# Outline for February 22, 2006

1. Greetings and felicitations!
1. Puzzle of the day
2. Public-Key Cryptography
1. Basic idea: 2 keys, one private, one public
2. Cryptosystem must satisfy:
1. Given public key, computationally infeasible to get private key;
2. Cipher withstands chosen plaintext attack;
3. Encryption, decryption computationally feasible [note: commutativity not required]
3. Benefits: can give confidentiality or authentication or both
3. Use of public key cryptosystem
1. Normally used as key interchange system to exchange secret keys (cheap)
2. Then use secret key system (too expensive to use PKC for this)
4. RSA
1. Provides both authenticity and confidentiality
2. Go through algorithm:
Idea: C = Me mod n, M = Cd mod n, with ed mod ϕ(n) = 1
Proof: Mϕ(n) mod n = 1 [by Fermat's theorem as generalized by Euler]; follows immediately from ed mod ϕ(n) = 1
Public key is (e, n); private key is d. Choose n = pq; then ϕ(n) = (p−1)(q−1).
3. Example: p = 5, q = 7; then n = 35, ϕ(n) = (5−1)(7−1) = 24. Pick d = 11. Then ed mod ϕ(n) = 1, so 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.
4. Example: p = 53, q = 61; then n = 3233, ϕ(n) = (53−1)(61−1) = 3120. Pick d = 791. Then e = 71.
To encipher M = RENAISSANCE, use the mapping A = 00, B = 01, ..., Z = 25, ␢ = 26. Then: M = RE NA IS SA NC E␢ = 1704 1300 0818 1800 1302 0426 C = (1704)71 mod 3233 = 3106; etc. = 3106 0100 0931 2691 1984 2927
5. Cryptographic Checksums
1. Function y = h(x): easy to compute y given x; computationally infeasible to compute x given y
2. Variant: given x and y, computationally infeasible to find a second x′ such that y = h(x′)
3. Keyed vs. keyless

Version of February 21, 2006 at 9:45 PM