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

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128 chapter 9 Ironically, if the cipher operators chose their indicators in a non-random manner, such as choosing all three letters from the same row of the keyboard, even more messages would be needed to get a full set of chains (M17). On the other hand the probability of finding two or more messages ‘in depth’ would be increased and the cryptanalysts might then be able to recover some plaintext, which might lead to the solution by a different route. Any kind of non-random feature in a cipher system or in its operational procedure might help the cryptanalysts. Aligning the chains The analysis that shows that chains of equal length occur in pairs also shows how to exploit this fact to discover the identity and setting of R1. To do this we must align pairs of chains of the same length, but one of the chains of such a pair must be reversed. Furthermore, we must try each possible alignment of the letters. Thus, for the two 10-letter chains above we can have 20 possible alignments, depending upon which chain we reverse and which letter comes first, viz: ABQNUMLPIR CHGYVTJSKD is one possible alignment, but there are 19 others, for example CDKSJTVYGH IPLMUNQBAR When we have the correct alignment of two related chains the vertical pairs of letters when encrypted through the correct R1 at its first setting will reveal pairs in the composite reflector whilst the NW–SE diagonal pairs when encrypted at the second setting of R1 will also reveal the same pairs. An incorrect wheel or incorrect settings will produce contradictions. This crucially important fact, a consequence of the operating procedure employed on the Enigma, was discovered by the Polish cryptanalysts in 1932; a proof is given in [9.1]. In this example we have some short chains and since there are fewer possibilities of alignment these would probably be tried first. Identifying R1 and its setting An example of the identification of R1 and its setting on a full size Enigma with known wheels but unknown plugboard would require a
great deal of data and many pages of analysis, but that was the problem faced daily by the cryptanalysts. The method can, however, be illustrated using the data from the 12-point mini-Enigma example above. We assume that the plugboard is known and that we have to see if any of the known wheels at one of 12 possible starting positions could be R1. Since there are many incorrect possibilities only two cases will be examined: one incorrect and the other correct. Example 9.3 The doublets in the 12-point mini-Enigma above are believed to have been enciphered at consecutive wheel positions on a common groundsetting with R1 having the encipherment table shown in Table 9.2. Table 9.2 Setting Input letter 1 2 3 4 5 6 7 8 9 10 11 12 A K A G L H F I C F D L E B F L B H A I G J D G E A C B G A C I B J H K E H F D G C H B D J C K I L F I E J H D I C E K D L J A G F H K I E J D F L E A K B G C I L J F K E G A F B L H A D J A K G L F H B G C I D B E K B L H A G I C H J I E C F L C A I B H J D K E J F D G A D B J C I K L L F K G E H B E C K D J Show that aligning the chains as D AICJK H BLFGE The Enigma cipher machine 129 (1) leads to contradictions if we assume that R1 is originally at Setting 1, but (2) produces a solution if we take R1 to be originally at Setting 2. Solution (1) We begin by enciphering vertical pairs from the chain at Setting 1 and diagonal (NW–SE) pairs at Setting 2. The 12 pairs that we obtain should be consistent and provide the 6 pairs of the composite reflector: Table 9.3.
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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
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
Page 135 and 136: The Enigma cipher machine 123 The m
Page 137 and 138: show how, in a typical encipherment
Page 139: interested reader can find it in [9
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
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
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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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