In my customary "Happy New Year" mail, I sent out a small note :

PS: Quiz : Is 2009 a prime number ? - Ans : No

Soon I started getting responses about that wisecrack of mine. The one that really stumped me came from a dear friend which said,

Enjoy the new year and happy computing. BTW is 200000000000000000000000000009 a prime number?

Soon enough found a piece of python code to find factors - Factor for Python which seemed to find all the factors for a number. Wasn't likely to work too fast, so quickly modified it to the following code which finds all the prime factors for a number. Here's the code

Update: The first version had a bug which did not show the largest factor. Has since been corrected.

    """
            Get the factors for a number
            (Note: Not optimised for tail recursion)
    """
    def factor(n):
            if n == 1: return [1]
            i = 2
            limit = n**0.5
            while i <= limit:
                    if n % i == 0:
                            ret = factor(n/i)
                            ret.append(i)
                            return ret
                    i += 1

            return [n]

    if __name__ == "__main__":
            import sys
            for index in xrange(1,len(sys.argv)):
                    print "Factors for %s : %s" %(sys.argv[index], str(factor(int(sys.argv[index]))))

Quickly gave me the following output (used all of 26.2 seconds to get there) :

    # python getfactors.py 200000000000000000000000000009
    Factors for 200000000000000000000000000009 : [13430577524641L, 2094523, 89, 47, 47, 43, 29, 29]

And in case you are wondering what are the factors for 2009

    #  python getfactors.py 2009
    Factors for 2009 : [41, 7, 7]

And which is the next year which is a prime number ?

    #  python getfactors.py 2011
    Factors for 2011 : [2011]

Pretty playful hacking for 10 mins I thought.

Update: The code above was written in 10 mins, but it kept bothering me .. just wasn't idiomatic python. So once I got some time, I got back to it and decided to write a generator instead. The results obviously stay the same but the time came down from 26 seconds to 4.7 seconds. I of course threw in an additional optimisation which had nothing to do with a generator - basically the value of i no longer restarts from 2, it resumes with the last factor (the same that was yielded and moves on from there). Here's the new code implementing a generator.

    """
      Generator for getting factors for a number
    """
    def factor(n):
      yield 1
      i = 2
      limit = n**0.5
      while i <= limit:
        if n % i == 0:
          yield i
          n = n / i
          limit = n**0.5
        else:
          i += 1
      if n > 1:
        yield n


    if __name__ == "__main__":  
      import sys  
      for index in xrange(1,len(sys.argv)):
        print "Factors for %s : %s" %(sys.argv[index], [i for i in factor(int(sys.argv[index]))])