Download euclid

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 = p1    pt = q1    qs  s = t and i j: pi= qj
• a = p1e1    ptet
• b = p1f1    ptft
• gcd(a,b) = p1min(e1,f1)    ptmin(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
a1 = a, a2 = b
a1 = q3  a2 + a3, 0  a3 < a2

ai = qi  ai+1 + ai+2, 0  ai+2 < ai+1

an= qn  an+1
gcd(a,b) = an+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
• F0 = 0, F1 = 1
• Fn+2 = Fn+1 + Fn
Solving the Fibonacci Recurrence
• Fn = 1/5(n + * n),
 = (1 + 5)/2, * = (1 - 5)/2
• Fn  1/5n+1
Solving the Fibonacci Recurrence
Solving the Fibonacci Recurrence
Maximum Number of Divisions
Theorem. The smallest pair of integers that
require n divisions to compute their gcd is Fn+2
and Fn+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
Correctness
Probability of Relative Primality
p/d2 = 1  p = 1/(2)
(z) = 1/nz
(2) = 2/6
Formal Proof
Let qn be the number of 1 a,b  n such that
gcd(a,b) = 1. Then limn qn/n2 = 6/2
Mobius Function
• (1) = 1
• (p1    pt) = -1t
• (n) = 0 if p2|n
(ab) = (a)(b) if gcd(a,b) = 1.
Mobius Inversion
d|n (d) = 0
(n (n)ns)(n 1/ns) = 1
Formal Proof
qn = n n/k2
limn qn/n2 = n (n)n2 = n 1/n2 = 6/2