// 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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//
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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::SizedCostFunction;
using ceres::Problem;
using ceres::Solver;
using ceres::Solve;
// 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:
virtual ~QuadraticCostFunction() {}
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.
//
// Since the Evaluate function can be called with the jacobians
// pointer equal to NULL, 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 NULL, 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 != NULL && jacobians[0] != NULL) {
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, NULL, &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;
}