ceres求解ICP--SLAM 十四讲第七章课后题

by jie 2018.8.10
参考：如何用ceres进行两帧之间的BA优化

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思路分析

之前用ceres求解了pnp问题，3d-2d构建cost fuction是最小重投影。那3d-3d呢？
也可以用最小重投影.思路是，将第一帧图像坐标系下的3d点经过旋转平移到第二帧图像下，然后通过相机内参求得其投影到图像坐标系下的坐标。第二帧观测到的与之匹配的3d点，也可以进行重投影得到图像坐标系下的坐标。两者就残差即可。

先来完整代码：

#include 
#include 
#include 
#include 

#include 
#include 
#include 
#include 
#include 
#include 

#include "common/rotation.h"
using namespace std;
using namespace cv;

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