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

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138 chapter 10 Note that the 19- and 17-wheels have made a complete cycle during the generation of these 20 key values so that their contributions to the total key begin to repeat, as indicated by the underlined values. Note also that since the key is interpreted by the print wheel (mod 26) the key value of 27 in the third position is effectively a key value of 1. Example 10.2 Use the 20 key values in the example above to encipher the following text using the Hagelin method of encipherment H A G E L I N X E N C I P H E R M E N T We first convert the text into numbers H A G E L I N X E N C I P H E R M E N T 7 0 6 4 11 8 13 23 4 13 2 8 15 7 4 17 12 4 13 19 we then subtract these numbers from the corresponding key values (mod 26) Key 21 15 27 15 2 16 1 7 13 20 12 19 22 12 9 18 3 13 16 15 Text 7 0 6 4 11 8 13 23 4 13 2 8 15 7 4 17 12 4 13 19 Cipher 14 15 21 11 17 8 14 10 9 7 10 11 7 5 5 1 17 9 3 22 Converting the cipher to letters and splitting into five-letter groups the cipher text would be transmitted as OPVLR IOKJH KLHFF BRJDW. Choosing the cage for the Hagelin In theory, the unoverlapped Hagelin cage may be any combination of six non-negative integers that add up to 27 or less. In practice, many of these cages would be very weak from a cryptographic point of view. Ideally, a cage should generate each of the possible key values, 0 to 25, equally often. This ideal is not achievable for, since each of the six wheels can be either active or inactive in any given position, there are 2 6 �64 possible combinations of the six pins and hence 64 possible key values. Since all key values are interpreted (mod 26) the 26 possible values cannot all occur equally often because 64 is not a multiple of 26. On average, a key
value can be expected to occur two or three times in a well-chosen cage. A cage such as (9, 9, 9, 0, 0, 0) is obviously very poor since the only key values that can be generated are 0, 1 (which occurs as 27), 9 and 18 and even these are non-uniformly represented, viz: Key value 0 1 19 18 Number of occurrences out of 64 possible 8 8 24 24 Solving a message sent with such a cage wouldn’t be very difficult since there are only four possible key values at any stage and two of these are much more likely than the other two. In the case of such a poor cage as this the process of solution is easily illustrated. Example 10.3 The following is the cipher text of a message which has been enciphered on a Hagelin with cage (0, 0, 0, 9, 9, 9). X is used for spacing/punctuation. Decrypt the message. ZCTAL BRDSV IBGDZ SMFVM. Solution With such a cage the only possible key values are 0, 1, 9 and 18. If we subtract each letter of the cipher from these four values and write the resultant texts in four rows the decrypt must lie somewhere within the four rows. Since 9 and 18 are three times more likely to occur than 0 or 1 we would expect the majority of the plaintext letters to lie in the third and fourth rows. X is used for spacing/punctuation and, to make it more obvious, we replace it by ^ wherever it occurs. We can save ourselves the tedium of subtracting the numerical equivalent of the cipher letters from the key values if we construct a table once and for all. This has the additional advantage that, unlike the book cipher tables of Chapter 7, the same table can be used for both encipherment and decipherment, because of the symmetry of the Hagelin encipherment/decipherment process mentioned above. When we have the table, given by Table 10.2, we simply look up the entries in the row corresponding to the cipher, or plaintext, letter and the column of the key value to obtain the plaintext, or cipher, letter, giving Table 10.3. The space marks are helpful and it is easy to pick out the plaintext, marked in bold in Table 10.3: THIS IS A BAD CAGE The Hagelin cipher machine 139
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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
Page 99 and 100: mistake can occur if the sender lea
Page 101 and 102: inks and ciphers were provided by h
Page 103 and 104: Example 7.6 Encipher the message AG
Page 105 and 106: Ciphers for spies 93 simply a ‘ra
Page 107 and 108: going into the mathematical criteri
Page 109 and 110: numbers, 0 to 31 inclusive, into bi
Page 111 and 112: Linear recurrences The sequences lo
Page 113 and 114: Having converted the characters of
Page 115 and 116: What about sequences of higher orde
Page 117 and 118: 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: of each cipher period, to which sid
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 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
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(1) X precedes M with information w
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ways, since choosing to pair, say,
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For example; if someone claims that
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depth, mentioned in Chapter 3. If w
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M9 Combining two biased streams of
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and so 2�4�6�12 �(4095)�4
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Verification Let the recurrence be
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the lettersatsetting2ofthewheel,and
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M16 Probability of a ‘depth’ in
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(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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