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How to implement rule1 that succeeds iff rule2 returns two or more results?

rule1(X) :-
    rule2(X, _).

How can I count the results, and then set a minimum for when to succeed?

Palec
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John Dorian
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    Should be easy to encode using [findall/3](http://stackoverflow.com/questions/1468150/how-do-i-find-all-solutions-to-a-goal-in-prolog). – Volker Stolz Nov 15 '12 at 08:35

2 Answers2

5

How can I count the results, and then set a minimum for when it's true?

It is not clear what you mean by results. So I will make some guesses. A result might be:

A solution. For example, the goal member(X,[1,2,1]) has two solutions. Not three. In this case consider using either setof/3 or a similar predicate. In any case, you should first understand setof/3 before addressing the problem you have.

An answer. The goal member(X,[1,2,1]) has three answers. The goal member(X,[Y,Z]) has two answers, but infinitely many solutions.

So if you want to ensure that there are at least a certain number of answers, define:

at_least(Goal, N) :-
   \+ \+ call_nth(Goal, N).

with call_nth/2 defined in another SO-answer.

Note that the other SO-answers are not correct: They either do not terminate or produce unexpected instantiations.

Community
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false
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2

you can use library(aggregate) to count solutions

:- use_module(library(aggregate)).

% it's useful to declare this for modularization
:- meta_predicate at_least(0, +).

at_least(Predicate, Minimum) :-
      aggregate_all(count, Predicate, N),
      N >= Minimum.

example:

?- at_least(member(_,[1,2,3]),3).
true.

?- at_least(member(_,[1,2,3]),4).
false.

edit here is a more efficient way, using SWI-Prolog facilities for global variables

at_least(P, N) :-
    nb_setval(at_least, 0),
    P,
    nb_getval(at_least, C),
    S is C + 1,
    ( S >= N, ! ; nb_setval(at_least, S), fail ).

with this definition, P is called just N times. (I introduce a service predicate m/2 that displays what it returns)

m(X, L) :- member(X, L), writeln(x:X).

?- at_least(m(X,[1,2,3]),2).
x:1
x:2
X = 2.

edit accounting for @false comment, I tried

 ?- call_nth(m(X,[1,2,3]),2).
x:1
x:2
X = 2 ;
x:3
false.

with call_nth from here.

From the practical point of view, I think nb_setval (vs nb_setarg) suffers the usual tradeoffs between global and local variables. I.e. for some task could be handly to know what's the limit hit to accept the condition. If this is not required, nb_setarg it's more clean.

Bottom line: the better way to do would clearly be using call_nth, with the 'trick' of double negation solving the undue variable instantiation.

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CapelliC
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    Your use of global variables is incorrect. And you produce unexpected substitutions (`X = 2`) in your query. – false Nov 15 '12 at 09:32