Pollard's rho algorithm for logarithms

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 such that , where belongs to a group generated by . The algorithm computes integers , , , and such that . Assuming, for simplicity, that the underlying group is cyclic of order , we can calculate as a solution of the equation .

To find the needed , , , and 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 into three subsets (not necessarily subgroups) of approximately equal size: , , and . If is in then double both and ; if then increment , if then increment .

Algorithm

Let be a cyclic group of order , 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 a^x = b, or failure

1. Initialise a₀ ← 0

   b₀ ← 0

   x₀ ← 1 ∈ G

   i ← 1

2. x_i ← f(x_i-1), a_i ← g(x_i-1,a_i-1), b_i ← h(x_i-1,b_i-1)
3. x_2i ← f(f(x_2i-2)), a_2i ← g(f(x_2i-2),g(x_2i-2,a_2i-2)), b_2i ← h(f(x_2i-2),h(x_2i-2,b_2i-2))
4. If x_i = x_2i then
   1. r ← b_i - b_2i
   2. If r = 0 return failure
   3. x ← r ^-1 (a_2i - a_i) mod p
   4. return x
5. If x_i ≠ x_2i then i ← i+1, and go to step 2.

Example

Consider, for example, the group generated by 2 modulo (the order of the group is , 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(void) {
   int x=1, a=0, b=0;
   int X=x, A=a, B=b;
   int i;
   for( 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 and so , for which is a solution as expected. As is not prime, there is another solution , for which holds.

Complexity

The running time is approximately O() where p is n's smallest prime factor.

References

• Pollard, J. M. (1978). "Monte Carlo methods for index computation (mod p)". Mathematics of Computation 32 (143): 918–924. JSTOR 2006496. 
• Menezes, Alfred J.; van Oorschot, Paul C.; Vanstone, Scott A. (2001). "Chapter 3". Handbook of Applied Cryptography. http://www.cacr.math.uwaterloo.ca/hac/about/chap3.pdf.