CS250: Discrete Math for Computer Science

Transcription

1 CS250: Discrete Math for Computer Science L6: Euclid s Algorithm & Multiplicative Inverses Mod m

2 Greatest Common Divisors, GCD If d a and d b then d is a common divisor of a and b. 1, 2, 3, and 6 are common divisors of 12,18. 1 is a common divisor of every pair of integers a,b. The greatest common divisor of a,b is denoted gcd(a,b). gcd(12, 18) = 6 gcd(5,11) = 1 5 and 11 are relatively prime gcd(17, 34) = 17 gcd(30, 100) = 10 gcd(98, 105) = 7

3 How do we efficiently compute gcd(a, b)? Easy if we know the prime factors of a and b: 12 = = gcd(12,18) = = = gcd(5,11) = = = gcd(17,34) = = = gcd(30,100) = = = gcd(98,105) = Prop: If a = p a 1 1 pa 2 2 pa k k and b = p b 1 for primes p 1 < p 2 < < p k, 1 pb 2 2 pb k Then gcd(a,b) = p min(a 1,b 1 ) 1 p min(a 2,b 2 ) 2 p min(a k,b k ) k k

4 But, factoring integers is computationally difficult To factor a thousand-bit integer, a, we would try all divisors up to a but that we would be about divisors! This is exponential time in terms of the size of the input, so it is not feasible. Today, we will see how over 2300 years ago, Euclid gave a very efficient algorithm to compute gcd(a,b), without factoring. This was in Euclid s Geometry text. He was thinking about line segments and wanted to be able to compute the length d of the longest line segment that evenly divided two given line segments, a and b.

5 Euclid s algorithm To compute: gcd(12, 18) = gcd(18, 12), Divide the bigger number, b, by the smaller, s, computing the remainder, r. Repeat until remainder = 0. Answer = the last postive remainder. 18 = Answer: gcd(18,12) = 6 12 =

6 Euclid s algorithm Compute: gcd(123, 42) 123 = = Answer: gcd(123,42) = 3 39 = Compute: gcd(13, 8) 13 = = = = Answer: gcd(13,8) = 1 2 =

7 Euclid s Algorithm: Why It Works Algorithm: GCD(b, s) Input: integers b > s 0 1. while (s 0) do { b := s; s := (b % s) } 2. return(b) Lemma [Euclid, 300 B.C.] If b > s > 0 Then gcd(b,s) = gcd(s,(b % s)). Each iteration of the while loop decreases s. Thus Euclid s Algorithm eventually halts. By Euclid s Lemma, each iteration preserves value of gcd(b,s). Correct answer at last step: because gcd(b,0) = b because b 0 (b 0 = 0) Thus, Euclid s Algorithm (GCD) correctly computes gcd.

8 Lemma [Euclid, 300 B.C.] If b > s > 0 Then gcd(b,s) = gcd(s,(b % s)). proof: Let r = b % s and b = q s + r Claim: d[(d b d s) (d s d r)] let d be arbitrary suppose (d b d s) Thus, d (b q s), i.e., d r. Conversely, suppose (d s d r) Thus, d (q s + r), i.e., d b. This proves the claims. Thus gcd(b, s) = gcd(s, r).

9 Euclid s Algorithm: How Long Does it Take? Algorithm: GCD(b, s) Input: integers b > s 0 1. while (s 0) do { b := s; s := (b % s) } 2. return(b) Claim: After two iterations: b b/2. b := s; s := (b % s) b := s ; s := (b % s ) proof: Suppose s b/2, then reduced by half after one iteration. Otherwise: s > b/2. Thus, s = b%s = b s so s < b/2. Thus b = s < b/2 Thus, Euclid s Algorithm takes at most 2 log b iterations. Thus linear, i.e., O(n) iterations where n = b is the number of bits to represent b.

10 Euclid s Algorithm Backwards Thm: ab xy(ax + by = gcd(a,b)) Remember that our universe of discourse is Z. 18 = gcd(18,12) = 6 Express gcd(a,b) in terms of previous values. 6 = ( 1)

11 ab xy(ax + by = gcd(a, b)) 123 = = gcd(123,42) = 3 Express gcd(a,b) in terms of previous values; regroup; repeat. 3 = ( 1) 3 = ( ( 2)) ( 1) 3 = 123 ( 1)

12 ab xy(ax + by = gcd(a, b)) 13 = = = = gcd(13,8) = 1 Express gcd(a,b) in terms of previous values; regroup; repeat. 1 = ( 1) 1 = (5 + 3 ( 1)) ( 1) 1 = 5 ( 1) = 5 ( 1) + (8 + 5 ( 1)) 2 1 = 8 (2) = 8 (2) + ( ( 1)) 3 1 = 13 ( 3) Don t forget to check: 1 =

13 ab xy(ax + by = gcd(a, b)) In fact, x and y can be computed very efficiently by running Euclid s Algorithm backwards. 13 ( 3) 1 (mod 8) 8 (5) 1 (mod 13) 1 = 13 ( 3) Thm: If gcd(a,b) = 1 then we can efficiently compute the multiplicative inverse of a (mod b). Proof. It s x in the above equation. ax + by = 1 ax 1(mod b)

14 Multipicative Inverses Mod m Thm: For all integers a,m s.t. m > 1, a has a multiplicative inverse mod m iff a and m are relatively prime. Proof: Recall a and m are relatively prime iff gcd(a,m) = 1. Let a and m be arbitrary with m > 1. Assume: gcd(a,m) = 1 Then xy(ax + my = 1) so x is mult. inv. of a (mod m). Assume: ax 1(mod m) Thus, y(ax + my = 1). If, d a and d m, then d (ax + my), i.e., d 1. Thus, gcd(a,m) = 1. Since a and m were arbitrary, Thm holds for all such a,m.

15 a, m > 1 (a has mult. inverse mod m iff gcd(a, m) = 1) mod