# lec7-4 - Admin Stuff E-mail Modern Cryptography Lecture 7...

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Modern Cryptography Lecture 7 Yongdae Kim 2 Admin Stuff E-mail Subject should have [5471] in front, e.g. “[5471] Project proposal” CC TA and PostDoc: [email protected], [email protected] Office hours Me: T 1:30 ~ 2:30, Th 10:00 ~ 11:00 (and by appointment) TA: M 1:15 PM ~ 2:15 PM Work on projects Interim Report due: Mar 30 (Firm Deadline: Read instruction) 4th assignment is due: 3/23 9:00 AM. Study Guide Come and talk to me and TA as much as possible. (Google chat is good!) Check Calendar 3 Recap Math… Proof techniques Direct/Indirect proof, Proof by contradiction, Proof by cases, Existential/Universal Proof, Forward/backward reasoning Divisibility: a divides b (a|b) if c such that b = ac GCD, LCM, relatively prime, existence of GCD Eucledean Algorithm d = gcd (a, b) x, y such that d = a x + b y. gcd(a, b) = gcd(a, b + ka) Modular Arithmetic a b (mod m ) iff | a-b iff a = b + mk for some k (mod ), c d (mod ) a+c ( b+d ) (mod ), ac bd (mod ) gcd(a, n) =1 a has an arithmetic inverse modulo n. Counting, probability, cardinality, … 4 Recap (cnt) Security Symmetric Key vs. Public Key, Hash function, MAC, Digital signature, Key management through SKE and PKE, certificate Block Cipher Modes of operation and their properties: ECB, OFB, CFB, CBC, CTR Meet-in-the-middle attack and the Double (triple) DES Feistal Cipher and DES Hash function and MAC Probability and Birthday paradox Merkle-Damgard Construction MD4: design and break MAC

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5 Z n, Z n * The integers modulo n denoted by Z n is the set of integers 0,1,2. ..n-1. Z 12 ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} a b (mod n) if n | a - b Let a Z n , the multiplicative inverse of a is an integer x Z n , s.t. ax 1 (mod n) 5 x ´ 1 (mod 12) x ´ 5 (mod 12) 5 x ´ 1 (mod 14) x ´ 11 (mod 14) a is invertible iff gcd(a,n) = 1 Z n * ={ a Z n | gcd(a,n)=1} Z 12 * ={1, 5, 7, 11}, Z 14 * ={1, 3, 5, 9, 11, 13} If n is a prime then Z n * ={ a Z n | 1 a n-1} 6 CRT Given r integers which are pairwise relatively prime , m 1 , m 2 ,…, m r , then x b 1 (mod m 1 ) x b 2 (mod m 2 ) x b 3 (mod m 3 ) …. x b r (mod m r ) has the unique solution : x = y 1 b 1 M 1 + … + y r b r M r mod M where M = Π m i , M i = M/m i , y i M i 1 (mod m i ).
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## This note was uploaded on 10/21/2011 for the course CSCI 5471 taught by Professor Staff during the Spring '08 term at Minnesota.

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lec7-4 - Admin Stuff E-mail Modern Cryptography Lecture 7...

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