write a program to find solution of linear equation using Gauss-Seidel method in c++.

write a program to find solution of linear equation using Gauss-Seidel method in c++.


Source Code:



/*************************************
    language: C
    Author : Tanveer Khan
    Rajasthan University, Jaipur
************************************************/


#include<iostream>

#include<conio.h>

 

using namespace std;

 

int main(void)

{

    float a[10][10], b[10], x[10], y[10];

    int n = 0, m = 0, i = 0, j = 0;

    cout << "Enter size of 2d array(Square matrix) : ";

    cin >> n;

    for (i = 0; i < n; i++)

    {

        for (j = 0; j < n; j++)

        {

            cout << "Enter values no :(" << i << ", " << j << ") ";

            cin >> a[i][j];

        }

    }

    cout << "\nEnter Values to the right side of equation\n";

    for (i = 0; i < n; i++)

    {

        cout << "Enter values no :(" << i << ", " << j << ") ";

        cin >> b[i];

    }

    cout << "Enter initial values of x\n";

    for (i = 0; i < n; i++)

    {

        cout << "Enter values no. :(" << i<<"):";

        cin >> x[i];

    }

    cout << "\nEnter the no. of iteration : ";

    cin >> m;

    while (m > 0)

    {

        for (i = 0; i < n; i++)

        {

            y[i] = (b[i] / a[i][i]);

            for (j = 0; j < n; j++)

            {

                if (j == i)

                    continue;

                y[i] = y[i] - ((a[i][j] / a[i][i]) * x[j]);

                x[i] = y[i];

            }

            printf("x%d = %f    ", i + 1, y[i]);

        }

        cout << "\n";

        m--;

    }

    return 0;

}









OUTPUT




$ g++ GaussSeidel.cpp
$ a.out
 
Enter size of 2d array(Square matrix) : 3
Enter values no :(0, 0) 2
Enter values no :(0, 1) 3
Enter values no :(0, 2) 1
Enter values no :(1, 0) 5
Enter values no :(1, 1) 4
Enter values no :(1, 2) 6
Enter values no :(2, 0) 8
Enter values no :(2, 1) 7
Enter values no :(2, 2) 9
 
Enter Values to the right side of equation
Enter values no :(0, 3) 2
Enter values no :(1, 3) 3
Enter values no :(2, 3) 4
 
Enter initial values of x
Enter values no. :(0): 0
Enter values no. :(1): 0
Enter values no. :(2): 0
 
Enter the no. of iteration : 4
x1 = 1.000000    x2 = -0.500000    x3 = -0.055556    
x1 = 1.777778    x2 = -1.388889    x3 = -0.055556    
x1 = 3.111111    x2 = -3.055555    x3 = 0.055555    
x1 = 5.555555    x2 = -6.277777    x3 = 0.388889