如何在序言中计算序言中父母的孩子数(不使用列表)? [英] How to count the number of children of parents in prolog (without using lists) in prolog?

问题描述

我有以下问题.我有一定数量的事实，例如:父母(简，迪克).父母(迈克尔，迪克).我想要一个谓词，例如:儿童数量(迈克尔，X)所以如果我这样称呼它，它会显示 X=1.

我在网上搜了一下，大家都把孩子放在列表里，有没有不使用列表的方法?

解决方案

计算解决方案的数量需要一些额外的逻辑工具(它本质上是非单调的).这是一个可能的解决方案:

<预><代码>:- 动态 count_solutions_store/1.count_solutions(目标，N):-断言(count_solutions_store(0))，重复，(呼叫(目标)，收回(count_solutions_store(SoFar))，更新的是 SoFar + 1，断言(count_solutions_store(更新))，失败;收回(count_solutions_store(T))),！，N = T.

I have the following problem. I have a certain number of facts such as: parent(jane,dick). parent(michael,dick). And I want to have a predicate such as: numberofchildren(michael,X) so that if I call it like that it shows X=1.

I've searched the web and everyone puts the children into lists, is there a way not to use lists?

解决方案

Counting number of solutions requires some extra logical tool (it's inherently non monotonic). Here a possible solution:

:- dynamic count_solutions_store/1.

count_solutions(Goal, N) :-
    assert(count_solutions_store(0)),
    repeat,
    (   call(Goal),
        retract(count_solutions_store(SoFar)),
        Updated is SoFar + 1,
        assert(count_solutions_store(Updated)),
        fail
    ;   retract(count_solutions_store(T))
    ),
    !, N = T.

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