Ceres-Solver学习笔记(2)

先从简单的开始学习。

1.继承SizedCostFunction，构造CostFunction

#include <vector>
#include "ceres/ceres.h"
#include "glog/logging.h"

using ceres::CostFunction;
using ceres::SizedCostFunction;
using ceres::Problem;
using ceres::Solver;
using ceres::Solve;

// class inheritance of SizedCostFunction
class QuadraticCostFunction
  : public SizedCostFunction<1 /* number of residuals */,
                             1 /* size of first parameter */> {
 public:
  virtual ~QuadraticCostFunction() {}

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

    // 残差函数
    residuals[0] = 10 - x;

    // f'(x) = -1. 只有一个参数，参数只有一个维度，jacobians只有一个元素
    //
    // 因为jacobians可以设置为NULL, Evaluate必须检查jacobians是否需要计算
    //
    // 对这个简单问题，检查jacobians[0]是否为空是足够的，但是通常的CostFunctions来说，
    // Ceres可能只需要参数的子集的导数
    if (jacobians != NULL && jacobians[0] != NULL) {
      jacobians[0][0] = -1;
    }
    return true;
  }
};
// main 
int main(int argc, char** argv) {
  google::InitGoogleLogging(argv[0]);

  // 变量初始值
  double x = 0.5;
  const double initial_x = x;

  // Build the problem.
  Problem problem;

  // 构建CostFunction（残差方程）
  CostFunction* cost_function = new QuadraticCostFunction;
  // CostFunction为QuadraticCostFunction，不使用LossFunction,只有一个优化参数x
  problem.AddResidualBlock(cost_function, NULL, &x);

  // Run the solver!
  Solver::Options options;
  options.minimizer_progress_to_stdout = true;
  Solver::Summary summary;
  Solve(options, &problem, &summary);

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

  return 0;
}

这个例子并没能解决我的疑惑，jacobians用在哪里？我们继续下一个例子。

2.曲线拟合与LossFunction

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

using ceres::AutoDiffCostFunction;
using ceres::CostFunction;
using ceres::Problem;
using ceres::Solver;
using ceres::Solve;

// 用于拟合曲线的点数量
const int kNumObservations = 67;
const double data[] = {
    0.000000e+00, 1.133898e+00, //第1个点
    7.500000e-02, 1.334902e+00, //第2个点
    //..., 这里省略64个点
    4.950000e+00, 4.669206e+00, //第67个点
    }

// 构建CostFunction结构体
struct ExponentialResidual {
  // 输入一对坐标x,y
  ExponentialResidual(double x, double y)
      : x_(x), y_(y) {}
  // 函数y = e^{mx + c}. 
  // 残差为 y_ - y
  template <typename T> bool operator()(const T* const m,
                                        const T* const c,
                                        T* residual) const {
    residual[0] = T(y_) - exp(m[0] * T(x_) + c[0]);
    return true;
  }

 private:
  const double x_;
  const double y_;
};
// main

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

  // m c 初始值
  double m = 0.0;
  double c = 0.0;

  Problem problem;
  for (int i = 0; i < kNumObservations; ++i) {
    problem.AddResidualBlock(
        // CostFunction为ExponentialResidual，
        // 残差个数为1，参数1(x)维度为1，参数2(y)维度为1，
        // 损失函数为CauchyLoss(0.5)
        new AutoDiffCostFunction<ExponentialResidual, 1, 1, 1>(
            new ExponentialResidual(data[2 * i], data[2 * i + 1])),
        new CauchyLoss(0.5),
        &m, &c);
  }

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

  Solver::Summary summary;
  Solve(options, &problem, &summary);
  std::cout << summary.BriefReport() << "\n";
  std::cout << "Initial m: " << 0.0 << " c: " << 0.0 << "\n";
  std::cout << "Final   m: " << m << " c: " << c << "\n";
  return 0;
}

介绍了两个简单的例子，下一节介绍更复杂一点的BA