recursion - count the number of calls of a clause

Question

I have a clause like following:

lock_open:-
        conditional_combination(X),
        equal(X,[8,6,5,3,6,9]),!,
        print(X).

this clause succeed. But I want to know how many times conditional_combination() is called before equal(X,[8,6,5,3,6,9]) is become true. the program is to generate a permutation by following some rules. And I need to how many permutation is need to generate to get a particular value like 865369.

Answer 1

What you actually want is something slightly different: You want to count the number of answers (so far) of a goal.

The following predicate call_nth(Goal_0, Nth) succeeds like call(Goal_0) but has an additional argument which indicates that the answer found is the n-th answer. This definition is highly specific to SWI or YAP. Do not use things like nb_setarg/3 in your general programs, but use them for well encapsulated cases as this one. Even within those two systems, the precise meaning of these constructs is not well defined for the general case. Here is a definition for SICStus.

call_nth(Goal_0, C) :-
   State = count(0,_), % note the extra argument which remains a variable
   Goal_0,
   arg(1, State, C1),
   C2 is C1+1,
   nb_setarg(1, State, C2),
   C = C2.

A more robust abstraction is provided by Eclipse:

call_nth(Goal_0, Nth) :-
   shelf_create(counter(0), CounterRef),
   call(Goal_0),
   shelf_inc(CounterRef, 1),
   shelf_get(CounterRef, 1, Nth).

?- call_nth(between(1,5,I),Nth).
I = Nth, Nth = 1 ;
I = Nth, Nth = 2 ;
I = Nth, Nth = 3 ;
I = Nth, Nth = 4 ;
I = Nth, Nth = 5.

So simply wrap it around:

lock_open :-
   call_nth(conditional_combination(X), Nth),
   X = [8,6,5,3,6,9],
   !,
   ....