General Unconstrained Minimization
Ceres 求解器除了能求解非线性最小二乘法问题外，还能使用目标函数值和梯度求解一般无约束问题。在本章中，我们将了解如何做到这一点。
Rosenbrock’s Function
考虑尝试最小化 Rosenbrock’s function。最简单方法是定义一个模板函数来评估误差函数，然后使用 Ceres Solver 的自动微分来计算其导数。我们首先定义一个模板化的函数，然后使用 AutoDiffFirstOrderFunction 构造一个 FirstOrderFunction 接口的实例。这个对象负责计算目标函数值和梯度（如果需要）。这与 Ceres 中定义非线性最小二乘法问题时的 CostFunction类似。
// f(x,y) = (1-x)^2 + 100(y - x^2)^2;
struct Rosenbrock {
template <typename T>
bool operator()(const T* parameters, T* cost) const {
const T x = parameters[0];
const T y = parameters[1];
cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * (y - x * x) * (y - x * x);
return true;
}
static ceres::FirstOrderFunction* Create() {
constexpr int kNumParameters = 2;
return new ceres::AutoDiffFirstOrderFunction<Rosenbrock, kNumParameters>();
}
};然后，只需构建一个 GradientProblem 对象并调用 Solve() 即可将其最小化。
double parameters[2] = {-1.2, 1.0};
ceres::GradientProblem problem(Rosenbrock::Create());
ceres::GradientProblemSolver::Options options;
options.minimizer_progress_to_stdout = true;
ceres::GradientProblemSolver::Summary summary;
ceres::Solve(options, problem, parameters, &summary);
std::cout << summary.FullReport() << "\n";执行该代码，使用 limited memory BFGS算法解决问题。
0: f: 2.420000e+01 d: 0.00e+00 g: 2.16e+02 h: 0.00e+00 s: 0.00e+00 e: 0 it: 1.19e-05 tt: 1.19e-05
1: f: 4.280493e+00 d: 1.99e+01 g: 1.52e+01 h: 2.01e-01 s: 8.62e-04 e: 2 it: 7.30e-05 tt: 1.72e-04
2: f: 3.571154e+00 d: 7.09e-01 g: 1.35e+01 h: 3.78e-01 s: 1.34e-01 e: 3 it: 1.60e-05 tt: 1.93e-04
3: f: 3.440869e+00 d: 1.30e-01 g: 1.73e+01 h: 1.36e-01 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 1.97e-04
4: f: 3.213597e+00 d: 2.27e-01 g: 1.55e+01 h: 1.06e-01 s: 4.59e-01 e: 1 it: 1.19e-06 tt: 2.00e-04
5: f: 2.839723e+00 d: 3.74e-01 g: 1.05e+01 h: 1.34e-01 s: 5.24e-01 e: 1 it: 9.54e-07 tt: 2.03e-04
6: f: 2.448490e+00 d: 3.91e-01 g: 1.29e+01 h: 3.04e-01 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 2.05e-04
7: f: 1.943019e+00 d: 5.05e-01 g: 4.00e+00 h: 8.81e-02 s: 7.43e-01 e: 1 it: 9.54e-07 tt: 2.08e-04
8: f: 1.731469e+00 d: 2.12e-01 g: 7.36e+00 h: 1.71e-01 s: 4.60e-01 e: 2 it: 2.15e-06 tt: 2.11e-04
9: f: 1.503267e+00 d: 2.28e-01 g: 6.47e+00 h: 8.66e-02 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 2.14e-04
10: f: 1.228331e+00 d: 2.75e-01 g: 2.00e+00 h: 7.70e-02 s: 7.90e-01 e: 1 it: 0.00e+00 tt: 2.16e-04
11: f: 1.016523e+00 d: 2.12e-01 g: 5.15e+00 h: 1.39e-01 s: 3.76e-01 e: 2 it: 1.91e-06 tt: 2.25e-04
12: f: 9.145773e-01 d: 1.02e-01 g: 6.74e+00 h: 7.98e-02 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 2.28e-04
13: f: 7.508302e-01 d: 1.64e-01 g: 3.88e+00 h: 5.76e-02 s: 4.93e-01 e: 1 it: 9.54e-07 tt: 2.30e-04
14: f: 5.832378e-01 d: 1.68e-01 g: 5.56e+00 h: 1.42e-01 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 2.33e-04
15: f: 3.969581e-01 d: 1.86e-01 g: 1.64e+00 h: 1.17e-01 s: 1.00e+00 e: 1 it: 1.19e-06 tt: 2.36e-04
16: f: 3.171557e-01 d: 7.98e-02 g: 3.84e+00 h: 1.18e-01 s: 3.97e-01 e: 2 it: 1.91e-06 tt: 2.39e-04
17: f: 2.641257e-01 d: 5.30e-02 g: 3.27e+00 h: 6.14e-02 s: 1.00e+00 e: 1 it: 1.19e-06 tt: 2.42e-04
18: f: 1.909730e-01 d: 7.32e-02 g: 5.29e-01 h: 8.55e-02 s: 6.82e-01 e: 1 it: 9.54e-07 tt: 2.45e-04
19: f: 1.472012e-01 d: 4.38e-02 g: 3.11e+00 h: 1.20e-01 s: 3.47e-01 e: 2 it: 1.91e-06 tt: 2.49e-04
20: f: 1.093558e-01 d: 3.78e-02 g: 2.97e+00 h: 8.43e-02 s: 1.00e+00 e: 1 it: 2.15e-06 tt: 2.52e-04
21: f: 6.710346e-02 d: 4.23e-02 g: 1.42e+00 h: 9.64e-02 s: 8.85e-01 e: 1 it: 8.82e-06 tt: 2.81e-04
22: f: 3.993377e-02 d: 2.72e-02 g: 2.30e+00 h: 1.29e-01 s: 4.63e-01 e: 2 it: 7.87e-06 tt: 2.96e-04
23: f: 2.911794e-02 d: 1.08e-02 g: 2.55e+00 h: 6.55e-02 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.00e-04
24: f: 1.457683e-02 d: 1.45e-02 g: 2.77e-01 h: 6.37e-02 s: 6.14e-01 e: 1 it: 1.19e-06 tt: 3.03e-04
25: f: 8.577515e-03 d: 6.00e-03 g: 2.86e+00 h: 1.40e-01 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.06e-04
26: f: 3.486574e-03 d: 5.09e-03 g: 1.76e-01 h: 1.23e-02 s: 1.00e+00 e: 1 it: 1.19e-06 tt: 3.09e-04
27: f: 1.257570e-03 d: 2.23e-03 g: 1.39e-01 h: 5.08e-02 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.12e-04
28: f: 2.783568e-04 d: 9.79e-04 g: 6.20e-01 h: 6.47e-02 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.15e-04
29: f: 2.533399e-05 d: 2.53e-04 g: 1.68e-02 h: 1.98e-03 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.17e-04
30: f: 7.591572e-07 d: 2.46e-05 g: 5.40e-03 h: 9.27e-03 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.20e-04
31: f: 1.902460e-09 d: 7.57e-07 g: 1.62e-03 h: 1.89e-03 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.23e-04
32: f: 1.003030e-12 d: 1.90e-09 g: 3.50e-05 h: 3.52e-05 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.26e-04
33: f: 4.835994e-17 d: 1.00e-12 g: 1.05e-07 h: 1.13e-06 s: 1.00e+00 e: 1 it: 1.19e-06 tt: 3.34e-04
34: f: 1.885250e-22 d: 4.84e-17 g: 2.69e-10 h: 1.45e-08 s: 1.00e+00 e: 1 it: 9.54e-07 tt: 3.37e-04
Solver Summary (v 2.2.0-eigen-(3.4.0)-lapack-suitesparse-(7.1.0)-metis-(5.1.0)-acceleratesparse-eigensparse)
Parameters 2
Line search direction LBFGS (20)
Line search type CUBIC WOLFE
Cost:
Initial 2.420000e+01
Final 1.955192e-27
Change 2.420000e+01
Minimizer iterations 36
Time (in seconds):
Cost evaluation 0.000000 (0)
Gradient & cost evaluation 0.000000 (44)
Polynomial minimization 0.000061
Total 0.000438
Termination: CONVERGENCE (Parameter tolerance reached. Relative step_norm: 1.890726e-11 <= 1.000000e-08.)
Initial x: -1.2 y: 1
Final x: 1 y: 1如果由于某种原因无法使用自动微分（例如因为需要调用外部库），那么可以使用数值微分。在这种情况下，函数的定义如下
// f(x,y) = (1-x)^2 + 100(y - x^2)^2;
struct Rosenbrock {
bool operator()(const double* parameters, double* cost) const {
const double x = parameters[0];
const double y = parameters[1];
cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * (y - x * x) * (y - x * x);
return true;
}
static ceres::FirstOrderFunction* Create() {
constexpr int kNumParameters = 2;
return new ceres::NumericDiffFirstOrderFunction<Rosenbrock,
ceres::CENTRAL,
kNumParameters>();
}
};最后，如果你更愿意手工计算导数（比如因为参数向量太大而无法自动微分）。那么就应该定义 FirstOrderFunction 的实例，它与 CostFunction 类似，适用于非线性最小二乘法问题
// f(x,y) = (1-x)^2 + 100(y - x^2)^2;
class Rosenbrock final : public ceres::FirstOrderFunction {
public:
bool Evaluate(const double* parameters,
double* cost,
double* gradient) const override {
const double x = parameters[0];
const double y = parameters[1];
cost[0] = (1.0 - x) * (1.0 - x) + 100.0 * (y - x * x) * (y - x * x);
if (gradient) {
gradient[0] = -2.0 * (1.0 - x) - 200.0 * (y - x * x) * 2.0 * x;
gradient[1] = 200.0 * (y - x * x);
}
return true;
}
int NumParameters() const override { return 2; }
};Footnotes
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