📄 d1r9.cpp
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#include "iostream.h"
#include "math.h"
void asub(double xin[], double xout[])
{
int i,n = 20;
xout[1] = xin[1] + 2.0 * xin[2];
xout[n] = -2.0 * xin[n - 1] + xin[n];
for (i = 2; i<=n-1; i++)
{
xout[i] = -2.0 * xin[i - 1] + xin[i] + 2.0 * xin[i + 1];
}
}
void atsub(double xin[], double xout[])
{
int i,n = 20;
xout[1] = xin[1] - 2.0 * xin[2];
xout[n] = 2.0 * xin[n - 1] + xin[n];
for (i = 2; i<=n-1; i++)
{
xout[i] = 2.0 * xin[i - 1] + xin[i] - 2.0 * xin[i + 1];
}
}
void sparse(double b[], int n, double x[], double rsq)
{
const int nmax = 500;
double eps2,eps = 0.000001;
double g[500], h[500], xi[500], xj[500];
int j;
eps2 = n * pow(eps, 2);
int irst = 0;
r1: irst = irst + 1;
asub(x, xi);
double rp = 0.0;
double bsq = 0.0;
for (j = 1; j<=n; j++)
{
bsq = bsq + pow(b[j],2);
xi[j] = xi[j] - b[j];
rp = rp + pow(xi[j],2);
}
atsub(xi, g);
for (j = 1; j<=n; j++)
{
g[j] = -g[j];
h[j] = g[j];
}
for (int iter = 1; iter<=10 * n; iter++)
{
asub(h, xi);
double anum = 0.0;
double aden = 0.0;
for (j = 1; j<=n; j++)
{
anum = anum + g[j] * h[j];
aden = aden + pow(xi[j] ,2);
}
if (aden == 0.0)
{
cout<<"very singular matrix"<<endl;
}
anum = anum / aden;
for (j = 1; j<=n; j++)
{
xi[j] = x[j];
x[j] = x[j] + anum * h[j];
}
asub(x, xj);
double rsq = 0.0;
for (j = 1; j<=n; j++)
{
xj[j] = xj[j] - b[j];
rsq = rsq + pow(xj[j], 2);
}
if (rsq == rp || rsq <= bsq * eps2)
{
return;
}
if (rsq > rp)
{
for (j = 1; j<=n; j++)
{
x[j] = xi[j];
}
if (irst >= 3)
{
return;
}
goto r1;
}
rp = rsq;
atsub(xj, xi);
double gg = 0.0;
double dgg = 0.0;
for (j = 1; j<=n; j++)
{
gg = gg + pow(g[j],2);
dgg = dgg + (xi[j] + g[j]) * xi[j];
}
if (gg == 0.0)
{
return;
}
double gam = dgg / gg;
for (j = 1; j<=n; j++)
{
g[j] = -xi[j];
h[j] = g[j] + gam * h[j];
}
}
cout<<"too many iterations"<<endl;
}
void main()
{
//program d1r9
//driver program for routine sparse
int i,j,n = 20;
double b[21], x[21], bcmp[21];
//输入已知的方程组的右端向量b
for (i = 1; i<=n; i++)
{
x[i] = 0.0;
b[i] = 1.0;
}
b[1] = 3.0;
b[n] = -1.0;
//利用下列子程序asub输入已知的方程组的系数矩阵
double rsq=0;
sparse(b, n, x, rsq);
cout<<endl;
cout<<"sum - squared residual:"<<rsq<<endl;
cout<<endl;
cout.setf(ios::fixed|ios::right);
cout.precision(5);
cout<<"计算出的方程组的解"<<endl;
for (i = 1; i<=4; i++)
{
for (j = i * 5 - 4; j<=i * 5; j++)
{
cout.width(12);
cout<<x[j];
}
cout<<endl;
}
//将计算出的方程组的解乘以系数矩阵,以验证计算结果正确
asub(x, bcmp);
cout<<endl;
cout<<"计算出的方程组的解乘以系数矩阵的结果"<<endl;
cout<<endl;
cout<<"解乘以系数矩阵"<<" "<<"方程组的右端向量"<<endl;
for (i = 1; i<=n; i++)
{
cout.width(10);
cout<<bcmp[i]<<" ";
cout.width(12);
cout<<b[i]<<endl;
}
}
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