Pollard's rho algorithm for logarithms is an algorithm introduced by John Pollard in 1978 to solve the discrete logarithm problem, analogous to Pollard's rho algorithm to solve the integer factorization problem. The goal is to compute \gamma such that, where \beta belongs to a cyclic group G generated by \alpha. The algorithm computes integers a, b, A, and B such that. If the underlying group is cyclic of order n, by substituting \beta as and noting that two powers are equal if and only if the exponents are equivalent modulo the order of the base, in this case modulo n, we get that \gamma is one of the solutions of the equation. Solutions to this equation are easily obtained using the extended Euclidean algorithm. To find the needed a, b, A, and B the algorithm uses Floyd's cycle-finding algorithm to find a cycle in the sequence, where the function is assumed to be random-looking and thus is likely to enter into a loop of approximate length after steps. One way to define such a function is to use the following rules: Partition G into three disjoint subsets S_0, S_1, and S_2 of approximately equal size using a hash function. If x_i is in S_0 then double both a and b; if x_i \in S_1 then increment a, if x_i \in S_2 then increment b.

Algorithm

Let G be a cyclic group of order n, and given, and a partition , let f:G\to G be the map and define maps and by input: a: a generator of G b: an element of G output: An integer x such that ax = b, or failure Initialise i ← 0, a0 ← 0, b0 ← 0, x0 ← 1 ∈ G loop i ← i + 1 xi ← f(xi−1), ai ← g(xi−1, ai−1), bi ← h(xi−1, bi−1) x2i−1 ← f(x2i−2), a2i−1 ← g(x2i−2, a2i−2), b2i−1 ← h(x2i−2, b2i−2) x2i ← f(x2i−1), a2i ← g(x2i−1, a2i−1), b2i ← h(x2i−1, b2i−1) while xi ≠ x2i r ← bi − b2i if r = 0 return failure return r−1(a2i − ai) mod n

Example

Consider, for example, the group generated by 2 modulo N=1019 (the order of the group is n=1018, 2 generates the group of units modulo 1019). The algorithm is implemented by the following C++ program: The results are as follows (edited): i x a b X A B -- 1 2 1 0 10 1 1 2 10 1 1 100 2 2 3 20 2 1 1000 3 3 4 100 2 2 425 8 6 5 200 3 2 436 16 14 6 1000 3 3 284 17 15 7 981 4 3 986 17 17 8 425 8 6 194 17 19 .............................. 48 224 680 376 86 299 412 49 101 680 377 860 300 413 50 505 680 378 101 300 415 51 1010 681 378 1010 301 416 That is and so, for which \gamma_1=10 is a solution as expected. As n=1018 is not prime, there is another solution, for which holds.

Complexity

The running time is approximately. If used together with the Pohlig–Hellman algorithm, the running time of the combined algorithm is, where p is the largest prime factor of n.