Prolog:成员谓词的真/假输出? [英] Prolog: true / false output of member predicate?

问题描述

我想知道为什么 Prolog 的 member/2 谓词提供了多种选择(通过回溯?！)，如果 true 已经统一用于输出.

I am wondering why the member/2 predicate of Prolog delivers multiple alternatives (via backtracking?!), if true was already unified for the output.

例如 member(1, [1,2,3]). 提供以下输出:

For example member(1, [1,2,3]). delivers the following output:

true ;
false.

为什么member在已经发现原子1确实是列表[1,2,3]的成员后返回false代码>?

Why does member return false after it already found out that the atom 1 is indeed a member of the list [1,2,3]?

更令我困惑的是以下输出:

Even more confusing to me is the following output:

?- member(1, [1,2,3,1]).
true ;
true.

推荐答案

在你的第一个例子中，你要求它证明 member(1,[1,2,3]).;因为它可以，所以它报告 true.当您输入 ; 时，您要求找到另一种方法来证明该查询；因为它不能，所以它报告 false.

In your first example, you ask it to prove member(1,[1,2,3]).; since it can, it reports true. When you enter ;, you are asking to find another way to prove that query; since it cannot, it reports false.

在第二种情况下，第一个 true 是因为它在列表中找到了 1 之一；第二个是因为它找到了第二个.如果您再次点击 ;，它会返回 false，因为它没有其他方法来证明查询.(注意:正如@WillNess 指出的那样，您实际上没有机会再次点击 ;；这可能是由于 member 的实现使得 Prolog 知道没有其他选择.如果列表没有以 1 结尾，您将能够再次点击 ;.)

In the second case, the first true is because it found one of the 1's in the list; the second is because it found the second. If you had hit ; again, it would have come back with false, as it had no other ways to prove the query. (Note: As @WillNess points out, you don't actually get the chance to hit ; again; this is probably due to the implementation of member being such that Prolog knows there are no remaining alternatives. If the list did not end with a 1, you would be able to hit ; again.)

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