Re: determinism

This may help :

- go/2 takes G and T. It copies G to G1 (no shared variable), calls G1
(hence G) and if success, next/3 will decide.

- next/3 retries G (hence G1) but if there is a solution different from
the first call (in go/2), then cuts and conludes on a failure.

Otherwise, next/3 takes the only solution into T.

- more_than_one is here to write some blabla.

Other 'write(...)' are also there for a trace.

Try it with :

go(p0(X), T).

go(p1(X), T).

go(p2(X), T).

go(p3(X), T).

Only go(p0(X), T). succeeds. Others will write things but fail.

PS : there is no try with an nonexistent predicat; you may set the
appropriate flag and try with that case also.

Cheers

Alex

------------------

go(G, T) :- copy_term(G,G1), call(G1), next(G,G1,T).
next(G,G1,T) :- G, G \== G1, !, more_than_one(G,G1,T).
next(_G,G1,T) :- T=G1, write('case one solution\n'). % succeed, only one solution.
more_than_one(_G,_G1,_T) :- write('case with more than one solution \n'), fail.

% ex for G with 0, 1 and more solutions.
p0(_):- fail. % zero solution for p0

p1(1). % one solution predicate

p2(1). % Two solutions for p2
p2(2).

p3(1). % 3 (or more) soultions
p3(2).
p3(3).

------------------------

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.

--
Aleksander S. Saidi
Maître de Conférences
Ecole Centrale de Lyon
Département Mathématiques-Informatique
Mél : [email protected]
Tél : 04.72.18.65.30, Fax : 04.78.33.16.15

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