Prolog 使用带有'length/2'的约束变量`
问题是：Prolog 使用带有'length/2'的约束变量`,prolog,clpfd,meta-predicate,prolog-coroutining,Prolog,Clpfd,Meta Predicate,Prolog Coroutining,问题是： $ 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 redistrib
$ 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
我想我需要在使用它之前标记
N长度/2?- length(L, N), N in 1..3.
NNfd_length(L, N) :-
N #>= 0,
fd_length(L, N, 0).
fd_length([], N, N0) :-
N #= N0.
fd_length([_|L], N, N0) :-
N1 is N0+1,
N #>= N1,
fd_length(L, N, N1).
（#）/11：严格地说，这“完全失败”只适用于SICStus、SWI和YAP。因为在这些系统中，不会由于耗尽当前表示而发生意外故障。也就是说，他们的失败总是可以被视为诚实的拒绝。 以下基于和的解决方案如何 :- ([, ]). 列表_FDlen（Xs，N）：- （（对），Xs，N）。 让我们提问 ?- N in 1..3, list_FDlen(Xs, N). N = 1, Xs = [_A] ; N = 2, Xs = [_A,_B] ; N = 3, Xs = [_A,_B,_C] ; false. % terminates universally ?- N in inf..2, list_FDlen(Xs, N). N = 0, Xs = [] ; N = 1, Xs = [_A] ; N = 2, Xs = [_A,_B] ; false. % terminates universally, too ？-N在1..3中，列出（Xs，N）。 N=1，Xs=[[u A] ; N=2，Xs=[[u A，u B] ; N=3，Xs=[[u A，[u B，[u C] ; 错.%普遍终止 -在inf..2中的-N，列表_FDlen（Xs，N）。 N=0，Xs=[] ; N=1，Xs=[[u A] ; N=2，Xs=[[u A，u B] ; 错.%也普遍终止 那么这个特殊的查询呢 ?- N in 2..sup, list_FDlen(Xs, N). N = 2, Xs = [_A,_B] ; N = 3, Xs = [_A,_B,_C] ; N = 4, Xs = [_A,_B,_C,_D] ... % does not terminate (as expected) ？-N在2..sup，list_FDlen（Xs，N）中。 N=2，Xs=[[u A，u B] ; N=3，Xs=[[u A，[u B，[u C] ; N=4，Xs=[[u A，[u B，[u C，[u D] ... % 不终止（如预期）
与不太确定的
长度/2len/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)
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)
长度/2fd_length([], N) :-
N #= 0.
fd_length([_|L], N0) :-
N0 #>= 1,
N1 #= N0-1,
fd_length(L, N1).
?- 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)
list\u FD\u size/3脚注1： 在这里，我们通过使用延迟目标来保持决定论（避免创建选择点）。
对于：SICStus，GNU，也有同样的问题……这是一个多么复杂的Prolog和扩展的好例子。。。谢谢你解决了这个问题…我很想看到这个功能变得生动起来@重复：我认为，一旦修复了语义，可能会非常简单。。。只需扩展长度/2处理属性变量…@capelical。听起来不错！但是，恕我直言，clpfd实现（在SWI中使用，也可以在yap中使用）是100%Prolog和0%C。当黑客攻击C实现时，例如，
length/2？-N#<0，length（L，N）。fd#length/2？-fd_长度（L，N）。G，falseGlength/2same_length/3suspend（len（Xs，L，N），0，[[Xs0，N]->inst，N->min]）length/2