How to express a disjunction of inequalities compactly without redundant answers/solutions

Question

Consider what I have tried:

dif_to_orto(A, B, C) :-
   (  dif(A, B)
   ;  dif(A, C)
   ).

While this definition is fine from a declarative viewpoint it contains many redundancies. Think of:

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 2.
   A = 1, B = 2, C = 2
;  A = 1, B = 2, C = 2.   % unexpected redundant solution

And not even in this case:

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 3.
   A = 1, B = 2, C = 3
;  A = 1, B = 2, C = 3.   % unexpected redundant solution

At least, here is a case without redundancy...

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 1.
   A = 1, B = 2, C = 1
;  false.                 % unexpected inefficient leftover choicepoint

...but with a resource-wasting leftover choicepoint.

Rare are the occasions where this definition is efficient:

?- dif_to_orto(A, B, C), A = 1, B = 1, C = 2.
   A = 1, B = 1, C = 2.

Also that the most general query produces two answers sounds very inefficient to me:

?- dif_to_orto(A, B, C).
   dif:dif(A,B)
;  dif:dif(A,C).

... which produces also the following redundancy:

?- dif_to_orto(1, B, B).
   dif:dif(1,B)
;  dif:dif(1,B).    % unexpected redundant answer

One dif/2 would be enough!

Is there a way to avoid all these redundancies and inefficiencies?

Answer 1

How about this one:

dif_to_orto(A, B, C) :-
   dif(A-A, B-C).

The test cases:

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 2.
A = 1,
B = C, C = 2.

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 3.
A = 1,
B = 2,
C = 3.

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 1.
A = C, C = 1,
B = 2.

?- dif_to_orto(A, B, C), A = 1, B = 1, C = 2.
A = B, B = 1,
C = 2.

?- dif_to_orto(A, B, C).
dif(f(B, A), f(A, C)).

?- dif_to_orto(1, B, B).
dif(B, 1).

Answer 2

This solution first awaits for 2 of the 3 variables to be comparable, then if it cannot decide whether the constraint should succeed adds a new constraint:

dif_to_orto(A, B, C) :-
    when((?=(A, B) ; ?=(A, C) ; ?=(B, C)),
         (   ?=(A, B) ->
              ( A\==B ->  true ; dif(A, C) )
         ;
             (
                ?=(A, C) ->
                 ( A\==C -> true ; dif(A, B) )
             ;
                 ( B\==C -> true ; dif(A, B) )
             )
         )).

Sample runs:

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 2.
A = 1,
B = C, C = 2.

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 3.
A = 1,
B = 2,
C = 3.

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 1.
A = C, C = 1,
B = 2.

?- dif_to_orto(A, B, C).
when((?=(A, C);?=(B, C);?=(A, B)),  (?=(A, B)->(A\==B->true;dif(A, C));?=(A, C)->(A\==C->true;dif(A, B));B\==C->true;dif(A, B))).

?- dif_to_orto(1, 2, Z).
true.

?- dif_to_orto(1, B, B).
dif(B, 1).

---

Reversing the checks:

dif_to_orto(A, B, C) :-
    when((?=(A, B) ; ?=(A, C) ; ?=(B, C)),
         (
           A==B -> dif(A, C)
           ;
           ((A==C ; B==C) -> dif(A, B) ; true)
         )).

Answer 3

Expanding on the definition of dif/2:

dif_to_orto(A, B, C):-
   when((?=(A,B), ?=(A, C)), (A \== B -> true ; A \== C)).

Sample runs:

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 2.
A = 1,
B = C, C = 2.

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 3.
A = 1,
B = 2,
C = 3.

?- dif_to_orto(A, B, C), A = 1, B = 2, C = 1.
A = C, C = 1,
B = 2.

?- dif_to_orto(A, B, C).
when((?=(A, B), ?=(A, C)),  (A\==B->true;A\==C)).

Answer 4

Here is one suggestion. As far as I can tell, it does not create choice points or redundant solutions:

dif_to_orto(A, B, C) :-
   when(?=(A,B),(A==B->dif(A,C);true)),
   when(?=(A,C),(A==C->dif(A,B);true)).

For each disjunct, wait until it is known to be true or false. Once known, check its truth and, if false, then post the other disjunct.