Euclidean algorithm and extended Euclidean algorithm

Euclid's algorithm is based on such a GCD recursive theorem:

$GCD (A, b) = gcd (b, a\bmod{b}) $

The proof is as follows:

Suppose $a > b$, $a = kb + R (0 <= R < b) $, i.e. $a \bmod{b} = R $.

If there is $d \mid a$ and $d \mid b$, there must be $d \mid a-kb$, that is, $d \mid R. It is learned that the number of $a $ with $b $ of all conventions must be $b $ with $r $ of the number of conventions.

If there is $d \mid R $ and $d \mid b$, there must be $d \mid KB + R $, which is $d \mid a$. It is learned that the number of $b $ with $r $ of all conventions must be $a $ with $b $ of the number of conventions.

So $GCD (A, b) = gcd (b, r) $, i.e. $GCD (A, b) = gcd (b, a\bmod{b}) $

The $GCD (A, b) $ can then be calculated by continually returning to the $b = 0$.

int gcd (intint  b) {    return b? gcd (b, a% b): A;}

The extended Euclidean algorithm utilizes some useful information from Euclid's algorithm to find a set of solutions with the Comodule equation $ax + by = gcd (A, b) $. The mathematical induction method can prove the correctness of the extended Euclidean algorithm.

1. When $b = 0$, $GCD (A, b) = a$, obviously $x = 1, y = 0$ is a set of valid solutions

2. When $b > 0$, suppose we have obtained the equation $bx + (a\bmod{b}) y = gcd (b, a\bmod{b}) $ for a set of solutions $ (x_0, y_0$), while setting $ (x_1, Y_1) $ is the equation $ax + by = GCD (A, b) $ of a set of solutions, then there are

$ax _1 + by_1 = bx_0 + (a\bmod{b}) y_0$

$ax _1 + by_1 = bx_1 + (A-\frac{a}{b}\cdot b) y_1$

$ax _1 + by_1 = B (X_0-\frac{a}{b}\cdot y_0) + ay_0$

correspond to equal, we can get

$x _1 = y_0$

$y _1 = x_0-\frac{a}{b}\cdot y_0$

void exgcd (intint int int. int int &y)    {if  (! b) {        = A;         1 ;         0 ;     Else {        % B, D, y, x);         -= (A/b) * x;    }}

(The code used a little bit of a trick: let $x $ with $y $ in recursive calls to reduce the amount of code, collation of the corresponding relationship can be found correct.)

Using Euclidean algorithm to get equation $ax + by = gcd (A, b) $ for a set of solutions $ (x_1, y_1) $ after assuming another set of equations for the solution of $ (x_2, y_2) $, then there are

$ax _1 + by_1 = ax_2 + by_2$

Organized to

$a (x_1-x_2) = B (y_2-y_1) $

Divide the two sides $g =GCD (A, b) $, get

$a ' (x_1-x_2) = B ' (y_2-y_1) $

Because $a ' $ with $b ' $ coprime, so $b ' \mid (x_1-x_2) $, you can set $x _1-x_2 = KB ' $, then there are

$a ' k = y_2-y_1$

That is $y _2 = Y_1+ka ' $

At the same time also got $x _2 = x_1+kb ' $

From the top, get a set of solutions $ (x_1, y_1) $ after you get another set of solutions $x _2 = X_1+kb ', y_2 = Y_1+ka ' $.

By extending Euclidean algorithm, we can solve the general indefinite equation $ax + by = C $ solution:

If $\GCD (A, B) \mid C $, the equation has a solution. Assuming $ (x_0, Y_0) $ is the equation $ax + by = gcd (A, b) $ for a set of solutions, then we can get the equation $ax + by = C $ for a set of solutions $ (\FRAC{C}{GCD (A, B)}\cdot{x_0}, \FRAC{C}{GCD (A, B)} \CDOT{Y_0}) $;

Otherwise, the equation has no integer solution.

In addition, we can use the extended Euclidean algorithm to solve the linear same comodule equation set. Like what:

$x \equiv m_1\pmod{a_1}$

$x \equiv m_2\pmod{a_2}$

...

$x \equiv m_n\pmod{a_n}$

We can merge it 22. For example, for the first two equations, suppose $m _1y+a_1=x$, $m _2z+a_2=x$, then

$m _1y-m_2z=a_2-a_1$

This is a general indefinite equation, using the extended Euclidean algorithm to get a $y _0$ value, we can calculate a feasible $x _0$ value, then obviously the general solution of X is

$x = X_0 + t[m_1, m_2]$

By this equation we can construct a new equation:

$x \equiv x_0\pmod{[m_1, m_2]}$

The equation is then coupled with the third equation to get a new equation. Repeat this process to get a solution to the equation set.

Euclidean algorithm and extended Euclidean algorithm