Prolog Out of stack error

Question

I'm working on Problem 26 from 99 Prolog Problems:

P26 (**) Generate the combinations of K distinct objects chosen from the N elements of a list

Example:

?- combination(3,[a,b,c,d,e,f],L).

L = [a,b,c] ;

L = [a,b,d] ;

L = [a,b,e] ;

So my program is:

:- use_module(library(clpfd)).

combination(0, _, []).
combination(Tot, List, [H|T]) :- 
    length(List, Length), Tot in 1..Length,
    append(Prefix, [H], Stem), 
    append(Stem, Suffix, List), 
    append(Prefix, Suffix, SubList),
    SubTot #= Tot-1,
    combination(SubTot, SubList, T).

My query result starts fine but then returns a Global out of stack error:

?- combination(3,[a,b,c,d,e,f],L).
L = [a, b, c] ;
L = [a, b, d] ;
L = [a, b, e] ;
L = [a, b, f] ;
Out of global stack

I can't understand why it works at first, but then hangs until it gives Out of global stack error. Happens on both SWISH and swi-prolog in the terminal.

Answer 1

if you try to input, at the console prompt, this line of your code, and ask for backtracking:

?- append(Prefix, [H], Stem).
Prefix = [],
Stem = [H] ;
Prefix = [_6442],
Stem = [_6442, H] ;
Prefix = [_6442, _6454],
Stem = [_6442, _6454, H] ;
...

maybe you have a clue about the (main) problem. All 3 vars are free, then Prolog keeps on generating longer and longer lists on backtracking. As Boris already suggested, you should keep your program far simpler... for instance

combination(0, _, []).
combination(Tot, List, [H|T]) :- 
    Tot #> 0,
    select(H, List, SubList),
    SubTot #= Tot-1,
    combination(SubTot, SubList, T).

that yields

?- aggregate(count,L^combination(3,[a,b,c,d,e],L),N).
N = 60.

IMHO, library(clpfd) isn't going to make your life simpler while you're moving your first steps into Prolog. Modelling and debugging plain Prolog is already difficult with the basic constructs available, and CLP(FD) is an advanced feature...

Answer 2

I can't understand why it works at first, but then hangs until it gives Out of global stack error.

The answers Prolog produces for a specific query are shown incrementally. That is, the actual answers are produced lazily on demand. First, there were some answers you expected, then a loop was encountered. To be sure that a query terminates completely you have to go through all of them, hitting SPACE/or ; all the time. But there is a simpler way:

Simply add false at the end of your query. Now, all the answers are suppressed:

?- combination(3,[a,b,c,d,e,f],L), false.
ERROR: Out of global stack

By adding further false goals into your program, you can localize the actual culprit. See below all my attempts: I started with the first attempt, and then added further false until I found a terminating fragment (failure-slice).

combination(0, _, []) :- false.                % 1st
combination(Tot, List, [H|T]) :-
    length(List, Length), Tot in 1..Length,    % 4th terminating
    append(Prefix, [H], Stem), false,          % 3rd loops
    append(Stem, Suffix, List), false,         % 2nd loops
    append(Prefix, Suffix, SubList),
    SubTot #= Tot-1, false,                    % 1st loops
    combination(SubTot, SubList, T).

To remove the problem with non-termination you have to modify something in the remaining visible part. Evidently, both Prefix and Stem occur here for the first time.

Answer 3

The use of library(clpfd) in this case is very suspicious. After length(List, Length), Length is definitely bound to a non-negative integer, so why the constraint? And your Tot in 1..Length is weird, too, since you keep on making a new constrained variable in every step of the recursion, and you try to unify it with 0. I am not sure I understand your logic overall :-(

If I understand what the exercise is asking for, I would suggest the following somewhat simpler approach. First, make sure your K is not larger than the total number of elements. Then, just pick one element at a time until you have enough. It could go something like this:

k_comb(K, L, C) :-
    length(L, N),
    length(C, K),
    K =< N,
    k_comb_1(C, L).

k_comb_1([], _).
k_comb_1([X|Xs], L) :-
    select(X, L, L0),
    k_comb_1(Xs, L0).

The important message here is that it is the list itself that defines the recursion, and you really don't need a counter, let alone one with constraints on it.

select/3 is a textbook predicate, I guess you should find it in standard libraries too; anyway, see here for an implementation.

This does the following:

?- k_comb(2, [a,b,c], C).
C = [a, b] ;
C = [a, c] ;
C = [b, a] ;
C = [b, c] ;
C = [c, a] ;
C = [c, b] ;
false.

And with your example:

?- k_comb(3, [a,b,c,d,e,f], C).
C = [a, b, c] ;
C = [a, b, d] ;
C = [a, b, e] ;
C = [a, b, f] ;
C = [a, c, b] ;
C = [a, c, d] ;
C = [a, c, e] ;
C = [a, c, f] ;
C = [a, d, b] ;
C = [a, d, c] ;
C = [a, d, e] ;
C = [a, d, f] ;
C = [a, e, b] ;
C = [a, e, c] . % and so on

Note that this does not check that the elements of the list in the second argument are indeed unique; it just takes elements from distinct positions.

This solution still has problems with termination but I don't know if this is relevant for you.