By Duong Hieu Phan, David Pointcheval (auth.), Chi-Sung Laih (eds.)

ISBN-10: 3540205926

ISBN-13: 9783540205920

This publication constitutes the refereed complaints of the ninth foreign convention at the conception and alertness of Cryptology and knowledge protection, ASIACRYPT 2003, held in Taipei, Taiwan in November/December 2003.

The 32 revised complete papers provided including one invited paper have been rigorously reviewed and chosen from 188 submissions. The papers are geared up in topical sections on public key cryptography, quantity conception, effective implementations, key administration and protocols, hash services, staff signatures, block cyphers, broadcast and multicast, foundations and complexity conception, and electronic signatures.

**Read or Download Advances in Cryptology - ASIACRYPT 2003: 9th International Conference on the Theory and Application of Cryptology and Information Security, Taipei, Taiwan, November 30 – December 4, 2003. Proceedings PDF**

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**Additional resources for Advances in Cryptology - ASIACRYPT 2003: 9th International Conference on the Theory and Application of Cryptology and Information Security, Taipei, Taiwan, November 30 – December 4, 2003. Proceedings**

**Sample text**

On the other hand, from the construction of B, it follows that z + zmN = Z + zM N mod N 2 . Thus, we can eﬃciently compute Z = z(1+(m−M )N ) mod N 2 . With the following theorem we make explicit the relation existing between the lift Diﬃe-Hellman problem and the partial Discrete Logarithm problem. Theorem 10. If the Partial Discrete Logarithm problem is hard then so is the Lift Diﬃe-Hellman problem. Proof. The proof goes by a standard reduction argument. Assume we are given an oracle O for the lift Diﬃe-Hellman problem that on input a triplet of the form (X, Y, Z) = (g x mod N 2 , g y mod N 2 , g xy mod N ) returns the value g xy mod N 2 with some non negligible probability .

Let c = ab + r mod ord(G), we can note that r is random and uniformly distributed in [1, ord(G)] and can be written as r1 + r2 λ(N )/2, with r1 , r2 ∈ ZN . The information received by the adversary (together with the public key) is of the form g b mod N 2 , g ab+r (1 + md N ) mod N 2 Let us concentrate on the second value (for the sake of simplicity let us assume that g λ(N )/2 = (1 + N ) mod N 2 ). g ab+r (1 + md N ) = g ab g r1 g r2 λ(N )/2 (1 + md N ) mod N 2 = g ab+r1 (1 + N )r2 (1 + md N ) mod N 2 = g ab+r1 (1 + (r2 + md )N ) mod N 2 .

Thus we have to compute: δ= Pr r1 ∈R Zλ/2 c∈G r ∈ Z 2 R N Pr = c∈G = c∈G r1 ∈R Zλ/2 g r1 +r2 λ/2 = c − [r1 = c1 ] Pr r1 ∈R Z(N +1)/4 r2 ∈R ZN Pr [r2 = c2 ] − r2 ∈R ZN g r1 (1 + r2 N ) = c Pr r1 ∈R Z(N +1)/4 r2 ∈R ZN g r1 (1 + r2 N ) = c 1 2 × − Pr g r1 (1 + r2 N ) = c λ N r1 ∈R Z(N +1)/4 r2 ∈R ZN Denoting g λ/2 = 1 + αN mod N 2 and β = α−1 mod N , we have g r1 (1 + r2 N ) = g r1 +r2 βλ/2 mod N 2 . Then we observe that for λ/2 ≤ r1 < N4+1 , we have the following “collision”: g r1 +r2 βλ/2 = g (r1 −λ/2)+(r2 β+1)λ/2 (mod N )2 Hence, two cases appear when summing up (of course, the probabilities that r2 or r2 β or r2 β + 1 equals a given c2 are all 1/N ): 4 1 N +1 λ 2 · N +1 × N if 0 ≤ c < 4 − 2 r1 +r2 βλ/2 c1 +c2 λ/2 Pr g = =g 1 · N4+1 × N1 if N4+1 − λ2 ≤ c < λ2 Consequently, we gets (recall that δ= N +1 4 8 p+q 2 − + 4 λN N (N + 1) ≤0 This is easily seen negligible.

### Advances in Cryptology - ASIACRYPT 2003: 9th International Conference on the Theory and Application of Cryptology and Information Security, Taipei, Taiwan, November 30 – December 4, 2003. Proceedings by Duong Hieu Phan, David Pointcheval (auth.), Chi-Sung Laih (eds.)

by Kevin

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