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

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78 chapter 7 Table 7.3 Decipher table for a book cipher Problem 7.2 A message has been enciphered using a book cipher. The book used was The Poems of Rupert Brooke and the key for the message was the passage beginning STANDSXTHEXCHURCHXCLOCKXATXTENXTOXTHREE The cipher text was LAEKV MPILG QZOUJ ZTLXP RZDLX EFOIE MHCIQ Decrypt the message. Key text Cipher A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A A Z Y X W V U T S R Q P O N M L K J I H G F E D C B B B A Z Y X W V U T S R Q P O N M L K J I H G F E D C C C B A Z Y X W V U T S R Q P O N M L K J I H G F E D D D C B A Z Y X W V U T S R Q P O N M L K J I H G F E E E D C B A Z Y X W V U T S R Q P O N M L K J I H G F F F E D C B A Z Y X W V U T S R Q P O N M L K J I H G G G F E D C B A Z Y X W V U T S R Q P O N M L K J I H H H G F E D C B A Z Y X W V U T S R Q P O N M L K J I I I H G F E D C B A Z Y X W V U T S R Q P O N M L K J J J I H G F E D C B A Z Y X W V U T S R Q P O N M L K K K J I H G F E D C B A Z Y X W V U T S R Q P O N M L L L K J I H G F E D C B A Z Y X W V U T S R Q P O N M M M L K J I H G F E D C B A Z Y X W V U T S R Q P O N N N M L K J I H G F E D C B A Z Y X W V U T S R Q P O O O N M L K J I H G F E D C B A Z Y X W V U T S R Q P P P O N M L K J I H G F E D C B A Z Y X W V U T S R Q Q Q P O N M L K J I H G F E D C B A Z Y X W V U T S R R R Q P O N M L K J I H G F E D C B A Z Y X W V U T S S S R Q P O N M L K J I H G F E D C B A Z Y X W V U T T T S R Q P O N M L K J I H G F E D C B A Z Y X W V U U U T S R Q P O N M L K J I H G F E D C B A Z Y X W V V V U T S R Q P O N M L K J I H G F E D C B A Z Y X W W W V U T S R Q P O N M L K J I H G F E D C B A Z Y X X X W V U T S R Q P O N M L K J I H G F E D C B A Z Y Y Y X W V U T S R Q P O N M L K J I H G F E D C B A Z Z Z Y X W V U T S R Q P O N M L K J I H G F E D C B A A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Letter frequencies in book ciphers The frequencies of the letters of the alphabet in the cipher produced by a book cipher when the key is a passage of English text will be different from those of unenciphered English. Whilst it will certainly be true that some letters, such as those letters used for ‘space’, E and T, will no longer occur very much more frequently than others such as Z or J, it is also true that the letters will not be equally represented. It is possible to estimate the frequencies of the cipher letters and we find that although the frequencies of the individual letters do not vary so much as they do in samples of normal (unenciphered) English they are still by no means uniform and observation of this type of variation would alert the cryptanalyst to the possibility that a book cipher was being used. This variation can be seen in Table 7.4. In the left-hand column are the frequencies of the 26 letters of the alphabet, and a 27th ‘letter’ which covers all punctuation marks and ‘space’, as they occur in a typical sample of normal English and, in the right-hand column, the frequencies of the same letters as they are predicted to occur in a passage enciphered by a book cipher using English texts. The sample size in both cases is 1000 so that ‘on average’ each letter should occur about 37 times. It will be seen that this is very far from being the case in the unenciphered text, and even in the enciphered text there is considerable variation in the frequencies, though not a lot more than we would expect at random (for further comments see M6). A book cipher might be regarded as an extreme case of a Vigenère cipher in which the key length is the same as the length of the message itself. Evidently a book cipher ought to be more secure than a Vigenère since the latter uses a key of fixed length. An alternative to using a key which is an English text is to use a key in a different language but which uses an alphabet of not more than 26 letters. Diacriticals, such as accents or umlauts, would be ignored if French or German were used. This would make life somewhat harder for the cryptanalyst at least until he realised what was going on. Solving a book cipher Ciphers for spies 79 Assuming that a cryptanalyst has realised that the cipher text is the result of enciphering an English text with an English book used as the key how might he go about trying to solve it? Although the letter frequencies in
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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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From Julius Caesar to simple substi
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Table 2.6 From Julius Caesar to sim
Page 39 and 40: From Julius Caesar to simple substi
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
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: Table 7.2 Encipher table for a book
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 and 140: 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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