1. Public Key Encryption

2. Why public key cryptology?
Solving key distribution
Two parties already share a key
Use a key distribution center (KDC)
Digital signature

3. Confidentiality

4. Authentication

5. Confidentiality & Authentication

6. Requirement for public-key cryptology
Proposed by Diffie & Hellman (1976)
It is computationally easy for a party B to generate a pair of public key KUb and private key KRb
It is computationally easy for a sender A, knowing the public key and the message M, to generate ciphertext
C = EKUb (M)
It is computationally easy for a receiver B to decrypt the ciphertext using the private key
M = DKRb (C) = DKRb (EKUb (M))
It is computationally infeasible for opponent, knowing public key KUb and ciphertext C to recover M
It is computationally infeasible for opponent, knowing public key Kub to determine the private key KRb

7. Knapsack Algorithm Based on the knapsack problem
What is the knapsack problem?
Determining which objects are in the knapsack
Proposed by Ralph Merkle (1978)
Message: n bits
Cargo vector a = (a1, a2, …, an), ai integer
Plaintext message block x = (x1, x2, …, xn), xi binary
Ciphertext S = a. x =  (ai . xi)
Encryption is easy
Decryption: recover x using S and a
It is difficult

8. Example a = (1, 3, 2, 5) S = 3 What is x? x = 1010 or x = 0100
Requirement 1: Unique inverse for each value of S
Requirement 2: Decryption is hard in general, BUT easy if specified knowledge is available

9. Merke’s method
Combine easy superincreasing knapsack with difficult knapsack
How?
Choose a random easy knapsack vector a’ with n elements
Select 2 integers w and m such that
m >  ai’
gcd (w, m) = 1
Construct a hard knapsack vector a, where
a = w . a’ mod m

10. Knapsack algorithm Choose a superincreasing vector a’ (private)
Choose an integer m larger than  ai’ (private)
Choose an integer w relatively prime to m (private)
Calculate w-1 the inverse of w (modulo m) (private)
Calculate a = w. a’ (mod m) (public)
Private key = {w-1, m, a’}
Public key = {a}

11. Knapsack algorithm cont…
Encryption
S = a . x
Decryption
Define S’ = w-1 S (mod m)
S’ = w-1 (w a’) x (mod m)
= a’. x

12. RSA (Rivest-Shamir-Adleman) Algorithm
Block cipher, plaintext and ciphertext are integers between 0 and (n-1)
Plaintext is encrypted in block, each block having a binary value less than some number n
Based on number theory and modular arithmetic
Depends on difficulty of determining prime factors of large numbers
Two keys are used (d,e)
Any of the key can be public
Remaining one should be private

14. Algorithm Description
C = Me mod n
M = Cd mod n = Med mod n
Both sender and receiver must know the value of n
The sender know the value of e
Only the receiver know the value of d
Public key: {e, n}
Private key: {d, n}

15. C = Me mod n M = Cd mod n = Med mod n
Knowing d, determining M is simple
Else, factoring Me is difficult

16. Requirements to be public-key encryption
Is it possible to find values e, d, n such that
Med = M mod n for all M < n
Is it easy to calculate Me and Cd for all M < n?
Is it infeasible to determine d given e and n?
The answer to these questions are explained on the white board.

17. Detail Algorithm Choose two prime numbers p and q (private)
Calculate n = pq (public)
Calculate d, with gcd(Φ(n), d) = 1, 1 < d < Φ(n) (private)
Chosen e = d-1 mod Φ(n) (public)
Public key: {e, n}
Private key: {d, n}

18. Two Issues Encryption and Decryption Key generation
The efficiency of exponentiation computation
Key generation
Determining two prime numbers p and q
Selecting either e or d and calculating the other

19. Efficient Exponentiation
Procedure exponentiation (a, m, n)
d := 1
for i = (k-1) down to 0 do
d = d2 mod n
if bi = 1 then d = d . a mod n
Return d

20. Demo RSA www-cs-students.stanford.edu/~tjw/jsbn/rsa2.html
islab.oregonstate.edu/koc/ece575/02Project/Mor/