Prolog: eliminate repetitions in query

Question

I\'ve been trying to write a simple code, that would behave in this manner:

| ?- hasCoppiesOf(X,[a,b,a,b,a,b,a,b]).
X = [a,b] ? ;
X = [a,b,a,b] ? ;
X = [a,b,a,b         


        

Answer 1

I think you just made your predicate more complex than it needs to be, probably just overthinking it. A given solution may succeed in multiple paths through the logic.

You can do this without append/3 by aligning the front end of the lists and keep the original list to "reset" on repeats:

% Empty list base cases
dups_list([], []).
dups_list([_|_], []).

% Main predicate, calling aux predicate
dups_list(L, Ls) :-
    dups_list(L, L, Ls).

% Recursive auxiliary predicate
dups_list([], [_|_], []).
dups_list([], [X|Xs], [X|Ls]) :-
    dups_list(Xs, [X|Xs], Ls).
dups_list([X|Xs], L, [X|Ls]) :-
    dups_list(Xs, L, Ls).

Here are some results:

| ?- dups_list(X,[a,b,a,b,a,b,a,b]).

X = [a,b] ? a

X = [a,b,a,b]

X = [a,b,a,b,a,b,a,b]

no
| ?- dups_list([a,b,a,b,a,b,a,b], X).

X = [] ? ;

X = [a,b,a,b,a,b,a,b] ? ;

X = [a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b] ? ;

X = [a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b] ?
...
| ?- dups_list(A, B).

A = []
B = [] ? ;

A = [_|_]
B = [] ? ;

A = [C]
B = [C] ? ;

A = [C]
B = [C,C] ? ;

A = [C,D]
B = [C,D] ? ;

A = [C]
B = [C,C,C] ? ;

A = [C,D,E]
B = [C,D,E] ? ;
...

There may be a way to simplify the solution just a bit more, but I haven't played with it enough to determine if that's the case.

Answer 2

Okay, I got your problem, you want to eliminate the repetitions.

hasCoppiesOf(A,[]).


hasCoppiesOf([H1|T1], [H1|T2]) :-
  append(T1, [H1], X),
  hasCoppiesOf([H1|T1], X, T2).


hasCoppiesOf(A, A, B) :-
  hasCoppiesOf(A, B),!. %Change here, place a cut after the termination.

hasCoppiesOf(A, [H1|T1], [H1|T2]) :-
  append(T1, [H1], X),
  hasCoppiesOf(A, X, T2).

This is the change that you need to make.

hasCoppiesOf(A, A, B) :-
      hasCoppiesOf(A, B),!.

A Cut '!' terminates the unwanted backtracking and thereby repetitions.

Answer 3

I think this is what you're trying for...

coppies(Z,Z,[]).
coppies(X,Z,[Y|Ys]):- \+member(Y,Z),coppies(X,[Y|Z],Ys).
coppies(X,Z,[Y|Ys]):- member(Y,Z),coppies(X,Z,Ys).


copies(M,[Y|Ys]):-coppies(M,[],[Y|Ys]).

Input:

copies(X,[1,2,1,2,1,2]).

Output:

X = [2, 1].

BTW I've used some different names instead..