使用Pollard rho算法计算两个大整数的最小公倍数

本文主要介绍基本思路和具体的python代码实现

Pollard rho算法基本思路

1、F(x)=x^2+a 中的a可以试情况而定，一般取1，但尽量不要取0和2；
2、种子默认取2，原则上不小于2；
3、Pollard算法可以快速获取数n的几个因子，但这些因子不是n的质因子，也不是n的所有因子；
4、通过限定x>n , 规避循环问题。

使用Pollard rho算法求最小公倍数的基本思路

本方法只经过有限验证，不保证一定正确，仅做参考。

具体python2.7代码如下:

# !/user/bin/env python
# -*-coding: utf-8 -*-

'''
Using Pollard Pho factoring to find LCM(a, b)
seed = 2， f(x) = x^2 + 1
'''

from fractions import gcd


# find the factors of n with large power
def Pollard(n):
    # generate the list(List_x) of x
    x = 2
    f = lambda x: x * x + 1
    List_x = [2]
    while x < n:
        x = f(x)
        List_x.append(x)
    List_x = [x % n for x in List_x]

    # calculate p = gcd(x(i) - x(j), n) and remove the same p
    List_p = [gcd((List_x[i] - List_x[j]), n) for i in range(1, len(List_x)) for j in range(i)]
    List_p = list(set(List_p))
    List_p.sort()

    # list factors of n
    List_factor = [p for p in List_p if p > 1]
    return  List_factor


print 'Using Pollard Pho factoring to find LCM(a, b)'
print 'seed = 2, f(x) = x^2 + 1'

a = int(raw_input('input a:'))
b = int(raw_input('input b:'))

#obtain the factors of a and b, remove one of the same factor of both
prime_a = [a / i for i in Pollard(a)] + Pollard(a)
prime_b = [b / i for i in Pollard(b)] + Pollard(b)
temp = list(set(prime_a + prime_b))
temp.sort()
prime = [i for i in temp if i != 1]

#calculate LCM(a, b)
LCM = a * b
for i in prime:
    while (LCM / i % a == 0) and (LCM / i % b == 0):
        LCM = LCM / i
print '\nLCM(a, b) = %d\n' % LCM
raw_input('Press Enter to exit()')