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

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7 Ciphers for spies A spy operating in country X on behalf of country Y has the problem of communicating with his controller in such a way as to protect both himself and the contents of his messages. No matter how he sends his messages they will have to be ‘modified’ somehow so that their true meaning is hidden from anyone but the intended recipient. There are methods, such as the use of microdots or ‘invisible’ ink, which do not, per se, involve encipherment although some ‘modification’ of the text even in such cases would probably be used to provide extra security. When we say that a text has been ‘modified’ we do not necessarily mean that it has been enciphered but that the ‘secret’ text is not simply sent in an unaltered form: it might, for example, be hidden inside an apparently innocuous message. Hiding a secret text inside an innocuous one has the advantage that, being apparently unenciphered, it will not automatically attract the interest of unintended recipients or interceptors, such as the security forces of country X. A disadvantage is that it may not be too easy to construct a realistic non-secret text in which to embed it. Here is a simple illustration. Example 7.1 (‘Part of a letter from Agent 63’) As I was walking through the centre of town yesterday morning at about eleven thirty I chanced to see Ron Kingston. He was alone, driving a newish-looking ultramarine car, a Ford Escort. Previously he’s had only second-hand cars, not often less than three years old. Perhaps he has had an inheritance from some rich relative who has recently died? Secret message If we take the first letter of the message and the first letter of every fourth word after that, taking hyphenated words as two separate words, we get [72] ATTACKDUEONTHIRD
Stencil ciphers The example above is a very simple, and insecure, case of a stencil cipher. In such a cipher certain letters on a page are part of the secret message and all the other letters are merely ‘fillers’ which are used to compose a mundane-looking communication. The ‘stencil’ in the example is far too regular to be satisfactory; a more suitable stencil would use letters which are separated from each other by irregular intervals and which are not necessarily the first letters of words. To add further security the letters of the secret text would probably not occur in their correct order in the overall text. If the overall text is typed in a regular format the sender and recipient may have identical cards with holes punched in them at the positions of the letters of the secret message. Hence the name of this type of cipher. Each hole would have a number underneath it giving the position of its corresponding letter in the message. Such a cipher would be much harder to solve unless the same stencil was used repeatedly in which case, given enough messages, it might be possible for a cryptanalyst to recover part of the stencil and from that gradually to recover the rest. If, however, the stencil changed regularly it would be extremely difficult, if not impossible without further information, to solve the system. To get some idea of the difficulty consider this. Example 7.2 (stencil cipher) Text of message: Some of Shakespeare’s plays such as SSSS 10SSSS 6SSSSSSSSS SSSSSS 8SS 1 Anthony and Cleopatra are performed SS2SSSS S11SSSS9SSS3SS SSS SS15 SSS7 less frequently than others, Macbeth SSSS SSSSSSS12SS SS4S SSSSSS SSS5SSS13 King Lear and Hamlet, in particular. S14SS SSSS SS16 Ciphers for spies 73 The numbers show where the holes in the stencil were placed and give the order in which the visible letters are to be read; where a two-digit number is needed the letter above the first digit is the one where the hole occurs; the letter has been underlined for clarity, though of course it would not
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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
Page 33 and 34: From Julius Caesar to simple substi
Page 37 and 38: Table 2.6 From Julius Caesar to sim
Page 41 and 42: and the key which provides the enci
Page 43 and 44: Further examination reveals that th
Page 45 and 46: ALTHOUGH I AM AN OLD MAN NIGHT IS G
Page 47 and 48: Polyalphabetic systems 35 messages
Page 49 and 50: Before leaving Vigenère try the fo
Page 51 and 52: Polyalphabetic systems 39 blocks of
Page 53 and 54: eginning at the top, but it is then
Page 55 and 56: first row above, for example, we fi
Page 57 and 58: Table 4.4 Key 3 1 5 2 4 1 2 3 4 5 6
Page 59 and 60: giving B G L D I N A F K E J O C H
Page 61 and 62: then the cipher text is HORUX SXSEO
Page 63 and 64: The plaintext digraphs are now sepa
Page 65 and 66: ox increases. By enumerating the po
Page 67 and 68: Example 5.1 HAPPY BIRTHDAY encipher
Page 69 and 70: Encryption The ‘plaintext’ is A
Page 71 and 72: plaintext digraphs according to som
Page 73 and 74: Cryptanalytic aspects of Playfair P
Page 75 and 76: OURXSITUATI ONXISXDESPE RATEXSENDXS
Page 77 and 78: 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: If we generate the same sequence (m
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 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
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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
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part of the story of what has been
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Public key cryptography 165 graphic
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(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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