🚀 go-pugleaf

In article <10588@topaz.RUTGERS.EDU> chomicki@topaz.RUTGERS.EDU (Jan Chomicki) writes:
>From: Francois-Michel Lang <lang%omega.PRC.Unisys.COM@cis.upenn.edu>
>
>Any ideas about how to hack either of these?
>
>1. A predicate xor(G1,G2) which has the following behavior:
> (...)

A "real" exclusive-or should be equivalent to:  G1, not(G2) ; not(G1), G2.
The xor defined in the question is rather like:  G1 ; \+ G1, G2.
So one solution is:

xor(G1,G2) :- G1 ; \+ G1, G2.

This has the disadvantage that G1 is called twice. A more efficient solution
is possible if (as Lee Naish pointed out) you have a retrying "if":

xor(G1,G2) :- if(G1, true, G2).   % if G1 then true else G2

>2. Given a predicate
>
>      P :- A, B, C, D.
>
>   Is it possible to rewrite P to obtain the following behavior:
>   (...) if a given solution of A allows B to be solved,
>     then don't try to resatisfy A.
>   But if a given solution of A does not allow B to be solved,
>     then do try to resatisfy A.

Assuming that P has no more clauses, a possible solution might be:

P :- A(X), B(X,_), !, B(X,Y), C, D.

The variable X represents the solutions of A. Here too we have the
disadvantage that B is called twice. We also have the requirement that the
variables that represent the solutions of B for a given solution of A
are easily identifiable. This may or may not be the case.

-- 
Bjorn Danielsson, ZYX, +46 8 653205, ...mcvax!enea!zyx!bd