在 Java 中实现科波菲尔算法
原文:https://www . geesforgeks . org/impering-copper Smith-freivalds-algorithm-in-Java/
概念:科波菲尔算法是检查矩阵 A 乘以矩阵 B 是否等于给定的矩阵 c，用于验证矩阵乘法。借助于 A(Br)-(C*r)=0 的方程进行了验证，其中 r 是仅由 0/1 组成的随机列向量。
插图:
Input:
Enter the dimensions of the matrices:
2
Enter the 1st matrix:
-2 1
0 4
Enter the 2st matrix:
6 5
-7 1
Enter the result matrix:
-19 9
-28 4
Output: Yes, The matrix multiplication is correct.
进场:
将矩阵的大小作为用户的输入。
目标:根据方程我们需要验证矩阵 A *矩阵 B =矩阵 c。
取矩阵 A(nn)矩阵 B(nn)和结果矩阵 C(n*n)的输入作为输入。
1)随机取数组 r[n][1]，该数组仅由 0/1 的元素组成。
2)计算矩阵 Br，矩阵 Cr，然后计算矩阵 A(矩阵 Br)来计算表达式矩阵 A(矩阵 B * r)–(矩阵 Cr)
3)检查方程矩阵 A(矩阵 B * r)–(矩阵 Cr)=0 与否。
4)如果为零，则打印“是”，否则打印“否”。
实现:输入要按照上面显示的顺序进行，否则会导致错误的结果。下面是考虑的例子
Java 语言(一种计算机语言，尤用于创建网站)
// Importing class to create objects
// generating pseudo random numbers
import java.util.Random;
// Importing class to take input from user
import java.util.Scanner;
public class GFG {
public static double[][] multiplyVector(double[][] a,
double[][] b,
int n)
// Method to check the result of the equation.
{
double result[][] = new double[n][1];
for (int i = 0; i < n; i++) {
for (int j = 0; j < 1; j++) {
for (int k = 0; k < n; k++) {
result[i][j]
= result[i][j] + a[i][k] * b[i][j];
}
}
}
return result;
}
public static void main(String args[])
{
// Driver main method
Scanner input = new Scanner(System.in);
System.out.println(
" Enter the dimensions of the matrix");
int n = input.nextInt();
// n- size or dimensions of matrix
System.out.println("Enter the 1st matrix:");
// Taking input for 1st matrix
double a[][] = new double[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
a[i][j] = input.nextDouble();
}
}
//
System.out.println("Enter the 2nd matrix");
double b[][] = new double[n][n];
// Taking input for second matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
b[i][j] = input.nextDouble();
}
}
// Covering up Resultant matrix
System.out.println("Enter the resultant matrix");
double c[][] = new double[n][n];
// the resultant matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c[i][j] = input.nextDouble();
}
}
// generating random matrix r consisting of 0/1 only
double[][] r = new double[n][1];
Random random = new Random();
for (int i = 0; i < n; i++) {
r[i][0] = random.nextInt(2);
}
// testing of the standard equation A*(B*r)-(C*r)=0
double br[][] = new double[n][1];
double cr[][] = new double[n][1];
double abr[][] = new double[n][1];
br = multiplyVector(b, r, n);
cr = multiplyVector(c, r, n);
abr = multiplyVector(a, br, n);
// check for all zeroes in abr
boolean flag = true;
// Setting flag with true
for (int i = 0; i < n; i++) {
if (abr[i][0] == 0)
continue;
else
// Set flag to false(change flag)
flag = false;
}
// Boolean comparison resulting in message printing
if (flag == true)
System.out.println(
"Yes,The matrix multiplication is correct");
else
System.out.println(
"No,The matrix multiplication is wrong");
input.close();
}
}
输出:2 阶随机矩阵的自定义输入
时间复杂度: O(kN^2)其中 n 是矩阵的大小。
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