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

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3 Polyalphabetic systems Strengthening Julius Caesar: Vigenère ciphers The weakness of the Julius Caesar system is that there are only 25 possible decrypts and so the cryptanalyst can try them all. Life can obviously be made more difficult for him if we increase the number of cases that must be tried before success can be assured. We can do this if, instead of shifting each letter by a fixed number of places in the alphabet, we shift the letters by a variable amount depending upon their position in the text. Of course there must be a rule for deciding the amount of the shift in each case otherwise even an authorised recipient won’t be able to decrypt the message. A simple rule is to use several fixed shifts in sequence. For example, if instead of a fixed shift of 19 as was used in the message COME AT ONCE in the last chapter and which enciphered to VHFX TM HGVX we use two shifts, say 19 and 5, alternately, so that the first, third, fifth etc. letters are shifted 19 places and the second, fourth etc. are shifted 5 places then the cipher now becomes VTFJ TY HSVJ. If we replace the space character by Z in the message and use three shifts, say 19, 5 and 11, in sequence the plaintext becomes COMEZATZONCE. The cipher is now [28] VTXXELMEZGHP
and the key which provides the encipherment is 19-5-11. To read the message the recipient must use the decipherment key in which each of these three numbers is replaced by its complement (mod 26), i.e. by 7-21-15. Even if a cryptanalyst suspected that a Julius Caesar system with three shifts being used sequentially was being employed he would have to try 75 or more combinations (one of the shifts might be 0). On such a short message as this there would be the possibility of more than one solution. If the message is too short to identify the three shifts independently, as we shall do in the example below, a ‘brute force’ method might have to be tried but, since this would involve 25�25�25�15 625 trials, it would only be used as a last resort. In the extreme case where the number of shifts used was equal to the number of letters in the message the message becomes ‘unbreakable’, unless there is some non-random feature to the sequence of shifts. Where there is no non-random feature, such as when the sequence of shifts has been generated by some ‘random number process’, we have what is known as a ‘one-time pad’ cipher, which we come to in Chapter 7. This approach to strengthening the Julius Caesar cipher by means of several shifts has been used for some hundreds of years. Such systems are known under the name of Vigenère ciphers. Since most people find it easier to remember words rather than arbitrary strings of letters or numbers Vigenère keys often take the form of a keyword. This reduces the number of possible keys of course but that is the price the cryptographer has to pay for easing the burden on his memory. The letters of the keyword are interpreted as numbers in the usual way (A�0, B�1, C�2, ..., Z�25) so that, for example, the keyword CHAOS would be equivalent to using the five shifts 2, 7, 0, 14 and 18 in sequence. The keyword or numerical key would be written repeatedly above the plaintext and each plaintext letter moved the appropriate number of places to give the cipher. Thus if we enciphered COMEZATZONCE using Vigenère with the keyword CHAOS the layout would be CHAOSCHAOSCH COMEZATZONCE and the resultant cipher is EVMSRCAZCFEL. Polyalphabetic systems 29
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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 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 35 and 36: From Julius Caesar to simple substi
Page 37 and 38: Table 2.6 From Julius Caesar to sim
Page 39: From Julius Caesar to simple substi
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
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
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where row F (the row of the cipher
Page 97 and 98:
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
Page 105 and 106:
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
Page 111 and 112:
Linear recurrences The sequences lo
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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
Page 163 and 164:
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
Page 181 and 182:
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
Page 221 and 222:
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
Page 233 and 234:
4.2 (Number of possible transpositi
Page 235 and 236:
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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