Resolution_Part_4

Resolution_Part_4 - Unification vs. Matching Are p(X) and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Unification vs. Matching Are p(X) and p(f(X)) unifiable? e=[X/f(X)] X=f(f(f(f(f(. .... ?! This is not allowed in unification. Proper unification requires occurs check: a variable X can not be substituted by a term t if X occur in t. This is not done in Prolog’s matching for efficiency reasons. Therefore it is referred to as ‘matching’ in Prolog, and not ‘unification’. York University- CSE 3401 22 04_Resolution
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Given two clauses in the form: A 0 ..A i ..A m :-B 1 ...B n . and C 1 ...C k :- D 1 ..D j ..D l . If e is a unifier of A i and D j (i.e. e(A i )=e(D j )) Then the resolvent of the above two clauses is: e(A 0 ).. e(A i-1 )e(A i+1 ).. e(A m ) e(C 1 )..e(C k ) :- e(B 1 ).. e(B n ) e(D 1 ).. e(D j-1 )e(D j+1 ).. e(D l ). Example: C 1 : p(Y):- r(X, Y), q(Y, Z). C 2 : :- p(f(1)). Unifier of p( f(1) ) and p( Y ): e=[Y/f(1)] The resolvent of C 1 and C 2 : :- r( X, f(1) ), q( f(1), Z ). York University- CSE 3401
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

Resolution_Part_4 - Unification vs. Matching Are p(X) and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online