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I used the extended Euclidean algorithm when I was studying the eighth lecture of the Acwing c++ Blue Bridge Cup Tutorial Course - AcWing 1299. Wuzhishan , here are the knowledge points.
The current article has been included in the blog file directory index: blog directory index (continuously updated)
1. Euclidean Algorithm
Introduction :
Euclidean algorithm: Euclidean algorithm , also known as the rolling and dividing method , is used to calculate the greatest common divisor of two integers a and b.
- gcd(a, b) = d, the d is the greatest common divisor.
code :
class Solution {
//欧几里得算法(辗转相除法)
public static int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static void main(String[] args) {
System.out.println(gcd(12, 5));
}
}
2. Extended Euclidean Algorithm
2.1. Understanding Pei Shu Theorem
Peishu's theorem, also known as Beizu's theorem (extended Euclidean): for any integer a, b and their greatest common divisor d, the linear indeterminate equation about the unknowns x and y (called Peishu's equation) , is a theorem about the greatest common divisor.
If a, b are integers, and gcd(a, b) = d, then there must be integers x, y such that ax+by=d holds true.
2.2. Deduce ax+by=gcd(a, b) to get x and y
2.2.1. Derivation process
For the equation: ax+by=gcd(a, b)
①When b = 0, ax + by = a, then x = 1, y = 0. [In fact, it is when the greatest common divisor is found to be critical]
②When b ≠ 0, namely gcd(a, b) = gcd(b, a % b), according to Pei Shu's theorem gcd(a, b) = d => ax+by=d
gcd(b, a % b) = bx' + (a % b)y', and a % b = a - (int)(a / b) * b, where this a/b is an integer, that is, a % b = a - a/b*b
gcd(b, a % b) = bx' + (a - a/b*b)y', and then disassemble the formula as ay' + b(x' - (a/b) * y)
At this point, the formula is obtained: gcd(b, a % b) = ay' + b(x' - (a/b) * y')
又gcd(a, b) = ax + by，又gcd(a, b) = gcd(b, a % b)
where ax + by = ay' + b(x' - (a/b) * y')
- x = y’
- y = x’ - (a/b) * y’
This y', x' refers to x, y at the current recursion level.
2.2.2, code implementation
class Solution {
/**
* x.a + b * y = d
* @param a
* @param b
* @param arr 存储x，y
*/
public static int exGcd(int a, int b, int[] arr) {
if (b == 0) {
//此时已经求得最大公约数即为a，此时默认x为1，y为0
arr[0] = 1;
arr[1] = 0;
return a;
}
//递归求得最大公约数
int d = exGcd(b, a % b, arr);
//通过公式去化解转为 x = y'，y = x‘ - a/b*y'
int temp = arr[0];
arr[0] = arr[1];
arr[1] = temp - a / b * arr[1];
return d;
}
public static void main(String[] args) {
//扩展欧几里得计算得到x,y
int[] arr = new int[2];
System.out.printf("最大公约数为：%d\n", exGcd(12, 5, arr));
System.out.printf("x = %d, y = %d", arr[0], arr[1]);
}
}
2.3. Deduce all solutions of ax+by=gcd(a, b) and the minimum value of a or b (conclusion + verification)
Using the extended Euclidean algorithm, we can derive ax+by=gcd(a, b) to obtain a set of x and y solutions.
in conclusion
Through such a set of x and y solutions we can get all x and y solutions !
Here we use gcd(a,b) to represent d, ax+by = d.
Definition : a' = ad \frac{a}{d}da，b‘ = b d \frac{b}{d} db
x and y formula definition conclusion : all x, y solutions are
x = x0 + kb'
y = y0 - ka'
The conclusion of the minimum positive integer value of x or y (here is the minimum positive integer value of y): y = y % a', at this time the minimum value of y can be obtained.
- We can directly substitute y of the first set of solutions or use the obtained y0 to get the smallest positive integer solution!
Related algorithm topic: AcWing 1299. Wuzhishan
proof derivation
①X and y formula definition conclusion derivation
Let's try to bring the final derivation formula of x and y into the ax+by = d formula:
ax+by = a(x0 + kb') + b(y0 - to') = ax0 + akb' + by0 - bka' = ax0 + by0 + akb' - bka'
At this point we will a' = ad \frac{a}{d}da，b‘ = b d \frac{b}{d} dbSubstitute into akb' - bka', akb' = ak.b / d, bka' = bk.a / d, that is to say akb' = bka'
So ax0 + by0 + akb' - bka' = ax0 + by0 = d, proved!
Let's take an example to prove whether the corresponding a' and b' and x and y general formulas are correct:
给出a = 12，b = 5，d = gcd(a, b) = 1
根据扩展欧几里得可以得到式子：ax + by = d
通过算法来得到一组x、y解，x = -2，y = 5 // 式子为：-2*12 + 5*5 = 1
接着我们根据①中的公式定理，a' = a / d，b' = b / d // a' = 12/1 = 12， b' = 5 / 1 = 5
对应x与y的所有解公式为：x = x0 + kb'，y = y0 - ka'
此时我们可以去尝试得到x0的解(假设k=1)，-2 = x0 + 1*5，得到x0 = -7
尝试得到y0的解(假设k=1)，5 = y0 - 1*12，得到y0 = 17
ok，此时x0 = -7，y0 = 17，我们来自己通过公式去尝试构造一组解，设置k = 2，
x = -7 + 2 * 5 = 3
y = 17 - 2 * 12 = -7
尝试代入ax+by=d中，即为12 * 3 + 5 * (-7) = 1，此时能够验证成功！
②Example of the minimum positive integer value of y: y = y % a'
Still using the example in ①, let's find the minimum value of y: a = 12, b = 5, a' = 12, b' = 5, x0 = -7, y0 = 17
//使用y0代入
17 % 12 = 5
//使用y代入(第一组解得到的值y = 5)
5 % 12 = 5
//使用y为负数情况代入(自己设置k=2时得到的y解)
-7 % 12 = -5
//对于该特殊情况我们需要使用(y mod a' + a') % a'
(-7 % 12 + 12) % 12 = (-5 + 12) % 12 = 17 % 12 = 5
Conclusion: We can directly substitute the y of the first set of solutions or use the obtained y0 to get the smallest positive integer solution!