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# Exam1Sol - CIS 3362 Exam#1(Chapters 1-4 Solution Date 1(10...

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CIS 3362 Exam #1 (Chapters 1-4) Solution Date: 9/20/07 1) (10 pts) Encrypt "KNIGHTS" using the affine cipher with the encryption function f(x) = (7x + 17)%26. f(K) = (7x10 + 17)%26 = 9 (J) f(N) = (7x13 + 17)%26 = 4 (E) f(I) = (7x8 + 17)%26 = 21 (V) f(G) = (7x6 + 17)%26 = 7 (H) f(H) = (7x7 + 17)%26 = 14 (O) f(T) = (7x19 + 17)%26 = 20 (U) f(S) = (7x18 + 17)%26 = 13 (N) Answer = JEVHOUN 2) (10 pts) Determine the corresponding decryption function for the encryption function given in question #1. x = (7f -1 (x) + 17) mod 26 7f -1 (x) = (x – 17) mod 26 15*7f -1 (x) = 15(x – 17) mod 26, since 15 is the modular inverse of 7 mod 26. f -1 (x) = (15x – 255) mod 26 f -1 (x) = (15x + 5) mod 26

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3) (15 pts) Determine the value of 29 -1 mod 85. 85 = 2x29 + 27 29 = 1x27 + 2 27 = 13x2 + 1 1 = 27 – 13x2 1 = 27 – 13x(29 – 27) 1 = 27 – 13x29 + 13x27 1 = 14x27 – 13x29 1 = 14x(85 – 2x29) – 13x29 1 = 14x85 – 28x29 – 13x29 1 = 14x85 – 41x29 Consider this equation mod 85 which yields: -41x29 = 1 mod 85 The desired inverse is -41, which maps to 44 mod 85. 4) (6 pts) The Enigma was a rather good encryption machine for its time. However, there were some weaknesses with how it was used that helped the Polish and British crack messages encrypted by it. Name two of these weaknesses. There are other answers, but here are three: The message key was repeated twice at the beginning of every message. German operators often reused "predictable" message keys, such as girlfriends' initials. The Germans sent out a weather message every morning at nearly the same time with the same format.
5) (9 pts) In order to thwart frequency analysis of the substitution cipher, people added several "add-ons" to the usual substitution cipher. Describe three techniques that people used in conjunction with the substitution cipher to attempt to make it more difficult to cryptanalyze with frequency analysis.

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