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Upcoming: Distinguished Lecturer! Upcoming: Distinguished Lecturer! Lecture: Self-Reference and Ucomputability. Upcoming: Distinguished Lecturer! Lecture: Self-Reference and Ucomputability. Christos Papadimitriou. Upcoming: Distinguished

  1. Is public key crypto possible? We don’t really know. ...but we do it every day!!! RSA (Rivest, Shamir, and Adleman) Pick two large primes p and q . Let N = pq . Choose e relatively prime to ( p − 1 )( q − 1 ) . 1 Compute d = e − 1 mod ( p − 1 )( q − 1 ) . Announce N (= p · q ) and e : K = ( N , e ) is my public key! 1 Typically small, say e = 3.

  2. Is public key crypto possible? We don’t really know. ...but we do it every day!!! RSA (Rivest, Shamir, and Adleman) Pick two large primes p and q . Let N = pq . Choose e relatively prime to ( p − 1 )( q − 1 ) . 1 Compute d = e − 1 mod ( p − 1 )( q − 1 ) . Announce N (= p · q ) and e : K = ( N , e ) is my public key! mod ( x e , N ) . Encoding: 1 Typically small, say e = 3.

  3. Is public key crypto possible? We don’t really know. ...but we do it every day!!! RSA (Rivest, Shamir, and Adleman) Pick two large primes p and q . Let N = pq . Choose e relatively prime to ( p − 1 )( q − 1 ) . 1 Compute d = e − 1 mod ( p − 1 )( q − 1 ) . Announce N (= p · q ) and e : K = ( N , e ) is my public key! mod ( x e , N ) . Encoding: mod ( y d , N ) . Decoding: 1 Typically small, say e = 3.

  4. Is public key crypto possible? We don’t really know. ...but we do it every day!!! RSA (Rivest, Shamir, and Adleman) Pick two large primes p and q . Let N = pq . Choose e relatively prime to ( p − 1 )( q − 1 ) . 1 Compute d = e − 1 mod ( p − 1 )( q − 1 ) . Announce N (= p · q ) and e : K = ( N , e ) is my public key! mod ( x e , N ) . Encoding: mod ( y d , N ) . Decoding: Does D ( E ( m )) = m ed = m mod N ? 1 Typically small, say e = 3.

  5. Is public key crypto possible? We don’t really know. ...but we do it every day!!! RSA (Rivest, Shamir, and Adleman) Pick two large primes p and q . Let N = pq . Choose e relatively prime to ( p − 1 )( q − 1 ) . 1 Compute d = e − 1 mod ( p − 1 )( q − 1 ) . Announce N (= p · q ) and e : K = ( N , e ) is my public key! mod ( x e , N ) . Encoding: mod ( y d , N ) . Decoding: Does D ( E ( m )) = m ed = m mod N ? Yes! 1 Typically small, say e = 3.

  6. Example: p = 7, q = 11.

  7. Example: p = 7, q = 11. N = 77 .

  8. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60

  9. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1.

  10. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60).

  11. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60

  12. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7

  13. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4

  14. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4 7 ( 9 )+ 60 ( − 1 ) = 3

  15. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4 7 ( 9 )+ 60 ( − 1 ) = 3 7 ( − 17 )+ 60 ( 2 ) = 1

  16. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4 7 ( 9 )+ 60 ( − 1 ) = 3 7 ( − 17 )+ 60 ( 2 ) = 1

  17. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4 7 ( 9 )+ 60 ( − 1 ) = 3 7 ( − 17 )+ 60 ( 2 ) = 1 Confirm:

  18. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4 7 ( 9 )+ 60 ( − 1 ) = 3 7 ( − 17 )+ 60 ( 2 ) = 1 Confirm: − 119 + 120 = 1

  19. Example: p = 7, q = 11. N = 77 . ( p − 1 )( q − 1 ) = 60 Choose e = 7, since gcd ( 7 , 60 ) = 1. egcd(7,60). 7 ( 0 )+ 60 ( 1 ) = 60 7 ( 1 )+ 60 ( 0 ) = 7 7 ( − 8 )+ 60 ( 1 ) = 4 7 ( 9 )+ 60 ( − 1 ) = 3 7 ( − 17 )+ 60 ( 2 ) = 1 Confirm: − 119 + 120 = 1 d = e − 1 = − 17 = 43 = ( mod 60 )

  20. Encryption/Decryption Techniques.

  21. Encryption/Decryption Techniques. Public Key: ( 77 , 7 )

  22. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } .

  23. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2!

  24. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 )

  25. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e

  26. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7

  27. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7 ≡ 128 ( mod 77 )

  28. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7 ≡ 128 ( mod 77 ) = 51 ( mod 77 )

  29. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7 ≡ 128 ( mod 77 ) = 51 ( mod 77 ) D ( 51 ) = 51 43 ( mod 77 )

  30. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7 ≡ 128 ( mod 77 ) = 51 ( mod 77 ) D ( 51 ) = 51 43 ( mod 77 ) uh oh!

  31. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7 ≡ 128 ( mod 77 ) = 51 ( mod 77 ) D ( 51 ) = 51 43 ( mod 77 ) uh oh! Obvious way: 43 multiplcations. Ouch.

  32. Encryption/Decryption Techniques. Public Key: ( 77 , 7 ) Message Choices: { 0 ,..., 76 } . Message: 2! E ( 2 ) = 2 e = 2 7 ≡ 128 ( mod 77 ) = 51 ( mod 77 ) D ( 51 ) = 51 43 ( mod 77 ) uh oh! Obvious way: 43 multiplcations. Ouch. In general, O ( N ) multiplications!

  33. Repeated squaring.

  34. Repeated squaring. 51 43

  35. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1

  36. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) .

  37. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of...

  38. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ?

  39. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 )

  40. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 )

  41. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 )

  42. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 )

  43. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 ) 51 16 = ( 51 8 ) ∗ ( 51 8 ) = 53 ∗ 53 = 2809 ≡ 37 ( mod 77 )

  44. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 ) 51 16 = ( 51 8 ) ∗ ( 51 8 ) = 53 ∗ 53 = 2809 ≡ 37 ( mod 77 ) 51 32 = ( 51 16 ) ∗ ( 51 16 ) = 37 ∗ 37 = 1369 ≡ 60 ( mod 77 )

  45. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 ) 51 16 = ( 51 8 ) ∗ ( 51 8 ) = 53 ∗ 53 = 2809 ≡ 37 ( mod 77 ) 51 32 = ( 51 16 ) ∗ ( 51 16 ) = 37 ∗ 37 = 1369 ≡ 60 ( mod 77 ) 5 more multiplications.

  46. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 ) 51 16 = ( 51 8 ) ∗ ( 51 8 ) = 53 ∗ 53 = 2809 ≡ 37 ( mod 77 ) 51 32 = ( 51 16 ) ∗ ( 51 16 ) = 37 ∗ 37 = 1369 ≡ 60 ( mod 77 ) 5 more multiplications. 51 32 · 51 8 · 51 2 · 51 1 = ( 60 ) ∗ ( 53 ) ∗ ( 60 ) ∗ ( 51 ) ≡ 2 ( mod 77 ) .

  47. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 ) 51 16 = ( 51 8 ) ∗ ( 51 8 ) = 53 ∗ 53 = 2809 ≡ 37 ( mod 77 ) 51 32 = ( 51 16 ) ∗ ( 51 16 ) = 37 ∗ 37 = 1369 ≡ 60 ( mod 77 ) 5 more multiplications. 51 32 · 51 8 · 51 2 · 51 1 = ( 60 ) ∗ ( 53 ) ∗ ( 60 ) ∗ ( 51 ) ≡ 2 ( mod 77 ) . Decoding got the message back!

  48. Repeated squaring. 51 43 = 51 32 + 8 + 2 + 1 = 51 32 · 51 8 · 51 2 · 51 1 ( mod 77 ) . 4 multiplications sort of... Need to compute 51 32 ... 51 1 . ? 51 1 ≡ 51 ( mod 77 ) 51 2 = ( 51 ) ∗ ( 51 ) = 2601 ≡ 60 ( mod 77 ) 51 4 = ( 51 2 ) ∗ ( 51 2 ) = 60 ∗ 60 = 3600 ≡ 58 ( mod 77 ) 51 8 = ( 51 4 ) ∗ ( 51 4 ) = 58 ∗ 58 = 3364 ≡ 53 ( mod 77 ) 51 16 = ( 51 8 ) ∗ ( 51 8 ) = 53 ∗ 53 = 2809 ≡ 37 ( mod 77 ) 51 32 = ( 51 16 ) ∗ ( 51 16 ) = 37 ∗ 37 = 1369 ≡ 60 ( mod 77 ) 5 more multiplications. 51 32 · 51 8 · 51 2 · 51 1 = ( 60 ) ∗ ( 53 ) ∗ ( 60 ) ∗ ( 51 ) ≡ 2 ( mod 77 ) . Decoding got the message back! Repeated Squaring took 9 multiplications

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