## 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 *g ^{l}* ≡

*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 (

*a*,

_{i}*b*) modulo (

_{i}*p*− 1) and a sequence of integers

*x*modulo

_{i}*p*such that

*x*=

_{i}*t*g

^{ai}^{bi}(mod

*p*), beginning with

*a*

_{0}=

*b*

_{0}= 0 and

*x*

_{0}= 1. Then the rule for deriving the terms of the various sequences is:

- If 0 <
*x*<_{i}*p*/3, then*a*_{i+1}= (*a*+ 1) mod (_{i}*p*− 1),*b*_{i+1}=*b*, and_{i}*x*_{i+1}=*t x*(mod_{i}*p*). - If
*p*/3 <*x*< 2_{i}*p*/3, then*a*_{i+1}= 2*a*mod (_{i}*p*− 1),*b*_{i+1}= 2*b*mod (_{i}*p*− 1), and*x*_{i+1}=*x*_{i}^{2}mod*p*. - If 2
*p*/3 <*x*<_{i}*p*, then*a*_{i+1}=*a*,_{i}*b*_{i+1}= (*b*+ 1) mod (_{i}*p*− 1), and*x*_{i+1}=*g x*mod_{i}*p*.

Splitting the computation into three pieces “randomizes” the calculation, since the interval in which *x _{i}* is found has nothing to do with the logarithm. The sequences are computed until some

*x*=

_{j}*x*, at which point we have

_{k}*t*

^{aj}*g*=

^{bj}*t*

^{ak}*g*. Then, if

^{bk}*a*−

_{j}*a*is coprime to

_{j}*p*− 1, we compute the discrete logarithm

*l*as (

*a*−

_{j}*a*)

_{k}*l*≡

*b*−

_{k}*b*(mod (

_{j}*p*− 1)). However, if the greatest common divisor of

*a*−

_{j}*a*with

_{j}*p*− 1 is

*d*> 1, then we compute (

*a*−

_{j}*a*)

_{k}*l*

_{0}≡

*b*−

_{k}*b*(mod (

_{j}*p*− 1) /

*d*), and

*l*=

*l*

_{0}+

*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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