问题描述
问题来了:
$ swipl
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?- use_module(library(clpfd)).
true.
?- N in 1..3, length(L, N).
N = 1,
L = [_G1580] ;
N = 2,
L = [_G1580, _G1583] ;
N = 3,
L = [_G1580, _G1583, _G1586] ;
ERROR: Out of global stack % after a while
(我可以切换子查询的顺序，结果是一样的).
(I can switch the order of the subqueries, the result is the same).
我想我需要标记 N 才能使用它，但我想知道问题是什么?我之前没能把 length/2 噎住.
I guess I need to label N before I can use it, but I wonder what the problem is? I have not managed to choke up length/2 before.
推荐答案
可能比稍微不确定的 length/2 更有用的是适当的列表长度约束.您可以在 这里ECLiPSe 实现>，称为 len/2.有了这个，你会得到以下行为:
What's probably more useful than a slightly less nondeterministic length/2 is a proper list-length constraint. You can find an ECLiPSe implementation of it here, called len/2. With this you get the following behaviour:
?- N :: 1..3, len(Xs, N).
N = N{1 .. 3}
Xs = [_431|_482] % note it must contain at least one element!
There is 1 delayed goal.
Yes (0.00s cpu)
然后您可以通过枚举 N 来枚举有效列表:
You can then enumerate the valid lists either by enumerating N:
?- N :: 1..3, len(Xs, N), indomain(N).
N = 1
Xs = [_478]
Yes (0.00s cpu, solution 1, maybe more)
N = 2
Xs = [_478, _557]
Yes (0.02s cpu, solution 2, maybe more)
N = 3
Xs = [_478, _557, _561]
Yes (0.02s cpu, solution 3)
或通过使用旧标准 length/2 生成列表:
or by generating lists with good old standard length/2:
?- N :: 1..3, len(Xs, N), length(Xs, _).
N = 1
Xs = [_488]
Yes (0.00s cpu, solution 1, maybe more)
N = 2
Xs = [_488, _555]
Yes (0.02s cpu, solution 2, maybe more)
N = 3
Xs = [_488, _555, _636]
Yes (0.02s cpu, solution 3)
这篇关于使用带有 `length/2` 的约束变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！