As I understand it, cuts are delimited by the relations they appear in.
So it would go something more like this (using a notation I just made up):
a(X)
---->
b(X) ; c(X)
---->
[delimit b(1) ; b(2) ] ; c(X)
---->
[delimit X=1, ! ; b(2) ] ; c(X)
---->
X=1 ; c(X)
---->
X=1 ; c(2)
---->
X=1 ; X=2
I've omitted the delimit brackets for a and c because they don't
interact with any cuts.
Ryan
On 08/13/2012 02:15 PM, Jay McCarthy wrote:
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.dominiku...@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:
Download 'a.pl' (attached).
Run 'prolog -f a.pl' (if using SWI Prolog).
Enter 'a(X).'.
Press ';' (semicolon) key until it prints false.
Download 'a.rkt' (attached).
Run 'racket a.rkt'.
Expectation:
Racklog gives x=1 and x=2.
Thank you.
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