Hi Christophe,
Yes. This is an infinite recursion.
...
So it seems it's not a syntax problem, but a conceptual one. I thought
about @x and @y being the same person, but it shouldn't.
Another hint?
I'm not a Prolog expert, but fact is that if the first three clauses
(be bigger (me her))
On Mon, Jul 9, 2012 at 11:12 PM, Doug Snead semaphore_2...@yahoo.com wrote:
Hmmm. Where do alix and sw originate? I do not understand that.
Sorry for the confusion, I was doing my tests with real names in my
family but thought afterwards I should use names that could be
understood by anybody.
On Tue, Jul 10, 2012 at 09:56:26AM +0200, Alexander Burger wrote:
(be bigger (@x @y) (isbigger @x @z) (isbigger @z @y))
No, I think the last rule must be
(be bigger (@x @y) (isbigger @x @z) (bigger @z @y))
Cheers,
- Alex
--
UNSUBSCRIBE:
On Tue, Jul 10, 2012 at 9:56 AM, Alexander Burger a...@software-lab.de wrote:
don't find a match, the fourth one
(be bigger (@x @y) (bigger @x @z) (bigger @z @y))
will always match and recurse infinitely.
Is it because it always match or because to know if it would succeed,
it must test
The pilog trace can be helpful in these situations.
: (be bigger (me her))
: (be bigger (her son))
: (be bigger (son daughter))
: (be bigger (@x @y) (bigger @x @z) (bigger @z @y))
List the assertions you want to trace before the clause to be proved, in this
case bigger :
: (? bigger (bigger @x
--- On Sun, 7/8/12, Christophe Gragnic christophegrag...@gmail.com wrote:
: (be bigger (me her))
- bigger
: (be bigger (her son))
- bigger
: (be bigger (son daughter))
- bigger
: (be bigger (@x @y) (bigger @x @z) (bigger @z @y))
- bigger
: bigger
- NIL
: (? (bigger @x meily))
@x=alix