For Friday Read Chapter 11, sections 1 and 2 Homework: –Chapter 10, exercises 1 and 5

For Friday

• Read Chapter 11, sections 1 and 2

• Homework:– Chapter 10, exercises 1 and 5

Program 2

• Any questions?

Issue 4: Categories

• How are categories defined?

• What makes reasoning about categories hard?

Semantic Networks

• Use graphs to represent concepts and the relations between them.

• Simplest networks are ISA hierarchies

• Must be careful to make a type/token distinction: Garfield isa Cat Cat(Garfield)

Cat isa Feline x (Cat (x) Feline(x))

• Restricted shorthand for a logical representation.

Semantic Nets/Frames

• Labeled links can represent arbitrary relations between objects and/or concepts.

• Nodes with links can also be viewed as frames with slots that point to other objects and/or concepts.

First Order Representation

Rel(Alive,Animals,T)

Rel(Flies,Animals,F)

Birds Animals

Mammals Animals

Rel(Flies,Birds,T)

Rel(Legs,Birds,2)

Rel(Legs,Mammals,4)

Penguins Birds

Cats Mammals

Bats Mammals

Rel(Flies,Penguins,F)

Rel(Legs,Bats,2)

Rel(Flies,Bats,T)

Opus Penguins

Bill Cats

Pat Bats

Name(Opus,"Opus")

Name(Bill,"Bill")

Friend(Opus,Bill)

Friend(Bill,Opus)

Name(Pat,"Pat")

Inheritance• Inheritance is a specific type of inference that allows properties of objects

to be inferred from properties of categories to which the object belongs. – Is Bill alive? – Yes, since Bill is a cat, cats are mammals, mammals are animals, and animals are

alive.

• Such inference can be performed by a simple graph traversal algorithm and implemented very efficiently.

• However, it is basically a form of logical inference x (Cat(x) Mammal(x))

x (Mammal(x) Animal(x))

x (Animal(x) Alive(x))

Cat(Bill)

|- Alive(Bill)

Backward or Forward

• Can work either way

• Either can be inefficient

• Usually depends on branching factors

Semantic of Links

• Must be careful to distinguish different types of links.

• Links between tokens and tokens are different than links between types and types and links between tokens and types.

Link Types

Link Type Semantics ExampleA subset B A B Cats Mammals

A member B A B Bill Cats

A R B R(A,B) Bill Age 12

A R B x, x A R(x,B)

Birds Legs 2

A R B x y, x A y B R(x,y)

Birds Parent Birds

Inheritance with Exceptions

• Information specified for a type gives the default value for a relation, but this may be over ridden by a more specific type. – Tweety is a bird. Does Tweety fly?

Birds fly. Yes. – Opus is a penguin. Does Opus fly?

Penguin's don't fly. No.

Multiple Inheritance

• If hierarchy is not a tree but a directed acyclic graph (DAG) then different inheritance paths may result in different defaults being inherited.

• Nixon Diamond

Nonmonotonicity

• In normal monotonic logic, adding more sentences to a KB only entails more conclusions. if KB |- P then KB {S} |- P

• Inheritance with exceptions is not monotonic (it is nonmonotonic) – Bird(Opus) – Fly(Opus)? yes – Penguin(Opus) – Fly(Opus)? no

• Nonmonotonic logics attempt to formalize default reasoning by allow default rules of the form: – If P and concluding Q is consistent, then

conclude Q. – If Bird(X) then if consistent Fly(x)

Defaults with Negation as Failure• Prolog negation as failure can be used to implement

default inference. fly(X) : bird(X), not(ab(X)).

ab(X) : penguin(X).

ab(X) : ostrich(X).

bird(opus).

? fly(opus).

Yes

penguin(opus).

? fly(opus).

No

Planning

Search

• What are characteristics of good problems for search?

• What does the search know about the goal state?

• Consider the package problem on the exam:– How well would search REALLY work on that

problem?

Search vs. Planning

• Planning systems:– Open up action and goal representation to allow

selection– Divide and conquer by subgoaling– Relax the requirement for sequential

construction of solutions

Planning in Situation CalculusPlanResult(p,s) is the situation resulting from executing p in sPlanResult([],s) = s

PlanResult([a|p],s) = PlanResult(p,Result(a,s))

Initial state At(Home,S_0) Have(Milk,S_0) …

Actions as Successor State axiomsHave(Milk,Result(a,s)) [(a=Buy(Milk) At(Supermarket,s)) Have(Milk,s) a

...)]

Query s=PlanResult(p,S_0) At(Home,s) Have(Milk,s) …

Solution p = Go(Supermarket),Buy(Milk),Buy(Bananas),Go(HWS),…]

• Principal difficulty: unconstrained branching, hard to apply heuristics

The Blocks World

• We have three blocks A, B, and C

• We can know things like whether a block is clear (nothing on top of it) and whether one block is on another (or on the table)

• Initial State:

• Goal State:

A

B CA

B

C

Situation Calculus in Prologholds(on(A,B),result(puton(A,B),S)) :

holds(clear(A),S), holds(clear(B),S),

neq(A,B).

holds(clear(C),result(puton(A,B),S)) :

holds(clear(A),S), holds(clear(B),S),

holds(on(A,C),S),

neq(A,B).

holds(on(X,Y),result(puton(A,B),S)) :

holds(on(X,Y),S),

neq(X,A), neq(Y,A), neq(A,B).

holds(clear(X),result(puton(A,B),S)) :

holds(clear(X),S), neq(X,B).

holds(clear(table),S).

neq(a,table).

neq(table,a).

neq(b,table).

neq(table,b).

neq(c,table).

neq(table,c).

neq(a,b).

neq(b,a).

neq(a,c).

neq(c,a).

neq(b,c).

neq(c,b).

Situation Calculus Plannerplan([],_,_).

plan([G1|Gs], S0, S) :

holds(G1,S),

plan(Gs, S0, S),

reachable(S,S0).

reachable(S,S).

reachable(result(_,S1),S) :

reachable(S1,S). • However, what will happen if we try to make plans using

normal Prolog depth first search?

Stack of 3 Blocks

holds(on(a,b), s0).

holds(on(b,table), s0).

holds(on(c,table),s0).

holds(clear(a), s0).

holds(clear(c), s0).

| ? cpu_time(db_prove(6,plan([on(a,b),on(b,c)],s0,S)), T).

S = result(puton(a,b),result(puton(b,c),result(puton(a,table),s0)))

T = 1.3433E+01

Invert stack

holds(on(a,table), s0).

holds(on(b,a), s0).

holds(on(c,b),s0).

holds(clear(c), s0).

? cpu_time(db_prove(6,plan([on(b,c),on(a,b)],s0,S)),T).

S = result(puton(a,b),result(puton(b,c),result(puton(c,table),s0))),

T = 7.034E+00

Simple Four Block Stackholds(on(a,table), s0).

holds(on(b,table), s0).

holds(on(c,table),s0).

holds(on(d,table),s0).

holds(clear(c), s0).

holds(clear(b), s0).

holds(clear(a), s0).

holds(clear(d), s0).

| ? cpu_time(db_prove(7,plan([on(b,c),on(a,b),on(c,d)],s0,S)),T).

S = result(puton(a,b),result(puton(b,c),result(puton(c,d),s0))),

T = 2.765935E+04

7.5 hours!