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Unformatted text preview: e sends the decrypted message to Alice. DB(EB(M1)) = M1 or DB(EB(M2)) = M2 (6) Alice reads the result of the coin flip and verifies that the random string is correct. (7) Both Alice and Bob reveal their key pairs so that both can verify that the other did not cheat. 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)
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:
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----------- This protocol is self-enforcing. Either party can immediately detect cheating by the other, and no trusted third party is required to participate in either the actual protocol or any adjudication after the protocol has been completed. To see how this works, let’s try to cheat. If Alice wanted to cheat and force heads, she has three potential ways of affecting the outcome. First, she could encrypt two “heads” messages in step (2). Bob would discover this when Alice revealed her keys at step (7). Second, she could use some other key to decrypt the message in step (4). This would result in gibberish, which Bob would discover in step (5). Third, she could lie about the validity of the message in step (6). Bob would also discover this in step (7), when Alice could not prove that the message was not valid. Of course, Alice could refuse to participate in the protocol at any step, at which point Alice’s attempted deception would be obvious to Bob. If Bob wanted to cheat and force “tails, ” his options are just as poor. He could incorrec...
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