* Reading*: text, §9.3–9.4

- Greetings and felicitations!
- Puzzle of the day

- 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 PKC 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

Proof:*M*^{ϕ(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). - 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*C*= (1704)^{71}mod 3233 = 3106; etc. = 3106 0100 0931 2691 1984 2927

- Cryptographic Checksums
- Function
*y*=*h*(*x*): easy to compute*y*given*x*; computationally infeasible to compute*x*given*y* - Variant: given
*x*and*y*, computationally infeasible to find a second*x*′ such that*y*=*h*(*x*′) - Keyed vs. keyless

- Function

Version of February 21, 2006 at 9:45 PM

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