视觉SLAM十四讲CH7课后习题10_2
转载于：在ceres中实现ICP优化（仅优化位姿）_luo870604851的博客-CSDN博客一.仅优化位姿构造类和代价函数：struct ICPCeres{ ICPCeres ( Point3f uvw,Point3f xyz ) : _uvw(uvw),_xyz(xyz) {} // 残差的计算 template <typename T> bool operator() ( const T* const c...https://blog.csdn.net/luo870604851/article/details/82356394
10. *在Ceres 中实现PnP 和ICP 的优化。
在原作者的基础上我加了CMakeLists.txt文件，原文rotation.h和主要代码是分开写的，在Ubuntu上用cmake来编译时会报错，所以我将两个文件统一到了同一个cpp文件里，其次，借助原作者的思路将其用在了pnp求解上，这样能方便大家能够在Ubuntu上编译执行。具体的代码注释没有写，如果有大神写完了，到时候可以私信我，把他分享给我，我将感激不尽。
10icp.cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include "sophus/se3.hpp"
using namespace std;
using namespace Eigen;
using namespace cv;
#ifndef ROTATION_H
#define ROTATION_H
// math functions needed for rotation conversion.
// dot and cross production
template
inline T DotProduct(const T x[3], const T y[3]) {
return (x[0] * y[0] + x[1] * y[1] + x[2] * y[2]);
}
template
inline void CrossProduct(const T x[3], const T y[3], T result[3]){
result[0] = x[1] * y[2] - x[2] * y[1];
result[1] = x[2] * y[0] - x[0] * y[2];
result[2] = x[0] * y[1] - x[1] * y[0];
}
//
// Converts from a angle anxis to quaternion :
template
inline void AngleAxisToQuaternion(const T* angle_axis, T* quaternion){
const T& a0 = angle_axis[0];
const T& a1 = angle_axis[1];
const T& a2 = angle_axis[2];
const T theta_squared = a0 * a0 + a1 * a1 + a2 * a2;
if(theta_squared > T(std::numeric_limits::epsilon()) ){
const T theta = sqrt(theta_squared);
const T half_theta = theta * T(0.5);
const T k = sin(half_theta)/theta;
quaternion[0] = cos(half_theta);
quaternion[1] = a0 * k;
quaternion[2] = a1 * k;
quaternion[3] = a2 * k;
}
else{ // in case if theta_squared is zero
const T k(0.5);
quaternion[0] = T(1.0);
quaternion[1] = a0 * k;
quaternion[2] = a1 * k;
quaternion[3] = a2 * k;
}
}
template
inline void QuaternionToAngleAxis(const T* quaternion, T* angle_axis){
const T& q1 = quaternion[1];
const T& q2 = quaternion[2];
const T& q3 = quaternion[3];
const T sin_squared_theta = q1 * q1 + q2 * q2 + q3 * q3;
// For quaternions representing non-zero rotation, the conversion
// is numercially stable
if(sin_squared_theta > T(std::numeric_limits::epsilon()) ){
const T sin_theta = sqrt(sin_squared_theta);
const T& cos_theta = quaternion[0];
// If cos_theta is negative, theta is greater than pi/2, which
// means that angle for the angle_axis vector which is 2 * theta
// would be greater than pi...
const T two_theta = T(2.0) * ((cos_theta < 0.0)
? atan2(-sin_theta, -cos_theta)
: atan2(sin_theta, cos_theta));
const T k = two_theta / sin_theta;
angle_axis[0] = q1 * k;
angle_axis[1] = q2 * k;
angle_axis[2] = q3 * k;
}
else{
// For zero rotation, sqrt() will produce NaN in derivative since
// the argument is zero. By approximating with a Taylor series,
// and truncating at one term, the value and first derivatives will be
// computed correctly when Jets are used..
const T k(2.0);
angle_axis[0] = q1 * k;
angle_axis[1] = q2 * k;
angle_axis[2] = q3 * k;
}
}
template
inline void AngleAxisRotatePoint(const T angle_axis[3], const T pt[3], T result[3]) {
const T theta2 = DotProduct(angle_axis, angle_axis);
if (theta2 > T(std::numeric_limits::epsilon())) {
// Away from zero, use the rodriguez formula
//
// result = pt costheta +
// (w x pt) * sintheta +
// w (w . pt) (1 - costheta)
//
// We want to be careful to only evaluate the square root if the
// norm of the angle_axis vector is greater than zero. Otherwise
// we get a division by zero.
//
const T theta = sqrt(theta2);
const T costheta = cos(theta);
const T sintheta = sin(theta);
const T theta_inverse = 1.0 / theta;
const T w[3] = { angle_axis[0] * theta_inverse,
angle_axis[1] * theta_inverse,
angle_axis[2] * theta_inverse };
// Explicitly inlined evaluation of the cross product for
// performance reasons.
/*const T w_cross_pt[3] = { w[1] * pt[2] - w[2] * pt[1],
w[2] * pt[0] - w[0] * pt[2],
w[0] * pt[1] - w[1] * pt[0] };*/
T w_cross_pt[3];
CrossProduct(w, pt, w_cross_pt);
const T tmp = DotProduct(w, pt) * (T(1.0) - costheta);
// (w[0] * pt[0] + w[1] * pt[1] + w[2] * pt[2]) * (T(1.0) - costheta);
result[0] = pt[0] * costheta + w_cross_pt[0] * sintheta + w[0] * tmp;
result[1] = pt[1] * costheta + w_cross_pt[1] * sintheta + w[1] * tmp;
result[2] = pt[2] * costheta + w_cross_pt[2] * sintheta + w[2] * tmp;
} else {
// Near zero, the first order Taylor approximation of the rotation
// matrix R corresponding to a vector w and angle w is
//
// R = I + hat(w) * sin(theta)
//
// But sintheta ~ theta and theta * w = angle_axis, which gives us
//
// R = I + hat(w)
//
// and actually performing multiplication with the point pt, gives us
// R * pt = pt + w x pt.
//
// Switching to the Taylor expansion near zero provides meaningful
// derivatives when evaluated using Jets.
//
// Explicitly inlined evaluation of the cross product for
// performance reasons.
/*const T w_cross_pt[3] = { angle_axis[1] * pt[2] - angle_axis[2] * pt[1],
angle_axis[2] * pt[0] - angle_axis[0] * pt[2],
angle_axis[0] * pt[1] - angle_axis[1] * pt[0] };*/
T w_cross_pt[3];
CrossProduct(angle_axis, pt, w_cross_pt);
result[0] = pt[0] + w_cross_pt[0];
result[1] = pt[1] + w_cross_pt[1];
result[2] = pt[2] + w_cross_pt[2];
}
}
#endif // rotation.h
void find_feature_matches (
const Mat& img_1, const Mat& img_2,
std::vector& keypoints_1,
std::vector& keypoints_2,
std::vector< DMatch >& matches );
// 像素坐标转相机归一化坐标
Point2d pixel2cam ( const Point2d& p, const Mat& K );
void find_feature_matches (
const Mat& img_1, const Mat& img_2,
std::vector& keypoints_1,
std::vector& keypoints_2,
std::vector< DMatch >& matches );
// 像素坐标转相机归一化坐标
Point2d pixel2cam ( const Point2d& p, const Mat& K );
void pose_estimation_3d3d (
const vector& pts1,
const vector& pts2,
Mat& R, Mat& t
);
struct cost_function_define
{
cost_function_define(Point3f p1,Point3f p2):_p1(p1),_p2(p2){}
template
bool operator()(const T* const cere_r,const T* const cere_t,T* residual)const
{
T p_1[3];
T p_2[3];
p_1[0]=T(_p1.x);
p_1[1]=T(_p1.y);
p_1[2]=T(_p1.z);
AngleAxisRotatePoint(cere_r,p_1,p_2);
p_2[0]=p_2[0]+cere_t[0];
p_2[1]=p_2[1]+cere_t[1];
p_2[2]=p_2[2]+cere_t[2];
const T x=p_2[0]/p_2[2];
const T y=p_2[1]/p_2[2];
const T u=x*520.9+325.1;
const T v=y*521.0+249.7;
T p_3[3];
p_3[0]=T(_p2.x);
p_3[1]=T(_p2.y);
p_3[2]=T(_p2.z);
const T x1=p_3[0]/p_3[2];
const T y1=p_3[1]/p_3[2];
const T u1=x1*520.9+325.1;
const T v1=y1*521.0+249.7;
residual[0]=u-u1;
residual[1]=v-v1;
return true;
}
Point3f _p1,_p2;
};
int main ( int argc, char** argv )
{
if ( argc != 5 )
{
cout<<"usage: pose_estimation_3d3d img1 img2 depth1 depth2"< keypoints_1, keypoints_2;
vector matches;
find_feature_matches ( img_1, img_2, keypoints_1, keypoints_2, matches );
cout<<"一共找到了"< ( 3,3 ) << 520.9, 0, 325.1, 0, 521.0, 249.7, 0, 0, 1 );
vector pts1, pts2;
for ( DMatch m:matches )
{
ushort d1 = depth1.ptr ( int ( keypoints_1[m.queryIdx].pt.y ) ) [ int ( keypoints_1[m.queryIdx].pt.x ) ];
ushort d2 = depth2.ptr ( int ( keypoints_2[m.trainIdx].pt.y ) ) [ int ( keypoints_2[m.trainIdx].pt.x ) ];
if ( d1==0 || d2==0 ) // bad depth
continue;
Point2d p1 = pixel2cam ( keypoints_1[m.queryIdx].pt, K );
Point2d p2 = pixel2cam ( keypoints_2[m.trainIdx].pt, K );
float dd1 = float ( d1 ) /1000.0;
float dd2 = float ( d2 ) /1000.0;
pts1.push_back ( Point3f ( p1.x*dd1, p1.y*dd1, dd1 ) );
pts2.push_back ( Point3f ( p2.x*dd2, p2.y*dd2, dd2 ) );
}
cout<<"3d-3d pairs: "<(0,0);
cere_tranf[1]=t.at(1,0);
cere_tranf[2]=t.at(2,0);
// bundleAdjustment( pts1, pts2, R, t );
ceres::Problem problem;
for(int i=0;i(new cost_function_define(pts1[i],pts2[i]));
problem.AddResidualBlock(costfunction,NULL,cere_rot,cere_tranf);//注意，cere_rot不能为Mat类型
}
ceres::Solver::Options option;
option.linear_solver_type=ceres::DENSE_SCHUR;
//输出迭代信息到屏幕
option.minimizer_progress_to_stdout=true;
//显示优化信息
ceres::Solver::Summary summary;
//开始求解
ceres::Solve(option,&problem,&summary);
//显示优化信息
cout< ( 3,1 )< ( 3,1 )< descriptor = DescriptorExtractor::create ( "ORB" );
Ptr matcher = DescriptorMatcher::create("BruteForce-Hamming");
//-- 第一步:检测 Oriented FAST 角点位置
detector->detect ( img_1,keypoints_1 );
detector->detect ( img_2,keypoints_2 );
//-- 第二步:根据角点位置计算 BRIEF 描述子
descriptor->compute ( img_1, keypoints_1, descriptors_1 );
descriptor->compute ( img_2, keypoints_2, descriptors_2 );
//-- 第三步:对两幅图像中的BRIEF描述子进行匹配，使用 Hamming 距离
vector match;
// BFMatcher matcher ( NORM_HAMMING );
matcher->match ( descriptors_1, descriptors_2, match );
//-- 第四步:匹配点对筛选
double min_dist=10000, max_dist=0;
//找出所有匹配之间的最小距离和最大距离, 即是最相似的和最不相似的两组点之间的距离
for ( int i = 0; i < descriptors_1.rows; i++ )
{
double dist = match[i].distance;
if ( dist < min_dist ) min_dist = dist;
if ( dist > max_dist ) max_dist = dist;
}
printf ( "-- Max dist : %f \n", max_dist );
printf ( "-- Min dist : %f \n", min_dist );
//当描述子之间的距离大于两倍的最小距离时,即认为匹配有误.但有时候最小距离会非常小,设置一个经验值30作为下限.
for ( int i = 0; i < descriptors_1.rows; i++ )
{
if ( match[i].distance <= max ( 2*min_dist, 30.0 ) )
{
matches.push_back ( match[i] );
}
}
}
Point2d pixel2cam ( const Point2d& p, const Mat& K )
{
return Point2d
(
( p.x - K.at ( 0,2 ) ) / K.at ( 0,0 ),
( p.y - K.at ( 1,2 ) ) / K.at ( 1,1 )
);
}
void pose_estimation_3d3d (
const vector& pts1,
const vector& pts2,
Mat& R, Mat& t
)
{
Point3f p1, p2; // center of mass
int N = pts1.size();
for ( int i=0; i q1 ( N ), q2 ( N ); // remove the center
for ( int i=0; i 执行结果：
./10icp1 1.png 2.png 1_depth.png 2_depth.png
-- Max dist : 94.000000
-- Min dist : 4.000000
一共找到了79组匹配点
3d-3d pairs: 72
W= 271.775 -25.487 63.6929
-54.0082 96.3269 -144.436
98.6846 -144.995 240.551
U= 0.558087 -0.829399 -0.0252034
-0.428009 -0.313755 0.847565
0.710878 0.462228 0.530093
V= 0.617887 -0.784771 -0.0484805
-0.399894 -0.366747 0.839989
0.676979 0.499631 0.540434
ICP via SVD results:
R = [0.9969452356349386, 0.0598334680297723, -0.05020112795872303;
-0.05932606950498385, 0.9981719688174682, 0.01153855034871776;
0.0507997502148167, -0.008525067189780355, 0.9986724731399795]
t = [0.7207992665571041;
-0.3333924223878955;
-0.1504889019731461]
R_inv = [0.9969452356349386, -0.05932606950498385, 0.0507997502148167;
0.0598334680297723, 0.9981719688174682, -0.008525067189780355;
-0.05020112795872303, 0.01153855034871776, 0.9986724731399795]
t_inv = [-0.7307314580359445;
0.2883721227716854;
0.1903209253782324]
p1 = [-0.187062, -4.15408, 13.724]
p2 = [-0.0557393, -3.73382, 13.826]
(R*p2+t) = [-0.2522577364161404;
-3.897544243811159;
13.68615641608679]
p1 = [-1.21849, -0.588597, 7.924]
p2 = [-1.48606, -0.478307, 8.279]
(R*p2+t) = [-1.204950942925912;
-0.6271353782194089;
8.046107116892889]
p1 = [-3.13876, 0.800932, 6.698]
p2 = [-3.54823, 0.795163, 7.106]
(R*p2+t) = [-3.12573901212394;
0.7528119377320688;
6.759049858892065]
p1 = [-1.61722, 0.524364, 7.133]
p2 = [-1.9954, 0.602349, 7.419]
(R*p2+t) = [-1.604903085936805;
0.4718387339466705;
7.152161629899285]
p1 = [-3.13611, 0.50727, 6.558]
p2 = [-3.54854, 0.50108, 6.999]
(R*p2+t) = [-3.138279851397398;
0.4580510316850339;
6.65468298422584]
----------------------------------
calling bundle adjustment
iter cost cost_change |gradient| |step| tr_ratio tr_radius ls_iter iter_time total_time
0 2.028716e+05 0.00e+00 2.64e+06 0.00e+00 0.00e+00 1.00e+04 0 4.51e-05 1.87e-04
1 8.030585e+02 2.02e+05 1.40e+05 1.34e+00 9.97e-01 3.00e+04 1 8.68e-05 2.94e-04
2 1.474886e+02 6.56e+02 2.57e+02 1.31e-01 1.00e+00 9.00e+04 1 4.51e-05 3.47e-04
3 1.474823e+02 6.32e-03 1.04e-01 8.21e-04 1.00e+00 2.70e+05 1 4.39e-05 3.97e-04
Ceres Solver Report: Iterations: 4, Initial cost: 2.028716e+05, Final cost: 1.474823e+02, Termination: CONVERGENCE
-----------optional after---------------
cam_9d:
[0.9979033018242032, -0.05090521379356253, 0.03997073200461912;
0.049806825147126, 0.9983659841089123, 0.02801146092239384;
-0.04133134860026962, -0.0259619140834198, 0.9988081390537464]
cam_t:-0.63524 -0.0415229 0.301918
p1 = [-0.187062, -4.15408, 13.724]
p2 = [-0.0557393, -3.73382, 13.826]
(R*p1+t) = [-0.06188644477170713;
-3.813701172132307;
14.12513974193004]
p1 = [-1.21849, -0.588597, 7.924]
p2 = [-1.48606, -0.478307, 8.279]
(R*p1+t) = [-1.504485891696794;
-0.4678840063604472;
8.282116008489821]
p1 = [-3.13876, 0.800932, 6.698]
p2 = [-3.54823, 0.795163, 7.106]
(R*p1+t) = [-3.540469400381376;
0.7893890293101794;
7.100869992025547]
p1 = [-1.61722, 0.524364, 7.133]
p2 = [-1.9954, 0.602349, 7.419]
(R*p1+t) = [-1.990645496013816;
0.6012421905123817;
7.479644052702291]
p1 = [-3.13611, 0.50727, 6.558]
p2 = [-3.54854, 0.50108, 6.999]
(R*p1+t) = [-3.528465233005454;
0.4924174084735043;
6.968551213550437]10pnp1.cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include "sophus/se3.hpp"
#include
#include
#include
#include
using namespace std;
using namespace Eigen;
using namespace cv;
#ifndef ROTATION_H
#define ROTATION_H
// math functions needed for rotation conversion.
// dot and cross production
template
inline T DotProduct(const T x[3], const T y[3]) {
return (x[0] * y[0] + x[1] * y[1] + x[2] * y[2]);
}
template
inline void CrossProduct(const T x[3], const T y[3], T result[3]){
result[0] = x[1] * y[2] - x[2] * y[1];
result[1] = x[2] * y[0] - x[0] * y[2];
result[2] = x[0] * y[1] - x[1] * y[0];
}
//
// Converts from a angle anxis to quaternion :
template
inline void AngleAxisToQuaternion(const T* angle_axis, T* quaternion){
const T& a0 = angle_axis[0];
const T& a1 = angle_axis[1];
const T& a2 = angle_axis[2];
const T theta_squared = a0 * a0 + a1 * a1 + a2 * a2;
if(theta_squared > T(std::numeric_limits::epsilon()) ){
const T theta = sqrt(theta_squared);
const T half_theta = theta * T(0.5);
const T k = sin(half_theta)/theta;
quaternion[0] = cos(half_theta);
quaternion[1] = a0 * k;
quaternion[2] = a1 * k;
quaternion[3] = a2 * k;
}
else{ // in case if theta_squared is zero
const T k(0.5);
quaternion[0] = T(1.0);
quaternion[1] = a0 * k;
quaternion[2] = a1 * k;
quaternion[3] = a2 * k;
}
}
template
inline void QuaternionToAngleAxis(const T* quaternion, T* angle_axis){
const T& q1 = quaternion[1];
const T& q2 = quaternion[2];
const T& q3 = quaternion[3];
const T sin_squared_theta = q1 * q1 + q2 * q2 + q3 * q3;
// For quaternions representing non-zero rotation, the conversion
// is numercially stable
if(sin_squared_theta > T(std::numeric_limits::epsilon()) ){
const T sin_theta = sqrt(sin_squared_theta);
const T& cos_theta = quaternion[0];
// If cos_theta is negative, theta is greater than pi/2, which
// means that angle for the angle_axis vector which is 2 * theta
// would be greater than pi...
const T two_theta = T(2.0) * ((cos_theta < 0.0)
? atan2(-sin_theta, -cos_theta)
: atan2(sin_theta, cos_theta));
const T k = two_theta / sin_theta;
angle_axis[0] = q1 * k;
angle_axis[1] = q2 * k;
angle_axis[2] = q3 * k;
}
else{
// For zero rotation, sqrt() will produce NaN in derivative since
// the argument is zero. By approximating with a Taylor series,
// and truncating at one term, the value and first derivatives will be
// computed correctly when Jets are used..
const T k(2.0);
angle_axis[0] = q1 * k;
angle_axis[1] = q2 * k;
angle_axis[2] = q3 * k;
}
}
template
inline void AngleAxisRotatePoint(const T angle_axis[3], const T pt[3], T result[3]) {
const T theta2 = DotProduct(angle_axis, angle_axis);
if (theta2 > T(std::numeric_limits::epsilon())) {
const T theta = sqrt(theta2);
const T costheta = cos(theta);
const T sintheta = sin(theta);
const T theta_inverse = 1.0 / theta;
const T w[3] = { angle_axis[0] * theta_inverse,
angle_axis[1] * theta_inverse,
angle_axis[2] * theta_inverse };
// Explicitly inlined evaluation of the cross product for
// performance reasons.
/*const T w_cross_pt[3] = { w[1] * pt[2] - w[2] * pt[1],
w[2] * pt[0] - w[0] * pt[2],
w[0] * pt[1] - w[1] * pt[0] };*/
T w_cross_pt[3];
CrossProduct(w, pt, w_cross_pt);
const T tmp = DotProduct(w, pt) * (T(1.0) - costheta);
// (w[0] * pt[0] + w[1] * pt[1] + w[2] * pt[2]) * (T(1.0) - costheta);
result[0] = pt[0] * costheta + w_cross_pt[0] * sintheta + w[0] * tmp;
result[1] = pt[1] * costheta + w_cross_pt[1] * sintheta + w[1] * tmp;
result[2] = pt[2] * costheta + w_cross_pt[2] * sintheta + w[2] * tmp;
} else {
// Near zero, the first order Taylor approximation of the rotation
// matrix R corresponding to a vector w and angle w is
//
// R = I + hat(w) * sin(theta)
//
// But sintheta ~ theta and theta * w = angle_axis, which gives us
//
// R = I + hat(w)
//
// and actually performing multiplication with the point pt, gives us
// R * pt = pt + w x pt.
//
// Switching to the Taylor expansion near zero provides meaningful
// derivatives when evaluated using Jets.
//
// Explicitly inlined evaluation of the cross product for
// performance reasons.
/*const T w_cross_pt[3] = { angle_axis[1] * pt[2] - angle_axis[2] * pt[1],
angle_axis[2] * pt[0] - angle_axis[0] * pt[2],
angle_axis[0] * pt[1] - angle_axis[1] * pt[0] };*/
T w_cross_pt[3];
CrossProduct(angle_axis, pt, w_cross_pt);
result[0] = pt[0] + w_cross_pt[0];
result[1] = pt[1] + w_cross_pt[1];
result[2] = pt[2] + w_cross_pt[2];
}
}
#endif // rotation.h
void find_feature_matches(
const Mat &img_1, const Mat &img_2,
std::vector &keypoints_1,
std::vector &keypoints_2,
std::vector &matches);
// 像素坐标转相机归一化坐标
Point2d pixel2cam(const Point2d &p, const Mat &K);
//Ref:http://www.ceres-solver.org/nnls_tutorial.html#bundle-adjustment
struct PnPReprojectionError {
PnPReprojectionError(Point2f pts_2d, Point3f pts_3d)
: _pts_2d(pts_2d), _pts_3d(pts_3d) {}
template
bool operator()(const T* const rotation_vector,
const T* const translation_vector,
T* residuals) const {
T p_transformed[3], p_origin[3];
p_origin[0]=T(_pts_3d.x);
p_origin[1]=T(_pts_3d.y);
p_origin[2]=T(_pts_3d.z);
ceres::AngleAxisRotatePoint(rotation_vector, p_origin, p_transformed);
//旋转后加上平移向量
p_transformed[0] += translation_vector[0];
p_transformed[1] += translation_vector[1];
p_transformed[2] += translation_vector[2];
//归一化
T xp = p_transformed[0] / p_transformed[2];
T yp = p_transformed[1] / p_transformed[2];
double fx=520.9, fy=521.0, cx=325.1, cy=249.7;
// Compute final projected point position.
T predicted_x = fx * xp + cx;
T predicted_y = fy * yp + cy;
// The error is the difference between the predicted and observed position.
residuals[0] = T(_pts_2d.x) - predicted_x;
residuals[1] = T(_pts_2d.y) - predicted_y;
return true;
}
// 2，3，3指输出维度（residuals）为2
//待优化变量（rotation_vector，translation_vector）维度分别为3
static ceres::CostFunction* Create(const Point2f _pts_2d,
const Point3f _pts_3d) {
return (new ceres::AutoDiffCostFunction(
new PnPReprojectionError(_pts_2d, _pts_3d)));
}
Point2f _pts_2d;
Point3f _pts_3d;
};
// 通过引入Sophus库简化计算，并使用雅克比矩阵的解析解代替自动求导
class PnPSE3ReprojectionError : public ceres::SizedCostFunction<2, 6> {
public:
EIGEN_MAKE_ALIGNED_OPERATOR_NEW
PnPSE3ReprojectionError(Eigen::Vector2d pts_2d, Eigen::Vector3d pts_3d) :
_pts_2d(pts_2d), _pts_3d(pts_3d) {}
virtual ~PnPSE3ReprojectionError() {}
virtual bool Evaluate(
double const* const* parameters, double *residuals, double **jacobians) const {
Eigen::Map> se3(*parameters);
Sophus::SE3d T = Sophus::SE3d::exp(se3);
Eigen::Vector3d Pc = T * _pts_3d;
Eigen::Matrix3d K;
double fx = 520.9, fy = 521.0, cx = 325.1, cy = 249.7;
K << fx, 0, cx,
0, fy, cy,
0, 0, 1;
Eigen::Vector2d error = _pts_2d - (K * Pc).hnormalized();
residuals[0] = error[0];
residuals[1] = error[1];
if(jacobians != NULL) {
if(jacobians[0] != NULL) {
Eigen::Map> J(jacobians[0]);
double x = Pc[0];
double y = Pc[1];
double z = Pc[2];
double x2 = x*x;
double y2 = y*y;
double z2 = z*z;
//雅克比矩阵推导看书187页公式(7.46)
J(0,0) = -fx/z;
J(0,1) = 0;
J(0,2) = fx*x/z2;
J(0,3) = fx*x*y/z2;
J(0,4) = -fx-fx*x2/z2;
J(0,5) = fx*y/z;
J(1,0) = 0;
J(1,1) = -fy/z;
J(1,2) = fy*y/z2;
J(1,3) = fy+fy*y2/z2;
J(1,4) = -fy*x*y/z2;
J(1,5) = -fy*x/z;
}
}
return true;
}
private:
const Eigen::Vector2d _pts_2d;
const Eigen::Vector3d _pts_3d;
};
int main(int argc, char **argv){
if (argc != 5) {
cout << "usage: pose_estimation_3d2d img1 img2 depth1 depth2" << endl;
return 1;
}
//-- 读取图像
Mat img_1 = imread(argv[1], CV_LOAD_IMAGE_COLOR);
Mat img_2 = imread(argv[2], CV_LOAD_IMAGE_COLOR);
assert(img_1.data && img_2.data && "Can not load images!");
vector keypoints_1, keypoints_2;
vector matches;
find_feature_matches(img_1, img_2, keypoints_1, keypoints_2, matches);
cout << "一共找到了" << matches.size() << "组匹配点" << endl;
// 建立3D点
Mat d1 = imread(argv[3], CV_LOAD_IMAGE_UNCHANGED); // 深度图为16位无符号数，单通道图像
Mat K = (Mat_(3, 3) << 520.9, 0, 325.1, 0, 521.0, 249.7, 0, 0, 1);
vector pts_3d;
vector pts_2d;
vector pts_3d_eigen;
vector pts_2d_eigen;
for (DMatch m:matches) {
ushort d = d1.ptr(int(keypoints_1[m.queryIdx].pt.y))[int(keypoints_1[m.queryIdx].pt.x)];
if (d == 0) // bad depth
continue;
float dd = d / 5000.0;
Point2d p1 = pixel2cam(keypoints_1[m.queryIdx].pt, K);
pts_3d.push_back(Point3f(p1.x * dd, p1.y * dd, dd));//第一个相机观察到的3D点坐标
pts_2d.push_back(keypoints_2[m.trainIdx].pt);//特征点在第二个相机中的投影
pts_3d_eigen.push_back(Vector3d(p1.x * dd, p1.y * dd, dd));
pts_2d_eigen.push_back(Vector2d(keypoints_2[m.trainIdx].pt.x, keypoints_2[m.trainIdx].pt.y));
}
cout << "3d-2d pairs: " << pts_3d.size() << endl;
chrono::steady_clock::time_point t1 = chrono::steady_clock::now();
Mat r, t;
solvePnP(pts_3d, pts_2d, K, Mat(), r, t, false); // 调用OpenCV 的 PnP 求解，可选择EPNP，DLS等方法
Mat R;
cv::Rodrigues(r, R); // r为旋转向量形式，用Rodrigues公式转换为矩阵
chrono::steady_clock::time_point t2 = chrono::steady_clock::now();
chrono::duration time_used = chrono::duration_cast>(t2 - t1);
cout << "solve pnp in opencv cost time: " << time_used.count() << " seconds." << endl;
cout << "R=" << endl << R << endl;
cout << "t=" << endl << t << endl;
cout << endl;
//ceres求解PnP, 使用自动求导
cout << "以下是ceres求解（自动求导）" << endl;
double r_ceres[3]={0,0,0};
double t_ceres[3]={0,0,0};
ceres::Problem problem;
for (size_t i = 0; i < pts_2d.size(); ++i) {
ceres::CostFunction* cost_function =
PnPReprojectionError::Create(pts_2d[i],pts_3d[i]);
problem.AddResidualBlock(cost_function,
nullptr /* squared loss */,
r_ceres,
t_ceres);
}
t1 = chrono::steady_clock::now();
ceres::Solver::Options options;
options.linear_solver_type = ceres::DENSE_SCHUR;
options.minimizer_progress_to_stdout = true;
ceres::Solver::Summary summary;
ceres::Solve(options, &problem, &summary);
std::cout << summary.BriefReport() << "\n";
Mat r_ceres_cv=(Mat_(3, 1) <(3, 1) <>(t2 - t1);
cout << "solve pnp in ceres cost time: " << time_used.count() << " seconds." << endl;
return 0;
}
void find_feature_matches(const Mat &img_1, const Mat &img_2,
std::vector &keypoints_1,
std::vector &keypoints_2,
std::vector &matches) {
//-- 初始化
Mat descriptors_1, descriptors_2;
// used in OpenCV3
Ptr detector = ORB::create();
Ptr descriptor = ORB::create();
// use this if you are in OpenCV2
// Ptr detector = FeatureDetector::create ( "ORB" );
// Ptr descriptor = DescriptorExtractor::create ( "ORB" );
Ptr matcher = DescriptorMatcher::create("BruteForce-Hamming");
//-- 第一步:检测 Oriented FAST 角点位置
detector->detect(img_1, keypoints_1);
detector->detect(img_2, keypoints_2);
//-- 第二步:根据角点位置计算 BRIEF 描述子
descriptor->compute(img_1, keypoints_1, descriptors_1);
descriptor->compute(img_2, keypoints_2, descriptors_2);
//-- 第三步:对两幅图像中的BRIEF描述子进行匹配，使用 Hamming 距离
vector match;
// BFMatcher matcher ( NORM_HAMMING );
matcher->match(descriptors_1, descriptors_2, match);
//-- 第四步:匹配点对筛选
double min_dist = 10000, max_dist = 0;
//找出所有匹配之间的最小距离和最大距离, 即是最相似的和最不相似的两组点之间的距离
for (int i = 0; i < descriptors_1.rows; i++) {
double dist = match[i].distance;
if (dist < min_dist) min_dist = dist;
if (dist > max_dist) max_dist = dist;
}
printf("-- Max dist : %f \n", max_dist);
printf("-- Min dist : %f \n", min_dist);
//当描述子之间的距离大于两倍的最小距离时,即认为匹配有误.但有时候最小距离会非常小,设置一个经验值30作为下限.
for (int i = 0; i < descriptors_1.rows; i++) {
if (match[i].distance <= max(2 * min_dist, 30.0)) {
matches.push_back(match[i]);
}
}
}
Point2d pixel2cam(const Point2d &p, const Mat &K) {
return Point2d
(
(p.x - K.at(0, 2)) / K.at(0, 0),
(p.y - K.at(1, 2)) / K.at(1, 1)
);
} CMakeLists.txt
和上面一样
执行结果：
./10pnp1 1.png 2.png 1_depth.png 2_depth.png
-- Max dist : 94.000000
-- Min dist : 4.000000
一共找到了79组匹配点
3d-2d pairs: 75
solve pnp in opencv cost time: 0.000401286 seconds.
R=
[0.9979059095501266, -0.05091940089110201, 0.03988747043653948;
0.04981866254253315, 0.9983623157438158, 0.02812094175376485;
-0.04125404886078184, -0.0260749135288436, 0.998808391202765]
t=
[-0.1267821389557959;
-0.008439496817520764;
0.06034935748888207]
以下是ceres求解（自动求导）
iter cost cost_change |gradient| |step| tr_ratio tr_radius ls_iter iter_time total_time
0 2.025888e+04 0.00e+00 6.83e+05 0.00e+00 0.00e+00 1.00e+04 0 4.41e-05 1.86e-04
1 1.650410e+02 2.01e+04 1.47e+04 1.51e-01 9.99e-01 3.00e+04 1 9.20e-05 2.98e-04
2 1.498819e+02 1.52e+01 4.80e+01 7.30e-03 1.00e+00 9.00e+04 1 4.70e-05 3.52e-04
Ceres Solver Report: Iterations: 3, Initial cost: 2.025888e+04, Final cost: 1.498819e+02, Termination: CONVERGENCE
R=
[0.9979061714739172, -0.05092444873781777, 0.03987447121929216;
0.04982431246493198, 0.9983622113118902, 0.02811463874620321;
-0.04124088774099816, -0.02606905340019193, 0.9988090876804998]
t=
[-0.1267625259978171;
-0.008424980436889882;
0.06034877257668213]
solve pnp in ceres cost time: 0.000438856 seconds.
以下是ceres求解（雅克比矩阵给出解析解）
iter cost cost_change |gradient| |step| tr_ratio tr_radius ls_iter iter_time total_time
0 2.025888e+04 0.00e+00 6.83e+05 0.00e+00 0.00e+00 1.00e+04 0 8.70e-05 1.09e-04
1 2.052736e+02 2.01e+04 3.79e+04 1.52e-01 9.97e-01 1.00e+04 1 1.74e-04 2.92e-04
2 1.500357e+02 5.52e+01 2.38e+03 8.62e-03 9.97e-01 1.00e+04 1 9.20e-05 3.91e-04
3 1.498820e+02 1.54e-01 7.50e+01 2.41e-04 9.99e-01 1.00e+04 1 8.89e-05 4.86e-04
4 1.498819e+02 1.63e-04 4.05e+00 8.96e-06 9.97e-01 1.00e+04 1 8.70e-05 5.79e-04
Ceres Solver Report: Iterations: 5, Initial cost: 2.025888e+04, Final cost: 1.498819e+02, Termination: CONVERGENCE
estimated pose:
0.997906 -0.0509194 0.0398875 -0.126782
0.0498187 0.998362 0.0281209 -0.00843933
-0.041254 -0.0260749 0.998808 0.0603495
0 0 0 1
solve pnp in ceres cost time: 0.000701 seconds.
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.csdn.net/luo870604851/article/details/82356394