|prolog and backtracking firstname.lastname@example.org (James Sison) (2007-09-22)|
|Re: prolog and backtracking email@example.com (Markus Triska) (2007-09-24)|
|Re: prolog and backtracking firstname.lastname@example.org (2007-09-24)|
|From:||Markus Triska <email@example.com>|
|Date:||Mon, 24 Sep 2007 03:34:06 +0200|
|Posted-Date:||23 Sep 2007 22:31:51 EDT|
James Sison <firstname.lastname@example.org> writes:
> but jake has already been found and it should be different from the
> result of the first parent.
You can state this constraint in Prolog, using e.g. dif/2 or \==/2:
sibling(X,Y):- dif(X, Y), parent(bill,X), parent(bill,Y).
You could also omit symmetrical solutions using e.g. @</2:
sibling(X,Y):- parent(bill,X), parent(bill,Y), X @< Y.
This is hardly advisable though, since sibling/2 is after all a
symmetrical relation in reality, whereas you'd have for example:
%?- sibling(jake, ruby).
%?- sibling(ruby, jake).
> Is there a simpler way to address this problem?
Consider the built-in predicates setof/3 or findall/3, which you can
use to find all solutions; sort to eliminate duplicates if necessary:
%?- findall(X-Y, sibling(X, Y), XYs0), sort(XYs0, XYs).
%@ XYs0 = [jake-ruby, ruby-jake, jake-ruby, ruby-jake],
%@ XYs = [jake-ruby, ruby-jake]
>From this list you can easily generate your facts in Prolog (tested
with SWI-Prolog, where $XYs refers to the previous toplevel binding):
%?- forall(member(X-Y, $XYs), format("sibling(~w, ~w).\n", [X,Y])).
Return to the
Search the comp.compilers archives again.