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Exam1-Sol - CIS 3362 Test#1 Classical Cryptographic Schemes...

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CIS 3362 Test #1: Classical Cryptographic Schemes Date: 9/20/2010 SOLUTIONS 1) (10 pts) Encrypt “DOLPHINS” using the affine cipher with the encryption function f(x) = (15x+ 7) mod 26. f(D) = f(3) = 15(3) + 7 = 52 mod 26 = 0 = A f(O) = f(14) = 15(14) + 7 = 217 mod 26 = 9 = J f(L) = f(11) = 15(11) + 7 = 172 mod 26 = 16 = Q f(P) = f(15) = 15(15) + 7 = 232 mod 26 = 24 = Y f(H) = f(7) = 15(7) + 7 = 112 mod 26 = 8 = I f(I) = f(8) = 15(8) + 7 = 127 mod 26 = 23 = X f(N) = f(13) = 15(13) + 7 = 202 mod 26 = 20 = U f(S) = f(18) = 15(18) + 7 = 277 mod 26 = 17 = R Answer: AJQYIXUR Grading: -1 for first two mistakes, -2 for third mistake, -1 for next two mistakes, -2 for sixth mistake, -1 for remaining mistakes.
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2) (10 pts) Determine the corresponding decryption function for the encryption function given in question #1. x = 15f -1 (x) + 7 mod 26 (2 pts) 15f -1 (x) = (x – 7) mod 26 (2 pts) 7(15f -1 (x)) = 7(x – 7) mod 26 (3 pts) f -1 (x) = (7x – 49) mod 26 (1 pt) f -1 (x) = (7x + 3) mod 26 (2 pts)
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3) (15 pts) Determine the value of 31 -1 mod 86. 86 = 2 x 31 + 24 (1 pt) 31 = 1 x 24 + 7 (1 pt) 24 = 3 x 7 + 3 (1 pt) 7 = 2 x 3 + 1 (1 pt) 7 – 2 x 3 = 1 (2 pts) 7 – 2(24 – 3 x 7) = 1 7 – 2 x 24 + 6 x 7 = 1 7 x 7 – 2 x 24 = 1 (2 pts) 7(31 – 24) = 2 x 24 = 1 7 x 31 – 7 x 24 – 2 x 24 = 1 7 x 31 – 9 x 24 = 1 (3 pts) 7 x 31 – 9(86 – 2 x 31) = 1 7 x 31 – 9 x 86 + 18 x 31 = 1
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