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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 \(\vec{p} = (0.52, 0.73)\).

At the beginning of the program, we have to import the pysgpp library.

1 import pysgpp

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

1 f = lambda 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().

1 dim = 2
2 grid = pysgpp.Grid.createLinearGrid(dim)

Then we obtain 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.

1 gridStorage = grid.getStorage()
2 print("dimensionality: {}".format(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.

1 level = 3
2 grid.getGenerator().regular(level)
3 print("number of grid points: {}".format(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.)

1 alpha = pysgpp.DataVector(gridStorage.getSize())
2 alpha.setAll(0.0)
3 print("length of alpha vector: {}".format(len(alpha)))

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.

1 for i in range(gridStorage.getSize()):
2  gp = gridStorage.getPoint(i)
3  alpha[i] = f(gp.getStandardCoordinate(0), gp.getStandardCoordinate(1))
5 print("alpha before hierarchization: {}".format(alpha))

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

1 pysgpp.createOperationHierarchisation(grid).doHierarchisation(alpha)
2 print("alpha after hierarchization: {}".format(alpha))

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.

1 p = pysgpp.DataVector(dim)
2 p[0] = 0.52
3 p[1] = 0.73
4 opEval = pysgpp.createOperationEval(grid)
5 print("u(0.52, 0.73) = {}".format(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.