I have this predicate which returns true if S is equal to some equation say K + 2N + 3L = S. The money we have are 1, 5, and 10 respectively for K, N, L.
I don't want to use :- use_module(library(clpfd)), I want to solve this without it.
My intuition was to break this into subproblems like write a function breakMoney1(S,K) :- K is S. and create more helpers with one more parameter added however I am struggling with the problem of getting uninstantiated variables, when I compare.
breakMoney(S,K,N,L) :-
This is easier than you think, probably. A very naive solution following @Will Ness' suggestion would be:
break(Sum, K, N, L) :- integer(Sum), Sum >= 0,
% upper bounds for K, N, and L
K_Max is Sum div 1,
N_Max is Sum div 5,
L_Max is Sum div 10,
% enumerate possible values for K, N, and L
between(0, L_Max, L),
between(0, N_Max, N),
between(0, K_Max, K),
Sum =:= K + 5*N + 10*L.
This will "magically" turn into a clp(fd) solution with very little effort: for example, replace between with X in 0..Max, and the =:= with #=. Although, it should be enough to simply say that X #>= 0 for each of the denominations. It is a good exercise to see how much of the constraints you can remove and still get an answer:
break(Sum, K, N, L) :-
K #>= 0, N #>= 0, L #>= 0,
Sum #= K + 5*N + 10*L.
Depending on how you instantiate the arguments, you might immediately get a unique answer, or you might need to use label/1:
?- break(100, P, 8, 5).
P = 10.
?- break(10, K, N, L).
K in 0..10,
-1*K+ -5*N+ -10*L#= -10,
N in 0..2,
L in 0..1.
?- break(10, K, N, L), label([K, N, L]).
K = N, N = 0,
L = 1 ;
K = L, L = 0,
N = 2 ;
K = 5,
N = 1,
L = 0 ;
K = 10,
N = L, L = 0.
But as @lurker has pointed out, there is very little reason not to use constraint programming for this problem. Unless, of course, you have a very clever algorithm for solving this particular problem and you know for a fact that it will outsmart the generic clp(fd) solution. Even then, it might be possible to achieve the same effect by using the options to labelling/2.
