Comp 307 Lecture 4:4

Prolog and a small rule based system

How Prolog works?

Matching

Backtracking

Rule based system

Comp 307 Lecture 4:5

The first program% father(Father, Child)father(bob, liz). father(bob, jim).father(jim, pam).father(jim, pat).

mother(ann, jim).mother(pat, tom).

parent(X,Y) :- mother(X,Y).parent(X,Y) :- father(X,Y).

grandparent(X,Y) :- parent(X,Z), parent(Z,Y).

ancestor(X,Y) :- parent(X,Y).ancestor(X,Y) :- parent(X,Z), ancestor(Z,Y).

Comp 307 Lecture 4:6

|?-parent(X, Y).X=ann,Y=jim;X=pat,Y=tom;X=bob,Y=liz;X=bob,Y=jim;X=jim,Y=pam;X=jim,Y=pat;no.| ?- father(bob, X), mother(ann, X).X=jim?;no

Queries and Answers

Comp 307 Lecture 4:7

• Prolog answers a query by matching the query with FACTs and the HEADs of the rules

– If the query matches a fact, then succeed– If the query matches the head of a rule, then

the body of the rule is now treated as a query, and prolog tries to solve that

– Top-down, left-to-right, Depth-first

• If Prolog’s search fails, it backtracks to try alternative solutions, if possible.

How Prolog answers a query

Comp 307 Lecture 4:8

How Prolog answers a query|?- grandparent(bob, Y).

Comp 307 Lecture 4:9

• When– Arguments at the same place– The same variable in one clause– X = Y

• How– Propagate variable bindings forward– Once a variable has a binding, it cannot be

changed -- there is no destructive assignment

Matching

Comp 307 Lecture 4:10

Backtracking

• When– When a sub goal failed– Multiple answers, using “;”– ……., fail.

• How:– Undo variable bindings, try alternative clauses

Comp 307 Lecture 4:11

How Prolog answers a query|?-ancestor(bob, pat).

Comp 307

Simple Rule-based System in Prologmammal:-f(hair). mammal:-f(milk).bird:-f(feathers).bird:-f(eggs), f(flies).carnivore:-mammal, f(meat).classify(tiger):-carnivore, f(tawny), f(striped). classify(puma):-carnivore, f(black).

f(hair).f(black).f(meat).

|?-classify(X).X=puma.

Comp 307 Lecture 4:13

• Term Term Possible Instantiation a 6 no

a Z Z=a A C A=C date(2, 3, 2000) Z Z=date(2, 3, 2000) date(D, M, Y) Z Z=date(D, M, Y) date(X, Y, Z) date(Y, X, Z) X=Y date(D, 5, Y) date(D1, M, 2000) D=D1, M=5, Y=2000 date(6, 5, 2000) date(6, may, 2000) no date(D, M, Y) date1(D, M, Y) no

Matching (Unification)

Comp 307 Lecture 4:14

• Term Term Possible Instantiation [] [] yes

[a, b, c] [X, Y, c] X=a, Y=b

[1, [2, 3], 4] [X, Y, Z] X=1, Y=[2, 3], Z=4

[X,Y] [a,b,c] no[H|T] [a,b] H=a, T=[b][H|T] [1,[2,3],4] H=1, T=[[2,3],4]

Matching (Unification)

Comp 307 Lecture 4:15

Matching (Unification)• If S and T are constants, then S and T match only if they are the

same object• If S is a variable and T is anything, then they match, and S is

instantiated to T (and vice-versa)• If S and T are structures, then they match only if

– S and T have the same principal functor, and– all their corresponding components match (the recursive part)

• If there are more than one bindings, we always compute the most general unifier:

e.g. f(X,Y)=f(a,Z) X=a,Y=Z or X=a,Z=Y (Not X=a, Y=b, Z=b)

Comp 307 Lecture 4:16

terms: To match = or to compare ==

To match X=YTo compare X==Y, X\==Y

e.g: [a,b] == [a,b] yes [a,X] == [a,b] no [a,X] = [a,b] X=b

Note: X\==Y is always true if there are free variables (X, Y must not be free variable)