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

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12 chapter 1 not occasionally women, for indeed some are, but since the majority are men I use masculine pronouns, which may be interpreted as feminine everywhere. End matter At the end of the book are the following. First, the mathematical appendix is intended for those readers who would like to know something about the mathematics behind some of the systems, probabilities, analysis or problems mentioned in the text. A familiarity with pure mathematics up to about the standard of the English A-Level is generally sufficient but in a few cases some deeper mathematics would be required to give a full explanation and then I try to give a simplified account and refer the interested reader to a more advanced work. References to the mathematical appendix throughout the book are denoted by M1, M2 etc. Second, there are solutions to problems. Third, there are references; articles or books referred to in Chapter 5 for instance are denoted by [5.1], [5.2] etc.
2 From Julius Caesar to simple substitution Julius Caesar ciphers and their solution In the Julius Caesar cipher each letter of the alphabet was moved along 3 places circularly, that is A was replaced by D, B by E ... W by Z, X by A, Y by B and Z by C. Although Julius Caesar moved the letters 3 places he could have chosen to move them any number of places from 1 to 25. There are therefore 25 versions of the Julius Caesar cipher and this indicates how such a cipher can be solved: write down the cipher message and on 25 lines underneath it write the 25 versions obtained by moving each letter 1, 2, 3, .... , 25 places. One of these 25 lines will be the original message. Example 2.1 The text of a message enciphered by the Julius Caesar System is VHFX TM HGVX Decrypt the message. Solution We write out the cipher message and the 25 shifted versions, indicating the shift at the left of each line (see Table 2.1), and we see that the cipher used a shift of 19, for the cipher text is shifted 7 places to give the plain and this means that the plaintext has to be shifted (26�7)�19 places to give the cipher. It looks very likely, on the assumption that no other shift would have produced an intelligible message, that we have correctly decrypted the message and so there is no point in writing out the remaining lines. This assumption of uniqueness is reasonable when the cipher message is more than five or six characters in length but for very short [13]
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Page 5 and 6: R. F. Churchhouse Codes and ciphers
Page 7 and 8: Contents Preface ix 1 Introduction
Page 9 and 10: Cryptanalysisofalinearrecurrence 10
Page 11 and 12: Preface Virtually anyone who can re
Page 13 and 14: 1 Introduction Some aspects of secu
Page 15 and 16: Not a very sophisticated method, pa
Page 17 and 18: 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: When the modulus is 10 only the num
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 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
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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
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Table 7.2 Encipher table for a book
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Letter frequencies in book ciphers
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If then we try subtracting THE from
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where row F (the row of the cipher
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Ciphers for spies 85 that were nece
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mistake can occur if the sender lea
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inks and ciphers were provided by h
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Example 7.6 Encipher the message AG
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Ciphers for spies 93 simply a ‘ra
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going into the mathematical criteri
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numbers, 0 to 31 inclusive, into bi
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Linear recurrences The sequences lo
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Having converted the characters of
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What about sequences of higher orde
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combining the keys of two or more l
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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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