1// Ceres Solver - A fast non-linear least squares minimizer
2// Copyright 2010, 2011, 2012 Google Inc. All rights reserved.
3// http://code.google.com/p/ceres-solver/
4//
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6// modification, are permitted provided that the following conditions are met:
7//
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9//   this list of conditions and the following disclaimer.
10// * Redistributions in binary form must reproduce the above copyright notice,
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12//   and/or other materials provided with the distribution.
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16//
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28//
29// Author: keir@google.com (Keir Mierle)
30//
31// Minimize 0.5 (10 - x)^2 using jacobian matrix computed using
32// numeric differentiation.
33
34#include <vector>
35#include "ceres/ceres.h"
36#include "gflags/gflags.h"
37#include "glog/logging.h"
38
39using ceres::NumericDiffCostFunction;
40using ceres::CENTRAL;
41using ceres::CostFunction;
42using ceres::Problem;
43using ceres::Solver;
44using ceres::Solve;
45
46// A cost functor that implements the residual r = 10 - x.
47class QuadraticCostFunctor {
48 public:
49  bool operator()(const double* const x, double* residual) const {
50    residual[0] = 10.0 - x[0];
51    return true;
52  }
53};
54
55int main(int argc, char** argv) {
56  google::ParseCommandLineFlags(&argc, &argv, true);
57  google::InitGoogleLogging(argv[0]);
58
59  // The variable to solve for with its initial value.
60  double initial_x = 5.0;
61  double x = initial_x;
62
63  // Set up the only cost function (also known as residual). This uses
64  // numeric differentiation to obtain the derivative (jacobian).
65  CostFunction* cost =
66      new NumericDiffCostFunction<QuadraticCostFunctor, CENTRAL, 1, 1> (
67          new QuadraticCostFunctor);
68
69  // Build the problem.
70  Problem problem;
71  problem.AddResidualBlock(cost, NULL, &x);
72
73  // Run the solver!
74  Solver::Options options;
75  options.max_num_iterations = 10;
76  options.linear_solver_type = ceres::DENSE_QR;
77  options.minimizer_progress_to_stdout = true;
78  Solver::Summary summary;
79  Solve(options, &problem, &summary);
80  std::cout << summary.BriefReport() << "\n";
81  std::cout << "x : " << initial_x
82            << " -> " << x << "\n";
83  return 0;
84}
85