Advances in Cryptology - EUROCRYPT 2014: 33rd Annual by Phong Q. Nguyen, Elisabeth Oswald

By Phong Q. Nguyen, Elisabeth Oswald

This booklet constitutes the lawsuits of the thirty third Annual overseas convention at the concept and functions of Cryptographic ideas, EUROCRYPT 2014, held in Copenhagen, Denmark, in might 2014. The 38 complete papers integrated during this quantity have been conscientiously reviewed and chosen from 197 submissions. They take care of public key cryptanalysis, identity-based encryption, key derivation and quantum computing, secret-key research and implementations, obfuscation and multi linear maps, authenticated encryption, symmetric encryption, multi-party encryption, side-channel assaults, signatures and public-key encryption, practical encryption, foundations and multi-party computation.

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Additional resources for Advances in Cryptology - EUROCRYPT 2014: 33rd Annual International Conference on the Theory and Applications of Cryptographic Techniques, Copenhagen, Denmark, May 11-15, 2014, Proceedings

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Xq+1 0 n−1 spectively N(x) and Tr(x) with by definition N(x) = def Tr(x) = xq0 + x0 , . . , xqn−1 and + xn−1 . Shortening and Puncturing Codes. For a given code D ⊂ Fnq and a subset I ⊂ {0, . . , n − 1} the punctured code DI and shortened code D I are defined as: def DI = (ci )i∈I / | c ∈D ; D I = (ci )i∈I / | ∃c = (ci )i ∈ D such that ∀i ∈ I, ci = 0 . def Instead of writing D{j} and D {j} when I = {j} we rather use the notation Dj and D j . The following classical results will be used repeatedly.

To each vector x = (x0 , . . , xn−1 ) ∈ Fnq , we associate its locator polynomial denoted as πa def n−1 and defined as πx (z) = i=0 (z − xi ). Its first derivative is denoted as πx and one shows easily that its evaluation at the entries of x yields the vector πx (x) = . j=i (xi − xj ) 0 i

Table 3 summarizes this probability for other parameters proposed in [6] for m = 2 and t > 3. 32 A. Couvreur, A. –P. Tillich Table 3. 6 10−21 Step 3. First, notice that the vectors x0 and x1 punctured at the first position def are both equal to the vector x01 = (x2 , . . , xn−1 ) ∈ Fn−2 q2 . From the previous step, one can obtain the two following sets of vectors: def L0 = xq+1 ∪ 01 def (1 − a)q+1 xq+1 (x01 − a)−(q+1) 01 a ∈ Fq2 \ Lx aq+1 (x01 − 1)q+1 (x01 − a)−(q+1) a ∈ Fq2 \ Lx . (15) They are computed by puncturing the first entry of each solution vector and taking the inverse for the star product.

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