Predicate not working

Question

I'm new to prolog, and I really don't understand how things really work.

Giving the following knowledge base, from a simplified social network:

% user(nickname,email,city,friends_list)
user(nick0,mail0,lisbon,[nick1,nick8]).
user(nick1,mail1,lisbon,[nick0,nick3, nick5, nick9]).
user(nick2,mail2,london,[nick5,nick7,nick8]).
user(nick3,mail3,madrid,[nick1,nick4]).
user(nick4,mail4,paris,[nick1, nick3, nick2, nick5, nick6]).
user(nick5,mail5,madrid,[nick1, nick4, nick2]).
user(nick6,mail6,lisbon,[nick3,nick2]).
user(nick7,mail7,berlin,[nick3, nick5]).
user(nick8,mail8,berlin,[nick3, nick5, nick7]).
user(nick9,mail9,london,[nick1, nick4]).

I need a predicate places(NU,LC) that generates a list of citys (LC) with the citys where the number of users is greater or equal than (NU).

Example:

places(0,LC).

LC = [(lisbon,3), (london,2), (madrid,2), (paris,1), (berlin,2)]

I have the following code, and it doesn't work, I'd appreciate it if you guys could help me:

places(NU,LC):-
findall(X,user(_,_,X,_),List),
findall(Z,(user(_,_,Z,_),member(Z,List), Z>=NU),LC).

It may be a simple problem, but I'm still learning prolog.

CapelliC CapelliC · Accepted Answer · 2013-09-03T13:46:45

Of course, the answer will depend on what Prolog system you're using, because libraries could be available solving the query in single call. To stick with ISO predicates (well, apart member/2):

places(NU, LC) :-
    setof(City, User^Mail^Friends^user(User,Mail,City,Friends), Cities),
    findall((City, Count),
        (member(City, Cities),
         findall(_, user(_,_,City,_), L),
         length(L, Count),
         Count >= NU), LC).

Predicate not working

1 Answers