问题
Here is the problem:
$ 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
(I can switch the order of the subqueries, the result is the same).
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.
回答1:
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)
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)
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)
回答2:
Let's start with the most obvious one. If you switch the goals, you have:
?- length(L, N), N in 1..3.
which has the same termination properties as:
?- length(L, N), false,N in 1..3.
So obviously, this must not terminate with Prolog's execution mechanism.
However, if you put N in 1..3 in front, this might affect termination. To do so, it must be possible with finite means to prove that there is no N from 4 on. How can you prove this in a system without constraints - that is, only with syntactic unification present? Well, you can't. And length/2 is commonly defined just without constraints present.
With library(clpfd) things are trivial, for N #>= 4, N in 1..3 simply fails1. Note also that library(clpfd) does not collaborate much with library(clpq) which might be an interesting candidate, too.
As a consequence you would need to define your own length — for each constraint package you are interested in. That's a bit of a pity, but currently there is no generic way to do so in sight. ((That is, if you are interested and think a bit about it, you might come up with a nice API that every constraint system should adhere to. Alas, this will take some more decades, I suspect. Currently, there is much too much divergence.))
So here is a first naive way for fd_length/2:
fd_length([], N) :-
N #= 0.
fd_length([_|L], N0) :-
N0 #>= 1,
N1 #= N0-1,
fd_length(L, N1).
OK, this could be optimized to avoid the superfluous choicepoint. But there is a more fundamental problem: If you are determining the length of a list of length N, this will create N constraint variables! But we do need only one.
fd_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).
Again, this is not perfect for so many reasons: It could use Brent's algorithm like current systems do ; and combine it with all the fd properties. Also, arithmetic expressions are probably not a good idea to permit ; but I would have to wait for (#)/1 in SWI...
1: Strictly speaking, this "simply fails" only for SICStus, SWI, and YAP. For in those systems, there is no accidental failure due to exhaustion of the current representation. That is, their failure can always be taken as an honest no.
回答3:
How about the following baroque work-around based on clpfd and meta-predicate tcount/3?
:- use_module([library(clpfd), library(lambda)]). list_FDlen(Xs, N) :- tcount(\_^ =(true), Xs, N).
Let's query!
?- 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
What about this particular query?
?- 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)
回答4:
We present a clpfd-ish variant of length/2 that's tailored to @mat's clpfd implementation.
:- use_module(library(clpfd)). :- use_module(library(dialect/sicstus)). :- multifile clpfd:run_propagator/2.
The "exported" predicate lazy_len/2 is defined like this:
lazy_len(Es, N) :- N in 0..sup, % lengths are always non-negative integers lazylist_acc_len(Es, 0, N), create_mutable(Es+0, State), clpfd:make_propagator(list_FD_size(State,N), Propagator), clpfd:init_propagator(N, Propagator), clpfd:trigger_once(Propagator).
The global constraint handler list_FD_size/3 incrementally modifies its internal state as constraint propagation occurs. All modifications are trailed and are un-done upon backtracking.
clpfd:run_propagator(list_FD_size(State,N), _MState) :- get_mutable(Es0+Min0, State), fd_inf(N, Min), Diff is Min - Min0, length(Delta, Diff), append(Delta, Es, Es0), ( integer(N) -> Es = [] ; Delta = [] -> true % unchanged ; update_mutable(Es+Min, State) ).
lazy_len/2 tackles the problem from two sides; the clpfd constraint part of it was shown above. The tree side uses prolog-coroutining to walk down the list as far as the partial instantiation allows1:
lazylist_acc_len(_, _, N) :- integer(N), !. lazylist_acc_len(Es, N0, N) :- var(Es), !, when((nonvar(N);nonvar(Es)), lazylist_acc_len(Es,N0,N)). lazylist_acc_len([], N, N). lazylist_acc_len([_|Es], N0, N) :- N1 is N0+1, N in N1..sup, lazylist_acc_len(Es, N1, N).
Sample queries:
?- lazy_len(Xs, N). when((nonvar(N);nonvar(Xs)), lazylist_acc_len(Xs,0,N)), N in 0..sup, list_FD_size(Xs+0, N). ?- lazy_len(Xs, 3). Xs = [_A,_B,_C]. ?- lazy_len([_,_], L). L = 2. ?- lazy_len(Xs, L), L #> 0. Xs = [_A|_B], when((nonvar(L);nonvar(_B)), lazylist_acc_len(_B,1,L)), L in 1..sup, list_FD_size(_B+1, L). ?- lazy_len(Xs, L), L #> 2. Xs = [_A,_B,_C|_D], when((nonvar(L);nonvar(_D)), lazylist_acc_len(_D,3,L)), L in 3..sup, list_FD_size(_D+3, L). ?- lazy_len(Xs, L), L #> 0, L #> 2. Xs = [_A,_B,_C|_D], when((nonvar(L);nonvar(_D)), lazylist_acc_len(_D,3,L)), L in 3..sup, list_FD_size(_D+3, L).
And, at long last, one more query... well, actually two more: one going up—the other going down.
?- L in 1..4, lazy_len(Xs, L), labeling([up], [L]). L = 1, Xs = [_A] ; L = 2, Xs = [_A,_B] ; L = 3, Xs = [_A,_B,_C] ; L = 4, Xs = [_A,_B,_C,_D]. ?- L in 1..4, lazy_len(Xs, L), labeling([down], [L]). L = 4, Xs = [_A,_B,_C,_D] ; L = 3, Xs = [_A,_B,_C] ; L = 2, Xs = [_A,_B] ; L = 1, Xs = [_A].
Footnote 1: Here, we focus on preserving determinism (avoid the creation of choice-points) by using delayed goals.
来源：https://stackoverflow.com/questions/32478193/using-a-constrained-variable-with-length-2