By Alex Biryukov, Dmitry Khovratovich (auth.), Mitsuru Matsui (eds.)
This publication constitutes the refereed court cases of the fifteenth overseas convention at the concept and alertness of Cryptology and knowledge protection, ASIACRYPT 2009, held in Tokyo, Japan, in December 2009.
The forty-one revised complete papers awarded have been rigorously reviewed and chosen from 298 submissions. The papers are prepared in topical sections on block ciphers, quantum and post-quantum, hash capabilities I, encryption schemes, multi celebration computation, cryptographic protocols, hash funtions II, types and frameworks I, cryptoanalysis: sq. and quadratic, types and framework II, hash services III, lattice-based, and facet channels.
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Additional resources for Advances in Cryptology – ASIACRYPT 2009: 15th International Conference on the Theory and Application of Cryptology and Information Security, Tokyo, Japan, December 6-10, 2009. Proceedings
Ii) There exists a ﬁxed permutation S ∈ Perm(n) (represented by a deterministic stateless system) such that Δq (C(P), C (P)) ≤ Δq (C(S), C (S)). 42 P. Gaˇzi and U. Maurer Proof. The ﬁrst claim comes from , so here we only prove the second one. Since the random system P can be seen as a system that picks a permutation uniformly at random from Perm(n) and then realizes this permutation, we have: Δq (C(P), C (P)) ≤ 1 (2n )! Δq (C(S), C (S)). S∈Perm(n) If all the values Δq (C(S), C (S)) were smaller than Δq (C(P), C (P)) it would contradict the inequality above, hence there exists a permutation S ∈ Perm(n) such that Δq (C(P), C (P)) ≤ Δq (C(S), C (S)).
If the hypothesis is not rejected, perform exhaustive search for the remaining 25 key bits. 5-round attack to an attack on the 6-round variant of IDEA starting before the MA layer of the second round. 1 . 5round attack. 5-round attack without enlarging the time complexity. It is more diﬃcult to construct right plaintext pairs satisfying Proposition 1. Consider a pair of intermediate values X 3 and X 3 before the third round, which satisfy Proposition 1. If we partially decrypt X 3 and X 3 using any possible Z52 and Z62 , the only fact we know is that all the results have the same XOR of the ﬁrst and third words.
Y13 = ΔY33 = 0 b. ΔY23 = 8000x c. Y23 ⊕ Y43 = Y2 3 ⊕ Y4 3 then Δs3 = 0 and the probability of LSB(Δs4 ) = 0 can be determined by Equation (5). Proof. From Condition (a), ΔY13 = ΔY33 = 0, p3 is equal to p 3 . Then Δs3 = 0 is quite straightforward. From Condition (c), q 3 is equal to q 3 . If p3 and q 3 are ﬁxed, u3 and t3 are also ﬁxed with respect to any Z53 and Z63 . It indicates that X14 = Y13 ⊕ u3 = X14 . Note that Y14 and Y1 4 are the results of modular-multiplying X14 and X14 with the same Z14 , hence Y14 is equal to Y1 4 .
Advances in Cryptology – ASIACRYPT 2009: 15th International Conference on the Theory and Application of Cryptology and Information Security, Tokyo, Japan, December 6-10, 2009. Proceedings by Alex Biryukov, Dmitry Khovratovich (auth.), Mitsuru Matsui (eds.)