Pollard's rho algorithm for logarithms
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Pollard's rho algorithm for logarithms is an algorithm for solving the discrete logarithm problem analogous to Pollard's rho algorithm for solving the Integer factorization problem.
The goal is to compute γ such that αγ = β(modN), where β belongs to the group G generated by α. The algorithm computes integers a, b, A, and B such that αaβb = αAβB(modN). Assuming, for simplicity, that the underlying group is cyclic of order N and that n = φ(N), we can calculate γ as a solution of the equation (B − b)γ = (a − A)(mod n).
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 after approximately steps. One way to define such a function is to use the following rules: Divide G into three subsets (not necessarily subgroups) of approximately equal size: G0, G1, and G2. If xi is in G0 then double both a and b; if then increment a, if then increment b.
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[edit] Algorithm
Let G be a cyclic group of prime order p, and given , and a partition , let be a map
and define maps and by
- Inputs a a generator of G, b an element of G
- Output An integer x such that ax = b, or failure
- Initialise a0 ← 0
-
- b0 ← 0
- x0 ← 1 ∈ G
- i ← 1
-
- xi ← f(xi-1), ai ← g(xi-1,ai-1), bi ← h(xi-1,bi-1)
- x2i ← f(f(x2i-2)), a2i ← g(f(x2i-2),g(x2i-2,a2i-2)), b2i ← h(f(x2i-2),h(x2i-2,b2i-2))
- If xi = x2i then
- r ← bi - b2i
- If r = 0 return failure
- x ← r-1(a2i - ai) mod p
- return x
- If xi ≠ x2i then i ← i+1, and go to step 2.
- Initialise a0 ← 0
[edit] 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:
#include <stdio.h> const int n = 1018, N = n + 1; /* N = 1019 -- prime */ const int alpha = 2; /* generator */ const int beta = 5; /* 2^{10} = 1024 = 5 (N) */ void new_xab( int& x, int& a, int& b ) { switch( x%3 ) { case 0: x = x*x % N; a = a*2 % n; b = b*2 % n; break; case 1: x = x*alpha % N; a = (a+1) % n; break; case 2: x = x*beta % N; b = (b+1) % n; break; } } int main() { int x=1, a=0, b=0; int X=x, A=a, B=b; for( int i = 1; i < n; ++i ) { new_xab( x, a, b ); new_xab( X, A, B ); new_xab( X, A, B ); printf( "%3d %4d %3d %3d %4d %3d %3d\n", i, x, a, b, X, A, B ); if( x == X ) break; } return 0; }
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 26815378 = 1010 = 23015416(mod 1019) and so (416 − 378)γ = 681 − 301(mod 1018), for which γ1 = 10 is a solution as expected. As n = 1018 is not prime, there is another solution γ2 = 519, for which 2519 = 1014 = − 5(mod 1019) holds.
[edit] Complexity
The running time is approximately O() for a number n.
[edit] References
- J. Pollard, Monte Carlo methods for index computation mod p, Mathematics of Computation, Volume 32, 1978.
- Alfred J. Menezes, Paul C. van Oorschot, and Scott A. Vanstone, Handbook of Applied Cryptography, Chapter 3, 2001.