我试图使用LAPACK的C接口来求解一个通用带状矩阵,在Intel的MKL中称为LAPACKE。我试图调用的函数是*gbsv,其中*表示格式。不幸的是,我发现很难找到关于如何使用C接口格式化带状矩阵的工作示例。如果有人可以为所有的C用户提供一个工作的例子,我向你保证这将是有帮助的。
fortran布局给出了一个示例这里,但我不太确定如何将其格式化为LAPACKE的输入。我还应该指出,在我的问题中,我必须动态地建立带状矩阵。所以我有5个系数,A,B,C,D,E,每个I-节点,它们必须被放入一个带状矩阵形式,然后传递给LAPACKE。
发布于 2016-04-02 14:39:46
函数LAPACKE_dgbsv()的原型如下:
lapack_int LAPACKE_dgbsv( int matrix_layout, lapack_int n, lapack_int kl,
lapack_int ku, lapack_int nrhs, double* ab,
lapack_int ldab, lapack_int* ipiv, double* b,
lapack_int ldb )Lapack的函数dgbsv()的主要区别是参数matrix_layout,它可以是LAPACK_ROW_MAJOR (C序),也可以是LAPACK_COL_MAJOR (Fortran排序)。如果LAPACK_ROW_MAJOR,LAPACKE_dgbsv将转置矩阵,调用dgbsv(),然后将矩阵转回C序。
其他参数的含义与函数dgbsv()相同。如果使用LAPACK_ROW_MAJOR,那么dgbsv()的正确ldab将由LAPACKE_dgbsv()计算,参数ldab可以设置为n。但是,就像dgbsv()一样,必须为矩阵ab分配额外的空间来存储分解的细节。
下面的例子利用LAPACKE_dgbsv()用中心有限差分法求解一维平稳扩散问题。考虑了零温度边界条件,并以正弦波中的一个作为源项来检验其正确性。以下程序是由gcc main3.c -o main3 -llapacke -llapack -lblas -Wall编译的
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <lapacke.h>
int main(void){
srand (time(NULL));
//size of the matrix
int n=10;
// number of right-hand size
int nrhs=4;
int ku=2;
int kl=2;
// ldab is larger than the number of bands,
// to store the details of factorization
int ldab = 2*kl+ku+1;
//memory initialization
double *a=malloc(n*ldab*sizeof(double));
if(a==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
double *b=malloc(n*nrhs*sizeof(double));
if(b==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
int *ipiv=malloc(n*sizeof(int));
if(ipiv==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
int i,j;
double fact=1*((n+1.)*(n+1.));
//matrix initialization : the different bands
// are stored in rows kl <= j< 2kl+ku+1
for(i=0;i<n;i++){
a[(0+kl)*n+i]=0;
a[(1+kl)*n+i]=-1*fact;
a[(2+kl)*n+i]=2*fact;
a[(3+kl)*n+i]=-1*fact;
a[(4+kl)*n+i]=0;
//initialize source terms
for(j=0;j<nrhs;j++){
b[i*nrhs+j]=sin(M_PI*(i+1)/(n+1.));
}
}
printf("end ini \n");
int ierr;
// ROW_MAJOR is C order, Lapacke will compute ldab by himself.
ierr=LAPACKE_dgbsv(LAPACK_ROW_MAJOR, n, kl,ku,nrhs, a,n, ipiv, b,nrhs );
if(ierr<0){LAPACKE_xerbla( "LAPACKE_dgbsv", ierr );}
printf("output of LAPACKE_dgbsv\n");
for(i=0;i<n;i++){
for(j=0;j<nrhs;j++){
printf("%g ",b[i*nrhs+j]);
}
printf("\n");
}
//checking correctness
double norm=0;
double diffnorm=0;
for(i=0;i<n;i++){
for(j=0;j<nrhs;j++){
norm+=b[i*nrhs+j]*b[i*nrhs+j];
diffnorm+=(b[i*nrhs+j]-1./(M_PI*M_PI)*sin(M_PI*(i+1)/(n+1.)))*(b[i*nrhs+j]-1./(M_PI*M_PI)*sin(M_PI*(i+1)/(n+1.)));
}
}
printf("analical solution is 1/(PI*PI)*sin(x)\n");
printf("relative difference is %g\n",sqrt(diffnorm/norm));
free(a);
free(b);
free(ipiv);
return 0;
}https://stackoverflow.com/questions/36345090
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