Select N items from the list in Prolog

Question

I am trying to write a Prolog predicate (SWI) that will select N items from a list, for example:

selectn (+ N,? Elems,? List1,? List2) is true when List1 with all Elems removed results in List2.

selectn(N,Lps,L1s,[]) :- length(L1s,L), N >= L, permutation(L1s,Lps). selectn(0,[],L1s,Lps) :- permutation(L1s,Lps). selectn(N,[E|Es],L1s,L2s) :- select(E,L1s,L0s), N0 is N-1, selectn(N0,Es,L0s,L2s). 

My problem is that in some cases I get duplicate results, and I don't know how to avoid them:

 ?- findall(L,selectn(2,Es,[a,b,c],L),Ls),length(Ls,Solutions). Ls = [[c], [b], [c], [a], [b], [a]], Solutions = 6. 

This is not homework, but if you want to help me as if it were, I will be happy too.

Answer 1

this may answer your question (although I do not understand your first selectn / 4 clause, the permutation is already done using the "nested" select / 3)

 selectn(0, [], Rest, Rest). selectn(N, [A|B], C, Rest) :- append(H, [A|T], C), M is N-1, selectn(M, B, T, S), append(H, S, Rest). 

gives

 ?- findall(L,selectn(2,Es,[a,b,c],L),Ls),length(Ls,Solutions). Ls = [[c], [b], [a]], Solutions = 3.