CS 188: Artificial IntelligenceSpring 2007

Lecture 16: Review

3/8/2007

Srini Narayanan – ICSI and UC Berkeley

Midterm Structure

5 questions Search (HW1 and HW 2) CSP (Written 2) Games (HW 4) Logic (HW 3) Probability/BN (Today’s lecture)

One page cheat sheet and calculator allowed. Midterm weight: 15% of your total grade. Today Review 1: Mostly Probability/BN Sunday: Review 2: All topics review and Q/A

Probabilities

What you are expected to know Basics

Conditional and Joint Distributions Bayes Rule Converting from conditional to joint and vice-versa Independence and conditional independence

Graphical Models/Bayes Nets Building nets from descriptions of problems Conditional Independence Inference by Enumeration from the net Approximate inference (Prior sampling, rejection sampling,

likelihood weighting)

Review: Useful Rules

Conditional Probability (definition)

Chain Rule

Bayes Rule

Marginalization

Marginalization (or summing out) is projecting a joint distribution to a sub-distribution over subset of variables

T S P

warm sun 0.4

warm rain 0.1

cold sun 0.2

cold rain 0.3

T P

warm 0.5

cold 0.5

S P

sun 0.6

rain 0.4

The Product Rule

Sometimes joint P(X,Y) is easy to get Sometimes easier to get conditional P(X|Y)

Example: P(sun, dry)?

R P

sun 0.8

rain 0.2

D S P

wet sun 0.1

dry sun 0.9

wet rain 0.7

dry rain 0.3

D S P

wet sun 0.08

dry sun 0.72

wet rain 0.14

dry rain 0.06

Conditional Independence

Reminder: independence X and Y are independent ( ) iff

or equivalently,

X and Y are conditionally independent given Z ( ) iff

or equivalently,

(Conditional) independence is a property of a distribution

Conditional Independence

For each statement about distributions over X, Y, and Z, if the statement is not always true, state a conditional independence assumption which makes it true. P(x|y) = P(x, y) / p(y) P(x, y) = P(x)P(y) P(x, y, z) = P(x|z)P(y|z)P(z) P(x, y, z) = P(x)P(y)P(z|x, y) P(x, y) = Sumz (x, y, z)

Conditional Independence

For each statement about distributions over X, Y , and Z, if the statement is not always true, state a conditional independence assumption which makes it true. P(x|y) = P(x, y)/ P(y)

always true P(x, y) = P(x)P(y)

true if x and y are independent P(x, y, z) = P(x|z)P(y|z)P(z)

true if x and y and independent given z P(x, y, z) = P(x)P(y)P(z|x, y)

true if x and y are independent P(x, y) = Sumz (x, y, z)

always true

Conditional and Joint Distributions

Suppose I want to determine the joint distribution P (W,X,Y,Z).

Assume I know P(X,Y,Z) and P(W |X, Y)

What assumptions do I need to make to compute P(W,X,Y,Z)?

Conditional and Joint Distributions

Suppose I want to determine the joint distribution P (W,X,Y,Z).

Assume I know P(X,Y,Z) and P(W | X, Y)

What assumptions do I need to make to compute P(W,X,Y,Z)?

ANS: P(X,Y,Z,W) = P(X,Y,Z| W) P(W) = P(W |X,Y,Z) P(X,Y,Z) If I assume P(W | X,Y,Z) = P (W | X,Y) ie. W is

independent of Z given both X and Y, I can compute the joint.

Graphical Model Notation

Nodes: variables (with domains) Can be assigned (observed) or

unassigned (unobserved)

Arcs: interactions Similar to CSP constraints Indicate “direct influence” between

variables

arrows from a to b means b “depends on” a. Often the arrows indicate causation

Bayes’ Net Semantics

Let’s formalize the semantics of a Bayes’ net

A set of nodes, one per variable X A directed, acyclic graph A conditional distribution for each node

A distribution over X, for each combination of parents’ values

CPT: conditional probability table Description of a noisy “causal” process

A1

X

An

A Bayes net = Topology (graph) + Local Conditional Probabilities

Probabilities in BNs

Bayes’ nets implicitly encode joint distributions As a product of local conditional distributions To see what probability a BN gives to a full assignment, multiply

all the relevant conditionals together:

Example:

This lets us reconstruct any entry of the full joint Not every BN can represent every full joint

The topology enforces certain conditional independencies

Example: Alarm Network

.001 * .002 * .05 * .05 * .01 = 5x10-11

Analyzing Independence

Arc between nodes ==> (poss) dependence What if there is no direct arc? To answer this question in general, we only

need to understand 3-node graphs with 2 arcs

Cast of characters:

X Y Z

“Causal Chain” X

Y

Z

“Common Effect”

X

Y

Z

“Common Cause”

Example

R

T

B

D

L

T’

Yes

Yes

Yes

Example

Variables: R: Raining T: Traffic D: Roof drips S: I’m sad

Questions:

T

S

D

R

Yes

Question

Which nets guarantee each statement:

1

2

A

B

C A

B

CA B C

NET X NET Y NET Z

Approximate Inference: Prior Sampling

Cloudy

Sprinkler Rain

WetGrass

Cloudy

Sprinkler Rain

WetGrass

Example

We’ll get a bunch of samples from the BN:c, s, r, w

c, s, r, w

c, s, r, w

c, s, r, w

c, s, r, w

If we want to know P(W) We have counts <w:4, w:1> Normalize to get P(W) = <w:0.8, w:0.2> This will get closer to the true distribution with more samples Can estimate anything else, too What about P(C| r)? P(C| r, w)?

Cloudy

Sprinkler Rain

WetGrass

C

S R

W

Rejection Sampling

Let’s say we want P(C| s) Same thing: tally C outcomes,

but ignore (reject) samples which don’t have S=s

This is rejection sampling It is also consistent (correct in

the limit)

c, s, r, wc, s, r, wc, s, r, wc, s, r, wc, s, r, w

Cloudy

Sprinkler Rain

WetGrass

C

S R

W

Likelihood Weighting

Problem with rejection sampling: If evidence is unlikely, you reject a lot of samples You don’t exploit your evidence as you sample Consider P(B|a)

Idea: fix evidence variables and sample the rest

Problem: sample distribution not consistent! Solution: weight by probability of evidence given parents

Burglary Alarm

Burglary Alarm

Likelihood Sampling

Cloudy

Sprinkler Rain

WetGrass

Cloudy

Sprinkler Rain

WetGrass

Design of BN

When designing a Bayes net, why do we not make every variable depend on as many other variables as possible?

Design of a BN

You are considering founding a startup to make AI based robots to do household chores, and you want to reason about your future. There are three ways you can possibly get rich (R), either your company can go public via an IPO (I), it can be acquired (A), or you can win the lottery (L). Your company cannot go public if it gets acquired. Of course, in order for your company to either go public or get acquired, your robot has to actually work (W). You decide that if you do strike it rich then you will probably retire to Hawaii (H) to live the good life.

Draw a graphical model for the problem that reflects the causal structure as stated.

Bayes Net for the Question

Independence

Which of the following independence properties are true for your network? A ind I L ind I L ind I|R L ind W|H W ind H|L W ind H|R

Independence

Which of the following independence properties are true for your network? A ind I L ind I True L ind I|R L ind W|H W ind H|L W ind H|R True

Inference

Write out an expression for an entry P(a, h, i, l, r,w), of the joint distribution

encoded by your network, P(A,H, I, L,R,W) in terms of quantities provided by the network.

Inference

Write out an expression for an entry P(a, h, i, l, r,w), of the joint distribution

encoded by your network, P(A,H, I, L,R,W) in terms of quantities provided by the network.

P(a, h, i, l, r,w) = P(w)P(a|w)P(i|w, a)P(r|a, i, l)P(h|r)

The three prisoners Three prisoners A, B, and C have been tried for murder. Their verdicts will be read and sentence executed tomorrow. They know that only one of them will be declared guilty and hanged, the other two will

be set free. The identity of the guilty prisoner is not known to the prisoners, only to a prison guard. In the middle of the night, prisoner A calls the guard over and makes the following

request. A to Guard: Please take this letter to one of my friends, the one who is to be released.

You and I know that at least one of the others (B, C) will be freed. The guard agrees. An hour later, A calls the guard and asks “Can you tell me which person (B or C) you

gave the letter to. This should give me no clue about my chances since either of them had an equal chance of receiving the letter.”

The guard answers “I gave the letter to B. B will be released tomorrow.” A thinks “Before I talked to the guard, my chances of being executed were 1 in 3.

Now that he has told me that B will be released, only C and I remain, so my chances are 1 in 2. What did I do wrong? I made certain not to ask for any information relevant to my own fate..

Question: What is A’s chance of perishing at dawn. 1 in 2 or 1 in 3. Why?

Topic Review

Search CSP Games Logic

Search

Uninformed Search DFS, BFS Uniform Cost Iterative Deepening

Informed Search Best first greedy A*

Admissibility Consistency Coming up with admissible heuristics

relaxed problem

Local Search

CSP

Formulating problems as CSPs Basic solution with DFS with backtracking Heuristics (Min Remaining Value, LCV) Forward Checking Arc consistency for CSP

Games

Problem formulation Minimax and zero sum two player games Alpha-Beta pruning

Logic

Basics: Entailment, satisfiability, validity Prop Logic

Truth tables, enumeration converting propositional sentences to CNF Propositional resolution

First Order Logic Basics: Objects, relations, functions,

quantifiers Converting NL sentences into FOL

Search Review

Uninformed Search DFS, BFS Uniform Cost Iterative Deepening

Informed Search Best first greedy A*

Admissible Consistency Relaxed problem for heuristics

Local Search

Combining UCS and Greedy Uniform-cost orders by path cost, or backward cost g(n) Best-first orders by goal proximity, or forward cost h(n)

A* Search orders by the sum: f(n) = g(n) + h(n)

S a d

b

Gh=5

h=5

h=2

1

5

11

2

h=6 h=0

c

h=4

2

3

e h=11

Example: Teg Grenager

Admissible Heuristics

A heuristic is admissible (optimistic) if:

where is the true cost to a nearest goal

E.g. Euclidean distance on a map problem

Coming up with admissible heuristics is most of what’s involved in using A* in practice.

Trivial Heuristics, Dominance

Dominance:

Heuristics form a semi-lattice: Max of admissible heuristics is admissible

Trivial heuristics Bottom of lattice is the zero heuristic (what

does this give us?) Top of lattice is the exact heuristic

Constraint Satisfaction Problems

Standard search problems: State is a “black box”: any old data structure Goal test: any function over states Successors: any map from states to sets of states

Constraint satisfaction problems (CSPs): State is defined by variables Xi with values from a

domain D (sometimes D depends on i) Goal test is a set of constraints specifying

allowable combinations of values for subsets of variables

Simple example of a formal representation language

Allows useful general-purpose algorithms with more power than standard search algorithms

Constraint Graphs

Binary CSP: each constraint relates (at most) two variables

Constraint graph: nodes are variables, arcs show constraints

General-purpose CSP algorithms use the graph structure to speed up search. E.g., Tasmania is an independent subproblem!

Improving Backtracking

General-purpose ideas can give huge gains in speed: Which variable should be assigned next? In what order should its values be tried? Can we detect inevitable failure early? Can we take advantage of problem structure?

Minimum Remaining Values

Minimum remaining values (MRV): Choose the variable with the fewest legal values

Why min rather than max? Called most constrained variable “Fail-fast” ordering

Degree Heuristic

Tie-breaker among MRV variables Degree heuristic:

Choose the variable with the most constraints on remaining variables

Why most rather than fewest constraints?

Least Constraining Value

Given a choice of variable: Choose the least constraining

value The one that rules out the fewest

values in the remaining variables Note that it may take some

computation to determine this!

Why least rather than most?

Combining these heuristics makes 1000 queens feasible

Forward Checking Idea: Keep track of remaining legal values for

unassigned variables Idea: Terminate when any variable has no legal values

WASA

NT Q

NSW

V

Constraint Propagation Forward checking propagates information from assigned to

unassigned variables, but doesn't provide early detection for all failures:

NT and SA cannot both be blue! Why didn’t we detect this yet? Constraint propagation repeatedly enforces constraints (locally)

WASA

NT Q

NSW

V

Arc Consistency Simplest form of propagation makes each arc consistent

X Y is consistent iff for every value x there is some allowed y

If X loses a value, neighbors of X need to be rechecked! Arc consistency detects failure earlier than forward checking What’s the downside of arc consistency? Can be run as a preprocessor or after each assignment

WASA

NT Q

NSW

V

Conversion to CNF

B1,1 (P1,2 P2,1)

1. Eliminate , replacing α β with (α β)(β α).(B1,1 (P1,2 P2,1)) ((P1,2 P2,1) B1,1)

2. Eliminate , replacing α β with α β.(B1,1 P1,2 P2,1) ((P1,2 P2,1) B1,1)

3. Move inwards using de Morgan's rules and double-negation:(B1,1 P1,2 P2,1) ((P1,2 P2,1) B1,1)

4. Apply distributivity law ( over ) and flatten:(B1,1 P1,2 P2,1) (P1,2 B1,1) (P2,1 B1,1)

Resolution

Conjunctive Normal Form (CNF) conjunction of disjunctions of literals

E.g., (A B) (B C D) : Basic intuition, resolve B, B to get (A) (C D) (why?)

Resolution inference rule (for CNF):li … lk, m1 … mn

l1 … li-1 li+1 … lk m1 … mj-1 mj+1 ... mn

where li and mj are complementary literals. E.g., P1,3 P2,2, P2,2

P1,3

Resolution is sound and complete for propositional logic.

Basic Use: KB ╞ α iff (KB α) is unsatisfiable

Some examples of FOL sentences

How expressive is FOL? Some examples from natural language

Every gardener likes the sun. x gardener(x) => likes (x, Sun)

You can fool some of the people all of the time x (person(x) ^ ( t) (time(t) => can-fool(x,t)))

You can fool all of the people some of the time. x (person(x) => ( t) (time(t) ^ can-fool(x,t)))

No purple mushroom is poisonous. ~ x purple(x) ^ mushroom(x) ^ poisonous(x) or, equivalently,

x (mushroom(x) ^ purple(x)) => ~poisonous(x)