Remove duplicates in list (Prolog)

Question

I am completely new to Prolog and trying some exercises. One of them is:

Write a predicate set(InList,OutList) which takes as input an arbitrary list, and returns a list in which each element of the input list appears only once.

Here is my solution:

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

set([],[]).
set([H|T],[H|Out]) :-
    not(member(H,T)),
    set(T,Out).
set([H|T],Out) :-
    member(H,T),
    set(T,Out).

I'm not allowed to use any of built-in predicates (It would be better even do not use not/1). The problem is, that set/2 gives multiple same solutions. The more repetitions in the input list, the more solutions will result. What am I doing wrong? Thanks in advance.

Answer 1

You are getting multiple solutions due to Prolog's backtracking. Technically, each solution provided is correct, which is why it is being generated. If you want just one solution to be generated, you are going to have to stop backtracking at some point. This is what the Prolog cut is used for. You might find that reading up on that will help you with this problem.

Update: Right. Your member() predicate is evaluating as true in several different ways if the first variable is in multiple positions in the second variable.

I've used the name mymember() for this predicate, so as not to conflict with GNU Prolog's builtin member() predicate. My knowledge base now looks like this:

mymember(X,[X|_]).
mymember(X,[_|T]) :- mymember(X,T).

not(A) :- \+ call(A).

set([],[]).
set([H|T],[H|Out]) :-
    not(mymember(H,T)),
    set(T,Out).
set([H|T],Out) :-
    mymember(H,T),
    set(T,Out).

So, mymember(1, [1, 1, 1]). evaluates as true in three different ways:

| ?- mymember(1, [1, 1, 1]).

true ? a

true

true

no

If you want to have only one answer, you're going to have to use a cut. Changing the first definition of mymember() to this:

mymember(X,[X|_]) :- !.

Solves your problem.

Furthermore, you can avoid not() altogether, if you wish, by defining a notamember() predicate yourself. The choice is yours.

Answer 2

A simpler (and likely faster) solution is to use library predicate sort/2 which remove duplicates in O(n log n). Definitely works in Yap prolog and SWIPL