上下文：我正在编写代码，以便解决Lippmann-Schwinger方程并在使用不同电势后验证实验结果。 为了做到这一点，我必须计算一个稀疏和大矩阵的逆。我正在使用NR例程ludcmp和lubksb，但是有些不正确。在ludcmp之后，矩阵具有许多零，并且lubksb将向量b计算为b [i] / a [i] [j]，大部分时间生成-nan。我不明白我在做什么错。 感谢您的帮助！
这是我的代码中带有矩阵求逆的部分。
//lu decomp algorithm
void ludcmp(double **a, int n, int *indx, double d)
{
int i,imax,j,k;
double big,dum,sum,temp;
double *vv; //vv stores the implicit scaling of each row.
vv= new double[n];//vector(1,n);
d=1.0; //No row interchanges yet.
for (i=0;i<n;i++)
{//Loop over rows to get the implicit scaling information.
big=0.0;
for (j=0;j<n;j++)
if ((temp=fabs(a[i][j])) > big) big=temp;
//char const *p = "Singular matrix in routine ludcmp"; //
if (big == 0.0) std::cout<<("Singular matrix in routine ludcmp");
//No nonzero largest element.
vv[i]=1.0/big; //Save the scaling.
}
for (j=0;j<n;j++)
{//This is the loop over columns of Crout’s method.
for (i=0;i<j;i++)
{//This is equation (2.3.12) except for i=j.
sum=a[i][j];
for (k=0;k<i;k++) sum -= a[i][k]*a[k][j];
a[i][j]=sum;
}
big=0.0;
//Initialize for the search for largest pivot element.
for (i=j;i<n;i++)
{//This is i=j of equation (2.3.12) and i = j+1...N
sum=a[i][j];
for (k=0;k<j-1;k++)
sum -= a[i][k]*a[k][j];
a[i][j]=sum;
if ( (dum=vv[i]*fabs(sum)) >= big)
{//Is the figure of merit for the pivot
better than the best so far?
big=dum;
imax=i;
}
}
if (j != imax)
{//Do we need to interchange rows?
for (k=0;k<n;k++)
{//Yes, do so...
dum=a[imax][k];
a[imax][k]=a[j][k];
a[j][k]=dum;
}
d = -d; //...and change the parity of d.
vv[imax]=vv[j]; //Also interchange the scale factor.
}
indx[j]=imax;
if (a[j][j] == 0.0) a[j][j]=TINY;
/*If the pivot element is zero the matrix is singular
(at least to the precision of the algorithm).
For some applications on singular matrices,
it is desirable to substitute TINY for zero.*/
if (j != n)
{//Now, finally, divide by the pivot element.
dum=1.0/(a[j][j]);
for (i=j+1;i<n;i++) a[i][j] *= dum;
}
}
//Go back for the next column in the reduction.
delete[] vv;
}
void lubksb(double **a, int n, int *indx, double b[])
{
int i,ii=0,ip,j;
double sum;
//for (i=1;i<n;i++)
for (i=0;i<n;i++)
{//When ii is set to a positive value,
it will become the index of the first nonvanishing
element of b.
//We now do the forward substitution, equation (2.3.6).
//The only new wrinkle is to unscramble
//the permutation as we go.
ip=indx[i];
sum=b[ip];
b[ip]=b[i];
if (ii)
for (j=ii;j<i-1;j++) sum -= a[i][j]*b[j];
else if (sum) ii=i;
//A nonzero element was encountered,
//so from now on we will have to do the sums in the loop above.
b[i]=sum;
}
//for (i=n-2;i>(-1);i--)
for (i=n-1;i>=0;i--)
{//Now we do the backsubstitution, equation (2.3.7).
sum=b[i];
for (j=i+1;j<n;j++) sum -= a[i][j]*b[j];
b[i]=sum/a[i][i];
//Store a component of the solution vector X.
}
for(int i=0; i<n; i++) std::cout<<
"b["<<i<<"]= "<<b[i]<<std::endl;
}
//matrix inversion
void invmat(double **a, double **y, int N, int *indx, double *b, double d)
{
ludcmp(a,N,indx,d);
std::ofstream oo; oo.open("invmattttt.txt",std::ios::out);
oo<< "after lu dec" << std::endl;
for(int i=0; i< N; i++)
for(int j=0; j<N; j++) oo << "a["<<i<<"]["<<j<<"]= " << a[i][j] << std::endl;
//Decompose the matrix just once.
//for(int j=1;j<=N;j++)
for(int j=0;j<N;j++)
{
//Find inverse by columns.
//for(int i=1;i<=N;i++) b[i]=0.0;
for(int i=0;i<N;i++) b[i]=0.0;
b[j]=1.0;
lubksb(a,N,indx,b);
oo<< "\n\nafter lubksb" << std::endl;
for(int i=0; i< N; i++)
for(int j=0; j<N; j++) oo <<"a["<<i<<"]["<<j<<"]= " << a[i][j] << std::endl;
//for(int i=1;i<N;i++) {y[i][j]=b[i]; std::clog<<i<<std::endl;} std::clog<<"hey there"<<std::endl;
for(int i=0;i<(N);i++) {y[i][j]=b[i];}}
oo<< "\n\nafter all" << std::endl;
for(int i=0; i< N; i++)
for(int j=0; j<N; j++) oo << y[i][j] << std::endl;
}