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

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200 appendix Solution A linear recurrence of order 5 takes the form U n �aU (n�1) �bU (n�2) �cU (n�3) �dU (n�4) �eU (n�5) where a, b, c, d and e are unknown constants whose values are either 0 or 1, since all the arithmetic is (mod 2). We number the bits 1 to 15 from left to right and then put n�6, 7, 8, 9 and 10 in the recurrence to obtain five linear equations in the five unknowns a, b, c, d and e: a(1)�b(0)�c(1)�d(0)�e(1)�0, (A.1) a(0)�b(1)�c(0)�d(1)�e(0)�0, (A.2) a(0)�b(0)�c(1)�d(0)�e(1)�1, (A.3) a(1)�b(0)�c(0)�d(1)�e(0)�1, (A.4) a(1)�b(1)�c(0)�d(0)�e(1)�0. (A.5) From equations (A.1) and (A.3) we find that a�1 and then from equation (A.4) it follows that d�0. From equation (A.2) we then find that b�0 and from equation (A.5) that e�1 and, finally, from equation (A.1) that c�0. The solution to these five equations is therefore U n �U (n�1) �U (n�5) . We now need to confirm that this gives the correct values when n�11, 12, 13, 14 and 15 and it will be seen that this is the case. We have therefore verified that the recurrence of order 5 that we have just found does generate the given stretch of key. As mentioned in Chapter 8 it can happen that there are no solutions of order k or that there is more than one solution. In the former case the equations are inconsistent and in the latter the ambiguities can usually be resolved if extra key digits are available. The following examples illustrate these situations. Example (More than one solution) Verify that the 10-bit binary key 0110110110 can be generated by either of two binary linear recurrences of order 5.
Verification Let the recurrence be U n �aU (n�1) �bU (n�2) �cU (n�3) �dU (n�4) �eU (n�5) . We number the bits 1 to 10 from left to right and put n�6, 7, 8, 9 and 10 successively in the recurrence which yields the equations a � c � d � 1, a � b � d � e � 0, b � c � e � 1, a � c � d � 1, a � b � d � e � 0, which have two genuine sets of solutions of order 5 (that is, where e, the coefficient of U (n�5) , is not zero): and a�b�c�0, d�e�1 a�d�0, b�c�e�1. The two linear recurrences of order 5 are therefore and U n �U (n�4) �U (n�5) U n �U (n�2) �U (n�3) �U (n�5) . In addition, there are solutions where e�0, that is, solutions which are not of order 5, these include a�b�1, c�d�e�0 corresponding to the recurrence, which is of order 2, U n �U (n�1) �U (n�2) , which reveals the fact that the binary sequence above is just that of the Fibonacci numbers (mod 2). Example (No solution having the assumed order) Verify that the six-bit binary key 011010 cannot be generated by any binary linear recurrence of order 3. Mathematical aspects 201
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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
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Further examination reveals that th
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ALTHOUGH I AM AN OLD MAN NIGHT IS G
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Polyalphabetic systems 35 messages
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Before leaving Vigenère try the fo
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Polyalphabetic systems 39 blocks of
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eginning at the top, but it is then
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first row above, for example, we fi
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Table 4.4 Key 3 1 5 2 4 1 2 3 4 5 6
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giving B G L D I N A F K E J O C H
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then the cipher text is HORUX SXSEO
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The plaintext digraphs are now sepa
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ox increases. By enumerating the po
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Example 5.1 HAPPY BIRTHDAY encipher
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Encryption The ‘plaintext’ is A
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plaintext digraphs according to som
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Cryptanalytic aspects of Playfair P
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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
Page 161 and 162: Overlapping will obviously affect t
Page 163 and 164: Table 10.6 The Hagelin cipher machi
Page 165 and 166: 11 Beyond the Enigma The SZ42: a pr
Page 167 and 168: Description of the SZ42 machine The
Page 169 and 170: P1 41 43 Z1 (4) the five bits from
Page 171 and 172: which is approximately 1.6�10 19
Page 173 and 174: 12 Public key cryptography Historic
Page 175 and 176: part of the story of what has been
Page 177 and 178: Public key cryptography 165 graphic
Page 179 and 180: (4) Now (k x ) y �(k y ) x �m x
Page 181 and 182: Public key cryptography 169 Despite
Page 183 and 184: If the cryptographer were prepared
Page 185 and 186: number of tests increased by a fact
Page 187 and 188: In this case the remainder is 4, no
Page 189 and 190: key, d, we use the Euclidean Algori
Page 191 and 192: the decipher key, d, is used instea
Page 193 and 194: and and, since Table 13.1 n N�2 n
Page 195 and 196: The Data Encryption Standard (DES)
Page 197 and 198: the message were known to relate to
Page 199 and 200: whereas in block cipher systems, su
Page 201 and 202: (1) X precedes M with information w
Page 203 and 204: ways, since choosing to pair, say,
Page 205 and 206: For example; if someone claims that
Page 207 and 208: depth, mentioned in Chapter 3. If w
Page 209 and 210: M9 Combining two biased streams of
Page 211: and so 2�4�6�12 �(4095)�4
Page 215 and 216: the lettersatsetting2ofthewheel,and
Page 217 and 218: M16 Probability of a ‘depth’ in
Page 219 and 220: (2) In how many ways can N be repre
Page 221 and 222: Chapter 13 M21 (Rate of increase of
Page 223 and 224: Multiply each of these by M: M, 2M,
Page 225 and 226: If x 3 �0 divide x 2 by x 3 to gi
Page 227 and 228: then k�[ log 2 n], where [z] deno
Page 229 and 230: Mathematical aspects 217 problem un
Page 231 and 232: 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
Page 237 and 238: Chapter 8 8.1 (Recurrences of order
Page 239 and 240: columns in each case, are full of c
Page 241 and 242: Chapter 11 11.1 (Pin-setting errors
Page 243 and 244: [2.4] Moroney, M.J.: Facts from Fig
Page 245 and 246: [10.3] Almost any elementary book o
Page 247 and 248: Name index Adelman, L. 171, 234 And
Page 249 and 250: Subject index Abwehr Enigma 124, 13
Page 251 and 252: active pin 136 cage:‘good’ 141;
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Code and ciphers: Julius Caesar, the Enigma and the internet

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