为什么在定义转换两个原子关系的谓词时会出现堆栈限制超出错误？

Question

我想知道为什么在这些情况下程序会无限递归：

?- love(kay, amanda).

和

?- love(rob, amanda).

这是代码：

love(amanda, kay).
love(kay, geo).
love(geo, rob).

love(X, Y) :-
   love(X, Z),
   love(Z, Y).
love(X, Y) :-
   love(Y, X).

Answer 1

First, your program always goes into an infinite loop. No matter what names you are using. Even ?- loves(amanda, kay). loops. To better see this rather ask ?- loves(amanda, kay), false. Why am I so sure that your program always loops?

To see this, we need some falsework. By adding goals false into your program, we get a new program requiring less (or equal) inferences.

love(amanda, kay) :- false.
love(kay, geo) :- false.
love(geo, rob) :- false.
love(X, Y) :-
   love(X, Z), false,
   love(Z, Y).
love(X, Y) :-  false,
   love(Y, X).

Because this fragment (called a failure-slice) does not terminate, your original program will not terminate. As you can see, all the persons have been removed. And thus actual names cannot influence the outcome.

If you want to fix commutativity, rather introduce a further predicate:

love2(X, Y) :- love1(X, Y).
love2(X, Y) :- love1(Y, X).

And to include the transitive closure, use closure/3:

love3(X, Y) :-
   closure(love2, X, Y).