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? [英] Why do i get a stack limit exceeded error when defining a predicate that convert the relation of two atoms?

问题描述

我想知道为什么在这些情况下程序会无限递归:

I want to know why does the program goes in an infinite recursion in those cases:

?- 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).

推荐答案

首先，您的程序总是进入无限循环.无论您使用什么名称.甚至 ?- love(amanda, kay). 循环.为了更好地看到这一点，而不是问 ?-loves(amanda, kay), false. 为什么我这么确定你的程序总是循环?

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?

要看到这一点，我们需要做一些假动作.通过在您的程序中添加目标 false，我们得到了一个需要更少(或相等)推理的新程序.

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).

因为这个片段(称为 failure-slice) 不会终止，您的原始程序不会终止.如您所见，所有人员都已被删除.因此，真实姓名不会影响结果.

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.

如果你想修正可交换性，不如引入一个进一步的谓词:

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

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

要包含传递闭包，请使用 closure/3:

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

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

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