Briggs M.'s An Introduction to the General Number Field Sieve PDF

By Briggs M.

The final quantity box Sieve (GNFS) is the quickest recognized strategy for factoring "large" integers, the place huge is usually taken to intend over one hundred ten digits. This makes it the easiest set of rules for trying to unscramble keys within the RSA [2, bankruptcy four] public-key cryptography approach, the most typical equipment for transmitting and receiving mystery information. in truth, GNFS was once used lately to issue a 130-digit "challenge" quantity released through RSA, the most important variety of cryptographic value ever factored.

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Extra info for An Introduction to the General Number Field Sieve

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This basic idea leads to storing approximations to ln(a +bm) in the sieve array instead of the actual value a + bm. From the elementary theory of logarithms, dividing a + bm by a prime p is equivalent to subtracting ln(p) from ln(a + bm). Thus, for a fixed b and prime p, one subtracts ln(p) from the array location for a = −bm + kp for k ∈ Z with −u < a < u. After processing all primes in F for a fixed b, the sieve array is then scanned for values ≤ 0 = ln(1) instead of 1. Such a position yields a value for a with the value a + bm very “likely” to be smooth.

But there is only one Sylow 2-subgroup S2r of F ∗q so η s must generate S2r and the result follows. The method of Shanks and Tonelli improves even further upon this, requiring at most r steps instead of 2r . The idea is to produce a sequence of elements ωi and λi in F ∗q such that ωi2 = λi δ, with the order oi+1 of λi+1 strictly less than the order oi of λi and oi dividing 2r−1 for all i in the sequence. If such a sequence could be found then eventually λj = 1 for some j so ωj2 = a and a square root of δ has been found.

Any subfield of F pd contains pe elements where e is a divisor of d. Conversely, if e is a divisor of d then there is exactly one subfield of F pd containing pe elements. Proof. Suppose E is a subfield of F pd . Then d = [F pd : F p ] = [F pd : E][E : F p ] and hence E is a finite dimensional extension of F p and must therefore have pe elements for some positive Matthew E. Briggs Chapter 4. Filling in the Details 49 integer e. But this equation also implies that F pd is a finite dimensional extension of E and hence F pd must contain (pe )k elements for some positive integer k.

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An Introduction to the General Number Field Sieve by Briggs M.


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