Mercury nondet in det

Question

In Mercury, suppose you're in a det predicate, and you want to call a nondet predicate, as follows. If there are no solutions, you want Result = []; if there's one or more, you just want the first one like Result = [FirstSolution]. The nondet predicate may have an infinite number of solutions, so you can't enumerate them all and take the first one. The closest I've come is to use do_while and just stop after the first solution, but do_while is apparently cc_multi and I don't know how to coerce it back into a det context, or even back into a multi context so I can apply solutions to it.

Answer 1

While scanning through the builtin module for something else, I came across very some clear "if you want to use cc_multi in a det context" language that led me to this solution:

:- module nondetindet3.
:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module list, string, int, solutions.

:- pred numbers(int::out) is multi.
numbers(N) :-
    ( N = 1; N = 2; N = 3; N = 4 ),
    trace [io(!IO)] (io.format("number tried: %d\n", [i(N)], !IO)).

:- pred even_numbers(int::out) is nondet.
even_numbers(N) :-
    numbers(N),
    N rem 2 = 0.

:- pred first_number(int::out) is semidet.
first_number(N) :-
    promise_equivalent_solutions [N] (
        even_numbers(N)
    ).

main(!IO) :-
    ( if first_number(N1) then
        io.format("decided on: %d\n", [i(N1)], !IO)
    else true),
    ( if first_number(N2) then
        io.format("still want: %d\n", [i(N2)], !IO)
    else true),
    ( if promise_equivalent_solutions [N3] (even_numbers(N3), N3 > 2) then
        io.format("now want: %d\n", [i(N3)], !IO)
    else true).

I believe the meaning there is "hey, no need to keep searching for even_numbers/1 solutions, 'cause all those other solutions are going to be no better than the first one you get."

Output:

number tried: 1
number tried: 2
decided on: 2
number tried: 1
number tried: 2
still want: 2
number tried: 1
number tried: 2
number tried: 3
number tried: 4
now want: 4

Answer 2

Why exactly do you want to do this? If you're doing it to optimize some logical code, so that you do less unnecessary searching, there must be a better way. Something with solver types perhaps.

Anyway, this technically works:

:- module nondetindet.
:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module list, string, int, bool, solutions.

:- pred numbers(int::out) is multi.
numbers(N) :-
    ( N = 1; N = 2; N = 3; N = 4 ),
    trace [io(!IO)] (io.format("number tried: %d\n", [i(N)], !IO)).

:- pred even_numbers(int::out) is nondet.
even_numbers(N) :-
    numbers(N),
    N rem 2 = 0.

:- initialise(set_number/0).

:- mutable(number, int, 0, ground, [untrailed]).

:- impure pred set_number is cc_multi.
set_number :-
    do_while(even_numbers, (pred(N1::in, no::out, _::in, N1::out) is det), 0, N),
    impure set_number(N).

:- func first_number = int.
:- pragma promise_pure(first_number/0).
first_number = N :- semipure get_number(N).

main(!IO) :-
    io.format("decided on: %d\n", [i(first_number)], !IO),
    io.format("still want: %d\n", [i(first_number)], !IO).

And has output:

number tried: 1
number tried: 2
decided on: 2
still want: 2