How to add multiple heads to a list using recursion - list

Question

I want to remove every nth element from both list A and B that doesn't correspond with each other's nth element. So for example: I have list A containing [1,2,3,4,5] and list B containing [1,2,2,4,7]. The result should be list C containing [3,5] and list D containing [2,7]. I came up with this little piece of code, but it isn't working properly yet.
without_doubles([], [], [_], [_]).
without_doubles([H|T1], [H|T2], C, D):-
without_doubles(T1, T2, C, D).
without_doubles([H1|T1], [H2|T2], [H1|C], [H2|D]):-
without_doubles(T1, T2, C, D).
Instead of the [3,5] and [2,7], I'm getting [3,5,_G2442] and [2,7,_G2445]. I know it's probably easy to fix, but I can't seem to figure it out. Thanks for helping me in advance!

Answer 1

Your first rule should be:
without_doubles([], [], [], []).
the two anonymous variables _ that you had before were the cause of your problem.
There is however a second problem with your code:
?- without_doubles([1,2,3,4,5],[1,2,2,4,7],L,R).
L = [3, 5],
R = [2, 7] ;
L = [3, 4, 5],
R = [2, 4, 7] ;
L = [2, 3, 5],
R = [2, 2, 7] ;
L = [2, 3, 4, 5],
R = [2, 2, 4, 7] ;
L = [1, 3, 5],
R = [1, 2, 7] ;
L = [1, 3, 4, 5],
R = [1, 2, 4, 7] ;
L = [1, 2, 3, 5],
R = [1, 2, 2, 7] ;
L = [1, 2, 3, 4, 5],
R = [1, 2, 2, 4, 7].
Obviously we only want the first result, not the other ones. The reason why we get the other ones is because at no point do you state in your last rule that H1 and H2 should be different. This means that when Prolog can apply the second rule, it will also be able to apply the last rule and thus create all those choice points.
You can fix this by explicitely stating that H1 and H2 must be different in your last rule, using dif/2:
without_doubles([H1|T1], [H2|T2], [H1|C], [H2|D]):-
dif(H1, H2),
without_doubles(T1, T2, C, D).
Now we have:
?- without_doubles([1,2,3,4,5],[1,2,2,4,7],L,R).
L = [3, 5],
R = [2, 7] ;
false.

Answer 2

There are two problems with your code:
the basecase specifies singleton lists as third and fourth argument, whereas these should be empty lists;
nothing is preventing you from backtracking over the second clause, and thus take the third clause.
You can fix this as follows:
%% no [_], but []
without_doubles([], [], [], []).
without_doubles([H|T1], [H|T2], C, D):-
without_doubles(T1, T2, C, D).
%% guard the fact that H1 and H2 are different
without_doubles([H1|T1], [H2|T2], [H1|C], [H2|D]):-
dif(H1,H2),
without_doubles(T1, T2, C, D).