1. Asymmetric-Key Cryptography
GROUP MEMBER :-
SOURAV SHASHANK
SURAJ YADAV
SONAL RATHI
SUBHAM SINGHAL

2. Asymmetric key cryptography uses two separate keys: one private and one public.
Locking and unlocking in asymmetric-key cryptosystem

3. General idea of asymmetric-key cryptosystem

4. C = f (Kpublic , P) P = g(Kprivate , C)
Plaintext/Ciphertext
Unlike in symmetric-key cryptography, plaintext and ciphertext are treated as integers in asymmetric-key cryptography.
Encryption/Decryption
C = f (Kpublic , P) P = g(Kprivate , C)

5. C M Public Key (asymmetric): Requirements:
1. For every message M, encrypting with public key and then
decrypting resulting ciphertext with matching private key
results in M.
2. Encryption and Decryption can be efficiently applied to M
3. It is impractical to derive decryption key from encryption key.
Sender
Receiver
(encryption)
(public key
of Receiver) C M
(decryption)
(private key
of Receiver)

6. Combining Public/Private Key Systems
Public key encryption is more expensive than symmetric key encryption
For efficiency, combine the two approaches
(1) A B
(2)
Use public key encryption for authentication; once authenticated, transfer a shared secret symmetric key
(2) Use symmetric key for encrypting subsequent data transmissions

7. Rivest­Shamir­Adelman (RSA) Method
Named after the designers: Rivest, Shamir, and Adleman
Public-key cryptosystem and digital signature scheme.
Based on difficulty of factoring large integers
For large primes p & q, n = pq
Public key e and private key d calculated

8. RSA Key Generation
Every participant must generate a Public and Private key:
1. Let p and q be large prime numbers, randomly chosen from the set of all large prime numbers.
2. Compute n = pq.
3. Choose any large integer, d, so that:
GCD( d, ϕ(n)) = (where ϕ(n) = (p­1)(q­1) )
4. Compute e = d-1 (mod ϕ(n)).
5. Publish n and e. Keep p, q and d secret.
Note:
Step 4 can be written as:
Find e so that: e x d = 1 (modulo ϕ(n))
If we can obtain p and q, and we have (n, e), we can find d

9. Rivest­Shamir­Adelman (RSA) Method
Assume A wants to send something confidentially to B:
A takes M, computes C = Me mod n, where (e, n) is B’s public key. Sends C to B
B takes C, finds M = Cd mod n, where (d, n) is B’s private key
+ Confidentiality B A M M C
Me mod n
Cd mod n
(e, n)
(d, n)
Encryption Key for user B
(B’s Public Key)
Decryption Key for user B
(B’s PrivateKey)

10. RSA Method Example: 1. p = 5, q = 11 and n = 55.
(p­1)x(q­1) = 4 x 10 = 40
2. A valid d is 23 since GCD(40, 23) = 1
3. Then e = 7 since:
23 x 7 = 161 modulo 40 = 1
in other words
e = 23-1 (mod 40) = 7

11. Topics discussed in this section:
ELGAMAL CRYPTOSYSTEM
Besides RSA another public-key cryptosystem is ElGamal. ElGamal is based on the discrete logarithm problem discussed.
Topics discussed in this section:
ElGamal Cryptosystem
Procedure Proof
Analysis

12. Key generation, encryption, and decryption in ElGamal

13. Key Generation

15. Note
The bit-operation complexity of encryption or decryption in ElGamal cryptosystem is polynomial.

16. Proof of ElGamal Cryptosystem

17. Here is a trivial example. Bob chooses p = 11 and e1 = 2
Here is a trivial example. Bob chooses p = 11 and e1 = 2. and d = 3 e2 = e1d = 8. So the public keys are (2, 8, 11) and the private key is 3. Alice chooses r = 4 and calculates C1 and C2 for the plaintext 7.
Bob receives the ciphertexts (5 and 6) and calculates the plaintext.

18. Questions?