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