By Thomas Baigneres, Pascal Junod, Yi Lu, Jean Monnerat, Serge Vaudenay
This spouse workout and answer ebook to A Classical advent to Cryptography: purposes for Communications safeguard includes a rigorously revised model of educating fabric utilized by the authors and given as examinations to advanced-level scholars of the Cryptography and defense Lecture at EPFL from 2000 to mid-2005. A Classical creation to Cryptography workout Book covers a majority of the topics that make up modern day cryptology, together with symmetric or public-key cryptography, cryptographic protocols, layout, cryptanalysis, and implementation of cryptosystems. workouts don't require an intensive heritage in arithmetic, because the most crucial notions are brought and mentioned in lots of of the routines. The authors anticipate the readers to be ok with uncomplicated evidence of discrete chance concept, discrete arithmetic, calculus, algebra, and machine technological know-how. Following the version of A Classical advent to Cryptography: purposes for Communications safety, routines regarding the extra complex components of the textbook are marked with a celebrity.
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Additional resources for A Classical Introduction to Cryptography Exercise Book
The memory requirements is thus 2 . 256 . 27 . T3= 261 bytes. + Solution 6 *Exhaustive Search on 3DES 1 The algorithm successively tries each possible key. It does not stop until the last possible key is tried. 1), under the keys K1 and K2 Output: K1 and K2 Processing: 1: for all k E (0, 1)56do 2: 3: P~+DEs;~(o) store (Pk,k) in a table Tl (sorted according to Pk) DESK^, KZ ( P k ) 4: Ck 5: B~ + D E S ; ~ ( C ~ ) 6: 7: 8: 9: lo: 11: store (Bk,k) in a table T2 (sorted according to Bk) end for sort the table TI according to the Pk'svalues sort the table T2 according to the Bk7svalues Store the keys kl,k2 E (0, 1)56such that Bkl = Pk2in another table T.
The complexity corresponds to the expected number of blocks after which we can expect a collision (see Exercise 1, Chapter 3). , @ = 232. We note that the complexity of this attack is not increased by using 3DES instead of DES as the block size remains the same. 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.
3 In CBC mode of operation, the ith ciphertext block yi is where xi is the ith plaintext block. If it happens that yi = yj (which is a collision), we deduce that y+l$ xi = yj-1 $ x j which leads to yi-1 a3 yj-1 = xi a3 xj. Hence, we can thus deduce some plaintext information from the value yi-1 $ yj-1. The complexity corresponds to the expected number of blocks after which we can expect a collision (see Exercise 1, Chapter 3). , @ = 232. We note that the complexity of this attack is not increased by using 3DES instead of DES as the block size remains the same.