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

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58 chapter 5 French cryptanalyst, Georges Painvin, worked out a method for doing it given a number of messages with identical plaintext beginnings or endings. Although only a few days were solved the number of messages on those days was high and their contents were particularly valuable. It is said that ‘On one occasion the solution was so rapid that an important German operation disclosed by one message was completely frustrated’ [5.1]. Example 5.5 (A Japanese naval cipher) During World War II the Japanese Merchant Navy used a cipher, known as JN40, in which each Japanese kana symbol was replaced by two digits from a 10�10 square. The individual digits were then written into the columns of a rectangle, the ordering of the columns being given by a transposition. The digits were then taken out row by row, the ordering of the rows being given by a second transposition. The transpositions were changed every day. The cipher was solved in November 1942 when an operator left some details out of a message and then re-enciphered the full text using the same keys [5.2]. Problem 5.1 An MDTM system with keyword ABSOLUTE and transposition 3-1-5-2-4 has been applied to a message and the resulting cipher text is CFIGS FLTBC XKEEA EBHTB GLDPI Decrypt the message. Digraph to digraph Just as a simple substitution system replaces each individual letter of the alphabet by a single letter so one can construct a system in which every digraph is replaced by two letters. The ‘obvious’ way of doing this, as has been indicated before, is to have a list of all 676 (�26�26) possible digraphs and their cipher equivalents e.g. AA�TK, AB�LD, AC�ER,..., ZX�DW, ZY�HB, ZZ�MS, but this involves having two lists, one for encryption and one for decryption, each 676 long, and although this would provide a better level of security than simple substitution it would be tedious to use. An alternative method is to use a digraph square which contains just the 25 letters of the alphabet, excluding J say, and form the cipher digraphs from the
plaintext digraphs according to some rule. A nineteenth century system of this type is the method of Playfair encipherment The alphabet, ignoring J, is written into a 5�5 square in some order, either starting with a keyword, with the remaining letters following in order, or in a ‘random’ order. If a keyword has any repeated letters the second and later occurrences are ignored. Thus if TOMORROW is the keyword it would go into the square as TOMRW. Encipherment is then carried out digraph by digraph according to the following rules. (i) If the two letters of a digraph are at diagonally opposite corners of a rectangle in the 5�5 square they encipher to the pair at the other two corners. In the example below we adopt the convention that each letter is replaced by the corner letter in the same row as itself. The alternative convention is to replace each letter by the letter in the same column as itself and this is the usual practice with double Playfair ciphers, as we see below. (ii) If the two letters are in the same column of the 5�5 square they encipher as the letters below them, where row 1 is considered to be below row 5 if necessary. (iii) If the two letters are in the same row of the 5�5 square they encipher as the letters to the right of them, where column 1 is considered to be to the right of column 5 if necessary. (iv) If the two letters are identical a ‘dummy’ letter, such as Q, is inserted between them. (v) If necessary, a ‘dummy’ letter is inserted at the end of the plaintext. Example 5.6 Encipher the message SUPPORT NEEDED URGENTLY using Playfair encipherment with the keyword WALKING. Table 5.5 W A L K I N G B C D E F H M O P Q R S T U V X Y Z Two-letter ciphers 59
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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
Page 19 and 20: Another form of encryption is the u
Page 21 and 22: and so is valid. On the other hand
Page 23 and 24: When the modulus is 10 only the num
Page 25 and 26: 2 From Julius Caesar to simple subs
Page 27 and 28: Table 2.3 Shift Message 12 OUI 12 Y
Page 29 and 30: worry about but, on the other hand,
Page 31 and 32: 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: Encryption The ‘plaintext’ is A
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 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
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Problem 8.3 Starting with U 0 �1
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The Enigma cipher machine 111 syste
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The Enigma cipher machine 113 Plate
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The Enigma cipher machine 115 Plate
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Figure 9.2. wheel. For example, if
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wheel and then through the three wh
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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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