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FinalExamSol - CIS 3362 Final Exam solution Date 1(10 pts...

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CIS 3362 Final Exam solution Date: 12/5/06 1) (10 pts) Consider the affine function e(x) = (23x + 5) mod 52 for encryption, where both lower case and upper case letters are encoded. (A-Z = 0 – 25, and a-z = 26 – 51.) Determine the corresponding decryption function d(x). Show all of your steps and justify your work. x = 23d(x) + 5 mod 52 23d(x) = (x – 5) mod 52 Use the Extended Euclidean Algorithm to determine 23 -1 mod 52: 52 = 2x23 + 6 52 – 2x23 = 6 23 = 3x6 + 5 23 – 3x6 = 5 6 = 1x5 + 1 6 – 5 = 1, so 6 – (23 – 3x6) = 1 4x6 – 23 = 1 4x(52 – 2x23) – 23 = 1 4x52 – 8x23 – 23 = 1 4x52 – 9x23 = 1, So 23 -1 ≡ -9 mod 52 ≡ 43 mod 52 23d(x) = (x – 5) mod 52 (43)(23)d(x) = 43(x – 5) mod 52 d(x) = 43x – 215 (mod 52) d(x) = 43x + 45 (mod 52) 2) (10 pts) Alice and Bob have just learned about product ciphers and have decided to utilize the idea in implementing the "Double Hill Cipher". Instead of having one square matrix as the key, they would have two separate square matrices, M and N as the key. To encrypt a block of letters denoted by p, the product c = MNp would be calculated. To recover the plaintext, p = N -1 M -1 c would be calculated. Is the "Double Hill Cipher" more secure than the regular Hill Cipher? Give proper justification for your answer.
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