如何从 SWI-Prolog 中的某些查询产生最大结果的列表中找到输入？
我现在刚刚开始学习 Prolog，所以我不熟悉做大多数事情的正常方式。
本质上，我有一个从输入中给出值的规则：
ScoreFromInput(Input, Score) :- ...
我有一个输入列表，它们只是数字。我无法弄清楚如何找到产生最高分数的输入。
这就是我现在所拥有的，但我认为它会无限递归：
bestInput(BestInput) :-
%#Bind the list of valid inputs
legalInputs(ValidInputs),
%# -1000 is a dummy score, it should get replaced on the first call to bestInputHelper
bestInputHelper(ValidInputs,-1000,BestInput).
%#I think this rule should work if the first input in the list is not the best one
bestInputHelper([Input|RestOfInputs],BestScore,BestInput):-
bestInputHelper(RestOfInputs,RestBestScore,BestInput),
ScoreFromInput(Input,BestScore),
RestBestScore > BestScore.
%#And this one if it is the best input
bestInputHelper([Input|RestOfInputs],BestScore,Input):-
bestInputHelper(RestOfInputs,RestBestScore,_RestBestInput),
ScoreFromInput(Input,BestScore),
RestBestScore =< BestScore.
这是我到目前为止所拥有的，但我想有一种更直接的方法来做到这一点。任何帮助表示赞赏！谢谢！
I'm just picking up Prolog now, so I'm unfamiliar with the normal way of doing most things.
Essentially I have a rule which gives a value from an input:
ScoreFromInput(Input, Score) :- ...
And I have a list of inputs, which are just numbers. I'm having trouble figuring out how to find the input which yields the maximum score.
This is what I have right now, but I think it recurses infinitely:
bestInput(BestInput) :-
%#Bind the list of valid inputs
legalInputs(ValidInputs),
%# -1000 is a dummy score, it should get replaced on the first call to bestInputHelper
bestInputHelper(ValidInputs,-1000,BestInput).
%#I think this rule should work if the first input in the list is not the best one
bestInputHelper([Input|RestOfInputs],BestScore,BestInput):-
bestInputHelper(RestOfInputs,RestBestScore,BestInput),
ScoreFromInput(Input,BestScore),
RestBestScore > BestScore.
%#And this one if it is the best input
bestInputHelper([Input|RestOfInputs],BestScore,Input):-
bestInputHelper(RestOfInputs,RestBestScore,_RestBestInput),
ScoreFromInput(Input,BestScore),
RestBestScore =< BestScore.
This is what I have so far, but I imagine there is a much more straightforward way of doing it. Any help is appreciated! Thanks!
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一个简单的说法是，如果没有更好的输入，那么输入就是最好的：
要查看您自己的代码有什么问题，请尝试使用 SWI-Prolog 的图形跟踪器：
and please_use_read_names inSteadOfUnreadOnes。
A simple way to state it is that an input is best if there is no better input:
To see what is wrong with your own code, try for example SWI-Prolog's graphical tracer:
And please_use_readable_names inSteadOfUnreadableOnes.
尽管 Chris 对 Prolog 不太熟悉，但他概述的方法可能比 mat 的方法更有效地找到具有最大分数的输入。像克里斯这样的方法可以线性扫描可能的输入，而不是进行二次比较。
这里 maxScoreOfList/3 将返回有效输入列表的最佳项目 Z 和最佳分数 B 作为第三个参数。谓词将在空列表上失败。
需要一个“辅助”函数，如下所示，它说明了添加一些额外参数的“技巧”，以便当到达输入列表的末尾时，输出 Z 和 B 可以绑定到找到的最佳项目和分数“所以远的”：
Despite Chris's lack of familiarity with Prolog, the approach he outlined may be a more efficient way of finding the input with the maximum score than mat's. Instead of doing a quadratic number of comparisons, an approach like Chris's is possible that linearly scans the possible inputs.
Here maxScoreOfList/3 will return the best item Z and the best score B for a list of valid inputs as the third argument. The predicate will fail on an empty list.
A "helper" function is needed as follows, which illustrates the "trick" of adding some extra arguments so that when the end of the input list is reached, the outputs Z and B can be bound to the best item and score found "so far":