is_prime.py

def naive_is_prime(n):
  if n <= 1:
    return False
  if n == 2:
    return True
  if n % 2 == 0:
    return False
  for i in xrange(3,n):
    if n % i == 0:
      return False
  return True

import math

def is_prime(n):
  if n <= 1:
    return False
  if n == 2:
    return True
  if n % 2 == 0:
    return False
  max_possible = long(math.floor(math.sqrt(n)))	#the sqrt is the max possible unique factor
  for i in xrange(3,max_possible+1,2):
    if n % i == 0:
      return False
  return True

import re

#A clever, but inefficient way using regexes
def is_prime_regex(n):
  if re.match(r'^1?$|^(11+?)\1+$','1'*n) == None:
    return True
  else:
    return False