Extending Pollard’s P-1 Factorization Algorithm
March 19, 2010
We begin by restating the one-stage algorithm, in a somewhat different form than it appeared in the previous exercise:
(define (pollard1 n b1)
(let stage1 ((a 2) (i 2))
(if (< i b1)
(stage1 (expm a i n) (+ i 1))
(let ((d (gcd (- a 1) n)))
(if (< 1 d n) d #f)))))
The two-stage algorithm starts the same, but the continuation loops over the integers from B1 to B2, making the same computation as the first stage:
(define (pollard2 n b1 b2)
(let stage1 ((a 2) (i 2))
(if (< i b1)
(stage1 (expm a i n) (+ i 1))
(let ((d (gcd (- a 1) n)))
(if (< 1 d n) (list 'stage1 d)
(let stage2 ((j b1))
(if (= j b2) #f
(let ((d (gcd (- (expm a j n) 1) n)))
(if (< 1 d n) (list 'stage2 d)
(stage2 (+ j 1)))))))))))
We can see how this works using the sample problem given above:
> (pollard1 15770708441 150)
#f
> (pollard1 15770708441 180)
135979
> (pollard2 15770708441 150 180)
(stage2 135979)
We use expm from the Standard Prelude. You can run the program at http://programmingpraxis.codepad.org/VpaXvPEN.
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[...] Praxis – Extending Pollard’s P-1 Factorization Algorithm By Remco Niemeijer In today’s Programming Praxis exercise we need to write an improved version of a factorization algorithm. I [...]
My Haskell solution (see http://bonsaicode.wordpress.com/2010/03/19/programming-praxis-extending-pollard%E2%80%99s-p-1-factorization-algorithm/ for a version with comments):
import Data.Bits import Data.List expm :: Integer -> Integer -> Integer -> Integer expm b e m = 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 pollard :: (Integer -> t) -> (Integer -> t) -> Integer -> Integer -> t pollard found notFound n b1 = f 2 2 where f a i | i < b1 = f (expm a i n) (i + 1) | 1 < d && d < n = found d | otherwise = notFound a where d = gcd (a - 1) n pollard1 :: Integer -> Integer -> Maybe Integer pollard1 = pollard Just (const Nothing) pollard2 :: Integer -> Integer -> Integer -> Maybe (String, Integer) pollard2 n b1 b2 = pollard (Just . ((,) "stage1")) (f b1) n b1 where f j a | j == b2 = Nothing | 1 < d && d < n = Just ("stage2", d) | otherwise = f (j + 1) a where d = gcd (expm a j n - 1) nNice Blog… Going to go through one post a time and read this whole website like it was designed for a magazine.
[...] have studied John Pollard’s p−1 algorithm for integer factorization on two previous occasions, giving first the basic single-stage algorithm and later adding a second stage. In today’s [...]