Can you translate this to JAVA and show that it runs. #include <iostream> #inclu

Question

Can you translate this to JAVA and show that it runs.

#include <iostream> #include <math.h> using namespace std; void Naive_Gauss(int n, float a[5][5], float b[5], float x[5]) { float sum, xmult; // float a[n][n], b[n], x[n]; for(int k = 1; k <= n-1; k++) { for(int i = k + 1; i <= n; i++) { xmult = a[i][k]/a[k][k]; a[i][k] = xmult; for(int j = k + 1; j <= n; j++) { a[i][j] = a[i][j] - xmult * a[k][j]; } b[i] = b[i] - xmult * b[k]; } } x[n] = b[n] / a[n][n]; for(int i = n - 1; i <= 1; i--) { sum = b[i]; for(int j = i + 1; j <= n; j++) { sum = sum - a[i][j] * x[j]; } x[i] = sum / a[i][i]; } } int main(){ float a[][5] = { {0, 0, 0, 0, 0}, {0, 15, -2, -6, 0}, {0, -2, 12, -4, -1}, {0, -6, -4, 19, -9}, {0, 0, -1, -9, 2}}; float b[5]; float x[5]; for(int n = 4; n <= 5; n++) { for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { a[i][j] = pow((float) (i + 1), (float) (j-1)); cout << "a" << i << "," << j << " = " << a[i][j] << endl; } b[i] = (pow((float) (i + 1), (float) n) - 1) / i; cout << "b" << i << " = " << b[i] << endl; } Naive_Gauss(n, a, b, x); cout << "n = " << n << endl; } return 0; }

Explanation / Answer

public class Gaussian { static void Naive_Gauss(int n, float a[][], float b[], float x[]) { float sum, xmult; // float a[n][n], b[n], x[n]; for (int k = 1; k

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