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Public Key Crytography From: Introduction to Algorithms Cormen, Leiserson and Rivest

Public Key Crytography From: Introduction to Algorithms Cormen, Leiserson and Rivest. Purpose. Privacy: to send encrypted messages over an insecure channel. Authentication: To digitally sign messages. Public Key Cryptography was first introduced by Diffie and Hellman in 1976.

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Public Key Crytography From: Introduction to Algorithms Cormen, Leiserson and Rivest

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  1. Public Key Crytography From: Introduction to Algorithms Cormen, Leiserson and Rivest 95-712 OOP/Java

  2. Purpose Privacy: to send encrypted messages over an insecure channel. Authentication: To digitally sign messages. Public Key Cryptography was first introduced by Diffie and Hellman in 1976. 95-712 OOP/Java

  3. The RSA Public-Key Cryptosystem • Developed by Rivest, Shamir, and Aldeman in 1977 • Since then the field of cryptography has blossomed 95-712 OOP/Java

  4. The Cast of Characters • Eve - always peeking at messages she should not be seeing. • Mallory - always trying to manipulate messages she has no business manipulating. • Bob and Alice - always trying to communicate over insecure channels. • Trent - A trusted third party who tries to help Bob and Alice. 95-712 OOP/Java

  5. The RSA Public-Key Cryptosystem • Select at random two large prime numbers p and q. • These numbers would normally be about 500 digits • in length. • Compute n by the equation n = p X q. • Compute (n) = (p –1) X (q –1) • Select a small odd integer e that is relatively prime to • (n) 95-712 OOP/Java

  6. The RSA Public-Key Cryptosystem 5. Compute d as the multiplicative inverse of e modulo (n). A theorem in number theory asserts that d exists and is uniquely defined. 6. Publish the pair P = (e,n) as the RSA public key. 7. Keep secret the pair S = (d,n) as the RSA secret key. 95-712 OOP/Java

  7. The RSA Public-Key Cryptosystem 8. To encrypt a message M compute C = Me (mod n) 9. To decrypt a message C compute M = Cd (mod n) 95-712 OOP/Java

  8. Key Selection Phase • Select at random two large prime numbers p and q. • These numbers would normally be about 500 digits • in length. • p = 3 q = 11 • Compute n by the equation n = p X q. • n = 33 3. Compute (n) = (p –1) X (q –1) (n) = (2) X (10) = 20 95-712 OOP/Java

  9. Key Selection Phase p = 3 q = 11 n = 33 (n) = 20 4. Select a small odd integer e that is relatively prime to (n) e = 3 95-712 OOP/Java

  10. Key Selection Phase p = 3 q = 11 n = 33 (n) = 20 e = 3 5. Compute d as the multiplicative inverse of e, modulo (n). A theorem in number theory asserts that d exists and is uniquely defined (since e and (n) are relatively prime). We need a d so that ed mod  = 1 Let’s try 1. 3 X 1 mod 20 = 3 mod 20 = 3. Nope. 95-712 OOP/Java

  11. Key Selection Phase p = 3 q = 11 n = 33 (n) = 20 e = 3 We need a d so that ed mod  = 1 Let’s try 2. 3 X 2 mod 20 = 6 mod 20 = 6. Nope. Let’s try 7. 3 X 7 mod 20 = 21 mod 20 = 1. We found it! We need a better way….we’ll see this soon. 95-712 OOP/Java

  12. Key Selection Phase p = 3 q = 11 n = 33 (n) = 20 e = 3 d = 7 6. Publish the pair P = (e,n) as the RSA public key. “Hey everyone, my key pair is 3 and 33” 7. Keep secret the pair S = (d,n) as the RSA secret key. “I’m not telling anyone about 7 and 33!!” 95-712 OOP/Java

  13. Message encoding phase Bob’s public keys are e = 3 n = 33 Alice wants to send the letter ‘d’ to Bob. Suppose that we have a public code where ‘a’ = 0 ‘b’ = 1 ‘c’ = 2 ‘d’ = 3 and so on… Alice’s software knows that 8. To encrypt a message M compute C = Me (mod n) = 33 mod 33 = 27 mod 33 = 27 95-712 OOP/Java

  14. Message decoding phase Bob’s private keys are d = 7 n = 33 Bob receives a 27 from Alice: 9. To decrypt a message C compute M = Cd (mod n) = 277 mod 33 = 10460353203 mod 33 = 3 (which is ‘d’) 95-712 OOP/Java

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