# [racket] [bug?] Racklog cut differs from Prolog cut

 From: Jay McCarthy (jay.mccarthy at gmail.com) Date: Mon Aug 13 14:15:09 EDT 2012 Previous message: [racket] [bug?] Racklog cut differs from Prolog cut Next message: [racket] [bug?] Racklog cut differs from Prolog cut Messages sorted by: [date] [thread] [subject] [author]

Can you help me understand this example and why x should be 2? My
understanding of cut and prolog is that this should happen:

a(X)                [X = _ ]
---->
b(X) ; c(X)       [ X = _ ]
---->
b(1) ; b(2); c(X) [ X = _ ]
---->
X = 1; ! ; b(2) ; c(X)    [ X = _ ]
---->
! ; b(2) ; c(X)              [ X = 1 ]

! evaluates to succeed and then we get the X = 1 solution.

But because we cut, the change to the X logic variable won't be
undone, so when we try we get to...

b(2) ; c(X)
----->
1 = 2 ; c(X)
---->
fail ; c(X)
---->
c(X)
---->
c(2)
----->
1 = 2
---->
fail

and never see that X = 2

I must be wrong because you show that prologs actually give 2, but
could you help me see why?

Jay

On Sat, Aug 11, 2012 at 11:32 PM, Erik Dominikus
<erik.dominikus71 at gmail.com> wrote:
> Racket version:
>
> 5.2.
>
> Output of 'uname -a':
>
> Linux kire 2.6.32-41-generic #91-Ubuntu SMP Wed Jun 13 11:43:55 UTC 2012
> x86_64 GNU/Linux
>
> Symptom:
>
> In SWI Prolog (or any Prolog interpreter I think), querying a(X) gives
> X=1 and X=2. Racklog only gives x=1.
>
> How to reproduce:
>
> Run 'prolog -f a.pl' (if using SWI Prolog).
> Enter 'a(X).'.
> Press ';' (semicolon) key until it prints false.
>
> Run 'racket a.rkt'.
>
> Expectation:
>
> Racklog gives x=1 and x=2.
>
>
> Thank you.
>
> ____________________
>   Racket Users list:
>   http://lists.racket-lang.org/users
>

--
Jay McCarthy <jay at cs.byu.edu>
Assistant Professor / Brigham Young University
http://faculty.cs.byu.edu/~jay

"The glory of God is Intelligence" - D&C 93

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