Lecture 14 Outline
Reading: §10.2.5–10.3, F
Due: Homework 3, due on May 9, 2018 at 11:59pm; Lab 2, due on May 7, 2018 at 11:59pm
- Product ciphers: AES
- Public-Key Cryptography
- Basic idea: 2 keys, one private, one public
- Cryptosystem must satisfy:
- Given public key, computationally infeasible to get private key;
- Cipher withstands chosen plaintext attack;
- Encryption, decryption computationally feasible (note: commutativity not required)
- Benefits: can give confidentiality or authentication or both
- Use of public key cryptosystem
- Normally used as key interchange system to exchange secret keys (cheap)
- Then use secret key system (too expensive to use public key cryptosystem for this)
- RSA
- Provides both authenticity and confidentiality
- Go through algorithm:
Idea: C = M^{e} mod n, M = C^{d} mod n, with ed mod φ(n) = 1
Public key is (e, n); private key is d. Choose n = pq; then φ(n) = (p−1)(q−1).
- 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 = M^{e} mod n = 2^{11} mod 35 = 2048 mod 35 = 18, and M = C^{d} mod n = 18^{11} mod 35 = 2.
- 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
So: C = 1704^{71} mod 3233 = 3106; … = 3106 0100 0931 2691 1984 2927