Python Prime number checker

Chris picture Chris · Sep 16, 2013 · Viewed 253.5k times · Source

I have been trying to write a program that will take an inputed number, and check and see if it is a prime number. The code that I have made so far works perfectly if the number is in fact a prime number. If the number is not a prime number it acts strange. I was wondering if anyone could tell me what the issue is with the code.

a=2
num=13
while num > a :
  if num%a==0 & a!=num:
    print('not prime')
    a=a+1
  else:
    print('prime')
    a=(num)+1

the result given when 24 is inputed is: not prime not prime not prime prime

How would i fix the error with the reporting prime on every odd and not prime for every even

Answer

Steven Rumbalski picture Steven Rumbalski · Sep 16, 2013

You need to stop iterating once you know a number isn't prime. Add a break once you find prime to exit the while loop.

Making only minimal changes to your code to make it work:

a=2
num=13
while num > a :
  if num%a==0 & a!=num:
    print('not prime')
    break
  i += 1
else: # loop not exited via break
  print('prime')

Your algorithm is equivalent to:

for a in range(a, num):
    if a % num == 0:
        print('not prime')
        break
else: # loop not exited via break
    print('prime')

If you throw it into a function you can dispense with break and for-else:

def is_prime(n):
    for i in range(3, n):
        if n % i == 0:
            return False
    return True

Even if you are going to brute-force for prime like this you only need to iterate up to the square root of n. Also, you can skip testing the even numbers after two.

With these suggestions:

import math
def is_prime(n):
    if n % 2 == 0 and n > 2: 
        return False
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        if n % i == 0:
            return False
    return True

Note that this code does not properly handle 0, 1, and negative numbers.

We make this simpler by using all with a generator expression to replace the for-loop.

import math
def is_prime(n):
    if n % 2 == 0 and n > 2: 
        return False
    return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))