cse101_9_28_11 - show something simpler Lemma If a>= b...

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Modular Arithmetic 1. Greatest common divisor A. The GCD of a,b is the largest integer that divides both of them, eg. GCD (10,25) = 5. GCD(1650, 363) = ? Factoring is exponential B. We’d like an algorithm for computing the GCD of a,b. Idea: function gcd(a,b) Input: Integers a >= b >= 0. For c = b down to 1. If c| a and c | b: output c and halt. Running time if a,b are n-bit integers. O(n 2 2 n ). Inner loop: O(n 2 ) division, but could loop O(2 n ) times. C. Evelid’s algorithm: function Evelid(a,b) Input Integers a >= b >= 0. Output: gcd (a,b). if b = 0: return a. return Evelid (b, a mod b). D. Why does it work? Evelid’s rule: if a >=b then gcd (a,b) = gcd (a mod b, b) actually can
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Unformatted text preview: show something simpler. Lemma: If a >= b then gd (a,b) = gcd(a - b, b). Proof: Any number that divides a and b also divides a – b mod b. any number that divides a-b mod b also divides a and b. therefore, (a,b) and (a-b,b) have the same divisors, therefore, same GCD. Gcd (550, 121) = gcd (550 mod 121, 121) = gcd (121, 66) = gcd (55, 11) = gcd (11, 0) Gcd(231, 60) = gcd(51, 9) = gcd(9, 6) = gcd(6,3) = 3. E. How fast is it: Lemma: if a >=b then a mod b < a/2. Case(i) b <= a/2 case 2: b > a/2. . F....
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