Re: pilog in ersatz
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)) (be bigger (her son)) (be bigger (son daughter)) don't find a match, the fourth one (be bigger (@x @y) (bigger @x @z) (bigger @z @y)) will always match and recurse infinitely. I suspect this can be solved with a 'cut' (anybody out there who knows?), but another solution would be to separate it into two predicates: (be isbigger (me her)) (be isbigger (her son)) (be isbigger (son daughter)) (be bigger (@x @y) (isbigger @x @y)) (be bigger (@x @y) (isbigger @x @z) (isbigger @z @y)) This gives: : (? (bigger son daughter)) - T : (? (bigger me @A)) @A=her @A=son - NIL : (? (bigger @A @B)) @A=me @B=her @A=her @B=son @A=son @B=daughter @A=me @B=son @A=her @B=daughter - NIL Cheers, - Alex -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: pilog in ersatz
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. My handmade-search-and-replace thing is not very efficient. Someday I'll try to configure a vim hook for web forms, or a vimesque mail client. chri -- Envoyé de ma messagerie électronique. Merci de divulguer cette adresse mail au minimum (pour les envois en groupe, utilisez la copie invisible). -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: pilog in ersatz
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: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: pilog in ersatz
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 itself? Hence the recursion, now I got it. I suspect this can be solved with a 'cut' (anybody out there who knows?) I thought about the 'cut' before falling asleep, but I'm too weak at it. but another solution would be to separate it into two predicates: (be isbigger (me her)) (be isbigger (her son)) (be isbigger (son daughter)) (be bigger (@x @y) (isbigger @x @y)) (be bigger (@x @y) (isbigger @x @z) (isbigger @z @y)) This gives: : (? (bigger son daughter)) - T : (? (bigger me @A)) @A=her @A=son - NIL : (? (bigger @A @B)) @A=me @B=her @A=her @B=son @A=son @B=daughter @A=me @B=son @A=her @B=daughter - NIL It's a start, but it doesn't span enough: we can't prove «bigger me daughter». - search search - This it what is called a «transitive closure». Alex was right about using two predicates. The first is the base one, the second is its closure. So Alex's rule: (be bigger (@x @y) (isbigger @x @z) (isbigger @z @y)) should be: (be bigger (@x @y) (isbigger @x @z) (bigger @z @y)) Tests left as an exercise !!! Doug: Thanks for (trace). I thought about it but traced (?) instead. chri -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: pilog in ersatz
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 meily)) 4 (bigger @x meily) 1 (bigger me her) 4 (bigger her meily) 2 (bigger her son) 4 (bigger son meily) 3 (bigger son daughter) 4 (bigger daughter meily) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) 4 (bigger daughter @y) ... So, it is infinitely recurring when trying to prove (bigger daughter @y). The numbers in front of the pilog trace output lines correspond to the rule being proved at that stage. Hope that helps! Cheers, Doug --- On Sun, 7/8/12, Christophe Gragnic christophegrag...@gmail.com wrote: From: Christophe Gragnic christophegrag...@gmail.com Subject: pilog in ersatz To: picolisp@software-lab.de Date: Sunday, July 8, 2012, 11:36 AM Hi all, I'm currently investigating Picolisp to teach prefix notation, and Pilog to teach first order logic. I have a problem with a set of instructions in ersatz: : (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 @x=sw Then ersatz freezes, no prompt. Any idea? chri -- Envoyé de ma messagerie électronique. Merci de divulguer cette adresse mail au minimum (pour les envois en groupe, utilisez la copie invisible). -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: pilog in ersatz
--- 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 @x=sw Hmmm. Where do alix and sw originate? I do not understand that. (I was using full picolisp for my tracing before, I should have mentioned.) -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
