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From: | Gregory Bourassa |
Subject: | Re: determinism |
Date: | Mon, 12 Feb 2007 12:56:42 -0500 |
Daniel,
Does your solution not return one solution to the caller before it ascertains whether or not the goal is really deterministic?
This program is a bit simplistic and it actually tries G three times; but it only returns a solution if G is really deterministic.
det(G) :- retractall( success ), call(G), backtrack_only_once, !, fail.
det(G) :- success, call(G).
backtrack_only_once :- success.
backtrack_only_once :- not( success ), assertz( success ), !, fail.
Regards.
Gregory Bourassa
On Mon Feb 12 12:11 , Daniel Diaz sent:
michel levy a écrit :
> Could you help to write this program :
> det(T,G) succeeds if and only if G succeeds once and only once and give
> the T answer.
> 1) I don't want the solution below by findall, because I want to try at
> most two back tracks on G.
> det(T,G) :- findall(T,G,L),length(L,1)
>
> 2) I know already call_det but it's not the solution because G can have
> choice points, but only one answer.
>
You can do it using a global variable to count solutions. Stop when the
second is reached.
det(Goal) :-
det1(Goal).
det(_) :-
g_read(det_count, 1).
det1(Goal) :-
g_assign(det_count, 0),
call(Goal),
g_inc(det_count, Count), % increment and return the counter
Count = 2, % cut if 2nd solution is reached (else fail)
!,
fail.
Hope this helps
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