prolog , Predicate that pick elements which are on list twice not less not more

Predicate that pick elements which are on list twice not less not more


Tag: prolog

I'm trying to write a predicate twice(El,L) which will return true. when El is on list exactly twice. Here is what I have:

twice(El,L) :- select(El,L,L1), member(El,L1), \+ twice(El,L1).

It works nice for twice(2,[1,2,2,3,4]) but for twice(X,[1,1,2,2,3,3]) it doubles every number X = 1 ; X = 1 ; X = 2... How could I avoid this without using any accumulator?


You want to describe a sequence of elements. For such, there is a special formalism in Prolog called Definite Clause Grammars. Before using the formalism, let's try to figure out how a sequence with E occurring exactly twice looks like:

  1. First, is a possibly empty sequence which does not contain E
  2. then, there is one occurrence of E
  3. then again a possibly empty sequence without E
  4. then, there is the second occurrence of E
  5. then again a possibly empty sequence without E.

Now, to put this into the DCG formalism

twice(E, L) :-
   phrase(twice_occurring(E), L).  % Interface

twice_occurring(E) -->
   seq_without(E),    % 1.
   [E],               % 2.
   seq_without(E),    % 3.
   [E],               % 4.
   seq_without(E).    % 5.

seq_without(_E) -->
seq_without(E) -->

Or, more compactly by using all//1 and avoiding auxiliary definitions:

twice(E, L) :-
    phrase(( all(dif(E)), [E], all(dif(E)), [E], all(dif(E)) ), L).

There is essentially only one drawback with these definitions: On current systems, they are not optimally implemented. See this if you want to know more.


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