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

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98 chapter 8 signal which is used to switch a gate on or off, and this in turn is then interpreted as a 0 or a 1. If the circuits are carefully adjusted the binary stream so produced should be effectively random. If there is some residual bias, in that the probabilities of 0 and 1 occurring are slightly different from 0.5, the (mod 2) sum of two or more such streams will reduce it considerably. Two unrelated streams each of which has a bias of 0.51 to 0.49 in favour of 0, for example, will combine to produce a stream with a bias of only 0.5002 to 0.4998 (M9). Pseudo-random sequences We have already encountered the Fibonacci sequence in Chapter 6. This is an infinitely long sequence of integers generated by the simple rule that each number in the sequence is the sum of the two previous numbers. The sequence is traditionally started by taking the first two numbers as 0 and 1. The Fibonacci sequence unfortunately has many arithmetic properties, as was mentioned before, and so is quite unsuitable as a source of pseudo-random numbers. Suppose, however, that we modify the rule to, say, that each number is the sum of twice the previous number plus the number before that, would we get a better sequence for our purposes? If we begin with 0 and 1 as the first two terms, the first 10 terms of the sequence are 0,1, 2, 5, 12, 29, 70, 169, 408, 985. It will be noted that the terms are even and odd alternately and this, by itself, is sufficient to rule them out as a source of pseudo-random numbers. Of course we needn’t begin with 0 and 1 as the first two terms, we could start with any two numbers, but the flaw is fundamental and no sequence generated in this way would be satisfactory. The sequence, as might be expected after seeing the many features of the Fibonacci sequence, has many mathematical properties; for example, every third term is divisible by 5 and the ratio of consecutive terms rapidly approaches the fixed value which is 2.414 213 56... (1��2). (For more detail see M10).
Linear recurrences The sequences looked at above are examples of sequences generated by means of what are known as linear recurrences. Since each new term involved adding together multiples of the two preceding terms they are more specifically known as linear recurrences of order 2. More generally, a linear recurrence of order k is one in which each new term is the sum of multiples of the k preceding terms. So, for example, if we let U n denote the nth term of a sequence then U n �U (n�1) �2U (n�2) �U (n�3) is a linear recurrence of order 3 and U n �U (n�3) �U (n�5) is a linear recurrence of order 5. The fact that in the second case some of the preceding terms are not involved doesn’t matter; five preceding terms are required in order to find the next term but three of the terms, U (n�1) , U (n�2) and U (n�4) , have multipliers of 0. Had the term U (n�5) not been present however the recurrence would not have been of order 5. The multipliers, for our purposes, are always integers but may be positive, negative or zero. It is assumed that in a linear recurrence of order k the term U (n�k) is present, with either a positive or a negative multiplier, but not zero. The terms of linear recurrences usually grow very rapidly and although they often have interesting arithmetical properties they are only suitable for cryptographic purposes when the terms themselves are replaced by their values (mod 2), that is the terms are replaced by 0 if they are even and by 1 if they are odd, thus producing a binary sequence. Calculation of the terms of a linear recurrence (mod 2) is particularly easy, there is no need to compute the actual value of the terms and then replace them by 0 or 1. Each term is simply replaced by 0 or 1 as soon as it is calculated; we then only have to add up a number of 0s and 1s which is a lot easier than adding increasingly large integers. The resulting binary sequence is identical to the one which would be obtained by computing each term exactly and then replacing it by 0 or 1. So, for example, the linear recurrence of order 2 U n �3U (n�1) �2U (n�2) with the initial values U 0 �0, U 1 �1 continues 0, 1, 3, 7, 15, 31, 63, 127, 255, 511,..... Producing random numbers and letters 99
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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
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 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: numbers, 0 to 31 inclusive, into bi
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 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
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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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