A Python solution to Project Euler Problem 3

factors.py

def factors(num):
  """Generates the factors for num.
  By PEZ, from a description by BradC over at Stackoverflow.com
  http://stackoverflow.com/questions/439814#439850
  """

  from math import sqrt

  def _candidates():
    """Generates the sequence 2, then all odd ints starting with 3""" 
    c = 2
    yield c
    c += 1
    while True:
      yield c
      c += 2

  start = num
  candidates = _candidates()
  candidate = candidates.next()
  sqrt_num = sqrt(num)
  while candidate < sqrt_num:
    while num % candidate == 0:
      yield candidate
      num /= candidate
      sqrt_num = sqrt(num)
      if candidate > sqrt_num:
        yield num
        return
    candidate = candidates.next()
  if num < start:
    yield num

if __name__=='__main__':
  # The number 600851475143 takes 0.002 secs to factor on my MacBook
  from timeit import Timer
  t = Timer("print list(factors(600851475143))", "from __main__ import factors")
  print "%.5f" % t.timeit(1)