tutorial.m (Start Here)

To be able to quickly start with a toolkit, it is often advantageous (not only for the impatient users), to look at some code examples first.

In this tutorial, we give a short example program which interpolates a bivariate function on a regular sparse grid. Identical versions of the example are given in all languages currently supported by SG++: C++, Python, Java, and MATLAB.

In the example, we create a two-dimensional regular sparse grid of level 3 (with grid points \(\vec{x}_j \in [0, 1]^2\)) using piecewise bilinear basis functions \(\varphi_j\colon [0, 1]^2 \to \mathbb{R}\). We then interpolate the function

\[ f\colon [0, 1]^2 \to \mathbb{R},\quad f(x_0, x_1) := 16 (x_0 - 1) x_0 (x_1 - 1) x_1 \]


\[ u\colon [0, 1]^2 \to \mathbb{R},\quad u(x_0, x_1) := \sum_{j=0}^{N-1} \alpha_j \varphi_j(x_0, x_1) \]

by calculating the coefficients \(\alpha_j\) such that \(u(\vec{x}_j) = f(\vec{x}_j)\) for all \(j\). This process is called hierarchization in sparse grid contexts; the \(\alpha_j\) are called (hierarchical) surpluses. Note that \(f\) vanishes at the boundary of the domain \([0, 1]^2\); therefore, we don't have to spend sparse grid points on the boundary. Finally, we evaluate the sparse grid function \(u\) at a point \(\vec{p} = (0.52, 0.73)\).

Before starting, the function \(f\), which we want to interpolate, is defined.

f = @(x0, x1) (16.0 * (x0 - 1.0) * x0 * (x1 - 1.0) * x1);

First, we create a two-dimensional grid (type sgpp::base::Grid) with piecewise bilinear basis functions with the help of the factory method sgpp::base::Grid.createLinearGrid().

dim = 2;
grid = sgpp.Grid.createLinearGrid(dim);

Then we obtain a reference to the grid's sgpp::base::GridStorage object which allows us, e.g., to access grid points, to obtain the dimensionality (which we print) and the number of grid points.

gridStorage = grid.getStorage();
fprintf('dimensionality: %u\n', gridStorage.getDimension());

Now, we use a sgpp::base::GridGenerator to create a regular sparse grid of level 3. Thus, gridStorage.getSize() returns 17, the number of grid points of a two-dimensional regular sparse grid of level 3.

level = 3;
fprintf('number of grid points: %u\n', gridStorage.getSize());

We create an object of type sgpp::base::DataVector which is essentially a wrapper around a double array. The DataVector is initialized with as many entries as there are grid points. It serves as a coefficient vector for the sparse grid interpolant we want to construct. As the entries of a freshly created DataVector are not initialized, we set them to 0.0. (This is superfluous here as we initialize them in the next few lines anyway.)

alpha = sgpp.DataVector(gridStorage.getSize());
fprintf('length of alpha vector: %u\n', alpha.getSize());

The for loop iterates over all grid points: For each grid point gp, the corresponding coefficient \(\alpha_j\) is set to the function value at the grid point's coordinates which are obtained by getStandardCoordinate(dim). The current coefficient vector is then printed.

for i = 0:gridStorage.getSize()-1
gp = gridStorage.getPoint(i);
alpha.set(i, f(gp.getStandardCoordinate(0), gp.getStandardCoordinate(1)));
fprintf('alpha before hierarchization: %s\n', char(alpha.toString()));

An object of sgpp::base::OperationHierarchisation is created and used to hierarchize the coefficient vector, which we print.

fprintf('alpha after hierarchization: %s\n', char(alpha.toString()));

Finally, a second DataVector is created which is used as a point to evaluate the sparse grid function at. An object is obtained which provides an evaluation operation (of type sgpp::base::OperationEvaluation), and the sparse grid interpolant is evaluated at \(\vec{p}\), which is close to (but not exactly at) a grid point.

p = sgpp.DataVector(dim);
p.set(0, 0.52);
p.set(1, 0.73);
opEval = sgpp.createOperationEval(grid);
fprintf('u(0.52, 0.73) = %.4f\n', opEval.eval(alpha,p));

The example results in the following output:

dimensionality:         2
number of grid points:  17
length of alpha vector: 17
alpha before hierarchization: [1, 0.75, 0.75, 0.4375, 0.9375, 0.9375, 0.4375, 0.75, 0.75, 0.4375, 0.9375, 0.9375, 0.4375, 0.5625, 0.5625, 0.5625, 0.5625]
alpha after hierarchization:  [1, 0.25, 0.25, 0.0625, 0.0625, 0.0625, 0.0625, 0.25, 0.25, 0.0625, 0.0625, 0.0625, 0.0625, 0.0625, 0.0625, 0.0625, 0.0625]
u(0.52, 0.73) = 0.7696

It can be clearly seen that the surpluses decay with a factor of 1/4: On the first level, we obtain 1, on the second 1/4, and on the third 1/16 as surpluses.