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Modular Arithmetic & Cryptography

Modular Arithmetic & Cryptography. CSC2110 Tutorial 8 Darek Yung. Outline. Quick Review Examples Q & A. Quick Review. Prime Modular Arithmetic Multiplicative Inverse Turing’s Code RSA. Prime. If p is a prime, GCD(a, p) = 1 unless a is multiple of p

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Modular Arithmetic & Cryptography

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  1. Modular Arithmetic & Cryptography CSC2110 Tutorial 8 Darek Yung

  2. Outline • Quick Review • Examples • Q & A

  3. Quick Review • Prime • Modular Arithmetic • Multiplicative Inverse • Turing’s Code • RSA

  4. Prime • If p is a prime, GCD(a, p) = 1 unless a is multiple of p • If p is a prime, p | a1 * a2 * … * aN implies p | ai for some i • Every natural number n > 1 has a unique prime factorization

  5. Modular Arithmetic

  6. Multiplicative Inverse

  7. Multiplicative Inverse • Multiplicative inverse can be calculated by • Repeated squaring • Integer linear combination

  8. Turing’ Code (Version 1) • Secret: key k • Encryption: m’ = m * k • Decryption: m’ / k = (m * k) / k = m • Attacked by GCD(m1’, m2’) = GCD(m1k, m2k) = k

  9. Turing’ Code (Version 2) • Public: large prime p • Secret: key k, k < p • Encryption: m’ = rem(mk, p), m < p • Decryption: • multiplicative inverse of k (mod p) k’ • m’  mk (mod p) • m’k’  mkk’ (mod p)  m (mod p) • Plain text attack • Given m’, m, p • multiplicative inverse of m (mod p) m’’ • m’m’’  kmm’’ (mod p)  k (mod p)

  10. RSA • Generate two large primes p, q • Let n = pq, T = (p-1)(q-1) • Select e such that GCD(e,T) = 1 • Let d = multiplicative inverse of e (mod T)

  11. RSA • Public: n, public key e • Secret: private key d • Encryption: m’ = me (mod n), m < n • Decryption: (m’)d  med  m (mod n)

  12. Examples • Q20: False • Q21: False, modular arithmetic can be applied to INTEGER a, b • Q22: True, sum of digits is divisible by 9

  13. Examples • Q23: True, difference between sum of even digits and sum of odd digits is divisible by 11 • Q24: False, either of a, a’ can be negative. e.g. n = 5, a = 3, a’ = -3  k = -2 • Q25: True

  14. Examples • Q26: False, c may be multiple of n • Q27: False, k may be multiple of p

  15. Examples • Q28: True • Q29: False, k may be multiple of p • Q30: False, k may be multiple of p and there is no multiplicative inverse of k (mod p)

  16. Examples • Q31: True (false if p1 = p2) • Q32: True • Q33: False, Version 1.0 use simple multiplication

  17. Examples • Q34: True (will show in long question) • Q35: True • Q36: False, unique m with respect to a specific k. A large number of (m, k) pair.

  18. Examples • Q37: True • Q38: False, encrypted by public key • Q39: True • Q40: False, it is just to compute the power of an integer modulo another number

  19. Examples • It is now 5pm on Sunday, What’s the time and the day of the week after 17965 hours • 17965 = 758(24) + 13 • 758 = 108(7) + 2  Tuesday • 13 + (5+12) = 30 = 1(24) + 6  6am  Wed

  20. Examples • Calculate 579572 (mod 21) by repeated squaring • 72 = 64 + 8 • 5795  20 (mod 21) • 57952 1 (mod 21)  57954 57958 579516 1 (mod 21) 579532 579564 1 (mod 21) • 579572 579564 * 57958 20 (mod 21)

  21. Examples • Calculate multiplicative inverse of 47981 (mod 95963) • 95963 = 2(47981) + 1  1(95963) -2(47981) = 1  -2(47981)  1 (mod 95963)  required multiplicative inverse = -2

  22. Examples • Prove ac  bc (mod p) and GCD(c, p) = 1 imply a  b (mod p) for prime p • Since GCD(c, p) = 1, there exists c’ = multiplicative inverse of c (mod p) • ac  bc (mod p)  acc’  bcc’ (mod p)  a  b (mod p)

  23. Examples • Given a  b (mod p) and b  c (mod p) implies a  c (mod p) • And a  b (mod p) implies a+cb+c (mod p) • Prove a  b (mod p) and c  d (mod p) imply a + c  b + d (mod p) for prime p

  24. Examples • Prove a  b (mod p) and c  d (mod p) imply a + c  b + d (mod p) for prime p • a  b (mod p)  a + c  b + c (mod p) • c  d (mod p)  c + b  d + b (mod p)  a + c  b + c  c + b  d + b (mod p)  a + c  b + d (mod p)

  25. Examples • Explain why a number written in decimal is divisible by 9 if and only if the sum of its digits is a multiple of 9 • 10  1 (mod 9)  10k 1k  1 (mod 9) • All decimal number, d, can be written in form: dk(10k) + dk-1(10k-1) + … + d1(101) +d0  d  dk + dk-1 + … + d0 (mod 9)  divisible by 9 if and only if the sum of its digits is a multiple of 9

  26. Examples • Explain why a number written in decimal is divisible by 11 if and only if the difference between the sum of its odd digits and sum of its even digits is a multiple of 11 • 10  -1 (mod 11)  10k (-1)k (mod 11) • All decimal number, d, can be written in form: dk(10k) + dk-1(10k-1) + … + d1(101) +d0  d  (-1)kdk + (-1)k-1dk-1 + … + (-1)0d0 (mod 11)  d  dk - dk-1 + dk-2… - d1 + d0 (mod 11)  divisible by 11 if and only if the difference between the sum of its odd digits and sum of its even digits is a multiple of 11

  27. Examples • Under Turing’s Code (Version 1.0), given key k = 47 • Encrypt message m1 = 569 • Encrypt message m2 = 751 • Get k by having m1 and m2

  28. Example • Encrypt message m1 = 569 • m1’ = k * m1 = 26743 • Encrypt message m2 = 751 • m2’ = k * m2 = 35297 • How to crack?

  29. Examples • GCD(m1’, m2’) = GCD(m1k, m2k) = k • 35297 = 1(26743) + 8554 • 26743 = 3(8554) + 1081 • 8554 = 7(1081) + 987 • 1081 = 1(987) + 94 • 987 = 10(94) + 47 • 94 = 2(47) + 0  GCD(35297, 26743) = 47 = k

  30. Examples • Under Turing’s Code (Version 2.0), given prime p = 89, key k =48 • Encrypt message m = 78 • Compute multiplicative inverse of m (mod p) • Show how to perform plain-text attack

  31. Examples • Encrypt message m = 78 • m’ = rem (mk, p) = rem (3744, 89) = 6 • Compute m’’ = multiplicative inverse of m (mod p) • 89 = 78 + 11  11 = 89 - 78 • 78 = 7(11) + 1  1 =78 – 7(11)  1 = 8(78) – 7(89)  m’’ = 8

  32. Examples • Plain-text attack: given m’, m and p • m’  m * k (mod p)  m’’ * m’  m’’ * m * k  k (mod p) • m’’ * m’ = 6 * 8  48 (mod p)  k = 48

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