gauss_seidle.txt

By the proof of Lemma 2 of Section 5.2, this accomplished as follows:

# include <iostream>
# include <fstream>
# include <cmath>
using namespace std;
class Gauss_seidel{
private:
int N;
double**a,*x,*b;
public:
double precision;
Gauss_seidel();
~Gauss_seidel();
void strike_on();
};
Gauss_seidel::Gauss_seidel()
{   
	int i,j;
	cout<<"input the precision:";
	cin>>precision;
	cout<<"input the dimention:";
	cin>>N;
	a=new double*[N];
	for(i=0;i<N;i++)
		a[i]=new double[N];
	x=new double[N];
	b=new double[N];
	cout<<"input the matrix a[i][j]:"<<endl;
    ifstream in("e:\\numerical analysis\\gauss_seidel\\data.txt");
        for(i=0;i<N;i++)
          for(j=0;j<N;j++)
		   in>>a[i][j];
    cout<<"input the matrix b[i]:"<<endl; 
	    for(i=0;i<N;i++)
			in>>b[i];
}
Gauss_seidel::~Gauss_seidel()
{
	int i;
	for(i=0;i<N;i++)
	delete []a[i];
	delete[]a;
    delete[]x;
    delete[]b;
}
void Gauss_seidel::strike_on()
{
    int i,j,k;
	double temp,x0,max=0;
   for(i=0;i<N;i++)
		x[i]=0;
for(k=0;k<100;k++)
{
	for(i=0;i<N;i++)
	   {
		temp=0;
		x0=x[i];
		for(j=0;j<N;j++)
	      temp+=a[i][j]*x[j];
		x[i]+=(b[i]-temp)/a[i][i];
		if(fabs(x[i]-x0)>max)
		max=fabs(x[i]-x0);
		}
if(max<=precision) break;
}
	cout<<"the answer is:"<<endl;
	cout.setf(ios::fixed);
    cout.precision(6);
	for(i=0;i<N;i++)
	cout<<"x"<<i<<"="<<x[i]<<endl;
}


void main()
{
	Gauss_seidel a;
	a.strike_on();
}