Advances in Cryptology — CRYPTO 2000: 20th Annual by Arjen K. Lenstra, Eric R. Verheul (auth.), Mihir Bellare

By Arjen K. Lenstra, Eric R. Verheul (auth.), Mihir Bellare (eds.)

This booklet constitutes the refereed court cases of the twentieth Annual foreign Cryptology convention, CRYPTO 2000, held in Santa Barbara, CA, united states in August 2000. The 32 revised complete papers offered including one invited contribution have been rigorously reviewed and chosen from a hundred and twenty submissions. The papers are prepared in topical sections on XTR and NTRU, privateness for databases, safe disbursed computation, algebraic cryptosystems, message authentication, electronic signatures, cryptanalysis, traitor tracing and broadcast encryption, symmetric encryption, to devote or to not devote, protocols, and movement ciphers and Boolean capabilities.

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Example text

Of course, the parameters N , p, q are public too. The private key is the polynomial f , together with Fp . 3 Encryption and Decryption Procedure Encryption. The encryption works as follows. First, we select a message m from the set of plaintexts Lm . Next we choose randomly a polynomial φ ∈ Lφ and use the public key to compute: e ≡ pφ h+m (mod q). e is our encrypted message. Decryption. We have received an encrypted message e and we want to decrypt it using our private key f . 2. In order to decrypt e, we compute : a≡f e (mod q), where we choose the coefficients of a in the interval from −q/2 to q/2.

A Chosen-Ciphertext Attack against NTRU 33 3. Select a value of a polynomial P. P = xi1 + · · · + xin + h (xj1 + · · · + xjm ) (mod q) 4. Produce m corresponding to cP : m = [m + H(m, [cP ]p )X N −k + G([cP ]p )]p with m = [−G([cP ]p )]p mod X N −K . 5. Ask the decryption of cP + m . The answer should be m. If not, go back to 3. 6. For all i such that [[cxi + cP ]q ]p = [[cP ]q ]p , ask decryption of cxi + cP + m . If the answer is a decryption error, the (N-i)th coefficient of f is a 1, else we know it is not a 1.

One computes e ≡ pφ h]p )X N −K + G([pφ h + [m + H(m, [pφ h]p )]p (mod q). (1) To decrypt the message, the receiver uses his private key f and the standard NTRU decryption method to recover a polynomial n = [Fp [f e]q ]p ∈ Pp (N ). Next he computes b≡e−n (mod p) and c ≡ n − G(b) (mod p). and he writes c in the form c = c + c X N −K with deg(c ) < N − K and deg(c ) < K. Finally, he compares the quantities c and H(c , b). If they are the same, he accepts c as a valid decryption. Otherwise he rejects the message as invalid.

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