Outline for February 15/20, 2002 Reading: 9-9.3 1. Greetings and Felicitations! 2. Puzzle of the day 3. Classical a. monoalphabetic (simple substitution): f(a) = a + k mod n b. example: Caesar with k = 3, RENAISSANCE UHQDLVVDQFH c. polyalphabetic: Vigenre, fi(a) = (a + ki) mod n d. cryptanalysis: first do index of coincidence to see if it's monoalphabetic or polyalphabetic, then Kasiski method. e. problem: eliminate periodicity of key 4. Long key generation a. Running-key cipher: M=THETREASUREISBURIED; K=THESECONDCIPHERISAN; C=MOILVGOFXTMXZFLZAEQ; wedge is that (plaintext,key) letter pairs are not random (T/T, H/H, E/E, T/S, R/E, A/O, S/N, etc.) b. Perfect secrecy: when the probability of computing the plaintext message is the same whether or not you have the ciphertext c. Only cipher with perfect secrecy: one-time pads; C=AZPR; is that DOIT or DONT? 5. DES 6. Public-Key Cryptography a. Basic idea: 2 keys, one private, one public b. Cryptosystem must satisfy: i. given public key, CI to get private key; ii. cipher withstands chosen plaintext attack; iii. encryption, decryption computationally feasible [note: commutativity not required] c. Benefits: can give confidentiality or authentication or both 7. RSA a. Provides both authenticity and confidentiality b. Go through algorithm: Idea: C = Me mod n, M = Cd mod n, with ed mod f(n) = 1. Proof: Mf(n) mod n = 1 [by Fermat's theorem as generalized by Euler]; follows immediately from ed mod f(n) = 1. Public key is (e, n); private key is d. Choose n = pq; then f(n) = (p-1)(q-1). c. Example: p = 5, q = 7; n = 35, f(n) = (5-1)(7-1) = 24. Pick d = 11. Then de mod f(n) = 1, so choose e = 11. To encipher 2, C = Me mod n = 211 mod 35 = 2048 mod 35 = 18, and M = Cd mod n = 1811 mod 35 = 2. d. Example: p = 53, q = 61, n = 3233, f(n) = (53-1)(61-1) = 3120. Take d = 791; then e = 71. Encipher M = RENAISSANCE: A = 00, B = 01, -, Z = 25, blank = 26. Then: M = RE NA IS SA NC Eblank = 1704 1300 0818 1800 1302 0426 C = (1704)71 mod 3233 = 3106; etc. = 3106 0100 0931 2691 1984 2927