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Ceres 自动求导解析-从原理到实践

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Ceres 自动求导解析-从原理到实践

Ceres 自动求导解析-从原理到实践

目录
    Ceres 自动求导解析-从原理到实践
    • 1.0 前言
    • 2.0 Ceres求导简介
    • 3.0 Ceres 自动求导原理
      • 3.1 官方解释
      • 3.2 自我理解
    • 4.0 实践
        4.1 Jet 的实现
      • 4.2 多项式函数自动求导
      • 4.3 BA 问题中的自动求导
    • Reference

1.0 前言

精确的导数（或者雅克比矩阵），这极大节约了算法开发者的时间，但是笔者在使用的时候一直觉得这是个黑盒子，特别是之前在做深度学习的时候，神经网络本事是一个很盒模型了，再加上 pytorch 的自动求导，简直是黑上加黑。现在转入视觉SLAM方向，又碰到了 Ceres 的自动求导，是时候揭开其真实的面纱了。知其然并知其所以然才是一名算法工程师应有的基本素养。

2.0 Ceres求导简介

    解析求导，也就是手动计算出导数的解析形式。

    \[y = \frac{b_1}{(1+e^{b_2-b_3x}^{1/b_4}} \]

    构建误差函数：

    \[\begin{split}\begin{align} E(b_1, b_2, b_3, b_4 &= \sum_i f^2(b_1, b_2, b_3, b_4 ; x_i, y_i\\ &= \sum_i \left(\frac{b_1}{(1+e^{b_2-b_3x_i}^{1/b_4}} - y_i\right^2\\ \end{align}\end{split} \]
    \[\begin{split}\begin{align} D_1 f(b_1, b_2, b_3, b_4; x,y &= \frac{1}{(1+e^{b_2-b_3x}^{1/b_4}}\\ D_2 f(b_1, b_2, b_3, b_4; x,y &= \frac{-b_1e^{b_2-b_3x}}{b_4(1+e^{b_2-b_3x}^{1/b_4 + 1}} \\ D_3 f(b_1, b_2, b_3, b_4; x,y &= \frac{b_1xe^{b_2-b_3x}}{b_4(1+e^{b_2-b_3x}^{1/b_4 + 1}} \\ D_4 f(b_1, b_2, b_3, b_4; x,y & = \frac{b_1 \log\left(1+e^{b_2-b_3x}\right }{b_4^2(1+e^{b_2-b_3x}^{1/b_4}} \end{align}\end{split} \]

  • 数值求导，当对变量增加一个微小的增量，然后观察此时的残差和原先残差的下降比例即可，其实就是导数的定义。

    \[Df(x = \lim_{h \rightarrow 0} \frac{f(x + h - f(x}{h} \]
    \[Df(x \approx \frac{f(x + h - f(x}{h} \]

    第二种是 Central Differences：

    \[Df(x \approx \frac{f(x + h - f(x - h}{2h} \]

    这里有三种数值微分方法的效果对比，从右向左看：

    第三种则是今天要介绍的主角，自动求导。

3.0 Ceres 自动求导原理

3.1 官方解释

其实官方对自动求导做出了解释，但是笔者觉得写的不够直观，比较抽象，不过既然是官方出品，还是非常有必要去看一看的。http://ceres-solver.org/automatic_derivatives.html。

3.2 自我理解

举一个例子：

\[\mathrm{f}^{\prime}(\mathrm{x}=\mathrm{h}^{\prime}(\mathrm{x} \mathrm{g}(\mathrm{x}+\mathrm{h}(\mathrm{x} \mathrm{g}^{\prime}(\mathrm{x} \]

其中 \(h(x\, \(g(x\ 都是标量函数.
如果我们定义一种数据类型,

\[Data \{ double\ \ value, double\ \ derived \} \]
\[data1*data2=\begin{bmatrix} data1.value*data2.value \\ data1.derived*data2.value+data1.value*data2.derived \end{bmatrix} \]

令 \(h(x =[h(x,{h(x}' ] , g(x=[g(x,{g(x' }]\。\(f(x=h(x * g(x\, 那么f_x.derived 就是\(f(x\的导数，f_x.value 即为\(f(x\的数值。value 储存变量的函数值, derived 储存变量对 \(\mathrm{x}\ 的导数。类似，如果我们对数据类型 Data 重载所有可能用到的运算符. “\(+- * / \log , \exp , \cdots\” 。那么在变量 \(h(x,g(x\经过任意次运算后，\(result=h(x+g(x*h(x+exp(h(x…\, 任然能获得函数值 result.value 和他的导数值 result.derived，这就是Ceres 自动求导的原理。

并且对于数据类型 Data，乘法运算符重载为

\[data1*data2=\begin{bmatrix} data1.value*data2.value \\ derived[i]=data1.derived[i]*data2.value+data1.value*data2.derived[i] \end{bmatrix} \]

4.0 实践

4.1 Jet 的实现

这里我们模仿 Ceres 实现了 Jet，并准备了两个具体的示例程序，Jet 具体代码在 ceres_jet.hpp 中，包装成了一个头文件，在使用的时候进行调用即可。这里也包含了一个 ceres_rotation.hpp 的头文件，是为了我们的第二个例子实现。具体代码如下：

    ceres_jet.hpp 
#ifndef _CERES_JET_HPP__
#define _CERES_JET_HPP__
#include <math.h>
#include <stdio.h>

#include <eigen3/Eigen/Core>
#include <eigen3/Eigen/Dense>
#include <eigen3/Eigen/Sparse>
#include "eigen3/Eigen/Eigen"
#include "eigen3/Eigen/SparseQR"
#include <fstream>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
#include "ceres_rotation.hpp"

#include "algorithm"
#include "stdlib.h"

template <int N>
struct jet
{
  Eigen::Matrix<double, N, 1> v;
  double a;
  jet( : a(0.0 {}
  jet(const double& value : a(value { v.setZero(; }
  EIGEN_STRONG_INLINE jet(const double& value,
                          const Eigen::Matrix<double, N, 1>& v_
      : a(value, v(v_
  {
  }
  jet(const double value, const int index
  {
    v.setZero(;
    a = value;
    v(index, 0 = 1.0;
  }
  void init(const double value, const int index
  {
    v.setZero(;
    a = value;
    v(index, 0 = 1.0;
  }
};
/****************jet overload******************/
// for the camera BA,the autodiff only need overload the operator :jet+jet
// number+jet -jet jet-number jet*jet number/jet jet/jet sqrt(jet cos(jet
// sin(jet  +=(jet overload jet + jet
template <int N>
inline jet<N> operator+(const jet<N>& A, const jet<N>& B
{
  return jet<N>(A.a + B.a, A.v + B.v;
}  // end jet+jet

// overload number + jet
template <int N>
inline jet<N> operator+(double A, const jet<N>& B
{
  return jet<N>(A + B.a, B.v;
}  // end number+jet

template <int N>
inline jet<N> operator+(const jet<N>& B, double A
{
  return jet<N>(A + B.a, B.v;
}  // end number+jet

// overload jet-number
template <int N>
inline jet<N> operator-(const jet<N>& A, double B
{
  return jet<N>(A.a - B, A.v;
}
// overload number * jet because jet *jet need A.a *B.v+B.a*A.v.So the number
// *jet is required
template <int N>
inline jet<N> operator*(double A, const jet<N>& B
{
  return jet<N>(A * B.a, A * B.v;
}
template <int N>
inline jet<N> operator*(const jet<N>& A, double B
{
  return jet<N>(B * A.a, B * A.v;
}
// overload -jet
template <int N>
inline jet<N> operator-(const jet<N>& A
{
  return jet<N>(-A.a, -A.v;
}
template <int N>
inline jet<N> operator-(double A, const jet<N>& B
{
  return jet<N>(A - B.a, -B.v;
}
template <int N>
inline jet<N> operator-(const jet<N>& A, const jet<N>& B
{
  return jet<N>(A.a - B.a, A.v - B.v;
}
// overload jet*jet
template <int N>
inline jet<N> operator*(const jet<N>& A, const jet<N>& B
{
  return jet<N>(A.a * B.a, B.a * A.v + A.a * B.v;
}
// overload number/jet
template <int N>
inline jet<N> operator/(double A, const jet<N>& B
{
  return jet<N>(A / B.a, -A * B.v / (B.a * B.a;
}
// overload jet/jet
template <int N>
inline jet<N> operator/(const jet<N>& A, const jet<N>& B
{
  // This uses:
  //
  //   a + u   (a + u(b - v   (a + u(b - v
  //   ----- = -------------- = --------------
  //   b + v   (b + v(b - v        b^2
  //
  // which holds because v*v = 0.
  const double a_inverse = 1.0 / B.a;
  const double abyb = A.a * a_inverse;
  return jet<N>(abyb, (A.v - abyb * B.v * a_inverse;
}
// sqrt(jet
template <int N>
inline jet<N> sqrt(const jet<N>& A
{
  double t = std::sqrt(A.a;

  return jet<N>(t, 1.0 / (2.0 * t * A.v;
}
// cos(jet
template <int N>
inline jet<N> cos(const jet<N>& A
{
  return jet<N>(std::cos(A.a, -std::sin(A.a * A.v;
}
template <int N>
inline jet<N> sin(const jet<N>& A
{
  return jet<N>(std::sin(A.a, std::cos(A.a * A.v;
}
template <int N>
inline bool operator>(const jet<N>& f, const jet<N>& g
{
  return f.a > g.a;
}

#endif //_CERES_JET_HPP__
    ceres_rotation.hpp 
#ifndef CERES_ROTATION_HPP_
#define CERES_ROTATION_HPP_
#include <iostream>

template <typename T>
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 <typename T>
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<double>::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 = T(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]};
    const T tmp =
        (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]};

    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  // CERES_ROTATION_HPP_

4.2 多项式函数自动求导

这里我们准备了两个实践案例，一个是对下面的函数进行自动求导，求在 \(f(1,2\ 处的导数。

\[f(x,y=2x^2+3y^3+3 \]

#include <eigen3/Eigen/Core>
#include <eigen3/Eigen/Dense>

#include "ceres_jet.hpp"

int main(int argc, char const *argv[]
{
  /// f(x,y = 2*x^2 + 3*y^3 + 3
  /// 残差的维度，变量1的维度，变量2的维度
  const int N = 1, N1 = 1, N2 = 1;
  Eigen::Matrix<double, N, N1> jacobian_parameter1;
  Eigen::Matrix<double, N, N2> jacobian_parameter2;
  Eigen::Matrix<double, N, 1> jacobi_residual;

  /// 模板参数为向量的维度，一定要是 N1+N2
  /// 也就是总的变量的维度，因为要存储结果（残差）
  /// 对于每个变量的导数值
  /// 至于为什么有 N1 个 jet 表示 var_x
  /// 假设变量 1 的维度为 N1,则残差对该变量的导数的维度是一个 N*N1 的矩阵
  /// 一个 jet<N1 + N2> 只能表示变量中的某一个在当前点的导数和值
  jet<N1 + N2> var_x[N1];
  jet<N1 + N2> var_y[N2];
	jet<N1 + N2> residual[N];
	/// 假设我们求上面的方程在 (x,y->(1.0,2.0 处的导数值
	double var_x_init_value[N1] = {1.0};
	double var_y_init_value[N1] = {2.0};

  for (int i = 0; i < N1; i++
  {
    var_x[i].init(var_x_init_value[i], i;
  }
  for (int i = 0; i < N2; i++
  {
    var_y[i].init(var_y_init_value[i], i + N1;
  }
	/// f(x,y = 2*x^2 + 3*y^3 + 3
	/// f_x` = 4x
	/// f_y` = 9 * y^2
	residual[0] = 2.0 * var_x[0] * var_x[0]  + 3.0 * var_y[0] * var_y[0] * var_y[0] + 3.0;
	std::cout << "residual: " << residual[0].a << std::endl;
	std::cout << "jacobian: " << residual[0].v.transpose( << std::endl;
	/// residual: 29
	/// jacobian:  4 36
  return 0;
}

    输出结果，读者可以自己求导算一下，是正确的。

    residual: 29
 jacobian:  4 36

4.3 BA 问题中的自动求导

这里是用的 Bal 数据集中的某个观测构建的误差项求导

#include "ceres_jet.hpp"

class costfunction
{
 public:
  double x_;
  double y_;
  costfunction(double x, double y : x_(x, y_(y {}
  template <class T>
  void Evaluate(const T* camera, const T* point, T* residual
  {
    T result[3];
    AngleAxisRotatePoint(camera, point, result;
    result[0] = result[0] + camera[3];
    result[1] = result[1] + camera[4];
    result[2] = result[2] + camera[5];
    T xp = -result[0] / result[2];
    T yp = -result[1] / result[2];
    T r2 = xp * xp + yp * yp;
    T distortion = 1.0 + r2 * (camera[7] + camera[8] * r2;
    T predicted_x = camera[6] * distortion * xp;
    T predicted_y = camera[6] * distortion * yp;
    residual[0] = predicted_x - x_;
    residual[1] = predicted_y - y_;
  }
};

int main(int argc, char const* argv[]
{
  const int N = 2, N1 = 9, N2 = 3;
  Eigen::Matrix<double, N, N1> jacobi_parameter_1;
  Eigen::Matrix<double, N, N2> jacobi_parameter_2;
  Eigen::Matrix<double, N, 1> jacobi_residual;
  costfunction* costfunction_ = new costfunction(-3.326500e+02, 2.620900e+02;
  jet<N1 + N2> cameraJet[N1];
  jet<N1 + N2> pointJet[N2];
  double params_1[N1] = {
      1.5741515942940262e-02,  -1.2790936163850642e-02, -4.4008498081980789e-03,
      -3.4093839577186584e-02, -1.0751387104921525e-01, 1.1202240291236032e+00,
      3.9975152639358436e+02,  -3.1770643852803579e-07, 5.8820490534594022e-13};
  double params_2[N2] = {-0.612000157172, 0.571759047760, -1.847081276455};
  for (int i = 0; i < N1; i++
  {
    cameraJet[i].init(params_1[i], i;
  }
  for (int i = 0; i < N2; i++
  {
    pointJet[i].init(params_2[i], i + N1;
  }
  jet<N1 + N2>* residual = new jet<N1 + N2>[N];
  costfunction_->Evaluate(cameraJet, pointJet, residual;
  for (int i = 0; i < N; i++
  {
    jacobi_residual(i, 0 = residual[i].a;
  }
  for (int i = 0; i < N; i++
  {
    jacobi_parameter_1.row(i = residual[i].v.head(N1;
    jacobi_parameter_2.row(i = residual[i].v.tail(N2;
  }
  /* 
  real result:
  jacobi_parameter_1: 
   -283.512    -1296.34    -320.603     551.177 0.000204691    -471.095   -0.854706    -409.362    -490.465
    1242.05      220.93    -332.566 0.000204691     551.177       376.9     0.68381     327.511     392.397
  jacobi_parameter_2: 
  545.118 -5.05828 -478.067
  2.32675  557.047  368.163
  jacobi_residual: 
  -9.02023
    11.264
   */
  std::cout << "jacobi_parameter_1: \n" << jacobi_parameter_1 << std::endl;
  std::cout << "jacobi_parameter_2: \n" << jacobi_parameter_2 << std::endl;
  std::cout << "jacobi_residual: \n" << jacobi_residual << std::endl;
  delete (residual;
  return 0;
}

    输出结果

    jacobi_parameter_1: 
   -283.512    -1296.34    -320.603     551.177 0.000204691    -471.095   -0.854706    -409.362    -490.465
    1242.05      220.93    -332.566 0.000204691     551.177       376.9     0.68381     327.511     392.397
jacobi_parameter_2: 
 545.118 -5.05828 -478.067
 2.32675  557.047  368.163
jacobi_residual: 
-9.02023
  11.264

Reference

    http://ceres-solver.org/
  • https://blog.csdn.net/u012260559/article/details/105878468
  • https://www.ngui.cc/article/show-902862.html?action=onClick

编程笔记 » Ceres 自动求导解析-从原理到实践

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