Download PDF by Claude Crépeau, Joe Kilian (auth.), Shafi Goldwasser (eds.): Advances in Cryptology — CRYPTO’ 88: Proceedings

By Claude Crépeau, Joe Kilian (auth.), Shafi Goldwasser (eds.)

ISBN-10: 0387347992

ISBN-13: 9780387347998

ISBN-10: 0387971963

ISBN-13: 9780387971964

The papers during this quantity have been awarded on the CRYPTO '88 convention on concept and functions of cryptography, held in Santa Barbara, California, August 21-25, 1988. The papers have been selected for his or her perceived originality and infrequently signify initial studies on carrying on with reserach. the most sections take care of the subsequent issues: Zero-Knowledge, quantity concept, Pseudorandomness, Signatures, Complexity, Protocols, protection, Cryptoanalysis. As such, they'll provide the devoted reader a different perception into the very most recent advancements within the box.

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Let IP be the class of languages t h a t admit interactive proofs. Clearly N P 5 I P , for an NP-interaction is a special type of IF‘-interaction, in which the prover (Alice) sends the one and only message, and t h e verifier (Bob) never errs. However, IP may be a much larger class of languages. For example, there is a n interactive proof known for graph nonisomorphism, even though there are not known t o be succinct certificates for establishing that a pair of graphs are not isomorphic. S. ): Advances in Cryptology - CRYPT0 ’88, LNCS 403, pp.

To d o this, we could have P encrypt each message that it sends to V. That is, P uses a secure encryption function, E . On t h e i t h round, when P "would have" sent to V the string yi, P instead sends to the string E ( y i , di), a random encryption of y;. (We assume that E ( s ,s) = E ( y ,t ) implies 3 = y, and that from E(z,s) and s one can efficiently compute I. ) There are two immediate difficulties. First, how can V be expected to compute his responses to P , since he doesn't understand what P has sent?

If the tester flips the coin 1/S12times, even a very weak form of the law of large numbers would tell us that Eve can guess the bias of the coin at least 99% of the time. 1. m Notice the order of the quantifiers in the above result. We picked the protocol between Alice and Bob, then we picked the oracle (since the protocol is bound by definition to work with a random oracle). Then, we showed Eve can break the protocol. We prove a stronger result which reverses the quantifiers. First, we pick a random oracle; then a protocol for Alice and Bob (this time the protocol need not work properly on other oracles).

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Advances in Cryptology — CRYPTO’ 88: Proceedings by Claude Crépeau, Joe Kilian (auth.), Shafi Goldwasser (eds.)

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