By Thomas Baigneres, Pascal Junod, Yi Lu, Jean Monnerat, Serge Vaudenay

TO CRYPTOGRAPHY workout booklet Thomas Baignkres EPFL, Switzerland Pascal Junod EPFL, Switzerland Yi Lu EPFL, Switzerland Jean Monnerat EPFL, Switzerland Serge Vaudenay EPFL, Switzerland Springer - Thomas Baignbres Pascal Junod EPFL - I&C - LASEC Lausanne, Switzerland Lausanne, Switzerland Yi Lu Jean Monnerat EPFL - I&C - LASEC EPFL-I&C-LASEC Lausanne, Switzerland Lausanne, Switzerland Serge Vaudenay Lausanne, Switzerland Library of Congress Cataloging-in-Publication information A C.I.P. Catalogue checklist for this booklet is obtainable from the Library of Congress. A CLASSICAL advent TO CRYPTOGRAPHY workout ebook by means of Thomas Baignkres, Palcal Junod, Yi Lu, Jean Monnerat and Serge Vaudenay ISBN- 10: 0-387-27934-2 e-ISBN-10: 0-387-28835-X ISBN- thirteen: 978-0-387-27934-3 e-ISBN- thirteen: 978-0-387-28835-2 published on acid-free paper. O 2006 Springer Science+Business Media, Inc. All rights reserved. This paintings will not be translated or copied in complete or partially with out the written permission of the writer (Springer Science+Business Media, Inc., 233 Spring road, ny, manhattan 10013, USA), with the exception of short excerpts in reference to stories or scholarly research. Use in reference to any type of details garage and retrieval, digital model, software program, or by means of comparable or diverse method now be aware of or hereafter built is forbidden. The use during this booklet of exchange names, emblems, carrier marks and related phrases, whether the are usually not pointed out as such, isn't to be taken as an expression of opinion to whether or now not they're topic to proprietary rights. published within the u . s .

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Have values P, C E { O , I ) ~ We where the last sum simply is the number of permutations mapping P on C , which is the number of permutations of a set of cardinality 264 - 1. Finally, Pr[C*(P)= C] = 39 Conventional Cryptography 3 We assume that PrK[3DESK(P) = C] = Prc* [C*(P) = C] = 2-". Multiplying this probability by the number of tried keys, we obtain the number of keys that are displayed: All the displayed keys (except one) are wrong keys! 4 We consider Algorithm 6. The algorithm clearly displays k as we do A l g o r i t h m 6 Exhaustive key search algorithm on 3DES, using t plaintextlciphertext pairs I n p u t : t plaintext/ciphertext pairs (Pi,Ci), for i = 1,.

Therefore, from a known plaintext attack with only one known message, we can recover the key stream and decrypt any new ciphertext (of the same length or shorter). 2 The CFB mode is stronger against this issue, except for the first block. The first encrypted block is equal to the first plaintext block XORed with a value generated from IV and from the key only. The next values in the sequence depend on the plaintext. Similarly, note that if two plaintexts are equal on their first n blocks, the knowledge of one of the plaintexts allows to recover the ( n + 1)th block of the other plaintext.

Once K3 is found, how can K1 and K2 be recovered? What is the overall complexity of the attack? " - D Exercise 13 Solution on page 47 *A Variant of A511 I In stream ciphers, the prevailing encryption is a bitwise XOR operation between the m-bit plaintext and the m-bit keystream which is the output of a so-called keystream generator fed by the L-bit secret key, where m is much larger than !. An ideal assumption for good stream ciphers is that any &bit window of the m-bit keystream is eventually modified when the Gbit key is modified.