In today’s Programming Praxis problem we have to implement John Pollard’s factorization algorithm. Our target is 16 lines (I’m not counting the code for the primality test, since we did that already).
First, we’re going to need to reuse the code from the Miller-Rabin primality test exercise, since we need to determine whether or not the number is prime:
import Control.Arrow import Data.Bits import Data.List import System.Random isPrime :: Integer -> StdGen -> Bool isPrime n g = let (s, d) = (length *** head) . span even $ iterate (`div` 2) (n-1) xs = map (expm n d) . take 50 $ randomRs (2, n - 2) g in all (\x -> elem x [1, n - 1] || any (== n-1) (take s $ iterate (expm n 2) x)) xs expm :: Integer -> Integer -> Integer -> Integer expm m e b = foldl' (\r (b', _) -> mod (r * b') m) 1 . filter (flip testBit 0 . snd) . zip (iterate (flip mod m . (^ 2)) b) $ takeWhile (> 0) $ iterate (`shiftR` 1) e
There’s not much more to the factor function than simply writing down the algorithm. It’s just math with Haskell syntax, really. The only thing you have to take care of is not to calculate gcd (x-y) n when x and y are still 2, since that will give you an incorrect result.
factor :: Integer -> Integer -> Integer factor c n = factor' 2 2 1 where f x = mod (x * x + c) n factor' x y 1 = factor' x' y' (gcd (x' - y') n) where (x', y') = (f x, f $ f y) factor' _ _ d = if d == n then factor (c + 1) n else d
And factors does little more than recursively calling factor, while filtering out factors of two and making sure not to call factor on a prime (since the algorithm is not designed for that).
factors :: Integer -> StdGen -> [Integer] factors n g = sort $ fs n where fs x | even x = 2 : fs (div x 2) | isPrime x g = [x] | otherwise = f : fs (div x f) where f = factor 1 x
And to test:
main :: IO () main = print . factors (2^98 - 1) =<< getStdGen
The end result is 9 lines of code, which was to be expected, given that it’s just a question of writing math in your language’s syntax.