为什么我的谓词不回溯? [英] Why is my predicate not backtracking?

问题描述

我不明白为什么我的谓词不回溯并找到所有解决方案.

I don't understand why my predicate doesnt backtrack and find all solutions.

person(john).
person(erik).

allExceptSpider(person(Spider),T ):-
    setof(person(X),person(X),S),
    subtract(S,[person(Spider) ],T).

如果我用两个变量来调用该谓词:

If I call this predicate with Two variables:

allExceptSpider(person(Z),Q)

那么它只会给我答案Z = john，Q = [person(erik)] 但是找到Z = erik，Q = [person(john)]不会回溯 为什么?

Then it will only give me the answer Z = john, Q = [person(erik)] but it won't backtrack to find Z = erik ,Q = [person(john)] why?

推荐答案

TL; DR::如果您使用逻辑纯度.

person(john).
person(erik).

allExceptSpider(Spider, T) :-
   setof(X, person(X), S),
   subtract(S, [Spider], T).

保持纯度！怎么样?像这样使用 list_item_subtracted/3

Preserve purity! How? Use list_item_subtracted/3 like so:

allExceptSpiderNU(Spider, T) :-
   setof(X, person(X), S),
   list_item_subtracted(S, Spider, T).

直接查询示例:

?- allExceptSpider(Z, Q).
Q = [erik], Z = john.

?- allExceptSpiderNU(Z,Q).
   Q = [     erik]             ,     Z=john
;  Q = [john     ],     Z=erik
;  Q = [john,erik], dif(Z,erik), dif(Z,john).

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