Daily Algo: Primality of a given n

There are various questions one can ask while investigating primes algorithmically. One might be : Given a number n, is it prime ? A naive, brute force approach would be to iterate through all the numbers i from 0 to n-1, and test if n%i ==0. However, this approach would be naive in the extreme, and a more efficient approach can be devised by noting a few points.

  1. 0 and 1 are not primes
  2. 2 is a prime. However, all multiples of 2 are not prime. Hence we can reduce the numbers we have to inspect by 50%.
  3. If a number m divides n, then n/m also divides n, and n/m <=m or vice-versa. Hence all we need is to examine numbers up to sqrt(n) and no further.

The Python code is shown below:

In [15]: import math
def isPrime(n):
    if n <= 1 or n%2==0: return False 
    if n==2: return True
    ulim = int(math.sqrt(n))+1
    for i in range(3,ulim+1,2):
        if n%i ==0:
            return False
    return True

In [16]: isPrime(6767)
Out[16]: False
In [17]: isPrime(62417)
Out[17]: True

The code is available for download here : primes.py