CS 4100 Artificial Intelligence
Prof. C. Hafner
Class 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
–
–
–
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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 teams
2. 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 addresses
c. You can reserve a topic asap (first request)
3. Brief written project proposal due Fri March 23 10pm (email)
4. Each team will
a.
b.
c.
submit a written project report (due April 17, last day of class)
a running computer application (due April 17, last day of class)
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 domain
2. Model the domain
a.
b.
c.
“Logical/state model” : define an ontology w/ example world state
Implementation in Protégé – demo with some queries
“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.
b.
Specify planning axioms or STRIPS-type operators
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 vacations
Cooking domains: planning a meal, a dinner party, preparing a
recipe
Sports domains: One league or tournament?
Gaming domains: model a game that requires some strategy
Military mission planning
Exercise 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)
P(B)
or P(A, B) = P(A)
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 =
Cavity = true
Cavity = false
sunny rainy
0.144 0.02
0.576 0.08
cloudy snow
0.016 0.02
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
Review: Conditional probabilities and JPD (joint distribution)
Extend to P(A ^ B ^ C ^ …) = ?
Chain rule follows from this definition
• Product rule
P(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:
P( Y | E = e) = αP(Y ^ E = e) or αΣhP(Y ^ E= e ^ H = h)
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 calculated
from the joint distribution as: ΣhP(E= e ^ H = h)
Example (medical diagnosis)
Causal model:
D I S
(Y  H  E)
Cancer  anemia  fatigue
Kidney 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
P(Y | E = e) = αP(Y ^ E = e) = αΣhP(Y ^ E= e ^ H = h) [repeated]
• 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 arity
2. How to find the numbers to solve real problems?
(A solution to 1. : assume independence !!)
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(ab) = 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 independence
P(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