Ceres Tutorials 学习笔记（一）：非线性最小二乘

介绍

1.使用ceres求解非线性最小二乘的步骤
* 自定义误差项：CostFunctor
* 建立CostFunction
* 设定初值
* 构建problem
* 构建cost_function, 并通过problem.AddResidualBlock添加
* 运行solver: Solve(options, &problem, & summary)
-定义option：liner_solver_type, minimizer_progress_to_stdout,
-定义summary
* 输出结果： cout << summary.BriefReport() << endl;

2.代码：

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

  // The variable to solve for with its initial value.
  double initial_x = 5.0;
  double x = initial_x;

  // Build the problem.
  Problem problem;

  // Set up the only cost function (also known as residual). This uses
  // auto-differentiation to obtain the derivative (jacobian).
  CostFunction* cost_function =
      new AutoDiffCostFunction1, 1>(new CostFunctor);
  problem.AddResidualBlock(cost_function, NULL, &x);

  // Run the solver!
  Solver::Options options;
  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 << "x : " << initial_x
            << " -> " << x << "\n";
  return 0;
}

导数

数值导数

理论导数

1.自定义参差结构体，用来计算参差，以f(x) = 10 - x 为例：

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

2.通过自定义的结构体，构建CostFunction， 并加入到problem:

CostFunction* cost_function =
  new NumericDiffCostFunction<NumericDiffCostFunctor, ceres::CENTRAL, 1, 1>(
      new NumericDiffCostFunctor);
problem.AddResidualBlock(cost_function, NULL, &x);

注意：这里使用的是NumericDiffCostFunction，而前面用的是AutoDiffFunction。更推荐使用AutoDiffFunction, 因为更加高效。

解析导数

在一些场合（不太懂），需要使用解析导数，这就需要自定义参差和Jacobian。可以通过构建一个CostFunction或者SizedCostFuntion的子类来实现。
但是，除非必须使用解析导数，一般更加推荐使用AutoDiffCostFuntion 或者 NumericDiffCosFunction

讨论

求导是使用Ceres中最复杂的过程，这里只是一个介绍而已。

例：鲍威尔方程（Powell’s Function)

问题

第一步：定义参差项（以f4为例）：

struct F4 {
  template <typename T>
  bool operator()(const T* const x1, const T* const x4, T* residual) const {
    residual[0] = T(sqrt(10.0)) * (x1[0] - x4[0]) * (x1[0] - x4[0]);
    return true;
  }
};

第二步：添加误差项

double x1 =  3.0; double x2 = -1.0; double x3 =  0.0; double x4 = 1.0;

Problem problem;

// Add residual terms to the problem using the using the autodiff
// wrapper to get the derivatives automatically.
problem.AddResidualBlock(
  new AutoDiffCostFunction1, 1, 1>(new F1), NULL, &x1, &x2);
problem.AddResidualBlock(
  new AutoDiffCostFunction1, 1, 1>(new F2), NULL, &x3, &x4);
problem.AddResidualBlock(
  new AutoDiffCostFunction1, 1, 1>(new F3), NULL, &x2, &x3)
problem.AddResidualBlock(
  new AutoDiffCostFunction1, 1, 1>(new F4), NULL, &x1, &x4);