Prolog: Arrangements of k elements with sum of elements S - list

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.