Calculating whether number is prime in Prolog - prolog

Question

I am trying to calculate if the input is a prime number but something goes wrong... here's my code:
primeNumber(X):-
prime_prime(A, 1).
prime_prime(A, B):-
R is A mod B,
R =:= 1,
R =:= A.
prime_prime(X, B):-
B < A,
Next is B + 1,
prime_prime(A, Next).
It gives me false every time. Anyone got any clues or idea on what I am doing wrong?

Answer 1

See http://www.swi-prolog.org/pldoc/man?function=mod/2:
+IntExpr1 mod +IntExpr2
Modulo, defined as Result = IntExpr1 - (IntExpr1 div IntExpr2) × IntExpr2, where div is floored division.
So R should be 0. mod has only one result.
A working solution would be:
primeNumber(A) :-
A > 1, % Negative numbers, 0 and 1 are not prime.
prime_prime(A, 2). % Begin iteration:
prime_prime(A, B) :- % Test if A divides by B without remainder
B >= A % The limit was reached?
-> true % Then it's prime.
; 0 is A mod B % B divides A without a remainder?
-> false % Then it's not prime.
; succ(B, C), % Otherwise: C is B + 1
prime_prime(A, C). % Test if C divides A.
By the way, primeNumber/1 (a predicate named primeNumber, with one argument) is a totally separate predicate from primeNumber/2 (same name, two arguments). A "subfunction" that only gets an extra argument for the start value, is usually given the same name. So instead of prime_prime you should just use primeNumber, though in Prolog you normally don't use camelCase.
Using the optimization that Sergei Lodyagin proposed in the comments:
primeNumber(A) :-
A > 1, % Negative numbers, 0 and 1 are not prime.
sqrt(A, L), % A prime factor of A is =< the square root of A.
prime_prime(A, 2, L). % Begin iteration:
prime_prime(A, B, L) :- % Test if A divides by B without remainder
B >= L % The limit was reached?
-> true % Then it's prime.
; 0 is A mod B % B divides A without a remainder?
-> false % Then it's not prime.
; succ(B, C), % Otherwise: C is B + 1
prime_prime(A, C, L). % Test if C divides A.
And if you use the predefined predicate between(+Low, +High, ?Value):
primeNumber(A) :-
L is floor(sqrt(A)),
\+ (between(2, L, X),
0 is A mod X).
And to reduce the number of iterations even further, you only need to test for odd modules:
primeNumber(2).
primeNumber(A) :-
A > 2,
\+ 0 is A mod 2,
L is floor(sqrt(A) / 2),
\+ (between(1, L, X),
0 is A mod (1 + 2*X)).

Answer 2

Kay already provided a working modification of the broken program. I'll provide a simple analysis of what's broken.
When solving a problem in Prolog, it's good to be able to write out logically what it is you want first. In this case, it appears that you want to declare that:
A number, A, is prime if, for each number B < A, the value of A mod B is non-zero.
There are probably a couple of ways to render this directly into Prolog, of which Kay shows one.
However, the way your original rules are written, they say:
A number, A, is prime if:
(Rule 1) The value of A mod B, for a given value of B, is 1 and is also A.
OR (Rule 2) B < A and Rule 1 is satisfied with A and B+1.
As you can see, the rules as defined have a few issues:
The rules don't match the logical definition of prime described in terms of the modulo relationship between the original number and all the numbers less than itself.
The first rule expects an impossible mathematical condition when A is not equal to 1 (remember, the comma [,] in Prolog is a conjunction)
The rules are initiated with starting divisor of 1, which is probably bad since 1 divides everything and is likely to become an exception to any rules that work
EDIT
Getting back to the first definition of a prime using the modulo operator, we can translate that into Prolog as follows:
is_prime(N) :- % N is prime if...
N > 1, % N > 1, and
non_divisible_from(N, 2). % N is non-divisible by everything from 2 to N-1
non_divisible_from(N, D) :- % N is non-divisible by D through N-1 if...
N =< D. % D >= N
% --OR--
non_divisible_from(N, D) :- % N is non-divisible from D to N-1 if...
N > D, % N > D, and
N mod D =\= 0, % N is non-divisible by D, and
D1 is D + 1, % N is non-divisible by D+1 to N-1
non_divisible_from(N, D1).
This logic is basically the same as Kay's except he's using a Prolog if-then-else construct.