使用带有`length/2`的约束变量

Question

这是问题所在:

$ swipl
Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 7.3.6-5-g5aeabd5)
Copyright (c) 1990-2015 University of Amsterdam, VU Amsterdam
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software,
and you are welcome to redistribute it under certain conditions.
Please visit http://www.swi-prolog.org for details.

For help, use ?- help(Topic). or ?- apropos(Word).

?- 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).

我想N在使用之前我需要贴标签,但我想知道问题是什么？我之前没有成功length/2.

Answer 1

什么可能比略微不确定的更有用的length/2是适当的列表长度约束.您可以在此处找到它的ECLiPSe实现,称为.有了这个,你会得到以下行为:len/2

?- 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,您可以通过枚举来枚举有效列表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)

或者通过生成具有良好旧标准的列表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)