File: examples.rst

package info (click to toggle)
amgcl 1.4.4-1
  • links: PTS, VCS
  • area: contrib
  • in suites: sid
  • size: 5,676 kB
  • sloc: cpp: 34,043; python: 747; pascal: 258; f90: 196; makefile: 20
file content (240 lines) | stat: -rw-r--r-- 8,076 bytes parent folder | download | duplicates (3)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
Examples
========

Solving Poisson's equation
--------------------------

The easiest way to solve a problem with AMGCL is to use the
:cpp:class:`amgcl::make_solver` class. It has two template parameters: the
first one specifies a :doc:`preconditioner <components/preconditioners>` to
use, and the second chooses an :doc:`iterative solver
<components/iter_solvers>`.  The class constructor takes the system matrix in
one of supported :doc:`formats <components/adapters>` and parameters for the
chosen algorithms and for the :doc:`backend <components/backends>`.

Let us consider a simple example of `Poisson's equation`_ in a unit square.
Here is how the problem may be solved with AMGCL. We will use BiCGStab solver
preconditioned with smoothed aggregation multigrid with SPAI(0) for relaxation
(smoothing). First, we include the necessary headers. Each of those brings in
the corresponding component of the method:

.. _Poisson's equation: https://en.wikipedia.org/wiki/Poisson%27s_equation

.. code-block:: cpp

    #include <amgcl/make_solver.hpp>
    #include <amgcl/solver/bicgstab.hpp>
    #include <amgcl/amg.hpp>
    #include <amgcl/coarsening/smoothed_aggregation.hpp>
    #include <amgcl/relaxation/spai0.hpp>
    #include <amgcl/adapter/crs_tuple.hpp>

Next, we assemble sparse matrix for the Poisson's equation on a uniform
1000x1000 grid. See below for the definition of the :cpp:func:`poisson`
function:

.. code-block:: cpp

    std::vector<int>    ptr, col;
    std::vector<double> val, rhs;
    int n = poisson(1000, ptr, col, val, rhs);

For this example, we select the :cpp:class:`builtin <amgcl::backend::builtin>`
backend with double precision numbers as value type:

.. code-block:: cpp

    typedef amgcl::backend::builtin<double> Backend;

Now we can construct the solver for our system matrix. We use the convenient
adapter for :cpp:class:`std::tuple` here and just tie together the matrix size
and its CRS components:

.. code-block:: cpp

    typedef amgcl::make_solver<
        // Use AMG as preconditioner:
        amgcl::amg<
            Backend,
            amgcl::coarsening::smoothed_aggregation,
            amgcl::relaxation::spai0
            >,
        // And BiCGStab as iterative solver:
        amgcl::solver::bicgstab<Backend>
        > Solver;

    Solver solve( std::tie(n, ptr, col, val) );

Once the solver is constructed, we can apply it to the right-hand side to
obtain the solution. This may be repeated multiple times for different
right-hand sides. Here we start with a zero initial approximation. The solver
returns a boost tuple with number of iterations and norm of the achieved
residual:

.. code-block:: cpp

    std::vector<double> x(n, 0.0);
    int    iters;
    double error;
    std::tie(iters, error) = solve(rhs, x);

That's it! Vector ``x`` contains the solution of our problem now.

Input formats
-------------

We used STL vectors to store the matrix components in the above axample. This
may seem too restrictive if you want to use AMGCL with your own types.  But the
`crs_tuple` adapter will take anything that the Boost.Range_ library recognizes
as a random access range. For example, you can wrap raw pointers to your data
into a `boost::iterator_range`_:

.. _Boost.Range: http://www.boost.org/doc/libs/release/libs/range/
.. _`boost::iterator_range`: http://www.boost.org/doc/libs/release/libs/range/doc/html/range/reference/utilities/iterator_range.html

.. code-block:: cpp

    Solver solve( boost::make_tuple(
        n,
        boost::make_iterator_range(ptr.data(), ptr.data() + ptr.size()),
        boost::make_iterator_range(col.data(), col.data() + col.size()),
        boost::make_iterator_range(val.data(), val.data() + val.size())
        ) );

Same applies to the right-hand side and the solution vectors. And if that is
still not general enough, you can provide your own adapter for your matrix
type. See :doc:`components/adapters` for further information on this.

Setting parameters
------------------

Any component in AMGCL defines its own parameters by declaring a ``param``
subtype. When a class wraps several subclasses, it includes parameters of its
children into its own ``param``. For example, parameters for the
:cpp:class:`amgcl::make_solver\<Precond, Solver>` are declared as

.. code-block:: cpp

    struct params {
        typename Precond::params precond;
        typename Solver::params solver;
    };

Knowing that, we can easily set the parameters for individual components. For
example, we can set the desired tolerance for the iterative solver in the above
example like this:

.. code-block:: cpp

    Solver::params prm;
    prm.solver.tol = 1e-3;
    Solver solve( std::tie(n, ptr, col, val), prm );

Parameters may also be initialized with a `boost::property_tree::ptree`_. This
is especially convenient when the runtime interface is used, and the exact
structure of the parameters is not known at compile time:

.. code-block:: cpp

    boost::property_tree::ptree prm;
    prm.put("solver.tol", 1e-3);
    Solver solve( std::tie(n, ptr, col, val), prm );

.. _`boost::property_tree::ptree`: http://www.boost.org/doc/libs/release/doc/html/property_tree.html


Assembling matrix for Poisson's equation
----------------------------------------

The section provides an example of assembling the system matrix and the
right-hand side for a Poisson's equation in a unit square
:math:`\Omega=[0,1]\times[0,1]`:

.. math::

    -\Delta u = 1, \; u \in \Omega \quad u = 0, \; u \in \partial \Omega

The solution to the problem looks like this:

.. plot::

    from pylab import *
    from numpy import *
    h = linspace(-1, 1, 100)
    x,y = meshgrid(h, h)
    u = 0.5 * (1-x**2)
    for k in range(1,20,2):
        u -= 16/pi**3 * (sin(k*pi*(1+x)/2) / (k**3 * sinh(k * pi))
                * (sinh(k * pi * (1 + y) / 2) + sinh(k * pi * (1 - y)/2)))
    figure(figsize=(5,5))
    imshow(u, extent=(0,1,0,1))
    show()

Here is how the problem may be discretized on a uniform :math:`n \times n`
grid:

.. note: The CRS_ format [Saad03]_ is used for the discretized matrix.

.. _CRS: http://netlib.org/linalg/html_templates/node91.html

.. code-block:: cpp

    #include <vector>

    // Assembles matrix for Poisson's equation with homogeneous
    // boundary conditions on a n x n grid.
    // Returns number of rows in the assembled matrix.
    // The matrix is returned in the CRS components ptr, col, and val.
    // The right-hand side is returned in rhs.
    int poisson(
        int n,
        std::vector<int>    &ptr,
        std::vector<int>    &col,
        std::vector<double> &val,
        std::vector<double> &rhs
        )
    {
        int    n2 = n * n;        // Number of points in the grid.
        double h = 1.0 / (n - 1); // Grid spacing.

        ptr.clear(); ptr.reserve(n2 + 1); ptr.push_back(0);
        col.clear(); col.reserve(n2 * 5); // We use 5-point stencil, so the matrix
        val.clear(); val.reserve(n2 * 5); // will have at most n2 * 5 nonzero elements.

        rhs.resize(n2);

        for(int j = 0, k = 0; j < n; ++j) {
            for(int i = 0; i < n; ++i, ++k) {
                if (i == 0 || i == n - 1 || j == 0 || j == n - 1) {
                    // Boundary point. Use Dirichlet condition.
                    col.push_back(k);
                    val.push_back(1.0);

                    rhs[k] = 0.0;
                } else {
                    // Interior point. Use 5-point finite difference stencil.
                    col.push_back(k - n);
                    val.push_back(-1.0 / (h * h));

                    col.push_back(k - 1);
                    val.push_back(-1.0 / (h * h));

                    col.push_back(k);
                    val.push_back(4.0 / (h * h));

                    col.push_back(k + 1);
                    val.push_back(-1.0 / (h * h));

                    col.push_back(k + n);
                    val.push_back(-1.0 / (h * h));

                    rhs[k] = 1.0;
                }

                ptr.push_back(col.size());
            }
        }

        return n2;
    }