Re: prolog and backtracking

James Sison <the_evil_trembles@yahoo.com> 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).
      %@ Yes


      %?- sibling(ruby, jake).
      %@ No


> 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])).


Output:


      sibling(jake, ruby).
      sibling(ruby, jake).

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