March 19, 2010
We studied John Pollard’s p-1 factorization algorithm in a previous exercise. You may recall that the algorithm finds factors of a number n by calculating the least common multiple of the integers up to some bound B, call it k, then calculates the greatest common divisor of 2k-1 and n; if the greatest common divisor is between 1 and n, it is a factor of n.
What is happening mathematically is that we are trying to find a factor p|n (that’s “p divides n“, meaning that p is a factor of n, for those not familiar with the mathematical notation), for which we know the factorization of p-1. Consider the number 15770708441 = 135979 × 115979. If we apply Pollard’s p-1 algorithm with a bound of 150, no factors are found, but if we apply Pollard’s p-1 algorithm with a bound of 180 the 135979 factor is found, because 135979 – 1 = 2 × 3 × 131 × 173; increasing the bound to include the factor 173 makes Pollard’s p-1 algorithm work. The 135979 factor is found first because 115979 – 1 = 2 × 103 × 563, and 563 is out-of-bounds.
An alternative to increasing the bound is to call a second stage that looks for a p-1 for which all factors are less than the first-stage bound except one factor that is between the first-stage bound and the second-stage bound. That is, instead of calculating lcm(1..180), we calculate lcm(1..150) × j, where j ranges from 151 to 180. For small numbers like 150 and 180, the difference doesn’t matter, but for larger numbers like B1 = 106 and B2 = 109, the difference in the computational cost is noticeable.
Your task is to write a two-stage version of Pollard’s p-1 algorithm. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.