在 SWI-Prolog 中聚合谓词

问题描述

我需要计算 some_predicate(X) 所持有的所有 X，而且确实有很多这样的 X.最好的方法是什么?

I need to count all X for which some_predicate(X) holds, and there really a lot of such X.What is the best way to do that?

第一条线索是findall，累加成一个列表，返回列表的长度.

First clue is to findall, accumulate to a list and return the length of the list.

countAllStuff( X ) :-
    findall( Y
           , permutation( [1,2,3,4,5,6,7,8,9,10], Y )
           , List
           ),
    length( List, X ).

(permutation/2 只是一个虚拟的占位符，表明有很多结果，而且计算计数的方法很糟糕)

(permutation/2 is only a dummy placeholder demonstrating that there are many results and that it's bad way to compute the count)

显然，对于真实数据，会有堆栈溢出.

Obviously, with real data, there will be a stack overflow.

?- countAllStuff( X ).
ERROR: Out of global stack

然后，我试图用 setof 替换 findall，但无济于事.

Then, I'm trying to replace findall with setof, to no avail.

最后，我找到了 [aggregate][1](可点击)系列谓词，并尝试使用 aggregate/3 和 aggregate/4:

At last, I've found the [aggregate][1] (clickable) family of predicates, and trying to use aggregate/3 and aggregate/4:

?- aggregate(count, permutation([1,2,3,4], X), Y ).
X = [1, 2, 3, 4],
Y = 1 .

?- aggregate(count, [1,2,3,4], permutation([1,2,3,4], X), Y ).
X = [1, 2, 3, 4],
Y = 1 ;
X = [1, 2, 4, 3],
Y = 1 ;

我想，这一切都错了.我需要得到这样的东西:

It's all wrong, I think. I need to get something like this:

?- aggregate(count, permutation([1,2,3,4], X), Y ).
Y = 24 .

1. 我做错了什么?

1. What am I doing wrong?

我如何声明一个谓词来计算正确的答案?[1]:http://www.swi-prolog.org/pldoc/doc/home/vnc/prolog/lib/swipl/library/aggregate.pl

How can I declare a predicate to conpute the right answer?[1]: http://www.swi-prolog.org/pldoc/doc/home/vnc/prolog/lib/swipl/library/aggregate.pl

推荐答案

使用存在量化的变量，就像使用 setof 一样:

Use an existentially quantified variable, as you would with setof:

?- aggregate(count, X^permutation([1,2,3,4], X), N).
N = 24.

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