我正在使用RFP格式进行矩阵存储,并试图解决方程组。但结果是错误的。 =(请有人帮忙。
我明白了 b = {5.5,10,5.5}
但应该 b = {2.875,4.75,3.5}
我没有得到,我犯了错误。只需简单地使用标准函数:分解,然后求解分解矩阵。
谢谢。
#include "mkl.h"
#include "mkl_lapacke.h"
#include <math.h>
#include <iostream>
using namespace std;
#define NI 3
#define NJ 1
int main(int argc, char* argv[])
{
double a[NI][NI];
double b[NI][NJ];
a[0][0] = 2; a[0][1] = -1; a[0][2] = 0;
a[1][0] = 0; a[1][1] = 2; a[1][2] = -1;
a[2][0] = 0; a[2][1] = 0; a[2][2] = 2;
b[0][0] = 1;
b[0][1] = 6;
b[0][2] = 7;
cout << "A1 = \n";
for(int i = 0; i < NI; i++) {
for(int j = 0; j < NI; j++) {
cout << a[i][j] << "\t";
}
cout << "\n";
}
cout << "\n";
cout << "B1 = \n";
for(int i = 0; i < NI; i++) {
for(int j = 0; j < NJ; j++) {
cout << b[i][j] << "\t";
}
cout << "\n";
}
cout << "\n";
char transr = 'N';
char uplo = 'U';
lapack_int n = NI;
lapack_int lda = NI; //LDA is used to define the distance in memory between elements of two consecutive columns which have the same row index.
double * arf = new double[ n * ( n + 1 ) / 2 ];
lapack_int info = -1;
将一般矩阵转换为RFP格式
info = LAPACKE_dtrttf(LAPACK_ROW_MAJOR, transr, uplo, n, *a, lda, arf);
//lapack_int LAPACKE_<?>trttf( int matrix_order, char transr, char uplo, lapack_int n, const <datatype>* a, lapack_int lda, <datatype>* arf );
cout << "LAPACKE_dtrttf = " << info << "\n";
cout << "Rectangular full packed: \n";
//cout.setf(std::ios::scientific);
for(int i = 0; i < NI; i++) {
for(int j = 0; j < (NI+1)/2; j++) {
cout << arf[i * (NI+1)/2 + j] << "\t";
//cout << arf[i] << "\t";
}
cout << "\n";
}
cout << "\n";
分解矩阵
int matrix_order = LAPACK_ROW_MAJOR;
transr = 'N';
uplo = 'U';
n = NI;
info = LAPACKE_dpftrf( matrix_order, transr, uplo, n, arf );
//lapack_int LAPACKE_<?>pftrf( int matrix_order, char transr, char uplo, lapack_int n, <datatype>* a );
cout << "INFO LAPACKE_dpftrf = " << info << "\n";
cout << "Factorized matrix: " << endl;
for(int i = 0; i < NI; i++) {
for(int j = 0; j < (NI+1)/2; j++) {
cout << arf[i*(NI+1)/2+j] << "\t";
}
cout << "\n";
}
cout << "\n";
解决系统
lapack_int nrhs = NJ;
lapack_int ldb = NJ;
info = LAPACKE_dpftrs( matrix_order, transr, uplo, n, nrhs, arf, &b[0][0], ldb );
//lapack_int LAPACKE_<?>pftrs( int matrix_order, char transr, char uplo, lapack_int n, lapack_int nrhs, const <datatype>* a, <datatype>* b, lapack_int ldb );
cout << "INFO LAPACKE_dpftrs = " << info << "\n";
结果
cout << "Solved = \n";
for(int i = 0; i < NI; i++) {
for(int j = 0; j < NJ; j++) {
cout << b[i][j] << "\t";
}
cout << "\n";
}
cout << "\n";
delete [] arf;
char ch;
cin.get(ch);
return 0;
}
答案 0 :(得分:1)
您确定NJ的值对于以下内容是否正确:
b[NI][NJ];
b[0][0] = 1;
b[0][1] = 6;
b[0][2] = 7;
其中NI = 3且NJ = 1 ......
我认为你错误地将行的值放在了列上。如果是问题,那么你可能想知道为什么它没有给出任何错误,这又是一个已经存在的问题: Accessing an array out of bounds gives no error, why?