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)
139
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)))))
Hello, what is the ilog function?
@Ruslan: The integer logarithm (ilog) of n to the base b is the largest integer e such that be ≤ n. You can see an implementation by clicking to run the code.
Hi, when I run this code Python says” name ‘count’ is not defined”, where is the mistake?