在Prolog中建立递归列表

Question

I'm trying to write a rule in prolog that requires me to build a list recursively. I'm being able to build that list but the issue is that I dont know how to stop that process for building the list. So, I need to write a rule that would stop building that list further.

This is what my rule looks like:

%a few rules that are supplementing the rule popular:

member(H,[H|T]).
member(X,[H|T]) :- member(X,T).

notmember(H,A) :- member(H,A), !, fail.
notmember(H,A).

size([],0).
size([H|T],N) :- size(T,N1), N is N1+1.

listhasx(X,A) :- article(A,_,_,_,R), member(X,R).

notlisthasx(X,A) :- listhasx(X,A), !, fail.
notlisthasx(X,A).

makelist(A,L) :- ________________________________________________________.
makelist(A,L) :- listhasx(X,B), notmember(B,L), N2 is B, makelist(N2,L2), L=[A|L2].

popular(X,0).
popular(X,N) :- listhasx(X,A), makelist(A,L), size(L,N1), N-1<N1.

So, basically, I need to write a rule for popular, which I have written already. But, it uses the makelist rule which is partially incomplete. To be specific, the first part of makelist needs to be completed, which would imply when the list should be stopped from building.

I would be really thankful if someone could help me out on this..

Answer 1

The easiest way to build a list of X such that they satisfy a goal G, is to use the findall predicate. Here is an example:

?- [user].
p(3).
p(1).
p(2).
^D
Yes
?- p(X).
X = 3 ;
X = 1
?- findall(X, p(X), L).
L = [3, 1, 2]

The predicate finall/3 is part of the ISO core standard. So should be found in most Prolog systems. Related predicates are bagof/3 and setof/3.

Also useful are predicates from a module aggregate found in many Prolog systems, or solution counting and limiting predicates also found in many Prolog systems.

Bye