3))
| (+ 1 (foldr + 0 '(2 3)))
| (+ 1 (+ 2 (foldr + 0 '(3))))
| (+ 1 (+ 2 (+ 3 (foldr + 0 '()))))
| (+ 1 (+ 2 (+ 3 0)))
| (+ 1 (+ 2 3))
| (+ 1 5)
| 6
|
|%END
Remark:
As you can see, in each step, the second argument
for function f is the recursive function application.
The calculation starts when the function reaches the
end of the list (which returns the init value z), and
fold from right towards left.
Thm:
Because of this, we can rewrite foldr like this:
|%Racket_file
|
| (foldr f z '(e1 e2 e3 e4 ... eN))
|
|%END
Step1: Expand the foldr
|%STEPPER
|
| (foldr f z '(e1 e2 e3 e4 ... eN))
| (f e1 (foldr f z '(e2 e3 e4 ... eN)))
13

CS145 Midterm II Review Notes V3.0

820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882

| (f e1
| (f e1
| (f e1
| ...
| (f e1
|
|%END

(f e2 (foldr f z '(e3 e4 ... eN))))
(f e2 (f e3 (foldr f z '(e4 ... eN)))))
(f e2 (f e3 (f e4 (foldr f z '(... eN))))))
(f e2 (f e3 (f e4 (...(f eN (foldr f z '())))))))

Remark:
Note that (foldr f z '()) returns z, so we can start
doing calculations now.
Step2: Evaluate expressions.
Let rk be the result when applying f onto ek, r(k+1).
Note that when k = N, e(k+1) = z.
|%STEPPER
|
| (f e1 (f e2
| (f e1 (f e2
| ...
| (f e1 (f e2
| (f e1 (f e2
| (f e1 (f e2
| (f e1 r2)
| r1
|
|%END

(f e3 (f e4 (...(f eN (foldr f z '())))))))
(f e3 (f e4 (...(f eN z))))))
(f e3 (f e4 r5))))
(f e3 r4)))
r3))

-------------------------------------------------------------------------C4: foldl
Thm:
Main characteristic of foldl: ACCUMULATIVE RECURSION
Implementation:
|%Racket_file
|
| (define (foldl f z ls)
|
(define (calc ls acc)
|
(cond
|
[(empty? ls) acc]
|
[else (calc (rest ls)
|
(f (first ls) acc))]))
|
(calc ls z))
|
|%END
Ex.
|%Racket_file
|
| (foldl cons '() '(1 2 3))
|
| (foldl + 0 '(1 2 3))
|
|%END
|%STEPPER
|
| (foldl cons '() '(1 2 3)
14

CS145 Midterm II Review Notes V3.0

883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945

| (calc '(1 2 3) '())
| (calc '(2 3) (cons 1 '()))
| (calc '(3) (cons 2 (cons 1 '())))
| (calc '() (cons 3 (cons 2 (cons 1 '()))))
| (cons 3 (cons 2 (cons 1 '())))
| (cons 3 (cons 2 '(1)))
| (cons 3 '(2 1))
| '(3 2 1)
|
| (foldl + 0 '(1 2 3))
| (calc '(1 2 3) 0)
| (calc '(2 3) (+ 1 0)
| (calc '(3) (+ 2 (+ 1 0)))
| (calc '() (+ 3 (+ 2 (+ 1 0))))
| (+ 3 (+ 2 (+ 1 0)))
| (+ 3 (+ 2 1))
| (+ 3 3)
| 6
|
|%END
Remark:
In foldl, the second argument for each function application
of f is the accumulative result.
Thm:
We can rewrite foldl like this:
|%Racket_file
|
| (foldl f z '(e1 e2 e3 e4 ... eN))
|
|%END
Step1: Expand the foldl / Use the helper
|%STEPPER
|
| (foldl f z '(e1 e2 e3 e4 ... eN))
| (calc '(e1 e2 e3 e4 ... eN) z)
| (calc '(e2 e3 e4 ... eN) (f e1 z))
| (calc '(e3 e4 ... eN) (f e2 (f e1 z))
| (calc '(e4 ... eN) (f e3 (f e2 (f e1 z))
| (calc '(... eN) (f e4 (f e3 (f e2 (f e1 z))
| (calc '() (f eN (... (f e4 (f e3 (f e2 (f e1 z)))))))
|
|%END
Remark:
Note that (calc '()  ) returns , so we can start
doing calculations now.
Step2: Evaluate expressions.
Let rk be the result when applying f onto ek, r(k-1).
Note that when k = 1, e(k-1) = e0 = z.
|%STEPPER
|
| (f eN (...
| (f eN (...
| (f eN (...
| (f eN (...

(f
(f
(f
(f

e4
e4
e4
e4

(f e3 (f e2 (f e1 z)))))))
(f e3 (f e2 r1)))))
(f e3 r2))))
r3)))
15

CS145 Midterm II Review Notes V3.0

946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
1001
1002
1003
1004
1005
1006
1007
1008

| (f eN r(N-1))
| rN
|
|%END
-------------------------------------------------------------------------C5: foldr vs. foldl
Thm: Equivalence
For most simple pure-functional functions, (foldr f z ls) is
equivalent to (foldl f z (reverse ls)).
Ex.
|%Racket_file
|
| (foldr f z '(e1 e2 e3 e4 e5))
| (foldl f z '(e5 e4 e3 e2 e1))
|
|%END
|%STEPPER
|
| (foldr f z '(e1 e2 e3 e4 e5))
| (f e1 (foldr f z '(e2 e3 e4 e5)))
| (f e1 (f e2 (foldr f z '(e3 e4 e5))))
| (f e1 (f e2 (f e3 (foldr f z '(e4 e5)))))
| (f e1 (f e2 (f e3 (f e4 (foldr f z '(e5))))))
| (f e1 (f e2 (f e3 (f e4 (f e5 (foldr f z '()))))))
| (f e1 (f e2 (f e3 (f e4 (f e5 z)))))
|
| (foldl f z '(e5 e4 e3 e2 e1))
| (calc '(e5 e4 e3 e2 e1) z)
| (calc '(e4 e3 e2 e1) (f e5 z))
| (calc '(e3 e2 e1) (f e4 (f e5 z)))
| (calc '(e2 e1) (f e3 (f e4 (f e5 z))))
| (calc '(e1) (f e2 (f e3 (f e4 (f e5 z)))))
| (calc '() (f e1 (f e2 (f e3 (f e4 (f e5 z))))))
| (f e1 (f e2 (f e3 (f e4 (f e5 z)))))
|
|%END
Remark:
They produce the same outcome!!
Thm: Memory cost
In terms of memory costs, foldl <= foldr.
The main reason is that we use an accumulator for foldl,
where in foldr we need memory to store everything.
-------------------------------------------------------------------------C6: or/list
Note:
Suppose we want to define a function which takes the "or" value of
all elements in a list.
Implementation:
;; Type contract:
;; (or/list (listof Bool) -> Bool)
;;
16

CS145 Midterm II Review Notes V3.0

1009
1010
1011
1012
1013
1014
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
1061
1062
1063
1064
1065
1066
1067
1068
1069
1070
1071

;; Example:
;; (or/list '(true false false)) -> true
|%Racket_file
|
| (define or/list
|
(lambda (bool-list)
|
(foldr
|
(lambda (bool acc-val)
|
(or bool acc-val))
|
false
|
bool-list)))
|
|%END
Remark:
For main function or/list
- Argument:
bool-list: listof b : Bool
- Return:
b : Bool created by foldr
Remark:
For function foldr
- Argument:
A function
- Argument:
bool : Bool, an element from bool-list
acc-val : Bool, accumulator
- Return:
(or bool acc-val) : Bool
false : Bool
Remark:
For or function, the base case is false.
bool-list : listof b : Bool
- Return:
c : Bool
-------------------------------------------------------------------------C7: disjoin/list
Note:
We want to disjoin all functions in a list.
Implementation:
|%Racket_file
|
| (define disjoin/list
|
(lambda (list-of-func)
|
(foldr
|
disjoin
|
(lambda (x) false)
|
list-of-func)))
|
|%END
Remark:
For main function disjoin/list
- Argument:
listof f : A -> Bool
17

CS145 Midterm II Review Notes V3.0

1072
1073
1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
1105
1106
1107
1108
1109
1110
1111
1112
1113
1114
1115
1116
1117
1118
1119
1120
1121
1122
1123
1124
1125
1126
1127
1128
1129
1130
1131
1132
1133
1134

- Returns:
A function created by foldr
Remark:
For function foldr
- Arguments:
disjoin : (f : A -> Bool) (g : A -> Bool) -> (h : A -> Bool)
(lambda (x) false)
Remark:
Recall that the base case for or is false, so the initial value
of foldr inside the or is false. In this case, the base case
for disjoin is also false, but since we are working with
functions instead of Booleans, we need a function version of
"False". In a sense, (lambda (x) false) can be seens as the
function version of false.
listof f : A -> Bool
- Return:
A function h : A -> Bool
-------------------------------------------------------------------------C8: f^n: repetitive application
Note:
If you want to apply one function many times to one list, you can
let this f^n function help you.
Implementation:
|%Racket_file
|
| (define f^n
|
(lambda (f n)
|
(foldr
|
compose
|
identity
|
(make-list n f))))
|
|%END
Remark:
For main function f^n
- Argument:
f : A -> B
n : Nat
- Return:
A function g : A -> B created by foldr.
Remark:
For foldr
- Argument:
compose : (f : B -> C) (g : A -> B) -> (A -> C)
identity: (f : X -> X)
Remark:
Recall that in disjoin/list we used (lambda (x) false) as the
function version of false. Here, we want to compose a list of
functions together, and we need a start point. This is very
much alike to + and *. For those two functions, when we fold them,
we used their identities as the initial value inside foldr
(0 and 1, respectively). Now we want to find the functional
version of identity -- the function identity returns the same
18

CS145 Midterm II Review Notes V3.0

1135
1136
1137
1138
1139
1140
1141
1142
1143
1144
1145
1146
1147
1148
1149
1150
1151
1152
1153
1154
1155
1156
1157
1158
1159
1160
1161
1162
1163
1164
1165
1166
1167
1168
1169
1170
1171
1172
1173
1174
1175
1176
1177
1178
1179
1180
1181
1182
1183
1184
1185
1186
1187
1188
1189
1190
1191
1192
1193
1194
1195
1196
1197

thing it consumes, which is the ideal function version of
identity we are looking for.
(make-list n f) : listof f : A -> B, with length n.
Ex.
|%Racket_file
|
| > (f^n rest 2)
| #
|
|%END
|%STEPPER
|
| (f^n rest 2)
|
| ((lambda (f n)
|
(foldr
|
compose
|
identity
|
(make-list n f))) rest 2)
|
| (foldr
|
compose
|
identity
|
(make-list 2 rest))
|
| (foldr
|
compose
|
identity
|
'(rest rest))
|
| (compose rest (foldr compose identity '(rest)))
|
| (compose rest (compose rest (foldr compose identity '())))
|
| (compose rest (compose rest identity))
|
| (compose rest rest)
|
| ((lambda (f g)
|
(lambda (x)
|
(f (g x)))) rest rest)
|
| (lambda (x) (rest (rest x)))
|
|%END
Ex.
|%Racket_file
|
| > ((f^n rest 2) '(1 2 3 4 5))
| '(3 4 5)
|
|%END
|%STEPPER
|
| ((f^n rest 2) '(1 2 3 4 5))
|
19

CS145 Midterm II Review Notes V3.0

1198
1199
1200
1201
1202
1203
1204
1205
1206
1207
1208
1209
1210
1211
1212
1213
1214
1215
1216
1217
1218
1219
1220
1221
1222
1223
1224
1225
1226
1227
1228
1229
1230
1231
1232
1233
1234
1235
1236
1237
1238
1239
1240
1241
1242
1243
1244
1245
1246
1247
1248
1249
1250
1251
1252
1253
1254
1255
1256
1257
1258
1259
1260

| ...
|
| ((lambda (x) (rest (rest x))) '(1 2 3 4 5))
|
| (rest (rest '(1 2 3 4 5)))
|
| (rest '(2 3 4 5))
|
| '(3 4 5)
|
|%END
Ex.
|%Racket_file
|
| > (define cons-7
|
(lambda (lst)
|
(cons 7 lst)))
|
| > ((f^n cons-7 3) '(1 2 3))
| '(7 7 7 1 2 3)
|
|%END
Look at (f^n cons-7 3) first.
|%STEPPER
|
| (f^n con-7 3)
|
| ((lambda (f n)
|
(foldr
|
compose
|
identity
|
(make-list n f))) cons-7 3)
|
| (foldr
|
compose
|
identity
|
(make-list 3 cons-7))
|
| (foldr
|
compose
|
identity
|
'(cons-7 cons-7 cons-7))
|
| (compose cons-7 (compose cons-7 cons-7))
|
| ((lambda (x) (cons-7 (cons-7 (cons-7 x)))
|
|%END
Now back to ((f^n cons-7 3) '(1 2 3))
|%STEPPER
|
| ((f^n cons-7 3) '(1 2 3))
|
| ((lambda (x) (cons-7 (cons-7 (cons-7 x)))) '(1 2 3))
|
| (cons-7 (cons-7 '(7 1 2 3)))
|
20

CS145 Midterm II Review Notes V3.0

1261
1262
1263
1264
1265
1266
1267
1268
1269
1270
1271
1272
1273
1274
1275
1276
1277
1278
1279
1280
1281
1282
1283
1284
1285
1286
1287
1288
1289
1290
1291
1292
1293
1294
1295
1296
1297
1298
1299
1300
1301
1302
1303
1304
1305
1306
1307
1308
1309
1310
1311
1312
1313
1314
1315
1316
1317
1318
1319
1320
1321
1322
1323

| (cons-7 '(7 7 1 2 3))
|
| '(7 7 7 1 2 3)
|
|%END
-------------------------------------------------------------------------C9: currying
Note:
We can curry a 2-arg function into a "wrapped" 1-arg function.
Note:
Before: (f : A -> B -> C)
After: (f' : A -> (B -> C))
Note:
Before: (A -> B -> C)
After: (A -> (B -> C))
Implementation:
|%Racket_file
|
| (define curry
|
(lambda (f)
|
(lambda (a)
|
(lambda (b)
|
(f a b)))))
|
|%END
Remark:
- Argument:
f : (A -> B -> C)
- Return:
A function
- Argument
a : A
- Return
A function
- Argument
b : B
- Return
(f a b) : C
Ex.
|%Racket_file
|
| > (+ 1 2)
| 3
| > (+ 1)
| 1
| > (curry +)
| #
| ;; At this point, we have only provided "f", so we get a
| ;; procedure
| ;;
(lambda (a)
| ;;
(lambda (b)
| ;;
(lambda (+ a b))))
|
| > ((curry +) 1)
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| #
| ;; Now we have provided both f and a, so we have a
| ;; "partially applied" function. Our output lambda function
| ;; looks like this:
| ;;
(lambda (b)
| ;;
(lambda (+ 1 b)))
|
| > (((curry +) 1) 2)
| 3
| ;; We have finally provided all three arguments, so we get
| ;; a value back in return. In fact, ((curry +) 1) is equivalent
| ;; to the function add1.
|
|%END
Note:
Let's look at the stepper.
|%STEPPER
|
| ((curry +) 1)
|
| (((lambda (f)
|
(lambda (a)
|
(lambda (b)
|
(f a b)))) +) 1)
|
| ((lambda (a)
|
(lambda (b)
|
(+ a b))) 1)
|
| (lambda (b) (+ 1 b))
|
|%END
Remark:
You can see how at every step, one more argument gets
accumulated into the final expression. (f a b) eventually
turns into (lambda (b) (+ 1 b)) in a few steps.
Note:
In a sense, what we've done is "storing" some inputs
inside a lambda function. Consider the following example:
Implementation:
|%Racket_file
|
| (define curry*
|
(lambda (a)
|
(lambda (b)
|
(lambda (f)
|
(f a b)))))
|
|%END
Note:
What if we want to uncurry a function?
Let's start with type contract. We want to reverse the curry process.
Note:
For curry:
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Before: (f : A -> B -> C)
After: (f' : A -> (B -> C))
Before: (A -> B -> C)
After: (A -> (B -> C))
Then for uncurry:
Before: (f' : A -> (B -> C))
After: (f : A -> B -> C)
Before: (A -> (B -> C))
After: (A -> B -> C)
Implementation:
|%Racket_file
|
| (define uncurry
|
(λ (f)
|
(λ (a b)
|
((f a) b))))
|
|%END
Remark:
- Arguments:
Warning: f must be a curried function.
A function
- Arguments:
a : A
- Return:
A function
- Arguments:
b : B
- Return:
(f a b) : C
- Return:
A function
- Arguments:
a : A
b : B
- Return:
((f a) b) : C
Ex.
|%Racket_file
|
| (define curried-add
|
(curry +))
|
| (uncurry curried-add)
|
|%END
|%STEPPER
|
| (uncurry curried-add)
|
| ;; rewrite the uncurry function
| ((lambda (f)
23

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|
(lambda (a b)
|
((f a) b))) curried-add)
|
| ;; rewrite the curried-add function
| ((lambda (f)
|
(lambda (a b)
|
((f a) b)))
|
((lambda (f)
|
(lambda (a)
|
(lambda (b)
|
(f a b)))) +))
|
| ;; pass in +
| (lambda (f)
|
(lambda (a b)
|
((f a) b))
|
(lambda (a)
|
(lambda (b)
|
(+ a b))))
|
| ;; pass in the lambda function as f
| (lambda (a b)
|
(((lambda (a)
|
(lambda (b)
|
(+ a b)))
|
a)
|
b))
|
|%END
Remark:
- Argument:
a : A
b : B
- Return:
((A function
- Argument:
a : A
- Return:
A function
- Argument:
b : B
- Return:
(+ a b) : C) applied onto a and b) : C
Warning:
This part is very messy. Please read carefully
and make sure you understand it fully.
-------------------------------------------------------------------------C10. map
Implementation:
|%Racket_file
|
| (define map
|
(lambda (f ls)
|
(foldr
|
(lambda (x acc) (cons (f x) acc))
|
'()
|
ls))))
|
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|%END
Ex.
| (map add1 '(1 2 3)) => '(2 3 4)
-------------------------------------------------------------------------C11. filter
Implementation:
|%Racket_file
|
| (define filter
|
(lambda (f ls)
|
(foldr
|
(lambda (x acc) (if (f x) (cons x acc) acc))
|
'()
|
ls)))
|
|%END
==========================================================================
==========================================================================
Part D. Other Minor Topics.
Table of contents
D1. Modules
D2. Smart helpers
D3. Parameterized by total order
-------------------------------------------------------------------------D1. Modules
Theory: Modules
|%Racket_file
|
| #lang racket
|
| ;; (sumto n) sums the integers from 0 to n, where n
| ;;
is a non-negative integer.
| ;; running time: O(n)
|
| (provide sumto)
|
| (define (sumto n)
|
(if (zero? n) 0 (+ n (sumto (sub1 n)))))
|
|%END
Note: Provide statement
This provide statement is called a "directive", which
tells Racket that other files may use this.
|%Racket_file
|
| #lang racket
|
| ;; running time: O(n^2)
|
| (require "sum.rkt")
25

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|
| (define (sumsumto m)
|
(if (zero? m) 0 (+ (sumto m) (sumsumto (sub1 m)))))
|
| ;; Don't panic. This is just a double summation.
|
|%END
Note: Requirement statement
The require statement in Racket is just like import in Python
in other imperative langauges (Python, Java, etc.)
-------------------------------------------------------------------------D2. Smart helpers
Note:
You can use helper functions in a smart way when dealing with the
following situations:
1. a calculation costs too much so you want to avoid doing it
more than once, or
2. you want to record the value produced by the current function
application.
In short, you want to create a "variable" to store some values
for later use.
Implementation:
|%Racket_file
|
| (define (f x)
|
(h (g x) (g x)))
|
| ;; Let h be an arbitrary 2-arg function that you don't care about.
| ;; (For example, h can be + or *).
| ;; Let f, g be arbitrary 1-arg functions.
| ;; Suppose (g x) costs too much and thus
| ;; you want to avoid performing it twice.
| ;; Then this function can be rewritten as:
|
| (define (f' x)
|
(helper (g x)))
|
| (define (helper y)
|
(h y y))
|
| ;; or use local helper function:
|
| (define (f'' x)
|
(define (helper y)
|
(h y y))
|
(helepr (g x)))
|
|%END
Ex.
Now consider the function (drawcard n).
Suppose you want to do the following:
1. Given a list of cards called list-of-cards.
2. Draw a new card (drawcard n)
3. If the new card's secret number is in the list,
discard it. Otherwise cons the new card into
the list.
26

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If you use the old approach like this:
|%Racket_file
|
| ;; (drawcard n) produces a card which is represented
| ;;
by a two-elemtn list.
| ;; (first (drawcard n)) returns the secret number of the card.
| ;; (second (drawcard n)) returns n
|
| ;; Suppose we have defined a helper function called unique?
| ;;
which can determine if the an identical card (a card
| ;;
with the same secret number is in the list or not.
| ;; (unique : (c : Card) -> (l : listof Card) -> Bool)
|
| ;; (main : (n : Nat) -> (lst : listof Cards) -> (l : lstof Cards))
| (define (main n lst)
|
(cond
|
[(unique? (drawcard n) lst) ;; [Line #1]
|
(cons (drawcard n) lst)]
;; [Line #2]
|
[else lst]))
|
|%END
You would soon realize that the two (drawcard n) applications in line
#1 and line #2 actually produces different cards and there's a very
high chance these two cards have different secret numbers. Therefore
your entire algorithm would be incorrect.
To fix this, you can use the above helper:
|%Racket_file
|
| ;; main : (n : Nat) -> (lst : listof Card) -> (l : listof Card)
| (define (main n lst)
|
;; helper : (c : Card) -> (ls : listof Card) -> (l : listof Card)
|
(define (helper c ls)
|
[(unique? c ls) (cons c ls)]
|
[else ls])
|
(helper (drawcard n) lst))
|
|%END
Note:
This way, the c inside the helper function stores your value for
(drawcard n) and your program is correct.
-------------------------------------------------------------------------D3. Parameterized by total order
Note:
In short, we can define our own rules for ordering, and we can
also pass in > and < as arguments.
Note:
When we are defining our own struct, sometimes we need to define our
own rules for order. To determine which element is "greater", we
can use the following function to consume a function for comparison:
Implementation:
|%Racket_file
|
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| (define (less-than? f)
|
(lambda (a b)
|
(cond
|
[(< (f a) (f b)) true]
|
[(< (f b) (f a)) false]
|
[else (< a b]))))
|
|%END
Ex.
|%Racket_file
|
| > ((less-than? abs) -1 -2)
| #true
|
|%END
|%STEPPER
|
| ((less-than? abs) -1 -2)
|
| ;; rewrite less-than?
| ;; plug in abs as f
| ((lambda (a b)
|
(cond
|
[(< (abs a) (abs b)) true]
|
[(< (abs b) (abs a)) false]
|
[else (< a b])))
|
-1 -2)
|
| ;; plug in -1 and -2 as a and b
|
(cond
|
[(< (abs -1) (abs -2)) true]
|
[(< (abs -2) (abs -1)) false]
|
[else (< -1 -2]))))
|
| #true
|
|%END
==========================================================================
==========================================================================
Part E. Appendix
Table of contents
E1. Practice from tutorial Oct.25 by Ashish.
E2. Practice from tutorial Nov.1 by Ashish.
E3. Assignment 6~8 Recap.
E4. Functions you should 100% memorize.
E5. Functions that may help you on the test.
E6. Possible questions on the test.
Ex.
1. Rewrite the function in an unsugared form:
[Level of difficulty: easy]
Q1.rkt
|%Racket_file
|
| (define (make-identity)
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|
(lambda (x) x))
|
|%END
S1.rkt
|%Racket_file
|
| (define (make-identity x) x)
|
|%END
Ex.
2. Define "negate", which takes a boolean function
f, and returns a function that produces the
"not" of the output.
[Level of difficulty: easy]
;; Type contract:
;; (negate : (A -> Bool) -> (A -> Bool))
S2.rkt
|%Racket_file
|
| (define negate
|
(lambda (f)
|
(lambda (x)
|
(not (f x)))))
|
|%END
Ex.
3. Define equal-to?, which is a higher-order function.
It should take a value that can be compared using
equal? and return a function, that when applied
to another argument, returns the equal? of arguments.
[Level of difficulty: easy]
;; Type contract:
;; equal-to? : (a1 : A) -> ((a2 : A) -> Bool)
S3.rkt
|%Racket_file
|
| (define equal-to?
|
(lambda (a1)
|
(lambda (a2)
|
(equal? a1 a2))))
|
|%END
Ex.
4. Define a function "dedup" which takes an ordered
list and removes any duplicate elements. Write
this function using foldr.
[Level of difficulty: medium]
;; Type contract:
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;; dedup : (listof A) -> (listof A)
S4.rkt
|%Racket_file
|
| (define dedup
|
(lambda (lst)
|
(foldr
|
(lambda (x acc) (if (not (member x acc))
|
(cons x acc) acc))
|
'()
|
lst)))
|
|%END
Ex.
5. Define "my-map", which takes a function and a list,
and produces a list with f applied to all elements of
the list. Use foldr.
[Level of difficulty: medium]
;; Type contract:
;; my-map : (A -> B) -> (listof A) -> (listof B)
S5.rkt
|%Racket_file
|
| (define my-map
|
(lambda (f lst)
|
(foldr
|
(lambda (x acc) (cons (f x) acc))
|
'()
|
lst)))
|
|%END
Ex.
6. Create a function "plus" that adds two natural numbers
together. Use only above "f^n" and "add1", do not use
recursion.
[Level of difficulty: medium]
;; Type contract:
;; plus: Nat -> Nat -> Nat
S6.rkt
|%Racket_file
|
| (define plus
|
(lambda (a b)
|
((f^n add1 a) b)))
|
|%END
Remark:
Here is an example and the step-by-step process.
|%Racket_file
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|
| (plus 3 15)
|
|%END
|%STEPPER
|
| (plus 3 15)
|
| ;; rewrite plus
| ((lambda (a b) ((f^n add1 a) b)) 3 15)
|
| ;; pass in a, b
| ((f^n add1 3) 15)
|
| ;; rewrite f^n
| (((lambda (f n)
|
(foldr compose identity
|
(make-list n f))) add1 3) 15)
|
| ;; pass in add1, 3
| ((foldr compose identity '(add1 add1 add1)) 15)
|
| ;; rewrite foldr
| ((lambda (x)
|
(add1 (add1 (add1 x)))) 15)
|
| ;; pass in 15
| (add1 (add1 (add1 15)))
|
| 18
|
|%END
Ex.
7. Create a function "mult" that multiplies two natural
numbers together. Use only the above "f^n" and "plus",
do not use recursion.
[Level of difficulty: hard]
;; Type contract:
;; mult: Nat -> Nat -> Nat
S7.rkt
|%Racket_file
|
| (define mult
|
(lambda (a b)
|
((f^n ((curry plus) a) b) 0))
|
|%END
Remark:
This is very very confusing.
I'll provide an example and step-by-step analysis.
|%Racket_file
|
| (mult 3 5)
|
|%END
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|%STEPPER
|
| (mult 3 5)
|
| ;; rewrite mult
| ((lambda (a b) ((f^n ((curry plus) a) b) 0)) 3 5)
|
| ;; plus in 3, 5
| (((f^n (curry plus) 3) 5) 0)
|
| ;; rewrite curry
| ((f^n
|
((lambda (f)
|
(lambda (a)
|
(lambda (b)
|
(f a b)))) plus 3)
|
5) 0)
|
| ;; plug in plus, 3
| (((f^n
|
(lamdba (b) (plus 3 b))) 5) 0)
|
| ;; in the step above, we get a partial application of 3.
| ;; now, rewrite f^n
| (((lambda (f n) compose identity (make-list f n))
|
(lambda (b) (plus 3 b)) ;; this thing is the f we are passing
|
;; into the outer lambda.
|
5) 0)
|
| ;; plug in 5 as n and (lambda (b) (plus 3 b))
| ;; as f in the outer lambda
| ((compose identity (make-list (lambda (b) (plus 3 b)) 5)) 0)
|
| ;; compose
| ;; now it's the beauty of partial application.
| (plus 3 (plus 3 (plus 3 (plus 3 (plus 3 0)))))
|
| 15
|
|%END
Remark:
This process is very challenging. Read it carefully.
-------------------------------------------------------------------------E2. Practice from tutorial Nov.1 by Ashish
Note:
The following two functions are equivalent:
|%Racket_file
|
| (define map-add1
|
(lambda (lst)
|
(map add1 lst)))
|
| (define map-adder
|
((curry map) add1))
|
|%END
Note that ((curry map) add1) creates the
32

CS145 Midterm II Review Notes V3.0

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function (lambda (lst) (map add1 lst)).
Note:
The following two functions are equivalent:
|%Racket_file
|
| (define sum-list
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(lambda (lst)
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(foldr + 0 lst)))
|
| (define sum-list-curried
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(((curry foldr) + 0)))
|
|%END
Currying and partial application too op.
-------------------------------------------------------------------------E3. Assignment 6~8 Recap
Asst.6 a~e:
Practice on time complexity.
Read A1 and A6 and you should be good.
Asst.6 f:
Given AVL, define a new ADT set.
Remark:
Try to provide a wrapper struct.
Asst.7:
Create a game to draw cards and collect prizes.
Nothing too crazy but remember to check your running time.
Asst.8:
Use generate function to write (prime? n),
(my-build-list n f), (my-foldl f z l), (my-insert e l <),
(my-insertion-sort l <). This is an exercise to help us
what fold does.
Also there is a merge sort exercise. By now we should be
comfortable dealing with sorting and searching algorithms.
-------------------------------------------------------------------------E4. Functions you should 100% memorize.
|%Racket_file
|
| (define compose
|
(lambda (f g)
|
(lambda (x)
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(f (g x)))))
|
| (define map
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(lambda (f lst)
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(foldr
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(lambda (x z) (cons (f x) z))
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'()
|
lst)))
|
| (define filter
|
(lambda (f lst)
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CS145 Midterm II Review Notes V3.0

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|
(foldr
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(lambda (x z) (if (f x) (cons x z) z))
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'()
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lst)))
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| (define (foldr f z l)
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(cond
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[(empty? l) z]
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[else (f (first l) (foldr f z (rest l)))]))
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| (define (foldl f z l)
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(define (calc l z)
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(cond
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[(empty? l) z]
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[else (calc (rest l) (f (first l) z))]))
|
(calc l z))
|
| (define (f'' x)
|
(define (helper y)
|
(h y))
|
(helper (g x)))
|
|%END
-------------------------------------------------------------------------E5. Functions that may help you on the test.
|%Racket_file
|
| (define curry
|
(lambda (f)
|
(lambda (a)
|
(lambda (b)
|
(f a b)))))
|
| (define f^n
|
(lambda (f n)
|
(foldr
|
compose
|
identity
|
(make-list n f))))
|
|%END
-------------------------------------------------------------------------E6. Possible questions on the test.
1. Relatively short problems
1.1 Theory
- One question about big-O
- One question about ADT
- One question about modules
1.2 Short programs
- One question about big-O
- One question about ADT
- One question about writing a lambda function
- One question about application of a given lambda function
2. Topic: Big-O
2.1 Prove why f(x) is O(g(x))
2.2 Prove why h(x) is not O(g(x))
3. Topic: Lambda
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3.1 Write a function to ...
3.2 Write a function to ...
3.3 Given this function, do ...
4. Topic: Stuff from midterm I
4.1 Use tree to define a new ADT
4.2 Use tree to do some operations
5. Prove that merge sort / insertion sort is correct using induction.
-------------------------------------------------------------------------V3.0 Complete. 2017, Nov.3 by David Duan

35



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