Computer Science: An Overview: Global Edition (12th Edition)

Published by Pearson Higher Education
ISBN 10: 1292061162
ISBN 13: 978-1-29206-116-0

Chapter 6 - Programming Languages - Section 6.7 - Declarative Programming - Questions & Exercises - Page 323: 4

Answer

Prolog will conclude that Carol is her own sibling. To solve this problem, the rule needs to include the fact that X cannot be equal to Y, which in Prolog is written X \ = Y. Thus an improved version of the rule would be sibling (X, Y) :-X \= Y, parent(Z, X), parent(Z, Y). which says that X is Y’s sibling if X and Y are not equal and have a common parent. The following version would insist that X and Y are siblings only if they have both parents in common: sibling (X, Y) :− X \= Y, Z \= W parent (Z, X), parent (Z, Y), parent (W, X), parent (W, Y).
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Work Step by Step

Prolog will conclude that Carol is her own sibling. To solve this problem, the rule needs to include the fact that X cannot be equal to Y, which in Prolog is written X \ = Y. Thus an improved version of the rule would be sibling (X, Y) :-X \= Y, parent(Z, X), parent(Z, Y). which says that X is Y’s sibling if X and Y are not equal and have a common parent. The following version would insist that X and Y are siblings only if they have both parents in common: sibling (X, Y) :− X \= Y, Z \= W parent (Z, X), parent (Z, Y), parent (W, X), parent (W, Y).
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