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]))])

## Comments !