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. > > ____________________ > Racket Users list: > http://lists.racket-lang.org/users > -- Jay McCarthy <j...@cs.byu.edu> Assistant Professor / Brigham Young University http://faculty.cs.byu.edu/~jay "The glory of God is Intelligence" - D&C 93 ____________________ Racket Users list: http://lists.racket-lang.org/users
