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CS 312: Algorithm Analysis

This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License . CS 312: Algorithm Analysis. Lecture #4: Primality Testing, GCD. Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, Sean Warnick. Announcements. HW #2 Due Now

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CS 312: Algorithm Analysis

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  1. This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License. CS 312: Algorithm Analysis Lecture #4: Primality Testing, GCD Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, Sean Warnick

  2. Announcements • HW #2 Due Now • Homework: Required to show your work • Project #1 • Today we’ll work through the rest of the math • Early: this Wednesday, 1/16 • Due: Friday, 1/18 • Bonus help session on Proof by Induction • Tuesday, 4pm, 1066 TMCB

  3. Practice • Key points: • Represent exponent in binary • Break up the problem into factors (one per binary digit) • Use the substitution rule

  4. Objectives • Part 1: • Introduce Fermat’s Little Theorem • Understand and analyze the Fermat primality tester • Part 2: • Discuss GCD and Multiplicative Inverses, modulo N • Prepare to Introduce Public Key Cryptography • This adds up to a lot of ideas!

  5. Part 1: Primality Testing

  6. Fermat’s Little Theorem If p is prime, then a p-1  1 (mod p) for any a such that Examples: How do you wish you could use this theorem? p = 3, a = 2 p = 7, a = 4

  7. Logic Review a  b (a implies b) Which is equivalent to the above statement? • b  a • ~a  ~b • ~b  ~a

  8. Logic Review a  b (a implies b) Which is equivalent to the above statement? • b  a The Converse • ~a  ~b The Inverse • ~b  ~a The Contrapositive

  9. Contrapositive of Fermat’s Little Theorem If pand a are integers such that and a p-1 mod p 1, then p is not prime.

  10. First Prime Number Test function primality(N) Input: Positive integer N Output: yes/no // a is random positive integer between 1 and N-1 a = uniform(1..N-1) // if (modexp(a, N-1, N) == 1): return “possibly prime” else: return “not prime” // certain

  11. False Witnesses • If primality(N) returns “possibly prime”, then N might or might not be prime, as the answer indicates • Consider 15: • 414 mod 15 = 1 • but 15 clearly isn’t prime! • 4 is called a false witness of 15 • Given a non-prime N, we call a number a where an-1 mod N = 1 a “false witness” (to the claim that N is prime)

  12. Relatively Prime • Two numbers, a and N, are relatively prime if their greatest common divisor is 1. • 3 and 5? • 4 and 8? • 4 and 9? • Consider the Carmichael numbers: • They pass the test (i.e., an-1 mod N = 1) for all a relatively prime to N.

  13. False Witnesses • Ignoring Carmichael numbers, • How common are false witnesses? • Lemma: If an-1 mod n = 1 for some a relatively prime to n, then it must hold for at least half the choices of a < n

  14. State of Affairs • Summary: • If n is prime, then an-1 mod n = 1 for all a < n • If n is not prime, then an-1 mod n = 1 for at most half the values of a < n • Allows us to put a bound on how often our primality() function is wrong.

  15. Correctness • Question #1: Is the “Fermat test” correct? • No • Question #1’: How correct is the Fermat test? • The algorithm is ½-correct with one-sided error. • The algorithm has 0.5 probability of saying “yes N is prime” when N is not prime. • But when the algorithm says “no N is not prime”, then N must not be prime (by contrapositive of Fermat's Little Theorem)

  16. Amplification • Repeat the test • Decrease the probability of error: C = Composite; P = Prime • Amplification of stochastic advantage

  17. P(Correct) • k trials gives 1/(2k) probability of being incorrect when the answer is "prime“ P(Correct) =

  18. Modified Primality Test function primality2(N) Input: Positive integer N Output: yes/no for i = 1 to k do: a = uniform(1..N-1) if (modexp(a, N-1, N) == 1): // possibly prime; do nothing else: return “not prime” return yes

  19. 2. Greatest Common Divisor

  20. Greatest Common Divisor • Euclid’s rule: • gcd(x, y) = gcd (x mod y, y) = gcd(y, x mod y) • Can compute gcd(x,y) for large x, y by modular reduction until we reach the base case! function Euclid (a,b) Input: Two integers a and b with a  b  0 (n-bit integers) Output: gcd(a,b) if b=0: return a return Euclid(b, a mod b)

  21. Example • gcd(25, 11)

  22. 3 Questions • 1. Is it Correct? • 2. How long does it take? • 3. Can we do better?

  23. Analysis function Euclid (a,b) Input: Two integers a and b with a  b  0 (n-bit integers) Output: gcd(a,b) if b=0: return a return Euclid(b, a mod b)

  24. Bezout’s Identity • For two integers a, band their GCD d, there exist integers x and y such that:

  25. Extended Euclid Algorithm function extended-Euclid (a, b) Input: Two positive integers a & b with a  b  0 (n-bits) Output: Integers x, y, d such that d = gcd(a, b)and ax + by = d if b=0: return (1,0,a) (x’, y’, d) = extended-Euclid(b, a mod b) return (y’, x’ – floor(a/b)y’, d)

  26. Example • Note: there’s a great worked example of how to use the extended-Euclid algorithm on Wikipedia here: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm • And another linked from the reading column on the schedule for today

  27. Multiplicative Inverses • In the rationals, what’s the multiplicative inverse of a? • In modular arithmetic (modulo N), what is the multiplicative inverse?

  28. Multiplicative Inverses • In the rationals, what’s the multiplicative inverse of a? • In modular arithmetic (modulo N), what is the multiplicative inverse?

  29. Finding Multiplicative Inverses(Modulo N) • Modular division theorem: For any a mod N, a has a multiplicative inverse modulo N if and only if it is relatively prime to N. • Significance of extended-Euclid algorithm: • When two numbers, and , are relatively prime • extended-Euclid algorithm produces and such that • Thus, • Because for all integers • Then is the multiplicative inverse of modulo • i.e., I can use extended-Euclid to compute the multiplicative inverse of mod

  30. Multiplicative Inverses • The multiplicative inverse mod is exactly what we will need for RSA key generation • Notice also: when and are relatively prime, we can perform modular division in this way

  31. Next • RSA

  32. Assignment • Read and Understand: Section 1.4 • HW #3: 1.9, 1.18, 1.20 • Finish your project #1 early! • Submit on Wednesday by 5pm

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