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

ISBN-10: 0387279342

ISBN-13: 9780387279343

ISBN-10: 038728835X

ISBN-13: 9780387288352

TO CRYPTOGRAPHY workout e-book 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 facts A C.I.P. Catalogue checklist for this publication is on the market from the Library of Congress. A CLASSICAL creation TO CRYPTOGRAPHY workout booklet via 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 is probably not translated or copied in complete or partially with no the written permission of the writer (Springer Science+Business Media, Inc., 233 Spring highway, ny, new york 10013, USA), apart from short excerpts in reference to studies or scholarly research. Use in reference to any kind of info garage and retrieval, digital model, software program, or via related or distinct technique now understand or hereafter built is forbidden. The use during this booklet of exchange names, emblems, provider marks and related phrases, whether the will not be pointed out as such, isn't to be taken as an expression of opinion as to if or now not they're topic to proprietary rights. published within the united states.

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**Additional resources for A Classical Introduction to Cryptography: Exercise Book**

**Example text**

2e = 2e+1. In that case, the storage complexity is 2e. 2 As in Solution 6, we can prove that Pr[C*(x) = y] = 2Tn. Assuming that EK roughly behaves like a random permutation when K is randomly chosen among all possible wrong keys, we estimate Pr[EK(P) = C] x 2-n. For a cascade cipher, a total of wrong keys are displayed. 3 Algorithm 7 exploits the t pairs at disposal. , t O u t p u t : key candidate(s) for k Processing: 1: for each possible key K d o 2: i f C i = E K ( P ! , t t h e n 3: display K 4: e n d if 5: e n d for roughly behaves like a random permutation when K is chosen among all possible wrong keys, we obtain t &x ) 2-tn Pr[EK(Pi) = Ci for all i = 1 , .

More details about cascade ciphers and their security can be found in [29]. 11. 11. 4) holds then 3: display K3 4: end if 5: end for it does not yield any wrong key (with high probability). Once ks is found, the adversary can peel the third layer off, and do a meet-in-themiddle attack on the last two layers. Note that we typically need both plaintext blocks A and B in order to eliminate wrong key candidates during the meet-in-the-middle. The complexity of this part of the attack is ~ ( 2 ' )in time and ~ ( 2 ' )in storage.

So, as long as the leftmost bits of two non-zero LFSRs are equal and the clocking taps are both one, the variant A511 generates the all-zero keystream. 5 We consider the following four different cases: a Case where the three LFSRs all stop forever: we have 264-2-1 = 261 different initial states that satisfy two linear relations: one clocking constraint and one output constraint. a For R1 = 0: In this case, if R2[10] = R3[10] = 1 and R2[21] = R3[22] we know that we obtain the all-zero keystream.

### A Classical Introduction to Cryptography: Exercise Book by Thomas Baigneres, Pascal Junod, Yi Lu, Jean Monnerat, Serge Vaudenay

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