深蓝学院视觉slam十四讲第2章作业

首先给出有关四元数讲解一份比较好的pdf

2.1 设线性⽅程 Ax=b\bm Ax = bAx=b，则有解充要条件为：系数矩阵An×n\bm A_{n \times n}An×n与其增广矩阵Bn×(n+1)\bm B_{n \times (n+1)}Bn×(n+1)的秩相，即R(A)=R(B)R(\bm A)=R(\bm B)R(A)=R(B)；且解唯一的条件为R(A)=R(B)=nR(\bm A)=R(\bm B)=nR(A)=R(B)=n。

2.2 高斯消元法的原理是通过初等行变换把AAA变换为上三角矩阵，然后从最下面一行（仅有一个未知量）开始求解，逐行回代依次解除每个未知量

2.3 实数矩阵A\bm AA的QR分解是把A\bm AA分解为 A=QR\bm A = \bm Q \bm RA=QR ，这里的Q\bm QQ是正交矩阵，而R\bm RR是上三角矩阵，如果A\bm AA是非奇异的，且限定R\bm RR的对角线元素为正，则这个分解是唯一的。计算有很多方法，例如Givens旋转、Householder变换，以及Gram-Schmidt正交化等等。
用QR分解建立了计算矩阵特征值的QR方法

2.4 摘自wiki百科：Cholesky分解

2.5

#include <iostream>
#include <ctime>
// Eigen 核心部分
#include <Eigen/Core>
// 稠密矩阵的代数运算（逆，特征值等）
#include <Eigen/Dense>using namespace std;
using namespace Eigen;#define MATRIX_SIZE 100int main(int argc, char **argv) {MatrixXd matrix_NN = MatrixXd::Random(MATRIX_SIZE, MATRIX_SIZE);matrix_NN = matrix_NN * matrix_NN.transpose();  // 对称阵，保证半正定
//    cout << "matrix_NN:\n" << matrix_NN << endl;// 实对称矩阵可以正交相似对角化，保证对角化成功SelfAdjointEigenSolver<MatrixXd> eigen_solver(matrix_NN);cout << "Eigen values = \n" << eigen_solver.eigenvalues() << endl;
//    The eigenvalues are sorted in increasing order. 所有特征值均大于0，保证正定cout << "Eigen values min = \n" << eigen_solver.eigenvalues()(0) << endl;clock_t time_stt = clock(); // 计时Matrix<double, MATRIX_SIZE, 1> v_Nd = MatrixXd::Random(MATRIX_SIZE, 1);Matrix<double, MATRIX_SIZE, 1> x;// 通常用矩阵分解来求，例如QR分解，速度会快很多time_stt = clock();x = matrix_NN.colPivHouseholderQr().solve(v_Nd);cout << "time of Qr decomposition is "<< 1000 * (clock() - time_stt) / (double) CLOCKS_PER_SEC << "ms" << endl;cout << "x = " << x.transpose() << endl;// 对于正定矩阵，还可以用cholesky分解来解方程time_stt = clock();x = matrix_NN.ldlt().solve(v_Nd);cout << "time of ldlt decomposition is "<< 1000 * (clock() - time_stt) / (double) CLOCKS_PER_SEC << "ms" << endl;cout << "x = " << x.transpose() << endl;return 0;
}

运行结果：

#include<iostream>
#include<cmath>
using namespace std;
#include<Eigen/Core>
#include<Eigen/Dense>
using namespace Eigen;int main(int argc,char **argv)
{Quaterniond q1(0.55,0.3,0.2,0.2),q2(-0.1,0.3,-0.7,0.2);Vector3d t1(0.7,1.1,0.2),t2(-0.1,0.4,0.8);q1.normalize();q2.normalize();Vector3d p1(0.5,-0.1,0.2);//记得初始化为单位阵Isometry3d T1 = Isometry3d::Identity();Isometry3d T2 = Isometry3d::Identity();T1.rotate(q1);T2.rotate(q2);T1.pretranslate(t1);T2.pretranslate(t2);Vector3d p2=T2*T1.inverse()*p1;cout<<endl<<p2.transpose()<<endl;return 0;
}

运行结果：

4.1
坐标系[e1,e2,e3]\left[\bm e_{1}, \bm e_{2}, \bm e_{3}\right][e1,e2,e3]发生了旋转，变成[e1′,e2′,e3′]\left[\bm e_{1}^{\prime}, \bm e_{2}^{\prime}, \bm e_{3}^{\prime}\right][e1′,e2′,e3′]对应坐标变换如下：
[e1,e2,e3][a1a2a3]=[e1′,e2′,e3′][a1′a2′a3′]\left[\bm e_{1}, \bm e_{2}, \bm e_{3}\right] \left[ \begin{array}{c}a_{1} \\ a_{2} \\ a_{3}\end{array}\right]= \left[\bm e_{1}^{\prime}, \bm e_{2}^{\prime}, \bm e_{3}^{\prime}\right] \left[\begin{array}{c}a_{1}^{\prime} \\ a_{2}^{\prime} \\ a_{3}^{\prime}\end{array}\right] [e1,e2,e3]⎣⎡a1a2a3⎦⎤=[e1′,e2′,e3′]⎣⎡a1′a2′a3′⎦⎤
上式同时左乘[e1Te2Te3T]\left[\begin{array}{c}\bm e_{1}^{T} \\ \bm e_{2}^{T} \\ \bm e_{3}^{T}\end{array}\right]⎣⎡e1Te2Te3T⎦⎤得：
a≜[a1a2a3]=[e1Te1′e1Te2′e1Te3′e2Te1′e2Te2′e2Te3′e3Te1′e3Te2′e3Te3′][a1′a2′a3′]≜Ra′\bm a \triangleq \left[\begin{array}{l} a_{1} \\a_{2} \\a_{3} \end{array}\right]= \left[\begin{array}{lll} \bm e_{1}^{T} \bm{e}_{1}^{\prime} & \bm e_{1}^{T} \bm{e}_{2}^{\prime} & \bm{e}_{1}^{T} \bm{e}_{3}^{\prime} \\ \bm e_{2}^{T} \bm{e}_{1}^{\prime} & \bm e_{2}^{T} \bm{e}_{2}^{\prime} & \bm{e}_{2}^{T} \bm{e}_{3}^{\prime} \\ \bm e_{3}^{T} \bm{e}_{1}^{\prime} & \bm e_{3}^{T} \bm{e}_{2}^{\prime} & \bm{e}_{3}^{T} \bm{e}_{3}^{\prime} \end{array}\right] \left[\begin{array}{l} a_{1}^{\prime} \\a_{2}^{\prime} \\a_{3}^{\prime} \end{array}\right] \triangleq \bm R \bm a^{\prime} a≜⎣⎡a1a2a3⎦⎤=⎣⎡e1Te1′e2Te1′e3Te1′e1Te2′e2Te2′e3Te2′e1Te3′e2Te3′e3Te3′⎦⎤⎣⎡a1′a2′a3′⎦⎤≜Ra′

同理若上式同时左乘[e1′Te2′Te3′T]\left[\begin{array}{c}\bm e_{1}^{\prime T} \\ \bm e_{2}^{\prime T} \\ \bm e_{3}^{\prime T}\end{array}\right]⎣⎡e1′Te2′Te3′T⎦⎤得：
R−1a≜[e1′Te1e1′Te2e1′Te3e2′Te1e2′Te2e2′Te3e3′Te1′e3′Te2e3′Te3][a1a2a3]=[a1′a2′a3′]≜a′\bm R^{-1} \bm a \triangleq \left[\begin{array}{lll} \bm e_{1}^{\prime T} \bm{e}_{1} & \bm e_{1}^{\prime T} \bm{e}_{2} & \bm{e}_{1}^{\prime T} \bm{e}_{3} \\ \bm e_{2}^{\prime T} \bm{e}_{1} & \bm e_{2}^{\prime T} \bm{e}_{2} & \bm{e}_{2}^{\prime T} \bm{e}_{3} \\ \bm e_{3}^{\prime T} \bm{e}_{1}^{\prime} & \bm e_{3}^{\prime T} \bm{e}_{2} & \bm{e}_{3}^{\prime T} \bm{e}_{3} \end{array}\right] \left[\begin{array}{l} a_{1} \\a_{2} \\a_{3} \end{array}\right]= \left[\begin{array}{l} a_{1}^{\prime} \\a_{2}^{\prime} \\a_{3}^{\prime} \end{array}\right] \triangleq \bm a^{\prime} R−1a≜⎣⎡e1′Te1e2′Te1e3′Te1′e1′Te2e2′Te2e3′Te2e1′Te3e2′Te3e3′Te3⎦⎤⎣⎡a1a2a3⎦⎤=⎣⎡a1′a2′a3′⎦⎤≜a′
即
R−1=[e1′Te1e1′Te2e1′Te3e2′Te1e2′Te2e2′Te3e3′Te1′e3′Te2e3′Te3]\bm R^{-1} = \left[\begin{array}{lll} \bm e_{1}^{\prime T} \bm{e}_{1} & \bm e_{1}^{\prime T} \bm{e}_{2} & \bm{e}_{1}^{\prime T} \bm{e}_{3} \\ \bm e_{2}^{\prime T} \bm{e}_{1} & \bm e_{2}^{\prime T} \bm{e}_{2} & \bm{e}_{2}^{\prime T} \bm{e}_{3} \\ \bm e_{3}^{\prime T} \bm{e}_{1}^{\prime} & \bm e_{3}^{\prime T} \bm{e}_{2} & \bm{e}_{3}^{\prime T} \bm{e}_{3} \end{array}\right] R−1=⎣⎡e1′Te1e2′Te1e3′Te1′e1′Te2e2′Te2e3′Te2e1′Te3e2′Te3e3′Te3⎦⎤
而
RT=[e1Te1′e2Te1′e3Te1′e1Te2′e2Te2′e3Te2′e1Te3′e2Te3′e3Te3′]\bm R^{T} =\left[\begin{array}{lll} \bm e_{1}^{T} \bm{e}_{1}^{\prime} & \bm e_{2}^{T} \bm{e}_{1}^{\prime} & \bm e_{3}^{T} \bm{e}_{1}^{\prime} \\ \bm e_{1}^{T} \bm{e}_{2}^{\prime} & \bm e_{2}^{T} \bm{e}_{2}^{\prime} & \bm e_{3}^{T} \bm{e}_{2}^{\prime}\\ \bm{e}_{1}^{T} \bm{e}_{3}^{\prime}& \bm{e}_{2}^{T} \bm{e}_{3}^{\prime} & \bm{e}_{3}^{T} \bm{e}_{3}^{\prime} \end{array}\right] RT=⎣⎡e1Te1′e1Te2′e1Te3′e2Te1′e2Te2′e2Te3′e3Te1′e3Te2′e3Te3′⎦⎤
同时
[ei1,ei2,ei3][ej1′ej2′ej3′]=eiTej′=ej′Tei=[ej1′,ej2′,ej3′][ei1ei2ei3]\left[ e_{i1}, e_{i2}, e_{i3}\right] \left[ \begin{array}{c}e_{j1}^{\prime} \\e_{j2}^{\prime} \\e_{j3}^{\prime}\end{array}\right]= \bm e_{i}^{T} \bm{e}_{j}^{\prime} = \bm e_{j}^{\prime T} \bm{e}_{i}= \left[ e_{j1}^{\prime }, e_{j2}^{\prime }, e_{j3}^{\prime } \right] \left[ \begin{array}{c}e_{i1}\\ e_{i2}\\ e_{i3}\end{array}\right] [ei1,ei2,ei3]⎣⎡ej1′ej2′ej3′⎦⎤=eiTej′=ej′Tei=[ej1′,ej2′,ej3′]⎣⎡ei1ei2ei3⎦⎤
故
R−1=RT\bm R^{-1}=\bm R^{T}R−1=RT

∣R∣=∣e1Te2Te3T∣∣e1′,e2′,e3′∣=1|\bm R| =\left|\begin{array}{c}\bm e_{1}^{T} \\ \bm e_{2}^{T} \\ \bm e_{3}^{T}\end{array}\right| \left|\bm e_{1}^{\prime}, \bm e_{2}^{\prime}, \bm e_{3}^{\prime}\right| =1 ∣R∣=∣∣∣∣∣∣e1Te2Te3T∣∣∣∣∣∣∣e1′,e2′,e3′∣=1

4.2 四元数的实部η\etaη是一维，虚部ε\bm \varepsilonε是三维

4.3
记
q1=[ε1,η1]T,q2=[ε2,η2]T\bm {q}_{1}=[\bm \varepsilon_1,\eta_1]^T, \quad \bm {q}_{2}=[\bm \varepsilon_2,\eta_2]^Tq1=[ε1,η1]T,q2=[ε2,η2]T

或原始四元数表示法：
q1=η1+x1i+y1j+z1k,q2=η2+x2i+y2j+z2k\bm {q}_{1}=\eta_1+x_1 \bm i + y_1 \bm j + z_1\bm k, \quad \bm {q}_{2}=\eta_2+x_2 \bm i + y_2 \bm j + z_2\bm kq1=η1+x1i+y1j+z1k,q2=η2+x2i+y2j+z2k
有
q1q2=η1η2−x1x2−y1y2−z1z2+(η1x2+x1η2+y1z2−z1y2)i+(η1y2−x1z2+y1η2+z1x2)j+(η1z2+x1y2−y1x2+z1η2)k\begin{aligned} \bm {q}_{1}\bm {q}_{2}= &\eta_{1} \eta_{2}-x_{1} x_{2}-y_{1} y_{2}-z_{1} z_{2} \\ &+ \left(\eta_{1} x_{2}+x_{1} \eta_{2}+y_{1} z_{2}-z_{1} y_{2}\right)\bm i \\ &+\left(\eta_{1} y_{2}-x_{1} z_{2}+y_{1} \eta_{2}+z_{1} x_{2}\right)\bm j \\ &+\left(\eta_{1} z_{2}+x_{1} y_{2}-y_{1} x_{2}+z_{1} \eta_{2}\right)\bm k \end{aligned}q1q2=η1η2−x1x2−y1y2−z1z2+(η1x2+x1η2+y1z2−z1y2)i+(η1y2−x1z2+y1η2+z1x2)j+(η1z2+x1y2−y1x2+z1η2)k
q1q2=η1η2−x1x2−y1y2−z1z2+(x1η2+η1x2−z1y2+y1z2)i+(y1η2+z1x2+η1y2−x1z2)j+(z1η2−y1x2+x1y2+η1z2)k\begin{aligned} \bm {q}_{1}\bm {q}_{2}= &\eta_1 \eta_2-x_1 x_2-y_1 y_2-z_1 z_2\\ &+(x_1 \eta_2+\eta_1 x_2-z_1 y_2+y_1 z_2)\bm i \\ &+(y_1 \eta_2+z_1 x_2+\eta_1 y_2-x_1 z_2) \bm j \\ &+(z_1 \eta_2-y_1 x_2+x_1 y_2+\eta_1 z_2)\bm k \end{aligned}q1q2=η1η2−x1x2−y1y2−z1z2+(x1η2+η1x2−z1y2+y1z2)i+(y1η2+z1x2+η1y2−x1z2)j+(z1η2−y1x2+x1y2+η1z2)k

写成向量形式并利用内外积运算表示:
q1q2=[η1ε2+η2ε1+ε1×ε2,η1η2−ε1Tε2]T\bm {q}_{1}\bm {q}_{2}=[\eta_1 \bm \varepsilon_2 + \eta_2 \bm \varepsilon_1 + \bm \varepsilon_1 \times \bm \varepsilon_2, \eta_1 \eta_2 - \bm \varepsilon_1^T \bm \varepsilon_2]^Tq1q2=[η1ε2+η2ε1+ε1×ε2,η1η2−ε1Tε2]T
记
q1+=[η11+ε1×ε1−ε1Tη1],q2⊕=[η21−ε2×ε2−ε2Tη2]\bm {q}^{+}_1=\left[\begin{array}{cc} \eta_1 \bm {1}+ \bm {\varepsilon}^{\times}_1 & \bm {\varepsilon}_1 \\ -\bm \varepsilon^{T}_1 & \eta_1 \end{array}\right] , \quad \bm {q}^{\oplus}_2=\left[\begin{array}{cc} \eta_2 \bm 1-\bm \varepsilon^{\times}_2 & \bm \varepsilon_2 \\ -\bm \varepsilon^{T}_2 & \eta_2 \end{array}\right]q1+=[η11+ε1×−ε1Tε1η1],q2⊕=[η21−ε2×−ε2Tε2η2]
则
q1+q2=[η11+ε1×ε1−ε1Tη1][ε2η2]=[η1ε2+ε1×ε2+η2ε1−ε1Tε2+η1η2]=q1q2\bm {q}^{+}_1\bm {q}_{2}=\left[\begin{array}{cc} \eta_1 \bm {1}+ \bm {\varepsilon}^{\times}_1 & \bm {\varepsilon}_1 \\ -\bm \varepsilon^{T}_1 & \eta_1 \end{array}\right] \left[\begin{array}{c} \bm \varepsilon_2 \\ \eta_2 \end{array}\right]= \left[\begin{array}{c} \eta_1 \bm \varepsilon_2 + \bm \varepsilon_1 \times \bm \varepsilon_2 +\eta_2 \bm \varepsilon_1 \\ -\bm \varepsilon_1^T \bm \varepsilon_2 +\eta_1 \eta_2 \end{array}\right] = \bm {q}_{1}\bm {q}_{2} q1+q2=[η11+ε1×−ε1Tε1η1][ε2η2]=[η1ε2+ε1×ε2+η2ε1−ε1Tε2+η1η2]=q1q2
q2⊕q1=[η21−ε2×ε2−ε2Tη2][ε1η1]=[η2ε1−ε2×ε1+η1ε2−ε2Tε1+η1η2]=q1q2\bm {q}^{\oplus}_2\bm {q}_{1}= \left[\begin{array}{cc} \eta_2 \bm 1-\bm \varepsilon^{\times}_2 & \bm \varepsilon_2 \\ -\bm \varepsilon^{T}_2 & \eta_2 \end{array}\right] \left[\begin{array}{c} \bm \varepsilon_1 \\ \eta_1 \end{array}\right]= \left[\begin{array}{c} \eta_2 \bm \varepsilon_1 - \bm \varepsilon_2 \times \bm \varepsilon_1 +\eta_1 \bm \varepsilon_2 \\ -\bm \varepsilon_2^T \bm \varepsilon_1 +\eta_1 \eta_2 \end{array}\right] = \bm {q}_{1}\bm {q}_{2} q2⊕q1=[η21−ε2×−ε2Tε2η2][ε1η1]=[η2ε1−ε2×ε1+η1ε2−ε2Tε1+η1η2]=q1q2

5. 摘自维基百科 Rodrigues’ rotation formula

基本原理是将向量v\bm vv绕转轴k\bm kk的旋转分解为平行于转轴的v∥\bm v_{\parallel }v∥和垂直于转轴的v⊥\bm v _{\perp }v⊥的旋转，并用v\bm vv和k\bm kk表示出来；v∥\bm v_{\parallel }v∥旋转后的v∥rot\bm{v}_{\parallel\mathrm{rot}}v∥rot保持不变，v⊥\bm v_{\perp }v⊥旋转后的v⊥rot\bm{v}_{\perp \mathrm{rot}}v⊥rot在由v⊥\bm v_{\perp }v⊥和w\bm ww（与v⊥\bm v_{\perp }v⊥垂直）张成的二维平面内表示出来；最终获得vrot\bm{v}_{\mathrm{rot}}vrot

记
q=[s,v]T,p=[0,u]T\bm {q}=[s,\bm v]^{\mathrm{T}} , \bm {p}=[0,\bm u]^{\mathrm{T}} q=[s,v]T,p=[0,u]T
有
q+(q−1)⊕=[s−vTvsI+v∧][svT−vsI+v∧]=[100TvvT+s2I+2sv∧+(v∧)2]\bm q^{+}\left(\bm q^{-1}\right)^{\oplus}=\left[\begin{array}{cc} s & -\bm v^{\mathrm{T}} \\ \bm v & s \bm I+\bm v^{\wedge} \end{array}\right]\left[\begin{array}{cc} s & \bm v^{\mathrm{T}} \\ -\bm v & s\bm I+\bm v^{\wedge} \end{array}\right]= \left[\begin{array}{cc} 1 & \bm 0 \\ \bm 0^{\mathrm{T}} & \bm v\bm v^{\mathrm{T}}+s^{2} \bm I+2 s \bm v^{\wedge}+\left(\bm v^{\wedge}\right)^{2} \end{array}\right]q+(q−1)⊕=[sv−vTsI+v∧][s−vvTsI+v∧]=[10T0vvT+s2I+2sv∧+(v∧)2]
p′=qpq−1=q+(q−1)⊕p=[10T0vvT+s2I+2sv∧+(v∧)2][0u]=[0((vvT+s2I+2sv∧+(v∧)2)u]\bm p^{\prime} =\bm q \bm p \bm q^{-1}=\bm q^{+}\left(\bm q^{-1}\right)^{\oplus}\bm p= \left[\begin{array}{cc} 1 & \bm 0^{\mathrm{T}} \\ \bm 0 & \bm v\bm v^{\mathrm{T}}+s^{2} \bm I+2 s \bm v^{\wedge}+\left(\bm v^{\wedge}\right)^{2} \end{array}\right] \left[\begin{array}{c} 0 \\ \bm u \end{array}\right]= \left[\begin{array}{c} 0 \\ \left((\bm v\bm v^{\mathrm{T}}+s^{2} \bm I+2 s \bm v^{\wedge}+\left(\bm v^{\wedge}\right)^{2}\right) \bm u \end{array}\right] p′=qpq−1=q+(q−1)⊕p=[100TvvT+s2I+2sv∧+(v∧)2][0u]=[0((vvT+s2I+2sv∧+(v∧)2)u]
由上式结果看p′\bm p^{\prime}p′实部为000即p′\bm p^{\prime}p′为虚四元数
若记q=s+xi+yj+zk\bm q =s+x\bm i+y\bm j+z\bm kq=s+xi+yj+zk则旋转矩阵Q\bm QQ可表示为：
Q=q+(q−1)⊕=[100TvvT+s2I+2sv∧+(v∧)2]=[100001−2y2−2z22xy−2sz2sy+2xz02xy+2sz1−2x2−2z22yz−2sx02xz−2sy2sx+2yz1−2x2−2y2]\bm Q =\bm q^{+}\left(\bm q^{-1}\right)^{\oplus}= \left[\begin{array}{cc} 1 & \bm 0 \\ \bm 0^{\mathrm{T}} & \bm v\bm v^{\mathrm{T}}+s^{2} \bm I+2 s \bm v^{\wedge}+\left(\bm v^{\wedge}\right)^{2} \end{array}\right]= \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1-2 y^{2}-2 z^{2} & 2 x y-2 s z & 2 s y+2 x z \\ 0 & 2 x y+2 s z & 1-2 x^{2}-2 z^{2} & 2 y z-2 s x \\ 0 & 2 x z-2 s y & 2 s x+2 y z & 1-2 x^{2}-2 y^{2} \end{array}\right]Q=q+(q−1)⊕=[10T0vvT+s2I+2sv∧+(v∧)2]=⎣⎢⎢⎡100001−2y2−2z22xy+2sz2xz−2sy02xy−2sz1−2x2−2z22sx+2yz02sy+2xz2yz−2sx1−2x2−2y2⎦⎥⎥⎤

方法二：摘自 A brief introduction to the quaternions and its applications in 3D geometry

7.
所用的新特性已写在注释中

#include <iostream>
#include <vector>
#include <algorithm>using namespace std;class A {public:A(const int &i) : index(i) {}int index = 0;
};int main() {A a1(3), a2(5), a3(9);vector<A> avec{a1, a2, a3}; // 初始化列表// 运用lambda表达式 []// 提供了一个类似匿名函数的特性，而匿名函数则是在需要一个函数，但是又不想费力去命名一个函数的情况下去使用std::sort(avec.begin(), avec.end(), [](const A&a1, const A&a2) {return a1.index<a2.index;});// 范围for循环：用a遍历容器avec中的每个量  auto自动类型推导：根据a获得的值，自动推断出a的类型for ( auto& a: avec ) cout<<a.index<<" ";cout<<endl;return 0;
}

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