Prolog: dividing a number

Question

I wanted to make a predicate that returns a list of a number dividers. Example: 72 = 2*2*2*3*3.

prdel(A,[],_):- 
      A is 1.
prdel(P,[D|L],D):-
      0 is mod(P,D),
      P1 is P/D,
      prdel(P1,L,D).
prdel(P,L,D):-
      D1 is D+1,
      prdel(P,L,D1).

This works and returns the right list. The problem is that it does not stop after that but returns the same list over and over again if I press space (I am sorry I don't know the term in English when you use the same predicate to get different answer). I want it to stop after the first time.

I tried to edit the last one like that,

prdel(P,L,D):-
      D1 is D+1,
      D1<P,
      prdel(P,L,D1).

but now it returns only false and not the list.

EDIT:

I am looking for an answer without cut.

Answer 1

One problem in your code is that it keeps trying to divide the number P by D even when it is clear that the division is not going to succeed because D is too high. This lets D "run away" without a limit.

Adding a check for D1 to be below or equal to P fixes this problem:

prdel(1,[],_).

prdel(P,[D|L],D):-
      0 is mod(P,D),
      P1 is P/D,
      prdel(P1,L,D).

prdel(P,L,D):-
      D1 is D+1,
      D1 =< P,
      prdel(P,L,D1).

This produces all combinations of divisors, including non-prime ones (demo).

[[2, 2, 2, 3, 3], [2, 2, 2, 9], [2, 2, 3, 6],
 [2, 2, 18], [2, 3, 3, 4], [2, 3, 12], [2, 4, 9],
 [2, 6, 6], [2, 36], [3, 3, 8], [3, 4, 6], [3, 24],
 [4, 18], [6, 12], [8, 9], [72]]

If you do not want that, add the condition that mod(P,D) > 0 in the last clause:

prdel(1,[],_).

prdel(P,[D|L],D):-
    0 is mod(P,D),
    P1 is P/D,
    prdel(P1,L,D).

prdel(P,L,D):-
    mod(P,D) > 0,
    D1 is D+1,
    D1 =< P,
    prdel(P,L,D1).

This produces only [2, 2, 2, 3, 3] (demo).