Advances in Cryptology - CRYPTO 2009: 29th Annual by Shai Halevi

By Shai Halevi

This e-book constitutes the refereed complaints of the twenty ninth Annual foreign Cryptology convention, CRYPTO 2009, held in Santa Barbara, CA, united states in August 2009. The 38 revised complete papers offered have been rigorously reviewed and chosen from 213 submissions. Addressing all present foundational, theoretical and study points of cryptology, cryptography, and cryptanalysis in addition to complicated functions, the papers are equipped in topical sections on key leakage, hash-function cryptanalysis, privateness and anonymity, interactive proofs and zero-knowledge, block-cipher cryptanalysis, modes of operation, elliptic curves, cryptographic hardness, merkle puzzles, cryptography within the actual global, assaults on signature schemes, mystery sharing and safe computation, cryptography and game-theory, cryptography and lattices, identity-based encryption and cryptographers’ toolbox.

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Example text

Based on Lemma 4, we can obviously construct a π-OBDD Qm with w(Qm ) ≤ 2 that performs the consistency test for the observed keystream z such that the generator fulfills the BDD Assumption. Since each keystream bit involves 3 new bits, we can expect the Independence Assumption to hold. By plugging α, γ and p into the statement of Theorem 2, we obtain: Theorem 3. 65625n ≈ 2189 for n = 288. 30 D. 5 keystream bits [17] or in around 2135 operations from O(1) keystream bits [9]. 2 Grain-128 The regularly clocked stream cipher Grain-128 was proposed in [12] and supports key size of 128 bits and IV size of 96 bits.

Xn } is a directed, acyclic graph G = (V, E) with E ⊆ V × V × {0, 1}. Each inner node v has exactly two outgoing edges, a 0-edge (v, v0 , 0) and a 1-edge (v, v1 , 1) leading to the 0-successor v0 and the 1-successor v1 , respectively. A BDD contains exactly two nodes with outdegree 0, the sinks s0 and s1 . label = 1. There is exacly one node with indegree 0, the root of the BDD. , |G| := |V |. Each node v ∈ V represents a Boolean Function fv ∈ Bn = {f |f : {0, 1}n → {0, 1}} in the following manner: For an input a = (a1 , .

Lemma 5. For a Fibonacci FCSR R with q bits of memory, an integer m > 0, and π the canonical reading order, we can construct a π-OBDD Sm of width at most 2q+1 that tests for the internal bitstream w ∈ {0, 1}m of the modified FCSR with m = n − 1 + t(q + 1) whether the last q + 1 bits fulfill the update relation. Proof. In order to check whether σt = (σt−1 div 2) + equivalently test if q σt = i=1 i σt−1 n i=1 wt−i · ci , we can n 0 σt−i · ci . + i=1 This test can be performed in a π-OBDD as follows.

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