For Tuesday Read chapter 12, sections 1-4 Homework: Chapter 10, - - PowerPoint PPT Presentation

For Tuesday

• Read chapter 12, sections 1-4
• Homework:

– Chapter 10, exercise 9 – Chapter 13, exercise 8

Program 2

• Any questions?
• Due Friday night

Constructing the planning graph

• Level P1: all literals from the initial state
• Add an action in level Ai if all its

preconditions are present in level Pi

• Add a precondition in level Pi if it is the

effect of some action in level Ai-1 (including no-ops)

• Maintain a set of exclusion relations to

eliminate incompatible propositions and actions (thus reducing the graph size)

Planning graph

… … …

Mutual Exclusion relations

• Two actions (or literals) are mutually

exclusive (mutex) at some stage if no valid plan could contain both.

• Two actions are mutex if:

– Interference: one clobbers others‟ effect or precondition – Competing needs: mutex preconditions

• Two propositions are mutex if:

– All ways of achieving them are mutex

Dinner Date example

• Initial Conditions: garbage ˄ cleanHands ˄ quiet
• Goal: dinner ^ present ^ ¬garbage
• Actions:

– Cook precondition: cleanHands effect: dinner – Wrap precondition: quiet effect present – Carry precondition: effect: ¬garbage ^ ¬cleanHands – Dolly precondition: effect: ¬garbage ^ ¬quiet

Dinner Date example

Dinner Date example

Observation 1

Propositions monotonically increase

(always carried forward by no-ops) p ¬q ¬r p q ¬q ¬r p q ¬q r ¬r p q ¬q r ¬r A A B A B

Observation 2

Actions monotonically increase

p ¬q ¬r p q ¬q ¬r p q ¬q r ¬r p q ¬q r ¬r A A B A B

Observation 3

Proposition mutex relationships monotonically decrease

p q r … A p q r … p q r …

Observation 4

Action mutex relationships monotonically decrease

p q … B p q r s … p q r s … A C B C A p q r s … B C A

Observation 5

Planning Graph „levels off‟.

• After some time k all levels are identical
• Because it‟s a finite space, the set of literals

never decreases and mutexes don‟t reappear.

Valid plan

A valid plan is a planning graph where:

• Actions at the same level don‟t interfere
• Each action‟s preconditions are made true

by the plan

• Goals are satisfied

GraphPlan algorithm

• Grow the planning graph (PG) until all

goals are reachable and not mutex. (If PG levels off first, fail)

• Search the PG for a valid plan
• If none is found, add a level to the PG and

try again

Searching for a solution plan

• Backward chain on the planning graph
• Achieve goals level by level
• At level k, pick a subset of non-mutex actions to

achieve current goals. Their preconditions become the goals for k-1 level.

• Build goal subset by picking each goal and

choosing an action to add. Use one already selected if possible. Do forward checking on remaining goals (backtrack if can‟t pick non- mutex action)

Plan Graph Search

If goals are present & non-mutex:

Choose action to achieve each goal Add preconditions to next goal set

Termination for unsolvable problems

• Graphplan records (memoizes) sets of unsolvable

goals:

– U(i,t) = unsolvable goals at level i after stage t.

• More efficient: early backtracking
• Also provides necessary and sufficient conditions

for termination:

– Assume plan graph levels off at level n, stage t > n – If U(n, t-1) = U(n, t) then we know we‟re in a loop and can terminate safely.

Dinner Date example

• Initial Conditions: garbage ˄ cleanHands ˄ quiet
• Goal: dinner ^ present ^ ¬garbage
• Actions:

– Cook precondition: cleanHands effect: dinner – Wrap precondition: quiet effect present – Carry precondition: effect: ¬garbage ^ ¬cleanHands – Dolly precondition: effect: ¬garbage ^ ¬quiet

Dinner Date example

Dinner Date example

Dinner Date example

Shopping Example

Op( Action: Go(there); Precond: At(here); Effects: At(there), ¬At(here) ) Op( Action: Buy(x), Precond: At(store), Sells(store,x); Effects: Have(x) )

• A0:

– At(Home) Sells(SM,Banana) Sells(SM,Milk) Sells(HWS,Drill)

• A

– Have(Drill) Have(Milk) Have(Banana) At(Home)

Uncertainty

• Everyday reasoning and decision making is

based on uncertain evidence and inferences.

• Classical logic only allows conclusions to

be strictly true or strictly false

• We need to account for this uncertainty and

the need to weigh and combine conflicting evidence.

Coping with Uncertainty

• Straightforward application of probability theory

is impractical since the large number of conditional probabilities required are rarely, if ever, available.

• Therefore, early expert systems employed fairly

ad hoc methods for reasoning under uncertainty and for combining evidence.

• Recently, methods more rigorously founded in

probability theory that attempt to decrease the amount of conditional probabilities required have flourished.

Probability

• Probabilities are real numbers 0-1 representing

the a priori likelihood that a proposition is true.

P(Cold) = 0.1 P(¬Cold) = 0.9

• Probabilities can also be assigned to all values
• f a random variable (continuous or discrete)

with a specific range of values (domain), e.g. low, normal, high.

P(temperature=normal)=0.99 P(temperature=98.6) = 0.99

Probability Vectors

• The vector form gives probabilities for all

values of a discrete variable, or its probability distribution.

P(temperature) = <0.002, 0.99, 0.008>

• This indicates the prior probability, in

which no information is known.

Conditional Probability

• Conditional probability specifies the

probability given that the values of some

• ther random variables are known.

P(Sneeze | Cold) = 0.8 P(Cold | Sneeze) = 0.6

• The probability of a sneeze given a cold is

80%.

• The probability of a cold given a sneeze is

60%.

• Cond. Probability cont.
• Assumes that the given information is all that is

known, so all known information must be given.

P(Sneeze | Cold  Allergy) = 0.95

• Also allows for conditional distributions

P(X |Y) gives 2-D array of values for all P(X=xi|Y=yj)

• Defined as

P (A | B) = P (A  B) P(B)

Axioms of Probability Theory

• All probabilities are between 0 and 1.

0  P(A)  1

• Necessarily true propositions have probability 1,

necessarily false have probability 0.

P(true) = 1 P(false) = 0

• The probability of a disjunction is given by

P(A  B) = P(A) + P(B) - P(A  B)

Joint Probability Distribution

• The joint probability distribution for a set of random

variables X1…Xn gives the probability of every combination of values (an n-dimensional array with vn values if each variable has v values)

P(X1,...,Xn) Sneeze ¬Sneeze Cold 0.08 0.01 ¬Cold 0.01 0.9

• The probability of all possible cases (assignments of values

to some subset of variables) can be calculated by summing the appropriate subset of values from the joint distribution.

• All conditional probabilities can therefore also be

calculated

Bayes Theorem

P(H | e) = P(e | H) P(H) P(e)

• Follows from definition of conditional

probability: P (A | B) = P (A  B) P(B)

Other Basic Theorems

• If events A and B are independent then:

P(A  B) = P(A)P(B)

• If events A and B are incompatible then:

P(A  B) = P(A) + P(B)

Simple Bayesian Reasoning

• If we assume there are n possible disjoint

diagnoses, d1 … dn P(di | e) = P(e | di) P(di) P(e)

• P(e) may not be known but the total

probability of all diagnoses must always be 1, so all must sum to 1

• Thus, we can determine the most probable

without knowing P(e).

Efficiency

• This method requires that for each disease

the probability it will cause any possible combination of symptoms and the number

• f possible symptom sets, e, is exponential

in the number of basic symptoms.

• This huge amount of data is usually not

available.

Bayesian Reasoning with Independence (“Naïve” Bayes)

• If we assume that each piece of evidence (symptom) is

independent given the diagnosis (conditional independence), then given evidence e as a sequence {e1,e2,…,ed} of

• bservations, P(e | di) is the product of the probabilities of the
• bservations given di.
• The conditional probability of each individual symptom for

each possible diagnosis can then be computed from a set of data

• r estimated by the expert.
• However, symptoms are usually not independent and frequently

correlate, in which case the assumptions of this simple model are violated and it is not guaranteed to give reasonable results.

Bayes Independence Example

• Imagine there are diagnoses ALLERGY, COLD,

and WELL and symptoms SNEEZE, COUGH, and FEVER

Prob Well Cold Allergy P(d) 0.9 0.05 0.05 P(sneeze|d) 0.1 0.9 0.9 P(cough | d) 0.1 0.8 0.7 P(fever | d) 0.01 0.7 0.4

Calculations

• If symptoms sneeze & cough & no fever, what is

the diagnosis?

Problems with Probabilistic Reasoning

• If no assumptions of independence are made,

then an exponential number of parameters is needed for sound probabilistic reasoning.

• There is almost never enough data or

patience to reliably estimate so many very specific parameters.

• If a blanket assumption of conditional

independence is made, efficient probabilistic reasoning is possible, but such a strong assumption is rarely warranted.

Practical Naïve Bayes

• We‟re going to assume independence, so

what numbers do we need?

• Where do the numbers come from?
• What about zeroes?

Program 3