// Ceres Solver - A fast non-linear least squares minimizer
// Copyright 2010, 2011, 2012 Google Inc. All rights reserved.
// http://code.google.com/p/ceres-solver/
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// Author: keir@google.com (Keir Mierle)
//
// Minimize 0.5 (10 - x)^2 using jacobian matrix computed using
// numeric differentiation.

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

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

class ResidualWithNoDerivative
  : public SizedCostFunction<1 /* number of residuals */,
                             1 /* size of first parameter */> {
 public:
  virtual ~ResidualWithNoDerivative() {}
  virtual bool Evaluate(double const* const* parameters,
                        double* residuals,
                        double** jacobians) const {
    (void) jacobians;  // Ignored; filled in by numeric differentiation.

    // f(x) = 10 - x.
    residuals[0] = 10 - parameters[0][0];
    return true;
  }
};

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

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

  // Set up the only cost function (also known as residual). This uses
  // numeric differentiation to obtain the derivative (jacobian).
  CostFunction* cost =
      new NumericDiffCostFunction<ResidualWithNoDerivative, CENTRAL, 1, 1> (
          new ResidualWithNoDerivative, ceres::TAKE_OWNERSHIP);

  // Build the problem.
  Problem problem;
  problem.AddResidualBlock(cost, 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 : " << initial_x
            << " -> " << x << "\n";
  return 0;
}