Pollard’s Rho Algorithm For Discrete Logarithms
May 27, 2016
We studied discrete logarithms in two previous exercises. Today we look at a third algorithm for computing discrete algorithms, invented by John Pollard in the mid 1970s. Our presentation follows that in the book Prime Numbers: A Computational Perspective by Richard Crandall and Carl Pomerance, which differs somewhat from other sources.
Our goal is to compute l (some browsers mess that up; it’s a lower-case ell, for “logarithm”) in the expression gl ≡ t (mod p); here p is a prime greater than 3, g is an integer generator on the range 1 ≤ g < p, and t is an integer target on the range 1 ≤ g < p. Pollard takes a sequence of integer pairs (ai, bi) modulo (p − 1) and a sequence of integers xi modulo p such that xi = tai gbi (mod p), beginning with a0 = b0 = 0 and x0 = 1. Then the rule for deriving the terms of the various sequences is:
- If 0 < xi < p/3, then ai+1 = (ai + 1) mod (p − 1), bi+1 = bi, and xi+1 = t xi (mod p).
- If p/3 < xi < 2p/3, then ai+1 = 2 ai mod (p − 1), bi+1 = 2 bi mod (p − 1), and xi+1 = xi2 mod p.
- If 2p/3 < xi < p, then ai+1 = ai, bi+1 = (bi + 1) mod (p − 1), and xi+1 = g xi mod p.
Splitting the computation into three pieces “randomizes” the calculation, since the interval in which xi is found has nothing to do with the logarithm. The sequences are computed until some xj = xk, at which point we have taj gbj = tak gbk. Then, if aj − aj is coprime to p − 1, we compute the discrete logarithm l as (aj − ak) l ≡ bk − bj (mod (p − 1)). However, if the greatest common divisor of aj − aj with p − 1 is d > 1, then we compute (aj − ak) l0 ≡ bk − bj (mod (p − 1) / d), and l = l0 + m (p − 1) / d for some m = 0, 1, …, d − 1, which must all be checked until the discrete logarithm is found.
Thus, Pollard’s rho algorithm consists of iterating the sequences until a match is found, for which we use Floyd’s cycle-finding algorithm, just as in Pollard’s rho algorithm for factoring integers. Here are outlines of the two algorithms, shown side-by-side to highlight the similarities:
# find d such that d | n # find l such that g**l = t (mod p) function factor(n) function dlog(g, t, p) func f(x) := (x*x+c) % n func f(x,a,b) := ... as above ... t, h, d := 1, 1, 1 j := (1,0,0); k := f(1,0,0) while d == 1 while j.x <> k.x t = f(t) j(x,a,b) := f(j.x, j.a, j.b) h = f(f(h)) k(x,a,b) := f(f(k.x, k.a, k.b)) d = gcd(t-h, n) d := gcd(j.a-k.a, p-1) return d return l ... as above ...
Please pardon some abuse of notation; I hope the intent is clear. In the factoring algorithm, it is possible that d is the trivial factor n, in which case you must try again with a different constant in the f function; the logarithm function has no such possibility. Most of the time consumed in the computation is the modular multiplications in the calculations of the x sequence; the algorithm itself is O(sqrt p), the same as the baby-steps, giant-steps algorithm of a previous exercise, but the space requirement is only a small constant, rather than the O(sqrt p) space required of the previous algorithm. In practice, the random split is made into more than 3 pieces, which complicates the code but speeds the computation, as much as a 25% improvement on average.
Your task is to write a program that computes discrete logarithms using Pollard’s rho 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.
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