1.优化问题:
(y=exp(ax^{2}+bx+c)+w),由y和x,求解a,b,c
误差为:(e_{i}=y_{i}-exp(ax_{i}^{2}+bx_{i}+c))
误差项对每一个待估计量进行求导:
(frac{partial e_{i}}{partial a}=-x^{2}_{i}exp(ax^{2}_{i}+bx_{i}+c))
(frac{partial e_{i}}{partial b}=-x_{i}exp(ax^{2}_{i}+bx_{i}+c))
(frac{partial e_{i}}{partial c}=-exp(ax^{2}_{i}+bx_{i}+c))
雅可比矩阵(J_{i}=[frac{partial e_{i}}{partial a},frac{partial e_{i}}{partial b},frac{partial e_{i}}{partial c}]^{T}),
高斯的增量方程为: ((displaystyle sum^{100}_{i=1} J_{i}(sigma^{2})^{-1} J_{i}^{T})Delta x_{k}=displaystyle sum^{100}_{i=1} -J_{i}(sigma^{2})^{-1} e_{i})
(HDelta x_{k}=b)
噪声满足 w ~ ( 0,(sigma^{2}) )
#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; //真实参考值
double ae = 20.0, be = -10.0, ce = 10.0; //初始值,不能太大,初始化很重要
int N = 100; //数据总点数
double w_sigma = 1.0; //噪声sigma值
double inv_sigma = 1.0 / w_sigma;
cv::RNG rng; // opencv随机数产生
vector<double> x_data, y_data; //数据,生成真值数据加上随机数模拟实际采样值
for(int i=0; i<N; i++){
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) );
}
int iterations = 100; //迭代次数
double cost = 0, lastCost= 0; //每次迭代的误差平方和,用于判断退出迭代次数
chrono::steady_clock::time_point t1 = chrono::steady_clock::now();
for ( int iter=0; iter<iterations; iter++ ){
Matrix3d H = Matrix3d::Zero();
Vector3d b = Vector3d::Zero();
cost = 0;
for(int i=0; i<N; i++){
double xi = x_data[i], yi = y_data[i];
double error = yi - exp( ae*xi*xi + be*xi + ce );
Vector3d J; //雅克比矩阵
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();
b += -inv_sigma * inv_sigma * 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) { //误差变大,找到最小值,退出迭代
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;
}
CMakelists.txt:
cmake_minimum_required(VERSION 2.8)
project(gaussnewton)
set(CMAKE_CXX_FLAGS "${CMAKE_CXX_FLAGS} -std=c++11")
find_package(OpenCV REQUIRED)
include_directories(${OpenCV_INCLUDE_DIRS})
include_directories("/usr/include/eigen3")
set(SOURCE_FILES main.cpp)
add_executable(gaussnewton ${SOURCE_FILES})
target_link_libraries(gaussnewton ${OpenCV_LIBS})