Study Notes--Extended Euclid

foreword

I had some understanding of this algorithm before noip, but noipD1T1 did not expect to expand Europe at all, which means that what I learned before was not solid at all, so now I come to fill the hole

Related concepts:

You can take a look at this dalao's blog first, and at the same time, the proofs of many theorems below are also given in that blog:

https://blog.csdn.net/yoer77/article/details/69568676

Term Deduction & Euclidean Algorithm

Let's first introduce a more subtractive method that appeared in "Nine Chapters of Arithmetic" :

For any positive integer $a>=b$ , there are $gcd(a,b)=gcd(b,ab)=gcd(a,ab)$
Prove:

For any common divisor d of a and b, because $d\mid b,d|a$ , so $d\mid (ab)$ , so the set of common divisors of a,b is the same as $b,ab\ ) has the same set of common divisors, and the same is true for \(a, ab$ .

Then there is the famous Euclidean algorithm :

For any positive integer $a,b(b \ne 0) $yes,$ gcd(a,b)=gcd(b,a\mod b)$

When performing high-precision operations, the term deduction method can be considered instead.

Pei Shu's Theorem ( $Bézout's$ $lemma$ )

For any integer $a,b$ , there is a pair of integers that satisfy $ax+by=gcd(a,b)$
Prove:

I think the proof given in the Advanced Guide to Algorithm Competition is more helpful:

In the last step of the Euclidean algorithm, that is, $b=0$ , there is obviously a pair of integers $x=1, y=0$ , satisfying $a*1+b*0=gcd(a, 0)$

If $b > 0$ , then $gcd(a,b)=gcd(b,a \mod b)$ . Fake

Suppose there is a pair of integers $x,y$ , satisfying $b*x+(a \mod b)*y=gcd(b,a \mod b)$ ,

Because $bx+(a \mod b)y$

$=bx+(a-b\lfloor a/b \rfloor)y$

$=ay-b(x-\lfloor a/b \rfloor y)$，

So make $x'=x,y'=x-\lfloor a/b \rfloor y$ to get $ax'+by'=gcd(a,b)$

Applying mathematical induction (if you don't know what this is, you can think of it as a recursive or recursive process), the theorem holds.

The above proof process also gives the calculation process of $x,y$ , which is written in code as

ll ex_gcd(ll a,ll b,ll &x,ll &y){
    if(b==0){
        x=1,y=0;
        return a;
    } 
    ll d=ex_gcd(b,a%b,x,y);
    ll z=x;x=y,y=z-(a/b)*y;
    return d;
}

ex_gcd returns $gcd(a,b)$ and finds a set of solutions $x,y$

For $ax+by=c$ , it has a solution if and only if $gcd(a,b)\mid c$

Then, the general solution of this equation can be expressed as:

$x=(c/d)*x0+k*(b/d)$ ，$y=(c/d)*y0-k*(a/d)$

Where $k$ takes any integer, $d$ is $gcd(a,b)$ .

Why is this? I have searched various blogs without a clear explanation. Driven by my perseverance, I finally found a concise and easy-to-understand proof in the Luogu online class:

Prove:

Let's first consider the case of $ax+by=gcd(a,b)$ ,

Obviously $a*b+b*(-a)=0$ is divided by $gcd(a,b)\ ) at the same time , it is easy to know that this is feasible

So \(a*(b/gcd(a,b))+b*(-a/gcd(a,b))=0$

So if it is known that $x0,y0$ is a set of special solutions, that is, $a*x0+b*y0=gcd(a,b)$ , let $x=x0+k*b/gcd( a,b)$ , $y=y0-k*(a/gcd(a,b))$ ,

Bring in $ax+by$ to get $a(x0+k*b/gcd(a,b))+b*(y0-k*(a/gcd(a,b)))$

$=ax0+by0+k*(a*b/gcd(a,b)-b*a/gcd(a,b))$

$=ax0+by0=gcd(a,b)$

Then $ax+by=c$ is also solved very well, so I won't go into details here

Solving Linear Congruence Equations

For the equation $a*x \equiv b \pmod m$ , it can be transformed into $a*xb \equiv 0 \pmod m$ , you might as well set $(a*x)-b$ as $m \ ) times \(-y$ , that is $a*xb=-y*m$

get $a*x+m*y=b$

Then see above
find the inverse element

What is an inverse element? See this blog of mine: https://www.luogu.org/blog/Rye-Catcher/solution-p3811

$a*x \equiv 1 \pmod b$ is also converted to \(ax+by=1\ )

Then see above
example:

Congruence equation: https://www.luogu.org/problemnew/show/P1082

Frog Dating: https://www.luogu.org/problemnew/show/P1516