Get a limited amount of solutions from a prolog Goal?

[sbe...@gmail.com]

Hello,

First I would like to say that I'm a complete newbie in prolog so
maybe there is a better way to do what I'm trying to accomplish and a
bit of enlightment would be very welcomed.

Lets imagine that I have a possibly infinite predicate that represents
the integer numbers set.

positive_integer(X)

In this way positive_integer(X) would render 1,2,3,4,5,6.. each time a
new solution to the goal is demanded, but a findall would hang the
interpreter.

Is there any simple way to either have a new predicate built upon the
previous one that fails when a certain number of solutions have been
found as could be:

is_limited(Count),integer(X)

Which would only allow to REDO Count times before failing.

Or either build a findlimited(Term,Goal,N,LIST), which would only
return the first N Terms that satisfy the goal?

Up to now the only ways I have thought of, are either using a
combination of bag_create, bag_dissolve which I dont like, or using
assert and retract which I like even less.

Any ideas?

Thanks for your time

[Chip Eastham]

While it is possible to wrap the "generator"
predicate positive_integer(X) of your specific
illustration in an "outer" predicate,
is_limited(Limit,X):

is_limited(Limit,X) :- positive_integer(X),
( (X > Limit, !, fail) ; X <= Limit ).

so that the wrapper fails once X exceeds
Count, I don't think the general effect you
are looking for can be obtained without
side-effects, eg. assert/retract or
some other type of persistent accumulator.

regards, chip

[sbe...@gmail.com]

Thanks for the tips chip. Your solution would be perfect if not
because I can't know beforehand without exploring which is the
explicit limit for the interval im exploring. Maybe a better expample
wold have been the following:

is_limited(Count),integer(X), check_condition(X).

Here only integers that satisfay a certain condition would be
returned. The problem is that the constraint i need to impose is not
over the explored domain, but over the number of Terms allowed to
satisfy the predicate.

Thanks, Sergio.

[Joachim Schimpf]

As Chip, said, you need some persistent counter. Since you seem to be
using ECLiPSe, the simplest is to use a "shelf":

shelf_create(count(Limit), Count),
<your goal>,
( shelf_dec(Count, 1) -> true ; !, fail ).

-- Joachim

[sbe...@gmail.com]

Joachim,

My infinite gratitude to chip and you for both tips, it works like a
charm right now using the shelf_create solution.

Regards.

Sergio.