The simplest matrix is divided into the following steps
Simplify from row i=0 of the matrix
1. Find the longest element row: starting from the i-th row, look for the first row whose element is not 0 at the beginning, and use row swap to change the row to the i-th row of the matrix.
2. Simplify the i-th row: all the elements in the i-th row are the same except the first non-zero element of the row.
3. Matrix row simplification: Start from the j=0th row of the matrix. The element in the same column as the first non-zero element in the i-th row is the multiple of the row simplification. Each row subtracts this multiple * the value of the i-th row to get the result of a row simplification. Go back to step 1, i++.
/**
* 矩阵行变换成最简矩阵
* @tparam ElemType
* @return
*/
template<typename ElemType>
SmartDongLib::Matrix<ElemType> SmartDongLib::Matrix<ElemType>::simplyTransform() {
Matrix<ElemType> ret(*this);
//表示将要化1的矩阵行
for (int transferRow = 0; transferRow < theRows_; ++transferRow) {
//按列循环每行找首个非0元素，从transferRow行开始寻找最长元素行
int firstNotNullCol = 0;
for (; firstNotNullCol < ret.theCols_; ++firstNotNullCol) {
int row = transferRow;
bool isfind= false;
for (; row < ret.theRows_; ++row) {
ElemType firstelem=ret(row, firstNotNullCol);
if (!close(firstelem, (ElemType)0 )) {
isfind = true;
break;
}
}
if (isfind){
//如果找到最长元素行，则和transferRow行进行行交换
ret.rowSwap(transferRow,row);
break;
}
}
if (firstNotNullCol >= ret.theCols_){
//零矩阵或者后继行都是0
return ret;
}
//当前行首元素化1
Real factor = ret(transferRow,firstNotNullCol);
for (int col = firstNotNullCol; col < ret.theCols_; ++col) {
ret(transferRow,col) = ret(transferRow,col) / factor;
}
//同列元素行化0
for (int i = 0 ; i < ret.theRows_; ++i) {
if ( i ==transferRow )
continue;
Real factor2 = ret(i,firstNotNullCol);
for (int j = firstNotNullCol; j < ret.theCols_; ++j) {
ret(i,j) = ret(i,j) - ret(transferRow,j) * factor2;
}
}
}
return ret;
}Note: This code has operator overloading on parentheses and only provides programming ideas. If you have suggestions for code optimization, you can comment.