Hi!
In the below code, I have declared one matrix and one vector: A[3][3] and B[3]. The matrix A is symmetric from main diagonal. I converted the matrix A from dense format to CSR format, and from CSR format to coordinate format using the function mkl_dcsrcoo. All conversions are ok, but I can not solve correctly the system AxY=B. I obtained a wrong solution. The correct solution is: Y=[0.5 0.5 0.5]
How can I obtain the correct solution?
The code is following:
#include "stdafx.h" #include <stdlib.h> #include <stdio.h> #include<conio.h> #include<math.h> #include "mkl.h" #include "mkl_lapacke.h" int _tmain(int argc, _TCHAR* argv[]) { double A[3][3] = {{2.5,-0.5,0.},{-0.5,5,-0.5},{0,-0.5,2.5}}; double B[3]={1.0,2.0,1.0}; MKL_INT rows = 3, cols = 3, nnz = 7,info; MKL_INT job[6]={0,0,0,2,nnz,2}; //for conversion of matrix A to CSR format double *arrV = new double[nnz]; MKL_INT *Columns = new MKL_INT[nnz]; MKL_INT *Rows = new MKL_INT[rows+1]; mkl_ddnscsr(job,&rows, &cols, A[0], &rows, arrV, Columns, Rows,&info); printf("\Values\n"); for(int i=0;i<nnz;i++) printf("%.3f ",arrV[i]); printf("\nColumns\n"); for(int i=0;i<nnz;i++) printf("%d ",Columns[i]); printf("\nRows\n"); for(int i=0;i<rows+1;i++) printf("%d ",Rows[i]); printf("\n"); //convert to column format double *acoo = new double[nnz]; int *rowind = new int[nnz]; int *colind = new int[nnz]; MKL_INT job1[6]={0,0,0,0,nnz,3}; mkl_dcsrcoo(job1,&rows,arrV,Columns,Rows,&nnz,acoo,rowind,colind,&info); printf("info mkl_dcsrcoo=%d\nacoo\n",info); for(int i=0;i<nnz;i++) printf("%.3f ",acoo[i]); printf("\n"); for(int i=0;i<nnz;i++) printf("%d ",rowind[i]); printf("\n"); for(int i=0;i<nnz;i++) printf("%d ",colind[i]); MKL_INT colsC=3; double alpha=1; double *Y=new double[3]; mkl_dcoosm ("n",&rows,&colsC,&alpha,"TUNC",acoo,rowind,colind,&nnz,B,&rows,Y,&colsC); printf("\n"); for(int i=0;i<3;i++) printf("%.3f ",Y[i]); delete[] arrV; delete[] Columns; delete[] Rows; delete[] acoo; delete[] rowind; delete[] colind; _getch(); return 0; }