// Ceres Solver - A fast non-linear least squares minimizer
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// http://code.google.com/p/ceres-solver/
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// Author: keir@google.com (Keir Mierle)
//
// A simple example of using the Ceres minimizer.
//
// Minimize 0.5 (10 - x)^2 using analytic jacobian matrix.

#include <vector>

#include "ceres/ceres.h"

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

class SimpleCostFunction
  : public SizedCostFunction<1 /* number of residuals */,
                             1 /* size of first parameter */> {
 public:
  virtual ~SimpleCostFunction() {}
  virtual bool Evaluate(double const* const* parameters,
                        double* residuals,
                        double** jacobians) const {
    double x = parameters[0][0];

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

    // f'(x) = -1. Since there's only 1 parameter and that parameter
    // has 1 dimension, there is only 1 element to fill in the
    // jacobians.
    if (jacobians != NULL && jacobians[0] != NULL) {
      jacobians[0][0] = -1;
    }
    return true;
  }
};

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

  // The variable with its initial value that we will be solving for.
  double x = 5.0;

  // Build the problem.
  Problem problem;
  // Set up the only cost function (also known as residual).
  problem.AddResidualBlock(new SimpleCostFunction, NULL, &x);

  // Run the solver!
  Solver::Options options;
  options.max_num_iterations = 10;
  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 : 5.0 -> " << x << "\n";
  return 0;
}