Ceres-Solver 从入门到上手视觉SLAM位姿优化问题


[TOC]

概述

Ceres Solver is an open source C++ library for modeling and solving large, complicated optimization problems.

使用 Ceres Solver 求解非线性优化问题,主要包括以下几部分:

  • 构建代价函数(cost function) 或 残差(residual)
  • 构建优化问题(ceres::Problem):通过 AddResidualBlock 添加代价函数(cost function)、损失函数(loss function) 和 待优化状态量
    • LossFunction: a scalar function that is used to reduce the influence of outliers on the solution of non-linear least squares problems.
  • 配置求解器(ceres::Solver::Options)
  • 运行求解器(ceres::Solve(options, &problem, &summary))

注意

Ceres Solver 只接受最小二乘优化,也就是 $\min r^T r$;若要对残差加权重,使用马氏距离,即 $\min r^T P^{-1} r$,则要对 信息矩阵$P^{-1}$ 做 Cholesky分解,即 $LL^T=P^{−1}$,则 $d = r^T (L L^T) r = (L^T r)^T (L^T r)$,令 $r’ = (L^T r)$,最终 $\min r’^T r’$。

代码cggos/state_estimation

入门

先以最小化下面的函数为例,介绍 Ceres Solver 的基本用法

\[\frac{1}{2} (10 - x)^2\]

Part 1: Cost Function

(1)AutoDiffCostFunction(自动求导)

  • 构造 代价函数结构体(例如:struct CostFunctor),在其结构体内对 模板括号() 重载,定义残差
  • 在重载的 () 函数 形参 中,最后一个为残差,前面几个为待优化状态量
struct CostFunctor {
    template<typename T>
    bool operator()(const T *const x, T *residual) const {
        residual[0] = 10.0 - x[0]; // r(x) = 10 - x
        return true;
    }
};
  • 创建代价函数的实例,对于模板参数的数字,第一个为残差的维度,后面几个为待优化状态量的维度
ceres::CostFunction *cost_function;
cost_function = new ceres::AutoDiffCostFunction<CostFunctor, 1, 1>(new CostFunctor);

(2) NumericDiffCostFunction

  • 数值求导法 也是构造 代价函数结构体,但在重载 括号() 时没有用模板
struct CostFunctorNum {
    bool operator()(const double *const x, double *residual) const {
        residual[0] = 10.0 - x[0]; // r(x) = 10 - x
        return true;
    }
};
  • 并且在实例化代价函数时也稍微有点区别,多了一个模板参数 ceres::CENTRAL
ceres::CostFunction *cost_function;
cost_function =
new ceres::NumericDiffCostFunction<CostFunctorNum, ceres::CENTRAL, 1, 1>(new CostFunctorNum);

(3) 自定义 CostFunction

  • 构建一个 继承自 ceres::SizedCostFunction<1,1> 的类,同样,对于模板参数的数字,第一个为残差的维度,后面几个为待优化状态量的维度
  • 重载 虚函数virtual bool Evaluate(double const* const* parameters, double *residuals, double **jacobians) const,根据 待优化变量,实现 残差和雅克比矩阵的计算
class SimpleCostFunctor : public ceres::SizedCostFunction<1,1> {
public:
    virtual ~SimpleCostFunctor() {};

    virtual bool Evaluate(
      double const* const* parameters, double *residuals, double** jacobians) const {
        const double x = parameters[0][0];

        residuals[0] = 10 - x; // r(x) = 10 - x

        if(jacobians != NULL && jacobians[0] != NULL) {
            jacobians[0][0] = -1; // r'(x) = -1
        }

        return true;
    }
};

Part 2: AddResidualBlock

  • 声明 ceres::Problem problem;
  • 通过 AddResidualBlock代价函数(cost function)、损失函数(loss function) 和 待优化状态量 添加到 problem
ceres::CostFunction *cost_function;
cost_function = new ceres::AutoDiffCostFunction<CostFunctor, 1, 1>(new CostFunctor);

ceres::Problem problem;
problem.AddResidualBlock(cost_function, NULL, &x);

Part 3: Config & Solve

配置求解器,并计算,输出结果

ceres::Solver::Options options;
options.max_num_iterations = 25;
options.linear_solver_type = ceres::DENSE_QR;
options.minimizer_progress_to_stdout = true;

ceres::Solver::Summary summary;
ceres::Solve(options, &problem, &summary);
std::cout << summary.BriefReport() << "\n";

Simple Example Code

#include "ceres/ceres.h"
#include "glog/logging.h"

struct CostFunctor {
    template<typename T>
    bool operator()(const T *const x, T *residual) const {
        residual[0] = 10.0 - x[0]; // f(x) = 10 - x
        return true;
    }
};

struct CostFunctorNum {
    bool operator()(const double *const x, double *residual) const {
        residual[0] = 10.0 - x[0]; // f(x) = 10 - x
        return true;
    }
};

class SimpleCostFunctor : public ceres::SizedCostFunction<1,1> {
public:
    virtual ~SimpleCostFunctor() {};

    virtual bool Evaluate(
      double const* const* parameters, double *residuals, double **jacobians) const {
        const double x = parameters[0][0];

        residuals[0] = 10 - x; // f(x) = 10 - x

        if(jacobians != NULL && jacobians[0] != NULL) {
            jacobians[0][0] = -1; // f'(x) = -1
        }

        return true;
    }
};

int main(int argc, char** argv) {
    google::InitGoogleLogging(argv[0]);

    double x = 0.5;
    const double initial_x = x;

    ceres::Problem problem;

    // Set up the only cost function (also known as residual)
    ceres::CostFunction *cost_function;

    // auto-differentiation
//    cost_function = new ceres::AutoDiffCostFunction<CostFunctor, 1, 1>(new CostFunctor);

     //numeric differentiation
//    cost_function =
//    new ceres::NumericDiffCostFunction<CostFunctorNum, ceres::CENTRAL, 1, 1>(
//      new CostFunctorNum);

    cost_function = new SimpleCostFunctor;

    // 添加代价函数cost_function和损失函数NULL,其中x为状态量
    problem.AddResidualBlock(cost_function, NULL, &x);

    ceres::Solver::Options options;
    options.minimizer_progress_to_stdout = true;

    ceres::Solver::Summary summary;
    ceres::Solve(options, &problem, &summary);

    std::cout << summary.BriefReport() << "\n";
    std::cout << "x : " << initial_x << " -> " << x << "\n";

    return 0;
}

应用: 基于李代数的视觉SLAM位姿优化

下面以 基于李代数的视觉SLAM位姿优化问题 为例,介绍 Ceres Solver 的使用。

(1)残差(预测值 - 观测值)

\[r(\xi) = K \exp({\xi}^{\wedge}) P - u\]

(2)雅克比矩阵

\[\begin{aligned} J &= \frac{\partial r(\xi)}{\partial \xi} \\ &= \begin{bmatrix} \frac{f_x}{Z'} & 0 & -\frac{X'f_x}{Z'^2} & -\frac{X'Y'f_x}{Z'^2} & f_x+\frac{X'^2f_x}{Z'^2} & -\frac{Y'f_x}{Z'} \\ 0 & \frac{f_y}{Z'} & -\frac{Y'f_y}{Z'^2} & -f_y-\frac{Y'^2f_y}{Z'^2} & \frac{X'Y'f_y}{Z'^2} & \frac{X'f_y}{Z'} \end{bmatrix} \in \mathbb{R}^{2 \times 6} \end{aligned}\]

(3)核心代码

代价函数的构造:

class BAGNCostFunctor : public ceres::SizedCostFunction<2, 6> {
public:
    EIGEN_MAKE_ALIGNED_OPERATOR_NEW

    BAGNCostFunctor(Eigen::Vector2d observed_p, Eigen::Vector3d observed_P) :
            observed_p_(observed_p), observed_P_(observed_P) {}

    virtual ~BAGNCostFunctor() {}

    virtual bool Evaluate(
      double const* const* parameters, double *residuals, double **jacobians) const {

        Eigen::Map<const Eigen::Matrix<double,6,1>> T_se3(*parameters);

        Sophus::SE3 T_SE3 = Sophus::SE3::exp(T_se3);

        Eigen::Vector3d Pc = T_SE3 * observed_P_;

        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 residual =  (K * Pc).hnormalized() - observed_p_;

        residuals[0] = residual[0];
        residuals[1] = residual[1];

        if(jacobians != NULL) {

            if(jacobians[0] != NULL) {

                Eigen::Map<Eigen::Matrix<double, 2, 6, Eigen::RowMajor>> 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;

                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 observed_p_;
    const Eigen::Vector3d observed_P_;
};

构造优化问题,并求解相机位姿:

Sophus::Vector6d se3;

ceres::Problem problem;
for(int i=0; i<n_points; ++i) {
    ceres::CostFunction *cost_function;
    cost_function = new BAGNCostFunctor(p2d[i], p3d[i]);
    problem.AddResidualBlock(cost_function, NULL, se3.data());
}

ceres::Solver::Options options;
options.dynamic_sparsity = true;
options.max_num_iterations = 100;
options.sparse_linear_algebra_library_type = ceres::SUITE_SPARSE;
options.minimizer_type = ceres::TRUST_REGION;
options.linear_solver_type = ceres::SPARSE_NORMAL_CHOLESKY;
options.trust_region_strategy_type = ceres::DOGLEG;
options.minimizer_progress_to_stdout = true;
options.dogleg_type = ceres::SUBSPACE_DOGLEG;

ceres::Solver::Summary summary;
ceres::Solve(options, &problem, &summary);
std::cout << summary.BriefReport() << "\n";

std::cout << "estimated pose: \n" << Sophus::SE3::exp(se3).matrix() << std::endl;



^