By Valérie Nachef, Emmanuel Volte, Jacques Patarin (auth.), Michel Abdalla, Cristina Nita-Rotaru, Ricardo Dahab (eds.)

This publication constitutes the refereed court cases of the twelfth overseas convention on Cryptology and community safety, CANS 2013, held in Paraty, Brazil, in November 2013.

The 18 revised complete papers offered including 4 invited talks have been conscientiously reviewed and chosen from fifty seven submissions. The papers are prepared in topical sections on cryptanalysis, zero-knowledge protocols, allotted protocols, community safeguard and purposes, complicated cryptographic primitives, and verifiable computation.

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Extra info for Cryptology and Network Security: 12th International Conference, CANS 2013, Paraty, Brazil, November 20-22. 2013. Proceedings

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5 and OAEP respectively. 5 is included only for compatibility with existing applications. Padding Oracle Attack. 5. Padding Oracle Attack is a type of chosen ciphertext attack, which takes advantage of whether cryptographic operation is successfully executed. Usually, we assume the attacker can M. Abdalla, C. Nita-Rotaru, and R. ): CANS 2013, LNCS 8257, pp. 39–56, 2013. c Springer International Publishing Switzerland 2013 40 S. Gao, H. Chen, and L. Fan trick an honest user to decrypt the ciphertext he chose.

GM (x) = i=1 Then the function hH (M ) = gM (H) is called evaluation hash. The hash function outputs are masked by block cipher encryptions to produce the authentication tags, such as EK (N ) ⊕ hH (M ) and EK (hH (M )). Poly1305-AES [2], and the MAC schemes in GCM and SGCM [19] are all within this framework. We summarize the main observation by Procter and Cid in [17] as follows. For the convenience of the readers, we include a short proof of their result. Result 1 ([17]). With the same notation as above, if there exists a polynomial f (x) ∈ F[x] without a constant term, such that f (H) = 0, then forgeries of MAC schemes based on the evaluation hash hH (x) can be made.

Finally E(δij ) = p1 p2 + 21n p1 (1 − p2 ), and E(Ntype1 ) = m(m−1) p1 p2 + 21n p1 (1 − p2 ) . We have p1 2 1 in O( 22n ). Indeed, according to Lemma 24 k X (j)] ≤ k−1 2n . Using the same arguments, 1 k−1 ≤ p ≤ n 1 2 2n . 1 2n . In p2 the dominant term is of [8], we have we obtain 1 22n 1 2n ≤ P r[X k (i) = ≤ p2 ≤ (k−1)2 22n and We want to show that the variance behaves like the mean value. For this, we will use the covariance formula: V (Ntype1 ) = [E(δij δqv ) − E(δij )E(δqv )] V (δij ) + i

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