Extending Pollard’s P-1 Factorization Algorithm

March 19, 2010

We studied John Pollard’s p-1 factorization algorithm in a previous exercise. You may recall that the algorithm finds factors of a number n by calculating the least common multiple of the integers up to some bound B, call it k, then calculates the greatest common divisor of 2k-1 and n; if the greatest common divisor is between 1 and n, it is a factor of n.

What is happening mathematically is that we are trying to find a factor p|n (that’s “p divides n“, meaning that p is a factor of n, for those not familiar with the mathematical notation), for which we know the factorization of p-1. Consider the number 15770708441 = 135979 × 115979. If we apply Pollard’s p-1 algorithm with a bound of 150, no factors are found, but if we apply Pollard’s p-1 algorithm with a bound of 180 the 135979 factor is found, because 135979 – 1 = 2 × 3 × 131 × 173; increasing the bound to include the factor 173 makes Pollard’s p-1 algorithm work. The 135979 factor is found first because 115979 – 1 = 2 × 103 × 563, and 563 is out-of-bounds.

An alternative to increasing the bound is to call a second stage that looks for a p-1 for which all factors are less than the first-stage bound except one factor that is between the first-stage bound and the second-stage bound. That is, instead of calculating lcm(1..180), we calculate lcm(1..150) × j, where j ranges from 151 to 180. For small numbers like 150 and 180, the difference doesn’t matter, but for larger numbers like B1 = 106 and B2 = 109, the difference in the computational cost is noticeable.

Your task is to write a two-stage version of Pollard’s p-1 algorithm. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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5 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 […]

  5. Lucas A. Brown said
    def primegen():
        Generates primes lazily via the sieve of Eratosthenes.
        Code shamelessly stolen from someone else's work.  No idea where I got it.
        Input: none
            Sequence of integers
        yield 2; yield 3; yield 5; yield 7; yield 11; yield 13
        ps = primegen() # yay recursion
        p = ps.next() and ps.next()
        q, sieve, n = p**2, {}, 13
        while True:
            if n not in sieve:
                if n < q: yield n
                    next, step = q + 2*p, 2*p
                    while next in sieve: next += step
                    sieve[next] = step
                    p = ps.next()
                    q = p**2
                step = sieve.pop(n)
                next = n + step
                while next in sieve: next += step
                sieve[next] = step
            n += 2
    def ispower(n):
        If n is a perfect power, return the largest integer (in terms of absolute value) that, when squared/cubed/etc, yields n.
        If n is not a perfect power, return 0.
            n -- an integer
            An integer
        >>> [ispower(n) for n in [64, 25, -729, 1729]]
        [8, 5, -9, 0]
        for p in primegen():
            r = introot(n, p)
            if r is None: continue
            if r ** p == n: return r
            if r == 1: return 0
    def pollard_pm1(n, B1=100, B2=1000):       # TODO: What are the best default bounds and way to increment them?
        Integer factoring function.  Uses Pollard's p-1 algorithm.  Note that this is only efficient iff the number to be factored
        has a prime factor p such that p-1's largest prime factor is "small".  In this implementation, that tends to mean less than
        10,000,000 or so.
            n -- number to factor
            B1 -- Natural number.  Bound for phase 1.  Default == 100.
            B2 -- Natural number.  Bound for phase 2.  Must be > B1.  Default == 100.
            A factor of n.
        >>> pollard_pm1(1275683450258216989546041841)
        if isprime(n): return n
        m = ispower(n)
        if m: return m
        while True:
            pg = primegen()
            q = 2           # TODO: what about other initial values of q?
            p = pg.next()
            while p <= B1: q, p = pow(q, p**ilog(B1, p), n), pg.next()
            g = gcd(q-1, n)
            if 1 < g < n: return g
            while p <= B2: q, p = pow(q, p, n), pg.next()
            g = gcd(q-1, n)
            if 1 < g < n: return g
            # These bounds failed.  Increase and try again.
            B1 *= 10
            B2 *= 10

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