解决Prolog中的Kakuro拼图（5x5）

Question

Assuming that:

A+B+C=24
E+F+G=11
J+K+L=22
N+O+P=14
A+E=17
B+F+J+N=26
C+G+K+O=15
L+P=13

How could i find a possible solution to the problem, given the constraints above, using the predicate solve/1 ? My first attempt was below, with no result. Thanks in advance!

solve(L1) :-
    L1 = [A,B,C,E,F,G,J,K,L,N,O,P],
    A is 24-B-C,
    B is 26-F-J-N,
    C is 15-G-K-O,
    E is 11-F-G,
    E is 17-A,
    J is 22-K-L,
    N is 14-O-P,
    L is 13-P,
    write(L1).

Answer 1

As @lurker already said in his comment, use CLP(FD) constraints .

In addition, I recommend:

1. instead of solve/1 , use a declarative name like solution/1 . You should describe what holds for a solution, so that the relation makes sense in all directions, also for example if the solution is already given and you want to validate it.
2. By convention, it makes sense to let variables that stand for lists end with an s .
3. Separate side-effects from pure code. In fact, remove side-effects altogether. Let the prolog-toplevel do the printing for you!

For example:

:- use_module(library(clpfd)).

solution(Ls) :-
    Ls = [A,B,C,E,F,G,J,K,L,N,O,P],
    A #= 24-B-C,
    B #= 26-F-J-N,
    C #= 15-G-K-O,
    E #= 11-F-G,
    E #= 17-A,
    J #= 22-K-L,
    N #= 14-O-P,
    L #= 13-P.

This already works for queries like:

?- solution(Ls), Ls ins 0..sup, label(Ls).
Ls = [6, 3, 15, 11, 0, 0, 9, 0, 13, 14, 0, 0] ;
Ls = [6, 3, 15, 11, 0, 0, 10, 0, 12, 13, 0, 1] ;
Ls = [6, 3, 15, 11, 0, 0, 11, 0, 11, 12, 0, 2] ;
etc.

I leave completing this as an easy exercise.