math470 - SOLUTIONS MANUAL for INTRODUCTION TO CRYPTOGRAPHY...

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SOLUTIONS MANUAL for INTRODUCTION TO CRYPTOGRAPHY with Coding Theory, 2nd edition Wade Trappe Wireless Information Network Laboratory and the Electrical and Computer Engineering Department Rutgers University Lawrence C. Washington Department of Mathematics University of Maryland August 26, 2005
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Contents Exercises Chapter 2 - Exercises 1 Chapter 3 - Exercises 6 Chapter 4 - Exercises 14 Chapter 5 - Exercises 17 Chapter 6 - Exercises 19 Chapter 7 - Exercises 23 Chapter 8 - Exercises 25 Chapter 9 - Exercises 27 Chapter 10 - Exercises 28 Chapter 11 - Exercises 29 Chapter 12 - Exercises 31 Chapter 13 - Exercises 33 Chapter 14 - Exercises 34 Chapter 15 - Exercises 36 Chapter 16 - Exercises 40 Chapter 17 - Exercises 44 Chapter 18 - Exercises 46 -2
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-1 Chapter 19 - Exercises 51 Mathematica problems Chapter 2 52 Chapter 3 63 Chapter 6 66 Chapter 7 72 Chapter 8 74 Chapter 9 75 Chapter 12 78 Chapter 16 79 Chapter 18 81 Maple problems Chapter 2 84 Chapter 3 98 Chapter 6 102 Chapter 7 109 Chapter 8 112 Chapter 9 113 Chapter 12 116 Chapter 16 118 Chapter 18 121
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0 MATLAB problems Chapter 2 124 Chapter 3 147 Chapter 6 151 Chapter 7 161 Chapter 8 164 Chapter 9 165 Chapter 12 167 Chapter 16 169 Chapter 18 174
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Chapter 2 - Exercises 1. Among the shifts of EVIRE , there are two words: arena and river . Therefore, Anthony cannot determine where to meet Caesar. 2. The inverse of 9 mod 26 is 3. Therefore, the decryption function is x =3( y - 2) = 3 y - 2 (mod 26). Now simply decrypt letter by letter as follows. U = 20 so decrypt U by calculating 3 * 20 - 6 (mod 26) = 2, and so on. The decrypted message is ’cat’. 3. Changing the plaintext to numbers yields 7, 14, 22, 0, 17, 4, 24, 14, 20. Applying 5 x +7 to each yields 5 · 7+7 =42 16 (mod 26), 5 · 14+7 = 77 25, etc. Changing back to letters yields QZNHOBXZD as the ciphertext. 4. Let mx + n be the encryption function. Since h = 7 and N =13 ,we have m · 7+ n 13 (mod 26). Using the second letters yields m · 0+ n 14. Therefore n = 14. The Frst congruence now yields 7 m ≡- 1 (mod 26). This yields m = 11. The encryption function is therefore 11 x +14. 5. Let the decryption function be x = ay + b . The Frst letters tell us that 7 a · 2+ b (mod 26). The second letters tell us that 0 a · 17 + b .Subtracting yields 7 a · ( - 15) 11 a . Since 11 - 1 19 (mod 26), we have a 19 · 7 3 (mod 26). The Frst congruence now tells us that 7 3 · b ,so b = 1. The decryption function is therefore x 3 y + 1. Applying this to CRWWZ yields happy for the plaintext. 6. Let mx + n be one a±ne function and ax + b be another. Applying the Frst then the second yields the function a ( mx + n )+ b =( am ) x +( an + b ), which is an a±ne function. Therefore, successively encrypting with two a±ne functions is the same as encrypting with a single a±ne function. There is therefore no advantage of doing double encryption in this case. (Technical point: Since gcd( a, 26) = 1 and gcd( m, 26) = 1, it follows that gcd( am, 26) = 1, so the a±ne function we obtained is still of the required form.) 7. ²or an a±ne cipher mx + n (mod 27), we must have gcd(27 ,m )=1 , and we can always take 1 m 27. So we must exclude all multiples of 3, which leaves 18 possibilities for m . All 27 values of n are possible, so we have 18 · 27 = 486 keys. When we work mod 29, all values 1
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This note was uploaded on 02/14/2012 for the course MATH 470 taught by Professor Staff during the Spring '08 term at Texas A&M.

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math470 - SOLUTIONS MANUAL for INTRODUCTION TO CRYPTOGRAPHY...

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