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applied cryptography - protocols, algorithms, and source code in c

With a normal hash function she can try common

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Unformatted text preview: g, RA, encrypts it with K, and sends Bob EK(RA) (4) Bob decrypts the message to obtain RA. He generates another random string, RB, encrypts both strings with K, and sends Alice the result. EK(RA, RB) (5) Alice decrypts the message to obtain RA and RB. Assuming the RA she received from Bob is the same as the one she sent to Bob in step (3), she encrypts RB with K and sends it to Bob. EK(RB) (6) Bob decrypts the message to obtain RB. Assuming the RB he received from Alice is the same one he sent to Alice in step (4), the protocol is complete. Both parties now communicate using K as the session key. Previous Table of Contents Next Products | Contact Us | About Us | Privacy | Ad Info | Home Use of this site is subject to certain Terms & Conditions, Copyright © 1996-2000 EarthWeb Inc. All rights reserved. Reproduction whole or in part in any form or medium without express written permission of EarthWeb is prohibited. Read EarthWeb's privacy statement. To access the contents, click the chapter and section titles. Applied Cryptography, Second Edition: Protocols, Algorthms, and Source Code in C (cloth) Go! Keyword Brief Full Advanced Search Search Tips (Publisher: John Wiley & Sons, Inc.) Author(s): Bruce Schneier ISBN: 0471128457 Publication Date: 01/01/96 Search this book: Go! Previous Table of Contents Next ----------- At step (3), both Alice and Bob know K´ and K. K is the session key and can be used to encrypt all other messages between Alice and Bob. Eve, sitting between Alice and Bob, only knows EP(K´), EP(EK´(K)), and some messages encrypted with K. In other protocols, Eve could make guesses at P (people choose bad passwords all the time, and if Eve is clever she can make some good guesses) and then test her guesses. In this protocol, Eve cannot test her guess without cracking the public-key algorithm as well. And if both K´ and K are chosen randomly, this can be an insurmountable problem. The challenge-response portion of the protocol, steps (3) through (6), provides validation. Steps (3) through (5) prove to Alice that Bob knows K; steps (4) through (6)...
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