Williams’ P+1 Factorization Algorithm

June 4, 2010

The code is far simpler to write than to describe. Here is the basic arithmetic over the Lucas sequences:

(define (add a b a-b n) (modulo (- (* a b) a-b) n))

(define (double a n) (modulo (- (* a a) 2) n))

(define (mult k v n)
  (let loop ((ks (cdr (digits k 2))) (x v) (y (double v n)))
    (cond ((null? ks) x)
          ((odd? (car ks)) (loop (cdr ks) (add x y v n) (double y n)))
          (else (loop (cdr ks) (double x n) (add x y v n))))))

Then the first stage of the p+1 algorithm is the same as the p-1 algorithm, with v the random starting point, but the second stage changes somewhat:

(define (pplus1-factor n b1 b2 v)
  (let stage1 ((p 2) (v v))
    (let ((a (ilog p b1)))
      (if (< p b1) (stage1 (next-prime p) (mult (expt p a) v n))
        (let ((g (gcd (- v 2) n))) (if (< 1 g n) g
          (let* ((vr v) (c (+ (* (quotient b1 6) 6) 12))
                 (v6 (mult 6 vr n)) (v12 (mult 12 vr n))
                 (z (modulo (* (- v6 vr) (- v12 vr)) n)))
            (let stage2 ((c c) (x v6) (y v12) (z z))
              (if (< c b2)
                  (let ((x+y (add y v6 x n)))
                    (stage2 (+ c 6) y x+y (modulo (* z (- x+y vr)) n)))
                  (let ((g (gcd z n))) (if (< 1 g n) g #f)))))))))))

We used ilog and digits from the Standard Prelude and next-prime and prime? from previous exercises. You can run the program at http://programmingpraxis.codepad.org/gAMB9T4D. Here’s an example:

> (pplus1-factor 451889 10 50 7)

If you wish, you can add Williams’ p+1 method to the integer factorization program of a previous exercise. A one-stage version of the algorithm, parameters for the factors function, and the calling code are shown below. The p+1 method fits best between the p-1 method and the elliptic curve method:

(define (pplus1-factor n b v)
  (define (m x) (modulo x n))
  (define (add a b a-b) (m (- (* a b) a-b)))
  (define (double a) (m (- (* a a) 2)))
  (define (mult k v)
    (let loop ((ks (cdr (digits k 2))) (x v) (y (double v)))
      (cond ((null? ks) x)
            ((odd? (car ks)) (loop (cdr ks) (add x y v) (double y)))
            (else (loop (cdr ks) (double x) (add x y v))))))
  (let loop ((p 2) (v v) (k 0))
    (cond ((< b p) (let ((g (gcd (- v 2) n))) (if (< 1 g n) g #f)))
          ((zero? (modulo k 10 0)) (let ((g (gcd (- v 2) n))) (if (< 1 g n) g
            (loop (next-prime p) (mult (expt p (ilog p b)) v) (+ k 1)))))
          (else (loop (next-prime p) (mult (expt p (ilog p b)) v) (+ k 1))))))

  (define pplus1-limit 100000) ; iteration limit
  (define pplus1-trials 7) ; number of p+1 constants to try

      ; williams pplus1
      (let loop ((k pplus1-trials) (v (randint 3 n)))
        (when (positive? k)
          (msg "Williams p+1: bound=" pplus1-limit ", constant=" v)
          (if (factor? 'p+1 (pplus1-factor n pplus1-limit v))
              (loop k (randint n))
              (loop (- k 1) (randint n)))))

Pages: 1 2

3 Responses to “Williams’ P+1 Factorization Algorithm”

  1. Lucas A. Brown said
    def primegen():
        Generates primes lazily via the sieve of Eratosthenes
        Input: none
            Sequence of integers
        >>> list(takewhile(lambda x: x < 100, primegen()))
        [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
        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 mlucas(v, a, n):
        """ Helper function for williams_pp1().  Multiplies along a Lucas sequence modulo n. """
        v1, v2 = v, (v**2 - 2) % n
        for bit in bin(a)[3:]: v1, v2 = ((v1**2 - 2) % n, (v1*v2 - v) % n) if bit == "0" else ((v1*v2 - v) % n, (v2**2 - 2) % n)
        return v1
    def williams_pp1(n):
        Williams' p+1 integer factoring algorithm
            n -- integer to factor
            Integer.  A nontrivial factor of n.
        >>> williams_pp1(315951348188966255352482641444979927)
        for v in count(1):
            for p in primegen():
                e = ilog(sqrt(n), p)
                if e == 0: break
                for _ in xrange(e): v = mlucas(v, p, n)
                g = gcd(v - 2, n)
                if 1 < g < n: return g
                if g == n: break
  2. Ruslan Khamidullin said

    Hello, what is the ilog function?

  3. programmingpraxis said

    @Ruslan: The integer logarithm (ilog) of n to the base b is the largest integer e such that ben. You can see an implementation by clicking to run the code.

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