I am trying to print all solutions of the n-fractions problem for n=4:
:- lib(ic).
fractions(Digits) :-
Digits = [A,B,C,D,E,F,G,H,I,J,K,L],
Digits #:: 1..9,
ic:alldifferent(Digits),
X #= 10*B+C,
Y #= 10*E+F,
Z #= 10*H+I,
V #= 10*K+L,
A*Y*Z*V + D*X*Z*V + G*X*Y*V + J*X*Y*Z #= X*Y*Z*V,
A*Y #=< D*X,
D*Z #=< G*Y,
G*V #=< J*Z,
search(Digits,0,input_order,indomain,complete,[]).
When I run the query:
?- findall(Digits,fractions(Digits),List).
I get the following exception:
*** Overflow of the local/control stack!
You can use the "-l kBytes" (LOCALSIZE) option to have a larger stack.
Peak sizes were: local stack 105728 kbytes, control stack 25344 kbytes
I am thinking if there is a way to loop inside the program and print one solution each time, or I can't do that because the problem has too many solutions?
As has been pointed out, your code fails because the alldifferent(Digits) constraint is too restrictive. The digits must be allowed to occur between 1 and 2 times. In eclipse-clp, you can use constraints such as atleast/3, atmost/3, occurrences/3 or gcc/2 to express this.
Slightly off-topic: as you are using ECLiPSe's ic-solver (which can handle continuous domains), you can actually use a model much closer to the original specification, without introducing lots of multiplications:
:- lib(ic).
:- lib(ic_global).
fractions4(Digits) :-
Digits = [A,B,C,D,E,F,G,H,I,J,K,L],
Digits #:: 1..9,
A/(10*B+C) + D/(10*E+F) + G/(10*H+I) + J/(10*K+L) $= 1,
( for(I,1,9), param(Digits) do
occurrences(I, Digits, NOcc), NOcc #:: 1..2
),
lex_le([A,B,C], [D,E,F]), % lex-ordering to eliminate symmetry
lex_le([D,E,F], [G,H,I]),
lex_le([G,H,I], [J,K,L]),
labeling(Digits).
Apart from the main equality constraint (using $= instead of #= because we don't want to require integrality here), I've used occurrences/3 for the occurrence restrictions, and lexicographic ordering constraints as a more standard way of eliminating symmetry. Result:
?- findall(Ds, fractions4(Ds), Dss), length(Dss, NSol).
Dss = [[1, 2, 4, 3, 5, 6, 8, 1, 4, 9, 2, 7], [1, 2, 6, 5, 3, 9, 7, 1, 4, 8, 2, 4], [1, 2, 6, 5, 3, 9, 7, 8, 4, 9, 1, 2], [1, 2, 6, 7, 3, 9, 8, 1, 3, 9, 5, 4], [1, 2, 6, 8, 7, 8, 9, 1, 3, 9, 5, 4], [1, 3, 4, 5, 4, 6, 8, 1, 7, 9, 2, 3], [1, 3, 4, 7, 5, 6, 8, 1, 7, 9, 2, 4], [1, 3, 4, 8, 1, 7, 8, 5, 2, 9, 2, ...], [1, 3, 5, 6, 2, 8, 7, 1, 4, 9, ...], [1, 3, 6, 5, 2, 4, 7, 1, 8, ...], [1, 3, 6, 5, 3, 6, 7, 8, ...], [1, 3, 6, 5, 4, 5, 8, ...], [1, 3, 6, 5, 6, 3, ...], [1, 3, 6, 6, 5, ...], [1, 3, 6, 7, ...], [1, 3, 9, ...], [1, 3, ...], [1, ...], [...], ...]
NSol = 1384
Yes (82.66s cpu)
An added advantage of this model is that it can be quite easily turned into a generic model for arbitrary N.