我正在尝试使用从南加州大学CACS网站获得的TQLI算法计算特征值。我的测试脚本如下所示：
#include <stdio.h>
int main()
{
int i;
i = rand();
printf("My random number: %d\n", i);
float d[4] = {
{1, 2, 3, 4}
};
float e[4] = {
{0, 0, 0, 0}
};
float z[4][4] = {
{1.0, 0.0, 0.0, 0.0} ,
{0.0, 1.0, 0.0, 0.0} ,
{0.0, 0.0, 1.0, 0.0},
{0.0, 0.0, 0.0, 1.0}
};
double *zptr;
zptr = &z[0][0];
printf("Element [2][1] of identity matrix: %f\n", z[2][1]);
printf("Element [2][2] of identity matrix: %f\n", z[2][2]);
tqli(d, e, 4, zptr);
printf("First eigenvalue: %f\n", d[0]);
return 0;
}
如我在here中所见，当我尝试运行此脚本时，出现了分段错误错误。我的代码在什么位置产生此分段错误。因为我相信USC的代码没有错误，所以我很确定错误一定出在我调用该函数时。但是我看不到我在设置数组时出错的地方，因为我认为我遵循了说明。
最佳答案
使用TQLI算法的特征值计算因分割而失败
故障
分割错误来自于越过提供的阵列边界。 tqli需要特定的数据准备。
1）eigen code from CACS基于Fortran，从1开始计算索引。
2）tqli期望double指针为其矩阵和double向量。
/******************************************************************************/
void tqli(double d[], double e[], int n, double **z)
/*******************************************************************************
d和e应该声明为double。3）该程序需要针对上述功能的数据准备进行修改。
必须创建基于Helper 1索引的向量才能为
tqli提供正确格式的数据： double z[NP][NP] = { {2, 0, 0}, {0, 4, 0}, {0, 0, 2} } ;
double **a;
double *d,*e,*f;
d=dvector(1,NP); // 1-index based vector
e=dvector(1,NP);
f=dvector(1,NP);
a=dmatrix(1,NP,1,NP); // 1-index based matrix
for (i=1;i<=NP;i++) // loading data from zero besed `ze` to `a`
for (j=1;j<=NP;j++) a[i][j]=z[i-1][j-1];
下面提供了完整的测试程序。它使用eigen code from CACS：
/*******************************************************************************
Eigenvalue solvers, tred2 and tqli, from "Numerical Recipes in C" (Cambridge
Univ. Press) by W.H. Press, S.A. Teukolsky, W.T. Vetterling, and B.P. Flannery
*******************************************************************************/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define NR_END 1
#define SIGN(a,b) ((b) >= 0.0 ? fabs(a) : -fabs(a))
double **dmatrix(int nrl, int nrh, int ncl, int nch)
/* allocate a double matrix with subscript range m[nrl..nrh][ncl..nch] */
{
int i,nrow=nrh-nrl+1,ncol=nch-ncl+1;
double **m;
/* allocate pointers to rows */
m=(double **) malloc((size_t)((nrow+NR_END)*sizeof(double*)));
m += NR_END;
m -= nrl;
/* allocate rows and set pointers to them */
m[nrl]=(double *) malloc((size_t)((nrow*ncol+NR_END)*sizeof(double)));
m[nrl] += NR_END;
m[nrl] -= ncl;
for(i=nrl+1;i<=nrh;i++) m[i]=m[i-1]+ncol;
/* return pointer to array of pointers to rows */
return m;
}
double *dvector(int nl, int nh)
/* allocate a double vector with subscript range v[nl..nh] */
{
double *v;
v=(double *)malloc((size_t) ((nh-nl+1+NR_END)*sizeof(double)));
return v-nl+NR_END;
}
/******************************************************************************/
void tred2(double **a, int n, double d[], double e[])
/*******************************************************************************
Householder reduction of a real, symmetric matrix a[1..n][1..n].
On output, a is replaced by the orthogonal matrix Q effecting the
transformation. d[1..n] returns the diagonal elements of the tridiagonal matrix,
and e[1..n] the off-diagonal elements, with e[1]=0. Several statements, as noted
in comments, can be omitted if only eigenvalues are to be found, in which case a
contains no useful information on output. Otherwise they are to be included.
*******************************************************************************/
{
int l,k,j,i;
double scale,hh,h,g,f;
for (i=n;i>=2;i--) {
l=i-1;
h=scale=0.0;
if (l > 1) {
for (k=1;k<=l;k++)
scale += fabs(a[i][k]);
if (scale == 0.0) /* Skip transformation. */
e[i]=a[i][l];
else {
for (k=1;k<=l;k++) {
a[i][k] /= scale; /* Use scaled a's for transformation. */
h += a[i][k]*a[i][k]; /* Form sigma in h. */
}
f=a[i][l];
g=(f >= 0.0 ? -sqrt(h) : sqrt(h));
e[i]=scale*g;
h -= f*g; /* Now h is equation (11.2.4). */
a[i][l]=f-g; /* Store u in the ith row of a. */
f=0.0;
for (j=1;j<=l;j++) {
/* Next statement can be omitted if eigenvectors not wanted */
a[j][i]=a[i][j]/h; /* Store u/H in ith column of a. */
g=0.0; /* Form an element of A.u in g. */
for (k=1;k<=j;k++)
g += a[j][k]*a[i][k];
for (k=j+1;k<=l;k++)
g += a[k][j]*a[i][k];
e[j]=g/h; /* Form element of p in temporarily unused element of e. */
f += e[j]*a[i][j];
}
hh=f/(h+h); /* Form K, equation (11.2.11). */
for (j=1;j<=l;j++) { /* Form q and store in e overwriting p. */
f=a[i][j];
e[j]=g=e[j]-hh*f;
for (k=1;k<=j;k++) /* Reduce a, equation (11.2.13). */
a[j][k] -= (f*e[k]+g*a[i][k]);
}
}
} else
e[i]=a[i][l];
d[i]=h;
}
/* Next statement can be omitted if eigenvectors not wanted */
d[1]=0.0;
e[1]=0.0;
/* Contents of this loop can be omitted if eigenvectors not
wanted except for statement d[i]=a[i][i]; */
for (i=1;i<=n;i++) { /* Begin accumulation of transformation matrices. */
l=i-1;
if (d[i]) { /* This block skipped when i=1. */
for (j=1;j<=l;j++) {
g=0.0;
for (k=1;k<=l;k++) /* Use u and u/H stored in a to form P.Q. */
g += a[i][k]*a[k][j];
for (k=1;k<=l;k++)
a[k][j] -= g*a[k][i];
}
}
d[i]=a[i][i]; /* This statement remains. */
a[i][i]=1.0; /* Reset row and column of a to identity matrix for next iteration. */
for (j=1;j<=l;j++) a[j][i]=a[i][j]=0.0;
}
}
/******************************************************************************/
void tqli(double d[], double e[], int n, double **z)
/*******************************************************************************
QL algorithm with implicit shifts, to determine the eigenvalues and eigenvectors
of a real, symmetric, tridiagonal matrix, or of a real, symmetric matrix
previously reduced by tred2 sec. 11.2. On input, d[1..n] contains the diagonal
elements of the tridiagonal matrix. On output, it returns the eigenvalues. The
vector e[1..n] inputs the subdiagonal elements of the tridiagonal matrix, with
e[1] arbitrary. On output e is destroyed. When finding only the eigenvalues,
several lines may be omitted, as noted in the comments. If the eigenvectors of
a tridiagonal matrix are desired, the matrix z[1..n][1..n] is input as the
identity matrix. If the eigenvectors of a matrix that has been reduced by tred2
are required, then z is input as the matrix output by tred2. In either case,
the kth column of z returns the normalized eigenvector corresponding to d[k].
*******************************************************************************/
{
double pythag(double a, double b);
int m,l,iter,i,k;
double s,r,p,g,f,dd,c,b;
for (i=2;i<=n;i++) e[i-1]=e[i]; /* Convenient to renumber the elements of e. */
e[n]=0.0;
for (l=1;l<=n;l++) {
iter=0;
do {
for (m=l;m<=n-1;m++) { /* Look for a single small subdiagonal element to split the matrix. */
dd=fabs(d[m])+fabs(d[m+1]);
if ((double)(fabs(e[m])+dd) == dd) break;
}
if (m != l) {
if (iter++ == 30) printf("Too many iterations in tqli");
g=(d[l+1]-d[l])/(2.0*e[l]); /* Form shift. */
r=pythag(g,1.0);
g=d[m]-d[l]+e[l]/(g+SIGN(r,g)); /* This is dm - ks. */
s=c=1.0;
p=0.0;
for (i=m-1;i>=l;i--) { /* A plane rotation as in the original QL, followed by Givens */
f=s*e[i]; /* rotations to restore tridiagonal form. */
b=c*e[i];
e[i+1]=(r=pythag(f,g));
if (r == 0.0) { /* Recover from underflow. */
d[i+1] -= p;
e[m]=0.0;
break;
}
s=f/r;
c=g/r;
g=d[i+1]-p;
r=(d[i]-g)*s+2.0*c*b;
d[i+1]=g+(p=s*r);
g=c*r-b;
/* Next loop can be omitted if eigenvectors not wanted */
for (k=1;k<=n;k++) { /* Form eigenvectors. */
f=z[k][i+1];
z[k][i+1]=s*z[k][i]+c*f;
z[k][i]=c*z[k][i]-s*f;
}
}
if (r == 0.0 && i >= l) continue;
d[l] -= p;
e[l]=g;
e[m]=0.0;
}
} while (m != l);
}
}
/******************************************************************************/
double pythag(double a, double b)
/*******************************************************************************
Computes (a2 + b2)1/2 without destructive underflow or overflow.
*******************************************************************************/
{
double absa,absb;
absa=fabs(a);
absb=fabs(b);
if (absa > absb) return absa*sqrt(1.0+(absb/absa)*(absb/absa));
else return (absb == 0.0 ? 0.0 : absb*sqrt(1.0+(absa/absb)*(absa/absb)));
}
#define NP 3
#define TINY 1.0e-6
double sqrt(double x)
{
union
{
int i;
double x;
} u;
u.x = x;
u.i = (1<<29) + (u.i >> 1) - (1<<22);
return u.x;
}
int main()
{
int i,j,k;
double ze[NP][NP] = { {2, 0, 0}, {0, 4, 0}, {0, 0, 2} } ;
double **a;
double *d,*e,*f;
d=dvector(1,NP);
e=dvector(1,NP);
f=dvector(1,NP);
a=dmatrix(1,NP,1,NP);
for (i=1;i<=NP;i++)
for (j=1;j<=NP;j++) a[i][j]=ze[i-1][j-1];
tred2(a,NP,d,e);
tqli(d,e,NP,a);
printf("\nEigenvectors for a real symmetric matrix:\n");
for (i=1;i<=NP;i++) {
for (j=1;j<=NP;j++) {
f[j]=0.0;
for (k=1;k<=NP;k++)
f[j] += (ze[j-1][k-1]*a[k][i]);
}
printf("%s %3d %s %10.6f\n","\neigenvalue",i," =",d[i]);
printf("%11s %14s %9s\n","vector","mtrx*vect.","ratio");
for (j=1;j<=NP;j++) {
if (fabs(a[j][i]) < TINY)
printf("%12.6f %12.6f %12s\n",
a[j][i],f[j],"div. by 0");
else
printf("%12.6f %12.6f %12.6f\n",
a[j][i],f[j],f[j]/a[j][i]);
}
}
//free_dmatrix(a,1,NP,1,NP);
//free_dvector(f,1,NP);
//free_dvector(e,1,NP);
//free_dvector(d,1,NP);
return 0;
}
输出：
Eigenvectors for a real symmetric matrix:
eigenvalue 1 = 2.000000
vector mtrx*vect. ratio
1.000000 2.000000 2.000000
0.000000 0.000000 div. by 0
0.000000 0.000000 div. by 0
eigenvalue 2 = 4.000000
vector mtrx*vect. ratio
0.000000 0.000000 div. by 0
1.000000 4.000000 4.000000
0.000000 0.000000 div. by 0
eigenvalue 3 = 2.000000
vector mtrx*vect. ratio
0.000000 0.000000 div. by 0
0.000000 0.000000 div. by 0
1.000000 2.000000 2.000000
我希望它最终有助于澄清有关
tqli数据准备的困惑。关于c - 使用TQLI算法的特征值计算失败并出现段错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/48945001/