C 2 C E T ANSWERS TO QUESTIONS

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2 8 The Caesar cipher involves replacing each letter of the alphabet with. the letter standing k places further down the alphabet for k in the. range 1 through 25, 2 9 A monoalphabetic substitution cipher maps a plaintext alphabet to a. ciphertext alphabet so that each letter of the plaintext alphabet maps. to a single unique letter of the ciphertext alphabet. 2 10 The Playfair algorithm is based on the use of a 5 5 matrix of. letters constructed using a keyword Plaintext is encrypted two letters at. a time using this matrix, 2 11 A polyalphabetic substitution cipher uses a separate. monoalphabetic substitution cipher for each successive letter of. plaintext depending on a key, 2 12 1 There is the practical problem of making large quantities of random. keys Any heavily used system might require millions of random. characters on a regular basis Supplying truly random characters in. this volume is a significant task, 2 Even more daunting is the problem of key distribution and. protection For every message to be sent a key of equal length is. needed by both sender and receiver Thus a mammoth key. distribution problem exists, 2 13 A transposition cipher involves a permutation of the plaintext.
2 14 Steganography involves concealing the existence of a message. ANSWERS TO PROBLEMS, 2 1 a No A change in the value of b shifts the relationship between. plaintext letters and ciphertext letters to the left or right uniformly. so that if the mapping is one to one it remains one to one. b 2 4 6 8 10 12 13 14 16 18 20 22 24 Any value of a larger. than 25 is equivalent to a mod 26, c The values of a and 26 must have no common positive integer factor. other than 1 This is equivalent to saying that a and 26 are relatively. prime or that the greatest common divisor of a and 26 is 1 To see. this first note that E a p E a q 0 p q 26 if and only if. a p q is divisible by 26 1 Suppose that a and 26 are relatively. prime Then a p q is not divisible by 26 because there is no way. to reduce the fraction a 26 and p q is less than 26 2 Suppose. Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. that a and 26 have a common factor k 1 Then E a p E a q if. 2 2 There are 12 allowable values of a 1 3 5 7 9 11 15 17 19 21 23. 25 There are 26 allowable values of b from 0 through 25 Thus the. total number of distinct affine Caesar ciphers is 12 26 312. 2 3 Assume that the most frequent plaintext letter is e and the second most. frequent letter is t Note that the numerical values are e 4 B 1 t. 19 U 20 Then we have the following equations,1 4a b mod 26. 20 19a b mod 26, Thus 19 15a mod 26 By trial and error we solve a 3. Then 1 12 b mod 26 By observation b 15, 2 4 A good glass in the Bishop s hostel in the Devil s seat twenty one.
degrees and thirteen minutes northeast and by north main branch. seventh limb east side shoot from the left eye of the death s head a. bee line from the tree through the shot fifty feet out from The Gold. Bug by Edgar Allan Poe, 2 5 a The first letter t corresponds to A the second letter h corresponds to. B e is C s is D and so on Second and subsequent occurrences of a. letter in the key sentence are ignored The result,ciphertext SIDKHKDM AF HCRKIABIE SHIMC KD LFEAILA. plaintext basilisk to leviathan blake is contact, b It is a monoalphabetic cipher and so easily breakable. c The last sentence may not contain all the letters of the alphabet If. the first sentence is used the second and subsequent sentences may. also be used until all 26 letters are encountered, 2 6 The cipher refers to the words in the page of a book The first entry. 534 refers to page 534 The second entry C2 refers to column two. The remaining numbers are words in that column The names DOUGLAS. and BIRLSTONE are simply words that do not appear on that page. Elementary from The Valley of Fear by Sir Arthur Conan Doyle. Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. 2 8 10 7 9 6 3 1 4 5,C R Y P T O G A H I,B E A T T H E T H I.
R D P I L L A R F R,O M T H E L E F T O,U T S I D E T H E L. Y C E U M T H E A T,R E T O N I G H T A,T S E V E N I F Y O. U A R E D I S T R U,S T F U L B R I N G,T W O F R I E N D S. 4 2 8 10 5 6 3 7 1 9,N E T W O R K S C U,T R F H E H F T I N. B R O U Y R T U S T,E A E T H G I S R E,H F T E A T Y R N D.
I R O L T A O U G S,H L L E T I N I B I,T I H I U O V E U F. E D M T C E S A T W,T L E D M N E D L R,A P T S E T E R F O. ISRNG BUTLF RRAFR LIDLP FTIYO NVSEE TBEHI HTETA,EYHAT TUCME HRGTA IOENT TUSRU IEADR FOETO LHMET. NTEDS IFWRO HUTEL EITDS, b The two matrices are used in reverse order First the ciphertext is. laid out in columns in the second matrix taking into account the. order dictated by the second memory word Then the contents of. the second matrix are read left to right top to bottom and laid out in. columns in the first matrix taking into account the order dictated by. the first memory word The plaintext is then read left to right top to. c Although this is a weak method it may have use with time sensitive. information and an adversary without immediate access to good. cryptanalysis e g tactical use Plus it doesn t require anything. more than paper and pencil and can be easily remembered. 2 8 SPUTNIK, Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695.
2 9 PT BOAT ONE OWE NINE LOST IN ACTION IN BLACKETT STRAIT TWO. MILES SW MERESU COVE X CREW OF TWELVE X REQUEST ANY. INFORMATION,F H I J K M,G H I J K L,2 11 a UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZ. b UZTBDLGZPNNWLGTGTUEROVLDBDUHFPERHWQSRZ, c A cyclic rotation of rows and or columns leads to equivalent. substitutions In this case the matrix for part a of this problem is. obtained from the matrix of Problem 2 10a by rotating the columns. by one step and the rows by three steps,2 12 a 25 284. b Given any 5x5 configuration any of the four row rotations is. equivalent for a total of five equivalent configurations For each of. these five configurations any of the four column rotations is. equivalent So each configuration in fact represents 25 equivalent. configurations Thus the total number of unique keys is 25 25. 2 13 A mixed Caesar cipher The amount of shift is determined by the. keyword which determines the placement of letters in the matrix. Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. 2 14 a We need an even number of letters so append a q to the end. of the message Then convert the letters into the corresponding. alphabetic positions,M e e t m e a t t h e u s u a l. 13 5 5 20 13 5 1 20 20 8 5 21 19 21 1 12,P l a c e a t t e n r a t h e r.
16 12 1 3 5 1 20 20 5 14 18 1 20 8 5 18,T h a n e i g h t o c l o c k q. 20 8 1 14 5 9 7 8 20 15 3 12 15 3 11 17, The calculations proceed two letters at a time The first pair. 1 9 4 13 mod 26 137 mod 26 7,C2 5 7 5 100 22, The first two ciphertext characters are alphabetic positions 7 and 22. which correspond to GV The complete ciphertext,GVUIGVKODZYPUHEKJHUZWFZFWSJSDZMUDZMYCJQMFWWUQRKR. b We first perform a matrix inversion Note that the determinate of the. encryption matrix is 9 7 4 5 43 Using the matrix,inversion formula from the book.
9 4 1 7 4 7 4 161 92 5 12,mod 26 23 mod 26 mod 26,5 7 43 5 9 5 9 115 9 15 25. Here we used the fact that 43 1 23 in Z26 Once the inverse. matrix has been determined decryption can proceed Source. Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. 2 15 Consider the matrix K with elements kij to consist of the set of column. vectors Kj where, The ciphertext of the following chosen plaintext n grams reveals the. columns of K,B A A A A K1,A B A A A K2,A A A A B Kn. 2 16 a 7 134,f 24 132 1 13,2 17 a 80 10 mod 26 18,b 1 9 5 7 2 1 22 4 2 22 9 1 2 2 1. 5 7 4 mod 26,45 14 176 198 4 140 mod 26,107 mod 26 23.
2 18 We label the matrices as A and B respectively. a det A 44 3 mod 26 15,det A 1 7 using Table E 1 of Appendix E. cof A cof A,A 1 det A 1,cof A cof A,22 3 154 21 24 5. 7 mod 26 mod 26,1 2 7 14 19 14, Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. b det B 6 16 15 24 10 20 1 13 17,1 16 20 10 17 6 15 24 13 mod 26. 1440 4800 221 320 1020 4680 mod 26,441 mod 26 25,We use the formulas from Appendix E.
b ij mod 26 17 cof K mod 26,b11 25mod 26 16 15 10 17 25mod 26 5100 mod 26 8. b12 25mod 26 24 15 1 17 25mod 26 8575mod 26 5,b13 25mod 26 24 10 1 16 25mod 26 5600 mod 26 10. b21 25mod 26 13 15 10 20 25mod 26 125mod 26 21,b22 25mod 26 6 15 1 20 25mod 26 1750 mod 26 8. b23 25mod 26 6 10 1 13 25mod 26 1175mod 26 21,b31 25mod 26 13 17 16 20 25mod 26 2475mod 26 21. b32 25mod 26 6 17 24 20 25mod 26 9450 mod 26 12,b33 25mod 26 6 16 24 13 25mod 26 5400 mod 26 8.
B 1 21 8 21,2 19 key legleglegle,plaintext explanation. ciphertext PBVWETLXOZR, Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. s e n d m o r e m o n e y,18 4 13 3 12 14 17 4 12 14 13 4 24. 9 0 1 7 23 15 21 14 11 11 2 8 9,1 4 14 10 9 3 12 18 23 25 15 12 7. B E C K J D M S X Z P M H,c a s h n o t n e e d e d.
2 0 18 7 13 14 19 13 4 4 3 4 3,25 4 22 3 22 15 19 5 19 21 12 8 4. 1 4 14 10 9 3 12 18 23 25 15 12 7,B E C K J D M S X Z P M H. 2 21 your package ready Friday 21st room three Please destroy this. immediately, Full file at http AplusTestbank eu Solution Manual for Cryptography and Network Security Principles and Practice 6 E 6th Edition 133354695. 9 CHAPTER 2 CLASSICAL ENCRYPTION TECHNIQUES ANSWERS TO QUESTIONS 2 1 Plaintext encryption algorithm secret key ciphertext decryption algorithm 2 2 Permutation and substitution 2 3 One key for symmetric ciphers two keys for asymmetric ciphers

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