examples/helloworld_analytic_diff.cc

// Ceres Solver - A fast non-linear least squares minimizer
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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"
#include "glog/logging.h"

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

// A CostFunction implementing analytically derivatives for the
// function f(x) = 10 - x.
class QuadraticCostFunction
    : public SizedCostFunction<1 /* number of residuals */,
                               1 /* size of first parameter */> {
 public:
  bool Evaluate(double const* const* parameters,
                double* residuals,
                double** jacobians) const override {
    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.
    //
    // Since the Evaluate function can be called with the jacobians
    // pointer equal to nullptr, the Evaluate function must check to see
    // if jacobians need to be computed.
    //
    // For this simple problem it is overkill to check if jacobians[0]
    // is nullptr, but in general when writing more complex
    // CostFunctions, it is possible that Ceres may only demand the
    // derivatives w.r.t. a subset of the parameter blocks.
    if (jacobians != nullptr && jacobians[0] != nullptr) {
      jacobians[0][0] = -1;
    }

    return true;
  }
};

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

  // The variable to solve for with its initial value. It will be
  // mutated in place by the solver.
  double x = 0.5;
  const double initial_x = x;

  // Build the problem.
  Problem problem;

  // Set up the only cost function (also known as residual).
  CostFunction* cost_function = new QuadraticCostFunction;
  problem.AddResidualBlock(cost_function, nullptr, &x);

  // Run the solver!
  Solver::Options options;
  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;
}