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Therefore, 324 = 37 (mod 101) and we have solved this particular discrete log problem. In general, for the giant-step phase, we compute about fi values and store these in our table. Then in the baby-step phase, after trying about J r/2 choices for we expect to have found a solution. Assuming the table lookups do not carry any cost (the table could be sorted or a hash table used), this algorithm requires about 1.5J r niultiplications and it also requires storage (or space) of about &. Assuming the storage requirement is feasible, this is A significant improvement over the nai've method of trial multiplication, but still an cxponential work factor.
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As above, we are given the generator 9.a prime p , and x = ga (mod p ) , and we want to find a. Analogous to Dixon's factoring algorithm, we first select a bound B and a factor base of primes less than B . Then we pre-compute the discrete logarithms to the base y of the elements in the factor base and we save the logarithms of these elements. These logarithms of the factor base can he found efficiently by solving a system of linear equations; see Problem 11. Once the logarithms of the elements in the factor base are known, the attack is straightforward. We have 11: = g" (mod p ) , and we want to find a. Let { p ~ . p l ,. .. ,p,-l} be the set of primes in the factor base. We randomly select k E ( 0 , l . 2 , . . . , p - 2 } and compute y = J:. gk (mod p ) until we find a y that factors completely over the factor base. Given such a y we have
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Taking log, on both sides and simplifying, we find
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(7.11)
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Assuming that the logarithms of the elements in the factor base are known, we have determined a. Note that the mod p - 1 in (7.11) follows from Ferniat's Little Theorem, which is given in Appendix A-2 and, for example, in [as]. An example should clarify the algorithm. Suppose g = 3 , p = 101 and we are given z = 3" = 94 (mod 101) and we want to determine a. As our factor base we choose the set of primes { 2 , 3 , 5 , 7 } and by solving a system of linear equations (see Problem l l ) , we determine log3 2 = 29, log;3 3 = 1, log, 5 = 96, log3 7 = 61. Next, we randomly select k until we obtain a value 9 4 . 3' (mod 101) which factors completely over the factor base. For example, if we choose k = 10 tl.1er1 9 4 . 3"' = 50 (mod 101). (7.12)
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7.3 DISCRETE LOG ALGORITHMS
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and 50 = 2 . 52 factors over our factor base. Taking logarithms of both sides of (7.12) yields log, 94 = log, 2 + 2 log, 5 - 10 (mod 100). Substituting the values for the logarithms of the primes in the factor base, we have log, 94 = 29 192 - 10 = 11 (mod loo),
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which implies that 311 = 94 (mod 101). Then a = 11 and we have solved this particular discrete log problem. The index calculus provides the most efficient method for solving a general discrete log problem. The work factor for the index calculus is subexponential; more precisely, the work is on the order of e(1np)'/2('n 1np)'/2. The form of this work factor is the same as that given for the quadratic sieve factoring method in Section 7.2.4, which is, perhaps, not surprising since both rely on finding smooth integers. For more information on the index calculus, a good source is [99].
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