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.

About these ads

Pages: 1 2

4 Responses to “Extending Pollard’s P-1 Factorization Algorithm”

  1. [...] 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 [...]

  2. Remco Niemeijer said

    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) n
    
  3. Nice Blog… Going to go through one post a time and read this whole website like it was designed for a magazine.

  4. [...] 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 [...]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 574 other followers

%d bloggers like this: