// 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: sameeragarwal@google.com (Sameer Agarwal)
//
// An example program that minimizes Powell's singular function.
//
//   F = 1/2 (f1^2 + f2^2 + f3^2 + f4^2)
//
//   f1 = x1 + 10*x2;
//   f2 = sqrt(5) * (x3 - x4)
//   f3 = (x2 - 2*x3)^2
//   f4 = sqrt(10) * (x1 - x4)^2
//
// The starting values are x1 = 3, x2 = -1, x3 = 0, x4 = 1.
// The minimum is 0 at (x1, x2, x3, x4) = 0.
//
// From: Testing Unconstrained Optimization Software by Jorge J. More, Burton S.
// Garbow and Kenneth E. Hillstrom in ACM Transactions on Mathematical Software,
// Vol 7(1), March 1981.

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

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

struct F1 {
  template <typename T> bool operator()(const T* const x1,
                                        const T* const x2,
                                        T* residual) const {
    // f1 = x1 + 10 * x2;
    residual[0] = x1[0] + T(10.0) * x2[0];
    return true;
  }
};

struct F2 {
  template <typename T> bool operator()(const T* const x3,
                                        const T* const x4,
                                        T* residual) const {
    // f2 = sqrt(5) (x3 - x4)
    residual[0] = T(sqrt(5.0)) * (x3[0] - x4[0]);
    return true;
  }
};

struct F3 {
  template <typename T> bool operator()(const T* const x2,
                                        const T* const x4,
                                        T* residual) const {
    // f3 = (x2 - 2 x3)^2
    residual[0] = (x2[0] - T(2.0) * x4[0]) * (x2[0] - T(2.0) * x4[0]);
    return true;
  }
};

struct F4 {
  template <typename T> bool operator()(const T* const x1,
                                        const T* const x4,
                                        T* residual) const {
    // f4 = sqrt(10) (x1 - x4)^2
    residual[0] = T(sqrt(10.0)) * (x1[0] - x4[0]) * (x1[0] - x4[0]);
    return true;
  }
};

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

  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. The parameters, x1 through
  // x4, are modified in place.
  problem.AddResidualBlock(new AutoDiffCostFunction<F1, 1, 1, 1>(new F1),
                           NULL,
                           &x1, &x2);
  problem.AddResidualBlock(new AutoDiffCostFunction<F2, 1, 1, 1>(new F2),
                           NULL,
                           &x3, &x4);
  problem.AddResidualBlock(new AutoDiffCostFunction<F3, 1, 1, 1>(new F3),
                           NULL,
                           &x2, &x3);
  problem.AddResidualBlock(new AutoDiffCostFunction<F4, 1, 1, 1>(new F4),
                           NULL,
                           &x1, &x4);

  // Run the solver!
  Solver::Options options;
  options.max_num_iterations = 30;
  options.linear_solver_type = ceres::DENSE_QR;
  options.minimizer_progress_to_stdout = true;

  Solver::Summary summary;

  std::cout << "Initial x1 = " << x1
            << ", x2 = " << x2
            << ", x3 = " << x3
            << ", x4 = " << x4
            << "\n";

  Solve(options, &problem, &summary);

  std::cout << summary.BriefReport() << "\n";
  std::cout << "Final x1 = " << x1
            << ", x2 = " << x2
            << ", x3 = " << x3
            << ", x4 = " << x4
            << "\n";
  return 0;
}