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.
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, [])
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].
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 ?-
