Prime number check Python

Question

I've written this very simple prime number check:

prime = int(input())
if prime % prime == 0 and prime % 2 != 0 and prime % 3 != 0 or prime == 2 or prime == 3:
    print("true")
else:
    print("false")

... which seems to work somehow, but i'm not certain if its the correct way, can someone please confirm?

I don't understand this part: `prime % prime == 0` ? Isn't that the modulo operator, which would return 0 every time — keyser, May 19 '12 at 14:47
This code is not useful at all. You should not "hard code" these kinds of checks — keyser, May 19 '12 at 14:49
If you were able to come up with such a short and fast algorithm to check for a prime number, then it means you have solved one of the hardest and most ancient problems in mathematics. — jn1kk, May 19 '12 at 14:52

score 6 · Answer 1 · edited May 23 '17 at 12:26

6

As simple as it gets:

def isprime(n):
    """check if integer n is a prime"""
    # range starts with 2 and only needs to go up the squareroot of n
    for x in xrange(2, int(n**0.5)+1):
        if n % x == 0:
            return False
    return True

For an impressive prime-number generator, see here

edited May 23 '17 at 12:26

Community

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answered May 19 '12 at 14:50

Fredrik Pihl

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I think this fails whith n = 1 – Marcos Jul 30 '13 at 22:35

NPE · Accepted Answer · 2012-05-19T16:17:56.017

5

i'm not certain if its the correct way

It isn't. To give one counterexample, it thinks that 25 is a prime number. To make matters worse, there are infinitely many such counterexamples.

Wikipedia is worth of a read for various (correct) methods of doing this.

edited May 19 '12 at 16:17

answered May 19 '12 at 14:48

NPE

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Edwin Dalorzo · Answer 3 · 2017-07-15T10:25:54.930

The Wikipedia article on primality can help you design a better algorithm. There are many of them, but the basic ones are not that complicated.

First, depart from the fact that a prime number must be a positive integer bigger than 1. This invariant implies that if n < 2 you could return false immediatelly. In your code, n=0 fails.
In a naive approach, you can then move to check all divisors of n from 1 to n. If you just find two, then you know it's a prime.
A more intuitive approach could be to conclude that every number is divisible by 1 and itself, and so, you could check for divisors only between 2 and n-1. And in the moment that you find a divisor of n, you can conclude n is not a prime.
A improved approach recognizes that all even numbers are divisible by 2, and so, if n is not divisible by 2 then, from there on you can only check for odd divisors.
Finally, you do not need to check for all the divisors up to n. It should suffice to check divisor up to the square root of n. If you haven't found a divisor when you reach that threshold, then you can conclude n is a prime.

Prime number check Python

3 Answers3