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

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118 chapter 9 The ‘dots’ indicate places where we do not yet have enough information to give the cipher letter. The full 26�26 encipherment table can be filled in when we know the encipherment of all 26 letters at position 1. Note the important feature of the encipherment table: each diagonal from North-West to South-East constitutes a full alphabet, in normal order, starting from the letter in column 1. Alternatively, if we know the encipherment of A, or any other letter, at all 26 positions of the wheel we can equally well work out the encipherment of any letter at any position. For example, suppose that we wish to know the encipherment of N at position 11. The wire that has N as its entry point at position 11 is the wire that had A as its entry point 13 positions earlier, since N is 13 places after A in the alphabet. Now 11�13��2, and position �2 is the same as position 26�2, i.e. 24.We therefore look up what letter A enciphers to at position 24. If this is, say, G then N at position 11 will encipher to the letter which is13 places after G in the alphabet at position 11, that is, to T. Readers who are familiar with matrices will recognise that what we are doing, in effect, is representing the encipherment provided by a wheel as a 26 �26 matrix. The first column gives the encipherment of the complete alphabet at position 1 and the first row gives the encipherment of A at each of the 26 positions. The matrix can then be completely filled in from either its first row or its first column by using the ‘diagonal property’ explained above. A cryptographic feature of some importance is that whereas any column will contain all 26 letters of the alphabet, since two letters cannot encipher to the same letter at the same wheel position, the rows may contain one or more letters twice or more, since there is nothing to prevent a letter enciphering to the same letter at two or more positions. In fact, with a wheel of size 26, or any even number of contacts, it is certain that each row will contain at least one repeated letter. In the 6-letter example above this already occurs, C goes to A at positions 1 and 3. With an odd number of contact points it is possible that the rows will not contain any repeated letters. From a cryptographic point of view the fewer repeated letters in a row the better. (For an explanation of this, and further details see M14.) Encipherment by the Enigma We have just seen how a single wired wheel enciphers a letter. In the Enigma the current from the keyboard letter passes through the entry
wheel and then through the three wheels R1, R2 and R3, after which it is ‘turned around’ by the reflector, U, and then goes back through the three wheels in the order R3, R2 and R1 before finally passing back through the entry wheel to light up the lamp which indicates the cipher letter. The original plaintext letter thus undergoes 9 changes before it finally emerges as a cipher letter; in fact, as we shall see later, in most military versions of the Enigma there were 2 further changes, making 11 in all. If all the wheels were fixed the Enigma would merely provide a complex way of generating a simple substitution cipher, but they are not fixed. When a keyboard letter is pressed the rightmost wheel, R1, immediately turns one position and the current then passes through the machine. After 26 letters have been enciphered R1 will be back in its original position. Unless R2 or R3 had moved in the mean time the Enigma would be equivalent to 26 simple substitution ciphers; R2, however, will have moved. The notch ring on R1 moves with the wheel and so, some time during the 26 encipherments, the V-shaped notch will have reached the position immediately in front of a lever at the back of the machine opposite R1, this will allow the lever to engage with the V-shaped notch and this in turn will allow a lever opposite R2 to cause R2 to turn one position. Since R2 has now moved the encipherment alphabets will all be different from what they were 26 encipherments previously. R2 thus moves at least once in every 26 letter encipherments; in fact it moves slightly more frequently than that, for R2 also has a notch ring on it, and when its notch is opposite a lever behind R2 the third wheel, R3, is caused to move one position and R2 itself is turned as well. The consequence of all this is that the three wheels will not all have returned to their original positions until 26 �25�26�16 900 The Enigma cipher machine 119 letters have been enciphered. Thus the Enigma machine provides an automatic way of using 16 900 simple substitution ciphers in succession. So, for example, if the notch ring on R1 were so fixed that its notch caused R2 to turn when R1 was at setting Z as shown in its window, and likewise with the notch ring on R2, the successive positions of the three wheels when they are started at positions A, Y, Y (reading from left to right) will be A Y Y, A Y Z, A Z A, B A B.
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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
Page 79 and 80: From a cryptographic point of view
Page 81 and 82: 3 7 0 7 7 4 1 5 6 1 7 8 5 3 8 1 9 0
Page 83 and 84: If we generate the same sequence (m
Page 85 and 86: Stencil ciphers The example above i
Page 87 and 88: Book ciphers A spy must avoid arous
Page 89 and 90: Table 7.2 Encipher table for a book
Page 91 and 92: Letter frequencies in book ciphers
Page 93 and 94: If then we try subtracting THE from
Page 95 and 96: where row F (the row of the cipher
Page 97 and 98: 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: Figure 9.2. wheel. For example, if
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 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
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Public key cryptography 169 Despite
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If the cryptographer were prepared
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number of tests increased by a fact
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In this case the remainder is 4, no
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key, d, we use the Euclidean Algori
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the decipher key, d, is used instea
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and and, since Table 13.1 n N�2 n
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The Data Encryption Standard (DES)
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the message were known to relate to
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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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