返回两个或多个结果时成功的谓词 [英] Predicate that succeeds if two or more results are returned

问题描述

如果rule2返回两个或多个结果，如何实现成功的rule1?

rule1(X) :-规则 2(X, _).

如何计算结果，然后设置何时成功的最小值?

解决方案

如何计算结果，然后设置为真的最小值?

不清楚您所说的结果是什么意思.所以我会做一些猜测.结果可能是:

一个解决方案.例如，目标 member(X,[1,2,1]) 有两个解决方案.不是三个.在这种情况下，请考虑使用 setof/3 或类似的谓词.无论如何，在解决您遇到的问题之前，您应该先了解setof/3.

答案.目标 member(X,[1,2,1]) 有三个答案.目标 member(X,[Y,Z]) 有两个答案，但有无穷多个解.

因此，如果您想确保至少有一定数量的答案，请定义:

<预>至少(目标，N):-\+ \+ call_nth(目标，N).

使用 call_nth/2 在另一个 SO-answer 中定义.

请注意，其他 SO 答案是不正确的:它们要么不会终止，要么不会产生意外的实例.

How to implement rule1 that succeeds iff rule2 returns two or more results?

rule1(X) :-
    rule2(X, _).

How can I count the results, and then set a minimum for when to succeed?

解决方案

How can I count the results, and then set a minimum for when it's true?

It is not clear what you mean by results. So I will make some guesses. A result might be:

A solution. For example, the goal member(X,[1,2,1]) has two solutions. Not three. In this case consider using either setof/3 or a similar predicate. In any case, you should first understand setof/3 before addressing the problem you have.

An answer. The goal member(X,[1,2,1]) has three answers. The goal member(X,[Y,Z]) has two answers, but infinitely many solutions.

So if you want to ensure that there are at least a certain number of answers, define:

at_least(Goal, N) :-
   \+ \+ call_nth(Goal, N).

with call_nth/2 defined in another SO-answer.

Note that the other SO-answers are not correct: They either do not terminate or produce unexpected instantiations.

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