Code and ciphers: Julius Caesar, the Enigma and the internet

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214 appendix so but so Hence 1�6�1�(17�2�6)�3�6�17 6�91�5�17 1�3�(91�5�17)�17�3�91�16�17. m�3 and n�16. (Check: 3�91�273 and 16�17�272.) An alternative method The values of m and n can also be found by using continued fractions [6.7]. Although it looks different the method is essentially the same as that of the Euclidean Algorithm. To illustrate the method we again find the values of m and n such that 91m�17n�1. 91 6 �5� , 17 17 17 5 �2� , 6 6 6 1 �1� . 5 5 The partial quotients of the continued fraction are therefore (5, 2, 1, 5) and its convergents are 5 11 16 91 , , , and . 1 2 3 17 The numbers m and n are the numerator and denominator of the penultimate convergent: 16 and 3, as we found before. M26 Efficiency of finding powers by repeated squaring Given a number, X, that we wish to raise to the power n we could compute X n by multiplying X by itself (n�1) times. If n is small this is reasonable but if n is large it is very inefficient. Let k be such that 2 k �n�2 (k�1) ;
then k�[ log 2 n], where [z] denotes, as usual in mathematics, the integer part of z. If we compute X 2 , X 4 , X 8 , ....by repeated squaring we will need to carry out k squarings, that is k multiplications, to reach the power 2 k . The binary representation of n contains at most (k�1) 1s and so X n can be computed by multiplying together at most (k�1) of the numbers X, X 2 , X 4 , ... and this means that at most k further multiplications are required, giving a total of 2k multiplications in all. Since k� (log 2 n�1) we see that computing X n by repeated squaring involves less than 2(log 2 n�1) multiplications whereas the brute force method requires (n�1). If n is small the difference is not too great. When n�7, for example, the brute force method requires 6 multiplications and the repeated squaring method requires 4. As n increases however the difference rapidly becomes very significant. When n�127, for example, the brute force method requires 126 multiplications whereas repeated squaring needs only 12. For the really large exponents which are likely to occur in RSA encipherment/decipherment astronomical numbers of multiplications are replaced by a few hundred. M27 Expected number of false hits in the ‘meet-in-themiddle’ attack on the DES When we encipher a text using 2 56 different keys we will obtain 2 56 different encipherments. Since there are 2 64 different 64-bit binary vectors there is only one vector in 256 (�2 8 ) that will appear in the list of encipherments. The same is true when we decipher a text using 2 56 different keys. If we now compare the two lists the chance that a vector in the encipherment list also occurs in the decipherment list is one in 256. There are 2 56 vectors in the encipherment list and one in 256 of them would be expected to appear in the decipherment list. We therefore expect 2 48 agreements in all. All but one of these will be false, and one or more further tests must be applied to find the true solution. M28 Elliptic Curve Cryptography Despite the name the curves in question are not ellipses but are of the type Y 2 �X 3 �aX�b Mathematical aspects 215 where a and b are integers. We are interested in pairs (X, Y) which are also integers; all arithmetic being carried out (mod p) for some (very large) prime p. Curves of this type can be parametrised by Weierstrass elliptic functions, hence the name.
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R. F. Churchhouse Codes and ciphers
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Contents Preface ix 1 Introduction
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Cryptanalysisofalinearrecurrence 10
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Preface Virtually anyone who can re
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1 Introduction Some aspects of secu
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Not a very sophisticated method, pa
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change. As an example, anticipating
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Another form of encryption is the u
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and so is valid. On the other hand
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When the modulus is 10 only the num
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2 From Julius Caesar to simple subs
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Table 2.3 Shift Message 12 OUI 12 Y
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worry about but, on the other hand,
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then it is probably THE and the unk
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From Julius Caesar to simple substi
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Table 2.6 From Julius Caesar to sim
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and the key which provides the enci
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Further examination reveals that th
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ALTHOUGH I AM AN OLD MAN NIGHT IS G
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Polyalphabetic systems 35 messages
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Before leaving Vigenère try the fo
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Polyalphabetic systems 39 blocks of
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eginning at the top, but it is then
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first row above, for example, we fi
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Table 4.4 Key 3 1 5 2 4 1 2 3 4 5 6
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giving B G L D I N A F K E J O C H
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then the cipher text is HORUX SXSEO
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The plaintext digraphs are now sepa
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ox increases. By enumerating the po
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Example 5.1 HAPPY BIRTHDAY encipher
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Encryption The ‘plaintext’ is A
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plaintext digraphs according to som
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Cryptanalytic aspects of Playfair P
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OURXSITUATI ONXISXDESPE RATEXSENDXS
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the alphabet are represented by up
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From a cryptographic point of view
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3 7 0 7 7 4 1 5 6 1 7 8 5 3 8 1 9 0
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If we generate the same sequence (m
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Stencil ciphers The example above i
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Book ciphers A spy must avoid arous
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Table 7.2 Encipher table for a book
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Letter frequencies in book ciphers
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If then we try subtracting THE from
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where row F (the row of the cipher
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Ciphers for spies 85 that were nece
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mistake can occur if the sender lea
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inks and ciphers were provided by h
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Example 7.6 Encipher the message AG
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Ciphers for spies 93 simply a ‘ra
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going into the mathematical criteri
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numbers, 0 to 31 inclusive, into bi
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Linear recurrences The sequences lo
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Having converted the characters of
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What about sequences of higher orde
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combining the keys of two or more l
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Example 8.3 (1) Use the mid-square
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Problem 8.3 Starting with U 0 �1
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The Enigma cipher machine 111 syste
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The Enigma cipher machine 113 Plate
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The Enigma cipher machine 115 Plate
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Figure 9.2. wheel. For example, if
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wheel and then through the three wh
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first day of usage and so the asses
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The Enigma cipher machine 123 The m
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show how, in a typical encipherment
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interested reader can find it in [9
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great deal of data and many pages o
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It should be realised, of course, t
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10 The Hagelin cipher machine Histo
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ut there was no cryptographic advan
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of each cipher period, to which sid
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value can be expected to occur two
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Since some cages are obviously very
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2, there are 26! possible simple su
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The Hagelin cipher machine 145 Exam
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values are repeating at the appropr
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Overlapping will obviously affect t
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Table 10.6 The Hagelin cipher machi
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11 Beyond the Enigma The SZ42: a pr
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Description of the SZ42 machine The
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P1 41 43 Z1 (4) the five bits from
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which is approximately 1.6�10 19
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12 Public key cryptography Historic
Page 175 and 176: part of the story of what has been
Page 177 and 178: Public key cryptography 165 graphic
Page 179 and 180: (4) Now (k x ) y �(k y ) x �m x
Page 181 and 182: Public key cryptography 169 Despite
Page 183 and 184: If the cryptographer were prepared
Page 185 and 186: number of tests increased by a fact
Page 187 and 188: In this case the remainder is 4, no
Page 189 and 190: key, d, we use the Euclidean Algori
Page 191 and 192: the decipher key, d, is used instea
Page 193 and 194: and and, since Table 13.1 n N�2 n
Page 195 and 196: The Data Encryption Standard (DES)
Page 197 and 198: the message were known to relate to
Page 199 and 200: whereas in block cipher systems, su
Page 201 and 202: (1) X precedes M with information w
Page 203 and 204: ways, since choosing to pair, say,
Page 205 and 206: For example; if someone claims that
Page 207 and 208: depth, mentioned in Chapter 3. If w
Page 209 and 210: M9 Combining two biased streams of
Page 211 and 212: and so 2�4�6�12 �(4095)�4
Page 213 and 214: Verification Let the recurrence be
Page 215 and 216: the lettersatsetting2ofthewheel,and
Page 217 and 218: M16 Probability of a ‘depth’ in
Page 219 and 220: (2) In how many ways can N be repre
Page 221 and 222: Chapter 13 M21 (Rate of increase of
Page 223 and 224: Multiply each of these by M: M, 2M,
Page 225: If x 3 �0 divide x 2 by x 3 to gi
Page 229 and 230: Mathematical aspects 217 problem un
Page 231 and 232: RHAPSODY and SYMPHONY agree in posi
Page 233 and 234: 4.2 (Number of possible transpositi
Page 235 and 236: Table S.5 R H A P S O D Y B C E F G
Page 237 and 238: Chapter 8 8.1 (Recurrences of order
Page 239 and 240: columns in each case, are full of c
Page 241 and 242: Chapter 11 11.1 (Pin-setting errors
Page 243 and 244: [2.4] Moroney, M.J.: Facts from Fig
Page 245 and 246: [10.3] Almost any elementary book o
Page 247 and 248: Name index Adelman, L. 171, 234 And
Page 249 and 250: Subject index Abwehr Enigma 124, 13
Page 251 and 252: active pin 136 cage:‘good’ 141;
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Code and ciphers: Julius Caesar, the Enigma and the internet

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