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By Michael Lehmann, Willi Meier (auth.), Josef Pieprzyk, Ahmad-Reza Sadeghi, Mark Manulis (eds.)

This e-book constitutes the refereed court cases of the eleventh foreign convention on Cryptology and community protection, CANS 2012, held in Darmstadt, Germany, in December 2012. The 22 revised complete papers, provided have been conscientiously reviewed and chosen from ninety nine submissions. The papers are geared up in topical sections on cryptanalysis; community safeguard; cryptographic protocols; encryption; and s-box theory.

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Extra resources for Cryptology and Network Security: 11th International Conference, CANS 2012, Darmstadt, Germany, December 12-14, 2012. Proceedings

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2 Lm (t1 , t2 , · · · , tρ ) = Lm (t1 , t2 , · · · , tρ ) Cryptanalysis of a Lattice-Knapsack Mixed Public Key Cryptosystem 39 It can be successfully solved by linearization technique provided m ≥ ρ2 + 2ρ, or equivalently, provided l ≥ log2 2m+1 . 2. As an example, for m = 500, this number is 5, we need only 5 encryptions to recover the secret message. 5 Discussion and Open Problem The broadcast and multiple transmission attacks presented in the previous two sections do not use the inherited information about the structure of the public key H except that it is invertible.

Karakoç, and Ö. Boztaş We overview the biclique technique in Section 3. In Section 4 and 5 we present the attacks on TWINE-80 and TWINE-128, respectively. We conclude the paper in Section 6. 1 Notation and a Short Description of TWINE Notation Throughout the paper, we use the following notations: A : a bit string A(i) : i-th nibble of A. The left most nibble is A(0). , k) : concatenation of i, j, ... , k-th nibbles of A. , j-th nibbles of A where i ≤ j. A[i] : i-th bit of A. The left most bit of A is A[0].

This scheme is like the problem of Learning With Errors (LWE) [25]. It can be treated as the problem to solve a specific system of 2m quadratic equations over Fp in m variables of the form: ⎧ 2 x1 − x1 = 0 ⎪ ⎪ ⎪ ⎪ .. ⎪ ⎪ . ⎪ ⎪ ⎨ 2 xm − xm = 0 (c1 − h11 x1 − · · · − h1m xm )2 − (c1 − h11 x1 − · · · − h1m xm ) = 0 ⎪ ⎪ ⎪ ⎪ .. ⎪ ⎪ ⎪ . ⎪ ⎩ (cm − hm1 x1 − · · · − hmm xm )2 − (cm − hm1 x1 − · · · − hmm xm ) = 0 (3) 40 J. Xu et al. We do not know how to efficiently solve (if possibly) this seemingly very specific nonlinear system for large m.

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