Artificial Intelligence Logic Part 2

1. Artificial IntelligenceLogic Part 2 L. Manevitz L. Manevitz Lecture 5

2. Unification P(x,f(y),B) • Alphabetic variant. • Term substitution. • “Ground”. S1={z/x,w/y} P(z,f(w),B) S2={g(z)/x,A/y} P(g(z),f(A),B) S3={C/x,A/y} P(C,f(A),B) L. Manevitz Lecture 5

3. P(a) P(a) P(a) Q(a) P(x) Q(x) Q(a) Example x [P(x) Q(x)] P(a) Q(a) x [ P(x) Q(x)] Substitution L. Manevitz Lecture 5

4. {A(x),A(z)} {A(x),A(z)} {A(a)} {A(x)} {Ei} S1 = {Ei}g S2 Unification cont. m.g.u g: s s’ {Ei}s = {Ei}g s’ S1={a/x,a/z} S2={x/z} L. Manevitz Lecture 5

5. P(A) P[f(x),x,g(x)] P[f(x,g(A,y)),g(A,y)] 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]} L. Manevitz Lecture 5

6. 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: L. Manevitz Lecture 5

7. Unification – Algorithm cont. • Begin: • If E1 and E2 are identical, return NIL. • If E1 is a variable, do: • Begin: • If E1 occurs in E2, return FAIL. • Return {E2 / E1}. • End. • If E2 is a variable, return {E1 / E2}. • Return FAIL. • End. L. Manevitz Lecture 5

8. Unification – Algorithm cont. • F1the first element of E1, T1 the rest of E1. • F2 the first element of E2, T2 the rest of E2. • Z1UNIFY(F1,F2). • If Z1 = FAIL, return FAIL. L. Manevitz Lecture 5

9. Unification – Algorithm cont. • G1result of applying Z1 to T1. • G2result of applying Z1 to T2. • Z2UNIFY(G1,G2). • If Z2 = FAIL, return FAIL. • Return the composition of Z1and Z2. L. Manevitz Lecture 5

10. 4Trace of Unification • Unify ((fxx), (fAy)) • 4. F1 = f; T1 = (xx) ; F2 = f; T2 =(Ay) • 6. Unify (f, f) • 1. f = f so return NIL 8. 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/y Result is fAA. L. Manevitz Lecture 5

11. x [R(x) L(x)] x [D(x) L(x)] x [D(x) I(x)] Example • Whoever can read is literate. • Dolphins are not literate. • Some dolphins are intelligent. L. Manevitz Lecture 5

12. x [I(x) R(x)] Example cont. To prove: some who are intelligent can not read. L. Manevitz Lecture 5

13. Example cont. Clause Form: • R(x) L(x) • D(y) L(y) • Divided to two: A is Skolem constant • D(A) • I(A) • Negation of theory: I(z) R(z) L. Manevitz Lecture 5

14. I(z) R(z) I(A) R(A) L(A) D(A) D(A) R(x) L(x) D(y) L(y) 3.2 4 1 2 3.1 Example cont. NIL L. Manevitz Lecture 5

15. Control Methods • Breadth First. • Set of Support. • Linear Input. L. Manevitz Lecture 5

16. I(z) R(z) I(z) L(z) I(A) R(A) L(A) D(A) L(A) I(A) R(A) L(A) D(A) R(x) L(x) I(z) D(z) R(y) D(y) D(y) L(y) NIL Breadth First - Example L. Manevitz Lecture 5

17. I(z) R(z) I(z) L(z) I(A) R(A) L(A) D(A) I(A) D(A) L(A) D(A) D(A) R(x) L(x) I(y) D(y) D(y) L(y) Set of Support - Example L. Manevitz Lecture 5

18. I(z) R(z) I(z) L(z) I(A) L(A) L(A) R(A) L(A) R(A) I(A) D(A) R(x) L(x) I(z) D(z) I(y) D(y) R(y) D(y) D(y) L(y) Linear Input - Example L. Manevitz Lecture 5

19. Q(y) P(y) P(A) P(x) Q(w) Q(w) P(w) Q(u) P(A) Q(x) P(x) NIL A refutation tree(note: no linear input tree) L. Manevitz Lecture 5

20. 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 L. Manevitz Lecture 5

21. Q(y) P(y) P(A) P(x) Q(w) Q(w) P(w) Q(u) P(A) Q(x) P(x) NIL A refutation tree(note: no linear input tree) L. Manevitz Lecture 5

22. 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) L. Manevitz Lecture 5

23. 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. L. Manevitz Lecture 5

24. AT(FIDO,x) Negation of Goal AT(JOHN,x) y x AT(JOHN,SCHOOL) Axiom 2 x SCHOOL NIL Example no.1 cont. AT(JOHN,y) AT(FIDO,y) Axiom 1 L. Manevitz Lecture 5

25. AT(JOHN,x) AT(FIDO,x) AT(FIDO,x) AT(FIDO,x) AT(JOHN,y) AT(FIDO,y) y x AT(JOHN,SCHOOL) x SCHOOL AT(FIDO,SCHOOL) Example no.1-Green’s Trick L. Manevitz Lecture 5

26. 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)] L. Manevitz Lecture 5

27. Example no.2 cont. Clause form: • On(u,v) Above(u,v) • Above(x,y) Above(y,z) Above(x,z) • On(B,A) • On(A,Table) • Above(B,Table) L. Manevitz Lecture 5

28. Above(B,y) Above(y,Table) On(y,Table) Above(B,y) 2 5 1 1 3 4 NIL On(B,y) On(y,Table) On(A,Table) Example no.2 – Resolution L. Manevitz Lecture 5

29. 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 L. Manevitz Lecture 5

30. Frame Problem B A A B • Operator: • PutonTable(x): • If x not on table, it puts x on table. How should we represent this in logic ? L. Manevitz Lecture 5

31. 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) L. Manevitz Lecture 5

32. 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 L. Manevitz Lecture 5

33. Frame Problem cont. When will B be on the Table? Apply Green’s Trick Goal: 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 L. Manevitz Lecture 5

34. Frame Problem cont. Clause form: • On(B,A,S0) • On(A,Table,S0) • On(x,Table, S3) On(x,Table,PutonTable(x,S3)) • On(y,z,S4) z=Table On(y,Table,S4) • (A=B) • (B=Table) • (A=Table) • On(u,v,w) On(v,u,w) • On(B,Table,s) Goal Formula L. Manevitz Lecture 5

35. On(y,z,S4) z=Table On(y,Table,S4) On(B,Table, S3) On(B,Table, PutonTable(B,S3)) S3 S0 S4 S3 On(B,z,S3) z=Table 9 3 4 7 1 On(B,A,S0) On(B,A,S3) NIL On(B,Table, PutonTable(B,S3)) On(B,Table,S) On(B,Table, PutonTable(B,S3)) On(B,Table, PutonTable(B,S0)) S PutonTable(B,S3) x B z A y B On(x,Table, S3)On(x,Table,PutonTable(x,S3)) On(B,Table,S) (A=Table) Green’s Trick L. Manevitz Lecture 5

36. 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 L. Manevitz Lecture 5

37. Slide Missing on Start of Proofof A and B on table • Show where you get stuck L. Manevitz Lecture 5

38. 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 L. Manevitz Lecture 5

39. q Table On(A,Table,S0) A=B (A=B) 10 5 2 9 On(A,Table,S0) On(A,Table,S0) On(A,Table, PutonTable(B,S0)) NIL p A r B t S0 On(p,q,t) p=r On(p,q,PutonTable(r,t)) L. Manevitz Lecture 5