Artificial Intelligence Logic Part 2

All rights reserved L. Manevitz Lecture 5 1

Artificial IntelligenceLogic Part 2

L. Manevitz


Unification

• Alphabetic variant.

• Term substitution.

• “Ground”.

P(x,f(y),B)

S1={z/x,w/y}

S2={g(z)/x,A/y}

S3={C/x,A/y}

P(z,f(w),B)

P(g(z),f(A),B)

P(C,f(A),B)


Example

x [P(x) Q(x)]

P(a)

Q(a)

x [ P(x) Q(x)]

P(x) Q(x)

P(a) Q(a)

P(a)

P(a)

Q(a)

Substitution


Unification cont.

m.g.u g: s s’

{Ei}s = {Ei}g s’

{A(x),A(z)} {A(x),A(z)}

{A(a)} {A(x)}S1={a/x,a/

z}S2={x/z}

{Ei} S1 = {Ei}g S2


Example of m.g.u

{P(x),P(A)}

{P[f(x),y,g(y)],P[f(x),x,g(z)]}

{P[f(x,g(A,y)),g(A,y)],P[f(x,z),z]}

P(A)

P[f(x),x,g(x)]

P[f(x,g(A,y)),g(A,y)]


Unification – Algorithm

Unify(E1,E2)

• If E1 or E2 is an atom (that is, a predicate symbol, a function symbol, a constant symbol, a negation symbol or a variable),

interchange the arguments E1 and E2 (if necessary) so that E1 is an atom, and do:


Unification – Algorithm cont.

2. Begin:1. If E1 and E2 are identical, return NIL.2. If E1 is a variable, do:3. Begin:

2. If E1 occurs in E2, return FAIL.3. Return {E2 / E1}.

4. End.5. If E2 is a variable, return {E1 / E2}.6. Return FAIL.

3. End.



4. F1the first element of E1, T1 the rest of E1.

5. F2 the first element of E2, T2 the rest of E2.

6. Z1UNIFY(F1,F2).

7. If Z1 = FAIL, return FAIL.



8. G1result of applying Z1 to T1.

9. G2result of applying Z1 to T2.

10. Z2UNIFY(G1,G2).

11. If Z2 = FAIL, return FAIL.

12. Return the composition of Z1and Z2.


4Trace of Unification

• Unify ((fxx), (fAy))– 4. F1 = f; T1 = (xx) ; F2 = f; T2 =(Ay)– 6. Unify (f, f)

• 1. f = f so return NIL8. G1 = NIL(T1) = (xx) G2 = (Ay)10. Unify ((xx, Ay) 4. F1 = x F2 = A etc 6. Unify (x, A) 3.3 Return A/x

8. Apply A/x to (x) (y) get (A y) 10. Unify (A, y) get A/y Return composition of A/x and A/yResult is fAA.


Example

1. Whoever can read is literate.

2. Dolphins are not literate.

3. Some dolphins are intelligent.

x [R(x) L(x)]

x [D(x) L(x)]

x [D(x) I(x)]


Example cont.

To prove:

some who are intelligent can not read.

x [I(x) R(x)]


Example cont.

Clause Form:

1. R(x) L(x)

2. D(y) L(y)

3. Divided to two: A is Skolem constant1. D(A)

2. I(A)

4. Negation of theory:

I(z) R(z)


Example cont.

I(z) R(z) I(A)

R(A) R(x) L(x)

L(A) D(y) L(y)

D(A) D(A)

NIL

3.2 4

1

2

3.1


Control Methods

• Breadth First.

• Set of Support.

• Linear Input.


Breadth First - Example I(z) R(z)I(A)

R(A)

R(x) L(x)

L(A)

D(y) L(y)

L(A)

D(A)

NIL

I(z) L(z) R(y) D(y)

I(A) R(A) I(z) D(z) D(A)L(A)


Set of Support - Example I(z) R(z) I(A)

R(A)

R(x) L(x)

L(A)

D(y) L(y) D(A)

I(A)

I(y) D(y)

D(A)

L(A)

D(A) D(A)

I(z) L(z)


Linear Input - ExampleI(A)

R(A)

R(x) L(x)

L(A)

D(y) L(y) D(A)

I(A)

I(y) D(y)L(A) R(A)

I(z) L(z)

I(z) R(z)

R(y) D(y)

I(z) D(z)

L(A)


A refutation tree(note: no linear input tree)

Q(x) P(x) Q(y) P(y)

Q(w) P(w)

P(A)

P(x)

Q(w)

NIL

Q(u) P(A)


Ancestor Filtered

• Like linear input but also …

• Additional resolutions where one parent is the ancestor of the other one.

• Surprisingly, this is a complete strategy


A refutation tree(note: no linear input tree)

Q(x) P(x) Q(y) P(y)

Q(w) P(w)

P(A)

P(x)

Q(w)

NIL

Q(u) P(A)


Example no.1

x [AT(JOHN,x) AT(FIDO,x)]

AT(JOHN,School)

Where is FIDO ?

x [AT(FIDO,x)]

x [ AT(FIDO,x)]

AT(FIDO,x)


Example no.1 cont.

Answer:

• Append negation of each clause that arises from negation of theorem to each leaf in derivation tree in which it appears.

• Follow structure of derivation tree.

• Use clause at root as answer.


Example no.1 cont.AT(FIDO,x)Negation of Goal

AT(JOHN,y) AT(FIDO,y)Axiom 1

NIL

AT(JOHN,x)AT(JOHN,SCHOOL)

Axiom 2

y x

x SCHOOL


Example no.1-Green’s Trick

AT(JOHN,y) AT(FIDO,y)

AT(FIDO,SCHOOL)

AT(JOHN,SCHOOL)

y x

x SCHOOL

AT(FIDO,x) AT(FIDO,x)

AT(JOHN,x) AT(FIDO,x)


Example no.2

On (B,A)

On (A,Table)

T.S. Above (B,Table)

The axioms :

x y[On(x,y) Above(x,y)]

x y z[Above(x,y) Above(y,z) Above(x,z)]


Example no.2 cont.

Clause form:

1. On(u,v) Above(u,v)

2. Above(x,y) Above(y,z) Above(x,z)

3. On(B,A)

4. On(A,Table)

5. Above(B,Table)


Example no.2 – Resolution

Above(B,y) Above(y,Table)

On(y,Table) Above(B,y)

On(B,y) On(y,Table)

2 5

1

1

3

4 On(A,Table)

NIL


Example no.3

Predicates:

On(A,Table)

On(B,A)

Function: PutonTable?

x On(x,Table) How to use the function?

Need to add situation parameter


Frame Problem

AA

BB

BBAA

Operator:

•PutonTable(x):

•If x not on table, it puts x on table.

How should we represent this in logic ?How should we represent this in logic ?


Making Situation Part of Logic

S0 – initial situation.

On(A,Table,S0)

On(B,A,S0)

PutonTable axiom:

z x [ On(x,Table,z) On(x,Table,PutonTable(x,z)]

(PutonTable is function P:Object X Situation Situation)


Making Situation Part of Logic

Other axioms (how do we find them ?)

x y z [On(x,y,z) On(y,x,z)]

A = B

A = Table

B = Table


Frame Problem cont.

When will B be on the Table? Apply Green’s TrickGoal: z On(B,Table,z)

z On(B,Table,z)Axioms: On(B,A,S0)

On(A,Table,S0) z x[ On(x,Table,z) On(x,Table,PutonTable(x,z))] z y x[On(y,x,z) x=Table On(y,Table,z)] x y z[On(y,x,z) On(x,y,z)]

negated


Frame Problem cont.Clause form:1. On(B,A,S0)2. On(A,Table,S0)3. On(x,Table, S3) On(x,Table,PutonTable(x,S3))4. On(y,z,S4) z=Table On(y,Table,S4)5. (A=B)6. (B=Table)7. (A=Table)8. On(u,v,w) On(v,u,w)9. On(B,Table,s)

Goal Formula


On(x,Table, S3) On(x,Table,PutonTable(x,S3)) On(B,Table,S)

On(B,Table, S3)

On(B,z,S3) z=Table (A=Table)

On(B,A,S3)

NIL

93

On(y,z,S4) z=Table On(y,Table,S4)4

7

On(B,A,S0)1

x B

y B S4 S3

z A

S3 S0

S PutonTable(B,S3)

Green’s Trick

On(B,Table,S)

On(B,Table, PutonTable(B,S3))





Frame Problem cont.

Do we know a state where both A and B

on table?

Goal: s[On(B,Table,s) On(A,Table,s)]

Negate + Clause:

On(B,Table,s) On(A,Table,s)

New formula no.9


Slide Missing on Start of Proofof A and B on table

• Show where you get stuck


Frame Problem cont.

The knowledge that A stays put

wasn’t put into logic !

Needs a “frame axiom” :

s x y z[On(x,y,s) x=z On(x,y,PutonTable(z,s))]

Clause form:

On(p,q,t) p=r On(p,q,PutonTable(r,t))

Formula no.10


On(p,q,t) p=r On(p,q,PutonTable(r,t))

On(A,Table,S0) A=B (A=B)

On(A,Table,S0)

NIL

10

5

On(A,Table,S0)2

p A r B

t S0

On(A,Table, PutonTable(B,S0))

9

q Table

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Artificial Intelligence Logic Part 2