Solution: Modular Arithmetic and Chapter
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
Contents
Exercises
Chapter 2 - Exercises Chapter 3 - Exercises Chapter 4 - Exercises Chapter 5 - Exercises Chapter 6 - Exercises Chapter 7 - Exercises Chapter 8 - Exercises Chapter 9 - Exercises Chapter 10 - Exercises Chapter 11 - Exercises Chapter 12 - Exercises Chapter 13 - Exercises Chapter 14 - Exercises Chapter 15 - Exercises Chapter 16 - Exercises Chapter 17 - Exercises Chapter 18 - Exercises -2 1 6 14 17 19 23 25 27 28 29 31 33 34 36 40 44 46
-1 Chapter 19 - Exercises 51
Mathematica problems
Chapter 2 Chapter 3 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 12 Chapter 16 Chapter 18 52 63 66 72 74 75 78 79 81
Maple problems
Chapter 2 Chapter 3 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 12 Chapter 16 Chapter 18 84 98 102 109 112 113 116 118 121
0
MATLAB problems
Chapter 2 Chapter 3 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 12 Chapter 16 Chapter 18 124 147 151 161 164 165 167 169 174
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) = 3y − 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 5x + 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
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
Contents
Exercises
Chapter 2 - Exercises Chapter 3 - Exercises Chapter 4 - Exercises Chapter 5 - Exercises Chapter 6 - Exercises Chapter 7 - Exercises Chapter 8 - Exercises Chapter 9 - Exercises Chapter 10 - Exercises Chapter 11 - Exercises Chapter 12 - Exercises Chapter 13 - Exercises Chapter 14 - Exercises Chapter 15 - Exercises Chapter 16 - Exercises Chapter 17 - Exercises Chapter 18 - Exercises -2 1 6 14 17 19 23 25 27 28 29 31 33 34 36 40 44 46
-1 Chapter 19 - Exercises 51
Mathematica problems
Chapter 2 Chapter 3 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 12 Chapter 16 Chapter 18 52 63 66 72 74 75 78 79 81
Maple problems
Chapter 2 Chapter 3 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 12 Chapter 16 Chapter 18 84 98 102 109 112 113 116 118 121
0
MATLAB problems
Chapter 2 Chapter 3 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 12 Chapter 16 Chapter 18 124 147 151 161 164 165 167 169 174
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) = 3y − 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 5x + 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