Prolog: Arrangements of k elements with sum of elements S

Question

I am trying to compute arrangements of K elements in Prolog, where the sum of their elements is equal to a given S. So, I know that arrangements can be computed by finding the combinations and then permute them. I know how to compute combinations of K elements, something like:

comb([E|_], 1, [E]).
comb([_|T], K, R) :-
   comb(T, K, R).
comb([H|T], K, [H|R]) :-
   K > 1,
   K1 is K-1,
   comb(T, K1, R).

The permutations of a list, having the property that the sum of their elements is equal to a given S, I know to compute like this:

insert(E, L, [E|L]).
insert(E, [H|T], [H|R]) :-
   insert(E, T, R).

perm([], []).
perm([H|T], P) :-
   perm(T, R),
   insert(H, R, P).

sumList([], 0).
sumList([H], H) :-
   number(H).
sumList([H|Tail], R1) :-
   sumList(Tail, R),
   R1 is R+H.

perms(L, S, R) :-
   perm(L, R),
   sumList(R, S1),
   S = S1.

allPerms(L, LP) :-
   findall(R, perms(L,R), LP).

The problem is that I do not know how to combine them, in order to get the arrangements of K elements, having the sum of elements equal to a given S. Any help would be appreciated.

Answer 1

Use clpfd!

:- use_module(library(clpfd)).

Using SWI-Prolog 7.3.16 we query:

?- length(Zs,4), Zs ins 1..4, sum(Zs,#=,7), labeling([],Zs).
   Zs = [1,1,1,4]
;  Zs = [1,1,2,3]
;  Zs = [1,1,3,2]
;  Zs = [1,1,4,1]
;  Zs = [1,2,1,3]
;  Zs = [1,2,2,2]
;  Zs = [1,2,3,1]
;  Zs = [1,3,1,2]
;  Zs = [1,3,2,1]
;  Zs = [1,4,1,1]
;  Zs = [2,1,1,3]
;  Zs = [2,1,2,2]
;  Zs = [2,1,3,1]
;  Zs = [2,2,1,2]
;  Zs = [2,2,2,1]
;  Zs = [2,3,1,1]
;  Zs = [3,1,1,2]
;  Zs = [3,1,2,1]
;  Zs = [3,2,1,1]
;  Zs = [4,1,1,1].

To eliminate "redundant modulo permutation" solutions use chain/2:

?- length(Zs,4), Zs ins 1..4, chain(Zs,#=<), sum(Zs,#=,7), labeling([],Zs).
   Zs = [1,1,1,4]
;  Zs = [1,1,2,3]
;  Zs = [1,2,2,2]
;  false.

Answer 2

I use SWI-Prolog. You can write that

:- use_module(library(lambda)).

arrangement(K, S, L) :-
    % we have a list of K numbers
    length(L, K),
    % these numbers are between 1 (or 0) and S
    maplist(between(1, S), L),
    % the sum of these numbers is S
    foldl(\X^Y^Z^(Z is X+Y), L, 0, S).

The result

 ?- arrangement(5, 10, L).
L = [1, 1, 1, 1, 6] ;
L = [1, 1, 1, 2, 5] ;
L = [1, 1, 1, 3, 4] ;
L = [1, 1, 1, 4, 3] .

You can use also a CLP(FD) library.

Edited after the remark of @repeat.

Answer 3

This response is similar to response of @repeat

predicates that below are implemented using the SICStus 4.3.2 tool

after simple modification of gen_list(+,+,?)

edit Code

gen_list(Length,Sum,List) :-    length(List,Length), 
                                domain(List,0,Sum), 
                                sum(List,#=,Sum), 
                                labeling([],List),

                                % to avoid duplicate results
                                ordered(List).

Test

| ?- gen_list(4,7,L).
L = [0,0,0,7] ? ;
L = [0,0,1,6] ? ;
L = [0,0,2,5] ? ;
L = [0,0,3,4] ? ;
L = [0,1,1,5] ? ;
L = [0,1,2,4] ? ;
L = [0,1,3,3] ? ;
L = [0,2,2,3] ? ;
L = [1,1,1,4] ? ;
L = [1,1,2,3] ? ;
L = [1,2,2,2] ? ;
no

Answer 4

I don't think that permutations could be relevant for your problem. Since the sum operation is commutative, the order of elements should be actually irrelevant. So, after this correction

sumList([], 0).
%sumList([H], H) :-
%   number(H).
sumList([H|Tail], R1) :-
   sumList(Tail, R),
   R1 is R+H.

you can just use your predicates

'arrangements of K elements'(Elements, K, Sum, Arrangement) :-
    comb(Elements, K, Arrangement),
    sumList(Arrangement, Sum).

test:

'arrangements of K elements'([1,2,3,4,5,6],3,11,A).
A = [2, 4, 5] ;
A = [2, 3, 6] ;
A = [1, 4, 6] ;
false.

You already know how to use findall/3 to get all lists at once, if you need them.