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

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178 chapter 13 so 1�7�2�3 but, from the second line above, 3�17�2�7 and so 1�7�2�(17�2�7)�5�7�2�17 but, from the first line above, 7�3016�177�17 and so or 1�5�(3016�177�17)�2�17�5�3016�887�17 887�17��1 (mod 3016). We need a value of d that, when multiplied by 17, leaves remainder �1 when divided by 3016 and this is therefore �887, not �887. Now �887� 2129 (mod 3016) so, finally, the decipherment key, d, is 2129. (Check: 2129�17�36 193�12�3016�1.) The encipherment and decipherment processes in the RSA system To encipher a message using the RSA method the would-be user must know the values of the modulus, n (�pq), and the encipherment key, e; these are publicly available. Only the ‘owner’ of the system which is being used knows the values of p and q and the decipherment key, d. To encipher a message which is to be sent to the ‘owner’, and which he alone can decipher, the user must: (1) convert the letters of his message into numbers such as those given in Table 1.1, so that, for example A�00, B�01, ..., Z�25; (2) if the modulus n contains no more than D digits break the number version of the message into blocks of no more than D digits; denote these blocks by B 1 , B 2 , ...; (3) encipher the blocks in order and independently by computing (4) (B I ) e (mod n), I�1, 2, ..., giving cipher blocks C 1 , C 2 , ... The cipher message is C 1 C 2 ... (a string of numbers, not letters). To decipher a message the procedure is exactly the same as for encipherment except that the blocks to be handled now are the cipher blocks C I and
the decipher key, d, is used instead of the encipher key, e, at stage (4). It is a consequence of the way that d has been determined that (C I ) d �B I or, in other words, that the original message is recovered. Example 13.3 Encipher the message COMEXATXNOON using the RSA System with n�3127 and encipherment key, e,�17. Encipherment We convert the text into numbers using Table 1.1 in the usual way. Since the modulus, 3127, contains four digits we break the text up into pairs of letters,viz: CO ME XA TX NO ON 0214 1204 2300 1923 1314 1413 We now have to compute the value of each of the six four-digit numbers raised to the power 17 with respect to the modulus 3127. The computation is laborious by hand and so is only given in detail for the first number, 0214. The other values are easily obtained by a simple computer program. Our task then is to compute (0214) 17 (mod 3127). Whenever possible in modular computations of this kind we use repeated squarings to get powers of the number and we then multiply together appropriate powers. Since 16�2 4 we can get close to the required power and so we proceed as follows: so and hence (214) 2 �45 796�14�3127�2018�2018 (mod 3127) (214) 4 �(2018) 2 �4 072 324�1302�3127�970�970 (mod 3127) (214) 8 �(970) 2 �940 900�300�3127�2800�2800 (mod 3127), (214) 16 �(2800) 2 �7, 840, 000�2507�3127�611 �611 (mod 3127). Encipherment and the internet 179
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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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Page 141 and 142: great deal of data and many pages o
Page 143 and 144: It should be realised, of course, t
Page 145 and 146: 10 The Hagelin cipher machine Histo
Page 147 and 148: ut there was no cryptographic advan
Page 149 and 150: of each cipher period, to which sid
Page 151 and 152: value can be expected to occur two
Page 153 and 154: Since some cages are obviously very
Page 155 and 156: 2, there are 26! possible simple su
Page 157 and 158: The Hagelin cipher machine 145 Exam
Page 159 and 160: values are repeating at the appropr
Page 161 and 162: Overlapping will obviously affect t
Page 163 and 164: Table 10.6 The Hagelin cipher machi
Page 165 and 166: 11 Beyond the Enigma The SZ42: a pr
Page 167 and 168: Description of the SZ42 machine The
Page 169 and 170: P1 41 43 Z1 (4) the five bits from
Page 171 and 172: which is approximately 1.6�10 19
Page 173 and 174: 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: key, d, we use the Euclidean Algori
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 and 226: If x 3 �0 divide x 2 by x 3 to gi
Page 227 and 228: then k�[ log 2 n], where [z] deno
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
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Chapter 11 11.1 (Pin-setting errors
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[2.4] Moroney, M.J.: Facts from Fig
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[10.3] Almost any elementary book o
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Name index Adelman, L. 171, 234 And
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Subject index Abwehr Enigma 124, 13
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active pin 136 cage:‘good’ 141;
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Code and ciphers: Julius Caesar, the Enigma and the internet

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