Prolog no_duplicate function

Question

I'm trying to write a simple procedure that checks if a list has any duplicates. This is what I have tried so far:

% returns true if the list has no duplicate items.
no_duplicates([X|XS]) :- member(X,XS) -> false ; no_duplicates(XS).
no_duplicates([]) :- true. 

If I try no_duplicates([1,2,3,3]). It says true. Why is this? I'm probably misunderstanding Prolog here, but any help is appreciated.

Answer 1

To answer your questions: your solution actually fails as expected for no_duplicates([1,2,3,3]). So there is no problem.

Now take the queries:

?- A = 1, no_duplicates([A, 2]).
A = 1.
?-        no_duplicates([A, 2]), A = 1.

They both mean the same, so we should expect that Prolog will produce the same answer. (_{To be more precise we expect the same ignoring errors and non-termination}).

However, four proposed solutions differ! And the one that does not, differs for:

?- A = 2, no_duplicates([A, 2]).
false.
?-        no_duplicates([A, 2]), A = 2.

Note that it is always the second query that makes troubles. To solve this problem we need a good answer for no_duplicates([A, 2]). It cannot be false, since there are some values for A to make it true. Like A = 1. Nor can it be true, since some values do not fit, like A = 2.

Another possibility would be to issue an instantiation_error in this case. Meaning: I have not enough information so I better stop than mess around with potentially incorrect information.

Ideally, we get one answer that covers all possible solutions. This answer is dif(A, 2) which means that all A that are different to 2 are solutions.

dif/2 is one of the oldest built-in predicates, already Prolog 0 did possess it. Unfortunately, later developments discarded it in Prolog I and thus Edinburgh Prolog and thus ISO Prolog.

However, current systems including SICStus, YAP, SWI all offer it. And there is a safe way to approximate dif/2 safely in ISO-Prolog

no_duplicates(Xs) :-
   all_different(Xs). % the common name

all_different([]).
all_different([X|Xs]) :-
   maplist(dif(X),Xs).
   all_different(Xs).

See: prolog-dif

Answer 2

Duplicates in a list are same elements not at the same place in the list, so no_duplicates can be written :

no_duplicates(L) :-
    \+((nth0(Id1, L, V), nth0(Id2, L, V), Id1 \= Id2)).

Answer 3

Here's yet another approach, which works because sort/2 removes duplicates:

no_duplicates(L) :-
    length(L, N),
    sort(L, LS),
    length(LS, N).

Answer 4

I'd go at the problem more descriptively:

no_duplicates( []     ) .  % the empty list is unique
no_duplicates( [X|Xs] ) :- % a list of length 1+ is unique
  \+ member(X,Xs) ,        % - if its head is not found in the tail,
  no_duplicates(Xs)        % - and its tail is itself unique.
  .                        %

Thinking on this, since this is a somewhat expensive operation — O(n²)? — it might be more efficient to use sort/2 and take advantage of the fact that it produces an ordered set, removing duplicates. You could say something like

no_duplicates( L ) :-
  sort(L,R)   , % sort the source list, removing duplicates
  length(L,N) , % determine the length of the source list
  length(R,N) . % check that against the result list

Or you could use msort/3 (which doesn't remove duplicates), might be a bit faster, too:

no_duplicates( L ) :-
  msort(L,R),            % order the list
  \+ append(_,[X,X|_],R) % see if we can find two consecutive identical members
  .

Answer 5

Jay already noted that your code is working. An alternative, slightly less verbose

no_duplicates(L) :- \+ (append(_, [X|XS], L), memberchk(X, XS)).