#include <iostream> #include <chrono> #include <opencv2/opencv.hpp> #include <Eigen/Core> #include <Eigen/Dense> using namespace std; using namespace Eigen; int main(int argc, char **argv) { double ar = 1.0, br = 2.0, cr = 1.0; // 真实参数值 double ae = 2.0, be = -1.0, ce = 5.0; // 估计参数值初始值 int N = 100; // 数据点 double w_sigma = 2.0; // 噪声Sigma值 double inv_sigma = 1.0 / w_sigma; cv::RNG rng; // OpenCV随机数产生器 vector<double> x_data, y_data; // 真实数据 for (int i = 0; i < N; i++) { //生成数据 double x = i / 100.0; x_data.push_back(x); y_data.push_back(exp(ar * x * x + br * x + cr) + rng.gaussian(w_sigma * w_sigma)); } // 开始Gauss-Newton迭代 int iterations = 1000; // 迭代次数 double cost = 0, lastCost = 0; // 本次迭代的cost和上一次迭代的cost chrono::steady_clock::time_point t1 = chrono::steady_clock::now(); for (int iter = 0; iter < iterations; iter++) { Matrix3d H = Matrix3d::Zero(); // Hessian=J*J^T in Gauss-Newton Vector3d b = Vector3d::Zero(); // bias 书本中的g(x) cost = 0; for (int i = 0; i < N; i++) { double xi = x_data[i], yi = y_data[i]; // 第i个数据点 double error = yi - exp(ae * xi * xi + be * xi + ce); Vector3d J; // 雅可比矩阵 J[0] = -xi * xi * exp(ae * xi * xi + be * xi + ce); // de/da J[1] = -xi * exp(ae * xi * xi + be * xi + ce); // de/db J[2] = -exp(ae * xi * xi + be * xi + ce); // de/dc // H += inv_sigma * inv_sigma * J * J.transpose();//不知道为什么要乘inv_sigma * inv_sigma // b += -inv_sigma * inv_sigma * error * J; //两边同乘一个不为0的常数,实际对结果没有影响 H += J * J.transpose(); b += -error * J; cost += error * error; } // 求解线性方程 Hx=b Vector3d dx = H.ldlt().solve(b); if (isnan(dx[0])) { cout << "result is nan!" << endl; break; } if (iter > 0 && cost >= lastCost) { cout << "cost: " << cost << ">= last cost: " << lastCost << ", break." << endl; break; } ae += dx[0]; be += dx[1]; ce += dx[2]; lastCost = cost; cout << "total cost: " << cost << ", update: " << dx.transpose() << " estimated params: " << ae << "," << be << "," << ce << endl; } chrono::steady_clock::time_point t2 = chrono::steady_clock::now(); chrono::duration<double> time_used = chrono::duration_cast<chrono::duration<double>>(t2 - t1); cout << "solve time cost = " << time_used.count() << " seconds. " << endl; cout << "estimated abc = " << ae << ", " << be << ", " << ce << endl; return 0; }