# Euclidean Algorithm Applied Symbolic Computation CS 567 Jeremy Johnson.

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Euclidean Algorithm Applied Symbolic Computation CS 567 Jeremy Johnson

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

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)

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

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

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

Correctness

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

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

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

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).

Bounding Number of Divisions

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

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

Solving the Fibonacci Recurrence

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.

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).

Maximum Number of Divisions

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;

Correctness

Probability of Relative Primality  p/d 2 = 1  p = 1/  (2)  (z) =  1/n z  (2) =  2 /6

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

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

Mobius Inversion  d|n  (d) = 0 (  n  (n)n s )  (  n 1/n s ) = 1

Formal Proof q n =  n  n/k  2 lim n  q n /n 2 =  n  (n)n 2 =  n 1/n 2 = 6/  2