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

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158 chapter 11 Solution Plaintext letter (S) 1 0 1 0 0 Set A pin values 0 1 0 1 1 (mod 2) sum 1 1 1 1 1 Set C pin values 1 0 0 1 0 (mod 2) sum 0 1 1 0 1�P. So the cipher letter will be P. If we now start with the cipher letter, P, then, with the same pin values the decipherment process gives Cipher letter (P) 0 1 1 0 1 Set A pin values 0 1 0 1 1 (mod 2) sum 0 0 1 1 0 Set C pin values 1 0 0 1 0 (mod 2) sum 1 0 1 0 0�S, confirming the original plaintext letter. Breaking and setting the SZ42 Assuming that the cryptanalyst knew the design details of the SZ42, as given in Figure 11.1, and that it was being used, how many possibilities would he have to consider before he could be certain of being able to decipher a message? The answer is easily obtained. The motion of the wheels and the encipherment are completely determined by the pins on the 12 wheels. Since there are 501 pins in all and each of them can be placed in either of two states, ‘active’ or ‘inactive’, the number of possibilities is 2 501 � 10 151 . This is the ‘breaking work factor’ and it is so large that if every particle in the Universe was a computer and had been assigned full time ever since the ‘Big Bang’ to trying all the possibilities the solution would still not have been found. Clearly, a ‘brute force attack’ approach to breaking the SZ42 is hopeless. If the cryptanalyst knew all of the pin patterns and a new message appeared he would have the ‘easier’ task of finding the settings of the 12 wheels at the start of the message. The number of possibilities that he would have to try is the product of the wheel lengths, viz: 23�26�29�31�37�41�43�47�51�53�59�61
which is approximately 1.6�10 19 . This is the ‘setting work factor’ and although it looks insignificant compared to the work factor for breaking it would still rule out a brute force attack even on today’s most powerful computers, and in 1942 there were none. The story of how SZ42 messages were nevertheless decrypted is described in some detail in [11.3]. The basic attack was to find the settings of the wheels of set A, using statistical analysis to find ‘likely candidates’. Since even this involved more than 22 000 000 possibilities (the product of the wheel lengths of set A) a machine (known as Colossus) was designed and built for the purpose. Although intended for one very special purpose Colossus was in some respects an early electronic computer. Problem 11.1 A cipher operator made a single error in setting one of the pins on one of the wheels in each of the following cases and the cipher messages were resent using the corrected pin settings. What would the cryptanalyst notice when comparing the two cipher messages in each case? (1) The messages were enciphered on a Hagelin machine and the incorrect pin was on the 23-wheel; (2) the messages were enciphered on an SZ42 and the incorrect pin was on the 31-wheel; (3) the messages were enciphered on an SZ42 and the incorrect pin was on the 61-wheel. With 131 pins on the Hagelin and 501 on the SZ42 such errors could easily occur. Modifications to the SZ42 Beyond the Enigma 159 In order to increase the complexity of the SZ42 and so make life even more difficult for the cryptanalysts three modifications were introduced at different times. All of these were intended to make the stepping of the wheels of set B more unpredictable. Instead of the motion of the wheels of set B being controlled exclusively by the current pin value on the 37-wheel: (1) the current pin value on the 31-wheel was added (mod 2) to the current pin value on the 37-wheel; or (2) the current pin value on the 43-wheel was added (mod 2) to the current pin value on the 37-wheel; or (3) the value of P 5 two positions back was added (mod 2) to the current pin value on the 37-wheel.
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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
Page 119 and 120: Example 8.3 (1) Use the mid-square
Page 121 and 122: Problem 8.3 Starting with U 0 �1
Page 123 and 124: The Enigma cipher machine 111 syste
Page 125 and 126: The Enigma cipher machine 113 Plate
Page 127 and 128: The Enigma cipher machine 115 Plate
Page 129 and 130: Figure 9.2. wheel. For example, if
Page 131 and 132: wheel and then through the three wh
Page 133 and 134: first day of usage and so the asses
Page 135 and 136: The Enigma cipher machine 123 The m
Page 137 and 138: show how, in a typical encipherment
Page 139 and 140: interested reader can find it in [9
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: P1 41 43 Z1 (4) the five bits from
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 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
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Chapter 13 M21 (Rate of increase of
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Multiply each of these by M: M, 2M,
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If x 3 �0 divide x 2 by x 3 to gi
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then k�[ log 2 n], where [z] deno
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Mathematical aspects 217 problem un
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RHAPSODY and SYMPHONY agree in posi
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4.2 (Number of possible transpositi
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Table S.5 R H A P S O D Y B C E F G
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Chapter 8 8.1 (Recurrences of order
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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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