Extended Euclidean algorithm

0 , Euclid's theorem

The basis of all, nature is Euclidean theorem. Its form is very simple (sometimes naive)

gcd (A, a) =gcd (b,a mod b)

Prove:

Suppose a , the number of conventions for B is g, and $${a}={bx+y}{(x,y\in Z)}$$
There is obviously $${g \mid A},\qquad {g \mid b},\ Qquad{{a}\ mod\ {b}}={y}$$
$${\because g \mid b}$$
$${\therefore g \mid bx}$$
and $${\ Because G \mid a}$$
$${\therefore g \mid {(A-BX)}}$$
i.e. $${g \mid {{a}\ mod\ {b}}}$$
so g is b and a mod b the number of conventions
that is (A, B) The number of conventions for and (b,a mod b) is the same, so their greatest common divisor are the same.

2. Euclidean algorithm

   aka "the division of the Law", is the current general algorithm for greatest common divisor, to achieve a simple function powerful. (The code can be said to be very beautiful)

    int  gcd ( int  x, int  y) { return  y? GCD (Y, x%y): x;}  

    

   a mod b is necessarily less than, the last B becomes a, the last a mod B becomes B, so the worst case is O (log n) .

4. Bezout theorem

   Before introducing the extended Euclidean theorem, we first need to introduce the bezout theorem (Bézout theorem, bezout theorem).

   Bezout theorem: If ${ax+by=z}$, then there is $${GCD (A, b)}\mid{z}$$

   Bezout theorem Expansion: If ${a_1x_1+a_2x_2+...+a_nx_n}=z$, there is $${GCD (a_1,a_2,..., a_n)}\mid{z}$$

5. extended Euclidean algorithm

   The extended Euclidean algorithm is very powerful and can be used to find the general solution of a two-yuan equation.

   Obviously when $b =0$, $GCD (A, b) =a$. At this point $x =1,y=0$
   When $a>b>0$
   Set $ $ax _1+BY_1=GCD (A, b) $$
   $${bx_2}+{({a}\ mod\ {b}) y_2}={gcd (b,{a}\ mod\ {b})}$$
   By Euclidean theorem, $ $gcd (A, B) =gcd (b,{{a}\ mod\}) $$
   Then: $ $ax _1+by_1=bx_2+{({a}\ mod\ {b}) y_2}$$
   That is, $ $ax _1+by_1=ay_2+b (X_2-{[{a}/{b}]}{\times} y_2) $$
   According to the identity theorem: $${x_1=y_2},\qquad{y_1=x_2-{[{a}/{b}]}{\times} y_2}$$
   This gives us a way to solve ${x_1},{y_1}$: The value of ${x_1},{y_1}$ is based on ${x_2},{y_2}$
   By Bezout theorem We know: ${ax+by=z}$,z is gcd (A, B) several times, so we first solve ${AX+BY={GCD (A, B)}}$, and then multiply the solution by ${{Z}/{GCD (A, b)}}$. (The above '/' means divisible)

   CPP Code:

   int exgcd (intint int int y) {    int d = A;     if 0         {= EXGCD (b, a%B, y, x);         -= (A/b) *x;    }     Else {        10;    }     return D;}

Extended Euclidean algorithm