Advances in Cryptology — EUROCRYPT 2002: International by Rosario Gennaro, Daniele Micciancio (auth.), Lars R. Knudsen

By Rosario Gennaro, Daniele Micciancio (auth.), Lars R. Knudsen (eds.)

This e-book constitutes the refereed lawsuits of the overseas convention at the thought and alertness of Cryptographic recommendations, EUROCRYPT 2002, held in Amsterdam, The Netherlands, in April/May 2002.
The 33 revised complete papers provided have been conscientiously reviewed and chosen from a complete of 122 submissions. The papers are equipped in topical sections on cryptanalysis, public-key encryption, details conception and new versions, implementational research, move ciphers, electronic signatures, key trade, modes of operation, traitor tracing and id-based encryption, multiparty and multicast, and symmetric cryptology.

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Extra resources for Advances in Cryptology — EUROCRYPT 2002: International Conference on the Theory and Applications of Cryptographic Techniques Amsterdam, The Netherlands, April 28 – May 2, 2002 Proceedings

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Cr ) ∈ (Bn )r . Initialization: x0 = e(identity braid), ci = ci for all i. Loop: STEP 1: If inf(ci ) ≥ inf(ai ) for all i, then STOP STEP 2: Choose k such that inf(ck ) < inf(ak ). Compute the permutation braid H by applying Proposition 1 to (ak , ck ). STEP 3: x0 ← Hx0 , ci ← Hci H −1 for all i. GO TO STEP 1 Output: x0 and (c1 , . . , cr ). Because H in Proposition 1 is a suffix of x, so is x0 at each step in the above algorithm. Whenever Proposition 1 is applied in the loop, the word-length of x0 strictly increases and its final length is bounded above by |x|.

C be the left canonical forms of conjugate braids. It is natural to say that a is simpler than c if (i) the word-length of A1 . . Ak is smaller than that of C1 . . C , or (ii) they have same word-length but k is smaller than . The former is equivalent to u = inf(a) > v = inf(c) and the latter is equivalent to inf(a) = inf(c) and sup(a) < sup(b). Mathematical Algorithm for the Conjugacy Problem. The conjugacy problem in braid groups is: given (a, c) ∈ (Bn )2 , decide whether they are conjugate and if so, find x ∈ Bn such that x−1 ax = c.

Ar ) and (c1 , . . , cr ) in (Bn )r , find x ∈ Bn such that ci = x−1 ai x for all i simultaneously. We start with some discussions. Uniqueness of the Solution to the MSCP in Braid Groups. For generic choice of a1 , . . , ar , the solution x is unique up to a power of ∆2 (See Appendix B). That is, if x is another solution such that x −1 ai x = ci for all i, then x = ∆2k x for some integer k. Note that x −1 y −1 x y = x−1 y −1 xy for any y, because ∆2 is a central element. Therefore, it suffices to find ∆2k x for any k.

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