Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!uunet!stretch.cs.mun.ca!leif!jgarland
From: jgarland@kean.ucs.mun.ca
Newsgroups: comp.lang.prolog
Subject: Re: Deterministic predicate question
Message-ID: <129469@kean.ucs.mun.ca>
Date: 28 Aug 90 21:28:04 GMT
References: <90239.224751F0O@psuvm.psu.edu> <3631@goanna.cs.rmit.oz.au>
Organization: Memorial University. St.John's Nfld, Canada
Lines: 74

In article <3631@goanna.cs.rmit.oz.au>, ok@goanna.cs.rmit.oz.au (Richard A. O'Keefe) writes:
> In article <90239.224751F0O@psuvm.psu.edu>, F0O@psuvm.psu.edu writes:
>>     I'm curious as to why length is deterministic where both of the
>> build_board predicates are not.  Also, is there a way to make one of the
>> build_board predicates deterministic, without using the cut?
> 
> It would help if you indicated the language you are using.
> From the extremely confusing use of '=' for arithmetic, I
> infer this is Turbo/PDC ``Prolog''.
  !!!!!

Don't be so cute, you *know* it is--you've seen, and answered, Tim's queries 
before.
 
>>     build_board1(10).
>>     build_board1(SquareNum) :-
>>          NextSquare = SquareNum + 1,
>>          build_board1(NextSquare),
>>          assert(board(SquareNum," ")).
> 
> This predicate is non-determinate for the very simple reason that
> any query (such as ?- build_board1(10)) which matches the first
> clause will *also* match the second clause.  That's what it _means_
> to be non-determinate.  If you call
> 
> 	?- build_board1(0), fail.
        ^^^^^^^^^^^^^^^^^^^^^^^^^

I'm not sure exactly what T/PDC Prolog will do here...I intend to 
check it out the second I'm free...it's an interesting point.  I *think* it 
will try the first clause, then the next and then quit, but check it 
out for yourself.
 
> or however you do that in Turbo/PDC Prolog, it will loop forever.
> 
> The answer is simple:  say what you *really* mean.
                         ^^^^^^^^^^^^^^^^^^^^^^^^^^

About the best Prolog advice ever.  Or turn it around:  Prolog does 
what you *really* say...often in a way that is non-obvious to 
procedural thinking.

Anyway, this is a perfect place to use a cut.  You're essentially 
trying to implement a while-do loop.  And given the order of the 
clauses, tail recursion optimization takes care of any stack problems. 
[In terms of the basic loop, that is.  Not asserts.]

********listing******************************************

predicates
  
  while_do(integer)
  
clauses

  while_do(0) :- !.

  while_do(X) :-
     write("Countdown: ",X),nl,
     X1=X-1,
     while_do(X1).
     
goal

   while_do(10).
 
***********end of listing*************************************** 

Your stuff is looking much more elegant.

JG

jgarland@mun                                Bitnet
jgarland@kean.ucs.mun.ca                     Internet