Extended Euclidean algorithm, Euclidean Algorithm

Extended Euclidean algorithm, Euclidean Algorithm

2016.1.25

(Not updated)

1. Euclidean Algorithm (moving phase Division)

1. Purpose: quickly calculate the maximum number of common approx.

2. essence: gcd (a, B) = gcd (B, a % B)

3. Proof: If r = a mod B is set, we need to prove that gcd (a, B) = gcd (B, r)

If gcd (a, B) is set to c, a = mc, B = nc, and m, n are mutually qualitative. So r = a-kb = (m-kn) c, so c is also the factor of r.

If gcd (B, r)> c is set to d, B = m1 * d, r = n1 * d, so a = (m1 + n1) * d, then d is the factor of a, so gcd (a, B) = d, which is inconsistent with the question.

Therefore, gcd (a, B) = gcd (B, r)

4. Time Complexity: Obviously, after two recursion, the first parameter is reduced by at least half.

Therefore, the time complexity is roughly O (log max (a, B ))

5. Typical Example: Number of vertices on a line segment

 

Ii. Extended Euclidean Algorithm:

1. Purpose: quickly calculate the integer x and y so that ax + by = gcd (a, B)

2. essence code:

void extgcd(int a,int b,int &x,int &y){    if(!b)    {        x=1;y=0;        return a;    }    else    {        extgcd(b,a%b,y,x);        y-=(a/b)*x;     }}

 

3. Proof: Since gcd (a, B) = gcd (B, a % B)

So B * x1 + (a % B) * y1 = gcd (a, B)

A % B = a-(a/B) * B

Therefore, gcd (a, B) = B * x1 + (a-(a/B) * y1

= B * x1 + a * y1-(a/B) * B * y1

= A * y1 + B * (x1-a/B * y1)

For the x, y we want to make ax + by = gcd (a, B)

X = y1

Y = x1-a/B * y1

Certificate completion

As for the termination condition, when the Euclidean Algorithm terminates, a = gcd, B = 0, then when x = 1, y = 0, the target formula is required.

In addition, for the mandatory and sufficient condition of ax + by = c, why is c = gcd (a, B), you can Baidu's Yi Shu theorem.

4. Time Complexity: Consistent with euclidean Algorithm

5. Typical Example: Double Six