《视觉SLAM十四讲》手写高斯牛顿—笔记记录

《视觉SLAM十四讲》手写高斯牛顿—笔记记录

在这里插入图片描述
在这里插入图片描述

我们的最终目的:使用高斯牛顿法,拟合参数abc
我们的实际小目标:求解增量方程得到ΔX(有了Δx就可以不停的迭代Eabc使得拟合Rabc啦)

三个步骤:
1、先根据模型生成x,y的真值,并在真值中添加高斯分布的噪声
2、使用高斯牛顿法进行迭代
3、求解高斯牛顿法的增量方程

首先定义真实的a,b,c参数值为ar,br,cr这三个也是我们要拟合的最终目标,最终的拟合结果跟这三个值越接近越好
double ar = 1.0, br = 2.0, cr = 1.0;
然后定义初始的估计参数值ae,be,ce这三个就是用于迭代的,最后效果最好是ae,be,ce迭代到与ar,br,cr一样。
 double ae = 2.0, be = -1.0, ce = 5.0; 
然后我们假设有100个数据点,设置噪声值sigma

第一步,根据模型生成X,Y的真值,并添加高斯分布的噪声

  vector<double> x_data, y_data;      // 数据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));}

第二步,使用高斯牛顿法进行迭代
迭代次数设为100次
定义误差:

     double error = yi - exp(ae * xi * xi + be * xi + ce);

然后分别对每个误差项对a,b,c进行求导,然后生成雅可比矩阵

      Vector3d J; // 雅可比矩阵J[0] = -xi * xi * exp(ae * xi * xi + be * xi + ce);  // de/daJ[1] = -xi * exp(ae * xi * xi + be * xi + ce);  // de/dbJ[2] = -exp(ae * xi * xi + be * xi + ce);  // de/dc

然后求解海塞矩阵H和偏移量b

     H += inv_sigma * inv_sigma * J * J.transpose();b += -inv_sigma * inv_sigma * error * J;

第三步,求解高斯牛顿法的增量方程(相当于梯度下降的Δx,就是向前迈一小步~)
这里的关键语句是Vector3d dx = H.ldlt().solve(b);
涉及一个eigen库中的LDLT分解,其中L为一下三角形单位矩阵(即主对角线元素皆为1),D为一对角矩阵(只在主对角线上有元素,其余皆为零),L^T为L的转置矩阵。

    // 求解线性方程 Hx=bVector3d 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;   //更新损失

完整代码如下:

#include 
#include 
#include 
#include 
#include using namespace std;
using namespace Eigen;//先根据模型生成x,y的真值
//在真值中添加高斯分布的噪声
//使用高斯牛顿法从带噪声的数据拟合参数模型int main(int argc, char **argv) {double ar = 1.0, br = 2.0, cr = 1.0;         // 真实参数值  最后就是要迭代到跟这个真值接近double ae = 2.0, be = -1.0, ce = 5.0;        // 估计参数值  随便给定的一个初值,用于迭代到真值附近int N = 100;                                 // 数据点double w_sigma = 1.0;                        // 噪声Sigma值double inv_sigma = 1.0 / w_sigma;cv::RNG rng;                                 // OpenCV随机数产生器vector<double> x_data, y_data;      // 数据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));}// 开始Gauss-Newton迭代int iterations = 100;    // 迭代次数double cost = 0, lastCost = 0;  // 本次迭代的cost和上一次迭代的costchrono::steady_clock::time_point t1 = chrono::steady_clock::now();for (int iter = 0; iter < iterations; iter++){Matrix3d H = Matrix3d::Zero();             // Hessian = J^T W^{-1} J in Gauss-NewtonVector3d b = Vector3d::Zero();             // biascost = 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/daJ[1] = -xi * exp(ae * xi * xi + be * xi + ce);  // de/dbJ[2] = -exp(ae * xi * xi + be * xi + ce);  // de/dcH += inv_sigma * inv_sigma * J * J.transpose();b += -inv_sigma * inv_sigma * error * J;cost += error * error;}// 求解线性方程 Hx=bVector3d 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 << ", \t\tupdate: " << dx.transpose() <<"\t\testimated 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;
}

最终结果:

/home/yjc/SLAM_Study/ch6/cmake-build-debug/gaussNewton
total cost: 3.19575e+06, 		update: 0.0455771  0.078164 -0.985329		estimated params: 2.04558,-0.921836,4.01467
total cost: 376785, 		update:  0.065762  0.224972 -0.962521		estimated params: 2.11134,-0.696864,3.05215
total cost: 35673.6, 		update: -0.0670241   0.617616  -0.907497		estimated params: 2.04432,-0.0792484,2.14465
total cost: 2195.01, 		update: -0.522767   1.19192 -0.756452		estimated params: 1.52155,1.11267,1.3882
total cost: 174.853, 		update: -0.537502  0.909933 -0.386395		estimated params: 0.984045,2.0226,1.00181
total cost: 102.78, 		update: -0.0919666   0.147331 -0.0573675		estimated params: 0.892079,2.16994,0.944438
total cost: 101.937, 		update: -0.00117081  0.00196749 -0.00081055		estimated params: 0.890908,2.1719,0.943628
total cost: 101.937, 		update:   3.4312e-06 -4.28555e-06  1.08348e-06		estimated params: 0.890912,2.1719,0.943629
total cost: 101.937, 		update: -2.01204e-08  2.68928e-08 -7.86602e-09		estimated params: 0.890912,2.1719,0.943629
cost: 101.937>= last cost: 101.937, break.
solve time cost = 0.000237057 seconds. 
estimated abc = 0.890912, 2.1719, 0.943629进程已结束,退出代码为 0


本文来自互联网用户投稿,文章观点仅代表作者本人,不代表本站立场,不承担相关法律责任。如若转载,请注明出处。 如若内容造成侵权/违法违规/事实不符,请点击【内容举报】进行投诉反馈!

相关文章

立即
投稿

微信公众账号

微信扫一扫加关注

返回
顶部