Solution Manual Artificial Intelligence A Modern Approach

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Instructor’s Manual:

Exercise Solutions

for

Artiﬁcial Intelligence

A Modern Approach

Second Edition

Stuart J. Russell and Peter Norvig

Upper Saddle River, New Jersey 07458

Library of Congress Cataloging-in-Publication Data

Russell, Stuart J. (Stuart Jonathan)

Instructor’s solution manual for artiﬁcial intelligence : a modern approach

(second edition) / Stuart Russell, Peter Norvig.

Includes bibliographical references and index.

1. Artiﬁcial intelligence I. Norvig, Peter. II. Title.

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Preface

This Instructor’s Solution Manual provides solutions (or at least solution sketches) for

almost all of the 400 exercises in Artiﬁcial Intelligence: A Modern Approach (Second Edi-

tion). We only give actual code for a few of the programming exercises; writing a lot of code

would not be that helpful, if only because we don’t know what language you prefer.

In many cases, we give ideas for discussion and follow-up questions, and we try to

explain why we designed each exercise.

There is more supplementary material that we want to offer to the instructor, but we

have decided to do it through the medium of the World Wide Web rather than through a CD

or printed Instructor’s Manual. The idea is that this solution manual contains the material that

must be kept secret from students, but the Web site contains material that can be updated and

added to in a more timely fashion. The address for the web site is:

http://aima.cs.berkeley.edu

and the address for the online Instructor’s Guide is:

http://aima.cs.berkeley.edu/instructors.html

There you will ﬁnd:

Instructions on how to join the aima-instructors discussion list. We strongly recom-

mend that you join so that you can receive updates, corrections, notiﬁcation of new

versions of this Solutions Manual, additional exercises and exam questions, etc., in a

timely manner.

Source code for programs from the text. We offer code in Lisp, Python, and Java, and

point to code developed by others in C++ and Prolog.

Programming resources and supplemental texts.

Figures from the text; for overhead transparencies.

Terminology from the index of the book.

Other courses using the book that have home pages on the Web. You can see example

syllabi and assignments here. Please do not put solution sets for AIMA exercises on

public web pages!

AI Education information on teaching introductory AI courses.

Other sites on the Web with information on AI. Organized by chapter in the book; check

this for supplemental material.

We welcome suggestions for new exercises, new environments and agents, etc. The

book belongs to you, the instructor, as much as us. We hope that you enjoy teaching from it,

that these supplemental materials help, and that you will share your supplements and experi-

ences with other instructors.

iii

Solutions for Chapter 1

Introduction

1.1

a. Dictionary deﬁnitions of intelligence talk about “the capacity to acquire and apply

knowledge” or “the faculty of thought and reason” or “the ability to comprehend and

proﬁt from experience.” These are all reasonable answers, but if we want something

quantiﬁable we would use something like “the ability to apply knowledge in order to

perform better in an environment.”

b. We deﬁne artiﬁcial intelligence as the study and construction of agent programs that

perform well in a given environment, for a given agent architecture.

c. We deﬁne an agent as an entity that takes action in response to percepts from an envi-

ronment.

1.2 See the solution for exercise 26.1 for some discussion of potential objections.

The probability of fooling an interrogator depends on just how unskilled the interroga-

tor is. One entrant in the 2002 Loebner prize competition (which is not quite a real Turing

Test) did fool one judge, although if you look at the transcript, it is hard to imagine what

that judge was thinking. There certainly have been examples of a chatbot or other online

agent fooling humans. For example, see See Lenny Foner’s account of the Julia chatbot

at foner.www.media.mit.edu/people/foner/Julia/. We’d say the chance today is something

like 10%, with the variation depending more on the skill of the interrogator rather than the

program. In 50 years, we expect that the entertainment industry (movies, video games, com-

mercials) will have made sufﬁcient investments in artiﬁcial actors to create very credible

impersonators.

1.3 The 2002 Loebner prize (www.loebner.net) went to Kevin Copple’s program ELLA. It

consists of a prioritized set of pattern/action rules: if it sees a text string matching a certain

pattern, it outputs the corresponding response, which may include pieces of the current or

past input. It also has a large database of text and has the Wordnet online dictionary. It is

therefore using rather rudimentary tools, and is not advancing the theory of AI. It is provid-

ing evidence on the number and type of rules that are sufﬁcient for producing one type of

conversation.

1.4 No. It means that AI systems should avoid trying to solve intractable problems. Usually,

this means they can only approximate optimal behavior. Notice that humans don’t solve NP-

complete problems either. Sometimes they are good at solving speciﬁc instances with a lot of

2 Chapter 1. Introduction

structure, perhaps with the aid of background knowledge. AI systems should attempt to do

the same.

1.5 No. IQ test scores correlate well with certain other measures, such as success in college,

but only if they’re measuring fairly normal humans. The IQ test doesn’t measure everything.

A program that is specialized only for IQ tests (and specialized further only for the analogy

part) would very likely perform poorly on other measures of intelligence. See The Mismea-

sure of Man by Stephen Jay Gould, Norton, 1981 or Multiple intelligences: the theory in

practice by Howard Gardner, Basic Books, 1993 for more on IQ tests, what they measure,

and what other aspects there are to “intelligence.”

1.6 Just as you are unaware of all the steps that go into making your heart beat, you are

also unaware of most of what happens in your thoughts. You do have a conscious awareness

of some of your thought processes, but the majority remains opaque to your consciousness.

The ﬁeld of psychoanalysis is based on the idea that one needs trained professional help to

analyze one’s own thoughts.

1.7

a. (ping-pong) A reasonable level of proﬁciency was achieved by Andersson’s robot (An-

dersson, 1988).

b. (driving in Cairo) No. Although there has been a lot of progress in automated driving,

all such systems currently rely on certain relatively constant clues: that the road has

shoulders and a center line, that the car ahead will travel a predictable course, that cars

will keep to their side of the road, and so on. To our knowledge, none are able to avoid

obstacles or other cars or to change lanes as appropriate; their skills are mostly conﬁned

to staying in one lane at constant speed. Driving in downtown Cairo is too unpredictable

for any of these to work.

c. (shopping at the market) No. No robot can currently put together the tasks of moving in

a crowded environment, using vision to identify a wide variety of objects, and grasping

the objects (including squishable vegetables) without damaging them. The component

pieces are nearly able to handle the individual tasks, but it would take a major integra-

tion effort to put it all together.

d. (shopping on the web) Yes. Software robots are capable of handling such tasks, par-

ticularly if the design of the web grocery shopping site does not change radically over

time.

e. (bridge) Yes. Programs such as GIB now play at a solid level.

f. (theorem proving) Yes. For example, the proof of Robbins algebra described on page

309.

g. (funny story) No. While some computer-generated prose and poetry is hysterically

funny, this is invariably unintentional, except in the case of programs that echo back

prose that they have memorized.

h. (legal advice) Yes, in some cases. AI has a long history of research into applications

of automated legal reasoning. Two outstanding examples are the Prolog-based expert

systems used in the UK to guide members of the public in dealing with the intricacies of

the social security and nationality laws. The social security system is said to have saved

the UK government approximately $150 million in its ﬁrst year of operation. However,

extension into more complex areas such as contract law awaits a satisfactory encoding

of the vast web of common-sense knowledge pertaining to commercial transactions and

agreement and business practices.

i. (translation) Yes. In a limited way, this is already being done. See Kay, Gawron and

Norvig (1994) and Wahlster (2000) for an overview of the ﬁeld of speech translation,

and some limitations on the current state of the art.

j. (surgery) Yes. Robots are increasingly being used for surgery, although always under

the command of a doctor.

1.8 Certainly perception and motor skills are important, and it is a good thing that the ﬁelds

of vision and robotics exist (whether or not you want to consider them part of “core” AI).

But given a percept, an agent still has the task of “deciding” (either by deliberation or by

reaction) which action to take. This is just as true in the real world as in artiﬁcial micro-

worlds such as chess-playing. So computing the appropriate action will remain a crucial part

of AI, regardless of the perceptual and motor system to which the agent program is “attached.”

On the other hand, it is true that a concentration on micro-worlds has led AI away from the

really interesting environments (see page 46).

1.9 Evolution tends to perpetuate organisms (and combinations and mutations of organ-

isms) that are succesful enough to reproduce. That is, evolution favors organisms that can

optimize their performance measure to at least survive to the age of sexual maturity, and then

be able to win a mate. Rationality just means optimizing performance measure, so this is in

line with evolution.

1.10 Yes, they are rational, because slower, deliberative actions would tend to result in

more damage to the hand. If “intelligent” means “applying knowledge” or “using thought

and reasoning” then it does not require intelligence to make a reﬂex action.

1.11 This depends on your deﬁnition of “intelligent” and “tell.” In one sense computers only

do what the programmers command them to do, but in another sense what the programmers

consciously tells the computer to do often has very little to do with what the computer actually

does. Anyone who has written a program with an ornery bug knows this, as does anyone

who has written a successful machine learning program. So in one sense Samuel “told” the

computer “learn to play checkers better than I do, and then play that way,” but in another

sense he told the computer “follow this learning algorithm” and it learned to play. So we’re

left in the situation where you may or may not consider learning to play checkers to be s sign

of intelligence (or you may think that learning to play in the right way requires intelligence,

but not in this way), and you may think the intelligence resides in the programmer or in the

computer.

1.12 The point of this exercise is to notice the parallel with the previous one. Whatever you

decided about whether computers could be intelligent in 1.9, you are committed to making the

4 Chapter 1. Introduction

same conclusion about animals (including humans), unless your reasons for deciding whether

something is intelligent take into account the mechanism (programming via genes versus

programming via a human programmer). Note that Searle makes this appeal to mechanism

in his Chinese Room argument (see Chapter 26).

1.13 Again, the choice you make in 1.11 drives your answer to this question.

Solutions for Chapter 2

Intelligent Agents

2.1 The following are just some of the many possible deﬁnitions that can be written:

Agent: an entity that perceives and acts; or, one that can be viewed as perceiving and

acting. Essentially any object qualiﬁes; the key point is the way the object implements

an agent function. (Note: some authors restrict the term to programs that operate on

behalf of a human, or to programs that can cause some or all of their code to run on

other machines on a network, as in mobile agents.)

Agent function: a function that speciﬁes the agent’s action in response to every possible

percept sequence.

Agent program: that program which, combined with a machine architecture, imple-

ments an agent function. In our simple designs, the program takes a new percept on

each invocation and returns an action.

Rationality: a property of agents that choose actions that maximize their expected util-

ity, given the percepts to date.

Autonomy: a property of agents whose behavior is determined by their own experience

rather than solely by their initial programming.

Reﬂex agent: an agent whose action depends only on the current percept.

Model-based agent: an agent whose action is derived directly from an internal model

of the current world state that is updated over time.

Goal-based agent: an agent that selects actions that it believes will achieve explicitly

represented goals.

Utility-based agent: an agent that selects actions that it believes will maximize the

expected utility of the outcopme state.

Learning agent: an agent whose behavior improves over time based on its experience.

2.2 A performance measure is used by an outside observer to evaluate how successful an

agent is. It is a function from histories to a real number. A utility function is used by an agent

itself to evaluate how desirable states or histories are. In our framework, the utility function

may not be the same as the performance measure; furthermore, an agent may have no explicit

utility function at all, whereas there is always a performance measure.

2.3 Although these questions are very simple, they hint at some very fundamental issues.

Our answers are for the simple agent designs for static environments where nothing happens

6 Chapter 2. Intelligent Agents

while the agent is deliberating; the issues get even more interesting for dynamic environ-

ments.

a. Yes; take any agent program and insert null statements that do not affect the output.

b. Yes; the agent function might specify that the agent print when the percept is a

Turing machine program that halts, and otherwise. (Note: in dynamic environ-

ments, for machines of less than inﬁnite speed, the rational agent function may not be

implementable; e.g., the agent function that always plays a winning move, if any, in a

game of chess.)

c. Yes; the agent’s behavior is ﬁxed by the architecture and program.

d. There are agent programs, although many of these will not run at all. (Note: Any

given program can devote at most bits to storage, so its internal state can distinguish

among only past histories. Because the agent function speciﬁes actions based on per-

cept histories, there will be many agent functions that cannot be implemented because

of lack of memory in the machine.)

2.4 Notice that for our simple environmental assumptions we need not worry about quanti-

tative uncertainty.

a. It sufﬁces to show that for all possible actual environments (i.e., all dirt distributions and

initial locations), this agent cleans the squares at least as fast as any other agent. This is

trivially true when there is no dirt. When there is dirt in the initial location and none in

the other location, the world is clean after one step; no agent can do better. When there

is no dirt in the initial location but dirt in the other, the world is clean after two steps; no

agent can do better. When there is dirt in both locations, the world is clean after three

steps; no agent can do better. (Note: in general, the condition stated in the ﬁrst sentence

of this answer is much stricter than necessary for an agent to be rational.)

b. The agent in (a) keeps moving backwards and forwards even after the world is clean.

It is better to do once the world is clean (the chapter says this). Now, since

the agent’s percept doesn’t say whether the other square is clean, it would seem that

the agent must have some memory to say whether the other square has already been

cleaned. To make this argument rigorous is more difﬁcult—for example, could the

agent arrange things so that it would only be in a clean left square when the right square

was already clean? As a general strategy, an agent can use the environment itself as

a form of external memory—a common technique for humans who use things like

appointment calendars and knots in handkerchiefs. In this particular case, however, that

is not possible. Consider the reﬂex actions for and . If either of

these is , then the agent will fail in the case where that is the initial percept but

the other square is dirty; hence, neither can be and therefore the simple reﬂex

agent is doomed to keep moving. In general, the problem with reﬂex agents is that they

have to do the same thing in situations that look the same, even when the situations

are actually quite different. In the vacuum world this is a big liability, because every

interior square (except home) looks either like a square with dirt or a square without

dirt.

Agent Type Performance

Measure Environment Actuators Sensors

Robot soccer

player

Winning game,

goals for/against

Field, ball, own

team, other team,

own body

Devices (e.g.,

legs) for

locomotion and

kicking

Camera, touch

sensors,

accelerometers,

orientation

sensors,

wheel/joint

encoders

Internet

book-shopping

agent

Obtain re-

quested/interesting

books, minimize

expenditure

Internet Follow link,

enter/submit data

in ﬁelds, display

to user

Web pages, user

requests

Autonomous

Mars rover Terrain explored

and reported,

samples gathered

and analyzed

Launch vehicle,

lander, Mars

Wheels/legs,

sample collection

device, analysis

devices, radio

transmitter

Camera, touch

sensors,

accelerometers,

orientation

sensors, ,

wheel/joint

encoders, radio

receiver

Mathematician’s

theorem-proving

assistant

Figure S2.1 Agent types and their PEAS descriptions, for Ex. 2.5.

c. If we consider asymptotically long lifetimes, then it is clear that learning a map (in

some form) confers an advantage because it means that the agent can avoid bumping

into walls. It can also learn where dirt is most likely to accumulate and can devise

an optimal inspection strategy. The precise details of the exploration method needed

to construct a complete map appear in Chapter 4; methods for deriving an optimal

inspection/cleanup strategy are in Chapter 21.

2.5 Some representative, but not exhaustive, answers are given in Figure S2.1.

2.6 Environment properties are given in Figure S2.2. Suitable agent types:

a. A model-based reﬂex agent would sufﬁce for most aspects; for tactical play, a utility-

based agent with lookahead would be useful.

b. A goal-based agent would be appropriate for speciﬁc book requests. For more open-

ended tasks—e.g., “Find me something interesting to read”—tradeoffs are involved and

the agent must compare utilities for various possible purchases.

8 Chapter 2. Intelligent Agents

Task Environment Observable Deterministic Episodic Static Discrete Agents

Robot soccer Partially Stochastic Sequential Dynamic Continuous Multi

Internet book-shopping Partially Deterministic Sequential Static Discrete Single

Autonomous Mars rover Partially Stochastic Sequential Dynamic Continuous Single

Mathematician’s assistant Fully Deterministic Sequential Semi Discrete Multi

Figure S2.2 Environment properties for Ex. 2.6.

c. A model-based reﬂex agent would sufﬁce for low-level navigation and obstacle avoid-

ance; for route planning, exploration planning, experimentation, etc., some combination

of goal-based and utility-based agents would be needed.

d. For speciﬁc proof tasks, a goal-based agent is needed. For “exploratory” tasks—e.g.,

“Prove some useful lemmata concerning operations on strings”—a utility-based archi-

tecture might be needed.

2.7 The ﬁle "agents/environments/vacuum.lisp" in the code repository imple-

ments the vacuum-cleaner environment. Students can easily extend it to generate different

shaped rooms, obstacles, and so on.

2.8 A reﬂex agent program implementing the rational agent function described in the chap-

ter is as follows:

(defun reflex-rational-vacuum-agent (percept)

(destructuring-bind (location status) percept

(cond ((eq status ’Dirty) ’Suck)

((eq location ’A) ’Right)

(t ’Left))))

For states 1, 3, 5, 7 in Figure 3.20, the performance measures are 1996, 1999, 1998, 2000

respectively.

2.9 Exercises 2.4, 2.9, and 2.10 may be merged in future printings.

a. No; see answer to 2.4(b).

b. See answer to 2.4(b).

c. In this case, a simple reﬂex agent can be perfectly rational. The agent can consist of

a table with eight entries, indexed by percept, that speciﬁes an action to take for each

possible state. After the agent acts, the world is updated and the next percept will tell

the agent what to do next. For larger environments, constructing a table is infeasible.

Instead, the agent could run one of the optimal search algorithms in Chapters 3 and 4

and execute the ﬁrst step of the solution sequence. Again, no internal state is required,

but it would help to be able to store the solution sequence instead of recomputing it for

each new percept.

2.10

Figure S2.3 An environment in which random motion will take a long time to cover all

the squares.

a. Because the agent does not know the geography and perceives only location and local

dirt, and canot remember what just happened, it will get stuck forever against a wall

when it tries to move in a direction that is blocked—that is, unless it randomizes.

b. One possible design cleans up dirt and otherwise moves randomly:

(defun randomized-reflex-vacuum-agent (percept)

(destructuring-bind (location status) percept

(cond ((eq status ’Dirty) ’Suck)

(t (random-element ’(Left Right Up Down))))))

This is fairly close to what the Roomba vacuum cleaner does (although the Roomba

has a bump sensor and randomizes only when it hits an obstacle). It works reasonably

well in nice, compact environments. In maze-like environments or environments with

small connecting passages, it can take a very long time to cover all the squares.

c. An example is shown in Figure S2.3. Students may also wish to measure clean-up time

for linear or square environments of different sizes, and compare those to the efﬁcient

online search algorithms described in Chapter 4.

d. A reﬂex agent with state can build a map (see Chapter 4 for details). An online depth-

ﬁrst exploration will reach every state in time linear in the size of the environment;

therefore, the agent can do much better than the simple reﬂex agent.

The question of rational behavior in unknown environments is a complex one but it is

worth encouraging students to think about it. We need to have some notion of the prior

10 Chapter 2. Intelligent Agents

probaility distribution over the class of environments; call this the initial belief state.

Any action yields a new percept that can be used to update this distribution, moving

the agent to a new belief state. Once the environment is completely explored, the belief

state collapses to a single possible environment. Therefore, the problem of optimal

exploration can be viewed as a search for an optimal strategy in the space of possible

belief states. This is a well-deﬁned, if horrendously intractable, problem. Chapter 21

discusses some cases where optimal exploration is possible. Another concrete example

of exploration is the Minesweeper computer game (see Exercise 7.11). For very small

Minesweeper environments, optimal exploration is feasible although the belief state

update is nontrivial to explain.

2.11 The problem appears at ﬁrst to be very similar; the main difference is that instead of

using the location percept to build the map, the agent has to “invent” its own locations (which,

after all, are just nodes in a data structure representing the state space graph). When a bump

is detected, the agent assumes it remains in the same location and can add a wall to its map.

For grid environments, the agent can keep track of its location and so can tell when it

has returned to an old state. In the general case, however, there is no simple way to tell if a

state is new or old.

2.12

a. For a reﬂex agent, this presents no additional challenge, because the agent will continue

to as long as the current location remains dirty. For an agent that constructs a

sequential plan, every action would need to be replaced by “ until clean.”

If the dirt sensor can be wrong on each step, then the agent might want to wait for a

few steps to get a more reliable measurement before deciding whether to or move

on to a new square. Obviously, there is a trade-off because waiting too long means

that dirt remains on the ﬂoor (incurring a penalty), but acting immediately risks either

dirtying a clean square or ignoring a dirty square (if the sensor is wrong). A rational

agent must also continue touring and checking the squares in case it missed one on a

previous tour (because of bad sensor readings). it is not immediately obvious how the

waiting time at each square should change with each new tour. These issues can be

clariﬁed by experimentation, which may suggest a general trend that can be veriﬁed

mathematically. This problem is a partially observable Markov decision process—see

Chapter 17. Such problems are hard in general, but some special cases may yield to

careful analysis.

b. In this case, the agent must keep touring the squares indeﬁnitely. The probability that

a square is dirty increases monotonically with the time since it was last cleaned, so the

rational strategy is, roughly speaking, to repeatedly execute the shortest possible tour of

all squares. (We say “roughly speaking” because there are complications caused by the

fact that the shortest tour may visit some squares twice, depending on the geography.)

This problem is also a partially observable Markov decision process.

Solutions for Chapter 3

Solving Problems by Searching

3.1 Astate is a situation that an agent can ﬁnd itself in. We distinguish two types of states:

world states (the actual concrete situations in the real world) and representational states (the

abstract descriptions of the real world that are used by the agent in deliberating about what to

do).

Astate space is a graph whose nodes are the set of all states, and whose links are

actions that transform one state into another.

Asearch tree is a tree (a graph with no undirected loops) in which the root node is the

start state and the set of children for each node consists of the states reachable by taking any

action.

Asearch node is a node in the search tree.

Agoal is a state that the agent is trying to reach.

An action is something that the agent can choose to do.

Asuccessor function described the agent’s options: given a state, it returns a set of

(action, state) pairs, where each state is the state reachable by taking the action.

The branching factor in a search tree is the number of actions available to the agent.

3.2 In goal formulation, we decide which aspects of the world we are interested in, and

which can be ignored or abstracted away. Then in problem formulation we decide how to

manipulate the important aspects (and ignore the others). If we did problem formulation ﬁrst

we would not know what to include and what to leave out. That said, it can happen that there

is a cycle of iterations between goal formulation, problem formulation, and problem solving

until one arrives at a sufﬁciently useful and efﬁcient solution.

3.3 In Python we have:

#### successor_fn defined in terms of result and legal_actions

def successor_fn(s):

return [(a, result(a, s)) for a in legal_actions(s)]

#### legal_actions and result defined in terms of successor_fn

def legal_actions(s):

return [a for (a, s) in successor_fn(s)]

def result(a, s):

12 Chapter 3. Solving Problems by Searching

for (a1, s1) in successor_fn(s):

if a == a1:

return s1

3.4 From http://www.cut-the-knot.com/pythagoras/ﬁfteen.shtml, this proof applies to the

ﬁfteen puzzle, but the same argument works for the eight puzzle:

Deﬁnition: The goal state has the numbers in a certain order, which we will measure as

starting at the upper left corner, then proceeding left to right, and when we reach the end of a

row, going down to the leftmost square in the row below. For any other conﬁguration besides

the goal, whenever a tile with a greater number on it precedes a tile with a smaller number,

the two tiles are said to be inverted.

Proposition: For a given puzzle conﬁguration, let denote the sum of the total number

of inversions and the row number of the empty square. Then is invariant under any

legal move. In other words, after a legal move an odd remains odd whereas an even

remains even. Therefore the goal state in Figure 3.4, with no inversions and empty square in

the ﬁrst row, has , and can only be reached from starting states with odd , not from

starting states with even .

Proof: First of all, sliding a tile horizontally changes neither the total number of in-

versions nor the row number of the empty square. Therefore let us consider sliding a tile

vertically.

Let’s assume, for example, that the tile is located directly over the empty square.

Sliding it down changes the parity of the row number of the empty square. Now consider the

total number of inversions. The move only affects relative positions of tiles , , , and .

If none of the , , caused an inversion relative to (i.e., all three are larger than ) then

after sliding one gets three (an odd number) of additional inversions. If one of the three is

smaller than , then before the move , , and contributed a single inversion (relative to

) whereas after the move they’ll be contributing two inversions - a change of 1, also an odd

number. Two additional cases obviously lead to the same result. Thus the change in the sum

is always even. This is precisely what we have set out to show.

So before we solve a puzzle, we should compute the value of the start and goal state

and make sure they have the same parity, otherwise no solution is possible.

3.5 The formulation puts one queen per column, with a new queen placed only in a square

that is not attacked by any other queen. To simplify matters, we’ll ﬁrst consider the –rooks

problem. The ﬁrst rook can be placed in any square in column 1, the second in any square in

column 2 except the same row that as the rook in column 1, and in general there will be

elements of the search space.

3.6 No, a ﬁnite state space does not always lead to a ﬁnite search tree. Consider a state

space with two states, both of which have actions that lead to the other. This yields an inﬁnite

search tree, because we can go back and forth any number of times. However, if the state

space is a ﬁnite tree, or in general, a ﬁnite DAG (directed acyclic graph), then there can be no

loops, and the search tree is ﬁnite.

3.7

a. Initial state: No regions colored.

Goal test: All regions colored, and no two adjacent regions have the same color.

Successor function: Assign a color to a region.

Cost function: Number of assignments.

b. Initial state: As described in the text.

Goal test: Monkey has bananas.

Successor function: Hop on crate; Hop off crate; Push crate from one spot to another;

Walk from one spot to another; grab bananas (if standing on crate).

Cost function: Number of actions.

c. Initial state: considering all input records.

Goal test: considering a single record, and it gives “illegal input” message.

Successor function: run again on the ﬁrst half of the records; run again on the second

half of the records.

Cost function: Number of runs.

Note: This is a contingency problem; you need to see whether a run gives an error

message or not to decide what to do next.

d. Initial state: jugs have values .

Successor function: given values , generate , , (by ﬁll-

ing); , , (by emptying); or for any two jugs with current values

and , pour into ; this changes the jug with to the minimum of and the

capacity of the jug, and decrements the jug with by by the amount gained by the ﬁrst

jug.

Cost function: Number of actions.

2 3

4 5 6 7

8 9 10 1211 13 14 15

Figure S3.1 The state space for the problem deﬁned in Ex. 3.8.

3.8

a. See Figure S3.1.

b. Breadth-ﬁrst: 1 2 3 4 5 6 7 8 9 10 11

Depth-limited: 1 2 4 8 9 5 10 11

Iterative deepening: 1; 1 2 3; 1 2 4 5 3 6 7; 1 2 4 8 9 5 10 11

14 Chapter 3. Solving Problems by Searching

c. Bidirectional search is very useful, because the only successor of in the reverse direc-

tion is . This helps focus the search.

d. 2 in the forward direction; 1 in the reverse direction.

e. Yes; start at the goal, and apply the single reverse successor action until you reach 1.

3.9

a. Here is one possible representation: A state is a six-tuple of integers listing the number

of missionaries, cannibals, and boats on the ﬁrst side, and then the seond side of the

river. The goal is a state with 3 missionaries and 3 cannibals on the second side. The

cost function is one per action, and the successors of a state are all the states that move

1 or 2 people and 1 boat from one side to another.

b. The search space is small, so any optimal algorithm works. For an example, see the

ﬁle "search/domains/cannibals.lisp". It sufﬁces to eliminate moves that

circle back to the state just visited. From all but the ﬁrst and last states, there is only

one other choice.

c. It is not obvious that almost all moves are either illegal or revert to the previous state.

There is a feeling of a large branching factor, and no clear way to proceed.

3.10 For the 8 puzzle, there shouldn’t be much difference in performance. Indeed, the ﬁle

"search/domains/puzzle8.lisp" shows that you can represent an 8 puzzle state as

a single 32-bit integer, so the question of modifying or copying data is moot. But for the

puzzle, as increases, the advantage of modifying rather than copying grows. The

disadvantage of a modifying successor function is that it only works with depth-ﬁrst search

(or with a variant such as iterative deepening).

3.11 a. The algorithm expands nodes in order of increasing path cost; therefore the ﬁrst

goal it encounters will be the goal with the cheapest cost.

b. It will be the same as iterative deepening, iterations, in which nodes are

generated.

d. Implementation not shown.

3.12 If there are two paths from the start node to a given node, discarding the more ex-

pensive one cannot eliminate any optimal solution. Uniform-cost search and breadth-ﬁrst

search with constant step costs both expand paths in order of -cost. Therefore, if the current

node has been expanded previously, the current path to it must be more expensive than the

previously found path and it is correct to discard it.

For IDS, it is easy to ﬁnd an example with varying step costs where the algorithm returns

a suboptimal solution: simply have two paths to the goal, one with one step costing 3 and the

other with two steps costing 1 each.

3.13 Consider a domain in which every state has a single successor, and there is a single goal

at depth . Then depth-ﬁrst search will ﬁnd the goal in steps, whereas iterative deepening

search will take steps.

3.14 As an ordinary person (or agent) browsing the web, we can only generarte the suc-

cessors of a page by visiting it. We can then do breadth-ﬁrst search, or perhaps best-search

search where the heuristic is some function of the number of words in common between the

start and goal pages; this may help keep the links on target. Search engines keep the complete

graph of the web, and may provide the user access to all (or at least some) of the pages that

link to a page; this would allow us to do bidirectional search.

3.15

a. If we consider all points, then there are an inﬁnite number of states, and of paths.

b. (For this problem, we consider the start and goal points to be vertices.) The shortest

distance between two points is a straight line, and if it is not possible to travel in a

straight line because some obstacle is in the way, then the next shortest distance is a

sequence of line segments, end-to-end, that deviate from the straight line by as little

as possible. So the ﬁrst segment of this sequence must go from the start point to a

tangent point on an obstacle – any path that gave the obstacle a wider girth would be

longer. Because the obstacles are polygonal, the tangent points must be at vertices of

the obstacles, and hence the entire path must go from vertex to vertex. So now the state

space is the set of vertices, of which there are 35 in Figure 3.22.

c. Code not shown.

d. Implementations and analysis not shown.

3.16 Code not shown.

3.17

a. Any path, no matter how bad it appears, might lead to an arbitraily large reward (nega-

tive cost). Therefore, one would need to exhaust all possible paths to be sure of ﬁnding

the best one.

b. Suppose the greatest possible reward is . Then if we also know the maximum depth of

the state space (e.g. when the state space is a tree), then any path with levels remaining

can be improved by at most , so any paths worse than less than the best path can be

pruned. For state spaces with loops, this guarantee doesn’t help, because it is possible

to go around a loop any number of times, picking up rewward each time.

c. The agent should plan to go around this loop forever (unless it can ﬁnd another loop

with even better reward).

d. The value of a scenic loop is lessened each time one revisits it; a novel scenic sight

is a great reward, but seeing the same one for the tenth time in an hour is tedious, not

rewarding. To accomodate this, we would have to expand the state space to include a

memory—a state is now represented not just by the current location, but by a current

location and a bag of already-visited locations. The reward for visiting a new location

is now a (diminishing) function of the number of times it has been seen before.

e. Real domains with looping behavior include eating junk food and going to class.

3.18 The belief state space is shown in Figure S3.2. No solution is possible because no path

leads to a belief state all of whose elements satisfy the goal. If the problem is fully observable,

16 Chapter 3. Solving Problems by Searching

L R

S SS

Figure S3.2 The belief state space for the sensorless vacuum world under Murphy’s law.

the agent reaches a goal state by executing a sequence such that is performed only in a

dirty square. This ensures deterministic behavior and every state is obviously solvable.

3.19 Code not shown, but a good start is in the code repository. Clearly, graph search

must be used—this is a classic grid world with many alternate paths to each state. Students

will quickly ﬁnd that computing the optimal solution sequence is prohibitively expensive for

moderately large worlds, because the state space for an world has states. The

completion time of the random agent grows less than exponentially in , so for any reasonable

exchange rate between search cost ad path cost the random agent will eventually win.

Solutions for Chapter 4

Informed Search and Exploration

4.1 The sequence of queues is as follows:

L[0+244=244]

M[70+241=311], T[111+329=440]

L[140+244=384], D[145+242=387], T[111+329=440]

D[145+242=387], T[111+329=440], M[210+241=451], T[251+329=580]

C[265+160=425], T[111+329=440], M[210+241=451], M[220+241=461], T[251+329=580]

T[111+329=440], M[210+241=451], M[220+241=461], P[403+100=503], T[251+329=580], R[411+193=604],

D[385+242=627]

M[210+241=451], M[220+241=461], L[222+244=466], P[403+100=503], T[251+329=580], A[229+366=595],

R[411+193=604], D[385+242=627]

M[220+241=461], L[222+244=466], P[403+100=503], L[280+244=524], D[285+242=527], T[251+329=580],

A[229+366=595], R[411+193=604], D[385+242=627]

L[222+244=466], P[403+100=503], L[280+244=524], D[285+242=527], L[290+244=534], D[295+242=537],

T[251+329=580], A[229+366=595], R[411+193=604], D[385+242=627]

P[403+100=503], L[280+244=524], D[285+242=527], M[292+241=533], L[290+244=534], D[295+242=537],

T[251+329=580], A[229+366=595], R[411+193=604], D[385+242=627], T[333+329=662]

B[504+0=504], L[280+244=524], D[285+242=527], M[292+241=533], L[290+244=534], D[295+242=537], T[251+329=580],

A[229+366=595], R[411+193=604], D[385+242=627], T[333+329=662], R[500+193=693], C[541+160=701]

4.2 gives . This behaves exactly like uniform-cost search—the factor

of two makes no difference in the ordering of the nodes. gives A search. gives

, i.e., greedy best-ﬁrst search. We also have

which behaves exactly like A search with a heuristic . For , this is always

less than and hence admissible, provided is itself admissible.

4.3

a. When all step costs are equal, , so uniform-cost search reproduces

breadth-ﬁrst search.

b. Breadth-ﬁrst search is best-ﬁrst search with ; depth-ﬁrst search is

best-ﬁrst search with ; uniform-cost search is best-ﬁrst search with

18 Chapter 4. Informed Search and Exploration

c. Uniform-cost search is A search with .

SAG

h=7

h=5

h=1 h=0

Figure S4.1 A graph with an inconsistent heuristic on which GRAPH-SEARCH fails to

return the optimal solution. The successors of are with and with . is

expanded ﬁrst, so the path via will be discarded because will already be in the closed

list.

4.4 See Figure S4.1.

4.5 Going between Rimnicu Vilcea and Lugoj is one example. The shortest path is the

southern one, through Mehadia, Dobreta and Craiova. But a greedy search using the straight-

line heuristic starting in Rimnicu Vilcea will start the wrong way, heading to Sibiu. Starting

at Lugoj, the heuristic will correctly lead us to Mehadia, but then a greedy search will return

to Lugoj, and oscillate forever between these two cities.

4.6 The heuristic (adding misplaced tiles and Manhattan distance) sometimes

overestimates. Now, suppose (as given) and let be a goal that is

suboptimal by more than , i.e., . Now consider any node on a path to an

optimal goal. We have

so will never be expanded before an optimal goal is expanded.

4.7 A heuristic is consistent iff, for every node and every successor of generated by

any action ,

One simple proof is by induction on the number of nodes on the shortest path to any goal

from . For , let be the goal node; then . For the inductive

case, assume is on the shortest path steps from the goal and that is admissible by

hypothesis; then

so at steps from the goal is also admissible.

4.8 This exercise reiterates a small portion of the classic work of Held and Karp (1970).

a. The TSP problem is to ﬁnd a minimal (total length) path through the cities that forms

a closed loop. MST is a relaxed version of that because it asks for a minimal (total

length) graph that need not be a closed loop—it can be any fully-connected graph. As

a heuristic, MST is admissible—it is always shorter than or equal to a closed loop.

b. The straight-line distance back to the start city is a rather weak heuristic—it vastly

underestimates when there are many cities. In the later stage of a search when there are

only a few cities left it is not so bad. To say that MST dominates straight-line distance

is to say that MST always gives a higher value. This is obviously true because a MST

that includes the goal node and the current node must either be the straight line between

them, or it must include two or more lines that add up to more. (This all assumes the

triangle inequality.)

c. See "search/domains/tsp.lisp" for a start at this. The ﬁle includes a heuristic

based on connecting each unvisited city to its nearest neighbor, a close relative to the

MST approach.

d. See (Cormen et al., 1990, p.505) for an algorithm that runs in time, where

is the number of edges. The code repository currently contains a somewhat less

efﬁcient algorithm.

4.9 The misplaced-tiles heuristic is exact for the problem where a tile can move from square

A to square B. As this is a relaxation of the condition that a tile can move from square A to

square B if B is blank, Gaschnig’s heuristic canot be less than the misplaced-tiles heuristic.

As it is also admissible (being exact for a relaxation of the original problem), Gaschnig’s

heuristic is therefore more accurate.

If we permute two adjacent tiles in the goal state, we have a state where misplaced-tiles

and Manhattan both return 2, but Gaschnig’s heuristic returns 3.

To compute Gaschnig’s heuristic, repeat the following until the goal state is reached:

let B be the current location of the blank; if B is occupied by tile X (not the blank) in the

goal state, move X to B; otherwise, move any misplaced tile to B. Students could be asked to

prove that this is the optimal solution to the relaxed problem.

4.11

a. Local beam search with is hill-climbing search.

b. Local beam search with : strictly speaking, this doesn’t make sense. (Exercise

may be modiﬁed in future printings.) The idea is that if every successor is retained

(because is unbounded), then the search resembles breadth-ﬁrst search in that it adds

one complete layer of nodes before adding the next layer. Starting from one state, the

algorithm would be essentially identical to breadth-ﬁrst search except that each layer is

generated all at once.

c. Simulated annealing with at all times: ignoring the fact that the termination step

would be triggered immediately, the search would be identical to ﬁrst-choice hill climb-

20 Chapter 4. Informed Search and Exploration

ing because every downward successor would be rejected with probability 1. (Exercise

may be modiﬁed in future printings.)

d. Genetic algorithm with population size : if the population size is 1, then the

two selected parents will be the same individual; crossover yields an exact copy of the

individual; then there is a small chance of mutation. Thus, the algorithm executes a

random walk in the space of individuals.

4.12 If we assume the comparison function is transitive, then we can always sort the nodes

using it, and choose the node that is at the top of the sort. Efﬁcient priority queue data

structures rely only on comparison operations, so we lose nothing in efﬁciency—except for

the fact that the comparison operation on states may be much more expensive than comparing

two numbers, each of which can be computed just once.

Arelies on the division of the total cost estimate into the cost-so-far and the cost-

to-go. If we have comparison operators for each of these, then we can prefer to expand a

node that is better than other nodes on both comparisons. Unfortunately, there will usually

be no such node. The tradeoff between and cannot be realized without numerical

values.

4.13 The space complexity of LRTA is dominated by the space required for ,

i.e., the product of the number of states visited ( ) and the number of actions tried per state

( ). The time complexity is at least for a naive implementation because for each

action taken we compute an value, which requires minimizing over actions. A simple

optimization can reduce this to . This expression assumes that each state–action pair

is tried at most once, whereas in fact such pairs may be tried many times, as the example in

Figure 4.22 shows.

4.14 This question is slightly ambiguous as to what the percept is—either the percept is just

the location, or it gives exactly the set of unblocked directions (i.e., blocked directions are

illegal actions). We will assume the latter. (Exercise may be modiﬁed in future printings.)

There are 12 possible locations for internal walls, so there are possible environ-

ment conﬁgurations. A belief state designates a subset of these as possible conﬁgurations;

for example, before seeing any percepts all 4096 conﬁgurations are possible—this is a single

belief state.

a. We can view this as a contingency problem in belief state space. The initial belief state

is the set of all 4096 conﬁgurations. The total belief state space contains belief

states (one for each possible subset of conﬁgurations, but most of these are not reach-

able. After each action and percept, the agent learns whether or not an internal wall

exists between the current square and each neighboring square. Hence, each reachable

belief state can be represnted exactly by a list of status values (present, absent, un-

known) for each wall separately. That is, the belief state is completely decomposable

and there are exactly reachable belief states. The maximum number of possible

wall-percepts in each state is 16 ( ), so each belief state has four actions, each with up

to 16 nondeterministic successors.

b. Assuming the external walls are known, there are two internal walls and hence

possible percepts.

c. The initial null action leads to four possible belief states, as shown in Figure S4.2. From

each belief state, the agent chooses a single action which can lead to up to 8 belief states

(on entering the middle square). Given the possibility of having to retrace its steps at

a dead end, the agent can explore the entire maze in no more than 18 steps, so the

complete plan (expressed as a tree) has no more than nodes. On the other hand,

there are just , so the plan could be expressed more concisely as a table of actions

indexed by belief state (a policy in the terminology of Chapter 17).

? ??

NoOp

Right

Figure S4.2 The maze exploration problem: the initial state, ﬁrst percept, and one

selected action with its perceptual outcomes.

4.15 Here is one simple hill-climbing algorithm:

Connect all the cities into an arbitrary path.

Pick two points along the path at random.

Split the path at those points, producing three pieces.

Try all six possible ways to connect the three pieces.

Keep the best one, and reconnect the path accordingly.

Iterate the steps above until no improvement is observed for a while.

4.1 4.16 Code not shown.

22 Chapter 4. Informed Search and Exploration

4.17 Hillclimbing is surprisingly effective at ﬁnding reasonable if not optimal paths for very

little computational cost, and seldom fails in two dimensions.

a. It is possible (see Figure S4.3(a)) but very unlikely—the obstacle has to have an unusual

shape and be positioned correctly with respect to the goal.

b. With convex obstacles, getting stuck is much more likely to be a problem (see Fig-

ure S4.3(b)).

c. Notice that this is just depth-limited search, where you choose a step along the best path

even if it is not a solution.

d. Set to the maximum number of sides of any polygon and you can always escape.

Current

position Goal

(a) (b)

Current

position

Goal

Figure S4.3 (a) Getting stuck with a convex obstacle. (b) Getting stuck with a nonconvex

obstacle.

4.18 The student should ﬁnd that on the 8-puzzle, RBFS expands more nodes (because

it does not detect repeated states) but has lower cost per node because it does not need to

maintain a queue. The number of RBFS node re-expansions is not too high because the

presence of many tied values means that the best path changes seldom. When the heuristic is

slightly perturbed, this advantage disappears and RBFS’s performance is much worse.

For TSP, the state space is a tree, so repeated states are not an issue. On the other hand,

the heuristic is real-valued and there are essentially no tied values, so RBFS incurs a heavy

penalty for frequent re-expansions.

Solutions for Chapter 5

Constraint Satisfaction Problems

5.1 Aconstraint satisfaction problem is a problem in which the goal is to choose a value

for each of a set of variables, in such a way that the values all obey a set of constraints.

Aconstraint is a restriction on the possible values of two or more variables. For exam-

ple, a constraint might say that is not allowed in conjunction with .

Backtracking search is a form depth-ﬁrst search in which there is a single representa-

tion of the state that gets updated for each successor, and then must be restored when a dead

end is reached.

A directed arc from variable to variable in a CSP is arc consistent if, for every

value in the current domain of , there is some consistent value of .

Backjumping is a way of making backtracking search more efﬁcient, by jumping back

more than one level when a dead end iss reached.

Min-conﬂicts is a heuristic for use with local search on CSP problems. The heuristic

says that, when given a variable to modify, choose the value that conﬂicts with the fewest

number of other variables.

5.2 There are 18 solutions for coloring Australia with three colors. Start with SA, which

can have any of three colors. Then moving clockwise, WA can have either of the other two

colors, and everything else is strictly determined; that makes 6 possibilities for the mainland,

times 3 for Tasmania yields 18.

5.3 The most constrained variable makes sense because it chooses a variable that is (all

other things being equal) likely to cause a failure, and it is more efﬁcient to fail as early

as possible (thereby pruning large parts of the search space). The least constraining value

heuristic makes sense because it allows the most chances for future assignments to avoid

conﬂict.

5.4 a. Crossword puzzle construction can be solved many ways. One simple choice is

depth-ﬁrst search. Each successor ﬁlls in a word in the puzzle with one of the words in the

dictionary. It is better to go one word at a time, to minimize the number of steps.

b. As a CSP, there are even more choices. You could have a variable for each box in

the crossword puzzle; in this case the value of each variable is a letter, and the constraints are

that the letters must make words. This approach is feasible with a most-constraining value

heuristic. Alternately, we could have each string of consecutive horizontal or vertical boxes

be a single variable, and the domain of the variables be words in the dictionary of the right

24 Chapter 5. Constraint Satisfaction Problems

length. The constraints would say that two intersecting words must have the same letter in the

intersecting box. Solving a problem in this formulation requires fewer steps, but the domains

are larger (assuming a big dictionary) and there are fewer constraints. Both formulations are

feasible.

5.5 a. For rectilinear ﬂoor-planning, one possibility is to have a variable for each of the

small rectangles, with the value of each variable being a 4-tuple consisting of the and

coordinates of the upper left and lower right corners of the place where the rectangle will

be located. The domain of each variable is the set of 4-tuples that are the right size for the

corresponding small rectangle and that ﬁt within the large rectangle. Constraints say that no

two rectangles can overlap; for example if the value of variable is , then no other

variable can take on a value that overlaps with the to rectangle.

b. For class scheduling, one possibility is to have three variables for each class, one with

times for values (e.g. MWF8:00, TuTh8:00, MWF9:00, ...), one with classrooms for values

(e.g. Wheeler110, Evans330, ...) and one with instructors for values (e.g. Abelson, Bibel,

Canny, ...). Constraints say that only one class can be in the same classroom at the same time,

and an instructor can only teach one class at a time. There may be other constraints as well

(e.g. an instructor should not have two consecutive classes).

5.6 The exact steps depend on certain choices you are free to make; here are the ones I

made:

a. Choose the variable. Its domain is .

b. Choose the value 1 for . (We can’t choose 0; it wouldn’t survive forward checking,

because it would force to be 0, and the leading digit of the sum must be non-zero.)

c. Choose , because it has only one remaining value.

d. Choose the value 1 for .

e. Now and Xare tied for minimum remaining values at 2; let’s choose .

f. Either value survives forward checking, let’s choose 0 for .

g. Now has the minimum remaining values.

h. Again, arbitrarily choose 0 for the value of .

i. The variable must be an even number (because it is the sum of less than 5

(because ). That makes it most constrained.

j. Arbitrarily choose 4 as the value of .

k. now has only 1 remaining value.

l. Choose the value 8 for .

m. now has only 1 remianing value.

n. Choose the value 7 for .

o. must be an even number less than 9; choose .

p. The only value for that survives forward checking is 6.

q. The only variable left is .

r. The only value left for is 3.

s. This is a solution.

This is a rather easy (under-constrained) puzzle, so it is not surprising that we arrive at a

solution with no backtracking (given that we are allowed to use forward checking).

5.7 There are implementations of CSP algorithms in the Java, Lisp, and Python sections

of the online code repository; these should help students get started. However, students will

have to add code to keep statistics on the experiments, and perhaps will want to have some

mechanism for making an experiment return failure if it exceeds a certain time limit (or

number-of-steps limit). The amount of code that needs to be written is small; the exercise is

more about running and analyzing an experiment.

5.8 We’ll trace through each iteration of the while loop in AC-3 (for one possible ordering

of the arcs):

a. Remove , delete from .

b. Remove , delete from , leaving only .

c. Remove , delete from .

d. Remove , delete from , leaving only .

e. Remove , delete from .

f. Remove , delete from , leaving only .

g. Remove , delete from .

h. Remove , delete from .

i. remove , delete from , leaving no domain for .

5.9 On a tree-structured graph, no arc will be considered more than once, so the AC-3

algorithm is , where is the number of edges and is the size of the largest do-

main.

5.10 The basic idea is to preprocess the constraints so that, for each value of , we keep

track of those variables for which an arc from to is satisﬁed by that particular value

of . This data structure can be computed in time proportional to the size of the problem

representation. Then, when a value of is deleted, we reduce by 1 the count of allowable

values for each arc recorded under that value. This is very similar to the forward

chaining algorithm in Chapter 7. See ? (?) for detailed proofs.

5.11 The problem statement sets out the solution fairly completely. To express the ternary

constraint on , and that , we ﬁrst introduce a new variable, . If the

domain of and is the set of numbers , then the domain of is the set of pairs of

numbers from , i.e. . Now there are three binary constraints, one between and

saying that the value of must be equal to the ﬁrst element of the pair-value of ; one

between and saying that the value of must equal the second element of the value

of ; and ﬁnally one that says that the sum of the pair of numbers that is the value of

must equal the value of . All other ternary constraints can be handled similarly.

Now that we can reduce a ternary constraint into binary constraints, we can reduce a

4-ary constraint on variables by ﬁrst reducing to binary constraints as

26 Chapter 5. Constraint Satisfaction Problems

shown above, then adding back in a ternary constraint with and , and then reducing

this ternary constraint to binary by introducing .

By induction, we can reduce any -ary constraint to an -ary constraint. We can

stop at binary, because any unary constraint can be dropped, simply by moving the effects of

the constraint into the domain of the variable.

5.12 A simple algorithm for ﬁnding a cutset of no more than nodes is to enumerate all

subsets of nodes of size , and for each subset check whether the remaining nodes

form a tree. This algorithm takes time , which is .

Becker and Geiger (1994; http://citeseer.nj.nec.com/becker94approximation.html) give

an algorithm called MGA (modiﬁed greedy algorithm) that ﬁnds a cutset that is no more than

twice the size of the minimal cutset, using time , where is the number of

edges and is the number of variables.

Whether this makes the cycle cutset approach practical depends more on the graph

involved than on the agorithm for ﬁnding a cutset. That is because, for a cutset of size , we

still have an exponential factor before we can solve the CSP. So any graph with a large

cutset will be intractible to solve, even if we could ﬁnd the cutset with no effort at all.

5.13 The “Zebra Puzzle” can be represented as a CSP by introducing a variable for each

color, pet, drink, country and cigaret brand (a total of 25 variables). The value of each variable

is a number from 1 to 5 indicating the house number. This is a good representation because it

easy to represent all the constraints given in the problem deﬁnition this way. (We have done

so in the Python implementation of the code, and at some point we may reimplement this

in the other languages.) Besides ease of expressing a problem, the other reason to choose a

representation is the efﬁciency of ﬁnding a solution. here we have mixed results—on some

runs, min-conﬂicts local search ﬁnds a solution for this problem in seconds, while on other

runs it fails to ﬁnd a solution after minutes.

Another representation is to have ﬁve variables for each house, one with the domain of

colrs, one with pets, and so on.

Solutions for Chapter 6

Adversarial Search

6.1 Figure S6.1 shows the game tree, with the evaluation function values below the terminal

nodes and the backed-up values to the right of the non-terminal nodes. The values imply that

the best starting move for X is to take the center. The terminal nodes with a bold outline are

the ones that do not need to be evaluated, assuming the optimal ordering.

x o x

−1 1 −2

1 −1 0 0 1 −1 −2 0 −1 0

1 2

Figure S6.1 Part of the game tree for tic-tac-toe, for Exercise 6.1.

6.2 Consider a MIN node whose children are terminal nodes. If MIN plays suboptimally,

then the value of the node is greater than or equal to the value it would have if MIN played

optimally. Hence, the value of the MAX node that is the MIN node’s parent can only be

increased. This argument can be extended by a simple induction all the way to the root. If

the suboptimal play by MIN is predictable, then one can do better than a minimax strategy.

For example, if MIN always falls for a certain kind of trap and loses, then setting the trap

guarantees a win even if there is actually a devastating response for MIN. This is shown in

Figure S6.2.

6.3

a. (5) The game tree, complete with annotations of all minimax values, is shown in Fig-

ure S6.3.

b. (5) The “?” values are handled by assuming that an agent with a choice between win-

ning the game and entering a “?” state will always choose the win. That is, min(–1,?)

is –1 and max(+1,?) is +1. If all successors are “?”, the backed-up value is “?”.

28 Chapter 6. Adversarial Search

MAX

MIN

B D

−101000 1000

−5−5−5

Figure S6.2 A simple game tree showing that setting a trap for MIN by playing is a win

if MIN falls for it, but may also be disastrous. The minimax move is of course , with value

(1,4)

(2,4)

(2,3)

(1,3)

(1,2)

(3,2)

(3,4)

(4,3)

(3,1)

(2,4)

(1,4)

−1

Figure S6.3 The game tree for the four-square game in Exercise 6.3. Terminal states are

in single boxes, loop states in double boxes. Each state is annotated with its minimax value

in a circle.

c. (5) Standard minimax is depth-ﬁrst and would go into an inﬁnite loop. It can be ﬁxed

by comparing the current state against the stack; and if the state is repeated, then return

a “?” value. Propagation of “?” values is handled as above. Although it works in this

case, it does not always work because it is not clear how to compare “?” with a drawn

position; nor is it clear how to handle the comparison when there are wins of different

degrees (as in backgammon). Finally, in games with chance nodes, it is unclear how to

compute the average of a number and a “?”. Note that it is not correct to treat repeated

states automatically as drawn positions; in this example, both (1,4) and (2,4) repeat in

the tree but they are won positions.

What is really happening is that each state has a well-deﬁned but initially unknown

value. These unknown values are related by the minimax equation at the bottom of page

163. If the game tree is acyclic, then the minimax algorithm solves these equations by

propagating from the leaves. If the game tree has cycles, then a dynamic programming

method must be used, as explained in Chapter 17. (Exercise 17.8 studies this problem in

particular.) These algorithms can determine whether each node has a well-determined

value (as in this example) or is really an inﬁnite loop in that both players prefer to stay

in the loop (or have no choice). In such a case, the rules of the game will need to deﬁne

the value (otherwise the game will never end). In chess, for eaxmple, a state that occurs

3 times (and hence is assumed to be desirable for both players) is a draw.

d. This question is a little tricky. One approach is a proof by induction on the size of the

game. Clearly, the base case is a loss for A and the base case is a win for

A. For any , the initial moves are the same: A and B both move one step towards

each other. Now, we can see that they are engaged in a subgame of size on the

squares , except that there is an extra choice of moves on squares and

. Ignoring this for a moment, it is clear that if the “ ” is won for A, then A

gets to the square before B gets to square (by the deﬁnition of winning) and

therefore gets to before B gets to , hence the “ ” game is won for A. By the same

line of reasoning, if “ ” is won for B then “ ” is won for B. Now, the presence of

the extra moves complicates the issue, but not too much. First, the player who is slated

to win the subgame never moves back to his home square. If the player

slated to lose the subgame does so, then it is easy to show that he is bound to lose the

game itself—the other player simply moves forward and a subgame of size is

played one step closer to the loser’s home square.

6.4 See "search/algorithms/games.lisp" for deﬁnitions of games, game-playing

agents, and game-playing environments. "search/algorithms/minimax.lisp" con-

tains the minimax and alpha-beta algorithms. Notice that the game-playing environment is

essentially a generic environment with the update function deﬁned by the rules of the game.

Turn-taking is achieved by having agents do nothing until it is their turn to move.

See "search/domains/cognac.lisp" for the basic deﬁnitions of a simple game

(slightly more challenging than Tic-Tac-Toe). The code for this contains only a trivial eval-

uation function. Students can use minimax and alpha-beta to solve small versions of the

game to termination (probably up to ); they should notice that alpha-beta is far faster

than minimax, but still cannot scale up without an evaluation function and truncated horizon.

Providing an evaluation function is an interesting exercise. From the point of view of data

structure design, it is also interesting to look at how to speed up the legal move generator by

precomputing the descriptions of rows, columns, and diagonals.

Very few students will have heard of kalah, so it is a fair assignment, but the game

is boring—depth 6 lookahead and a purely material-based evaluation function are enough

30 Chapter 6. Adversarial Search

to beat most humans. Othello is interesting and about the right level of difﬁculty for most

students. Chess and checkers are sometimes unfair because usually a small subset of the

class will be experts while the rest are beginners.

6.5 This question is not as hard as it looks. The derivation below leads directly to a deﬁni-

tion of and values. The notation refers to (the value of) the node at depth on the path

from the root to the leaf node . Nodes are the siblings of node .

a. We can write , giving

Then can be similarly replaced, until we have an expression containing itself.

b. In terms of the and values, we have

Again, can be expanded out down to . The most deeply nested term will be

c. If is a max node, then the lower bound on its value only increases as its successors

are evaluated. Clearly, if it exceeds it will have no further effect on . By extension,

if it exceeds it will have no effect. Thus, by keeping track of this

value we can decide when to prune . This is exactly what - does.

d. The corresponding bound for min nodes is .

6.7 The general strategy is to reduce a general game tree to a one-ply tree by induction on

the depth of the tree. The inductive step must be done for min, max, and chance nodes, and

simply involves showing that the transformation is carried though the node. Suppose that the

values of the descendants of a node are , and that the transformation is , where

is positive. We have

Hence the problem reduces to a one-ply tree where the leaves have the values from the original

tree multiplied by the linear transformation. Since if , the

best choice at the root will be the same as the best choice in the original tree.

6.8 This procedure will give incorrect results. Mathematically, the procedure amounts to

assuming that averaging commutes with min and max, which it does not. Intuitively, the

choices made by each player in the deterministic trees are based on full knowledge of future

dice rolls, and bear no necessary relationship to the moves made without such knowledge.

(Notice the connection to the discussion of card games on page 179 and to the general prob-

lem of fully and partially observable Markov decision problems in Chapter 17.) In practice,

the method works reasonably well, and it might be a good exercise to have students compare

it to the alternative of using expectiminimax with sampling (rather than summing over) dice

rolls.

6.9 Code not shown.

6.10 The basic physical state of these games is fairly easy to describe. One important thing

to remember for Scrabble and bridge is that the physical state is not accessible to all players

and so cannot be provided directly to each player by the environment simulator. Particularly

in bridge, each player needs to maintain some best guess (or multiple hypotheses) as to the

actual state of the world. We expect to be putting some of the game implementations online

as they become available.

6.11 One can think of chance events during a game, such as dice rolls, in the same way

as hidden but preordained information (such as the order of the cards in a deck). The key

distinctions are whether the players can inﬂuence what information is revealed and whether

there is any asymmetry in the information available to each player.

a. Expectiminimax is appropriate only for backgammon and Monopoly. In bridge and

Scrabble, each player knows the cards/tiles he or she possesses but not the opponents’.

In Scrabble, the beneﬁts of a fully rational, randomized strategy that includes reasoning

about the opponents’ state of knowledge are probably small, but in bridge the questions

of knowledge and information disclosure are central to good play.

b. None, for the reasons described earlier.

c. Key issues include reasoning about the opponent’s beliefs, the effect of various actions

on those beliefs, and methods for representing them. Since belief states for rational

agents are probability distributions over all possible states (including the belief states of

others), this is nontrivial.

function MAX-VALUE( ) returns

if TERMINAL-TEST( ) then return UTILITY( )

for in SUCCESSORS( ) do

if WINNER( ) = MAX

then MAX(v, MAX-VALUE( ))

else MAX(v, MIN-VALUE( ))

return

Figure S6.4 Part of the modiﬁed minimax algorithm for games in which the winner of the

previous trick plays ﬁrst on the next trick.

6.12 (In the ﬁrst printing, this exericse refers to WINNER( ); subsequent printings refer

to WINNER( ), denoting the winner of the trick just completed (if any), or null.) This question

is interpreted as applying only to the observable case.

a. The modiﬁcation to MAX-VALUE is shown in Figure S6.4. If MAX has just won a trick,

MAX gets to play again, otherwise play alternates. Thus, the successors of a MAX node

32 Chapter 6. Adversarial Search

MAX MIN

S 2

H 6 4

D 6

C 9,8 10,5

MAX MIN

S 2

H 4

D 6

C 9,8 10,5

MAX MIN

S 2

H 6 4

C 9,8 10,5

MAX MIN

S 2

H 6 4

D 6

C 9 10,5

MAX MIN

S 2

D 6

C 9,8 10,5

MAX MIN

S 2

D 6

C 9 10,5

MAX MIN

S 2

C 9,8 10,5

MAX MIN

C 9,8 10,5

MAX MIN

S 2

C 9,8 10

MAX MIN

S 2

C 9,8 5

MAX MIN

C 9 10,5

MAX MIN

S 2

C 9 10

MAX MIN

S 2

C 9 5

MAX MIN

C 9 5

MAX MIN

C 9 10

MAX MIN

C 9

MAX MIN

C 10

MAX MIN

S 2

C 9

MAX MIN

S 2

C 9

MAX MIN

C 9

MAX MIN

S 2

MAX MIN

S 2

D 6

C 9 10

MAX MIN

S 2

D 6

C 9 5

MAX MIN

S 2

C 9 10

MAX MIN

S 2

D 6

C 10

MAX MIN

C 9 10

MAX MIN

S 2

C 9

MAX MIN

C 10

MAX MIN

S 2

MAX MIN

D 6

MAX MIN

S 2

D 6

MAX MIN

D 6

C 9 5

MAX MIN

S 2

D 6

C 9

MAX MIN

C 9 5

MAX MIN

D 6

C 5

MAX MIN

C 9

MAX MIN

D 6

MAX MIN

S 2

D 6

MAX MIN

S 2

+2 +2 0 +4 +2 +4 0 0 −2 +2

0+4

+2 0 0 +2

Figure S6.5 Ex. 6.12: Part of the game tree for the card game shown on p.179.

can be a mixture of MAX and MIN nodes, depending on the various cards MAX can

play. A similar modiﬁcation is needed for MIN-VALUE.

b. The game tree is shown in Figure S6.5.

6.13 The naive approach would be to generate each such position, solve it, and store the

outcome. This would be enormously expensive—roughly on the order of 444 billion seconds,

or 10,000 years, assuming it takes a second on average to solve each position (which is

probably very optimistic). Of course, we can take advantage of already-solved positions

when solving new positions, provided those solved positions are descendants of the new

positions. To ensure that this always happens, we generate the ﬁnal positions ﬁrst, then their

predecessors, and so on. In this way, the exact values of all successors are known when each

state is generated. This method is called retrograde analysis.

6.14 The most obvious change is that the space of actions is now continuous. For example,

in pool, the cueing direction, angle of elevation, speed, and point of contact with the cue ball

are all continuous quantities.

The simplest solution is just to discretize the action space and then apply standard meth-

ods. This might work for tennis (modelled crudely as alternating shots with speed and direc-

tion), but for games such as pool and croquet it is likely to fail miserably because small

changes in direction have large effects on action outcome. Instead, one must analyze the

game to identify a discrete set of meaningful local goals, such as “potting the 4-ball” in pool

or “laying up for the next hoop” in croquet. Then, in the current context, a local optimization

routine can work out the best way to achieve each local goal, resulting in a discrete set of pos-

sible choices. Typically, these games are stochastic, so the backgammon model is appropriate

provided that we use sampled outcomes instead of summing over all outcomes.

Whereas pool and croquet are modelled correctly as turn-taking games, tennis is not.

While one player is moving to the ball, the other player is moving to anticipate the opponent’s

return. This makes tennis more like the simultaneous-action games studied in Chapter 17. In

particular, it may be reasonable to derive randomized strategies so that the opponent cannot

anticipate where the ball will go.

6.15 The minimax algorithm for non-zero-sum games works exactly as for multiplayer

games, described on p.165–6; that is, the evaluation function is a vector of values, one for

each player, and the backup step selects whichever vector has the highest value for the player

whose turn it is to move. The example at the end of Section 6.2 (p.167) shows that alpha-beta

pruning is not possible in general non-zero-sum games, because an unexamined leaf node

might be optimal for both players.

6.16 With 32 pieces, each needing 6 bits to specify its position on one of 64 squares, we

need 24 bytes (6 32-bit words) to store a position, so we can store roughly 20 million positions

in the table (ignoring pointers for hash table bucket lists). This is about one-ninth of the 180

million positions generated during a three-minute search.

Generating the hash key directly from an array-based representation of the position

might be quite expensive. Modern programs (see, e.g., Heinz, 2000) carry along the hash

key and modify it as each new position is generated. Suppose this takes on the order of 20

operations; then on a 2GHz machine where an evaluation takes 2000 operations we can do

roughly 100 lookups per evaluation. Using a rough ﬁgure of one millisecond for a disk seek,

we could do 1000 evaluations per lookup. Clearly, using a disk-resident table is of dubious

value, even if we can get some locality of reference to reduce the number of disk reads.

Solutions for Chapter 7

Agents that Reason Logically

7.1 The wumpus world is partially observable, deterministic, sequential (you need to re-

member the state of one location when you return to it on a later turn), static, discrete, and

single agent (the wumpus’s sole trick—devouring an errant explorer—is not enough to treat

it as an agent). Thus, it is a fairly simple environment. The main complication is the partial

observability.

7.2 To save space, we’ll show the list of models as a table rather than a collection of dia-

grams. There are eight possible combinations of pits in the three squares, and four possibili-

ties for the wumpus location (including nowhere).

We can see that because every line where is true also has true.

Similarly for .

7.3

a. There is a pl true in the Python code, and a version of ask in the Lisp code that

serves the same purpose. The Java code did not have this function as of May 2003, but

it should be added soon.)

b. The sentences , , and can all be determined to be true or false in

a partial model that does not specify the truth value for .

c. It is possible to create two sentences, each with variables that are not instantiated in

the partial model, such that one of them is true for all possible values of the variables,

while the other sentence is false for one of the values. This shows that in general one

must consider all possibilities. Enumerating them takes exponential time.

d. The python implementation of pl true returns true if any disjunct of a disjunction

is true, and false if any conjunct of a conjunction is false. It will do this even if other

disjuncts/conjuncts contains uninstantiated variables. Thus, in the partial model where

is true, returns true, and returns false. But the truth values of ,

, and are not detected.

e. Our version of tt entails already uses this modiﬁed pl true. It would be slower

if it did not.

7.4 Remember, iff in very model in which is true, is also true. Therefore,

a. A valid sentence is one that is true in all models. The sentence is also valid in all

models. So if is valid then the entailment holds (because both and hold

Model

, ,

,,,

, ,

,,,

, ,

,,,

Figure 7.1 A truth table constructed for Ex. 7.2. Propositions not listed as true on a given

line are assumed false, and only entries are shown in the table.

in every model), and if the entailment holds then must be valid, because it must be

true in all models, because it must be true in all models in which holds.

b. doesn’t hold in any model, so trivially holds in every model that holds

in.

c. holds in those models where holds or where holds. That is precisely the

case if is valid.

d. This follows from applying cin both directions.

36 Chapter 7. Agents that Reason Logically

e. This reduces to c, because is unsatisﬁable just when is valid.

7.5 These can be computed by counting the rows in a truth table that come out true. Re-

member to count the propositions that are not mentioned; if a sentence mentions only and

, then we multiply the number of models for by to account for and .

a. 6

b. 12

c. 4

7.6 A binary logical connective is deﬁned by a truth table with 4 rows. Each of the four

rows may be true or false, so there are possible truth tables, and thus 16 possible

connectives. Six of these are trivial ones that ignore one or both inputs; they correspond to

, , , , and . Four of them we have already studied: .

The remaining six are potentially useful. One of them is reverse implication ( instead of

), and the other ﬁve are the negations of and . (The ﬁrst two of these

are sometimes called nand and nor.)

7.7 We use the truth table code in Lisp in the directory logic/prop.lisp to show each

sentence is valid. We substitute P, Q, R for because of the lack of Greek letters in

ASCII. To save space in this manual, we only show the ﬁrst four truth tables:

> (truth-table "P ˆ Q <=> Q ˆ P")

-----------------------------------------

P Q P ˆ Q Q ˆ P (P ˆ Q) <=> (Q ˆ P)

-----------------------------------------

F F F F $true$

T F F F T

F T F F T

T T T T T

-----------------------------------------

NIL

> (truth-table "P | Q <=> Q | P")

-----------------------------------------

P Q P | Q Q | P (P | Q) <=> (Q | P)

-----------------------------------------

F F F F T

T F T T T

F T T T T

T T T T T

-----------------------------------------

NIL

> (truth-table "P ˆ (Q ˆ R) <=> (P ˆ Q) ˆ R")

-----------------------------------------------------------------------

P Q R Q ˆ R P ˆ (Q ˆ R) P ˆ Q ˆ R (P ˆ (Q ˆ R)) <=> (P ˆ Q ˆ R)

-----------------------------------------------------------------------

F F F F F F T

T F F F F F T

F T F F F F T

T T F F F F T

F F T F F F T

T F T F F F T

F T T T F F T

T T T T T T T

-----------------------------------------------------------------------

NIL

> (truth-table "P | (Q | R) <=> (P | Q) | R")

-----------------------------------------------------------------------

P Q R Q | R P | (Q | R) P | Q | R (P | (Q | R)) <=> (P | Q | R)

-----------------------------------------------------------------------

F F F F F F T

T F F F T T T

F T F T T T T

T T F T T T T

F F T T T T T

T F T T T T T

F T T T T T T

T T T T T T T

-----------------------------------------------------------------------

NIL

For the remaining sentences, we just show that they are valid according to the validity

function:

> (validity "˜˜P <=> P")

VALID

> (validity "P => Q <=> ˜Q => ˜P")

VALID

> (validity "P => Q <=> ˜P | Q")

VALID

> (validity "(P <=> Q) <=> (P => Q) ˆ (Q => P)")

VALID

> (validity "˜(P ˆ Q) <=> ˜P | ˜Q")

VALID

> (validity "˜(P | Q) <=> ˜P ˆ ˜Q")

VALID

> (validity "P ˆ (Q | R) <=> (P ˆ Q) | (P ˆ R)")

VALID

> (validity "P | (Q ˆ R) <=> (P | Q) ˆ (P | R)")

VALID

7.8 We use the validity function from logic/prop.lisp to determine the validity

of each sentence:

> (validity "Smoke => Smoke")

VALID

> (validity "Smoke => Fire")

SATISFIABLE

> (validity "(Smoke => Fire) => (˜Smoke => ˜Fire)")

SATISFIABLE

> (validity "Smoke | Fire | ˜Fire")

VALID

38 Chapter 7. Agents that Reason Logically

> (validity "((Smoke ˆ Heat) => Fire) <=> ((Smoke => Fire) | (Heat => Fire))")

VALID

> (validity "(Smoke => Fire) => ((Smoke ˆ Heat) => Fire)")

VALID

> (validity "Big | Dumb | (Big => Dumb)")

VALID

> (validity "(Big ˆ Dumb) | ˜Dumb")

SATISFIABLE

Many people are fooled by (e) and (g) because they think of implication as being cau-

sation, or something close to it. Thus, in (e), they feel that it is the combination of Smoke

and Heat that leads to Fire, and thus there is no reason why one or the other alone should

lead to ﬁre. Similarly, in (g), they feel that there is no necessary causal relation between Big

and Dumb, so the sentence should be satisﬁable, but not valid. However, this reasoning is

incorrect, because implication is not causation—implication is just a kind of disjunction (in

the sense that is the same as ). So is

equivalent to , which is equivalent to ,

which is true whether is true or false, and is therefore valid.

7.9 From the ﬁrst two statements, we see that if it is mythical, then it is immortal; otherwise

it is a mammal. So it must be either immortal or a mammal, and thus horned. That means it

is also magical. However, we can’t deduce anything about whether it is mythical. Using the

propositional reasoning code:

> (setf kb (make-prop-kb))

#S(PROP-KB SENTENCE (AND))

> (tell kb "Mythical => Immortal")

> (tell kb "˜Mythical => ˜Immortal ˆ Mammal")

> (tell kb "Immortal | Mammal => Horned")

> (tell kb "Horned => Magical")

> (ask kb "Mythical")

NIL

> (ask kb "˜Mythical")

NIL

> (ask kb "Magical")

> (ask kb "Horned")

7.10 Each possible world can be written as a conjunction of symbols, e.g. .

Asserting that a possible world is not the case can be written by negating that, e.g.

, which can be rewritten as . This is the form of a clause; a conjunction

of these clauses is a CNF sentence, and can list all the possible worlds for the sentence.

7.11

a. This is a disjunction with 28 disjuncts, each one saying that two of the neighbors are

true and the others are false. The ﬁrst disjunct is

The other 27 disjuncts each select two different to be true.

b. There will be disjuncts, each saying that of the symbols are true and the others

false.

c. For each of the cells that have been probed, take the resulting number revealed by the

game and construct a sentence with disjuncts. Conjoin all the sentences together.

Then use DPLL to answer the question of whether this sentence entails for the

particular pair you are interested in.

d. To encode the global constraint that there are mines altogether, we can construct

a disjunct with disjuncts, each of size . Remember, . So for

a Minesweeper game with 100 cells and 20 mines, this will be morre than , and

thus cannot be represented in any computer. However, we can represent the global

constraint within the DPLL algorithm itself. We add the parameter min and max to

the DPLL function; these indicate the minimum and maximum number of unassigned

symbols that must be true in the model. For an unconstrained problem the values 0 and

will be used for these parameters. For a mineseeper problem the value will be

used for both min and max. Within DPLL, we fail (return false) immediately if min is

less than the number of remaining symbols, or if max is less than 0. For each recursive

call to DPLL, we update min and max by subtracting one when we assign a true value

to a symbol.

e. No conclusions are invalidated by adding this capability to DPLL and encoding the

global constraint using it.

f. Consider this string of alternating 1’s and unprobed cells (indicated by a dash):

|-|1|-|1|-|1|-|1|-|1|-|1|-|1|-|

There are two possible models: either there are mines under every even-numbered

dash, or under every odd-numbered dash. Making a probe at either end will determine

whether cells at the far end are empty or contain mines.

7.12

a.is equivalent to by implication elimination (Figure 7.11), and

is equivalent to by de Morgan’s rule, so

is equivalent to .

b. A clause can have positive and negative literals; arrange them in the form

. Then, setting , we have

is equivalent to

40 Chapter 7. Agents that Reason Logically

c. For atoms where UNIFY :

SUBST

7.13

c. These formulae are the same as (7.7) and (7.8), except that the for pit is replaced by

for wumpus, and for breezy is replaced by for smelly.

7.14 Optimal behavior means achieving an expected utility that is as good as any other

agent program. The PL-WUMPUS-AGENT is clearly non-optimal when it chooses a random

move (and may be non-optimal in other braqnches of its logic). One example: in some cases

when there are many dangers (breezes and smells) but no safe move, the agent chooses at

random. A more thorough analysis should show when it is better to do that, and when it is

better to go home and exit the wumpus world, giving up on any chance of ﬁnding the gold.

Even when it is best to gamble on an unsafe location, our agent does not distinguish degrees

of safety – it should choose the unsafe square which contains a danger in the fewest number

of possible models. These reﬁnements are hard to state using a logical agent, but we will see

in subsequent chapters that a probabilistic agent can handle them. does

7.15 PL-WUMPUS-AGENT keeps track of 6 static state variables besides KB. The difﬁculty

is that these variables change—we don’t just add new information about them (as we do with

pits and breezy locations), we modify exisitng information. This does not sit well with logic,

which is designed for eternal truths. So there are two alternatives. The ﬁrst is to superscript

each proposition with the time (as we did with the circuit agents), and then we could, for

example, do TELL to say that the agent is at location at time 3. Then at time 4,

we would have to copy over many of the existing propositions, and add new ones. The second

possibility is to treat every proposition as a timeless one, but to remove outdated propositions

from the KB. That is, we could do RETRACT and then progTell to

indicate that the agent has moved from to . Chapter 10 describes the semantics and

implementation of RETRACT.

NOTE: Avoid assigning this problem if you don’t feel comfortable requiring students

to think ahead about the possibility of retraction.

7.16 It will take time proportional to the number of pure symbols plus the number of unit

clauses. We assume that is false, and prove a contradiction. is

equivalent to . From this sentence the algorithm will ﬁrst eliminate all the pure

symbols, then it will work on unit clauses until it chooses (which is a unit clause); at that

point it will immediately recognize that either choice (true or false) for leads to failure,

which means that the original non-negated assertion is true.

7.17 Code not shown.

Solutions for Chapter 8

First-Order Logic

8.1 This question will generate a wide variety of possible solutions. The key distinction

between analogical and sentential representations is that the analogical representation au-

tomatically generates consequences that can be “read off” whenever suitable premises are

encoded. When you get into the details, this distinction turns out to be quite hard to pin

down—for example, what does “read off” mean?—but it can be justiﬁed by examining the

time complexity of various inferences on the “virtual inference machine” provided by the

representation system.

a. Depending on the scale and type of the map, symbols in the map language typically

include city and town markers, road symbols (various types), lighthouses, historic mon-

uments, river courses, freeway intersections, etc.

b. Explicit and implicit sentences: this distinction is a little tricky, but the basic idea is that

when the map-drawer plunks a symbol down in a particular place, he says one explicit

thing (e.g. that Coit Tower is here), but the analogical structure of the map representa-

tion means that many implicit sentences can now be derived. Explicit sentences: there

is a monument called Coit Tower at this location; Lombard Street runs (approximately)

east-west; San Francisco Bay exists and has this shape. Implicit sentences: Van Ness

is longer than North Willard; Fisherman’s Wharf is north of the Mission District; the

shortest drivable route from Coit Tower to Twin Peaks is the following .

c. Sentences unrepresentable in the map language: Telegraph Hill is approximately coni-

cal and about 430 feet high (assuming the map has no topographical notation); in 1890

there was no bridge connecting San Francisco to Marin County (map does not repre-

sent changing information); Interstate 680 runs either east or west of Walnut Creek (no

disjunctive information).

d. Sentences that are easier to express in the map language: any sentence that can be

written easily in English is not going to be a good candidate for this question. Any

linguistic abstraction from the physical structure of San Francisco (e.g. San Francisco

is on the end of a peninsula at the mouth of a bay) can probably be expressed equally

easily in the predicate calculus, since that’s what it was designed for. Facts such as

the shape of the coastline, or the path taken by a road, are best expressed in the map

language. Even then, one can argue that the coastline drawn on the map actually consists

of lots of individual sentences, one for each dot of ink, especially if the map is drawn

using a digital plotter. In this case, the advantage of the map is really in the ease of

inference combined with suitability for human “visual computing” apparatus.

e. Examples of other analogical representations:

Analog audio tape recording. Advantages: simple circuits can record and repro-

duce sounds. Disadvantages: subject to errors, noise; hard to process in order to

separate sounds or remove noise etc.

Traditional clock face. Advantages: easier to read quickly, determination of how

much time is available requires no additional computation. Disadvantages: hard to

read precisely, cannot represent small units of time (ms) easily.

All kinds of graphs, bar charts, pie charts. Advantages: enormous data compres-

sion, easy trend analysis, communicate information in a way which we can in-

terpret easily. Disadvantages: imprecise, cannot represent disjunctive or negated

information.

8.2 The knowledge base does not entail . To show this, we must give a model

where and but is false. Consider any model with three domain elements,

where and refer to the ﬁrst two elements and the relation referred to by holds only for

those two elements.

8.3 The sentence is valid. A sentence is valid if it is true in every model. An

existentially quantiﬁed sentence is true in a model if it holds under any extended interpretation

in which its variables are assigned to domain elements. According to the standard semantics

of FOL as given in the chapter, every model contains at least one domain element, hence,

for any model, there is an extended interpretation in which and are assigned to the ﬁrst

domain element. In such an interpretation, is true.

8.4 stipulates that there is exactly one object. If there are two objects, then

there is an extended interpretation in which and are assigned to different objects, so the

sentence would be false. Some students may also notice that any unsatisﬁable sentence also

meets the criterion, since there are no worlds in which the sentence is true.

8.5 We will use the simplest counting method, ignoring redundant combinations. For the

constant symbols, there are assignments. Each predicate of arity is mapped onto a -ary

relation, i.e., a subset of the possible -element tuples; there are such mappings. Each

function symbol of arity is mapped onto a -ary function, which speciﬁes a value for each

of the possible -element tuples. Including the invisible element, there are choices

for each value, so there are functions. The total number of possible combinations

is therefore

Two things to note: ﬁrst, the number is ﬁnite; second, the maximum arity is the most

crucial complexity parameter.

8.6 In this exercise, it is best not to worry about details of tense and larger concerns with

consistent ontologies and so on. The main point is to make sure students understand con-

44 Chapter 8. First-Order Logic

nectives and quantiﬁers and the use of predicates, functions, constants, and equality. Let the

basic vocabulary be as follows:

: student takes course in semester ;

: student passes course in semester ;

: the score obtained by student in course in semester ;

: is greater than ;

and : speciﬁc French and Greek courses (one could also interpret these sentences as re-

ferring to any such course, in which case one could use a predicate meaning

that the subject of course is ﬁeld ;

: buys from (using a binary predicate with unspeciﬁed seller is OK but

less felicitous);

: sells to ;

: person shaves person

: person is born in country ;

: is a parent of ;

: is a citizen of country for reason ;

: is a resident of country ;

: person fools person at time ;

, , , , , , ,

, : predicates satisﬁed by members of the corresponding categories.

a. Some students took French in spring 2001.

b. Every student who takes French passes it.

c. Only one student took Greek in spring 2001.

d. The best score in Greek is always higher than the best score in French.

e. Every person who buys a policy is smart.

f. No person buys an expensive policy.

g. There is an agent who sells policies only to people who are not insured.

h. There is a barber who shaves all men in town who do not shave themselves.

i. A person born in the UK, each of whose parents is a UK citizen or a UK resident, is a

UK citizen by birth.

j. A person born outside the UK, one of whose parents is a UK citizen by birth, is a UK

citizen by descent.

k. Politicians can fool some of the people all of the time, and they can fool all of the people

some of the time, but they can’t fool all of the people all of the time.

8.7 The key idea is to see that the word “same” is referring to every pair of Germans. There

are several logically equivalent forms for this sentence. The simplest is the Horn clause:

8.8 . This axiom is no longer true in

certain states and countries.

8.9 This is a very educational exercise but also highly nontrivial. Once students have

learned about resolution, ask them to do the proof too. In most cases, they will discover

missing axioms. Our basic predicates are ( heard about event at time );

(event occurred at time ); ( is alive at time ).

8.10 It is not entirely clear which sentences need to be written, but this is one of them:

That is, a square is breezy if and only if there is a pit in a neighboring square. Generally

speaking, the size of the axiom set is independent of the size of the wumpus world being

described.

8.11 Make sure you write deﬁnitions with . If you use , you are only imposing con-

46 Chapter 8. First-Order Logic

straints, not writing a real deﬁnition.

A second cousin is a a child of one’s parent’s ﬁrst cousin, and in general an th cousin

is deﬁned as:

The facts in the family tree are simple: each arrow represents two instances of

(e.g., and ), each name represents a

sex proposition (e.g., or ), each double line indicates a

proposition (e.g. ). Making the queries of the logical

reasoning system is just a way of debugging the deﬁnitions.

8.12 . This does follow from the Peano axioms (although we

should write the ﬁrst axiom for + as ). Roughly

speaking, the deﬁnition of + says that , where is shorthand for

the function applied times. Similarly, . Hence, the axioms

imply that and are equal to syntactically identical expressions. This argument

can be turned into a formal proof by induction.

8.13 Although these axioms are sufﬁcient to prove set membership when is in fact a

member of a given set, they have nothing to say about cases where is not a member. For

example, it is not possible to prove that is not a member of the empty set. These axioms may

therefore be suitable for a logical system, such as Prolog, that uses negation-as-failure.

8.14 Here we translate to mean “proper list” in Lisp terminology, i.e., a cons structure

with as the “rightmost” atom.

8.15 There are several problems with the proposed deﬁnition. It allows one to prove, say,

but not ; so we need an additional symmetry

axiom. It does not allow one to prove that is false, so it needs to be

written as

Finally, it does not work as the boundaries of the world, so some extra conditions must be

added.

8.16 We need the following sentences:

8.17 There are three stages to go through. In the ﬁrst stage, we deﬁne the concepts of one-

bit and -bit addition. Then, we specify one-bit and -bit adder circuits. Finally, we verify

that the -bit adder circuit does -bit addition.

One-bit addition is easy. Let be a function of three one-bit arguments (the third

is the carry bit). The result of the addition is a list of bits representing a 2-bit binary

number, least signiﬁcant digit ﬁrst:

-bit addition builds on one-bit addition. Let be a function that takes

two lists of binary digits of length (least signiﬁcant digit ﬁrst) and a carry bit (initially

0), and constructs a list of length that represents their sum. (It will always be

exactly bits long, even when the leading bit is 0—the leading bit is the overﬂow

bit.)

The next step is to deﬁne the structure of a one-bit adder circuit, as given in Section ??.

Let be true of any circuit that has the appropriate components and

48 Chapter 8. First-Order Logic

connections:

Notice that this allows the circuit to have additional gates and connections, but they

won’t stop it from doing addition.

Now we deﬁne what we mean by an -bit adder circuit, following the design of Figure

8.6. We will need to be careful, because an -bit adder is not just an -bit adder

plus a one-bit adder; we have to connect the overﬂow bit of the -bit adder to the

carry-bit input of the one-bit adder. We begin with the base case, where :

Now, for the recursive case we specify that the ﬁrst connect the “overﬂow” output of

the -bit circuit as the carry bit for the last bit:

Now, to verify that a one-bit adder circuit actually adds correctly, we ask whether, given

any setting of the inputs, the outputs equal the sum of the inputs:

If this sentence is entailed by the KB, then every circuit with the design

is in fact an adder. The query for the -bit can be written as

where and are deﬁned appropriately to map bit

sequences to the actual terminals of the circuit. [Note: this logical formulation has not

been tested in a theorem prover and we hesitate to vouch for its correctness.]

8.18 Strictly speaking, the primitive gates must be deﬁned using logical equivalences to

exclude those combinations not listed as correct. If we are using a logic programming system,

we can simply list the cases. For example,

For the one-bit adder, we have

The veriﬁcation query is

It is not possible to ask whether particular terminals are connected in a given circuit, since

the terminals is not reiﬁed (nor is the circuit itself).

8.19 The answers here will vary by country. The two key rules for UK passports are given

in the answer to Exercsie 8.6.

Solutions for Chapter 9

Inference in First-Order Logic

9.1 We want to show that any sentence of the form entails any universal instantiation

of the sentence. The sentence is true if is true in all possible extended interpretations.

But replacing with any ground term must count as one of the interpretations, so if the

original sentence is true, then the instantiated sentence must also be true.

9.2 For any sentence containing a ground term and for any variable not occuring in

, we have

SUBST

where SUBST is a function that substitutes for a single occurrence of with .

9.3 Both b and c are valid; a is invalid because it introduces the previously-used symbol

Everest. Note that c does not imply that there are two mountains as high as Everest, be-

cause nowhere is it stated that BenNevis is different from Kilimanjaro (or Everest, for that

matter).

9.4 This is an easy exercise to check that the student understands uniﬁcation.

a.(or some permutation of this).

b. No uniﬁer ( cannot bind to both and ).

c. .

d. No uniﬁer (because the occurs-check prevents uniﬁcation of with ).

9.5 Employs(Mother(John), Father(Richard)) This page isn’t wide enough to draw the dia-

gram as in Figure 9.2, so we will draw it with indentation denoting children nodes:

[1] Employs(x, y)

[2] Employs(x, Father(z))

[3] Employs(x, Father(Richard))

[4] Employs(Mother(w), Father(Richard))

[5] Employs(Mother(John), Father(Richard))

[6] Employs(Mother(w), Father(z))

[4] ...

[7] Employs(Mother(John), Father(z))

[5] ...

[8] Employs(Mother(w), y)

[9] Employs(Mother(John), y)

[10] Employs(Mother(John), Father(z)

[5] ...

[6] ...

9.6 We will give the average-case time complexity for each query/scheme combination in

the following table. (An entry of the form “ ” means that it is to ﬁnd the ﬁrst solution

to the query, but to ﬁnd them all.) We make the following assumptions: hash tables give

access; there are people in the data base; there are people of any speciﬁed age;

every person has one mother; there are people in Houston and people in Tiny Town;

is much less than ; in Q4, the second conjunct is evaluated ﬁrst.

Q1 Q2 Q3 Q4

Anything that is can be considered “efﬁcient,” as perhaps can anything . Note

that S1 and S5 dominate the other schemes for this set of queries. Also note that indexing on

predicates plays no role in this table (except in combination with an argument), because there

are only 3 predicates (which is ). It would make a difference in terms of the constant

factor.

9.7 This would work if there were no recursive rules in the knowledge base. But suppose

the knowledge base contains the sentences:

Now take the query , with a backward chaining system. We unify the

query with the consequent of the implication to get the substitution .

We then substitute this in to the left-hand side to get and try to back

chain on that with the substitution . When we then try to apply the implication again, we

get a failure because cannot be both and . In other words, the failure to standardize

apart causes failure in some cases where recursive rules would result in a solution if we did

standardize apart.

9.8 Consider a 3-SAT problem of the form

We want to rewrite this as a single deﬁne clause of the form

along with a few ground clauses. We can do that with the deﬁnite clause

52 Chapter 9. Inference in First-Order Logic

along with the ground clauses

9.9 We use a very simple ontology to make the examples easier:

(Note we couldn’t do because that is not in the form

expected by Generalized Modus Ponens.)

f.(here is a Skolem function).

9.10 This questions deals with the subject of looping in backward-chaining proofs. A loop

is bound to occur whenever a subgoal arises that is a substitution instance of one of the goals

on the stack. Not all loops can be caught this way, of course, otherwise we would have a way

to solve the halting problem.

a. The proof tree is shown in Figure S9.1. The branch with and

repeats indeﬁnitely, so the rest of the proof is never reached.

b. We get an inﬁnite loop because of rule b, .

The speciﬁc loop appearing in the ﬁgure arises because of the ordering of the clauses—

it would be better to order before the rule from b. However, a loop

will occur no matter which way the rules are ordered if the theorem-prover is asked for

all solutions.

c. One should be able to prove that both Bluebeard and Charlie are horses.

d. Smith et al. (1986) recommend the following method. Whenever a “looping” goal

occurs (one that is a substitution instance of a supergoal higher up the stack), sus-

pend the attempt to prove that subgoal. Continue with all other branches of the proof

for the supergoal, gathering up the solutions. Then use those solutions (suitably in-

stantiated if necessary) as solutions for the suspended subgoal, continuing that branch

of the proof to ﬁnd additional solutions if any. In the proof shown in the ﬁgure, the

is a repeated goal and would be suspended. Since no other

way to prove it exists, that branch will terminate with failure. In this case, Smith’s

method is sufﬁcient to allow the theorem-prover to ﬁnd both solutions.

9.11 Surprisingly, the hard part to represent is “who is that man.” We want to ask “what

relationship does that man have to some known person,” but if we represent relations with

Parent(y,Bluebeard)

Horse(h)

Offspring(h,y)

Parent(y,h)

Horse(Bluebeard)

Yes, {y/Bluebeard,

h/Charlie}

Offspring(Bluebeard,y)

Figure S9.1 Partial proof tree for ﬁnding horses.

predicates (e.g., ) then we cannot make the relationship be a variable in ﬁrst-

order logic. So instead we need to reify relationships. We will use to say that the

family relationship holds between people and . Let denote me and denote

“that man.” We will also need the Skolem constants for the father of and for

the father of . The facts of the case (put into implicative normal form) are:

We want to be able to show that is the only son of my father, and therefore that is

father of , who is male, and therefore that “that man” is my son. The relevant deﬁnitions

from the family domain are:

and the query we want is:

We want to be able to get back the answer . Translating 1-9 and into INF

54 Chapter 9. Inference in First-Order Logic

(and negating and including the deﬁnition of ) we get:

Note that (1) is non-Horn, so we will need resolution to be be sure of getting a solution. It

turns out we also need demodulation (page 284) to deal with equality. The following lists the

steps of the proof, with the resolvents of each step in parentheses:

9.12 Here is a goal tree:

goals = [Criminal(West)]

goals = [American(West), Weapon(y), Sells(West, y, z), Hostile(z)]

goals = [Weapon(y), Sells(West, y, z), Hostile(z)]

goals = [Missle(y), Sells(West, y, z), Hostile(z)]

goals = [Sells(West, M1, z), Hostile(z)]

goals = [Missle(M1), Owns(Nono, M1), Hostile(Nono)]

goals = [Owns(Nono, M1), Hostile(Nono)]

goals = [Hostile(Nono)]

goals = []

9.13

a. In the following, an indented line is a step deeper in the proof tree, while two lines at

the same indentation represent two alternative ways to prove the goal that is unindented

above it. The P1 and P2 annotation on a line mean that the ﬁrst or second clause of P

was used to derive the line.

P(A, [1,2,3]) goal

P(1, [1|2,3]) P1 => solution, with A = 1

P(1, [1|2,3]) P2

P(2, [2,3]) P1 => solution, with A = 2

P(2, [2,3]) P2

P(3, [3]) P1 => solution, with A = 3

P(3, [3]) P2

P(2, [1, A, 3]) goal

P(2, [1|2, 3]) P1

P(2, [1|2, 3]) P2

P(2, [2|3]) P1 => solution, with A = 2

P(2, [2|3]) P2

P(2, [3]) P1

P(2, [3]) P2

b.Pcould better be called Member; it succeeds when the ﬁrst argument is an element of

the list that is the second argument.

9.14 The different versions of sort illustrate the distinction between logical and procedu-

ral semantics in Prolog.

a.sorted([]).

sorted([X]).

sorted([X,Y|L]) :- X<Y, sorted([Y|L]).

b.perm([],[]).

perm([X|L],M) :-

delete(X,M,M1),

perm(L,M1).

delete(X,[X|L],L). %% deleting an X from [X|L] yields L

delete(X,[Y|L],[Y|M]) :- delete(X,L,M).

member(X,[X|L]).

member(X,[_|L]) :- member(X,L).

c.sort(L,M) :- perm(L,M), sorted(M).

This is about as close to an executable formal speciﬁcation of sorting as you can

get—it says the absolute minimum about what sort means: in order for Mto be a sort of

L, it must have the same elements as L, and they must be in order.

d. Unfortunately, this doesn’t fare as well as a program as it does as a speciﬁcation. It

is a generate-and-test sort: perm generates candidate permutations one at a time, and

sorted tests them. In the worst case (when there is only one sorted permutation, and

it is the last one generated), this will take generations. Since each perm is

and each sorted is , the whole sort is in the worst case.

e. Here’s a simple insertion sort, which is :

isort([],[]).

isort([X|L],M) :- isort(L,M1), insert(X,M1,M).

56 Chapter 9. Inference in First-Order Logic

insert(X,[],[X]).

insert(X,[Y|L],[X,Y|L]) :- X=<Y.

insert(X,[Y|L],[Y|M]) :- Y<X, insert(X,L,M).

9.15 This exercise illustrates the power of pattern-matching, which is built into Prolog.

a. The code for simpliﬁcation looks straightforward, but students may have trouble ﬁnding

the middle way between undersimplifying and looping inﬁnitely.

simplify(X,X) :- primitive(X).

simplify(X,Y) :- evaluable(X), Y is X.

simplify(Op(X)) :- simplify(X,X1), simplify_exp(Op(X1)).

simplify(Op(X,Y)) :- simplify(X,X1), simplify(Y,Y1), simplify_exp(Op(X1,Y1)).

simplify_exp(X,Y) :- rewrite(X,X1), simplify(X1,Y).

simplify_exp(X,X).

primitive(X) :- atom(X).

b. Here are a few representative rewrite rules drawn from the extensive list in Norvig (1992).

Rewrite(X+0,X).

Rewrite(0+X,X).

Rewrite(X+X,2*X).

Rewrite(X*X,Xˆ2).

Rewrite(Xˆ0,1).

Rewrite(0ˆX,0).

Rewrite(X*N,N*X) :- number(N).

Rewrite(ln(eˆX),X).

Rewrite(XˆY*XˆZ,Xˆ(Y+Z)).

Rewrite(sin(X)ˆ2+cos(X)ˆ2,1).

c. Here are the rules for differentiation, using d(Y,X) to represent the derivative of ex-

pression Ywith respect to variable X.

Rewrite(d(X,X),1).

Rewrite(d(U,X),0) :- atom(U), U /= X.

Rewrite(d(U+V,X),d(U,X)+d(V,X)).

Rewrite(d(U-V,X),d(U,X)-d(V,X)).

Rewrite(d(U*V,X),V*d(U,X)+U*d(V,X)).

Rewrite(d(U/V,X),(V*d(U,X)-U*d(V,X))/(Vˆ2)).

Rewrite(d(UˆN,X),N*Uˆ(N-1)*d(U,X)) :- number(N).

Rewrite(d(log(U),X),d(U,X)/U).

Rewrite(d(sin(U),X),cos(U)*d(U,X)).

Rewrite(d(cos(U),X),-sin(U)*d(U,X)).

Rewrite(d(eˆU,X),d(U,X)*eˆU).

9.16 Once you understand how Prolog works, the answer is easy:

solve(X,[X]) :- goal(X).

solve(X,[X|P]) :- successor(X,Y), solve(Y,P).

We could render this in English as “Given a start state, if it is a goal state, then the path

consisting of just the start state is a solution. Otherwise, ﬁnd some successor state such that

there is a path from the successor to the goal; then a solution is the start state followed by that

path.”

Notice that solve can not only be used to ﬁnd a path Pthat is a solution, it can also be

used to verify that a given path is a solution.

If you want to add heuristics (or even breadth-ﬁrst search), you need an explicit queue.

The algorithms become quite similar to the versions written in Lisp or Python or Java or in

pseudo-code in the book.

9.17 This question tests both the student’s understanding of resolution and their ability to

think at a high level about relations among sets of sentences. Recall that resolution allows one

to show that by proving that is inconsistent. Suppose that in general the

resolution system is called using ASK . Now we want to show that a given sentence,

say is valid or unsatisﬁable.

A sentence is valid if it can be shown to be true without additional information. We

check this by calling ASK where is the empty knowledge base.

A sentence that is unsatisﬁable is inconsistent by itself. So if we the empty knowl-

edge base again and call ASK the resolution system will attempt to derive a con-

tradiction starting from . If it can do so, then it must be that , and hence , is

inconsistent.

9.18 This is a form of inference used to show that Aristotle’s syllogisms could not capture

all sound inferences.

(Here . comes from the ﬁrst sentence in a. while the others come from the second.

and are Skolem constants.)

c. Resolve and to yield . Resolve this with to give .

Resolve this with to obtain a contradiction.

9.19 This exercise tests the students understanding of models and implication.

a. (A) translates to “For every natural number there is some other natural number that is

smaller than or equal to it.” (B) translates to “There is a particular natural number that

is smaller than or equal to any natural number.”

b. Yes, (A) is true under this interpretation. You can always pick the number itself for the

“some other” number.

c. Yes, (B) is true under this interpretation. You can pick 0 for the “particular natural

number.”

d. No, (A) does not logically entail (B).

e. Yes, (B) logically entails (A).

f. We want to try to prove via resolution that (A) entails (B). To do this, we set our knowl-

edge base to consist of (A) and the negation of (B), which we will call (-B), and try to

58 Chapter 9. Inference in First-Order Logic

derive a contradiction. First we have to convert (A) and (-B) to canonical form. For (-B),

this involves moving the in past the two quantiﬁers. For both sentences, it involves

introducing a Skolem function:

(A)

(-B)

Now we can try to resolve these two together, but the occurs check rules out the uniﬁca-

tion. It looks like the substitution should be , but that is equivalent

to , which fails because is bound to an expression contain-

ing . So the resolution fails, there are no other resolution steps to try, and therefore (B)

does not follow from (A).

g. To prove that (B) entails (A), we start with a knowledge base containing (B) and the

negation of (A), which we will call (-A):

(-A)

(B)

This time the resolution goes through, with the substitution , thereby

yielding , and proving that (B) entails (A).

9.20 One way of seeing this is that resolution allows reasoning by cases, by which we can

prove by proving that either or is true, without knowing which one. With deﬁnite

clauses, we always have a single chain of inference, for which we can follow the chain and

instantiate variables.

9.21 No. Part of the deﬁnition of algorithm is that it must terminate. Since there can be

an inﬁnite number of consequences of a set of sentences, no algorithm can generate them

all. Another way to see thatthe answer is no is to remember that entailment for FOL is

semidecidable. If there were an algorithm that generates the set of consequences of a set of

sentences , then when given the task of deciding if is entailed by , one could just check

if is in the generated set. But we know that this is not possible, therefore generating the set

of sentences is impossible.

Solutions for Chapter 10

Knowledge Representation

10.1 Shooting the wumpus makes it dead, but there are no actions that cause it to come

alive. Hence the successor-state axiom for just has the second clause:

where is deﬁned appropriately in terms of the locations of and and the

orientation of . Possession of the arrow is lost by shooting, and again there is no way to

make it true:

10.2

Notice that the recursion needs no base case because we already have the axiom

10.3 This question takes the student through the initial stages of developing a logical rep-

resentation for actions that incorporates more and more realism. Implementing the reasoning

tasks in a theorem-prover is also a good idea. Although the use of logical reasoning for the

initial task—ﬁnding a route on a graph—may seem like overkill, the student should be im-

pressed that we can keep making the situation more complicated simply by describing those

added complications, with no additions to the reasoning system.

a. .

b. .

c. The successor-state axiom should be mechanical by now. :

60 Chapter 10. Knowledge Representation

d. To represent the amount of fuel the robot has in a given situation, use the function

. Let denote the distance between cities and , mea-

sured in units of fuel consumption. Let be a constant denoting the fuel capacity of

the tank.

e. The initial situation is described by .

The above axiom for location is extended as follows (note that we do not say what

happens if the robot runs out of gas). :

f. The simplest way to extend the representation is to add the predicate ,

which is true of cities with gas stations. The action is described by adding

another clause to the above axiom for , saying that when .

10.4 This question was inadvertently left in the exercises after the corresponding material

was excised from the chapter. Future printings may omit or replace this exercise.

10.5 Remember that we deﬁned substances so that is a category whose elements

are all those things of which one might say “it’s water.” One tricky part is that the English

language is ambiguous. One sense of the word “water” includes ice (“that’s frozen water”),

while another sense excludes it: (“that’s not water—it’s ice”). The sentences here seem to

use the ﬁrst sense, so we will stick with that. It is the sense that is roughly synonymous with

The other tricky part is that we are dealing with objects that change (freeze and melt)

over time. Thus, it won’t do to say , because (a mass of water) might be a

liquid at one time and a solid at another. For simplicity, we will use a situation calculus

representation, with sentences such as . There are many possible correct

answers to each of these. The key thing is to be consistent in the way that information is

represented. For example, do not use as a predicate on objects if is used as a

substance category.

a. “Water is a liquid between 0 and 100 degrees.” We will translate this as “For any

water and any situation, the water is liquid iff and only if the water’s temperature in the

situation is between 0 and 100 centigrade.”

b. “Water boils at 100 degrees.” It is a good idea here to do some tool-building. On

page 243 we used as a predicate applying to individual instances of

a substance. Here, we will deﬁne to denote the boiling point of all

instances of a substance. The basic meaning of boiling is that instances of the substance

becomes gaseous above the boiling point:

Then we need only say .

c. “The water in John’s water bottle is frozen.”

We will use the constant to represent the situation in which this sentence holds.

Note that it is easy to make mistakes in which one asserts that only some of the water

in the bottle is frozen.

d. “Perrier is a kind of water.”

e. “John has Perrier in his water bottle.”

f. “All liquids have a freezing point.”

Presumably what this means is that all substances that are liquid at room temper-

ature have a freezing point. If we use to denote this class of

substances, then we have

where is deﬁned similarly to . Note that this state-

ment is false in the real world: we can invent categories such as “blue liquid” which do

not have a unique freezing point. An interesting exercise would be to deﬁne a “pure”

substance as one all of whose instances have the same chemical composition.

g. “A liter of water weighs more than a liter of alcohol.”

10.6 This is a fairly straightforward exercise that can be done in direct analogy to the cor-

responding deﬁnitions for sets.

62 Chapter 10. Knowledge Representation

a. holds between a set of parts and a whole, saying

that anything that is a part of the whole must be a part of one of the set of parts.

b. holds between a set of parts and a whole when the set is disjoint and is

an exhaustive decomposition.

c. A set of parts is if when you take any two parts from the set, there

is nothing that is a part of both parts.

It is not the case that for any . A set may consist of

physically overlapping objects, such as a hand and the ﬁngers of the hand. In that case,

is equal to the hand, but is not a partition of it. We need to ensure that the

elements of are partwise disjoint:

10.7 For an instance of a substance with price per pound and weight pounds, the

price of will be , or in other words:

If is the set of tomatoes in a bag, then is the composite object consisting of

all the tomatoes in the bag. Then we have

10.8 In the scheme in the chapter, a conversion axiom looks like this:

“50 dollars” is just , the name of an abstract monetary quantity. For any measure func-

tion such as , we can extend the use of as follows:

Since the conversion axiom for dollars and cents has

it follows immediately that .

In the new scheme, we must introduce objects whose lengths are converted:

There is no obvious way to refer directly to “50 dollars” or its relation to “50 cents”. Again,

we must introduce objects whose monetary value is 50 dollars or 50 cents:

10.9 We will deﬁne a function that takes three arguments: a source cur-

rency, a time interval, and a target currency. It returns a number representing the exchange

rate. For example,

means that you can get 5.8677 Krone for a dollar on February 17th. This was the Federal

Reserve bank’s Spot exchange rate as of 10:00 AM. It is the mid-point between the buying and

selling rates. A more complete analysis might want to include buying and selling rates, and

the possibility for many different exchanges, as well as the commissions and fees involved.

Note also the distinction between a currency such as and a unit of measurement,

such as is used in the expression .

10.10 Another fun problem for clear thinkers:

a. Remember that means that some event of type occurs throughout the interval

Using as the question requests is not so easy, because the interval subsumes

all events within it.

b. A event is one in which both and occur throughout the duration of the

event. There is only one way this can happen: both and have to persist for the whole

interval. Another way to say it:

is logically equivalent to

whereas the same equivalence fails to hold for disjunction (see the next part).

c. means that a event occurs throughout or a event does:

On the other hand, holds if, at every point in , either a or a is

happening:

d. should mean that there is never an event of type going on in any

subinterval of , while should mean that there is no single event of type

that spans all of , even though there may be one or more events of type for subinter-

vals of :

64 Chapter 10. Knowledge Representation

One could also ask students to prove two versions of de Morgan’s laws using the two

types of negation, each with its corresponding type of disjunction.

10.11 Any object is an event, and is the event that for every subinterval of

time, refers to the place where is. For example, is the complex event

consisting of his home from midnight to about 9:00 today, then various parts of the road, then

his ofﬁce from 10:00 to 1:30, and so on. To say that an event is ﬁxed is to say that any two

moments of the event have the same spatial extent:

10.12 We will omit universally quantiﬁed variables:

10.13

“ ” “ “” “””

10.14 Here is an initial sketch of one approach. (Others are possible.) A given object to be

purchased may require some additional parts (e.g., batteries) to be functional, and there may

also be optional extras. We can represent requirements as a relation between an individual

object and a class of objects, qualiﬁed by the number of objects required:

We also need to know that a particular object is compatible, i.e., ﬁlls a given role appropri-

ately. For example,

Then it is relatively easy to test whether the set of ordered objects contains compatible re-

quired objects for each object.

10.15 Plurals can be handled by a relation between strings, e.g.,

“ ” “ ”

plus an assertion that the plural (or singular) of a name is also a name for the same category:

Conjunctions can be handled by saying that any conjunction string is a name for a category if

one of the conjuncts is a name for the category:

where is deﬁned appropriately in terms of concatenation. Probably it would be

better to redeﬁne instead.

10.16 Chapter 22 explains how to use logic to parse text strings and extract semantic infor-

mation. The outcome of this process is a deﬁnition of what objects are acceptable to the user

for a speciﬁc shopping request; this allows the agent to go out and ﬁnd offers matching the

user’s requirements. We omit the full deﬁnition of the agent, although a skeleton may appear

on the AIMA project web pages.

10.17 Here is a simple version of the answer; it can be elaborated ad inﬁnitum. Let the term

denote the event category of buyer buying object from seller for price .

We want to say about it that transfers the money to , and transfers ownership of to .

10.18 Let denote the class of events where person trades object to

person for object :

Now the only tricky part about deﬁning buying in terms of trading is in distinguishing a price

(a measurement) from an actual collection of money.

10.19 There are many possible approaches to this exercise. The idea is for the students to

think about doing knowledge representation for real; to consider a host of complications and

ﬁnd some way to represent the facts about them. Some of the key points are:

Ownership occurs over time, so we need either a situation-calculus or interval-calculus

approach.

There can be joint ownership and corporate ownership. This suggests the owner is a

group of some kind, which in the simple case is a group of one person.

Ownership provides certain rights: to use, to resell, to give away, etc. Much of this is

outside the deﬁnition of ownership per se, but a good answer would at least consider

how much of this to represent.

Own can own abstract obligations as well as concrete objects. This is the idea behind

the futures market, and also behind banks: when you deposit a dollar in a bank, you

are giving up ownership of that particular dollar in exchange for ownership of the right

66 Chapter 10. Knowledge Representation

to withdraw another dollar later. (Or it could coincidentally turn out to be the exact

same dollar.) Leases and the like work this way as well. This is tricky in terms of

representation, because it means we have to reify transactions of this kind. That is,

must be an object, not a predicate.

10.20 Most schools distinguish between required courses and elected courses, and between

courses inside the department and outside the department. For each of these, there may be re-

quirements for the number of courses, the number of units (since different courses may carry

different numbers of units), and on grade point averages. We show our chosen vocabulary by

example:

Student Jones’ complete course of study for the whole college career consists of Math1,

CS1, CS2, CS3, CS21, CS33 and CS34, and some other courses outside the major.

Jones meets the requirements for a major in Computer Science

Courses Math1, CS1, CS2, and CS3 are required for a Computer Science major.

A student must take at least 18 units in the CS department to get a degree in CS.

One can easily imagine other kinds of requirements; these just give you a ﬂavor.

In this solution we took “over an extended period” to mean that we should recommend

a set of courses to take, without scheduling them on a semester-by-semester basis. If you

wanted to do that, you would need additional information such as when courses are taught,

what is a reasonable course load in a semester, and what courses are prerequisites for what

others. For example:

The problem with ﬁnding the best program of study is in deﬁning what best means to the

student. It is easy enough to say that all other things being equal, one prefers a good teacher

to a bad one, or an interesting course to a boring one. But how do you decide which is best

when one course has a better teacher and is expected to be easier, while an alternative is more

interesting and provides one more credit? Chapter 16 uses utility theory to address this. If

you can provide a way of weighing these elements against each other, then you can choose a

best program of study; otherwise you can only eliminate some programs as being worse than

others, but can’t pick an absolute best one. Complexity is a further problem: witha general-

purpose theorem-prover it’s hard to do much more than enumerate legal programs and pick

the best.

10.21 This exercise and the following two are rather complex, perhaps suitable for term

projects. At this point, we want to strongly urge that you do assign some of these exercises

(or ones like them) to give your students a feeling of what it is really like to do knowl-

edge representation. In general, students ﬁnd classiﬁcation hierarchies easier than other rep-

resentation tasks. A recent twist is to compare one’s hierarchy with online ones such as

yahoo.com.

10.22 This is the most involved representation problem. It is suitable for a group project of

2 or 3 students over the course of at least 2 weeks.

10.23 Normally one would assign 10.22 in one assignment, and then when it is done, add

this exercise (posibly varying the questions). That way, the students see whether they have

made sufﬁcient generalizations in their initial answer, and get experience with debugging and

modifying a knowledge base.

10.24 In many AI and Prolog textbooks, you will ﬁnd it stated plainly that implications

sufﬁce for the implementation of inheritance. This is true in the logical but not the practical

sense.

a. Here are three rules, written in Prolog. We actually would need many more clauses on

the right hand side to distinguish between different models, different options, etc.

worth(X,575) :- year(X,1973), make(X,dodge), style(X,van).

worth(X,27000) :- year(X,1994), make(X,lexus), style(X,sedan).

worth(X,5000) :- year(X,1987), make(X,toyota), style(X,sedan).

To ﬁnd the value of JB, given a data base with year(jb,1973),make(jb,dodge)

and style(jb,van) we would call the backward chainer with the goal worth(jb,D),

and read the value for D.

b. The time efﬁciency of this query is , where in this case is the 11,000 entries in

the Blue Book. A semantic network with inheritance would allow us to follow a link

from JB to 1973-dodge-van, and from there to follow the worth slot to ﬁnd the

dollar value in time.

c. With forward chaining, as soon as we are told the three facts about JB, we add the new

fact worth(jb,575). Then when we get the query worth(jb,D), it is to

ﬁnd the answer, assuming indexing on the predicate and ﬁrst argument. This makes

logical inference seem just like semantic networks except for two things: the logical

68 Chapter 10. Knowledge Representation

inference does a hash table lookup instead of pointer following, and logical inference

explicitly stores worth statements for each individual car, thus wasting space if there

are a lot of individual cars. (For this kind of application, however, we will probably

want to consider only a few individual cars, as opposed to the 11,000 different models.)

d. If each category has many properties—for example, the speciﬁcations of all the replace-

ment parts for the vehicle—then forward-chaining on the implications will also be an

impractical way to ﬁgure out the price of a vehicle.

e. If we have a rule of the following kind:

worth(X,D) :- year-make-style(X,Yr,Mk,St),

year-make-style(Y,Yr,Mk,St), worth(Y,D).

together with facts in the database about some other speciﬁc vehicle of the same type

as JB, then the query worth(jb,D) will be solved in time with appropriate

indexing, regardless of how many other facts are known about that type of vehicle and

regardless of the number of types of vehicle.

10.25 When categories are reiﬁed, they can have properties as individual objects (such as

and ) that do not apply to their elements. Without the distinction be-

tween boxed and unboxed links, the sentence might mean

that every singleton set has one element, or that there si only one singleton set.

Solutions for Chapter 11

Planning

11.1 Both problem solver and planner are concerned with getting from a start state to a goal

using a set of deﬁned operations or actions. But in planning we open up the representation of

states, goals, and plans, whcihc allows for a wider variety of algorithms that decompose the

search space.

11.2 This is an easy exercise, the point of which is to understand that “applicable” means

satisfying the preconditions, and that a concrete action instance is one with the variables

replaced by constants. The applicable actions are:

A minor point of this is that the action of ﬂying nowhere—from one airport to itself—is

allowable by the deﬁnition of , and is applicable (if not useful).

11.3 For the regular schema we have:

When we add we get:

In general, we (1) created a predicate for each action, and then, for each ﬂuent

such as , we create a predicate that says the ﬂuent keeps its old value if either an irrelevant

action is taken, or an action whose precondition is not satisﬁed, and it takes on a new value

according to the effects of a relevant action if that action’s preconditions are satisﬁed.

70 Chapter 11. Planning

11.4 This exercise is intended as a fairly easy exercise in describing a domain. It is similar

to the Shakey problem (11.13), so you should probably assign only one of these two.

a. The initial state is:

b. The actions are:

ACTION: PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

ACTION:

PRECOND:

EFFECT:

ACTION:

PRECOND:

EFFECT:

ACTION:

PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

c. In situation calculus, the goal is a state such that:

In STRIPS, we can only talk about the goal state; there is no way of representing the fact

that there must be some relation (such as equality of location of an object) between two

states within the plan. So there is no way to represent this goal.

d. Actually, we did include the precondition. This is an example of the qualiﬁ-

cation problem.

11.5 Only positive literals are represented in a state. So not mentioning a literal is the same

as having it be negative.

11.6 Goals and preconditions can only be positive literals. So a negative effect can only

make it harder to achieve a goal (or a precondition to an action that achieves the goal). There-

fore, eliminating all negative effects only makes a problem easier.

11.7

a. It is feasible to use bidirectional search, because it is possible to invert the actions.

However, most of those who have tried have concluded that biderectional search is

generally not efﬁcient, because the forward and backward searches tend to miss each

other. This is due to the large state space. A few planners, such as PRODIGY (Fink and

Blythe, 1998) have used bidirectional search.

b. Again, this is feasible but not popular. PRODIGY is in fact (in part) a partial-order

planner: in the forward direction it keeps a total-order plan (equivalent to a state-based

planner), and in the backward direction it maintains a tree-structured partial-order plan.

c. An action can be added if all the preconditions of have been achieved by other

steps in the plan. When is added, ordering constraints and causal links are also added

to make sure that appears after all the actions that enabled it and that a precondition

is not disestablished before can be executed. The algorithm does search forward, but

it is not the same as forward state-space search because it can explore actions in parallel

when they don’t conﬂict. For example, if has three preconditions that can be satisﬁed

by the non-conﬂicting actions , , and , then the solution plan can be represented

as a single partial-order plan, while a state-space planner would have to consider all

permutations of , , and .

d. Yes, this is one possible way of implementing a bidirectional search in the space of

partial-order plans.

11.8 The drawing is actually rather complex, and doesn’t ﬁt well on this page. Some key

things to watch out for: (1) Both and actions are possible at level ; the planes

can still ﬂy when empty. (2) Negative effects appear in , and are mutex with their positive

counterparts.

11.9

a. Literals are persistent, so if it does not appear in the ﬁnal level, it never will and never

did, and thus cannot be achieved.

b. In a serial planning graph, only one action can occur per time step. The level cost (the

level at which a literal ﬁrst appears) thus represents the minimum number of actions in

a plan that might possibly achieve the literal.

11.10 A forward state-space planner maintains a partial plan that is a strict linear sequence

of actions; the plan reﬁnement operator is to add an applicable action to the end of the se-

quence, updating literals according to the action’s effects.

A backward state-space planner maintains a partial plan that is a reversed sequence of

actions; the reﬁnement operator is to add an action to the beginning of the sequence as long

as the action’s effects are compatible with the state at the beginning of the sequence.

11.11 The initial state is:

The goal is:

First we’ll explain why it is an anomaly for a noninterleaved planner. There are two subgoals;

suppose we decide to work on ﬁrst. We can clear off of and then move

72 Chapter 11. Planning

on to . But then there is no way to achieve without undoing the work we have

done. Similarly, if we work on the subgoal ﬁrst we can immediately achieve it in

one step, but then we have to undo it to get on .

Now we’ll show how things work out with an interleaved planner such as POP. Since

isn’t true in the initial state, there is only one way to achieve it: ,

for some . Similarly, we also need a step, for some . Now let’s look

at the step. We need to achieve its precondition . We could do

that either with or with . Let’s assume we choose the

latter. Now if we bind to , then all of the preconditions for the step

are true in the initial state, and we can add causal links to them. We then notice that there

is a threat: the step threatens the condition that is required by

the step. We can resolve the threat by ordering after the

step. Finally, notice that all the preconditions for are true in

the initial state. Thus, we have a complete plan with all the preconditions satisﬁed. It turns

out there is a well-ordering of the three steps:

11.12 The actions we need are the four from page 346:

ACTION: PRECOND: EFFECT:

ACTION: EFFECT:

ACTION: PRECOND: EFFECT:

ACTION: EFFECT:

One solution found by GRAPHPLAN is to execute RightSock and LeftSock in the ﬁrst time

step, and then RightShoe and LeftShoe in the second.

Now we add the following two actions (neither of which has preconditions):

ACTION: EFFECT:

The partial-order plan is shown in Figure S11.1. We saw on page 348 that there are 6 total-

order plans for the shoes/socks problem. Each of these plans has four steps, and thus ﬁve

arrow links. The next step, Hat could go at any one of these ﬁve locations, giving us

total-order plans, each with ﬁve steps and six links. Then the ﬁnal step, Coat, can go in

any one of these 6 positions, giving us total-order plans.

11.13 The actions are quite similar to the monkey and banannas problem—you should prob-

Start

Finish

Left

Sock Right

Sock

Left

Shoe Right

Shoe

Hat Coat

Figure S11.1 Partial-order plan including a hat and coat, for Exercise 11.1.

ably assign only one of these two problems. The actions are:

ACTION: PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

ACTION: PRECOND:

EFFECT:

The initial state is:

74 Chapter 11. Planning

A plan to achieve the goal is:

11.14 GRAPHPLAN is a propositional algorithm, so, just as we could solve certain FOL by

translating them into propositional logic, we can solve certain situation calculus problems by

translating into propositional form. The trick is how exactly to do that.

The Finish action in POP planning has as its preconditions the goal state. We can create

aFinish action for GRAPHPLAN, and give it the effect Done. In this case there would be a

ﬁnite number of instantiations of the Finish action, and we would reason with them.

11.15 (Figure (11.1) is a little hard to ﬁnd—it is on page 403.)

a. The point of this exercise is to consider what happens when a planner comes up with an

impossible action, such as ﬂying a plane from someplace where it is not. For example,

suppose is at and we give it the action of ﬂying from Bangalore to Brisbane.

By (11.1), was at and did not ﬂy away, so it is still there.

b. Yes, the plan will still work, because the ﬂuents hold from the situation before an inap-

plicable action to the state afterward, so the plan can continue from that state.

c. It depends on the details of how the axioms are written. Our axioms were of the form

Action is possible Rule. This tells us nothing about the case where the action is

not possible. We would need to reformulate the axioms, or add additional ones to say

what happens when the action is not possible.

11.16 Aprecondition axiom is of the form

There are of these axioms, where is the number of time steps, is

the number of planes and is the number of airports. More generally, if there are action

schemata of maximum arity , with objects, then there are axioms.

With symbol-splitting, we don’t have to describe each speciﬁc ﬂight, we need only say

that for a plane to ﬂy anywhere, it must be at the start airport. That is,

More generally, if there are action schemata of maximum arity , with objects, and each

precondition axiom depends on just two of the arguments, then there are

axioms, for a speedup of .

An action exclusion axiom is of the form

With the notation used above, there are axioms for . More generally,

there can be up to axioms.

With symbol-splitting, we wouldn’t gain anything for the axioms, but we would

gain in cases where there is another variable that is not relevant to the exclusion.

11.17

a. Yes, this will ﬁnd a plan whenever the normal SATPLAN ﬁnds a plan no longer than

b. No.

c. There is no simple and clear way to induce WALKSAT to ﬁnd short solutions, because

it has no notion of the length of a plan—the fact that the problem is a planning problem

is part of the encoding, not part of WALKSAT. But if we are willing to do some rather

brutal surgery on WALKSAT, we can achieve shorter solutions by identifying the vari-

ables that represent actions and (1) tending to randomly initialize the action variables

(particularly the later ones) to false, and (1) prefering to randomly ﬂip an earlier action

variable rather than a later one.

Solutions for Chapter 12

Planning and Acting in the Real

World

12.1

a. is eligible to be an effect because the action does have the effect of moving

the clock by . It is possible that the duration depends on the action outcome, so if

disjunctive or conditional effects are used there must be a way to associate durations

with outcomes, which is most easily done by putting the duration into the outcome

expression.

b. The STRIPS model assumes that actions are time points characterized only by their pre-

conditions and effects. Even if an action occupies a resource, that has no effect on the

outcome state (as explained on page 420). Therefore, we must extend the STRIPS for-

malism. We could do this by treating a RESOURCE: effect differently from other effects,

but the difference is sufﬁciently large that it makes more sense to treat it separately.

12.2 The basic idea here is to record the initial resource level in the precondition and the

change in resource level in the effect of each action.

a. Let denote the fact that there are screws. We need to add

to the initial state, and add a fourth argument to the predicate indicating the

number of screws required—i.e., and .

We add to the precondition of and add as a fourth argument

of the literal. Then add to the effect of .

b. A simple solution is to say that any action that consumes a resource is potentially in

conﬂict with any causal link protecting the same resource.

c. The planner can keep track of the resource requirements of actions added to the plan

and backtrack whenever the total usage exceeds the initial amount.

12.3 There is a wide range of possible answers to this question. The important point is that

students understand what constitutes a correct implementation of an action: as mentioned on

page 424, it must be a consistent plan where all the preconditions and effects are accounted

for. So the ﬁrst thing we need is to decide on the preconditions and effects of the high-level

actions. For GetPermit, assume the precondition is owning land, and the effect is having a

permit for that piece of land. For HireBuilder, the precondition is having the ability to pay,

and the effect is having a signed contract in hand.

One possible decomposition for GetPermit is the three-step sequence GetPermitForm,

FillOutForm, and GetFormApproved. There is a causal link with the condition HaveForm

between the ﬁrst two, and one with the condition HaveCompletedForm between the last two.

Finally, the GetFormApproved step has the effect HavePermit. This is a valid decomposition.

For HireBuilder, suppose we choose the three-step sequence InterviewBuilders,Choose-

Builder, and SignContract. This last step has the precondition AbleToPay and the effect Have-

ContractInHand. There are also causal links between the substeps, but they don’t affect the

correctness of the decomposition.

12.4 Consider the problem of building two adjacent walls of the house. Mostly these sub-

plans are independent, but they must share the step of putting up a common post at the corner

of the two walls. If that step was not shared, we would end up with an extra post, and two

unattached walls.

Note that tasks are often decomposed speciﬁcally so as to minimize the amount of step

sharing. For example, one could decompose the house building task into subtasks such as

“walls” and “ﬂoors.” However, real contractors don’t do it that way. Instead they have “rough

walls” and “rough ﬂoors” steps, followed by a “ﬁnishing” step.

12.5 In the HTN view, the space of possible decompositions may constrain the allowable

solutions, eliminating some possible sequences of primitive actions. For example, the de-

composition of the LAToNYRoundTrip action can stipulate that the agent should go to New

York. In a simple STRIPS formulation where the start and goal states are the same, the empty

plan is a solution. We can get around this problem by rethinking the goal description. The

goal state is not , but . We add as an effect of

. Then, the solution must be a trip that includes New York. There remains the prob-

lem of preventing the STRIPS plan from including other stops on its itinerary; ﬁxing this is

much more difﬁcult because negated goals are not allowed.

12.6 Suppose we have a STRIPS action description for with precondition and effect .

The “action” to be decomposed is . The decomposition has two steps:

and . This can be extended in the obvious way for conjunctive effects and precondi-

tions.

12.7 We need one action, , which assigns the value in the source register (or variable

if you prefer, but the term “register” makes it clearer that we are dealing with a physical

location) to the destination register :

ACTION:

PRECOND:

EFFECT:

Now suppose we start in an initial state with

and we have the goal . Unfortunately, there

is no way to solve this as is. We either need to add an explicit condition to the

initial state, or we need a way to create new registers. That could be done with an action for

78 Chapter 12. Planning and Acting in the Real World

allocating a new register:

ACTION:

EFFECT:

Then the following sequence of steps constitues a valid plan:

12.8 For the ﬁrst case, where one instance of action schema is in the plan, the reformula-

tion is correct, in the sense that a solution for the original disjunctive formulation is a solution

for the new formulation and vice versa. For the second case, where more than one instance

of the action schema may occur, the reformulation is incorrect. It assumes that the outcomes

of the instances are governed by a single hidden variable, so that if, for example, is the

outcome of one instance it must also be the outcome of the other. It is possible that a solution

for the reformulated case will fail in the original formulation.

12.9 With unbounded indeterminacy, the set of possible effects for each action is unknown

or too large to be enumerated. Hence, the space of possible actions sequences required to

handle all these eventualities is far too large to consider.

12.10 Using the second deﬁnition of in the chapter—namely, that there is a clear

space for a block—the only change is that the destination remains clear if it is the table:

PRECOND:

EFFECT:when

12.11 Let be true iff the robot’s current square is clean and be true iff the

other square is clean. Then is characterized by

PRECOND: EFFECT:

Unfortunately, moving affects these new literals! For we have

PRECOND:

EFFECT:when when

when when

with the dual for .

12.12 Here we borrow from the last description of the on page 433:

PRECOND:

EFFECT:when when

when when

12.13 The main thing to notice here is that the vacuum cleaner moves repeatedly over dirty

areas—presumably, until they are clean. Also, each forward move is typically short, followed

by an immediate reversing over the same area. This is explained in terms of a disjunctive

outcome: the area may be fully cleaned or not, the reversing enables the agent to check, and

the repetition ensures completion (unless the dirt is ingrained). Thus, we have a strong cyclic

plan with sensing actions.

12.14

a. “Lather. Rinse. Repeat.”

This is an unconditional plan, if taken literally, involving an inﬁnite loop. If the pre-

condition of is , and the goal is , then execution monitoring will

cause execution to terminate once is achieved because at that point the correct

repair is the empty plan.

b. “Apply shampoo to scalp and let it remain for several minutes. Rinse and repeat if

necessary.”

This is a conditional plan where “if necessary” presumably tests .

c. “See a doctor if problems persist.”

This is also a conditional step, although it is not speciﬁed here what problems are tested.

12.17 First, we need to decide if the precondition is satisﬁed. There are three cases:

a. If it is known to be unsatisﬁed, the new belief state is identical to the old (since we

assume nothing happens).

b. If it is known to be satisﬁed, the unconditional effects (which are all knowledge propo-

sitions) are added and deleted from the belief state in the usual STRIPS fashion. Each

conditional effect whose condition is known to be true is handled in the same way. For

each setting of the unknown conditions, we create a belief state with the appropriate

additions and deletions.

c. If the status of the precondition is unknown, each new belief state is effectively the

disjunction of the unchanged belief state from (a) with one of the belief states obtained

from (b). To enforce the “list of knowledge propositions” representation, we keep those

propositions that are identical in each of the two belief states being disjoined and discard

those that differ. This results in a weaker belief state than if we were to retain the

disjunction; on the other hand, retaining the disjunctions over many steps could lead to

exponentially large representations.

12.18 For we have the obvious dual version of Equation 12.2:

PRECOND:

EFFECT:when

when when

With , dirt is sometimes deposited when the square is clean. With automatic dirt sensing,

80 Chapter 12. Planning and Acting in the Real World

this is always detected, so we have a disjunctive conditional effect:

PRECOND:

EFFECT:when

when

12.19 The continuous planning agent described in Section 12.6 has at least one of the listed

abilities, namely the ability to accept new goals as it goes along. A new goal is simply added

as an extra open precondition in the step, and the planner will ﬁnd a way to satisfy

it, if possible, along with the other remaining goals. Because the data structures built by

the continuous planning agent as it works on the plan remain largely in place as the plan is

executed, the cost of replanning is usually relatively small unless the failure is catastrophic.

There is no speciﬁc time bound that is guaranteed, and in general no such bound is possible

because changing even a single state variable might require completely reconstructing the

plan from scratch.

12.20 Let be the proposition that the patient is dehydrated and be the side effect. We

have

PRECOND: EFFECT:

PRECOND: EFFECT:when

and the initial state is . The solution plan is .

There are two possible worlds, one where holds and one where holds. In the ﬁrst,

causes and has no effect; in the second, has no effect and

causes . In both cases, the ﬁnal state is .

12.21 One solution plan is if then .

Solutions for Chapter 13

Uncertainty

13.1 The “ﬁrst principles” needed here are the deﬁnition of conditional probability,

, and the deﬁnitions of the logical connectives. It is not enough to say that

if is “given” then must be true! From the deﬁnition of conditional probability, and

the fact that and that conjunction is commutative and associative, we have

13.2 The main axiom is axiom 3: . For the discrete

random variable , let be the event that , and be the event that has any other

value. Then we have

where we know that is 0 because a variable cannot take on two

distinct values. If we now break down the case of , we eventually get

But the left-hand side is equivalent to , which is 1 by axiom 2, so the sum of the

right-hand side must also be 1.

13.3 Probably the easiest way to keep track of what’s going on is to look at the probabil-

ities of the atomic events. A probability assignment to a set of propositions is consistent

with the axioms of probability if the probabilities are consistent with an assignment to the

atomic events that sums to 1 and has all probabilities between 0 and 1 inclusive. We call the

probabilities of the atomic events , , , and , as follows:

a b

c d

We then have the following equations:

82 Chapter 13. Uncertainty

From these, it is straightforward to infer that , , , and .

Therefore, . Thus the probabilities given are consistent with a rational

assignment, and the probability is exactly determined. (This latter fact can be seen

also from axiom 3 on page 422.)

If , then . Thus, even though the bet outlined in

Figure 13.3 loses if and are both true, the agent believes this to be impossible so the bet

is still rational.

13.4 ? (?, ?) argues roughly as follows: Suppose we present an agent with a choice: either

deﬁnitely receive monetary payoff , or choose a lottery that pays if event occurs

and 0 if does not occur. The number for which the agent is indifferent between the two

choices (assuming linear utility of money), is deﬁned to be the agent’s degree of belief in .

A set of degrees of belief speciﬁed for some collection of events will either be coherent,

which is deﬁned to mean that there is no set of bets based on these stated beliefs that will

guarantee that the agent will lose money, or incoherent, which is deﬁned to mean that there

is such a set of bets. De Finetti showed that coherent beliefs satisfy the axioms of probability

theory.

Axiom 1: , for any and . If an agent speciﬁes , then the agent

is offering to pay more than to enter a lottery in which the biggest prize is . If an agent

speciﬁes , then the agent is offering to pay either or 0 in exchange for a negative

amount. Either way, the agent is guaranteed to lose money if the opponent accepts the right

offer.

Axiom 2: when is and when is . Suppose the agent assigns

as the degree of belief in a known true event. Then the agent is indifferent between a payoff

of and one of . This is only coherent when . Similarly, a degree of belief of

for a known false event means the agent is indifferent between and 0. Only makes

this coherent.

Axiom 3: Given two mutually exclusive, exhaustive events, and , and respective

degrees of belief and and payoffs and , it must be that the degree of belief for

the combined event equals . The idea is that the beliefs and constitute

an agreement to pay in order to enter a lottery in which the prize is when

occurs. So the net gain is deﬁned as . To avoid the possibility

of the amounts being chosen to guarantee that every is negative, we have to assure that

the determinant of the matrix relating to is zero, so that the linear system can be solved.

This requires that . The result extends to the case with mutually exclusive,

exhaustive events rather than two.

13.5 This is a classic combinatorics question that could appear in a basic text on discrete

mathematics. The point here is to refer to the relevant axioms of probability: principally,

axiom 3 on page 422. The question also helps students to grasp the concept of the joint

probability distribution as the distribution over all possible states of the world.

a. There are = = 2,598,960 possible

ﬁve-card hands.

b. By the fair-dealing assumption, each of these is equally likely. By axioms 2 and 3, each

hand therefore occurs with probability 1/2,598,960.

c. There are four hands that are royal straight ﬂushes (one in each suit). By axiom 3, since

the events are mutually exclusive, the probability of a royal straight ﬂush is just the sum

of the probabilities of the atomic events, i.e., 4/2,598,960 = 1/649,740.

d. Again, we examine the atomic events that are “four of a kind” events. There are 13

possible “kinds” and for each, the ﬁfth card can be one of 48 possible other cards. The

total probability is therefore .

These questions can easily be augmented by more complicated ones, e.g., what is the proba-

bility of getting a full house given that you already have two pairs? What is the probability of

getting a ﬂush given that you have three cards of the same suit? Or you could assign a project

of producing a poker-playing agent, and have a tournament among them.

13.6 The main point of this exercise is to understand the various notations of bold versus

non-bold P, and uppercase versus lowercase variable names. The rest is easy, involving a

small matter of addition.

a. This asks for the probability that Toothache is true.

b. This asks for the vector of probability values for the random variable Cavity. It has two

values, which we list in the order . First add up

. Then we have

c. This asks for the vector of probability values for Toothache, given that Cavity is true.

d. This asks for the vector of probability values for Cavity, given that either Toothache or

Catch is true. First compute

. Then

13.7 Independence is symmetric (that is, and are independent iff and are indepen-

dent) so is the same as . So we need only prove that

is equivalent to . The product rule, , can

be used to rewrite as , which simpliﬁes to

13.8 We are given the following information:

84 Chapter 13. Uncertainty

and the observation . What the patient is concerned about is . Roughly

speaking, the reason it is a good thing that the disease is rare is that is propor-

tional to , so a lower prior for will mean a lower value for .

Roughly speaking, if 10,000 people take the test, we expect 1 to actually have the disease, and

most likely test positive, while the rest do not have the disease, but 1% of them (about 100

people) will test positive anyway, so will be about 1 in 100. More precisely,

using the normalization equation from page 428:

The moral is that when the disease is much rarer than the test accuracy, a positive test result

does not mean the disease is likely. A false positive reading remains much more likely.

Here is an alternative exercise along the same lines: A doctor says that an infant who

predominantly turns the head to the right while lying on the back will be right-handed, and

one who turns to the left will be left-handed. Isabella predominantly turned her head to the

left. Given that 90% of the population is right-handed, what is Isabella’s probability of being

right-handed if the test is 90% accurate? If it is 80% accurate?

The reasoning is the same, and the answer is 50% right-handed if the test is 90% accu-

rate, 69% right-handed if the test is 80% accurate.

13.9 The basic axiom to use here is the deﬁnition of conditional probability:

a. We have

and

P P P

hence

P P P

b. The derivation here is the same as the derivation of the simple version of Bayes’ Rule

on page 426. First we write down the dual form of the conditionalized product rule,

simply by switching and in the above derivation:

P P P

Therefore the two right-hand sides are equal:

P P P P

Dividing through by Pwe get

PP P

13.10 The key to this exercise is rigorous and frequent application of the deﬁnition of con-

ditional probability, P P P . The original statement that we are given

is:

P P P

We start by applying the deﬁnition of conditional probability to two of the terms in this

statement:

Pand PP

Now we substitute the right hand side of these deﬁnitions for the left hand sides in the original

statement to get:

PPP

Now we need the deﬁnition once more:

PP P

We substitute this right hand side for Pto get:

P P

PPP

Finally, we cancel the Pand Ps to get:

The second part of the exercise follows from by a similar derivation, or by noticing that

and are interchangeable in the original statement (because multiplication is commutative

and means the same as ).

In Chapter 14, we will see that in terms of Bayesian networks, the original statement

means that is the lone parent of and also the lone parent of . The conclusion is that

knowing the values of and is the same as knowing just the value of in terms of telling

you something about the value of .

13.11

a. There are ways to pick a coin, and 2 outcomes for each ﬂip (although with the fake

coin, the results of the ﬂip are indistinguishable), so there are total atomic events.

Of those, only 2 pick the fake coin, and result in heads. So the probability

of a fake coin given heads, , is .

b. Now there are atomic events, of which pick the fake coin, and result

in heads. So the probability of a fake coin given a run of heads, , is

. Note this approaches 1 as increases, as expected. If ,

for example, than .

c. There are two kinds of error to consider. Case 1: A fair coin might turn up heads

times in a row. The probability of this is , and the probability of a fair coin being

chosen is . Case 2: The fake coin is chosen, in which case the procedure

86 Chapter 13. Uncertainty

always makes an error. The probability of drawing the fake coin is . So the total

probability of error is

13.12 The important point here is that although there are often many possible routes by

which answers can be calculated in such problems, it is usually better to stick to systematic

“standard” routes such as Bayes’ Rule plus normalization. Chapter 14 describes general-

purpose, systematic algorithms that make heavy use of normalization. We could guess that

, or we could calculate it from the information already given (although the

idea here is to assume that is not known):

Normalization proceeds as follows:

13.13 The question would have been slightly more consistent if we had asked about the

calculation of Pinstead of . Showing that a given set of information

is sufﬁcient is relatively easy: ﬁnd an expression for Pin terms of the given

information. Showing insufﬁciency can be done by showing that the information provided

does not contain enough independent numbers.

a. Bayes’ Rule gives

PP P

Hence the information in (ii) is sufﬁcient—in fact, we don’t need Pbecause

we can use normalization. Intuitively, the information in (iii) is insufﬁcient because

Pand Pprovide no information about correlations between and

that might be induced by . Mathematically, suppose has possible values and

and have and possible respectively. Pcontains

independent numbers, whereas the information in (iii) contains

numbers—clearly insufﬁcient for large , , and . Similarly, the

information in (i) contains numbers—again

insufﬁcient.

b. If and are conditionally independent given , then

P P P

Using normalization, (i), (ii), and (iii) are each sufﬁcient for the calculation.

13.14 When dealing with joint entries, it is usually easiest to get everything into the form of

probabilities of conjunctions, since these can be expressed as sums of joint entries. Beginning

with the conditional independence constraint

P P P

we can rewrite it using the deﬁnition of conditional probability on each term to obtain

Hence we can write an expression for joint entries:

PP P

This gives us 8 equations constraining the 8 joint entries, but several of the equations are

redundant.

13.15 The relevant aspect of the world can be described by two random variables: means

the taxi was blue, and means the taxi looked blue. The information on the reliability of

color identiﬁcation can be written as

We need to know the probability that the taxi was blue, given that it looked blue:

Thus we cannot decide the probability without some information about the prior probability

of blue taxis, . For example, if we knew that all taxis were blue, i.e., , then

obviously . On the other hand, if we adopt Laplace’s Principle of Indifference,

which states that propositions can be deemed equally likely in the absence of any differenti-

ating information, then we have and . Usually we will have

some differentiating information, so this principle does not apply.

Given that 9 out of 10 taxis are green, and assuming the taxi in question is drawn

randomly from the taxi population, we have . Hence

13.16 This question is extremely tricky. It is a variant of the “Monty Hall” problem, named

after the host of the game show “Let’s Make a Deal” in which contestants are asked to choose

between the prize they have already won and an unknown prize behind a door. Several dis-

tinguished professors of statistics have very publically got the wrong answer. Certainly, all

such questions can be settled by repeated trials!

Let = “x will be freed”, = “x will be executed”. If the information provided by

the guard is expressed as then we get:

This would be quite a shock to —his chances of execution have increased! On the other

hand, if the information provided by the guard is expressed as = “The guard said that ”

then we get:

88 Chapter 13. Uncertainty

Thus the key thing that is missed by the naive approach is that the guard has a choice of whom

to inform in the case where A will be executed.

One can produce variants of the question that reinforce the intuitions behind the correct

approach. For example: Suppose there now a thousand prisoners on death row, all but one

of whom will be pardoned. Prisoner A ﬁnds a printout with the last page torn off. It gives

the names of 998 pardonees, not including A’s name. What is the probability that A will

be executed? Now suppose A is left-handed, and the program was printing the names of all

right-handed pardonees. What is the probability now? Clearly, in the ﬁrst case it is quite

reasonable for to get worried, whereas in the second case it is not—the names of right-

handed prisoners to be pardoned should not inﬂuence A’s chances. It is this second case that

applies in the original story, because the guard is precluded from giving information about

13.17 We can apply the deﬁnition of conditional independence as follows:

P e P e P e P e

Now, divide the effect variables into those with evidence, E, and those without evidence, Y.

We have

PeyP y e

yP P y P

P P yP P y

P P

where the last line follows because the summation over yis 1. Therefore, the algorithm

computes the product of the conditional probabilitites of the evidence variables given each

value of the cause, multiplies each by the prior probability of the cause, and normalizes the

result.

13.18 This question is essentially previewing material in Chapter 23 (page 842), but stu-

dents should have little difﬁculty in ﬁguring out how to estimate a conditional probability

from complete data.

a. The model consists of the prior probability Pand the conditional probabil-

ities P. For each category , Pis estimated as the

fraction of all documents that are of category . Similarly, P

is estimated as the fraction of documents of category that contain word .

b. See the answer for 13.17. Here, every evidence variable is observed, since we can tell

if any given word appears in a given document or not.

c. The independence assumption is clearly violated in practice. For example, the word pair

“artiﬁcial intelligence” occurs more frequently in any given document category than

would be suggested by multiplying the probabilities of “artiﬁcial” and “intelligence”.

13.19 This probability model is also appropriate for Minesweeper (Ex. 7.11). If the total

number of pits is ﬁxed, then the variables and are no longer independent. In general,

because learning that makes it less likely that there is a mine at (as there are

now fewer to spread around). The joint distribution places equal probability on all assign-

ments to that have exactly 3 pits, and zero on all other assignments. Since there

are 15 squares, the probability of each 3-pit assignment is .

To calculate the probabilities of pits in and , we start from Figure 13.7. We

have to consider the probabilities of complete assignments, since the probability of the “other”

region assignment does not cancel out. We can count the total number of 3-pit assignments

that are consistent with each partial assignment in 13.7(a) and 13.7(b).

In 13.7(a), there are three partial assignments with :

The ﬁrst ﬁxes all three pits, so corresponds to 1 complete assignment.

The second leaves 1 pit in the remaining 10 squares, so corresponds to 10 complete

assignments.

The third also corresponds to 10 complete assignments.

Hence, there are 21 complete assignments with .

In 13.7(b), there are two partial assignments with :

The ﬁrst leaves 1 pit in the remaining 10 squares, so corresponds to 10 complete assign-

ments.

The second leaves 2 pits in the remaining 10 squares, so corresponds to com-

plete assignments.

Hence, there are 55 complete assignments with . Normalizing, we obtain

With , there are four partial assignments with a total of

complete assignments. With , there is only one partial assignment

with complete assignments. Hence

Solutions for Chapter 14

Probabilistic Reasoning

14.1 Adding variables to an existing net can be done in two ways. Formally speaking,

one should insert the variables into the variable ordering and rerun the network construction

process from the point where the ﬁrst new variable appears. Informally speaking, one never

really builds a network by a strict ordering. Instead, one asks what variables are direct causes

or inﬂuences on what other ones, and builds local parent/child graphs that way. It is usually

easy to identify where in such a structure the new variable goes, but one must be very careful

to check for possible induced dependencies downstream.

a.is not caused by any of the car-related variables, so needs no parents.

It directly affects the battery and the starter motor. is an additional

precondition for . The new network is shown in Figure S14.1.

b. Reasonable probabilities may vary a lot depending on the kind of car and perhaps the

personal experience of the assessor. The following values indicate the general order of

Radio

Battery

Ignition Gas

Starts

Moves

IcyWeather

StarterMotor

Figure S14.1 Car network amended to include and

( ).

magnitude and relative values that make sense:

A reasonable prior for IcyWeather might be 0.05 (perhaps depending on location

and season).

, .

, other entries 0.0.

c. With 8 Boolean variables, the joint has = 255 independent entries.

d. Given the topology shown in Figure S14.1, the total number of independent CPT entries

is 1+2+2+2+2+1+8+2= 20.

e. The CPT for describes a set of nearly necessary conditions that are together

almost sufﬁcient. That is, all the entries are nearly zero except for the entry where all

the conditions are true. That entry will be not quite 1 (because there is always some

other possible fault that we didn’t think of), but as we add more conditions it gets closer

to 1. If we add a node as an extra parent, then the probability is exactly 1 when

all parents are true. We can relate noisy-AND to noisy-OR using de Morgan’s rule:

. That is, noisy-AND is the same as noisy-OR except that the

polarities of the parent and child variables are reversed. In the noisy-OR case, we have

where is the probability that the presence of the th parent fails to cause the child to

be true. In the noisy-AND case, we can write

where is the probability that the absence of the th parent fails to cause the child to

be false (e.g., it is magically bypassed by some other mechanism).

14.2 This question exercises many aspects of the student’s understanding of Bayesian net-

works and uncertainty.

a. A suitable network is shown in Figure S14.2. The key aspects are: the failure nodes are

parents of the sensor nodes, and the temperature node is a parent of both the gauge and

the gauge failure node. It is exactly this kind of correlation that makes it difﬁcult for

humans to understand what is happening in complex systems with unreliable sensors.

b. No matter which way the student draws the network, it should not be a polytree because

of the fact that the temperature inﬂuences the gauge in two ways.

c. The CPT for is shown below. The wording of the question is a little tricky because

and are deﬁned in terms of “incorrect” rather than “correct.”

92 Chapter 14. Probabilistic Reasoning

TGA

FGFA

Figure S14.2 A Bayesian network for the nuclear alarm problem.

d. The CPT for is as follows:

0 0 0 1

1 1 1 0

e. This part actually asks the student to do something usually done by Bayesian network

algorithms. The great thing is that doing the calculation without a Bayesian network

makes it easy to see the nature of the calculations that the algorithms are systematizing.

It illustrates the magnitude of the achievement involved in creating complete and correct

algorithms.

Abbreviating and by and , the probability of interest here

is . Because the alarm’s behavior is deterministic, we can reason

that if the alarm is working and sounds, must be . Because and are

d-separated from , we need only calculate .

There are several ways to go about doing this. The “opportunistic” way is to notice

that the CPT entries give us , which suggests using the generalized Bayes’

Rule to switch and with as background:

We then use Bayes’ Rule again on the last term:

A similar relationship holds for :

Normalizing, we obtain

The “systematic” way to do it is to revert to joint entries (noticing that the subgraph

of , , and is completely connected so no loss of efﬁciency is entailed). We have

Now we use the chain rule formula (Equation 15.1 on page 439) to rewrite the joint

entries as CPT entries:

which of course is the same as the expression arrived at above. Letting ,

, and , we get

14.3

a. Although (i) in some sense depicts the “ﬂow of information” during calculation, it is

clearly incorrect as a network, since it says that given the measurements and ,

the number of stars is independent of the focus. (ii) correctly represents the causal

structure: each measurement is inﬂuenced by the actual number of stars and the focus,

and the two telescopes are independent of each other. (iii) shows a correct but more

complicated network—the one obtained by ordering the nodes , , , , . If

you order before you would get the same network except with the arrow from

to reversed.

b. (ii) requires fewer parameters and is therefore better than (iii).

c. To compute P, we will need to condition on (that is, consider both possible

cases for , weighted by their probabilities).

P P P P P

P P P P

Let be the probability that the telescope is out of focus. The exercise states that this

will cause an “undercount of three or more stars,” but if = 3 or less the count will

be 0 if the telescope is out of focus. If it is in focus, then we will assume there is a

probability of of counting one two few, and of counting one too many. The rest of

the time , the count will be accurate. Then the table is as follows:

f + e(1-f) f f

(1-2e)(1-f) e(1-f) 0.0

e(1-f) (1-2e)(1-f) e(1-f)

0.0 e(1-f) (1-2e)(1-f)

0.0 0.0 e(1-f)

Notice that each column has to add up to 1. Reasonable values for and might be

0.05 and 0.002.

94 Chapter 14. Probabilistic Reasoning

d. This question causes a surprising amount of difﬁculty, so it is important to make sure

students understand the reasoning behind an answer. One approach uses the fact that

it is easy to reason in the forward direction, that is, try each possible number of stars

and see whether measurements and are possible. (This is a sort of

mental simulation of the physical process.) An alternative approach is to enumerate the

possible focus states and deduce the value of for each. Either way, the solutions are

, 4, or .

e. We cannot calculate the most likely number of stars without knowing the prior distribu-

tion . Let the priors be , , and . The posterior for is ;

for it is at most (at most, because with the out-of-focus telescope

could measure 0 instead of 1); for it is at most . If we assume that the

priors are roughly comparable, then is most likely because we are told that is

much smaller than .

For follow-up or alternate questions, it is easy to come up with endless variations on the

same theme involving sensors, failure nodes, hidden state. One can also add in complex

mechanisms, as for the variable in exercise 14.1.

14.5 This exercise is a little tricky and will appeal to more mathematicaly oriented students.

a. The basic idea is to multiply the two densities, match the result to the standard form for

a multivariate Gaussian, and hence identify the entries in the inverse covariance matrix.

Let’s begin by looking at the multivariate Gaussian. From page 982 in Appendix A we

have

xx x

where is the mean vector and is the covariance matrix. In our case, xis

and let the (as yet) unknown be . Suppose the inverse covariance matrix is

Then, if we multiply out the exponent, we obtain

x x

Looking at the distributions themselves, we have

and

hence

We can obtain equations for , , and by picking out the coefﬁcients of , , and

We can check these by comparing the normalizing constants.

from which we obtain the constraint

which is easily conﬁrmed. Similar calculations yield and , and plugging the re-

sults back shows that is indeed multivariate Gaussian. The covariance matrix

b. The induction is on , the number of variables. The base case for is trivial.

The inductive step asks us to show that if any constructed with linear–

Gaussian conditional densities is multivariate Gaussian, then any

constructed with linear–Gaussian conditional densities is also multivariate Gaussian.

Without loss of generality, we can assume that is a leaf variable added to a net-

work deﬁned in the ﬁrst variables. By the product rule we have

which, by the inductive hypothesis, is the product of a linear Gaussian with a multivari-

ate Gaussian. Extending the argument of part (a), this is in turn a multivariate Gaussian

of one higher dimension.

14.6

a. With multiple continuous parents, we must ﬁnd a way to map the parent value vector to

a single threshold value. The simplest way to do this is to take a linear combination of

the parent values.

b. For ordered values , consider the Boolean proposition deﬁned

by . The proposition is just for . Now we can

propose probit distributions for each , with means also in increasing order.

14.7 This question deﬁnitely helps students get a solid feel for variable elimination. Stu-

dents may need some help with the last part if they are to do it properly.

96 Chapter 14. Probabilistic Reasoning

b. Including the normalization step, there are 7 additions, 16 multiplications, and 2 divi-

sions. The enumeration algorithm has two extra multiplications.

c. To compute Pusing enumeration, we have to evaluate two complete

binary trees (one for each value of ), each of depth , so the total work is .

Using variable elimination, the factors never grow beyond two variables. For example,

the ﬁrst step is

P f f

P f

The last line is isomorphic to the problem with variables instead of ; the work

done on the ﬁrst step is a constant independent of , hence (by induction on , if you

want to be formal) the total work is .

d. Here we can perform an induction on the number of nodes in the polytree. The base

case is trivial. For the inductive hypothesis, assume that any polytree with nodes can

be evaluated in time proportional to the size of the polytree (i.e., the sum of the CPT

sizes). Now, consider a polytree with nodes. Any node ordering consistent with

the topology will eliminate ﬁrst some leaf node from this polytree. To eliminate any

leaf node, we have to do work proportional to the size of its CPT. Then, because the

network is a polytree, we are left with independent subproblems, one for each parent.

Each subproblem takes total work proportional to the sum of its CPT sizes, so the total

work for nodes is proportional to the sum of CPT sizes.

C1C3C4

0.50.50.5 0.50.5

A C D E

Figure S14.3 A Bayesian network corresponding to a SAT problem.

14.8 Consider a SAT problem such as the following:

The idea is to encode this as a Bayes net, such that doing inference in the Bayes net gives the

answer to the SAT problem.

a. Figure S14.3 shows the Bayes net corresponding to this SAT problem. The general

construction method is as follows:

The root nodes correspond to the logical variables of the SAT problem. They have

a prior probability of 0.5.

Each clause is a node. Its parents are the variables in the clause. The CPT is

deterministic and implements the disjunction given in the clause. (Negative literals

in the clause are indicated by negation symbols on the links in the ﬁgure.)

A single sentence node has all the clauses as parents and a CPT that implements

deterministic conjunction.

It is clear that iff the SAT problem is satisﬁable. Hence, we have reduced

SAT to Bayes net inference. Since SAT is NP-complete, we have shown that Bayes net

inference is NP-hard (even without evidence).

b. The prior probability of each complete assignment is . is therefore

where is the number of satisfying assignments. Hence, we can count the number

of satisfying assignments by computing . This amounts to a reduction of the

problem of counting satisfying assignments to Bayes net inference; since the former is

#P-complete, the latter is #P-hard.

14.9

a. To calculate the cumulative distribution of a discrete variable, we start from a vector

representation of the original distribution and a vector of the same dimension.

98 Chapter 14. Probabilistic Reasoning

Then, we loop through , adding up the values as we go along and setting to the

running sum, . To sample from the distribution, we generate a random number

uniformly in , and then return for the smallest such that . A naive

way to ﬁnd this is to loop through starting at 1 until . This takes time. A

more efﬁcient solution is binary search: start with the full range , choose at the

midpoint of the range. If , set the range from to the upper bound, otherwise set

the range from the lower bound to . After iterations, we terminate when the

bounds are identical or differ by 1.

b. If we are generating samples, we can afford to preprocess the cumulative

distribution. The basic insight required is that if the original distribution were uniform,

it would be possible to sample in time by returning . That is, we can index

directly into the correct part of the range (analog random access, one might say) instead

of searching for it. Now, suppose we divide the range into equal parts and

construct a -element vector, each of whose entries is a list of all those for which

is in the corresponding part of the range. The we want is in the list with index

. We retrieve this list in time and search through it in order (as in the naive

implementation). Let be the number of elements in list . Then the expected runtime

is given by

The variance of the runtime can be reduced by further subdividing any part of the range

whose list contains more than some small constant number of elements.

c. One way to generate a sample from a univariate Gaussian is to compute the discretized

cumulative distribution (e.g., integrating by Taylor’s rule) and use the algorithm de-

scribed above. We can compute the table once and for all for the standard Gaussian

(mean 0, variance 1) and then scale each sampled value to . If we had a

closed-form, invertible expression for the cumulative distribution , we could sam-

ple exactly, simply by returning . Unfortunately the Gaussian density is not

exactly integrable. Now, the density is exactly integrable, and there are cute

schemes for using two samples and this density to obtain an exact Gaussian sample. We

leave the details to the interested instructor.

d. When querying a continuous variable using Monte carlo inference, an exact closed-form

posterior cannot be obtained. Instead, one typically deﬁnes discrete ranges, returning

a histogram distribution simply by counting the (weighted) number of samples in each

range.

14.10 These proofs are tricky for those not accustomed to manipulating probability expres-

sions, and students may require some hints.

a. There are several ways to prove this. Probably the simplest is to work directly from the

global semantics. First, we rewrite the required probability in terms of the full joint:

Now, all terms in the product in the denominator that do not contain can be moved

outside the summation, and then cancel with the corresponding terms in the numerator.

This just leaves us with the terms that do mention , i.e., those in which is a child

or a parent. Hence, is equal to

Now, by reversing the argument in part (b), we obtain the desired result.

b. This is a relatively straightforward application of Bayes’ rule. Let Ybe the

children of and let Zbe the parents of other than . Then we have

P Y Z Z

P Z Z P Y Z Z

P P Y Z Z

where the derivation of the third line from the second relies on the fact that a node is

independent of its nondescendants given its children.

14.11

a. There are two uninstantiated Boolean variables ( and ) and therefore four

possible states.

b. First, we compute the sampling distribution for each variable, conditioned on its Markov

blanket.

PP P P

P P P P

P P P

Strictly speaking, the transition matrix is only well-deﬁned for the variant of MCMC in

which the variable to be sampled is chosen randomly. (In the variant where the variables

are chosen in a ﬁxed order, the transition probabilities depend on where we are in the

ordering.) Now consider the transition matrix.

100 Chapter 14. Probabilistic Reasoning

Entries on the diagonal correspond to self-loops. Such transitions can occur by

sampling either variable. For example,

Entries where one variable is changed must sample that variable. For example,

Entries where both variables change cannot occur. For example,

This gives us the following transition matrix, where the transition is from the state given

by the row label to the state given by the column label:

c.Qrepresents the probability of going from each state to each state in two steps.

d.Q(as ) represents the long-term probability of being in each state starting in

each state; for ergodic Qthese probabilities are independent of the starting state, so

every row of Qis the same and represents the posterior distribution over states given

the evidence.

e. We can produce very large powers of Qwith very few matrix multiplications. For

example, we can get Qwith one multiplication, Qwith two, and Qwith . Unfor-

tunately, in a network with Boolean variables, the matrix is of size , so each

multiplication takes operations.

14.12

a. The classes are , with instances , , and , and , with instances ,

, and . Each team has a quality and each match has a and and

an . The team names for each match are of course ﬁxed in advance. The prior

over quality could be uniform and the probability of a win for team 1 should increase

with .

b. The random variables are , , , , , and .

The network is shown in Figure S14.4.

c. The exact result will depend on the probabilities used in the model. With any prior on

quality that is the same across all teams, we expect that the posterior over

will show that is more likely to win than .

d. The inference cost in such a model will be because all the team qualities become

coupled.

e. MCMC appears to do well on this problem, provided the probabilities are not too

skewed. Our results show scaling behavior that is roughly linear in the number of

teams, although we did not investigate very large .

101

A.Q B.Q C.Q

AB.Outcome BC.Outcome CA.Outcome

Figure S14.4 A Bayesian network corresponding to the soccer model of Exercise 14.12.

Solutions for Chapter 15

Probabilistic Reasoning over Time

15.1 For each variable that appears as a parent of a variable , deﬁne an auxiliary

variable , such that is parent of and is a parent of . This gives us

a ﬁrst-order Markov model. To ensure that the joint distribution over the original variables

is unchanged, we keep the CPT for is unchanged except for the new parent name, and

we require that Pis an identity mapping, i.e., the child has the same value as the

parent with probability 1. Since the parameters in this model are ﬁxed and known, there is no

effective increase in the number of free parameters in the model.

15.2

a. For all , we the ﬁltering formula

P P P

At the ﬁxed point, we additionally expect that P P . Let the

ﬁxed-point probabilities be . This provides us with a system of linear equa-

tions:

Solving this system, we ﬁnd that .

b. The probability converges to as illustrated in Figure S15.1. This convergence

makes sense if we consider a ﬁxed-point equation for P:

That is, .

Notice that the ﬁxed point does not depend on the initial evidence.

15.3 This exercise develops the Island algorithm for smoothing in DBNs (?).

102

103

0 2 4 6 8 10 12 14 16 18 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Figure S15.1 A graph of the probability of rain as a function of time, forecast into the

future.

a. The chapter shows that P X e can be computed as

P X e P X e P e X f b

The forward recursion (Equation 15.3) shows that fcan be computed from fand

e, which can in turn be computed from fand e, and so on down to fand

e. Hence, fcan be computed from fand e. The backward recursion (Equation

15.7) shows that bcan be computed from band e, which in turn can be

computed from band e, and so on up to band e. Hence, bcan be

computed from band e. Combining these two, we ﬁnd that P X e can

be computed from f,b, and e.

b. The reasoning for the second half is essentially identical: for between and ,

P X e can be computed from f,b, and e.

c. The algorithm takes 3 arguments: an evidence sequence, an initial forward message,

and a ﬁnal backward message. The forward message is propagated to the halfway point

and the backward message is propagated backward. The algorithm then calls itself

recursively on the two halves of the evidence sequence with the appropriate forward

and backward messages. The base case is a sequence of length 1 or 2.

d. At each level of the recursion the algorithm traverses the entire sequence, doing

work. There are levels, so the total time is . The algorithm does

a depth-ﬁrst recursion, so the total space is proportional to the depth of the stack,

i.e., . With islands, the recursion depth is , so the total time is

but the space is .

104 Chapter 15. Probabilistic Reasoning over Time

15.4 This is a very good exercise for deepening intuitions about temporal probabilistic rea-

soning. First, notice that the impossibility of the sequence of most likely states cannot come

from an impossible observation because the smoothed probability at each time step includes

the evidence likelihood at that time step as a factor. Hence, the impossibility of a sequence

must arise from an impossible transition. Now consider such a transition from to

for some , , . For to be the most likely state at time , even though

it cannot be reached from the most likely state at time , we can simply have an -state system

where, say, the smoothed probability of is and the remaining states

have probability . The remaining states all transition deterministically to .

From here, it is a simple matter to work out a speciﬁc model that behaves as desired.

15.5

a. Looking at the fragment of the model containing just , X, and X, we have

P X xx P x

From the properties of the Kalman ﬁlter, we know that the integral gives a Gaussian

for each different value of . Hence, the prediction distribution is a mixture of

Gaussians, each weighted by .

b. The update equation for the switching Kalman ﬁlter is

P X e

P e X xP x e P X x

P e X e P xP x e P X x

We are given that P x e is a mixture of Gaussians. Each Gaussian is subject to

different linear–Gaussian projections and then updated by a linear-Gaussian observa-

tion, so we obtain a sum of Gaussians. Thus, after steps we have Gaussians.

c. Each weight represents the probability of one of the sequences of values for the

switching variable.

15.6 This is a simple exercise in algebra. We have

105

0 2 4 6 8 10 12

Variance of posterior distribution

Number of time steps

sx2=0.1, sz2=0.1

sx2=0.1, sz2=1.0

sx2=0.1, sz2=10.0

sx2=1.0, sz2=0.1

sx2=1.0, sz2=1.0

sx2=1.0, sz2=10.0

sx2=10.0, sz2=0.1

sx2=10.0, sz2=1.0

sx2=10.0, sz2=10.0

Figure S15.2 Graph for Ex. 15.7, showing the posterior variance as a function of for

various values of and .

15.7

a. See Figure S15.2.

b. We can ﬁnd a ﬁxed point by solving

for . Using the quadratic formula and requiring , we obtain

We omit the proof of convergence, which, presumably, can be done by showing that the

update is a contraction (i.e., after updating, two different starting points for become

closer).

c. As , we see that the ﬁxed point also. This is because implies

a deterministic path for the object. Each observation supplies more information about

this path, until its parameters are known completely.

106 Chapter 15. Probabilistic Reasoning over Time

As , the variance update gives immediately. That is, if we have an

exact observation of the object’s state, then the posterior is a delta function about that

observed value regardless of the transition variance.

15.8 There is one class, . Each element of the class has a attribute with

possible values and an attribute which may be discrete or continuous. The parent

of the attribute is the attribute of the same time step. Each time step has

relation to another time step, and the parent of the attribute is the

attribute of the time step.

15.9

a. The curve of interest is the one for . In the absence

of any useful sensor information from the battery meter, the posterior distribution for

the battery level is the same as the projection without evidence. The transition model

for the battery includes a small probability for downward transitions in the battery level

at each time step, but zero probability for upward transitions (there are no recharging

actions in the model). Thus, the stationary distribution towards which the battery level

tends has value 0 with probability 1. The curve for

will asymptote to 0.

b. See Figure S15.3. The CPT for has a probability of transient failure (i.e.,

reporting 0) that increases with temperature.

c. The agent can obviously calculate the posterior distribution over by ﬁltering

the observation sequence in the usual way. This posterior can be informative if the

effect of temperature on transient failure is non-negligible and transient failures occur

more frequently than do major changes in temperature. Essentially, the temperature is

estimated from the frequency of “blips” in the sequence of battery meter readings.

15.10 The process works exactly as on page 507. We start with the full expression:

P P P

Whichever order we push in the summations, the variable elimination process never creates

factors containing more than two variables, which is the same size as the CPTs in the original

network. In fact, given an HMM sequence of arbitrary length, we can eliminate the state

variables in any order.

15.11 The model is shown in Figure S15.4. The key point to note is that any phone variation

at one point (here, [aa] vs. [ey] in the fourth phone) results in variation at three points because

of the effects on the preceding and succeeding phones.

15.12 The [C1,C2] and [C6,C7] must come from the onset and end, respectively. The C3

could come from 2 sources, the onset or the mid, and the [C4, C4] combination could come

from 3 sources: mid-mid, mid-end, or end-end. You can’t go directly from onset to end, so

the only possible paths (along with their path and output probabilities) are as follows:

107

Battery

Battery 0

BMeter

BMBroken 1

BMBroken

Temp 1

Temp

Figure S15.3 Modiﬁcation of Figure 15.13(a) to include the effect of external temperature

on the battery meter.

t(_,ow) ow(t,m)

m(ow.ey)

m(ow,aa)

ey(m,t)

aa(m,t)

t(ey,ow)

ow(t,_)

t(aa,ow)

Figure S15.4 Transition diagram for the triphone pronunciation model in Ex. 15.11.

C1 C2 C3 C4 C4 C6 C7

OOOMMEE (* .3 .3 .7 .9 .1 .4 .6 .5 .2 .3 .7 .7 .5 .4) = 4.0e-6

OOOMEEE (* .3 .3 .7 .1 .4 .4 .6 .5 .2 .3 .7 .1 .5 .4) = 2.5e-7

OOMMMEE (* .3 .7 .9 .9 .1 .4 .6 .5 .2 .2 .7 .7 .5 .4) = 8.0e-6

OOMMEEE (* .3 .7 .9 .1 .4 .4 .6 .5 .2 .2 .7 .1 .5 .4) = 5.1e-7

OOMEEEE (* .3 .7 .1 .4 .4 .4 .6 .5 .2 .2 .1 .1 .5 .4) = 3.2e-8

So the most probable path is OOMMMEE (that is, onset-onset-mid-mid-mid-end-end), with

a probability of . The total probability of the sequence is the sum of the ﬁve paths,

or .

Solutions for Chapter 16

Making Simple Decisions

16.1 It is interesting to create a histogram of accuracy on this task for the students in the

class. It is also interesting to record how many times each student comes within, say, 10% of

the right answer. Then you get a proﬁle of each student: this one is an accurate guesser but

overly cautious about bounds, etc.

16.2 The expected monetary value of the lottery is

Although , it is not necessarily irrational to buy the ticket. First we will consider

just the utilities of the monetary outcomes, ignoring the utility of actually playing the lottery

game. Using to represent the utility to the agent of having dollars more than the

current state, and assuming that utility is linear for small values of money (i.e.,

for ), the utility of the lottery is:

This is more than when . Thus, for a purchase

to be rational (when only money is considered), the agent must be quite risk-seeking. This

would be unusual for low-income individuals, for whom the price of a ticket is non-trivial. It

is possible that some buyers do not internalize the magnitude of the very low probability of

winning—to imagine an event is to assign it a “non-trivial” probability, in effect. Apparently,

these buyers are better at internalizing the large magnitude of the prize. Such buyers are

clearly acting irrationally.

Some people may feel their current situation is intolerable, that is,

. Therefore the situation of having one dollar more or less would be equally intolerable,

and it would be rational to gamble on a high payoff, even if one that has low probability.

Gamblers also derive pleasure from the excitement of the lottery and the temporary

possession of at least a non-zero chance of wealth. So we should add to the utility of playing

the lottery the term to represent the thrill of participation. Seen this way, the lottery is just

another form of entertainment, and buying a lottery ticket is no more irrational than buying

108

109

a movie ticket. Either way, you pay your money, you get a small thrill , and (most likely)

you walk away empty-handed. (Note that it could be argued that doing this kind of decision-

theoretic computation decreases the value of . It is not clear if this is a good thing or a bad

thing.)

16.3

a. The probability that the ﬁrst heads appears on the th toss is , so

b. Typical answers range between $4 and $100.

c. Assume initial wealth (after paying to play the game) of ; then

Assume for simplicity. Then

d. The maximum amount is given by the solution of

For our simple case, we have

or .

16.4 This is an interesting exercise to do in class. Choose , = $100, $1000,

$10000, $1000000. Ask for a show of hands of those preferring the lottery at different values

of . Students will almost always display risk aversion, but there may be a wide spread in its

onset. A curve can be plotted for the class by ﬁnding the smallest yielding a majority vote

for the lottery.

16.5 The program itself is pretty trivial. But note that there are some studies showing you

get better answers if you ask subjects to move a slider to indicate a proportion, rather than

asking for a probability number. So having a graphical user interface is an advantage. The

main point of the exercise is to examine the data, expose inconsistent behavior on the part of

the subjects, and see how people vary in their choices.

16.6 The protocol would be to ask a series of questions of the form “which would you

prefer” involving a monetary gain (or loss) versus an increase (or decrease) in a risk of death.

110 Chapter 16. Making Simple Decisions

For example, “would you pay $100 for a helmet that would eliminate completely the one-in-

a-million chance of death from a bicycle accident.”

16.7 The complete proof is given by Keeney and Raiffa (1976).

16.8 This exercise can be solved using an inﬂuence diagram package such as IDEAL. The

speciﬁc values are not especially important. Notice how the tedium of encoding all the entries

in the utility table cries out for a system that allows the additive, multiplicative, and other

forms sanctioned by MAUT.

One of the key aspects of the fully explicit representation in Figure 16.5 is its amenabil-

ity to change. By doing this exercise as well as Exercise 16.9, students will augment their

appreciation of the ﬂexibility afforded by declarative representations, which can otherwise

seem tedious.

a. For this part, one could use symbolic values (high, medium, low) for all the variables

and not worry too much about the exact probability values, or one could use actual

numerical ranges and try to assess the probabilities based on some knowledge of the

domain. Even with three-valued variables, the cost CPT has 54 entries.

b. This part almost certainly should be done using a software package.

c. If each aircraft generates half as much noise, we need to adjust the entries in the

CPT.

d. If the noise attribute becomes three times more important, the utility table entries must

all be altered. If an appropriate (e.g., additive) representation is available, then one

would only need to adjust the appropriate constants to reﬂect the change.

e. This part should be done using a software package. Some packages may offer VPI

calculation already. Alternatively, one can invoke the decision-making package repeat-

edly to do all the what-if calculations of best actions and their utilities, as required in

the VPI formula. Finally, one can write general-purpose VPI code as an add-on to a

decision-making package.

16.9 The information associated with the utility node in Figure 16.6 is an action-value table,

and can be constructed simply by averaging out the , , and nodes in Fig-

ure 16.5. As explained in the text , modiﬁcations to aircraft noise levels or to the importance

of noise do not result in simple changes to the action-value table. Probably the easiest way

to do it is to go back to the original table in Figure 16.5. The exercise therefore illustrates the

tradeoffs involved in using compiled representations.

16.10 The answer to this exercise depends on the probability values chosen by the stu-

dent.

16.11 This question is a simple exercise in sequential decision making, and helps in making

the transition to Chapter 17. It also emphasizes the point that the value of information is

computed by examining the conditional plan formed by determining the best action for each

possible outcome of the test. It may be useful to introduce “decision trees” (as the term is

used in the decision analysis literature) to organize the information in this question. (See Pearl

(1988), Chapter 6.) Each part of the question analyzes some aspect of the tree. Incidentally,

111

Test

Buy U

Outcome

Quality

Figure S16.1 A decision network for the car-buying problem.

the question assumes that utility and monetary value coincide, and ignores the transaction

costs involved in buying and selling.

a. The decision network is shown in Figure S16.1.

b. The expected net gain in dollars is

c. The question could probably have been stated better: Bayes’ theorem is used to compute

, etc., whereas conditionalization is sufﬁcient to compute .

Using Bayes’ theorem:

d. If the car passes the test, the expected value of buying is

Thus buying is the best decision given a pass. If the car fails the test, the expected value

of buying is

Buying is again the best decision.

112 Chapter 16. Making Simple Decisions

e. Since the action is the same for both outcomes of the test, the test itself is worthless (if

it is the only possible test) and the optimal plan is simply to buy the car without the test.

(This is a trivial conditional plan.) For the test to be worthwhile, it would need to be

more discriminating in order to reduce the probability . The test would

also be worthwhile if the market value of the car were less, or if the cost of repairs were

more.

An interesting additional exercise is to prove the general proposition that if is the

best action for all the outcomes of a test then it must be the best action in the absence

of the test outcome.

16.12 Intuitively, the value of information is nonnegative because in the worst case one

could simply ignore the information and act as if it was not available. A formal proof therefore

begins by showing that this policy results in the same expected utility. The formula for the

value of information is

If the agent does given the information , its expected utility is

where the equality holds because the LHS is just the conditionalization of the RHS with

respect to . By deﬁnition,

hence .

16.13 This is relatively straightforward in the AIMA2e code release. We need to add node

types for action nodes and utility nodes; we need to be able to run standard Bayes net infer-

ence on the network given ﬁxed actions, in order to compute the posterior expected utility; and

we need to write an “outer loop” that can try all possible actions to ﬁnd the best. Given this,

adding VPI calculation is straightforward, as described in the answer to Exercise 16.8.

Solutions for Chapter 17

Making Complex Decisions

17.1 This question helps to bring home the difference between deterministic and stochastic

environments. Here, even a short sequence spreads the agent all over the place. The easiest

way to answer the question systematically is to draw a tree showing the states reached after

each step and the transition probabilities. Then the probability of reaching each leaf is the

product of the probabilities along the path, because the transition probabilities are Markovian.

If the same state appears at more than one leaf, the probabilities of the leaves are summed

because the events corresponding to the two paths are disjoint. (It is always a good idea to

ensure that students justify these kinds of probabilistic calculations, especially since some-

times the “naive” approach gets the wrong answer.) The states and probabilities are: (3,1),

0.01; (3,2), 0.08; (3,3), 0.09; (4,2), 0.18; (4,3), 0.64.

17.2 Stationarity requires the agent to have identical preferences between the sequence pair

and between the sequence pair . If the

utility of a sequence is its maximum reward, we can easily violate stationarity. For example,

but

We can still deﬁne as the expected maximum reward obtained by executing starting

in . The agent’s preferences seem peculiar, nonetheless. For example, if the current state

has reward max, the agent will be indifferent among all actions, but once the action is

executed and the agent is no longer in , it will suddenly start to care about what happens

next.

17.3 A ﬁnite search problem (see Chapter 3) is deﬁned by an initial state , a successor

function that returns a set of action–state pairs, a step cost function , and a goal

test. An optimal solution is a least-cost path from to any goal state.

To construct the corresponding MDP, deﬁne unless is a

goal state, in which case (see Ex. 17.5 for how to obtain the effect of a three-

argument reward function); deﬁne if ; and . An optimal

solution to this MDP is a policy that follows the least-cost path from each state to its nearest

goal state.

17.4

113

114 Chapter 17. Making Complex Decisions

a. Intuitively, the agent wants to get to state 3 as soon as possible, because it will pay a

cost for each time step it spends in states 1 and 2. However, the only action that reaches

state 3 (action ) succeeds with low probability, so the agent should minimize the cost

it incurs while trying to reach the terminal state. This suggests that the agent should

deﬁnitely try action in state 1; in state 2, it might be better to try action to get to state

1 (which is the better place to wait for admission to state 3), rather than aiming directly

for state 3. The decision in state 2 involves a numerical tradeoff.

b. The application of policy iteration proceeds in alternating steps of value determination

and policy update.

Initialization: , .

Value determination:

That is, and .

Policy update: In state 1,

while

so action is still preferred for state 1.

In state 2,

while

so action is preferred for state 1. We set and proceed.

Value determination:

Once more ; now, . Policy update: In state 1,

while

115

so action is still preferred for state 1.

In state 2,

while

so action is still preferred for state 1. remains , and we termi-

nate.

Note that the resulting policy matches our intuition: when in state 2, try to move to state

1, and when in state 1, try to move to state 3.

c. An initial policy with action in both states leads to an unsolvable problem. The initial

value determination problem has the form

and the ﬁrst two equations are inconsistent. If we were to try to solve them iteratively,

we would ﬁnd the values tending to .

Discounting leads to well-deﬁned solutions by bounding the penalty (expected dis-

counted cost) an agent can incur at either state. However, the choice of discount factor

will affect the policy that results. For small, the cost incurred in the distant future

plays a negligible role in the value computation, because is near 0. As a result,

an agent could choose action in state 2 because the discounted short-term cost of re-

maining in the non-terminal states (states 1 and 2) outweighs the discounted long-term

cost of action failing repeatedly and leaving the agent in state 2. An additional exer-

cise could ask the student to determine the value of at which the agent is indifferent

between the two choices.

17.5 This is a deceptively simple exercise that tests the student’s understanding of the for-

mal deﬁnition of MDPs. Some students may need a hint or an example to get started.

a. The key here is to get the max and summation in the right place. For we have

and for we have

b. There are a variety of solutions here. One is to create a “pre-state” for

every , , , such that executing in leads not to but to . In this state

is encoded the fact that the agent came from and did to get here. From the pre-state,

116 Chapter 17. Making Complex Decisions

there is just one action that always leads to . Let the new MDP have transition ,

reward , and discount . Then

c. In keeping with the idea of part (b), we can create states for every , , such

that

17.6 The framework for this problem is in "uncertainty/domains/4x3-mdp.lisp".

There is still some synthesis for the student to do for answer b. For c. some experimental de-

sign is necessary.

17.7 This can be done fairly simply by:

Call policy-iteration (from "uncertainty/algorithms/dp.lisp")on

the Markov Decision Processes representing the 4x3 grid, with values for the step cost

ranging from, say, 0.0 to 1.0 in increments of 0.02.

For any two adjacent policies that differ, run binary search on the step cost to pinpoint

the threshold value.

Convince yourself that you haven’t missed any policies, either by using too coarse an

increment in step size (0.02), or by stopping too soon (1.0).

One useful observation in this context is that the expected total reward of any ﬁxed

policy is linear in , the per-step reward for the empty states. Imagine drawing the total reward

of a policy as a function of —a straight line. Now draw all the straight lines corresponding

to all possible policies. The reward of the optimal policy as a function of is just the max of

all these straight lines. Therefore it is a piecewise linear, convex function of . Hence there

is a very efﬁcient way to ﬁnd all the optimal policy regions:

For any two consecutive values of that have different optimal policies, ﬁnd the optimal

policy for the midpoint. Once two consecutive values of give the same policy, then

the interval between the two points must be covered by that policy.

Repeat this until two points are known for each distinct optimal policy.

Suppose and are points for policy , and and

are the next two points, for policy . Clearly, we can draw straight lines through these

pairs of points and ﬁnd their intersection. This does not mean, however, that there is no

other optimal policy for the intervening region. We can determine this by calculating

117

the optimal policy for the intersection point. If we get a different policy, we continue

the process.

The policies and boundaries derived from this procedure are shown in Figure S17.1. The

ﬁgure shows that there are nine distinct optimal policies! Notice that as becomes more

negative, the agent becomes more willing to dive straight into the –1 terminal state rather

than face the cost of the detour to the +1 state.

The somewhat ugly code is as follows. Notice that because the lines for neighboring

policies are very nearly parallel, numerical instability is a serious problem.

(defun policy-surface (mdp r1 r2 &aux prev (unchanged nil))

"returns points on the piecewise-linear surface

defined by the value of the optimal policy of mdp as a

function of r"

(setq rvplist

(list (cons r1 (r-policy mdp r1)) (cons r2 (r-policy mdp r2))))

(do ()

(unchanged rvplist)

(setq unchanged t)

(setq prev nil)

(dolist (rvp rvplist)

(let* ((rest (cdr (member rvp rvplist :test #’eq)))

(next (first rest))

(next-but-one (second rest)))

(dprint (list (first prev) (first rvp)

’* (first next) (first next-but-one)))

(when next

(unless (or (= (first rvp) (first next))

(policy-equal (third rvp) (third next) mdp))

(dprint "Adding new point(s)")

(setq unchanged nil)

(if (and prev next-but-one

(policy-equal (third prev) (third rvp) mdp)

(policy-equal (third next) (third next-but-one) mdp))

(let* ((intxy (policy-vertex prev rvp next next-but-one))

(int (cons (xy-x intxy) (r-policy mdp (xy-x intxy)))))

(dprint (list "Found intersection" intxy))

(cond ((or (< (first int) (first rvp))

(> (first int) (first next)))

(dprint "Intersection out of range!")

(let ((int-r (/ (+ (first rvp) (first next)) 2)))

(setq int (cons int-r (r-policy mdp int-r))))

(push int (cdr (member rvp rvplist :test #’eq))))

((or (policy-equal (third rvp) (third int) mdp)

(policy-equal (third next) (third int) mdp))

(dprint "Found policy boundary")

(push (list (first int) (second int) (third next))

(cdr (member rvp rvplist :test #’eq)))

(push (list (first int) (second int) (third rvp))

(cdr (member rvp rvplist :test #’eq))))

(t (dprint "Found new policy!")

(push int (cdr (member rvp rvplist :test #’eq))))))

118 Chapter 17. Making Complex Decisions

(let* ((int-r (/ (+ (first rvp) (first next)) 2))

(int (cons int-r (r-policy mdp int-r))))

(dprint (list "Adding split point" (list int-r (second int))))

(push int (cdr (member rvp rvplist :test #’eq))))))))

(setq prev rvp))))

(defun r-policy (mdp r &aux U)

(set-rewards mdp r)

(setq U (value-iteration mdp

(copy-hash-table (mdp-rewards mdp) #’identity)

:epsilon 0.0000000001))

(list (gethash ’(1 1) U) (optimal-policy U (mdp-model mdp) (mdp-rewards mdp))))

(defun set-rewards (mdp r &aux (rewards (mdp-rewards mdp))

(terminals (mdp-terminal-states mdp)))

(maphash #’(lambda (state reward)

(unless (member state terminals :test #’equal)

(setf (gethash state rewards) r)))

rewards))

(defun policy-equal (p1 p2 mdp &aux (match t)

(terminals (mdp-terminal-states mdp)))

(maphash #’(lambda (state action)

(unless (member state terminals :test #’equal)

(unless (eq (caar (gethash state p1)) (caar (gethash state p2)))

(setq match nil))))

p1)

match)

(defun policy-vertex (rvp1 rvp2 rvp3 rvp4)

(let ((a (make-xy :x (first rvp1) :y (second rvp1)))

(b (make-xy :x (first rvp2) :y (second rvp2)))

(c (make-xy :x (first rvp3) :y (second rvp3)))

(d (make-xy :x (first rvp4) :y (second rvp4))))

(intersection-point (make-line :xy1 a :xy2 b)

(make-line :xy1 c :xy2 d))))

(defun intersection-point (l1 l2)

;;; l1 is line ab; l2 is line cd

;;; assume the lines cross at alpha a + (1-alpha) b,

;;; also known as beta c + (1-beta) d

;;; returns the intersection point unless they’re parallel

(let* ((a (line-xy1 l1))

(b (line-xy2 l1))

(c (line-xy1 l2))

(d (line-xy2 l2))

(xa (xy-x a)) (ya (xy-y a))

(xb (xy-x b)) (yb (xy-y b))

(xc (xy-x c)) (yc (xy-y c))

(xd (xy-x d)) (yd (xy-y d))

(q (- (* (- xa xb) (- yc yd))

(* (- ya yb) (- xc xd)))))

119

(unless (zerop q)

(let ((alpha (/ (- (* (- xd xb) (- yc yd))

(* (- yd yb) (- xc xd)))

q)))

(make-xy :x (float (+ (* alpha xa) (* (- 1 alpha) xb)))

:y (float (+ (* alpha ya) (* (- 1 alpha) yb))))))))

− 1

+ 1

− 1

+ 1

− 1

+ 1

− 1

+ 1

− 1

+ 1

− 1

+ 1

− 1

+ 1

− 1

+ 1

− 1

+ 1

r = [−1.6284 : −1.3702] r = [−1.3702 : −0.7083]

r = [−0.7083 : −0.4278] r = [−0.4278 : −0.0850] r = [−0.0850 : −0.0480]

r = [−0.0480 : −0.0274] r = [−0.0274 : −0.0218] r = [−0.0218 : 0.0000]

r = [− : −1.6284]

Figure S17.1 Optimal policies for different values of the cost of a step in the envi-

ronment, and the boundaries of the regions with constant optimal policy.

17.8

a. For we have

and for we have

120 Chapter 17. Making Complex Decisions

(1,4)

(1,3)

(1,2)

(2,4)

(2,3)

(2,1)

(3,4)

(3,2)

(3,1)

(4,3)

(4,2)

−1 −1

Figure S17.2 State-space graph for the game on page 190 (for Ex. 17.8).

b. To do value iteration, we simply turn each equation from part (a) into a Bellman update

and apply them in alternation, applying each to all states simultaneously. The process

terminates when the utility vector for one player is the same as the previous utility

vector for the same player (i.e., two steps earlier). (Note that typically and are

not the same in equilibrium.)

c. The state space is shown in Figure S17.2.

d. We mark the terminal state values in bold and initialize other values to 0. Value iteration

proceeds as follows:

(1,4) (2,4) (3,4) (1,3) (2,3) (4,3) (1,2) (3,2) (4,2) (2,1) (3,1)

00000+1 0 0 +1 –1 –1

0 0 0 0 –1 +1 0 –1 +1 –1 –1

0 0 0 –1 +1 +1 –1 +1 +1 –1 –1

–1 +1 +1 –1 –1 +1 –1 –1 +1 –1 –1

+1 +1 +1 –1 +1 +1 –1 +1 +1 –1 –1

–1 +1 +1 –1 –1 +1 –1 –1 +1 –1 –1

and the optimal policy for each player is as follows:

(1,4) (2,4) (3,4) (1,3) (2,3) (4,3) (1,2) (3,2) (4,2) (2,1) (3,1)

(2,4) (3,4) (2,4) (2,3) (4,3) (3,2) (4,2)

(1,3) (2,3) (3,2) (1,2) (2,1) (1,3) (3,1)

17.9 This question is simple a matter of examining the deﬁnitions. In a dominant strategy

equilibrium , it is the case that for every player , is optimal for every combi-

nation by the other players:

In a Nash equilibrium, we simply require that is optimal for the particular current combi-

nation by the other players:

Therefore, dominant strategy equilibrium is a special case of Nash equilibrium. The converse

does not hold, as we can show simply by pointing to the CD/DVD game, where neither of the

Nash equilibria is a dominant strategy equilibrium.

121

17.10 In the following table, the rows are labelled by A’s move and the columns by B’s

move, and the table entries list the payoffs to A and B respectively.

R P S F W

R 0,0 –1,1 1,-1 –1,1 1,-1

P 1,-1 0,0 –1,1 –1,1 1,-1

S –1,1 1,-1 0,0 –1,1 1,-1

F 1,-1 1,-1 1,-1 0,0 –1,1

W –1,1 –1,1 –1,1 1,-1 0,0

Suppose chooses a mixed strategy , where

. The payoff to A of B’s possible pure responses are as follows:

It is easy to see that no option is dominated over the whole region. Solving for the intersection

of the hyperplanes, we ﬁnd and . By symmetry, we will ﬁnd the

same solution when chooses a mixed strategy ﬁrst.

17.11 The payoff matrix for three-ﬁnger Morra is as follows:

O: one O: two O: three

E: one

E: two

E: three

Suppose chooses a mixed strategy , where .

The payoff to E of O’s possible pure responses are as follows:

It is easy to see that no option is dominated over the whole region. Solving for the intersection

of the hyperplanes, we ﬁnd . , . The expected value is 0.

17.12 Every game is either a win for one side (and a loss for the other) or a tie. With 2 for a

win, 1 for a tie, and 0 for a loss, 2 points are awarded for every game, so this is a constant-sum

game.

If 1 point is awarded for a loss in overtime, then for some games 3 points are awarded

in all. Therefore, the game is no longer constant-sum.

Suppose we assume that team A has probability of winning in regular time and team

B has probability of winning in regular time (assuming normal play). Furthermore, assume

122 Chapter 17. Making Complex Decisions

team B has a probability of winning in overtime (which occurs if there is a tie after regular

time). Once overtime is reached (by any means), the expected utilities are as follows:

In normal play, the expected utilities are derived from the probability of winning plus the

probability of tying times the expected utility of overtime play:

Hence A has an incentive to agree if , or

or or

and B has an incentive to agree if , or

or or

When both of these inequalities hold, there is an incentive to tie in regulation play. For any

values of and , there will be values of and such that both inequalities hold.

For an in-depth statistical analysis of the actual effects of the rule change and a more

sophisticated treatment of the utility functions, see “Overtime! Rules and Incentives in the

National Hockey League” by Stephen T. Easton and Duane W. Rockerbie, available at

http://people.uleth.ca/˜rockerbie/OVERTIME.PDF.

17.13 We apply iterated strict dominance to ﬁnd the pure strategy. First, Pol: do nothing

dominates Pol: contract, so we drop the Pol: contract row. Next, Fed: contract dominates

Fed: do nothing and Fed: expand on the remaining rows, so we drop those columns. Finally,

Pol: expand dominates Pol: do nothing on the one remaining column. Hence the only Nash

equilibrium is a dominant strategy equilibrium with Pol: expand and Fed: contract. This is

not Pareto optimal: it is worse for both players than the four strategy proﬁles in the top right

quadrant.

Solutions for Chapter 18

Learning from Observations

18.1 The aim here is to couch language learning in the framework of the chapter, not to

solve the problem! This is a very interesting topic for class discussion, raising issues of

nature vs. nurture, the indeterminacy of meaning and reference, and so on. Basic references

include Chomsky (1957) and Quine (1960).

The ﬁrst step is to appreciate the variety of knowledge that goes under the heading

“language.” The infant must learn to recognize and produce speech, learn vocabulary, learn

grammar, learn the semantic and pragmatic interpretation of a speech act, and learn strategies

for disambiguation, among other things. The performance elements for this (in humans) and

their associated learning mechanisms are obviously very complex and as yet little is known

about them.

A naive model of the learning environment considers just the exchange of speech sounds.

In reality, the physical context of each utterance is crucial: a child must see the context in

which “watermelon” is uttered in order to learn to associate “watermelon” with watermel-

ons. Thus, the environment consists not just of other humans but also the physical objects

and events about which discourse takes place. Auditory sensors detect speech sounds, while

other senses (primarily visual) provide information on the physical context. The relevant

effectors are the speech organs and the motor capacities that allow the infant to respond to

speech or that elicit verbal feedback.

The performance standard could simply be the infant’s general utility function, however

that is realized, so that the infant performs reinforcement learning to perform and respond to

speech acts so as to improve its well-being—for example, by obtaining food and attention.

However, humans’ built-in capacity for mimicry suggests that the production of sounds sim-

ilar to those produced by other humans is a goal in itself. The child (once he or she learns to

differentiate sounds and learn about pointing or other means of indicating salient objects) is

also exposed to examples of supervised learning: an adult says “shoe” or “belly button” while

indicating the appropriate object. So sentences produced by adults provide labelled positive

examples, and the response of adults to the infant’s speech acts provides further classiﬁcation

feedback.

Mostly, it seems that adults do not correct the child’s speech, so there are very few neg-

ative classiﬁcations of the child’s attempted sentences. This is signiﬁcant because early work

on language learning (such as the work of Gold, 1967) concentrated just on identifying the

set of strings that are grammatical, assuming a particular grammatical formalism. If there are

123

124 Chapter 18. Learning from Observations

only positive examples, then there is nothing to rule out the grammar . Some

theorists (notably Chomsky and Fodor) used what they call the “poverty of the stimulus” argu-

ment to say that the basic universal grammar of languages must be innate, because otherwise

(given the lack of negative examples) there would be no way that a child could learn a lan-

guage (under the assumptions of language learning as learning a set of grammatical strings).

Critics have called this the “poverty of the imagination” argument—I can’t think of a learning

mechanism that would work, so it must be innate. Indeed, if we go to probabilistic context

free grammars, then it is possible to learn a language without negative examples.

18.2 Learning tennis is much simpler than learning to speak. The requisite skills can be

divided into movement, playing strokes, and strategy. The environment consists of the court,

ball, opponent, and one’s own body. The relevant sensors are the visual system and propri-

oception (the sense of forces on and position of one’s own body parts). The effectors are

the muscles involved in moving to the ball and hitting the stroke. The learning process in-

volves both supervised learning and reinforcement learning. Supervised learning occurs in

acquiring the predictive transition models, e.g., where the opponent will hit the ball, where

the ball will land, and what trajectory the ball will have after one’s own stroke (e.g., if I hit

a half-volley this way, it goes into the net, but if I hit it that way, it clears the net). Rein-

forcement learning occurs when points are won and lost—this is particularly important for

strategic aspects of play such as shot placement and positioning (e.g., in 60% of the points

where I hit a lob in response to a cross-court shot, I end up losing the point). In the early

stages, reinforcement also occurs when a shot succeeds in clearing the net and landing in the

opponent’s court. Achieving this small success is itself a sequential process involving many

motor control commands, and there is no teacher available to tell the learner’s motor cortex

which motor control commands to issue.

18.3 This is a deceptively simple question, designed to point out the plethora of “excep-

tions” in real-world situations and the way in which decision trees capture a hierarchy of

exceptions. One possible tree is shown in Figure S18.1. One can, of course, imagine many

more exceptions. The qualiﬁcation problem, deﬁned originally for action models, applies a

fortiori to condition–action rules.

18.4 In standard decision trees, attribute tests divide examples according to the attribute

value. Therefore any example reaching the second test already has a known value for the

attribute and the second test is redundant. In some decision tree systems, however, all tests

are Boolean even if the attributes are multivalued or continuous. In this case, additional tests

of the attribute can be used to check different values or subdivide the range further (e.g., ﬁrst

check if , and then if it is, check if ).

18.5 The algorithm may not return the “correct” tree, but it will return a tree that is logi-

cally equivalent, assuming that the method for generating examples eventually generates all

possible combinations of input attributes. This is true because any two decision tree deﬁned

on the same set of attributes that agree on all possible examples are, by deﬁnition, logically

equivalent. The actually form of the tree may differ because there are many different ways to

represent the same function. (For example, with two attributes and we can have one tree

125

No Yes

NoYes

Yes

No Yes

FrontOfQueue?

IntersectionBlocked?CarAheadMoving?

No Yes

CrossTraffic?

Pedestrians?

NoTurning?

No RightLeft

OncomingTraffic? Cyclist?

No Yes

NoYes

Figure S18.1 A decision tree for deciding whether to move forward at a trafﬁc intersec-

tion, given a green light.

with at the root and another with at the root.) The root attribute of the original tree may

not in fact be the one that will be chosen by the information gain geuristic when applied to

the training examples.

18.6 This is an easy algorithm to implement. The main point is to have something to test

other learning algorithms against, and to learn the basics of what a learning algorithm is in

terms of inputs and outputs given the framework provided by the code repository.

18.7 If we leave out an example of one class, then the majority of the remaining examples

are of the other class, so the majority classiﬁer will always predict the wrong answer.

18.8 This question brings a little bit of mathematics to bear on the analysis of the learning

problem, preparing the ground for Chapter 20. Error minimization is a basic technique in

both statistics and neural nets. The main thing is to see that the error on a given training

set can be written as a mathematical expression and viewed as a function of the hypothesis

chosen. Here, the hypothesis in question is a single number returned at the leaf.

a. If is returned, the absolute error is

when

This is minimized by setting

126 Chapter 18. Learning from Observations

That is, is the majority value.

b. First calculate the sum of squared errors, and its derivative:

The fact that the second derivative, , is greater than zero means that

is minimized (not maximized) where , i.e., when .

18.9 Suppose that we draw examples. Each example has input features plus its classi-

ﬁcation, so there are distinct input/output examples to choose from. For each example,

there is exactly one contradictory example, namely the example with the same input features

but the opposite classiﬁcation. Thus, the probability of ﬁnding no contradiction is

number of sequences of mnon-contradictory examples

number of sequences of mexamples

For , with 2048 possible examples, a contradiction becomes likely with probability

after 54 examples have been drawn.

18.10 This result emphasizes the fact that any statistical ﬂuctuations caused by the random

sampling process will result in an apparent information gain.

The easy part is showing that the gain is zero when each subset has the same ratio of

positive examples. The gain is deﬁned as

Since and , if for all , then we must have

for all , and also . From this, we obtain

18.11 This is a fairly small, straightforward programming exercise. The only hard part is

the actual computation; you might want to provide your students with a library function

to do this.

18.12 This is another straightforward programming exercise. The follow-up exercise is to

run tests to see if the modiﬁed algorithm actually does better.

127

18.13 Let the prior probabilities of each attribute value be . (These prob-

abilities are estimated by the empirical fractions among the examples at the current node.)

From page 540, the intrinsic information content of the attribute is

Given this formula and the empirical estimates of , the modiﬁcation to the code is

straightforward.

18.14

18.15 Note: this is the only exercise to cover the material in section 18.6. Although the

basic ideas of computational learning theory are both important and elegant, it is not easy to

ﬁnd good exercises that are suitable for an AI class as opposed to a theory class. If you are

teaching a graduate class, or an undergraduate class with a strong emphasis on learning, it

might be a good idea to use some of the exercises from Kearns and Vazirani (1994).

a. If each test is an arbitrary conjunction of literals, then a decision list can represent

an arbitrary DNF (disjunctive normal form) formula directly. The DNF expression

, where is a conjunction of literals, can be represented by a

decision list in which is the th test and returns if successful. That is:

Since any Boolean function can be written as a DNF formula, then any Boolean function

can be represented by a decision list.

b. A decision tree of depth can be translated into a decision list whose tests have at most

literals simply by encoding each path as a test. The test returns the corresponding leaf

value if it succeeds. Since the decision tree has depth , no path contains more than

literals.

Solutions for Chapter 19

Knowledge in Learning

19.1 In CNF, the premises are as follows:

We can prove the desired conclusion directly rather than by refutation. Resolve the ﬁrst two

premises with to obtain

Resolve this with to obtain

which is the desired conclusion .

19.2 This question is tricky in places. It is important to see the distinction between the

shared and unshared variables on the LHS and RHS of the determination. The shared vari-

ables will be instantiated to the objects to be compared in an analogical inference, while the

unshared variables are instantiated with the objects’ observed and inferred properties.

a. Here we are talking about the zip codes and states of houses (or addresses or towns). If

two houses/addresses/towns have the same zip code, they are in the same state:

The determination is true because the US Postal Service never draws zipcode bound-

aries across state lines (perhaps for some reason having to do with interstate commerce).

b. Here the objects being reasoned about are coins, and design, denomination, and mass

are properties of coins. So we have

This is (very nearly exactly) true because coins of a given denomination and design are

stamped from the same original die using the same material; size and shape determine

volume; and volume and material determine mass.

128

129

c. Here we have to be careful. The objects being reasoned about are not programs but

runs of a given program. (This determination is also one often forgotten by novice

programmers.) We can use situation calculus to refer to the runs:

Here the captures the variable so that it does not participate in the determination

as one of the shared or unshared variables. The situation is the shared variable. The

determination expands out to the following Horn clause:

That is, if has the same input in two different situations it will have the same output

in those situations. This is generally true because computers operate on programs and

inputs deterministically; however, it is important that “input” include the entire state of

the computer’s memory, ﬁle system, and so on. Notice that the “naive” choice

expands out to

which says that if any two programs have the same input they produce the same output!

d. Here the objects being reasoned are people in speciﬁc time intervals. (The intervals

could be the same in each case, or different but of the same kind such as days, weeks,

etc. We will stick to the same interval for simplicity. As above, we need to quantify the

interval to “precapture” the variable.) We will use to mean that person

experiences climate in interval , and we will assume for the sake of variety that a

person’s metabolism is constant.

While the determinations seems plausible, it leaves out such factors as water intake,

clothing, disease, etc. The qualiﬁcation problem arises with determinations just as with

implications.

e. Let mean that person has baldness (which might be , ,

or , say). A ﬁrst stab at the determination might be

but this would only allow an inference when two people have the same mother and ma-

ternal grandfather because the and are the unshared variables on the LHS. Also, the

RHS has no unshared variable. Notice that the determination does not say speciﬁcally

that baldness is inherited without modiﬁcation; it allows, for example, for a hypothet-

ical world in which the maternal grandchildren of a bald man are all hairy, or vice

versa. This might not seem particularly natural, but consider other determinations such

as “Whether or not I ﬁle a tax return determines whether or not my spouse must ﬁle a

tax return.”

130 Chapter 19. Knowledge in Learning

The baldness of the maternal grandfather is the relevant value for prediction, so that

should be the unshared variable on the LHS. The mother and maternal grandfather are

designated by skolem functions:

If we use and as function symbols, then the meaning becomes clearer:

Just to check, this expands into

which has the intended meaning.

19.3 Because of the qualiﬁcation problem, it is not usually possible in most real-world

applications to list on the LHS of a determination all the relevant factors that determine

the RHS. Determinations will usually therefore be true to an extent—that is, if two objects

agree on the LHS there is some probability (preferably greater than the prior) that the two

objects will agree on the RHS. An appropriate deﬁnition for probabilistic determinations

simply includes this conditional probability of matching on the RHS given a match on the

LHS. For example, we could deﬁne to mean

that if two people have the same nationality, then there is a 90% chance that they have the

same language.

19.4 This exercise test the student’s understanding of resolution and uniﬁcation, as well as

stressing the nondeterminism of the inverse resolution process. It should help a lot in making

the inverse resolution operation less mysterious and more amenable to mathematical analysis.

It is helpful ﬁrst to draw out the resolution “V” when doing these problems, and then to do a

careful case analysis.

a. There is no possible value for here. The resolution step would have to resolve away

both the on the LHS of and the on the right, which is not possible.

(Resolution can remove more than one literal from a clause, but only if those literals

are redundant—i.e., one subsumes the other.)

b. Without loss of generality, let contain the negative (LHS) literal to be resolved away.

The LHS of therefore contains one literal , while the LHS of must be empty.

The RHS of must contain such that and unify with some uniﬁer . Now we

have a choice: on the RHS of could come from the RHS of or of .

Thus the two basic solution templates are

;

Within these templates, the choice of is entirely unconstrained. Suppose is

and is . Then could be (or or ) and

131

the solutions are

;

c. As before, let contain the negative (LHS) literal to be resolved away, with on the

RHS of . We now have four possible templates because each of the two literals in

could have come from either or :

;

Again, we have a fairly free choice for . However, since contains and , cannot

bind those variables (else they would not appear in ). Thus, if is , then

must be also and will be empty.

19.5 We will assume that Prolog is the logic programming language. It is certainly true that

any solution returned by the call to will be a correct inverse resolvent. Unfortunately,

it is quite possible that the call will fail to return because of Prolog’s depth-ﬁrst search. If

the clauses in and are infelicitously arranged, the proof tree might go down

the branch corresponding to indeﬁnitely nested function symbols in the solution and never

return. This can be alleviated by redesigning the Prolog inference engine so that it works

using breadth-ﬁrst search or iterative deepening, although the inﬁnitely deep branches will

still be a problem. Note that any cuts used in the Prolog program will also be a problem for

the inverse resolution.

19.6 This exercise gives some idea of the rather large branching factor facing top-down ILP

systems.

a. It is important to note that position is signiﬁcant— is very different from

! The ﬁrst argument position can contain one of the ﬁve existing variables

or a new variable. For each of these six choices, the second position can contain one of

the ﬁve existing variables or a new variable, except that the literal with two new vari-

ables is disallowed. Hence there are 35 choices. With negated literals too, the total

branching factor is 70.

b. This seems to be quite a tricky combinatorial problem. The easiest way to solve it

seems to be to start by including the multiple possibilities that are equivalent under

renaming of the new variables as well as those that contain only new variables. Then

these redundant or illegal choices can be removed later. Now, we can use up to

new variables. If we use new variables, we can write literals, so using

exactly variables we can write literals. Each of these

is functionally isomorphic under any renaming of the new variables. With variables,

132 Chapter 19. Knowledge in Learning

there are are renamings. Hence the total number of distinct literals (including those

illegal ones with no old variables) is

Now we just subtract off the number of distinct all-new literals. With new variables,

the number of (not necessarily distinct) all-new literals is , so the number with exactly

is . Each of these has equivalent literals in the set. This gives us

the ﬁnal total for distinct, legal literals:

which can doubtless be simpliﬁed. One can check that for and this gives

35.

c. If a literal contains only new variables, then either a subsequent literal in the clause

body connects one or more of those variables to one or more of the “old” variables,

or it doesn’t. If it does, then the same clause will be generated with those two literals

reversed, such that the restriction is not violated. If it doesn’t, then the literal is either

always true (if the predicate is satisﬁable) or always false (if it is unsatisﬁable), inde-

pendent of the “input” variables in the head. Thus, the literal would either be redundant

or would render the clause body equivalent to .

19.7 FOIL is available on the web at http://www-2.cs.cmu.edu/afs/cs/project/ai-repository-

/ai/areas/learning/systems/foil/0.html (and possibly other places). It is worthwhile to experi-

ment with it.

Solutions for Chapter 20

Statistical Learning Methods

20.1 The code for this exercise is a straightforward implementation of Equations 20.1 and

20.2. Figure S20.1 shows the results for data sequences generated from and . (Plots

0.2

0.4

0.6

0.8

0 20 40 60 80 100

Posterior probability of hypothesis

Number of samples in d

P(h1 | d)

P(h2 | d)

P(h3 | d)

P(h4 | d)

P(h5 | d)

0.4

0.5

0.6

0.7

0.8

0.9

0 20 40 60 80 100

Probability that next candy is lime

Number of samples in d

(a) (b)

0.2

0.4

0.6

0.8

0 20 40 60 80 100

Posterior probability of hypothesis

Number of samples in d

P(h1 | d)

P(h2 | d)

P(h3 | d)

P(h4 | d)

P(h5 | d)

0.4

0.5

0.6

0.7

0.8

0.9

0 20 40 60 80 100

Probability that next candy is lime

Number of samples in d

Figure S20.1 Graphs for Ex. 20.1. (a) Posterior probabilities over a

sample sequence of length 100 generated from (50% cherry + 50% lime). (b) Bayesian

prediction given the data in (a). (c) Posterior probabilities

over a sample sequence of length 100 generated from (25% cherry

+ 75% lime). (d) Bayesian prediction given the data in (c).

133

134 Chapter 20. Statistical Learning Methods

for and are essentially identical to those for and .) Results obtained by students

may vary because the data sequences are generated randomly from the speciﬁed candy dis-

tribution. In (a), the samples very closely reﬂect the true probabilities and the hypotheses

other than are effectively ruled out very quickly. In (c), the early sample proportions are

somewhere between 50/50 and 25/75; furthermore, has a higher prior than . As a result,

and vie for supremacy. Between 50 and 60 samples, a preponderance of limes ensures

the defeat of and the prediction quickly converges to 0.75.

20.2 Typical plots are shown in Figure S20.2. Because both MAP and ML choose exactly

one hypothesis for predictions, the prediction probabilities are all 0.0, 0.25, 0.5, 0.75, or 1.0.

For small data sets the ML prediction in particular shows very large variance.

20.3 This is a nontrivial sequential decision problem, but can be solved using the tools

developed in the book. It leads into general issues of statistical decision theory, stopping

rules, etc. Here, we sketch the “straightforward” solution.

0.2

0.4

0.6

0.8

0 20 40 60 80 100

Probability that next candy is lime

Number of samples in d

0.2

0.4

0.6

0.8

0 20 40 60 80 100

Probability that next candy is lime

Number of samples in d

(a) (b)

0.2

0.4

0.6

0.8

0 20 40 60 80 100

Probability that next candy is lime

Number of samples in d

0.2

0.4

0.6

0.8

0 20 40 60 80 100

Probability that next candy is lime

Number of samples in d

Figure S20.2 Graphs for Ex. 20.2. (a) Prediction from the MAP hypothesis given a sample

sequence of length 100 generated from (50% cherry + 50% lime). (b) Prediction from the

ML hypothesis given the data in (a). (c) Prediction from the MAP hypothesis given data from

(25% cherry + 75% lime). (d) Prediction from the ML hypothesis given the data in (c).

135

We can think of this problem as a simpliﬁed form of POMDP (see Chapter 17). The

“belief states” are deﬁned by the numbers of cherry and lime candies observed so far in the

sampling process. Let these be and , and let be the utility of the corresponding

belief state. In any given state, there are two possible decisions: and . There is a

simple Bellman equation relating and for the sampling case:

Let the posterior probability of each be , the size of the bag be , and the

fraction of cherries in a bag of type be . Then the value obtained by selling is given by

the value of the sampled candies (which Ann gets to keep) plus the price paid by Bob (which

equals the expected utility of the remaining candies for Bob):

and of course we have

Thus we can set up a dynamic program to compute given the obvious boundary conditions

for the case where . The solution of this dynamic program gives the optimal policy

for Ann. It will have the property that if she should sell at , then she should also sell

at for all positive . Thus, the problem is to determine, for each , the threshold

value of at or above which she should sell. A minor complication is that the formula for

should take into account the non-replacement of candies and the ﬁniteness of ,

otherwise odd things will happen when is close to .

20.4 The Bayesian approach would be to take both drugs. The maximum likelihood ap-

proach would be to take the anti- drug. In the case where there are two versions of ,

the Bayesian still recommends taking both drugs, while the maximum likelihood approach is

now to take the anti- drug, since it has a 40% chance of being correct, versus 30% for each

of the cases. This is of course a caricature, and you would be hard-pressed to ﬁnd a doctor,

even a rabid maximum-likelihood advocate who would prescribe like this. But you can ﬁnd

ones who do research like this.

20.5 Boosted naive Bayes learning is discussed by ? (?). The application of boosting to

naive Bayes is straightforward. The naive Bayes learner uses maximum-likelihood parameter

estimation based on counts, so using a weighted training set simply means adding weights

rather than counting. Each naive Bayes model is treated as a deterministic classiﬁer that picks

the most likely class for each example.

20.6 We have

hence the equations for the derivatives at the optimum are

136 Chapter 20. Statistical Learning Methods

and the solutions can be computed as

20.7 There are a couple of ways to solve this problem. Here, we show the indicator vari-

able method described on page 743. Assume we have a child variable with parents

and let the range of each variable be . Let the noisy-OR parameters be

. The noisy-OR model then asserts that

Assume we have complete-data samples with values for and for each . The

conditional log likelihood for is given by

The gradient with respect to each noisy-OR parameter is

20.8

a. By integrating over the range , show that the normalization constant for the dis-

tribution is given by where is the Gamma

function, deﬁned by and . (For integer , .)

137

We will solve this for positive integer and by induction over . Let be

the normalization constant. For the base cases, we have

and

For the inductive step, we assume for all that

Now we evaluate using integration by parts. We have

Hence

as required.

b. The mean is given by the following integral:

c. The mode is found by solving for :

d. tends to very large values close to and ,

i.e., it expresses the prior belief that the distribution characterized by is nearly deter-

ministic (either positively or negatively). After updating with a positive example we

138 Chapter 20. Statistical Learning Methods

obtain the distribution , which has nearly all its mass near (and the

converse for a negative example), i.e., we have learned that the distribution character-

ized by is deterministic in the positive sense. If we see a “counterexample”, e.g., a

positive and a negative example, we obtain , which is close to uniform,

i.e., the hypothesis of near-determinism is abandoned.

20.9 Consider the maximum-likelihood parameter values for the CPT of node in the orig-

inal network, where an extra parent will be added to . If we set the parameters for

in the new network to be identical to in the original

netowrk, regardless of the value , then the likelihood of the data is unchanged. Maxi-

mizing the likelihood by altering the parameters can then only increase the likelihood.

20.10

a. With three attributes, there are seven parameters in the model and the empirical data

give frequencies for classes, which supply 7 independent numbers since the 8

frequencies have to sum to the total sample size. Thus, the problem is neither under-

nor over-constrained. With two attributes, there are ﬁve parameters in the model and

the empirical data give frequencies for classes, which supply 3 independent num-

bers. Thus, the problem is severely underconstrained. There will be a two-dimensional

surface of equally good ML solutions and the original parameters cannot be recovered.

b. The calculation is sketched and partly completed on pages 729 and 730. Completing

it by hand is tedious; students should encouraged to implement the method, ideally in

combination with a bayes net package, and trace its calculations.

c. The question should also assume to simplify the analysis. With every parameter

identical and , the new parameter values for bag 1 will be the same as those for

bag 2, by symmetry. Intuitively, if we assume initially that the bags are identical, then

it is as if we had just one bag. Likelihood is maximized under this constraint by setting

the proportions of candy types within each bag to the observed proportions. (See proof

below.)

d. We begin by writing out for the data in the table:

Now we can differentiate with respect to each parameter. For example,

Now if , , and , the denominators simplify, everything

cancels, and we have

139

First, note that will have the same value except that and are reversed,

so the parameters for bags 1 and 2 will move in lock step if . Second, note that

when , i.e., exactly the observed proportion of

candies with holes. Finally, we can calculate second derivatives and evlauate them at

the ﬁxed point. For example, we obtain

which is negative (indicating the ﬁxed point is a maximum) only when .

Thus, in general the ﬁxed point is a saddle point as some of the second derivatives may

be positive and some negative. Nonetheless, EM can reach it by moving along the ridge

leading to it, as long as the symmetry is unbroken.

20.11 XOR (in fact any Boolean function) is easiest to construct using step-function units.

Because XOR is not linearly separable, we will need a hidden layer. It turns out that just one

hidden node sufﬁces. To design the network, we can think of the XOR function as OR with

the AND case (both inputs on) ruled out. Thus the hidden layer computes AND, while the

output layer computes OR but weights the output of the hidden node negatively. The network

shown in Figure S20.3 does the trick.

t = 0.5 t = 0.2

W = 0.3

W =− 0.6

Figure S20.3 A network of step-function neurons that computes the XOR function.

20.12 The examples map from to coordinates as follows:

(negative) maps to

(positive) maps to

(negative) maps to

Thus, the positive examples have and the negative examples have .

The maximum margin separator is the line , with a margin of 1. The separator

corresponds to the and axes in the original space—this can be thought of as the

limit of a hyperbolic separator with two branches.

20.13 The perceptron adjusts the separating hyperplane deﬁned by the weights so as to

minimize the total error. The question assumes that the perceptron is trained to convergence

(if possible) on the accumulated data set after each new example arrives.

There are two phases. With few examples, the data may remain linearly separable and

the hyperplane will separate the positive and negative examples, although it will not represent

140 Chapter 20. Statistical Learning Methods

the XOR function. Once the data become non-separable, the hyperplane will move around to

ﬁnd a local minimum-error conﬁguration. With parity data, positive and negative examples

are distributed uniformly throughout the input space and in the limit of large sample sizes the

minimum error is obtained by a weight conﬁguration that outputs approximately 0.5 for any

input. However, in this region the error surface is basically ﬂat, and any small ﬂuctuation in

the local balance of positive and negative examples due to the sampling process may cause

the minimum-error plane to move around drastically.

Here is some code to try out a given training set:

(setq *examples*

’(((T . 0) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 0))

((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 1))

((T . 0) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 1))

((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0))

((T . 0) (I1 . 0) (I2 . 0) (I3 . 1) (I4 . 1))

((T . 1) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0))

((T . 0) (I1 . 1) (I2 . 1) (I3 . 0) (I4 . 0))

((T . 1) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1))

((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0))))

(deftest ex20.13

((setq problem

(make-learning-problem

:attributes ’((I1 0 1) (I2 0 1) (I3 0 1) (I4 0 1))

:goals ’((T 0 1))

:examples (subseq *examples* 0 <n>)))) ;;;; vary <n> as needed

;; Normally we’d call PERCEPTRON-LEARNING here, but we need added control

;; to set all the weights to 0.1, so we build the perceptron ourselves:

((setq net (list (list (make-unit :parents (iota 5)

:children nil

:weights ’(-1.0 0.1 0.1 0.1 0.1)

:g #’(lambda (i) (step-function 0 i)))))))

;; Now we call NN-LEARNING with the perceptron-update method,

;; but we also make it print out the weights, and set *debugging* to t

;; so that we can see the epoch number and errors.

((let ((*debugging* t))

(nn-learning

problem net

#’(lambda (net inputs predicted target &rest args)

(format t ‘‘˜&˜A Weights = ˜{˜4,1F ˜}˜%’’

(if (equal target predicted) ‘‘YES’’ ‘‘NO ‘‘)

(unit-weights (first (first net))))

(apply #’perceptron-update net inputs predicted target args))))))

Up to the ﬁrst 5 examples, the network converges to zero error, and converges to non-zero

error thereafter.

20.14 According to ? (?), the number of linearly separable Boolean functions with inputs

141

For we have

so the fraction of representable functions vanishes as increases.

20.15 These examples were generated from a kind of majority or voting function, where

each input has a different number of votes: 10 for , 4 for to , 2 for , and 1 for . If

you assign this problem, it is probably a good idea to tell the students this. Figure S?? shows

a perceptron and a feed-forward net with logical nodes that represent this function. Our in-

tuition was that the function should have been easy to learn with a perceptron, but harder

with other representations such as a decision tree. However, it turns out that there are not

enough examples for even a perceptron to learn. In retrospect, that should not be too surpris-

ing, as there are over different Boolean functions of six inputs, and only 14 examples.

Running the following code (which makes use of the code in learning/nn.lisp and

learning/perceptron.lisp) we ﬁnd that the perceptron quickly converges to learn

all 14 training examples, but it performs at chance level on the ﬁve-example test set we made

up (getting 2 or 3 out of 5 right on most runs). (The student who did not know what the un-

derlying function was would have to keep out some of the examples to use as a test set.) The

weights vary widely from run to run, although in every run the weight for is the highest

(as it should be for the speciﬁed function), and the weights for and are usually low, but

sometimes is higher than other nodes. This may be a result of the fact that the examples

were chosen to represent some borderline cases where casts the deciding vote. So this

serves as a lesson: if you are “clever” in choosing examples, but rely on a learning algorithm

that assumes examples are chosen at random, you will run into trouble. Here is the code we

used:

(defun test-nn (net problem &optional

(examples (learning-problem-examples problem)))

(let ((correct 0))

(for-each example in examples do

(if (eql (cdr (first example))

(first (nn-output net (rest example)

(learning-problem-attributes problem)

nil)))

(incf correct)))

(values correct ’out-of (length examples))))

(deftest ex20.15

((setq examples

’(((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 0) (I5 . 0) (I6 . 0))

((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 0))

((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 0) (I5 . 1) (I6 . 0))

((T . 1) (I1 . 1) (I2 . 1) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 1))

((T . 1) (I1 . 1) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 0))

((T . 1) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 1))

((T . 0) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 0))

((T . 1) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 1))

((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0) (I5 . 1) (I6 . 1))

((T . 0) (I1 . 0) (I2 . 0) (I3 . 0) (I4 . 1) (I5 . 1) (I6 . 0))

142 Chapter 20. Statistical Learning Methods

((T . 0) (I1 . 0) (I2 . 1) (I3 . 0) (I4 . 1) (I5 . 0) (I6 . 1))

((T . 0) (I1 . 0) (I2 . 0) (I3 . 0) (I4 . 1) (I5 . 0) (I6 . 1))

((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0) (I5 . 1) (I6 . 1))

((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 0)))))

((setq problem

(make-learning-problem

:attributes ’((I1 0 1) (I2 0 1) (I3 0 1) (I4 0 1) (I5 0 1) (I6 0 1))

:goals ’((T 0 1))

:examples examples)))

((setq net (perceptron-learning problem)))

((setq weights (unit-weights (first (first net)))))

((test-nn net problem)

(= * 14))

((test-nn net problem

’(((T . 1) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 1) (I5 . 0) (I6 . 0))

((T . 0) (I1 . 0) (I2 . 0) (I3 . 1) (I4 . 1) (I5 . 1) (I6 . 1))

((T . 1) (I1 . 1) (I2 . 1) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 0))

((T . 0) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0) (I5 . 0) (I6 . 0))

((T . 1) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 1) (I6 . 1))))

(= * 5)))

20.16 The probability output by the perceptron is , where is the sigmoid

function. Since , we have

For a datum with actual value , the log likelihood is

so the gradient of the log likelihood with respect to each weight is

20.17 This exercise reinforces the student’s understanding of neural networks as mathemat-

ical functions that can be analyzed at a level of abstraction above their implementation as a

network of computing elements. For simplicity, we will assume that the activation function

is the same linear function at each node: . (The argument is the same (only

messier) if we allow different and for each node.)

a. The outputs of the hidden layer are

The ﬁnal outputs are

143

Now we just have to see that this is linear in the inputs:

Thus we can compute the same function as the two-layer network using just a one-layer

perceptron that has weights and an activation function

b. The above reduction can be used straightforwardly to reduce an -layer network to an

-layer network. By induction, the -layer network can be reduced to a single-

layer network. Thus, linear activation function restrict neural networks to represent only

linearly functions.

20.18 The implementation of neural networks can be approached in several different ways.

The simplest is probably to store the weights in an array. Everything can be calculated

as if all the nodes were in each layer, with the zero weights ensuring that only appropriate

changes are made as each layer is processed.

Particularly for sparse networks, it can be more efﬁcient to use a pointer-based im-

plementation, with each node represented by a data structure that contains pointers to its

successors (for evaluation) and its predecessors (for backpropagation). Weights are at-

tached to node . In both types of implementation, it is convenient to store the summed input

and the value . The code repository contains an implementation

of the pointer-based variety. See the ﬁle learning/algorithms/nn.lisp, and the

function nn-learning in that ﬁle.

20.19 This question is especially important for students who are not expected to implement

or use a neural network system. Together with 20.15 and 20.17, it gives the student a concrete

(if slender) grasp of what the network actually does. Many other similar questions can be

devised.

Intuitively, the data suggest that a probabilistic prediction is

appropriate. The network will adjust its weights to minimize the error function. The error is

The derivative of the error with respect to the single output is

Setting the derivative to zero, we ﬁnd that indeed . The student should spot the

connection to Ex. 18.8.

20.20 The application of cross-validation is straightforward—the methodology is the same

as that for any parameter selection problem. With 10-fold cross-validation, for example, each

size of hidden layer is evaluated by training on 90% subsets and testing on the remaining

10%. The best-performing size is then chosen, trained on all the training data, and the result

is returned as the system’s hypothesis. The purpose of this exercise is to have the student

144 Chapter 20. Statistical Learning Methods

understand how to design the experiment, run some code to see results, and analyze the result.

The higher-order purpose is to cause the student to question results that make unwarranted

assumptions about the representation used in a learning problem, whether that representation

is the number of hidden nodes in a neural net, or any other representational choice.

20.21 The main purpose of this exercise is to make concrete the notion of the capacity of a

function class (in this case, linear halfspaces). It can be hard to internalize this concept, but

the examples really help.

a. Three points in general position on a plane form a triangle. Any subset of the points

can be separated from the rest by a line, as can be seen from the two examples in

Figure S20.4(a).

b. Figure S20.4(b) shows two cases where the positive and negative examples cannot be

separated by a line.

c. Four points in general position on a plane form a tetrahedron. Any subset of the points

can be separated from the rest by a plane, as can be seen from the two examples in

Figure S20.4(c).

d. Figure S20.4(d) shows a case where a negative point is inside the tetrahedron formed

by four positive points; clearly no plane can separate the two sets.

(a) (b)

Figure S20.4 Illustrative examples for Ex. 20.21.

Solutions for Chapter 21

Reinforcement Learning

21.1 The code repository shows an example of this, implemented in the passive envi-

ronment. The agents are found under lisp/learning/agents/passive*.lisp and

the environment is in lisp/learning/domains/4x3-passive-mdp.lisp. (The

MDP is converted to a full-blown environment using the function mdp->environment

which can be found in lisp/uncertainty/environments/mdp.lisp.)

21.2 Consider a world with two states, and , with two actions in each state: stay still

or move to the other state. Assume the move action is non-deterministic—it sometimes fails,

leaving the agent in the same state. Furthermore, assume the agent starts in and that is a

terminal state. If the agent tries several move actions and they all fail, the agent may conclude

that is 0, and thus may choose a policy with , which is an

improper policy. If we wait until the agent reaches before updating, we won’t fall victim

to this problem.

21.3 This question essentially asks for a reimplementation of a general scheme for asyn-

chronous dynamic programming of which the prioritized sweeping algorithm is an exam-

ple (Moore and Atkeson, 1993). For a., there is code for a priority queue in both the Lisp and

Python code repositories. So most of the work is the experimentation called for in b.

21.4 When there are no terminal states there are no sequences, so we need to deﬁne se-

quences. We can do that in several ways. First, if rewards are sparse, we can treat any state

with a reward as the end of a sequence. We can use equation (21.2); the only problem is that

we don’t know the exact totl reward of the state at the end of the sequence, because it is not

a terminal state. We can estimate it using the current estimate. Another option is to

arbitrarily limit sequences to states, and then consider the next states, etc. A variation on

this is to use a sliding window of states, so that the ﬁrst sequence is states , the second

sequence is , etc.

21.5 The idea here is to calculate the reward that the agent will actually obtain using a

given estimate of and a given estimated model . This is distinct from the true utility of

the states visited. First, we compute the policy for the agent by calculating, for each state, the

action with the highest estimated utility:

145

146 Chapter 21. Reinforcement Learning

Then the expected values can be found by applying value determination with policy and

can then be compared to the optimal values.

21.6 The conversion of the vacuum world problem speciﬁcation into an environment suit-

able for reinforcement learning can be done by merging elements of mdp->environment

from lisp/uncertainty/environments/mdp.lisp with elements of the corre-

sponding function problem->environment from lisp/search/agents.lisp. The

point here is twofold: ﬁrst, that the reinforcement learning algorithms in Chapter 20 are lim-

ited to accessible environments; second, that the dirt makes a huge difference to the size of

the state space. Without dirt, the state space is with locations. With dirt, the state

space is because each location can be clean or dirty. For this reason, input general-

ization is clearly necessary for above about 10. This illustrates the misleading nature of

“navigation” domains in which the state space is proportional to the “physical size” of the

environment. “Real-world” environments typically have some combinatorial structure that

results in exponential growth.

21.7 Code not shown. Several reinforcement learning agents are given in the directory

lisp/learning/agents.

21.8 This utility estimation function is similar to equation (21.9), but adds a term to repre-

sent Euclidean distance on a grid. Using equation (21.10), the update equations are the same

for through , and the new parameter can be calculated by taking the derivative with

respect to :

21.9 Possible features include:

Distance to the nearest terminal state.

Number of adjacent terminal states.

Number of adjacent obstacles.

Number of obstacles that intersect with a path to the nearest terminal state.

21.10 This is a relatively time-consuming exercise. Code not shown to compute three-

dimensional plots. The utility functions are:

a. is the true utility, and is linear.

b. Same as in a, except that .

c. The exact utility depends on the exact placement of the obstacles. The best approxima-

tion is the same as in a. The features in exercise 21.9 might improve the approximation.

147

d. The optimal policy is to head straight for the goal from any point on the right side of

the wall, and to head for (5, 10) ﬁrst (and then for the goal) from any point on the left

of the wall. Thus, the exact utility function is:

(if )

Unfortunately, this is not linear in and , as stated. Fortunately, we can restate the

optimal policy as “head straight up to row 10 ﬁrst, then head right until column 10.”

This gives us the same exact utility as in a, and the same linear approximation.

e. is the true utility. This is also not linear in and ,

because of the absolute value signs. All can be ﬁxed by introducing the features

and .

21.11 The modiﬁcation involves combining elements of the environment converter for games

(game->environment in lisp/search/games.lisp) with elements of the function

mdp->environment. The reward signal is just the utility of winning/drawing/losing and

occurs only at the end of the game. The evaluation function used by each agent is the utility

function it learns through the TD process. It is important to keep the TD learning process

(which is entirely independent of the fact that a game is being played) distinct from the

game-playing algorithm. Using the evaluation function with a deep search is probably better

because it will help the agents to focus on relevant portions of the search space by improving

the quality of play. There is, however, a tradeoff: the deeper the search, the more computer

time is used in playing each training game.

21.12 Code not shown.

21.13 Reinforcement learning as a general “setting” can be applied to almost any agent

in any environment. The only requirement is that there be a distinguished reward signal.

Thus, given the signals of pain, pleasure, hunger, and so on, we can map human learning

directly into reinforcement learning—although this says nothing about how the “program”

is implemented. What this view misses out, however, is the importance of other forms of

learning that occur in humans. These include “speedup learning” (Chapter 21); supervised

learning from other humans, where the teacher’s feedback is taken as a distinguished input;

and the process of learning the world model, which is “supervised” by the environment.

21.14 DNA does not, as far as we know, sense the environment or build models of it. The

reward signal is the death and reproduction of the DNA sequence, but evolution simply mod-

iﬁes the organism rather than learning a or function. The really interesting problem is

deciding what it is that is doing the evolutionary learning. Clearly, it is not the individual (or

the individual’s DNA) that is learning, because the individual’s DNA gets totally intermingled

within a few generations. Perhaps you could say the species is learning, but if so it is learning

to produce individuals who survive to reproduce better; it is not learning anything to do with

the species as a whole rather than individuals. In The Selﬁsh Gene, Richard Dawkins (1976)

proposes that the gene is the unit that learns to succeed as an “individual” because the gene is

preserved with small, accumulated mutations over many generations.

Solutions for Chapter 22

Communication

22.1 No answer required; just read the passage.

Iyou

he she it

you

they

you

him her it

usyou

them

smell

smells

smell

Figure S22.1 A partial DCG for , modiﬁed to handle subject–verb number/person

agreement as in Ex. 22.2.

22.2 See Figure S22.1 for a partial DCG. We include both person and number annotation al-

though English really only differentiates the third person singular for verb agreement (except

for the verb be).

22.3 See Figure S22.2

148

149

a an the

the some many

Figure S22.2 A partial DCG for , modiﬁed to handle article–noun agreement as in

Ex. 22.3.

22.4 The purpose of this exercise is to get the student thinking about the properties of natural

language. There is a wide variety of acceptable answers. Here are ours:

Grammar and Syntax Java: formally deﬁned in a reference book. Grammaticality is

crucial; ungrammatical programs are not accepted. English: unknown, never formally

deﬁned, constantly changing. Most communication is made with “ungrammatical” ut-

terances. There is a notion of graded acceptability: some utterances are judged slightly

ungrammatical or a little odd, while others are clearly right or wrong.

Semantics Java: the semantics of a program is formally deﬁned by the language spec-

iﬁcation. More pragmatically, one can say that the meaning of a particular program

is the JVM code emitted by the compiler. English: no formal semantics, meaning is

context dependent.

Pragmatics and Context-Dependence Java: some small parts of a program are left

undeﬁned in the language speciﬁcation, and are dependent on the computer on which

the program is run. English: almost everything about an utterance is dependent on the

situation of use.

Compositionality Java: almost all compositional. The meaning of “A + B” is clearly

derived from the meaning of “A” and the meaning of “B” in isolation. English: some

compositional parts, but many non-compositional dependencies.

Lexical Ambiguity Java: a symbol such as “Avg” can be locally ambiguous as it might

refer to a variable, a class, or a function. The ambiguity can be resolved simply by

checking the declaration; declarations therefore fulﬁll in a very exact way the role

played by background knowledge and grammatical context in English. English: much

lexical ambiguity.

Syntactic Ambiguity Java: the syntax of the language resolves ambiguity. For exam-

ple, in “if (X) if (Y) A; else B;” one might think it is ambiguous whether the “else”

belongs to the ﬁrst or second “if,” but the language is speciﬁed so that it always belongs

to the second. English: much syntactic ambiguity.

Reference Java: there is a pronoun “this” to refer to the object on which a method was

invoked. Other than that, there are no pronouns or other means of indexical reference;

no “it,” no “that.” (Compare this to stack-based languages such as Forth, where the

stack pointer operates as a sort of implicit “it.”) There is reference by name, however.

Note that ambiguities are determined by scope—if there are two or more declarations

150 Chapter 22. Communication

of the variable “X”, then a use of X refers to the one in the innermost scope surrounding

the use. English: many techniques for reference.

Background Knowledge Java: none needed to interpret a program, although a local

“context” is built up as declarations are processed. English: much needed to do disam-

biguation.

Understanding Java: understanding a program means translating it to JVM byte code.

English: understanding an utterance means (among other things) responding to it appro-

priately; participating in a dialog (or choosing not to participate, but having the potential

ability to do so).

As a follow-up question, you might want to compare different languages, for example: En-

glish, Java, Morse code, the SQL database query language, the Postscript document descrip-

tion language, mathematics, etc.

22.5 This exercise is designed to clarify the relation between quasi-logical form and the

ﬁnal logical form.

a. Yes. Without a quasi-logical form it is hard to write rules that produce, for example,

two different scopes for quantiﬁers.

b. No. It just makes ambiguities and abstractions more concise.

c. Yes. You don’t need to explicitly represent a potentially exponential number of disjunc-

tions.

d. Yes and no. The form is more concise, and so easier to manipulate. On the other hand,

the quasi-logical form doesn’t give you any clues as to how to disambiguate. But if

you do have those clues, it is easy to eliminate a whole family of logical forms without

having to explicitly expand them out.

22.6 Assuming that is the interpretation of “is,” and is the interpretation of “it,” then

we get the following:

a. It is a wumpus:

b. The wumpus is dead:

c. The wumpus is in 2,2:

We should deﬁne what means—one reasonable axiom for one sense of “is” would be

. (This is the “is the same as” sense. There are others.) The

formula is a reasonable semantics for “It is a wumpus.” The problem is if

we use that formula, then we have nowhere to go for “It was a wumpus”—there is no event to

which we can attach the time information. Similarly, for “It wasn’t a wumpus,” we can’t use

, nor could we use . So it is best to have an explicit

event for “is.”

151

22.7 This is a very difﬁcult exercise—most readers have no idea how to answer the ques-

tions (except perhaps to remember that “too few” is better than “too many”). This is the

whole point of the exercise, as we will see in exercise 23.14.

22.8 The purpose of this exercise is to get some experience with simple grammars, and to

see how context-sensitive grammars are more complicated than context-free. One approach to

writing grammars is to write down the strings of the language in an orderly fashion, and then

see how a progression from one string to the next could be created by recursive application

of rules. For example:

a. The language : The strings are , , , . . . (where indicates the null string).

Each member of this sequence can be derived from the previous by wrapping an at

the start and a at the end. Therefore a grammar is:

b. The palindrome language: Let’s assume the alphabet is just , and . (In general, the

size of the grammar will be proportional to the size of the alphabet. There is no way to

write a context-free grammar without specifying the alphabet/lexicon.) The strings of

the language include , a, b, c, aa, bb, cc, aaa, aba, aca, bab, bbb, bcb, . . . . In general,

a string can be formed by bracketing any previous string with two copies of any member

of the alphabet. So a grammar is:

c. The duplicate language: For the moment, assume that the alphabet is just . (It is

straightforward to extend to a larger alphabet.) The duplicate language consists of the

strings: , , , , , , , . . . Note that all strings are of even length.

One strategy for creating strings in this language is this:

Start with markers for the front and middle of the string: we can use the non-

terminal for the front and for the middle. So at this point we have the string

Generate items at the front of the string: generate an followed by an , or a

followed by a . Eventually we get, say, . Then we no longer need

the marker and can delete it, leaving .

Move the non-terminals and down the line until just before the . We end

up with .

Hop the s and s over the , converting each to a terminal ( or ) as we go.

Then we delete the , and are left with the end result: .

152 Chapter 22. Communication

Here is a grammar to implement this strategy:

(starting markers)

(introduce symbols)

(delete the marker)

(move non-terminals down to the )

(hop over and convert to terminal)

(delete the marker)

Here is a trace of the grammar deriving :

22.9 Grammar (A) does not work, because there is no way for the verb “walked” followed

by the adverb “slowly” and the prepositional phrase “to the supermarket” to be parsed as a

verb phrase. A verb phrase in (A) must have either two adverbs or be just a verb. Here is the

parse under grammar (B):

S---NP-+-Pro---Someone

|-VP-+-V---walked

|-Vmod-+-Adv---slowly

|-Vmod---Adv---PP---Prep-+-to

|-NP-+-Det---the

|-NP---Noun---supermarket

153

Here is the parse under grammar (C):

S---NP-+-Pro---Someone

|-VP-+-V---walked

|-Adv-+-Adv---slowly

|-Adv---PP---Prep-+-to

|-NP-+-Det---the

|-NP---Noun---supermarket

22.10 Code not shown.

22.11 Code not shown.

22.12 This is the grammar suggested by the exercise. There are additional grammatical

constructions that could be covered by more ambitious grammars, without adding any new

vocabulary.

(for relative clause)

Running the parser function parses from the code repository with this grammar and with

strings of the form , and then just counting the number of results, we get:

N1 2 3 4 5 6 7 8 9 10

Number of parses 0 0 1 2 3 6 11 22 44 90

To count the number of sentences, de Marcken implemented a parser like the packed forest

parser of example 22.10, except that the representation of a forest is just an integer count of

the number of parses (and therefore the combination of adjacent forests is just the product

of the constituent forests). He then gets a single integer representing the parses for the whole

200-word sentence.

22.13 Here’s one way to draw the parse tree for the story on page 823. The parse tree of the

students’ stories will depned on their choice.

Segment(Evaluation)

Segment(1) ‘‘A funny thing happened’’

Segment(Ground-Figure)

Segment(Cause)

Segment(Enable)

Segment(2) ‘‘John went to a fancy restaurant’’

154 Chapter 22. Communication

Segment(3) ‘‘He ordered the duck’’

Segment(4) ‘‘The bill came to $50’’

Segment(Cause)

Segment(Enable)

Segment(Explanation)

Segment(5) ‘‘John got a shock...’’

Segment(6) ‘‘He had left his wallet at home’’

Segment(7) ‘‘The waiter said it was all right’’

Segment(8) ‘‘He was very embarrassed...’’

22.14 Now we can answer the difﬁcult questions of 22.7:

The steps are sorting the clothes into piles (e.g., white vs. colored); going to the washing

machine (optional); taking the clothes out and sorting into piles (e.g., socks versus

shirts); putting the piles away in the closet or bureau.

The actual running of the washing machine is never explicitly mentioned, so that is one

possible answer. One could also say that drying the clothes is a missing step.

The material is clothes and perhaps other washables.

Putting too many clothes together can cause some colors to run onto other clothes.

It is better to do too few.

So they won’t run; so they get thoroughly cleaned; so they don’t cause the machine to

become unbalanced.

Solutions for Chapter 23

Probabilistic Language Processing

23.1 Code not shown. The approach suggested here will work in some cases, for authors

with distinct vocabularies. For more similar authors, other features such as bigrams, average

word and sentence length, parts of speech, and punctuation might help. Accuracy will also

depend on how many authors are being distinguished. One interesting way to make the task

easier is to group authors into male and female, and try to distinguish the sex of an author not

previously seen. This was suggested by the work of Shlomo Argamon.

23.2 Code not shown. The distribution of words should fall along a Zipﬁan distribution: a

straight line on a log-log scale. The generated language should be similar to the examples in

the chapter.

23.3 Code not shown. There are now several open-source projects to do Bayesian spam

ﬁltering, so beware if you assign this exercise.

23.4 Doing the evaluation is easy, if a bit tedious (requiring 150 page evaluations for the

complete 10 documents 3 engines 5 queries). Explaining the differences is more difﬁ-

cult. Some things to check are whether the good results in one engine are even in the other

engines at all (by searching for unique phrases on the page); check whether the results are

commercially sponsored, are produced by human editors, or are algorithmically determined

by a search ranking algorithm; check whether each engine does the features mentioned in the

next exercise.

23.5 One good way to do this is to ﬁrst ﬁnd a search that yields a single page (or a few pages)

by searching for rare words or phrases on the page. Then make the search more difﬁcult by

adding a variant of one of the words on the page—a word with different case, different sufﬁx,

different spelling, or a synonym for one of the words on the page, and see if the page is still

returned. (Make sure that the search engine requires all terms to match for this technique to

work.)

23.6 Computations like this are given in the book Managing Gigabytes (Witten et al., 1999).

Here’s one way of doing the computation: Assume an average page is about 10KB (giving

us a 10TB corpus), and that index size is linear in the size of the corpus. Bahle et al. (2002)

show an index size of about 2GB for a 22GB corpus; so our billion page corpus would have

an index of about 1TB.

155

156 Chapter 23. Probabilistic Language Processing

23.7 Code not shown. The simplest approach is to look for a string of capitalized words,

followed by “Inc” or “Co.” or “Ltd.” or similar markers. A more complex approach is to

get a list of company names (e.g. from an online stock service), look for those names as

exact matches, and also extract patterns from them. Reporting recall and precision requires a

clearly-deﬁned corpus.

23.8 The main point of this exercise is to show that current translation software is far from

perfect. The mistakes made are often amusing for students.

23.9 Here is a start of a grammar:

time => hour ":" minute

| extendedhour

| extendedhour "o’clock"

| difference before_after extendedhour

hour => 1 | 2 | ... | 24 | "one" | ... | "twelve"

extendedhour => hour | "midnight" | "noon"

minute => 1 | 2 | ... | 60

before-after => "before" | "after" | "to" | "past"

difference => minute | "quarter" | "half"

23.10

a. “I have never seen a better programming language” is easy for most people to see.

b. “John loves mary” seems to be prefered to “Mary loves John” (on Google, by a margin

of 2240 to 499, and by a similar margin on a small sample of respondents), but both are

of course acceptable.

c. This one is quite difﬁcult. The ﬁrst sentence of the second paragraph of Chapter 22

is “Communication is the intentional exchange of information brought about by the

production and perception of signs drawn from a shared system of conventional signs.”

However, this cannot be reliably recovered from the string of words given here. Code

not shown for testing the probabilities of permutations.

23.11 In parlimentary debate, a standard expression of approval is “bravo” in French, and

“hear, hear” in English. That means that in going from French to English, “bravo” would

often have a fertility of 2, but for English to French, the fertility distribution of “hear” would

be half 0 and half 1 for this usage. For other usages, it would have various values, probably

centered closely around 1.

Solutions for Chapter 24

Perception

24.1 The small spaces between leaves act as pinhole cameras. That means that the circular

light spots you see are actually images of the circular sun. You can test this theory next time

there is a solar eclipse: the circular light spots will have a crescent bite taken out of them as

the eclipse progresses. (Eclipse or not, the light spots are easier to see on a sheet of paper

than on the rough forest ﬂoor.)

24.2 Given labels on the occluding edges (i.e., they are all arrows pointing in the clockwise

direction), there will be no backtracking at all if the order is ; each choice is forced

by the existing labels. With the order , the amount of backtracking depends on the

choices made. The ﬁnal labelling is shown in Figure S24.1.

24.3 Recall that the image brightness of a Lambertian surface (page 743) is given by

n s. Here the light source direction sis along the -axis. It is sufﬁcient to consider a

horizontal cross-section (in the – plane) of the cylinder as shown in Figure S24.2(a). Then,

the brightness for all the points on the right half of the cylinder. The left

+ +

+−

Figure S24.1 Labelling of the L-shaped object (Exercise 24.2).

157

158 Chapter 24. Perception

half is in shadow. As , we can rewrite the brightness function as which

reveals that the isobrightness contours in the lit part of the cylinder must be equally spaced.

The view from the -axis is shown in Figure S24.2(b).

illumination

viewer (a) (b)

Figure S24.2 (a) Geometry of the scene as viewed from along the -axis. (b) The scene

from the -axis, showing the evenly spaced isobrightness contours.

24.4 We list the four classes and give two or three examples of each:

a.depth: Between the top of the computer monitor and the wall behind it. Between the

side of the clock tower and the sky behind it. Between the white sheets of paper in the

foreground and the book and keyboard behind them.

b.surface normal: At the near corner of the pages of the book on the desk. At the sides of

the keys on the keyboard.

c.reﬂectance: Between the white paper and the black lines on it. Between the “golden”

bridge in the picture and the blue sky behind it.

d.illumination: On the windowsill, the shadow from the center glass pane divider. On the

paper with Greek text, the shadow along the left from the paper on top of it. On the

computer monitor, the edge between the white window and the blue window is caused

by different illumination by the CRT.

24.5 This exercise requires some basic algebra, and enough calculus to know that

. Students with freshman calculus as background should be able to handle it. Note

that all differentiation is with respect to . Crucially, this means that .

We work the solution for the discrete case; the continuous (integral) case is similar.

(deﬁnition of )

(derivative of a sum)

(since )

(deﬁnition of )

24.6 Before answering this exercise, we draw a diagram of the apparatus (top view), shown

in Figure S24.3. Notice that we make the approximation that the focal length is the distance

from the lens to the image plane; this is valid for objects that are far away. Notice that this

159

512x512

pixels

10cm

16cm

16m Object

0.5m

Figure S24.3 Top view of the setup for stereo viewing (Exercise 24.6).

question asks nothing about the coordinates of points; we might as well have a single line

of 512 pixels in each camera.

a. Solve this by constructing similar triangles: whose hypotenuse is the dotted line from

object to lens, and whose height is 0.5 meters and width 16 meters. This is similar

to a triangle of width 16cm whose hypotenuse projects onto the image plane; we can

compute that its height must be 0.5cm; this is the offset from the center of the image

plane. The other camera will have an offset of 0.5cm in the opposite direction. Thus the

total disparity is 1.0cm, or, at 512 pixels/10cm, a disparity of 51.2 pixels, or 51, since

there are no fractional pixels. Objects that are farther away will have smaller disparity.

Writing this as an equation, where is the disparity in pixels and is the distance to

the object, we have:

pixels

cm cm m

b. In other words, this question is asking how much further than 16m could an object be,

and still occupy the same pixels in the image plane? Rearranging the formula above by

swapping and , and plugging in values of 51 and 52 pixels for , we get values of

of 16.06 and 15.75 meters, for a difference of 31cm (a little over a foot). This is the

range resolution at 16 meters.

c. In other words, this question is asking how far away would an object be to generate a

disparity of one pixel? Objects farther than this are in effect out of range; we can’t say

where they are located. Rearranging the formula above by swapping and we get

51.2 meters.

24.7 In the 3-D case, the two-dimensional image projected by an arbitrary 3-D object can

vary greatly, depending on the pose of the object. But with ﬂat 2-D objects, the image is

always the “same,” except for its orientation and position. All feature points will always be

present because there can be no occlusion, so . Suppose we compute the center of

gravity of the image and model feature points. For the model this can be done ofﬂine; for

the image this is . Then we can take these two center of gravity points, along with an

160 Chapter 24. Perception

arbitrary image point and one of the model points, and try to verify the transformation for

each of the cases. Veriﬁcation is as before, so the whole process is .

A follow-up exercise is to look at some of the tricks used by Olson (1994) to see if they are

applicable here.

24.8 A, B, C can be viewed in stereo and hence their depths can be measured, allowing

the viewer to determine that B is nearest, A and C are equidistant and slightly further away.

Neither D nor E can be seen by both cameras, so stereo cannot be used. Looking at the

ﬁgure, it appears that the bottle occludes D from Y and E from X, so D and E must be further

away than A, B, C, but their relative depths cannot be determined. There is, however, another

possibility (noticed by Alex Fabrikant). Remember that each camera sees the camera’s-eye

view not the bird’s-eye view. X sees DABC and Y sees ABCE. It is possible that D is very

close to camera X, so close that it falls outside the ﬁeld of view of camera Y; similarly, E

might be very close to Y and be outside the ﬁeld of view of X. Hence, unless the cameras

have a 180-degree ﬁeld of view—probably impossible—there is no way to determine whether

D and E are in front of or behind the bottle.

24.9

a. False. This can be quite difﬁcult, particularly when some point are occluded from one

eye but not the other.

b. True. The grid creates an apparent texture whose distortion gives good information as

to surface orientation.

c. False. It applies only to trihedral objects, excluding many polyhedra such as four-sided

pyramids.

d. True. A blade can become a fold if one of the incident surfaces is warped.

e. True. The detectable depth disparity is inversely proportional to .

f. False.

g. False. A disk viewed edge-on appears as a straight line.

24.10 There are at least two reasons: (1) The leftmost car appears bigger and cars are

usually roughly similar in size, therefore it is closer. (2) It is assumed that the road surface is

an approximately horizontal ground plane, and that both cars are on it. In that case, because

the leftmost car appears lower in the image, it must be closer.

25 ROBOTICS

25.1 To answer this question, consider all possibilities for the initial samples before and

after resampling. This can be done because there are only ﬁnitely many states. The following

C++ program calculates the results for ﬁnite . The result for is simply the posterior,

calculated using Bayes rule.

int

main(int argc, char *argv[])

{

// parse command line argument

if (argc != 3){

cerr << "Usage: " << argv[0] << " <number of samples>"

<< " <number of states>" << endl;

exit(0);

}

int numSamples = atoi(argv[1]);

int numStates = atoi(argv[2]);

cerr << "number of samples: " << numSamples << endl

<< "number of states: " << numStates << endl;

assert(numSamples >= 1);

assert(numStates >= 1);

// generate counter

int samples[numSamples];

for (int i = 0; i < numSamples; i++)

samples[i] = 0;

// set up probability tables

assert(numStates == 4); // presently defined for 4 states

double condProbOfZ[4] = {0.8, 0.4, 0.1, 0.1};

double posteriorProb[numStates];

for (int i = 0; i < numStates; i++)

posteriorProb[i] = 0.0;

double eventProb = 1.0 / pow(numStates, numSamples);

//loop through all possibilities

for (int done = 0; !done; ){

// compute importance weights (is probability distribution)

double weight[numSamples], totalWeight = 0.0;

for (int i = 0; i < numSamples; i++)

totalWeight += weight[i] = condProbOfZ[samples[i]];

// normalize them

for (int i = 0; i < numSamples; i++)

weight[i] /= totalWeight;

// calculate contribution to posterior probability

for (int i = 0; i < numSamples; i++)

posteriorProb[samples[i]] += eventProb * weight[i];

// increment counter

for (int i = 0; i < numSamples && i != -1;){

samples[i]++;

if (samples[i] >= numStates)

samples[i++] = 0;

else

i = -1;

if (i == numSamples)

done = 1;

}

// print result

cout << "Result: ";

for (int i = 0; i < numStates; i++)

cout << " " << posteriorProb[i];

cout << endl;

// calculate asymptotic expectation

double totalWeight = 0.0;

for (int i = 0; i < numStates; i++)

totalWeight += condProbOfZ[i];

cout << "Unbiased:";

for (int i = 0; i < numStates; i++)

cout << " " << condProbOfZ[i] / totalWeight;

cout << endl;

// calculate KL divergence

double kl = 0.0;

for (int i = 0; i < numStates; i++)

kl += posteriorProb[i] * (log(posteriorProb[i]) -

log(condProbOfZ[i] / totalWeight));

cout << "KL divergence: " << kl << endl;

}

(a) (b)

Figure S25.1 Code to calculate answer to exercise 25.1.

a. The program (correctly) calculates the following posterior distributions for the four

states, as a function of the number of samples . Note that for , the measurement

is ignored entirely! The correct posterior for is calculated using Bayes rule.

161

162 Chapter 25. Robotics

sample at sample at sample at sample at

0.25 0.25 0.25 0.25

0.368056 0.304167 0.163889 0.163889

0.430182 0.314463 0.127677 0.127677

0.466106 0.314147 0.109874 0.109874

0.488602 0.311471 0.0999636 0.0999636

0.503652 0.308591 0.0938788 0.0938788

0.514279 0.306032 0.0898447 0.0898447

0.522118 0.303872 0.0870047 0.0870047

0.528112 0.30207 0.0849091 0.0849091

0.532829 0.300562 0.0833042 0.0833042

0.571429 0.285714 0.0714286 0.0714286

b. Plugging the posterior for into the deﬁnition of the Kullback Liebler Diver-

gence gives us:

c. The proof for is trivial, since the re-weighting ignores the measurement proba-

bility entirely. Therefore, the probability for generating a sample in any of the locations

in is given by the initial distribution, which is uniform.

For , a proof is easily obtained by considering all ways in which

initial samples are generated:

number samples probability weights probability of resampling

of sample set for each sample for each sample for each location in

1 0 0 0 0 0

2 0 1 0 0

3 0 2 0 0

4 0 3 0 0

5 1 0 0 0

6 1 1 0 0 0

7 1 2 0 0

8 1 3 0 0

9 2 0 0 0

10 2 1 0 0

11 2 2 0 0 0

12 2 3 0 0

13 3 0 0 0

14 3 1 0 0

15 3 2 0 0

16 3 3 0 0 0

sum of all probabilities

163

A quick check should convince you that these numbers are the same as above. Placing

this into the deﬁnition of the Kullback Liebler divergence with the correct posterior

distribution, gives us .

For we know that the sampler is unbiased. Hence, the probability of gen-

erating a sample is the same as the posterior distribution calculated by Bayes ﬁlters.

Those are given above as well.

d. Here are two possible modiﬁcations. First, if the initial robot location is known with

absolute certainty, the sampler above will always be unbiased. Second, if the sensor

measurement is equally likely for all states, that is

, it will also be unbiased. An invalid answer, which we frequently encountered

in class, pertains to the algorithm (instead of the problem formulation). For example,

replacing particle ﬁlters by the exact discrete Bayes ﬁler remedies the problem but is

not a legitimate answer to this question. Neither is the use of inﬁnitely many particles.

25.2 Implementing Monte Carlo localization requires a lot of work but is a premiere way to

gain insights into the basic workings of probabilistic algorithms in robotics, and the intricacies

inherent in real data. We have used this exercise in many courses, and students consistently

expressed having learned a lot. We strongly recommend this exercise!

The implementation is not as straightforward as it may appear at ﬁrst glance. Common

problems include:

The sensor model models too little noise, or the wrong type of noise. For example, a

simple Gaussian will not work here.

The motion model assumes too little or too much noise, or the wrong type of noise.

Here a Gaussian will work ﬁne though.

The implementation may introduce unnecessarily high variance in the resulting sam-

pling set, by sampling too often, or by sampling in the wrong way. This problem man-

ifests itself by diversity disappearing prematurely, often with the wrong samples sur-

viving. While the basic MCL algorithm, as stated in the book, suggests that sampling

should occur after each motion update, implementations that sample less frequently

tend to yield superior results. Further, drawing samples independently of each other

is inferior to so-called low variance samplers. Here is a version of low variance sam-

pling, in which denotes the particles and their importance weights. The resulting

resampled particles reside in the set .

function LOW-VARIANCE-WEIGHTED-SAMPLE-WITH-REPLACEMENT

( ):

for to do

add to

modulo

return

164 Chapter 25. Robotics

The parameter determines the speed at which we cycle through the sample set. While

each sample’s probability remains the same as if it were sampled independently, the re-

sulting samples are dependent, and the variance of the sample set is lower (assuming

). As a pleasant side effect, the low-variance samples is also easily implemented

in time, which is more difﬁcult for the independent sampler.

Samples are started in the occupied or unknown parts of the map, or are allowed into

those parts during the forward sampling (motion prediction) step of the MCL algorithm.

Too few samples are used. A few thousand should do the job, a few hundred will

probably not.

The algorithm can be sped up by pre-caching all noise-free measurements, for all - - poses

that the robot might assume. For that, it is convenient to deﬁne a grid over the space of all

poses, with 10 centimeters spatial and 2 degrees angular resolution. One might then compute

the noise-free measurements for the centers of those grid cells. The sensor model is clearly

just a function of those correct measurements; and computing those takes the bulk of time in

MCL.

25.3 Let be the shoulder and be the elbow angle. The coordinates of the end effector are

then given by the following expression. Here is the height and the horizontal displacement

between the end effector and the robot’s base (origin of the coordinate system):

Notice that this is only one way to deﬁne the kinematics. The zero-positions of the angles

and can be anywhere, and the motors may turn clockwise or counterclockwise. Here we

chose deﬁne these angles in a way that the arm points straight up at ; furthermore,

increasing and makes the corresponding joint rotate counterclockwise.

Inverse kinematics is the problem of computing and from the end effector coordi-

nates and . For that, we observe that the elbow angle is uniquely determined by the

Euclidean distance between the shoulder joint and the end effector. Let us call this distance

. The shoulder joint is located above the origin of the coordinate system; hence, the

distance is given by . An alternative way to calculate is by

recovering it from the elbow angle and the two connected joints (each of which is

long): . The reader can easily derive this from basic trigonomy, exploiting

the fact that both the elbow and the shoulder are of equal length. Equating these two different

derivations of with each other gives us

(25.1)

(25.2)

In most cases, can assume two symmetric conﬁgurations, one pointing down and one

pointing up. We will discuss exceptions below.

To recover the angle , we note that the angle between the shoulder (the base) and the

end effector is given by . Here is the common generalization

165

of the arcus tangens to all four quadrants (check it out—it is a function in C). The angle

is now obtained by adding , again exploiting that the shoulder and the elbow are of equal

length:

(25.3)

Of course, the actual value of depends on the actual choice of the value of . With the

exception of singularities, can take on exactly two values.

The inverse kinematics is unique if assumes a single value; as a consequence, so does

alpha. For this to be the case, we need that

(25.4)

This is the case exactly when the argument of the is 1, that is, when the distance

and the arm is fully stretched. The end points then lie on a circle deﬁned by

. If the distance , there is no solution to the inverse

kinematic problem: the point is simply too far away to be reachable by the robot arm.

Unfortunately, conﬁgurations like these are numerically unstable, as the quotient may

be slightly larger than one (due to truncation errors). Such points are commonly called singu-

larities, and they can cause major problems for robot motion planning algorithms. A second

singularity occurs when the robot is “folded up,” that is, . Here the end effector’s

position is identical with that of the robot elbow, regardless of the angle : and

. This is an important singularity, as there are inﬁnitely many solutions to the

inverse kinematics. As long as , the value of can be arbitrary. Thus, this sim-

ple robot arm gives us an example where the inverse kinematics can yield zero, one, two, or

inﬁnitely many solutions.

25.4 Code not shown.

25.5

a. The conﬁgurations of the robots are shown by the black dots in the following ﬁgures.

Figure S25.2 Conﬁguration of the robots.

166 Chapter 25. Robotics

b. The above ﬁgure answers also the second part of this exercise: it shows the conﬁgura-

tion space of the robot arm constrained by the self-collision constraint and the constraint

imposed by the obstacle.

c. The three workspace obstacles are shown in the following diagrams:

Figure S25.3 Workspace obstacles.

d. This question is a great mind teaser that illustrates the difﬁculty of robot motion plan-

ning! Unfortunately, for an arbitrary robot, a planar obstacle can decompose the workspace

into any number of disconnected subspaces. To see, imagine a 1-DOF rigid robot that

moves on a horizontal rod, and possesses upward-pointing ﬁngers, like a giant fork.

A single planar obstacle protruding vertically into one of the free-spaces between the

ﬁngers could effectively separate the conﬁguration space into disjoint subspaces.

A second DOF will not change this.

More interesting is the robot arm used as an example throughout this book. By

slightly extending the vertical obstacles protruding into the robot’s workspace we can

decompose the conﬁguration space into ﬁve disjoint regions. The following ﬁgures

show the conﬁguration space along with representative conﬁgurations for each of the

ﬁve regions.

Is ﬁve the maximum for any planar object that protrudes into the workspace of this

particular robot arm? We honestly do not know; but we offer a $1 reward for the ﬁrst

person who presents to us a solution that decomposes the conﬁguration space into six,

seven, eight, nine, or ten disjoint regions. For the reward to be claimed, all these regions

must be clearly disjoint, and they must be a two-dimensional manifold in the robot’s

conﬁguration space.

For non-planar objects, the conﬁguration space is easily decomposed into any num-

ber of regions. A circular object may force the elbow to be just about maximally

bent; the resulting workspace would then be a very narrow pipe that leave the shoulder

largely unconstrained, but conﬁnes the elbow to a narrow range. This pipe is then easily

chopped into pieces by small dents in the circular object; the number of such dents can

be increased without bounds.

167

25.6 A simple deliberate controller might work as follows: Initialize the robot’s map with

an empty map, in which all states are assumed to be navigable, or free. Then iterate the

following loop: Find the shortest path from the current position to the goal position in the

map using A*; execute the ﬁrst step of this path; sense; and modify the map in accordance

with the sensed obstacles. If the robot reaches the goal, declare success. The robot declares

failure when A* fails to ﬁnd a path to the goal. It is easy to see that this approach is both

complete and correct. The robot always ﬁnd a path to a goal if one exists. If no such path

exists, the approach detects this through failure of the path planner. When it declares failure,

it is indeed correct in that no path exists.

A common reactive algorithm, which has the same correctness and completeness prop-

erty as the deliberate approach, is known as the BUG algorithm. The BUG algorithm dis-

tinguishes two modes, the boundary-following and the go-to-goal mode. The robot starts in

go-to-goal mode. In this mode, the robot always advances to the adjacent grid cell closest to

the goal. If this is impossible because the cell is blocked by an obstacle, the robot switches to

the boundary-following mode. In this mode, the robot follows the boundary of the obstacle

until it reaches a point on the boundary that is a local minimum to the straight-line distance

to the goal. If such a point is reached, the robot returns to the go-to-goal mode. If the robot

reaches the goal, it declares success. It declares failure when the same point is reached twice,

which can only occur in the boundary-following mode. It is easy to see that the BUG algo-

rithm is correct and complete. If a path to the goal exists, the robot will ﬁnd it. When the

robot declares failure, no path to the goal may exist. If no such path exists, the robot will

Figure S25.4 Conﬁguration space for each of the ﬁve regions.

168 Chapter 25. Robotics

ultimately reach the same location twice and detect its failure.

Both algorithms can cope with continuous state spaces provides that they can accurately

perceive obstacles, plan paths around them (deliberative algorithm) or follow their boundary

(reactive algorithm). Noise in motion can cause failures for both algorithms, especially if the

robot has to move through a narrow opening to reach the goal. Similarly, noise in perception

destroys both completeness and correctness: In both cases the robot may erroneously con-

clude a goal cannot be reached, just because its perception was noise. However, a deliberate

algorithm might build a probabilistic map, accommodating the uncertainty that arises from

the noisy sensors. Neither algorithm as stated can cope with unknown goal locations; how-

ever, the deliberate algorithm is easily converted into an exploration algorithm by which the

robot always moves to the nearest unexplored location. Such an algorithm would be complete

and correct (in the noise-free case). In particular, it would be guaranteed to ﬁnd and reach

the goal when reachable. The BUG algorithm, however, would not be applicable. A common

reactive technique for ﬁnding a goal whose location is unknown is random motion; this algo-

rithm will with probability one ﬁnd a goal if it is reachable; however, it is unable to determine

when to give up, and it may be highly inefﬁcient. Moving obstacles will cause problems for

both the deliberate and the reactive approach; in fact, it is easy to design an adversarial case

where the obstacle always moves into the robot’s way. For slow-moving obstacles, a common

deliberate technique is to attach a timer to obstacles in the grid, and erase them after a certain

number of time steps. Such an approach often has a good chance of succeeding.

25.7 There are a number of ways to extend the single-leg AFSM in Figure 25.22(b) into a set

of AFSMs for controlling a hexapod. A straightforward extension—though not necessarily

the most efﬁcient one—is shown in the following diagram. Here the set of legs is divided into

two, named A and B, and legs are assigned to these sets in alternating sequence. The top level

controller, shown on the left, goes through six stages. Each stage lifts a set of legs, pushes the

ones still on the ground backwards, and then lowers the legs that have previously been lifted.

The same sequence is then repeated for the other set of legs. The corresponding single-

leg controller is essentially the same as in Figure 25.22(b), but with added wait-steps for

synchronization with the coordinating AFSM. The low-level AFSM is replicated six times,

once for each leg.

For showing that this controller is stable, we show that at least one leg group is on the

ground at all times. If this condition is fulﬁlled, the robot’s center of gravity will always be

above the imaginary triangle deﬁned by the three legs on the ground. The condition is easily

proven by analyzing the top level AFSM. When one group of legs in (or on the way to

from ), the other is either in or , both of which are on the ground. However, this proof

only establishes that the robot does not fall over when on ﬂat ground; it makes no assertions

about the robot’s performance on non-ﬂat terrain. Our result is also restricted to static stabil-

ity, that is, it ignores all dynamic effects such as inertia. For a fast-moving hexapod, asking

that its center of gravity be enclosed in the triangle of support may be insufﬁcient.

25.8 We have used this exercise in class to great effect. The students get a clearer picture of

why it is hard to do robotics. The only drawback is that it is a lot of fun to play, and thus the

students want to spend a lot of time on it, and the ones who are just observing feel like they

169

(a) (b)

Figure S25.5 Controller for a hexapod robot.

are missing out. If you have laboratory or TA sections, you can do the exercise there.

Bear in mind that being the Brain is a very stressful job. It can take an hour just to stack

three boxes. Choose someone who is not likely to panic or be crushed by student derision.

Help the Brain out by suggesting useful strategies such as deﬁning a mutually agreed Hand-

centric coordinate system so that commands are unambiguous. Almost certainly, the Brain

will start by issuing absolute commands such as “Move the Left Hand 12 inches positive y

direction” or “Move the Left Hand to (24,36).” Such actions will never work. The most useful

“invention” that students will suggest is the guarded motion discussed in Section 25.5—that

is, macro-operators such as “Move the Left Hand in the positive y direction until the eyes say

the red and green boxes are level.” This gets the Brain out of the loop, so to speak, and speeds

things up enormously.

We have also used a related exercise to show why robotics in particular and algorithm

design in general is difﬁcult. The instructor uses as props a doll, a table, a diaper and some

safety pins, and asks the class to come up with an algorithm for putting the diaper on the

baby. The instructor then follows the algorithm, but interpreting it in the least cooperative

way possible: putting the diaper on the doll’s head unless told otherwise, dropping the doll

on the ﬂoor if possible, and so on.

Solutions for Chapter 26

Philosophical Foundations

26.1 We will take the disabilities (see page 949) one at a time. Note that this exercise might

be better as a class discussion rather than written work.

a.be kind: Certainly there are programs that are polite and helpful, but to be kind requires

an intentional state, so this one is problematic.

b.resourceful: Resourceful means “clever at ﬁnding ways of doing things.” Many pro-

grams meet this criteria to some degree: a compiler can be clever making an optimiza-

tion that the programmer might not ever have thought of; a database program might

cleverly create an index to make retrievals faster; a checkers or backgammon program

learns to play as well as any human. One could argue whether the machines are “re-

ally” clever or just seem to be, but most people would agree this requirement has been

achieved.

c.beautiful: Its not clear if Turing meant to be beautiful or to create beauty, nor is it clear

whether he meant physical or inner beauty. Certainly the many industrial artifacts in

the New York Museum of Modern Art, for example, are evidence that a machine can

be beautiful. There are also programs that have created art. The best known of these

is chronicled in Aaron’s code: Meta-art, artiﬁcial intelligence, and the work of Harold

Cohen (McCorduck, 1991).

d.friendly This appears to fall under the same category as kind.

e.have initiative Interestingly, there is now a serious debate whether software should take

initiative. The whole ﬁeld of software agents says that it should; critics such as Ben

Schneiderman say that to achieve predictability, software should only be an assistant,

not an autonomous agent. Notice that the debate over whether software should have

initiative presupposes that it has initiative.

f.have a sense of humor We know of no major effort to produce humorous works. How-

ever, this seems to be achievable in principle. All it would take is someone like Harold

Cohen who is willing to spend a long time tuning a humor-producing machine. We note

that humorous text is probably easier to produce than other media.

g.tell right from wrong There is considerable research in applying AI to legal reasoning,

and there are now tools that assist the lawyer in deciding a case and doing research. One

could argue whether following legal precedents is the same as telling right from wrong,

and in any case this has a problematic conscious aspect to it.

170

171

h.make mistakes At this stage, every computer user is familiar with software that makes

mistakes! It is interesting to think back to what the world was like in Turing’s day,

when some people thought it would be difﬁcult or impossible for a machine to make

mistakes.

i.fall in love This is one of the cases that clearly requires consciousness. Note that while

some people claim that their pets love them, and some claim that pets are not conscious,

I don’t know of anybody who makes both claims.

j.enjoy strawberries and cream There are two parts to this. First, there has been little to

no work on taste perception in AI (although there has been related work in the food and

perfume industries; see http://198.80.36.88/popmech/tech/U045O.html for one such ar-

tiﬁcial nose), so we’re nowhere near a breakthrough on this. Second, the “enjoy” part

clearly requires consciousness.

k.make someone fall in love with it This criteria is actually not too hard to achieve; ma-

chines such as dolls and teddy bears have been doing it to children for centuries. Ma-

chines that talk and have more sophisticated behaviors just have a larger advantage in

achieving this.

l.learn from experience Part VI shows that this has been achieved many times in AI.

m.use words properly No program uses words perfectly, but there have been many natural

language programs that use words properly and effectively within a limited domain (see

Chapters 22-23).

n.be the subject of its own thought The problematic word here is “thought.” Many pro-

grams can process themselves, as when a compiler compiles itself. Perhaps closer to

human self-examination is the case where a program has an imperfect representation

of itself. One anecdote of this involves Doug Lenat’s Eurisko program. It used to run

for long periods of time, and periodically needed to gather information from outside

sources. It “knew” that if a person were available, it could type out a question at the

console, and wait for a reply. Late one night it saw that no person was logged on, so it

couldn’t ask the question it needed to know. But it knew that Eurisko itself was up and

running, and decided it would modify the representation of Eurisko so that it inherits

from “Person,” and then proceeded to ask itself the question!

o.have as much diversity of behavior as man Clearly, no machine has achieved this, al-

though there is no principled reason why one could not.

p.do something really new This seems to be just an extension of the idea of learning

from experience: if you learn enough, you can do something really new. “Really” is

subjective, and some would say that no machine has achieved this yet. On the other

hand, professional backgammon players seem unanimous in their belief that TDGam-

mon (Tesauro, 1992), an entirely self-taught backgammon program, has revolutionized

the opening theory of the game with its discoveries.

26.2 No. Searle’s Chinese room thesis says that there are some cases where running a

program that generates the right output for the Chinese room does not cause true understand-

ing/consciousness. The negation of this thesis is therefore that all programs with the right

172 Chapter 26. Philosophical Foundations

output do cause true understanding/consciousness. So if you were to disprove Searle’s the-

sis, then you would have a proof of machine consciousness. However, what this question is

getting at is the argument behind the thesis. If you show that the argument is faulty, then you

may have proved nothing more: it might be that the thesis is true (by some other argument),

or it might be false.

26.3 Yes, this is a legitimate objection. Remember, the point of restoring the brain to normal

(page 957) is to be able to ask “What was it like during the operation?” and be sure of

getting a “human” answer, not a mechanical one. But the skeptic can point out that it will not

do to replace each electronic device with the corresponding neuron that has been carefully

kept aside, because this neuron will not have been modiﬁed to reﬂect the experiences that

occurred while the electronic device was in the loop. One could ﬁx the argument by saying,

for example, that each neuron has a single activation energy that represents its “memory,” and

that we set this level in the electronic device when we insert it, and then when we remove it,

we read off the new activation energy, and somehow set the energy in the neuron that we put

back in. The details, of course, depend on your theory of what is important in the functional

and conscious functioning of neurons and the brain; a theory that is not well-developed so

far.

26.4 This exercise depends on what happens to have been published lately. The NEWS

and MAGS databases, available on many online library catalog systems, can be searched

for keywords such as Penrose, Searle, Chinese Room, Dreyfus, etc. We found about 90

reviews of Penrose’s books. Here are some excerpts from a fairly typical one, by Adam

Schulman (1995).

Roger Penrose, the distinguished mathematical physicist, has again entered the lists to rid

the world of a terrible dragon. The name of this dragon is ”strong artiﬁcial intelligence.”

Strong Al, as its defenders call it, is both a widely held scientiﬁc thesis and an ongoing

technological program. The thesis holds that the human mind is nothing but a fancy calcu-

lating machine-”-a computer made of meat”–and that all thinking is merely computation;

the program is to build faster and more powerful computers that will eventually be able to

do everything the human mind can do and more. Penrose believes that the thesis is false

and the program unrealizable, and he is conﬁdent that he can prove these assertions.

In Part I of Shadows of the Mind Penrose makes his rigorous case that human conscious-

ness cannot be fully understood in computational terms. How does Penrose prove that

there is more to consciousness than mere computation? Most people will already ﬁnd it

inherently implausible that the diverse faculties of human consciousness–self-awareness,

understanding, willing, imagining, feeling–differ only in complexity from the workings

of, say, an IBM PC.

Students should have no problem ﬁnding things in this and other articles with which to dis-

agree. The comp.ai Newsnet group is also a good source of rash opinions.

Dubious claims also emerge from the interaction between journalists’ desire to write

entertaining and controversial articles and academics’ desire to achieve prominence and to be

viewed as ahead of the curve. Here’s one typical result— Is Nature’s Way The Best Way?,

Omni, February 1995, p. 62:

173

Artiﬁcial intelligence has been one of the least successful research areas in computer

science. That’s because in the past, researchers tried to apply conventional computer

programming to abstract human problems, such as recognizing shapes or speaking in

sentences. But researchers at MIT’s Media Lab and Boston University’s Center for Adap-

tive Systems focus on applying paradigms of intelligence closer to what nature designed

for humans, which include evolution, feedback, and adaptation, are used to produce com-

puter programs that communicate among themselves and in turn learn from their mistakes.

Proﬁles In Artiﬁcial Intelligence, David Freedman.

This is not an argument that AI is impossible, just that it has been unsuccessful. The full

text of the article is not given, but it is implied that the argument is that evolution worked

for humans, therefore it is a better approach for programs than is “conventional computer

programming.” This is a common argument, but one that ignores the fact that (a) there are

many possible solutions to a problem; one that has worked in the past may not be the best in

the present (b) we don’t have a good theory of evolution, so we may not be able to duplicate

human evolution, (c) natural evolution takes millions of years and for almost all animals

does not result in intelligence; there is no guarantee that artiﬁcial evolution will do better (d)

artiﬁcial evolution (or genetic algorithms, ALife, neural nets, etc.) is not the only approach

that involves feedback, adaptation and learning. “Conventional” AI does this as well.

26.5 This also might make a good class discussion topic. Here are our attempts:

intelligence: a measure of the ability of an agent to make the right decisions, given the

available evidence. Given the same circumstances, a more intelligent agent will make better

decisions on average.

thinking: creating internal representations in service of the goal of coming to a conclu-

sion, making a decision, or weighing evidence.

consciousness: being aware of one’s own existence, and of one’s current internal state.

Here are some objections [with replies]:

For intelligence, too much emphasis is put on decision-making. Haven’t you ever

known a highly intelligent person who made bad decisions? Also no mention is made of

learning. You can’t be intelligent by using brute-force look-up, for example, could you? [The

emphasis on decision-making is only a liability when you are working at too coarse a gran-

ularity (e.g., “What should I do with my life?”) Once you look at smaller-grain decisions

(e.g., “Should I answer a, b, c or none of the above?), you get at the kinds of things tested by

current IQ tests, while maintaining the advantages of the action-oriented approach covered

in Chapter 1. As to the brute-force problem, think of intelligence in terms of an ecological

niche: an agent only needs to be as intelligent as is necessary to be successful. If this can be

accomplished through some simple mechanism, ﬁne. For the complex environments that we

humans are faced with, more complex mechanisms are needed.]

For thinking, we have the same objections about decision-making, but in general, think-

ing is the least controversial of the three terms.

For consciousness, the weakness is the deﬁnition of “aware.” How does one demon-

strate awareness? Also, it is not one’s true internal state that is important, but some kind of

abstraction or representation of some of the features of it.

174 Chapter 26. Philosophical Foundations

26.6 It is hard to give a deﬁnitive answer to this question, but it can provoke some interesting

essays. Many of the threats are actually problems of computer technology or industrial society

in general, with some components that can be magniﬁed by AI—examples include loss of

privacy to surveillance, and the concentration of power and wealth in the hands of the most

powerful. As discussed in the text, the prospect of robots taking over the world does not

appear to be a serious threat in the foreseeable future.

26.7 Biological and nuclear technologies provide mush more immediate threats of weapons,

yielded either by states or by small groups. Nanotechnlogy threatens to produce rapidly re-

producing threats, either as weapons or accidently, but the feasibility of this technology is still

quite hypothetical. As discussed in the text and in the previous exercise, computer technology

such as centralized databases, network-attached cameras, and GPS-guided weapons seem to

pose a more serious portfolio of threats than AI technology, at least as of today.

26.8 To decide if AI is impossible, we must ﬁrst deﬁne it. In this book, we’ve chosen a

deﬁnition that makes it easy to show it is possible in theory—for a given architecture, we

just enumerate all programs and choose the best. In practice, this might still be infeasible,

but recent history shows steady progress at a wide variety of tasks. Now if we deﬁne AI as

the production of agents that act indistinguishably form (or at least as intellgiently as) human

beings on any task, then one would have to say that little progress has been made, and some,

such as Marvin Minsky, bemoan the fact that few attempts are even being made. Others think

it is quite appropriate to address component tasks rather than the “whole agent” problem.

Our feeling is that AI is neither impossible nor a ooming threat. But it would be perfectly

consistent for someone to ffel that AI is most likely doomed to failure, but still that the risks

of possible success are so great that it should not be persued for fear of success.

Solutions for Chapter 27

AI: Present and Future

There are no exercises in this chapter. There are many topics that are worthy of class dis-

cussion, or of paper assignments for those who like to emphasize such things. Examples

are:

What are the biggest theoretical obstacles to successful AI systems?

What are the biggest practical obstacles? How are these different?

What is the right goal for rational agent design? Does the choice of a goal make all that

much difference?

What do you predict the future holds for AI?

175

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177

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