prolog - Using a constrained variable with `length/2`

Question

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.

Answer 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)