我不知道为什么,但是我的矩阵乘法非常慢,我需要优化它。而且矩阵(1000X1000)的打印也需要很长时间。该函数的目的是计算矩阵指数,但我的主要问题是,这两个操作对于像1000X1000这样的大型矩阵来说非常慢。
这两个动作分别在poweMat()函数和printeResult()函数中实现,代码如下:
#define M 1000
#define e 2.71828182845904523536;
//declaration of the functions
void sumMatrices(vector<vector<double> >& mat1, vector<vector<double> >& mat2);
void printResult(vector<vector<double> >&matRes);
void mulMatWithFactorial(long factorialValue);
long factorialCalculate(int n);
void initializeMatrix();
void initializeIdenticalMatrix();
void checkIfTheMatrixIsDiagonal();
void calculateExpoMatrixWithDiagonalMatrix();
void readMatrixFromFile();
void powerMat(vector<vector<double> >& mat, int powNum);
//declaration of the variables
vector<vector<double>> inputMatrix(M, vector<double>(M));
vector<vector<double>> sumMatrixResult(M, vector<double>(M));
vector<vector<double>> powerMatrixResult(M, vector<double>(M));
vector<vector<double>> mulFactorialMatrixResult(M, vector<double>(M));
vector<vector<double>> finalMatrixResult(M, vector<double>(M));
vector<vector<double>> identicalMatrix(M, vector<double>(M));
vector<vector<vector<double>>> listOfMatrices;
bool matrixIsNilpotent = false;
int diagonaMatrixlFlag = 1;
int main() {
//variables
long factorialValue;
initializeIdenticalMatrix();
readMatrixFromFile();
//check if the matrix is diagonal - so we will have easier and faster compute
checkIfTheMatrixIsDiagonal();
if (diagonaMatrixlFlag == 1) {
calculateExpoMatrixWithDiagonalMatrix();
goto endOfLoop;
}
//loop for taylor series
for (int i = 0; i < 5; i++) {
if (i == 0) { // first we add identical matrix when the power is 0
sumMatrices(finalMatrixResult, identicalMatrix); // summarize between this 2 matrices
finalMatrixResult = sumMatrixResult; //copy matrices
}
if (i == 1) { // we add the matrix itself because the power is 1
sumMatrices(finalMatrixResult, inputMatrix);
finalMatrixResult = sumMatrixResult; //copy matrices
}
if (i > 1 ) {
powerMat(inputMatrix, i);
if (matrixIsNilpotent) { // it means that A^i is 0 for some integer, so the series terminates after a finite number
goto endOfLoop;
}
factorialValue = factorialCalculate(i); // calculate the factorial of i
mulMatWithFactorial(factorialValue); // multiply (1/i) * matrix^i - like in the algorithm
sumMatrices(finalMatrixResult, mulFactorialMatrixResult); // summarize it with the previous result
finalMatrixResult = sumMatrixResult; //copy matrices
}
}
endOfLoop:
printResult(finalMatrixResult); // print the final result - e^M
return 0;
}
//Summarize matrices
void sumMatrices(vector<vector<double> >& mat1, vector<vector<double> >& mat2) {
for (int i = 0; i < M; i++)
for (int j = 0; j < M; j++)
sumMatrixResult[i][j] = mat1[i][j] + mat2[i][j];
}
//Print matrix
void printResult(vector<vector<double> >& matRes) {
for (int i = 0; i < M; i++) {
for (int j = 0; j < M; j++) {
printf("%f ", matRes[i][j]);
if (j == M - 1) {
printf("\n");
}
}
}
}
//Calculate the factorial of n
long factorialCalculate(int n) {
long factorial = 1.0;
for (int i = 1; i <= n; ++i) {
factorial *= i;
}
return factorial;
}
// mutiply the matrix with scalar
void mulMatWithFactorial(long factorialValue) {
for (int i = 0; i < M; i++) {
for (int j = 0; j < M; j++) {
mulFactorialMatrixResult[i][j] = powerMatrixResult[i][j] * 1/factorialValue;
}
}
}
//initialize matrix
void initializeMatrix() {
for (int i = 0; i < M; i++) {
for (int j = 0; j < M; j++) {
powerMatrixResult[i][j] = 0;
}
}
}
void checkIfTheMatrixIsDiagonal() {
for (int i = 0; i < M; i++) {
for (int j = 0; j < M; j++) {
if (i == j)
{
if (inputMatrix[i][j] == 0) {
diagonaMatrixlFlag = 0;
goto endOfLoop;
}
}
else
{
if (inputMatrix[i][j] != 0) {
diagonaMatrixlFlag = 0;
goto endOfLoop;
}
}
}
}
endOfLoop:
return;
}
void calculateExpoMatrixWithDiagonalMatrix() {
for (int i = 0; i < M; i++) {
for (int j = 0; j < M; j++) {
if (i == j)
{
for (int k = 0; k < inputMatrix[i][j]; ++k)// loop to calculate the pow of e^alpha
{
finalMatrixResult[i][j] *= e;
}
}
else
{
finalMatrixResult[i][j] = 0;
}
}
}
}
void readMatrixFromFile() {
ifstream f("inv_matrix(1000x1000).txt");
for (int i = 0; i < M; i++)
for (int j = 0; j < M; j++) {
f >> inputMatrix[i][j];
if (f.peek() == ',')
f.ignore();
}
listOfMatrices.push_back(inputMatrix);
}
void initializeIdenticalMatrix() {
for (int i = 0; i < M; i++) {
for (int k = 0; k < M; k++) {
if (i == k) {
identicalMatrix[i][k] = 1;
}
else {
identicalMatrix[i][k] = 0;
}
}
}
}
void powerMat(vector<vector<double> >& mat, int powNum) {
int counterForNilpotent = 0;
initializeMatrix();
auto start = high_resolution_clock::now();
for (int i = 0; i < M; i++) {
for (int k = 0; k < M; k++) {
for (int j = 0; j < M; j++) {
powerMatrixResult[i][j] += mat[i][k] * listOfMatrices[powNum-2][k][j];
}
}
}
auto stop = high_resolution_clock::now();
auto duration = duration_cast<seconds>(stop - start);
cout << duration.count() << " seconds" << endl; // checking run time
listOfMatrices.push_back(powerMatrixResult);
// check if after we we did A^i , the matrix is equal to 0
for (int i = 0; i < M; i++) {
for (int j = 0; j < M; j++) {
if (powerMatrixResult[i][j] == 0) {
counterForNilpotent++;
}
}
}
if (counterForNilpotent == M * M) {
matrixIsNilpotent = true;
}
}
2条答案
按热度按时间w8f9ii691#
遍历n个数组中的每个元素的计算效率为O(n^2),这意味着对于大型数组来说,这需要一段时间,但不会是“宇宙生命周期”的时间长度。
通常要对这样的大规模数组进行运算,首先要对它们进行某种形式的约简,这样计算就可以更接近O(n),或者更好地使用矩阵约简形式的一些真理。
因此,更快的矩阵乘法实现将从两个矩阵上的某个rref()函数开始,然后仅计算那些在列和行中具有对象的矩阵部分。
这里有一些很好的地方复习/学习(免费)线性代数:
“3b 1b(2016年):线性代数的本质”= https://www.youtube.com/watch?v=kjBOesZCoqc&list=PL0-GT3co4r2y2YErbmuJw2L5tW4Ew2O5B
“麻省理工学院开放式课程(2009):线性代数”= https://www.youtube.com/watch?v=ZK3O402wf1c&list=PL49CF3715CB9EF31D&index=1
yks3o0rb2#
使用SSE2。它不是一个库。它是一个使用cpu矢量硬件的方法。
将操作设置为并行运行。
https://en.wikipedia.org/wiki/SSE2