ORIGIN

Extended Euclid and Inverse element

Algorithm 2 mins395 words

Today’s importance is inverse element, however, inverse element calculation is based on the extended Euclid algorithm. At first, we need to know the Euclid’s algorithm.

A much more efficient method is the Euclidean algorithm, which uses a division algorithm such as long division in combination with the observation that the gcd of two numbers also divides their difference. –from Wikipedia

Extended Euclidean

$$
ax+by = gcd(a, b)
$$

From the Euclid Method of calculating gcd, we know that
$$
gcd(a, b) = gcd(b,a % b)
$$
We set another function
$$
bx_1 + (a%b)y_1 = gcd(b, a % b)
$$
So, combine the first function with the third function we got
$$
ax+by = bx_1+(a%b)y_1\
\Rightarrow ax+by=bx_1+(a-a /b \times b)y_1\
\Rightarrow ax+by=bx_1+ay_1-b\times(a/b)y_1\
\Rightarrow ax + by = ay_1+ b(x_1-a/b\times y_1)\
\Rightarrow \left{
\begin{aligned}
x & = y_1 \
y & = (x_1) - a / b \times y_1)
\end{aligned}
\right.
$$
We define that if
$$
b = 0:x = 1 , y = 0
$$
So we can write our function in this way

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typedef pair<int,int> P;
P gcdEx(int a, int b){
if(b == 0) return P(1, 0);
else{
P p = gcdEx(b, a % b);
return P(t.second, t.first - (a / b) * t.second);
}
}

The previous one is the direct one we can get from the mathematical function. Here is a better way of writing it.

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

Inverse Element

$$
ax \equiv 1 (mod \space b)
$$

There is a inverse element only if a and b are both prime numbers. The equation above means both side mod b at the same time is equal. x is a reverse element of a under mod b.

We can change the equation into this.
$$
ax \space mod \space m = 1\
ax -my = 1
$$
y is the middle element used to solve x.

So the code looks like this.

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int x, y;
gcdEx(a, m, x, y);
x = (x % m + m) % m;

x is the inverse element of a.

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