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>getting the reverse solution (sum(?, 10)) Doing an "all modes" predicate: sum_list(List,Sum)
...where... sum_list([1,2,3], 6)
...is "True" as expected would probably be a pretty interesting exercise if you wanted to be able to get all of the lists. You'd probably need to do an "enumerate the rationals" type of thing, since you need to go to infinity in a couple of different directions at the same time. That is, you can't do nested loops/recursing: for(x=-∞; x<∞; ++x) {
for(y=-∞; y<∞; ++y) {
i.e. sum_list([],0).
sum_list([L|Ls],BigSum) :- sum_list(Ls, LittleSum), add(L,LittleSum,BigSum).
...with an all-modes "add/3". Since there are an infinite supply of pairs that add up to, say 10: ⋮
[12,-2]
[11,-1]
[10, 0]
[ 9, 1]
[ 8, 2]
⋮
[ 0,10]
[-1,11]
[-2,12]
⋮
...and you can also go to arbitrary list lengths: [10]
[1,9]
[1,1,8]
[1,1,1,7]
⋮
[1,1,1,1,1,1,1,1,1,1]
⋮
[0,1,1,1,1,1,1,1,1,1,1]
[1,0,1,1,1,1,1,1,1,1,1]
⋮
[1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0]
⋮
[-100,-99,-98,...,-12,-11,-9,-8,...,0,1,2,3,4,5,6,7,8,9,10,11,...,99,100]
...your life probably gets easier if you limit the domain to positive integers (without zero) since then the list length doesn't diverge. Then a declarative looking solution probably just has a all-modes equivalent to "sort" somewhere within? Certainly interesting to think about.https://www.cs.ox.ac.uk/people/jeremy.gibbons/publications/r... https://prolog-lang.org/ImplementersForum/0103-modes.html https://www.swi-prolog.org/pldoc/man?predicate=permutation%2... |