getting out of local stack at prolog

Question

p(0,0).
p(0,1).
p(0,2).
p(0,3).
p(0,4).
p(1,1).
p(1,2).
p(1,3).
p(1,4).
p(1,0).
p(2,0).
p(2,1).
p(2,2).
p(2,3).
p(2,4).
p(3,0).
p(3,1).
p(3,2).
p(3,3).
p(3,4).
p(4,0).
p(4,1).
p(4,2).
p(4,3).
p(4,4).

adjacent(p(X,Y),p(X,Z)) :-
    p(X,Y),
    p(X,Z),
    Z is Y+1.
adjacent(p(X,Y),p(X,Z)) :-
    p(X,Y),
    p(X,Z),
    Z is Y-1.
adjacent(p(X,Y),p(Z,Y)) :-
    p(X,Y),
    p(X,Z),
    Z is X+1.
adjacent(p(X,Y),p(Z,Y)) :-
    p(X,Y),
    p(X,Z),
    Z is X-1.

adjacentC(X,Y) :-
    adjacent(X,Y).
adjacentC(X,Y) :-
    adjacent(X,Z),
    adjacentC(Z,Y).

I don't know why this code I wrote isn't working.

e.g.:

?- adjacentC((0,0),(4,4)). ERROR

What does the error say? Does it just say "ERROR"? Did you try tracing the execution to make sure the program does what you think it does (obviously, it doesn't....) — , May 08 '15 at 07:16
ive found the problem. im using the adjacent predicate wrong, i want adjacent(p(0,0),X) to return facts EG: (0,1) and (1,0) — joaquin montero, May 08 '15 at 07:29

score 1 · Answer 1 · edited May 23 '17 at 11:43

Quick answer: The following works and terminates always using closure/3 defined elsewhere.

adjacentD(X,Y) :-
   closure(adjacent,X,Y).

However, this approach is extremely slow, due to the inefficient definition of adjacent/3. Here is a better one / oh forget it, here is a more correct one, first:

adjacent2(p(X0,Y0),p(X,Y)) :-
   p(X0,Y0),
   (  X0 = X,
      p(X,Y),
      abs(Y0-Y) =:= 1
   ;  Y0 = Y,
      p(X,Y),
      abs(X0-X) =:= 1
   ).

getting out of local stack at prolog

1 Answers1