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. (AZ = 0 – 25, and az = 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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 Fall '08
 Staff
 Information Security, Cryptography, Encryption, pts, cipher text, Hill cipher, Double Hill Cipher

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