brainwagon "There is much pleasure in useless knowledge." — Bertrand Russell


On Factoring, by Jevons’ The Principles of Science

While reading the section on factorization of large numbers in Knuth's Seminumerical Algorithms, I encountered a reference to an interesting claim by William Stanley Jevons. I did a google search, and found it in google books:

Knuth was quick to point out that the number that he gave was easily factorable in short order (even by hand, even in Jevons' day) using Fermat's method. I of course just handed it to the little program I wrote, and it instantly returned

8,616,460,799 =
	89,681 *

Addendum: How does Fermat's method work? You basically try to express the prime as the difference of two squares. Let's say that the prime P = u * v. If P = x^2 - y^2, then P = (x + y) (x - y), so the two factors u = x + y and v = x - y. You start by setting x to be just above the sqrt(P), and compute x*x - P. If the remainder is a square, then you have have two factors. Knuth gives several hints on how you could avoid testing many possible squares, but even acting trivially, you would only have to try 55 numbers before finding that...

8616460799 = 92880**2 - 3199**2
8616460799 = 96079 * 89681

Addendum: Lehmer factored this number in 1903. Golomb apparently had a Cryptologia article showing how it could be easily factored with a hand calculator, but I don't want to pay $37 to get the 3 page reprint.

Filed under: Math No Comments

A Simple Python Program for Factoring…

Sometimes, your interests converge. Over on Programming Praxis, he had a coding challenge to implement Monte Carlo factorization. The last couple of days, I have been thinking a bit more about factorization, so it seemed like a good idea to give it a whack. I cribbed a version of the Rabin Miller primality tester off the web, and then coded the rest up in python.

#!/usr/bin/env python

# a basic implementation of the Pollard rho factorization
# Written by Mark VandeWettering <>

import sys
import locale
import random

class FactorError(Exception):
    def __init__(self, value):
	self.value = value 
    def __str__(self):
	return repr(self.value)

def miller_rabin_pass(a, n):
    d = n - 1
    s = 0
    while d % 2 == 0:
        d >>= 1
        s += 1

    a_to_power = pow(a, d, n)
    if a_to_power == 1:
        return True
    for i in xrange(s-1):
        if a_to_power == n - 1:
            return True
        a_to_power = (a_to_power * a_to_power) % n
    return a_to_power == n - 1

def isprime(n):
    for repeat in xrange(20):
        a = 0
        while a == 0:
            a = random.randrange(n)
        if not miller_rabin_pass(a, n):
            return False
    return True

def gcd(a, b):
    while b != 0:
	a, b = b, a%b
    return a 

def findfactor(n):
    for c in range(1, 50):
	x = y = random.randint(1, n-1)
	x = (x * x + c) % n
	y = (y * y + c) % n
	y = (y * y + c) % n
	while True:
	    t = gcd(n, abs(x-y))
	    if t == 1:
		x = (x * x + c) % n
		y = (y * y + c) % n
		y = (y * y + c) % n
	    elif t == n:
		return t
    raise FactorError("couldn't find a factor.")
def factor(n):
    r = []
    while True:
	if isprime(n):
	    f = findfactor(n)
	    n = n / f
	except FactorError, msg:
    return r 

def doit(n):
	flist = factor(n)
	print locale.format("%d", n, True), "="
	for f in flist:
		print "\t%s" % locale.format("%d", f, True)

locale.setlocale(locale.LC_ALL, "")


This is a fairly challenging factorization, and after a few seconds, it outputs:

316,912,650,057,057,350,374,175,801,343 =

Use a Wii Balance Board with Linux

An interesting part of the Nintendo Wii design is that they rely on bluetooth for connecting their joysticks and (as it turns out) their Balance Board peripheral. This makes them pretty darned hackable. I've seen lots about using the "WiiMote" with computers, but this is the first I've seen that interfaces with the Balance Board. Nifty.

Use a Wii Balance Board with Linux
Use a Wii Balance Board with Linux

Filed under: General 1 Comment