SG++-Doxygen-Documentation
optimization.cpp

On this page, we look at an example application of the sgpp::optimization module.

Versions of the example are given in all languages currently supported by SG++: C++, Python, Java, and MATLAB.

The example interpolates a bivariate test function with B-splines instead of piecewise linear basis functions to obtain a smoother interpolant. The resulting sparse grid function is then minimized with the method of steepest descent. For comparison, we also minimize the objective function with Nelder-Mead's method.

First, we include all the necessary headers, including those of the sgpp::base and sgpp::optimization module.

#include <sgpp_base.hpp>
#include <algorithm>
#include <iostream>
#include <iterator>

The function $$f\colon [0, 1]^d \to \mathbb{R}$$ to be minimized is called objective function and has to derive from sgpp::optimization::ScalarFunction. In the constructor, we give the dimensionality of the domain (in this case $$d = 2$$). The eval method evaluates the objective function and returns the function value $$f(\vec{x})$$ for a given point $$\vec{x} \in [0, 1]^d$$. The clone method returns a std::unique_ptr to a clone of the object and is used for parallelization (in case eval is not thread-safe).

class ExampleFunction : public sgpp::optimization::ScalarFunction {
public:
ExampleFunction() : sgpp::optimization::ScalarFunction(2) {}
double eval(const sgpp::base::DataVector& x) {
// minimum is f(x) = -2 for x[0] = 3*pi/16, x[1] = 3*pi/14
return std::sin(8.0 * x[0]) + std::sin(7.0 * x[1]);
}
virtual void clone(std::unique_ptr<sgpp::optimization::ScalarFunction>& clone) const {
clone = std::unique_ptr<sgpp::optimization::ScalarFunction>(new ExampleFunction(*this));
}
};

Now, we can start with the main function.

void printLine() {
std::cout << "----------------------------------------"
"----------------------------------------\n";
}
int main(int argc, const char* argv[]) {
(void)argc;
(void)argv;
std::cout << "sgpp::optimization example program started.\n\n";
// increase verbosity of the output

Here, we define some parameters: objective function, dimensionality, B-spline degree, maximal number of grid points, and adaptivity.

// objective function
ExampleFunction f;
// dimension of domain
const size_t d = f.getNumberOfParameters();
// B-spline degree
const size_t p = 3;
// maximal number of grid points
const size_t N = 30;
const double gamma = 0.95;

First, we define a grid with modified B-spline basis functions and an iterative grid generator, which can generate the grid adaptively.

With the iterative grid generator, we generate adaptively a sparse grid.

printLine();
std::cout << "Generating grid...\n\n";
if (!gridGen.generate()) {
std::cout << "Grid generation failed, exiting.\n";
return 1;
}

Then, we hierarchize the function values to get hierarchical B-spline coefficients of the B-spline sparse grid interpolant $$\tilde{f}\colon [0, 1]^d \to \mathbb{R}$$.

printLine();
std::cout << "Hierarchizing...\n\n";
sgpp::base::DataVector functionValues(gridGen.getFunctionValues());
sgpp::base::DataVector coeffs(functionValues.getSize());
// solve linear system
if (!sleSolver.solve(hierSLE, functionValues, coeffs)) {
std::cout << "Solving failed, exiting.\n";
return 1;
}

We define the interpolant $$\tilde{f}$$ and its gradient $$\nabla\tilde{f}$$ for use with the gradient method (steepest descent). Of course, one can also use other optimization algorithms from sgpp::optimization::optimizer.

printLine();
std::cout << "Optimizing smooth interpolant...\n\n";
double fX0;
double ftX0;

The gradient method needs a starting point. We use a point of our adaptively generated sparse grid as starting point. More specifically, we use the point with the smallest (most promising) function value and save it in x0.

{
sgpp::base::GridStorage& gridStorage = grid.getStorage();
// index of grid point with minimal function value
size_t x0Index =
std::distance(functionValues.getPointer(),
std::min_element(functionValues.getPointer(),
functionValues.getPointer() + functionValues.getSize()));
x0 = gridStorage.getCoordinates(gridStorage[x0Index]);
fX0 = functionValues[x0Index];
ftX0 = ft.eval(x0);
}
std::cout << "x0 = " << x0.toString() << "\n";
std::cout << "f(x0) = " << fX0 << ", ft(x0) = " << ftX0 << "\n\n";

We apply the gradient method and print the results.

const double fXOpt = f.eval(xOpt);
std::cout << "\nxOpt = " << xOpt.toString() << "\n";
std::cout << "f(xOpt) = " << fXOpt << ", ft(xOpt) = " << ftXOpt << "\n\n";

For comparison, we apply the classical gradient-free Nelder-Mead method directly to the objective function $$f$$.

printLine();
std::cout << "Optimizing objective function (for comparison)...\n\n";
const double ftXOptNM = ft.eval(xOptNM);
std::cout << "\nnxOptNM = " << xOptNM.toString() << "\n";
std::cout << "f(xOptNM) = " << fXOptNM << ", ft(xOptNM) = " << ftXOptNM << "\n\n";
printLine();
std::cout << "\nsgpp::optimization example program terminated.\n";
return 0;
}

The example program outputs the following results:

sgpp::optimization example program started.

--------------------------------------------------------------------------------
Generating grid...

100.0% (N = 29, k = 3)
Done in 3ms.
--------------------------------------------------------------------------------
Hierarchizing...

Solving linear system (automatic method)...
estimated nnz ratio: 59.8%
constructing matrix (100.0%)
nnz ratio: 58.0%
Done in 0ms.
Done in 1ms.
--------------------------------------------------------------------------------
Optimizing smooth interpolant...

x0 = [0.625, 0.75]
f(x0) = -1.81786, ft(x0) = -1.81786

9 steps, f(x) = -2.000780
Done in 1ms.

xOpt = [0.589526, 0.673268]
f(xOpt) = -1.99999, ft(xOpt) = -2.00078

--------------------------------------------------------------------------------
Optimizing objective function (for comparison)...


We see that both the gradient-based optimization of the smooth sparse grid interpolant and the gradient-free optimization of the objective function find reasonable approximations of the minimum, which lies at $$(3\pi/16, 3\pi/14) \approx (0.58904862, 0.67319843)$$.