zoukankan      html  css  js  c++  java
  • 经典ICP算法的问题

    最近可能要用三维点云实现一个三维场景重建的功能,从经典的ICP算法开始,啃了一些文档,对其原理也是一知半解。

    迭代最近点算法综述

    大致参考了这份文档之后,照流程用MATLAB实现了一个简单的ICP算法,首先是发现这份文档中一个明显的错误,

    公式6

    求两个点集的协方差,其中(Pi-p)和(Qi-p')分别求两个点集的各点与重心的差,都是(3*1)向量,这是不能相乘的,根据后文推断,此物的结果应为(3*3)矩阵,所以我大(zuo)胆(si)的改为(Pi-p)' * (Qi-p'),做一次尝试。

    Matlab代码如下:

    %%% ICP迭代最近点算法
    
    function [sourcePoint,aimPoint,distance] = ICPiterator( sourcePoint , targetPoint )
    %%% 获得匹配点集,重心
    aimPoint = getAimPoint(sourcePoint,targetPoint);
    
    sourcePointCentre = getCentre(sourcePoint);
    aimPointCentre = getCentre(aimPoint);
    
    %%% 平移矩阵
    T = getTranslation(aimPointCentre,sourcePointCentre);
    
    %%% 中心化
    midSourcePoint = centreTransform(sourcePoint, sourcePointCentre);
    midAimPoint = centreTransform(aimPoint, aimPointCentre);
    
    %%%旋转四元数
    quaternion = getRevolveQuaternion(midSourcePoint,midAimPoint);
    
    %%%旋转矩阵
    revolveMatrix = getRevolveMatrix(quaternion);
    
    %%%变换
    
    sourcePoint = midSourcePoint * revolveMatrix;
    sourcePoint = counterCentreTransform(sourcePoint,sourcePointCentre);
    
    range = length(sourcePoint);
    for i = 1:1:range
        sourcePoint(i,:) = sourcePoint(i,:) + T;
    end
    
    %%%阈值判定,欧拉距离和
    distance = getDistance(sourcePoint,aimPoint);
        
    end
    
    %%% 点对搜索匹配,得到匹配点集
    function [aimPoint] = getAimPoint( sourcePoint , targetPoint ) 
    rangeS = length(sourcePoint );
    rangeT = length(targetPoint);
    aimPoint = zeros(rangeS,3);
    
    for i = 1:1:rangeS
        minDistance = getDistance(sourcePoint(i,:),targetPoint(1,:));
        aimPoint(i,:) = targetPoint(1,:);
        for j = 1:1:rangeT
            distance = getDistance(sourcePoint(i,:),targetPoint(j,:));
            if distance < minDistance
                minDistance = distance;
                aimPoint(i,:) = targetPoint(j,:);
            end
        end
    end
    end
    
    %%%旋转四元数
    function [quaternion] = getRevolveQuaternion( sourcePoint , targetPoint )
        %%% 协方差
        pp = sourcePoint' * targetPoint;
        range = size(sourcePoint,1);
        pp = pp / range;
        
        %%% 反对称矩阵
        dissymmetryMatrix = pp - pp' ;
        
        %%% 列向量delta
        delta = [dissymmetryMatrix(2,3) ; dissymmetryMatrix(3,1) ; dissymmetryMatrix(1,2)];
        
        %%%对称矩阵Q
        Q = [ trace(pp) delta' ; delta   pp + pp' - trace(pp)*eye(3) ];
        
        %%%最大特征值,对应特征向量即为旋转四元数
        maxEigenvalues = max(eig(Q));
        quaternion = null(Q - maxEigenvalues*eye(length(Q)));
    
    end
    
    %%% 旋转矩阵
    function [revolveMatrix] = getRevolveMatrix(quaternion)
        revolveMatrix = [ quaternion(1,1)^2 + quaternion(2,1)^2 - quaternion(3,1)^2 - quaternion(4,1)^2    2 * (quaternion(2,1)*quaternion(3,1) - quaternion(1,1)*quaternion(4,1))  2 * (quaternion(2,1)*quaternion(4,1) + quaternion(1,1)*quaternion(3,1));
                            2 * (quaternion(2,1)*quaternion(3,1) + quaternion(1,1)*quaternion(4,1))    quaternion(1,1)^2 - quaternion(2,1)^2 + quaternion(3,1)^2 - quaternion(4,1)^2     2 * (quaternion(3,1)*quaternion(4,1) - quaternion(1,1)*quaternion(2,1));
                            2 * (quaternion(2,1)*quaternion(4,1) - quaternion(1,1)*quaternion(3,1))  2 * (quaternion(3,1)*quaternion(4,1) + quaternion(1,1)*quaternion(2,1))   quaternion(1,1)^2 - quaternion(2,1)^2 - quaternion(3,1)^2 + quaternion(4,1)^2  ];
    end
    
    %%% 点集重心
    function [centre] = getCentre( point )
        range = length(point);
        centre = sum(point)/range;
    end
    
    %%% 获取平移矩阵
    function [T] = getTranslation( aimPointCentre , sourcePointCentre )
        T = aimPointCentre - sourcePointCentre;
    end
    
    %%% 点集中心化
    function [point] = centreTransform(point,centre)
    range = size(point,1);
    for i = 1:1:range
        point(i,:) = point(i,:) - centre;    
    end
    end
    
    function [point] = counterCentreTransform(point,centre)
    range = size(point,1);
    for i = 1:1:range
        point(i,:) = point(i,:) + centre;    
    end
    end
    
    
    %%% 计算两点距离的平方,即欧拉距离和
    function [distance] = getDistance(point1,point2)
        distance = (point1(1,1) - point2(1,1))^2 + (point1(1,2) - point2(1,2))^2 + (point1(1,3) - point2(1,3))^2;
    end
    
        

    为了看到迭代过程,这段代码每次只是进行一次迭代,但是实际情况下需要不断迭代,直到两点集的方差收敛,达到拟合要求。

    用随机数生成了一个含一百个点的点集A,并对A进行一次随机的空间变化,得到B,这样A,B是完全可以拟合的两个点集;

    点集A:

    点集B:

    用A,B来验证算法能不能实现点集的拟合。

    试验了几次之后,发现无法收敛:

    问题应该出在旋转四元数和旋转矩阵求解上,这块是一直没能理解透彻的部分。

  • 相关阅读:
    进击Node.js基础(一)
    关于bootstrap两个模态框的问题
    系列博文-Three.js入门指南(张雯莉)-网格 setInterval方法 requestAnimationFrame方法 使用stat.js记录FPS
    系列博文-Three.js入门指南(张雯莉)-照相机
    系列博文-Three.js入门指南(张雯莉)-静态demo和three.js功能概览
    for循环执行效率
    c/c++多维数组动态分配与释放
    C/C++数组指针与指针数组详解
    C/C++语言参数传递----值传递、引用传递、指针传递、指针引用传递
    float类型最大值和最小值
  • 原文地址:https://www.cnblogs.com/moranBlogs/p/3798257.html
Copyright © 2011-2022 走看看