Week 5 Lecture Notes

# Week 5 Lecture Notes - February 4: Lecture 13 1 + 1 = 10 1...

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February 4: Lecture 13 1 + 1 = 10 1 + 1 + 1 = 11 1 1 0 1 0 = 26 x 1 0 1 = 5 1 1 0 1 0 0 1 1 0 1 0 1 0 0 0 0 0 1 0 = 130 ab (mod P) = (a (mod P) x b (mod P))(mod P) a b (mod P) = (a(mod P)) b (mod P) 3 6 (mod 5) = ((3x3)(mod 5) x (3x3)(mod 5) x (3x3)(mod 5))(mod 5) = (4 x 4 x 4)(mod 5) = [(4x4)(mod 5) x (4)(mod 5)] mod 5 = (1x4)(mod 5) = 4 Alice Exchanged Info. Bob A Y B Y A (mod P) = αP Y B (mod P) = β β A (mod P) = K α and β α B (mod P) = K K = Y AB (mod P) = (Y A ) B (mod P) = (Y A (mod P)) B (mod P)

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February 6: Lecture 14 Assignment 3 up until Feb. 12, Tuesday Assignment 2 Answers : up on WebCT once you click on your submission of Assignment 2 Asymmetric Cipher : key used to encrypt different from key used to decrypt Mathematical description of the mechanics of RSA encryption and decryption P, Q = prime numbers N = P x Q e e and N are public key C = M e (mod N) *Everything is given except for M but you cannot solve for M (ONE-WAY FUNCTION) 88 7 (mod 187) = [(88 4
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## This note was uploaded on 04/03/2008 for the course LIGN 17 taught by Professor Kehler during the Winter '08 term at UCSD.

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Week 5 Lecture Notes - February 4: Lecture 13 1 + 1 = 10 1...

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