CS 188: Artificial Intelligence
Spring 2007
Lecture 16: Review
3/8/2007
Srini Narayanan – ICSI and UC Berkeley
Midterm Structure
 5 questions
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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
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Conditional and Joint Distributions
Bayes Rule
Converting from conditional to joint and vice-versa
Independence and conditional independence
 Graphical Models/Bayes Nets
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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
P
P
warm
0.5
cold
0.5
T
S
warm
sun
0.4
warm
rain
0.1
cold
sun
0.2
S
cold
rain
0.3
sun
0.6
rain
0.4
P
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
D
S
P
wet
sun
0.1
wet
sun
0.08
dry
sun
0.9
dry
sun
0.72
wet
rain
0.7
wet
rain
0.14
dry
rain
0.3
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.
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

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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
A1
An
 A set of nodes, one per variable X
 A directed, acyclic graph
 A conditional distribution for each node
X
 A distribution over X, for each combination
of parents’ values
 CPT: conditional probability table
 Description of a noisy “causal” process
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:
“Common Effect”
X
Y
X
Y
Z
Z
“Causal Chain”
X
Z
“Common Cause”
Y
Example
L
Yes
R
Yes
D
B
T
Yes
T’
Example
 Variables:
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R: Raining
T: Traffic
D: Roof drips
S: I’m sad
R
T
 Questions:
D
S
Yes
Question
 Which nets guarantee each statement:
A
C
A
A
B
B
NET X
 1
 2
C
C
B
NET Y
NET Z
Approximate Inference:
Prior Sampling
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
Cloudy
C
Sprinkler
S
Rain
R
WetGrass
W
 If we want to know P(W)
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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)?
Rejection Sampling
Cloudy
C
 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)
Sprinkler
S
Rain
R
WetGrass
W
c, s, r, w
c, s, r, w
c, s, r, w
c, s, r, w
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)
Burglary
Alarm
 Idea: fix evidence variables and sample the rest
Burglary
Alarm
 Problem: sample distribution not consistent!
 Solution: weight by probability of evidence given parents
Likelihood Sampling
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?
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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?
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A ind I
L ind I
L ind I|R
L ind W|H
W ind H|L
W ind H|R
True
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
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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
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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
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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)
5
e
h=1
1
S
h=6
1
3
a
1
b
h=5
1
h=5
2
d
2
h=2
G
h=0
c
h=4
 A* Search orders by the sum: f(n) = g(n) + h(n)
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:
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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
NT
WA
SA
Q
NSW
V
 Idea: Keep track of remaining legal values for
unassigned variables
 Idea: Terminate when any variable has no legal values
Constraint Propagation
NT
WA
SA
Q
NSW
V
 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)
Arc Consistency
NT
WA
SA
Q
NSW
V
 Simplest form of propagation makes each arc consistent
 X  Y is consistent iff for every value x there is some allowed y
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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
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 doublenegation:
(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)