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

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82 chapter 7 There are a few plausible looking trigraphs such as MEE at position 1, ROW at position 10, ONY at position 15, BEE at position 39, PEN at position 41. We now try extending the crib from THE to THEX to see if it produces a plausible tetragraph in any of these five cases. We insert X at positions 4, 13, 18, 42 and 44; the cipher letters at these positions are Q, C, U, L and L respectively and we decrypt these by ‘subtracting’ X from them, which means in effect moving each of these letters three places forward in the alphabet, thus giving the decrypts T, F, X, O and H so that the tetragraphs produced are: MEET at position 1, ROWF at position 10, ONYX at position 15, BEEO at position 39, PENH at position 41. The first of these looks particularly promising so we immediately investigate that further, keeping the others for analysis later. Since the first ‘hit’ of THE was at position 1 we examine the cipher text immediately following, the first 10 places initially. What we tentatively have is Cipher FLIQT NYQFK Text1 THEX. Text2 MEET. The first word in Text 2 might be MEET in which case it should be followed by X, as the separator, or it might be a longer word such as MEE- TINGX. In the first case the fifth letter of Text 1 would be W; in the second case letters 5, 6, 7 and 8 of TEXT 1 would be (T�I), (N�N), (Y�G) and (Q�X), i.e L, A, S and T, which looks extremely good since it implies that Text 1 begins THE LAST. We would now expect that LAST would be followed by X in which case the ninth letter of Text 2 would be (F�X) which is I (we could also obtain this by looking up Table 7.3 (the decipher table above) at the point
where row F (the row of the cipher letter) intersects column X (the column of the assumed plaintext letter in Text 1). The first 15 letters of the cipher and the partial texts then read Cipher FLIQTNYQFKVACEH Text1 THE LAST.. Text2 MEETING I. The letter following the I in Text 2 is very likely to be N or S and, in either case, the letter after that is probably X. The corresponding letters in Text 1 would then be either (K�N) and (V�X) or (K�S) and (V�X) that is either XY or SY. The latter is much more likely since the former implies a double space. We tentatively accept SX in Text 2 and the resulting SY in Text 1, and our partial decrypt now reads Cipher FLIQTNYQFKVACEH Text1 THE LAST SY.. Text2 MEETING IS .. Our next task is to discover what letter follows SY in Text 1. There are not many candidates and B, L, M, N and S are the most common. Since the cipher letter in position 12 is A the corresponding letter in Text 2 in these five cases is (A�B), (A�L), (A�M), (A�N) or(A�S) that is, Z, P, O, N or I. Of these only Z looks unlikely so we turn our attention elsewhere, to see if we can find some additional clues. Going back to the five possible occurrences of THE we see that we have confirmed the first (at position 1) and ruled out the second (at position 10) so we look at the third (at position 15) which would give us THEX in one text (we can’t decide immediately in which of the two texts it occurs) and ONYX in the other. If THEX occurs at position 15 then X ought to be found at position 14 and since the cipher letter there is E the letter preceding ONYX would then be (E�X) which is H. The third word in Text 1 would then be SY..HONY which strongly suggests that it is SYMPHONY. If this is so then since the 12th and 13th cipher letters are A and C, the corresponding letters of Text 2 are (A�M) and (C�P), i.e. O and N, which looks promising, and our decrypt now reads Cipher FLIQTNYQFKVACEHUCUAC Text1 THE LAST SYMPHONY .. Text2 MEETING IS ON THE .. Ciphers for spies 83
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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
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: If then we try subtracting THE from
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
Page 141 and 142: great deal of data and many pages o
Page 143 and 144: 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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