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

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86 chapter 7 been enciphered using a book as key, a fact that is not immediately apparent. Indicators These have already been referred to in Chapter 3 but they are sufficiently important to warrant mention again, since provision of the indicator may prove to be the Achilles heel of any cipher system. When a book cipher is used the sender has to let the receiver know at what line on what page of the key book he is beginning the encipherment. If, for example, he begins with line 15 of page 216 he may preface the message with the number 15216. This is, however, rather too obvious and he would be more likely to disguise it. There are numerous ways of doing this such as: (1) transposing the digits according to an agreed rule so that 15216 becomes, say, 65121; (2) adding an agreed number, such as 59382, digit by digit, thus producing 64598; (3) converting the digits of the indicator to letters and hiding the resulting five-letter group at an agreed place within the cipher text of the message; thus 15216 becomes BFCBG; (4) a combination of any of these. Provision of an indicator is not, of course, unique to book ciphers; it is an essential part of many cipher systems, but the basic principle is the same: the sender must somehow communicate it to the receiver in such a way as to make it as difficult as possible for a cryptanalyst to find it. The cryptanalyst, conversely, will give high priority to identifying the location of the indicator and discovering its method of encryption . Disastrous errors in using a book cipher A significant risk for the sender of a book cipher message is that if he makes a mistake and has to re-send the message he may provide the cryptanalyst with a relatively simple route into decryption. Such a crucial
mistake can occur if the sender leaves out a letter of the text of the book key, for example: Example 7.5 (Spy makes a mistake and gives the game away) Two messages from the same person and with identical indicators were sent within hours of each other; the cipher texts were ZECBH MOPJO IIUXJ ELFDR WRSJX CQ. ZECSS HLIEL RVBCM CUAKA OLPBP PPP We shall begin the decryption on the assumption that, since the first message has one letter fewer than the second, the sender left out a letter of the book key in the first message and then re-enciphered the text correctly in the second message. Start of solution Since the first three letters of the cipher text are the same in both cases we assume that it was the fourth letter of the book key that was not used in the first message. If we were to try all 26 possibilities for this letter in the second cipher message we would obtain 26 possibilities for the fourth letter of the message and, in each case, this would then give us the fifth letter of the book key from the first cipher message. With the fifth letter of the book key we would then recover the fifth letter of the message from the second cipher message, which would then lead us to the sixth letter of the book key from the first cipher message; and so on. We are thus enabled to unravel both the book key and the message from the fourth letter onwards. Of course we initially have to try all 26 possibilities for the fourth key letter but we would quickly be able to see which was the right one when the book key and message texts began to appear. To save time and space we will simply look at what happens when we choose the right letter, which is F,as the fourth letter of the key. We shall refer to the cipher texts as CT1 and CT2; CT2 is the correct text and CT1 has the error. The plaintext letters can be worked out directly, as shown below, or by using Table 7.3. From CT2: (cipher – key) at position 4�plaintext letter at position 4; i.e. S�F�N. From CT1: (cipher – plain) at position 4�key letter at position 5; i.e. B�N�O. Ciphers for spies 87
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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
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 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: Ciphers for spies 85 that were nece
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
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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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