Kandan Venkataraman wrote:
Assume I have a predicate defined f(X,Y) taking two parameters.
If X is a variable, or a partially instantiated variable (i.e. it is a
structure), then assume X can have multiple solutions depending on the
value of Y.
How do I store the multiple solutions of X in a list. The problem I am
having is that once X is bound to a value, I am
unable to unbind it , to find the next solution to store in the list.
You might consider simply using bagof/3 or setof/3 depending on whether
you would mind duplicate solutions. If this is for a class assignment,
the classic solution is use recursion as in and a clever one is to use a
difference list . . .
more,
l8r,
v
--
"The future is here. It's just not evenly distributed yet."
-- William Gibson, quoted by Whitfield Diffie
_______________________________________________
Users-prolog mailing list
[email protected]
http://lists.gnu.org/mailman/listinfo/users-prolog
