在没有 findall 的情况下收集所有解决方案 [英] Gathering all solutions without findall

问题描述

因此，作为我工作的一部分，我的代码需要打印查询的所有解决方案，但不使用 findall/3 谓词.我已经做了一些阅读，有一些方法涉及将解决方案添加到列表等等.我自己尝试过这样做，但没有成功；因此我希望有人能够展示我如何在不使用 findall 的情况下打印所有解决方案.

So as part of my work my code needs to print all the solutions to a query but without using the findall/3 predicate. I've done some reading around and there are ways involving adding the solutions to a list and so on. I tried doing this myself but to with no success; hence I was hoping someone may be able to show how I would print all the solutions without using findall.

程序代码如下:

    solutions(Q, 100):-
        Q = [X, Y, S],
        between(2,50,X),
        between(2,50,Y),
        S is X+Y,
        Y > X,
        S =< 50.

Q 和 100 在那里是因为程序的另一部分需要它，所以现在忽略它.当我使用 ?-solutions(Q, 100) 查询时，我得到结果 [2,3,5], [2,4,6], [2,5,7] 等等，但显然我需要按 ; 来获得每个新结果.我需要在不需要按 ; 和不使用 findall 的情况下显示所有这些.

The Q and the 100 are there because it's needed for another part of the program so ignore that for now. When I query using ?- solutions(Q, 100) I get the results [2,3,5], [2,4,6], [2,5,7] and so on but obviously I need to press ; to get each new result. I need all these to be displayed without the need to press ; and without using findall.

推荐答案

一个基于断言的解决方案(这实际上是在 Prolog 教科书中如何实现 findall 的):假设 solution/2 根据您的代码找到每个解决方案.现在我们按照 Paulo 的建议使用故障驱动循环来构建解决方案列表，使用 assert/1 来缓存解决方案.

An assert-based solution (which is actually how findall was implemented in Prolog textbooks): Assume that solution/2 finds each single solution, as per your code. Now we use a fail-driven loop, as Paulo suggested, to build a list of solutions, using assert/1 to cache solutions.

solutions(_, N) :-
  solution(Q, N),
  (cache(Qs) -> retractall(cache(_)) ; Qs = []),
  assert(cache([Q|Qs])),
  fail.
solutions(Qs, _) :-
  retract(cache(Qs)).

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