Reputation: 59
prime_factors(N, [_:_]) :- prime_factors(N, [_:_], 2).
prime_factors(N, [_:_], D) :- N mod D == 0, N1 is N div D,
prime_factors(N1, [_:D], D).
prime_factors(N, [_:_], D) :- N mod D =\= 0, D1 is D+1, prime_factors(N, [_:_], D1).
This is my proposed solution to find the prime factors of an input N.
When I try to run it I am getting an error about such a predicate/2 not existing - is my syntax somehow wrong with the extended predicate/3?
Upvotes: 2
Views: 2078
Reputation: 476977
Using a second parameter that only seems to unify in the second case, does not seem to make much sense. Furthermore this is not the way you construct a list in Prolog anyway, since:
[H|T]
, so then it should be [_|_]
;findall/3
to later construct a list. This is usually better since that means that we can also query like prime_factor(1425, 3)
to check if 3
is a prime factor of 1425
.We can thus construct a predicate that looks like:
prime_factor(N, D) :-
find_prime_factor(N, 2, D).
find_prime_factor(N, D, D) :-
0 is N mod D.
find_prime_factor(N, D, R) :-
D < N,
(0 is N mod D
-> (N1 is N/D, find_prime_factor(N1, D, R))
; (D1 is D + 1, find_prime_factor(N, D1, R))
).
For example:
?- prime_factor(1425, R).
R = 3 ;
R = 5 ;
R = 5 ;
R = 19 ;
false.
?- prime_factor(1724, R).
R = 2 ;
R = 2 ;
R = 431 ;
false.
If we want a list of all prime factors, we can use findall/3
for that:
prime_factors(N, L) :-
findall(D, prime_factor(N, D), L).
For example:
?- prime_factors(1425, R).
R = [3, 5, 5, 19].
?- prime_factors(1724, R).
R = [2, 2, 431].
?- prime_factors(14, R).
R = [2, 7].
?- prime_factors(13, R).
R = [13].
Upvotes: 4