Download presentation

Presentation is loading. Please wait.

Published bySeth Craner Modified about 1 year ago

1
Euclidean Algorithm Applied Symbolic Computation CS 567 Jeremy Johnson

2
Greatest Common Divisors g = gcd(a,b) – g|a and g|b – e|a and e|b e|g

3
Unique Factorization p|ab p|a or p|b a = p 1 p t = q 1 q s s = t and i j: p i = q j a = p 1 e1 p t et b = p 1 f1 p t ft gcd(a,b) = p 1 min(e1,f1) p t min(et,ft)

4
Bezout’s Identity g = gcd(a,b) x,y: g = ax + by

5
Bezout’s Identity g = gcd(a,b) x,y: g = ax + by

6
Euclidean Algorithm g = gcd(a,b) if (b = 0) then return a; else return gcd(b,a mod b)

7
Correctness

8
Tail Recursion g = gcd(a,b) if (b = 0) then return a; else return gcd(b,a mod b)

9
Iterative Algorithm g = gcd(a,b) a1 = a; a2 = b; while (a2 0) a3 = a1 mod a2; a1 = a2; a2 = a3; } return a1;

10
Remainder Sequence a 1 = a, a 2 = b a 1 = q 3 a 2 + a 3, 0 a 3 < a 2 a i = q i a i+1 + a i+2, 0 a i+2 < a i+1 a n = q n a n+1 gcd(a,b) = a n+1

11
Bounding Number of Divisions Theorem. Let a b 0 and n = number of divisions required by the Euclidean algorithm to compute gcd(a,b). Then n < 2lg(a).

12
Bounding Number of Divisions

13
Fibonacci Numbers F 0 = 0, F 1 = 1 F n+2 = F n+1 + F n

14
Solving the Fibonacci Recurrence F n = 1/ 5( n + * n ), = (1 + 5)/2, * = (1 - 5)/2 F n 1/ 5 n+1

15
Solving the Fibonacci Recurrence

17
Maximum Number of Divisions Theorem. The smallest pair of integers that require n divisions to compute their gcd is F n+2 and F n+1.

18
Maximum Number of Divisions Theorem. Let a b 0 and n = number of divisions required by the Euclidean algorithm to compute gcd(a,b). Then n < 1.44 lg(a).

19
Maximum Number of Divisions

20
Extended Euclidean Algorithm g = gcd(a,b,*x,*y) a1 = a; a2 = b; x1 = 1; x2 = 0; y1 = 0; y2 = 1; while (a2 0) a3 = a1 mod a2; q = floor(a1/a2); x3 = x1 – q*x2; y3 = y1 – q*y2; a1 = a2; a2 = a3; x1 = x2; x2 = x3; y1 = y2; y2 = y3; } return a1;

21
Correctness

23
Probability of Relative Primality p/d 2 = 1 p = 1/ (2) (z) = 1/n z (2) = 2 /6

24
Formal Proof Let q n be the number of 1 a,b n such that gcd(a,b) = 1. Then lim n q n /n 2 = 6/ 2

25
Mobius Function (1) = 1 (p 1 p t ) = -1 t (n) = 0 if p 2 |n (ab) = (a) (b) if gcd(a,b) = 1.

26
Mobius Inversion d|n (d) = 0 ( n (n)n s ) ( n 1/n s ) = 1

27
Formal Proof q n = n n/k 2 lim n q n /n 2 = n (n)n 2 = n 1/n 2 = 6/ 2

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google