## 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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### 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

```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. […] 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 […]

4. 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
Output:
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
else:
next, step = q + 2*p, 2*p
while next in sieve: next += step
sieve[next] = step
p = ps.next()
q = p**2
else:
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.
Input:
n -- an integer
Output:
An integer
Examples:
>>> [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.
Input:
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.
Output:
A factor of n.
Examples:
>>> pollard_pm1(1275683450258216989546041841)
1224040923709997L
"""
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
```