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

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

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General idea of asymmetric-key cryptosystem

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

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5 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. (encryption) (public key of Receiver) C MM (decryption) SenderReceiver (private key of Receiver)

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

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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 7

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RSA Key Generation 8 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)) = 1 (where ϕ (n) = (p1)(q1) ) 4. Compute e = d -1 (mod ϕ (n)). 5. Publish n and e. Keep p, q and d secret. Every participant must generate a Public and Private key: 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

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

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10 RSA Method Example: 1. p = 5, q = 11 and n = 55. (p1)x(q1) = 4 x 10 = 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 = (mod 40) = 7

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

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Key generation, encryption, and decryption in ElGamal C2

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Key Generation

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

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Proof of ElGamal Cryptosystem

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10.17 Here is a trivial example. Bob chooses p = 11 and e 1 = 2. and d = 3 e 2 = e 1 d = 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.

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18 Questions?

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