Permutations: Difference between revisions

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(Mercury!)
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(%i1) permutations([1, 2, 3]);
(%i1) permutations([1, 2, 3]);
(%o1) {[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]}
(%o1) {[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]}
</lang>

=={{header|Mercury}}==
<lang mercury>
:- module permutations2.
:- interface.

:- import_module io.

:- pred main(io::di, io::uo) is det.


:- import_module list.
:- import_module set_ordlist.
:- import_module set.
:- import_module solutions.

%% permutations(List, Set) is true if List is a permutation of Set:
:- pred permutationSet(list(A)::out,set(A)::in) is nondet.

%% Two ways to compute all permutations of a given list (using backtracking):
:- func all_permutations1(list(int))=set_ordlist.set_ordlist(list(int)).
:- func all_permutations2(list(int))=set_ordlist.set_ordlist(list(int)).

:- implementation.


permutationSet([],set.init).
permutationSet([H|T], S) :- set.member(H,S), permutationSet(T,set.delete(S,H)).

all_permutations1(L) =
solutions_set(pred(X::out) is nondet:-permutationSet(X,set.from_list(L))).

all_permutations2(L) =
solutions_set(pred(X::out) is nondet:-perm(L,X)).

main(!IO) :-
print(all_permutations1([1,2,3,4,5]),!IO),
nl(!IO),
print(all_permutations2([1,2,3,4,5]),!IO).
</lang>
</lang>