SWI-Prolog 9.2.6 Reference Manual: Section 4.30

## 4.30 Finding all Solutions to a Goal

[ISO]findall(+Template, :Goal, -Bag)
  Create a list of the instantiations Template gets successively on backtracking over Goal and unify the result with Bag. Succeeds with an empty list if Goal has no solutions.

  findall/3 is equivalent to bagof/3 with all free variables appearing in Goal scoped to the Goal with an existential (caret) operator (`^`), except that bagof/3 fails when Goal has no solutions.

findall(+Template, :Goal, -Bag, +Tail)
  As findall/3, but returns the result as the difference list Bag-Tail. The 3-argument version is defined as

      findall(Templ, Goal, Bag) :-
              findall(Templ, Goal, Bag, [])

[nondet]findnsols(+N, @Template, :Goal, -List)
[nondet]findnsols(+N, @Template, :Goal, -List, ?Tail)
  As findall/3 and findall/4, but generates at most N solutions. If N solutions are returned, this predicate succeeds with a choice point if Goal has a choice point. Backtracking returns the next chunk of (at most) N solutions. In addition to passing a plain integer for N, a term of the form `count(N)` is accepted. Using `count(N)`, the size of the next chunk can be controlled using nb_setarg/3. The non-deterministic behaviour used to implement the chunk option in `library(pengines)`. Based on Ciao, but the Ciao version is deterministic. Portability can be achieved by wrapping the goal in once/1. Below are three examples. The first illustrates standard chunking of answers. The second illustrates that the chunk size can be adjusted dynamically and the last illustrates that no choice point is left if Goal leaves no choice-point after the last solution.

      ?- findnsols(5, I, between(1, 12, I), L).
      L = [1, 2, 3, 4, 5] ;
      L = [6, 7, 8, 9, 10] ;
      L = [11, 12].

      ?- State = count(2),
         findnsols(State, I, between(1, 12, I), L),
         nb_setarg(1, State, 5).
      State = count(5), L = [1, 2] ;
      State = count(5), L = [3, 4, 5, 6, 7] ;
      State = count(5), L = [8, 9, 10, 11, 12].

      ?- findnsols(4, I, between(1, 4, I), L).
      L = [1, 2, 3, 4].

[ISO]bagof(+Template, :Goal, -Bag)
  Unify Bag with the alternatives of Template. If Goal has free variables besides the one sharing with Template, bagof/3 will backtrack over the alternatives of these free variables, unifying Bag with the corresponding alternatives of Template. The construct `+Var`^`Goal` tells bagof/3 not to bind Var in Goal. bagof/3 fails if Goal has no solutions.

  The example below illustrates bagof/3 and the `^` operator. The variable bindings are printed together on one line to save paper.

      2 ?- listing(foo).
      foo(a, b, c).
      foo(a, b, d).
      foo(b, c, e).
      foo(b, c, f).
      foo(c, c, g).
      true.

      3 ?- bagof(C, foo(A, B, C), Cs).
      A = a, B = b, C = G308, Cs = [c, d] ;
      A = b, B = c, C = G308, Cs = [e, f] ;
      A = c, B = c, C = G308, Cs = [g].

      4 ?- bagof(C, A^foo(A, B, C), Cs).
      A = G324, B = b, C = G326, Cs = [c, d] ;
      A = G324, B = c, C = G326, Cs = [e, f, g].

      5 ?-

[ISO]setof(+Template, +Goal, -Set)
  Equivalent to bagof/3, but sorts the result using sort/2 to get a sorted list of alternatives without duplicates.