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

ISBN-10: 038728835X

ISBN-13: 9780387288352

This better half workout and answer booklet to A Classical advent to Cryptography: functions for Communications Security incorporates a rigorously revised model of educating fabric. It used to be utilized by the authors or given as examinations to undergraduate and graduate-level scholars of the Cryptography and protection Lecture at EPFL from 2000 to mid-2005.

A Classical creation to Cryptography workout publication for A Classical creation to Cryptography: functions for Communications protection covers a majority of the themes that make up today's cryptology, comparable to symmetric or public-key cryptography, cryptographic protocols, layout, cryptanalysis, and implementation of cryptosystems. routines don't require a wide heritage in arithmetic, because the most crucial notions are brought and mentioned in lots of of the exercises.

The authors count on the readers to be pleased with simple proof of discrete chance idea, discrete arithmetic, calculus, algebra, in addition to computing device technological know-how. Following the version of A Classical advent to Cryptography: functions for Communications protection, workouts concerning the extra complex components of the textbook are marked with a celebrity.

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**Extra resources for A Classical Introduction to Cryptography Exercise Book**

**Sample text**

5 We consider the variant of A5/1. We first note that in this case, either exactly one LFSR is clocked (when its clocking tap is different from the two others) or no LFSR is clocked at all (when all three clocking taps are equal). Using the notations of Question 1, we have - P r [ T 2 # T 3 ] + P r [ T l = T 2 = T 3 ]1= - +1- = -3. pfixed - 2 4 4 Consequently, by symmetry, ppd - 3 and 4 The probability that all three LFSRs stay still during next clock is Pr[Tl = T2 = T3]= and the probability that exactly one LFSR is shifted is pfhifted + p;hifted + pihifted = 3 Ti.

Following the same reasoning, we deduce the following lower bound on the number of possible initializations states in this case: R2 # 0 and R1 = R3 = 0: We similarly obtain a lower bound eaual to w For For R3 # 0 and R1 = R2 = 0: We similarly obtain a lower bound 50 EXERCISE BOOK Summing these values, we conclude that there are at least 222 such initialization states. 4 When the initial clocking taps of the three LFSRs are all equal, none of the three LFSRs will ever be shifted. Hence, provided that the XOR of the three LFSRs output bits is zero at some time, we will obtain the all-zero keystream.

In order to thwart this attack, we thus need to enlarge the block size. 9. 5 With XL = X R , we obtain yr, = y~ = 3DESKI,K2 (xL). So a circuit which computes this new scheme can be used to compute 3DES. Similarly, with K l = K2, we obtain compatibility with DES. 6 The previous question leads to the intuition that this new scheme is at least as strong as DES and 3DES. 9. A 128 bit extention of DES as the key size is increased and at least as secure as 3DES as the key size is the same. The advantage of this scheme is that it is protected against the collision attack in CBC mode.

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

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