CS 4100 Artificial Intelligence

Prof. C. HafnerClass Notes March 15and20, 2012

Outline• Midterm planning problem: solution

http://www.ccs.neu.edu/course/cs4100sp12/classnotes/midterm-planning.doc

• Discuss term projects• Continue uncertain reasoning in AI

– Probability distribution (review)– Conditional Probability and the Chain Rule (cont.)– Bayes’ Rule– Independence, “Expert” systems and the combinatorics of

joint probabilities– Bayes networks– Assignment 6

Term Projects – The Process

1. Form teams of 3 or 4 people – 10-12 teams2. Before next class (Mar 20) each team send an email

a. Name and a main contact person (email)b. All team members’ names and email addressesc. You can reserve a topic asap (first request)

3. Brief written project proposal due Fri March 23 10pm (email)4. Each team will

a. submit a written project report (due April 17, last day of class)b. a running computer application (due April 17, last day of class)c. make a presentation of 15 minutes on their project (April 12 & 17)

5. Attendance is required and will be taken on April 12 & 17

Term Projects – The Content

1. Select a domain2. Model the domain

a. “Logical/state model” : define an ontology w/ example world stateb. Implementation in Protégé – demo with some queriesc. “Dynamics model” (of how the world changes)

Using Situation Calculus formalism or STRIPS-type operators

3. Define and solve example planning problems: initial state goal state

a. Specify planning axioms or STRIPS-type operatorsb. Show (on paper) a proof or derivation of a trivial plan and then a

more challenging one using resolution or the POP algorithm

Term Projects – Choosing Domains

Travel domains: Boston T, other kinds of trips or vacationsCooking domains: planning a meal, a dinner party, preparing a

recipeSports domains: One league or tournament? Gaming domains: model a game that requires some strategyMilitary mission planningExercise session/program planning (including use of equipment)Making a movie

An issue is granularity: how fine a level of detail

Review: Inference by enumeration• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is true: P(φ) = Σω:ω φ╞ P(ω)

• P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2• P(toothache, catch) = ???

•

Inference by enumeration• Start with the joint probability distribution:

• Can also compute conditional probabilities:

P(cavity | toothache) = P(cavity toothache)P(toothache)

= 0.016+0.064 0.108 + 0.012 + 0.016 + 0.064

= 0.4

Conditional probability and Bayes Rule• Definition of conditional probability:

P(a | b) = P(a b) / P(b) if P(b) > 0

• Product rule gives an alternative formulation:P(a b) = P(a | b) P(b) = P(b | a) P(a)

• Combine these to derive: Bayes' rule: P(a | b) = P(b | a) P(a) / P(b)

• Useful for assessing diagnostic probability from causal probability:– P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect)

– E.g., let M be meningitis, S be stiff neck:P(m|s) = P(s|m) P(m) / P(s) = 0.8 × 0.0001 / 0.1 = 0.0008

– Note: posterior probability of meningitis still very small!

•

The Chain Rule

• Chain rule is derived by successive application of product rule:P(X1, …,Xn) = P(X1,...,Xn-1) P(Xn | X1,...,Xn-1) = P(X1,...,Xn-2) P(Xn-1 | X1,...,Xn-2) P(Xn | X1,...,Xn-1) = …

= P(X1) P(X2 | X1) P(X3 | X1, X2) . . . P(Xn | X1, . . ., Xn-1)

OR: πi= 1 to n P(Xi | X1, … ,Xi-1)

Independence• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)

P(Toothache, Catch, Cavity, Weather)= P(Toothache, Catch, Cavity) P(Weather)

• 32 entries reduced to 12; for n independent biased coins, O(2n) →O(n)

• Absolute independence powerful but rare

• Dentistry is a large field with hundreds of variables, none of which are independent. What to do?

•

Example: Expert Systems for Medical Diagnosis

• 100 diseases (assume only one at a time!)• 20 symptoms

• # of parameters needed to calculate P(Di) when a patient provides his/her symptoms

• Strategy to reduce the size: assume independence of all symptoms

• Recalculate number of parameters needed

In class exercise• Given the joint distribution shown below and the

definition P(a | b) = P(a b) / P(b): – What is P(Cavity = True) ?– What is P(Weather = Sunny) ?– What is P(Cavity = True | Weather = Sunny)

• Given the meta-equation:– P(Weather,Cavity) = P(Weather | Cavity) P(Cavity)

What are the 8 equations represented here?

Weather = sunny rainy cloudy snow Cavity = true 0.144 0.02 0.016 0.02Cavity = false 0.576 0.08 0.064 0.08

Bayes' Rule and conditional independence

P(Cavity | toothache catch) = αP(toothache catch | Cavity) P(Cavity) = αP(toothache | Cavity) P(catch | Cavity) P(Cavity)

• This is an example of a naïve Bayes model:

P(Cause,Effect1, … ,Effectn) = P(Cause) πiP(Effecti|Cause)

• Total number of parameters is linear in n––

Conditional independence• P(Toothache, Cavity, Catch) has 23 – 1 = 7 independent entries

• If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:(1) P(catch | toothache, cavity) = P(catch | cavity)

• The same independence holds if I haven't got a cavity:(2) P(catch | toothache,cavity) = P(catch | cavity)

• Catch is conditionally independent of Toothache given Cavity:P(Catch | Toothache,Cavity) = P(Catch | Cavity)

• Equivalent statements:P(Toothache | Catch, Cavity) = P(Toothache | Cavity)P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)

––

»

Bayesian networks

• A simple, graphical notation for conditional independence assertions and hence for compact specification of full joint distributions

• Syntax:– a set of nodes, one per variable– a directed, acyclic graph (link ≈ "directly influences")– a conditional distribution for each node given its parents:

P (Xi | Parents (Xi))

• In the simplest case, conditional distribution represented as a conditional probability table (CPT) giving the distribution over Xi for each combination of parent values

Extend to P(A ^ B ^ C ^ …) = ?

Review: Conditional probabilities and JPD (joint distribution)

Chain rule follows from this definition

• Product ruleP(a b) = P(a | b) P(b) = P(b | a) P(a)

• Chain rule is derived by successive application of product rule:P(X1, …,Xn) can also be written P(X1 ^ ... ^ Xn) = P([Xn ^ [X1 ,. . . Xn-1]) = P(X1,...Xn-1) P(Xn | X1,...,Xn-1) = P(X1,...,Xn-2) P(Xn-1 | X1,...,Xn-2) P(Xn | X1,...,Xn-1) = …

= P(X1) P(X2 | X1) P(X3 | X1, X2) . . . P(Xn | X1, . . ., Xn-1)

–

Conditional Prob. example

Example

Likes Football Dislikes Neutral

Male .25 .1 .15

Female .1 .3 .1

In-class exercise:Calculate:

P(Likes Football | Male )P( ~ Likes Football | Female)

Review the Joint Distribution (JPD)

What assumption can we make ?

Test your understanding: Fill in the table

Structure for CP-based AI Models Given a set of RV’s X, typically, we are interested in

the posterior joint distribution of the query variables Y given specific values e for the evidence variables E

Let the hidden variables be H = X - Y – E

Then the required calculation of P(Y | E) is done by summing out the hidden variables:

Note: what is α ?

Given the definition: P(a | b) = P(a b) / P(b)

α is the denominator 1/P(E=e). P(E=e) can be calculatedfrom the joint distribution as: ΣhP(E= e ^ H = h)P( Y | E = e) = αP(Y ^ E = e) or αΣhP(Y ^ E= e ^ H

= h)

Example (medical diagnosis)Causal model: D I S (Y H E)

Cancer anemia fatigueKidney disease anemia fatigue

P(Y=cancer | E=fatigue) = α [ P(Y=cancer ^ E=fatigue ^ anemia) + P(Y=cancer ^ E=fatigue ^ ~anemia) ]

α = 1/P(E = fatigue) or 1/[P(E=fatigue ^ anemia) + P(E=fatigue ^ ~anemia) ]

Analysis

• The terms in the summation are joint entries because Y, E and H together exhaust the set of random variables

• Obvious problems:1. Time and space complexity O(dn) where d is the largest arity2. How to find the numbers to solve real problems?

(A solution to 1. : assume independence !!)

• P(Y | E = e) = αP(Y ^ E = e) = αΣhP(Y ^ E= e ^ H = h) [repeated]

What is Independence ??• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)

P(Toothache, Catch, Cavity, Weather) JD entries are 2x2x2x4= P(Toothache, Catch, Cavity) P(Weather) entries are 2x2x2 + 4

• 32 entries reduced to 12• In general, total independence assumption reduces

exponential to linear complexity

•

What is Independence ??• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)• Toss 10 coins, different OUTCOMES are 2^10 = 2048• Biased coins whose behavior is independent of each other:

O(2n) →O(n) = can compute P(all outcomes) with 10 values• All coins have the same bias (includes the case of fair coins) ????

How many values are needed ?

Test your understanding:• Consider a “3 sided coin” (or die). How many entries needed to

show the probabilities of all outcomes?• If you toss 10 of those and:• All have the same bias?• Bias unknown, but independence is assumed?• Bias unknown, no independence assumed?

•

Example: Expert Systems for Medical Diagnosis• 10 diseases• 20 symptoms

• # of parameters needed to calculate P(D | S) for all combinations using a JPD

• Strategy to reduce the size of the model: assume mutual independence of symptoms and diseases - Recalculate number of parameters needed

• Absolute independence powerful but rare• Medicine is a large field with hundreds of variables,

many of which are not independent. What to do?

Problem 2: We still need to find the numbers

Assuming independence, doctors may be able to estimate:P(symptom | disease) for each S/D pair (causal reasoning)

While what we need to know s/he may not be able to estimate as easily:

P(disease | symptom)

Thus, the importance of Bayes rule in probabilistic AI

Bayes' Rule• Product rule P(ab) = P(a | b) P(b) = P(b | a) P(a)

Bayes' rule: P(a | b) = P(b | a) P(a) / P(b)• or in distribution form

P(Y|X) = P(X|Y) P(Y) / P(X) = αP(X|Y) P(Y)

• Useful for assessing diagnostic probability from causal probability:

P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect) P(Disease|Symptom) = P(Symptom|Diease) P(Symptom) / (Disease)

– E.g., let M be meningitis, S be stiff neck:P(m|s) = P(s|m) P(m) / P(s) = 0.8 × 0.0001 / 0.1 = 0.0008

– Note: posterior probability of meningitis still very small!•

•

Bayes' Rule and conditional independenceP(Cavity | toothache catch)

= αP(toothache catch | Cavity) P(Cavity) = αP(toothache | Cavity) P(catch | Cavity) P(Cavity)

• We say: “toothache and catch are independent, given cavity”. This is an example of a naïve Bayes model. We will study this later as our simplest machine learning application

P(Cause,Effect1, … ,Effectn) = P(Cause) πiP(Effecti|Cause)

• Total number of parameters is linear in n (number of symptoms). This is our first Bayesian inference net.

––

Conditional independence• P(Toothache, Cavity, Catch) has 23 – 1 = 7 independent entries

• If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:(1) P(catch | toothache, cavity) = P(catch | cavity)

• The same independence holds if I haven't got a cavity:(2) P(catch | toothache,cavity) = P(catch | cavity)

• Catch is conditionally independent of Toothache given Cavity:P(Catch | Toothache,Cavity) = P(Catch | Cavity)

• Equivalent statements (from original definitions of independence):P(Toothache | Catch, Cavity) = P(Toothache | Cavity)P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)

––

»

Conditional independence contd.• Write out full joint distribution using chain rule:

P(Toothache, Catch, Cavity)= P(Toothache | Catch, Cavity) P(Catch, Cavity)= P(Toothache | Catch, Cavity) P(Catch | Cavity) P(Cavity)= P(Toothache | Cavity) P(Catch | Cavity) P(Cavity)

I.e., 2 + 2 + 1 = 5 independent numbers

• In most cases, the use of conditional independence reduces the size of the representation of the joint distribution from exponential in n to linear in n.

• Conditional independence is our most basic and robust form of knowledge about uncertain environments.

–

Remember this examples

Example of conditional independence

Test your understanding of the Chain Rule

This is our second Bayesian inference net

How to construct a Bayes Net

Test your understanding: design a Bayes net with plausible numbers

Calculating using Bayes’ Nets