The 3-point stencil example.
This example depends on simple-solver, poisson-solver.
Introduction
This example solves a 1D Poisson equation:
![$ u : [0, 1] \rightarrow R\\ u'' = f\\ u(0) = u0\\ u(1) = u1 $](form_15.png)
using a finite difference method on an equidistant grid with K
discretization points (K
can be controlled with a command line parameter). The discretization is done via the second order Taylor polynomial:

For an equidistant grid with K "inner" discretization points
and step size
, the formula produces a system of linear equations

which is then solved using Ginkgo's implementation of the CG method preconditioned with block-Jacobi. It is also possible to specify on which executor Ginkgo will solve the system via the command line. The function
is set to
(making the solution
), but that can be changed in the main
function.
The intention of the example is to show how Ginkgo can be integrated into existing software - the generate_stencil_matrix
, generate_rhs
, print_solution
, compute_error
and main
function do not reference Ginkgo at all (i.e. they could have been there before the application developer decided to use Ginkgo, and the only part where Ginkgo is introduced is inside the solve_system
function.
About the example
The commented program
/ *****************************<DECSRIPTION>***********************************
This example solves a 1D Poisson equation:
u : [0, 1] -> R
u'' = f
u(0) = u0
u(1) = u1
using a finite difference method on an equidistant grid with `K` discretization
points (`K` can be controlled with a command line parameter). The discretization
is done via the second order Taylor polynomial:
u(x + h) = u(x) - u'(x)h + 1/2 u''(x)h^2 + O(h^3)
u(x - h) = u(x) + u'(x)h + 1/2 u''(x)h^2 + O(h^3) / +
---------------------------------------------
-u(x - h) + 2u(x) + -u(x + h) = -f(x)h^2 + O(h^3)
For an equidistant grid with K "inner" discretization points x1, ..., xk, and
step size h = 1 / (K + 1), the formula produces a system of linear equations
2u_1 - u_2 = -f_1 h^2 + u0
-u_(k-1) + 2u_k - u_(k+1) = -f_k h^2, k = 2, ..., K - 1
-u_(K-1) + 2u_K = -f_K h^2 + u1
which is then solved using Ginkgo's implementation of the CG method
preconditioned with block-Jacobi. It is also possible to specify on which
executor Ginkgo will solve the system via the command line.
The function `f` is set to `f(x) = 6x` (making the solution `u(x) = x^3`), but
that can be changed in the `main` function.
The intention of the example is to show how Ginkgo can be integrated into
existing software - the `generate_stencil_matrix`, `generate_rhs`,
`print_solution`, `compute_error` and `main` function do not reference Ginkgo at
all (i.e. they could have been there before the application developer decided to
use Ginkgo, and the only part where Ginkgo is introduced is inside the
`solve_system` function.
*****************************<DECSRIPTION>********************************** /
#include <ginkgo/ginkgo.hpp>
#include <iostream>
#include <map>
#include <string>
#include <vector>
Creates a stencil matrix in CSR format for the given number of discretization points.
void generate_stencil_matrix(int discretization_points, int *row_ptrs,
int *col_idxs, double *values)
{
int pos = 0;
const double coefs[] = {-1, 2, -1};
row_ptrs[0] = pos;
for (int i = 0; i < discretization_points; ++i) {
for (auto ofs : {-1, 0, 1}) {
if (0 <= i + ofs && i + ofs < discretization_points) {
values[pos] = coefs[ofs + 1];
col_idxs[pos] = i + ofs;
++pos;
}
}
row_ptrs[i + 1] = pos;
}
}
Generates the RHS vector given f
and the boundary conditions.
template <typename Closure>
void generate_rhs(int discretization_points, Closure f, double u0, double u1,
double *rhs)
{
const auto h = 1.0 / (discretization_points + 1);
for (int i = 0; i < discretization_points; ++i) {
const auto xi = (i + 1) * h;
rhs[i] = -f(xi) * h * h;
}
rhs[0] += u0;
rhs[discretization_points - 1] += u1;
}
Prints the solution u
.
void print_solution(int discretization_points, double u0, double u1,
const double *u)
{
std::cout << u0 << '\n';
for (int i = 0; i < discretization_points; ++i) {
std::cout << u[i] << '\n';
}
std::cout << u1 << std::endl;
}
Computes the 1-norm of the error given the computed u
and the correct solution function correct_u
.
template <typename Closure>
double calculate_error(int discretization_points, const double *u,
Closure correct_u)
{
const auto h = 1.0 / (discretization_points + 1);
auto error = 0.0;
for (int i = 0; i < discretization_points; ++i) {
using std::abs;
const auto xi = (i + 1) * h;
error +=
abs(u[i] - correct_u(xi)) /
abs(correct_u(xi));
}
return error;
}
void solve_system(const std::string &executor_string,
unsigned int discretization_points, int *row_ptrs,
int *col_idxs, double *values, double *rhs, double *u,
double accuracy)
{
Some shortcuts
using vec = gko::matrix::Dense<double>;
using mtx = gko::matrix::Csr<double, int>;
using cg = gko::solver::Cg<double>;
const auto &dp = discretization_points;
Figure out where to run the code
std::map<std::string, std::shared_ptr<gko::Executor>> exec_map{
{"omp", omp},
{"reference", gko::ReferenceExecutor::create()}};
executor where Ginkgo will perform the computation
const auto exec = exec_map.at(executor_string);
executor where the application initialized the data
const auto app_exec = exec_map["omp"];
Tell Ginkgo to use the data in our application
Matrix: we have to set the executor of the matrix to the one where we want SpMVs to run (in this case exec
). When creating array views, we have to specify the executor where the data is (in this case app_exec
).
If the two do not match, Ginkgo will automatically create a copy of the data on exec
(however, it will not copy the data back once it is done
- here this is not important since we are not modifying the matrix).
val_array::view(app_exec, 3 * dp - 2, values),
idx_array::view(app_exec, 3 * dp - 2, col_idxs),
idx_array::view(app_exec, dp + 1, row_ptrs));
RHS: similar to matrix
val_array::view(app_exec, dp, rhs), 1);
Solution: we have to be careful here - if the executors are different, once we compute the solution the array will not be automatically copied back to the original memory locations. Fortunately, whenever apply
is called on a linear operator (e.g. matrix, solver) the arguments automatically get copied to the executor where the operator is, and copied back once the operation is completed. Thus, in this case, we can just define the solution on app_exec
, and it will be automatically transferred to/from exec
if needed.
val_array::view(app_exec, dp, u), 1);
Generate solver
auto solver_gen =
cg::build()
.with_criteria(
gko::stop::Iteration::build().with_max_iters(dp).on(exec),
.with_reduction_factor(accuracy)
.on(exec))
.with_preconditioner(bj::build().on(exec))
.on(exec);
auto solver = solver_gen->generate(
gko::give(matrix));
Solve system
}
int main(int argc, char *argv[])
{
if (argc < 2) {
std::cerr << "Usage: " << argv[0] << " DISCRETIZATION_POINTS [executor]"
<< std::endl;
std::exit(-1);
}
const int discretization_points = argc >= 2 ? std::atoi(argv[1]) : 100;
const auto executor_string = argc >= 3 ? argv[2] : "reference";
problem:
auto correct_u = [](double x) { return x * x * x; };
auto f = [](double x) { return 6 * x; };
auto u0 = correct_u(0);
auto u1 = correct_u(1);
matrix
std::vector<int> row_ptrs(discretization_points + 1);
std::vector<int> col_idxs(3 * discretization_points - 2);
std::vector<double> values(3 * discretization_points - 2);
right hand side
std::vector<double> rhs(discretization_points);
solution
std::vector<double> u(discretization_points, 0.0);
generate_stencil_matrix(discretization_points, row_ptrs.data(),
col_idxs.data(), values.data());
looking for solution u = x^3: f = 6x, u(0) = 0, u(1) = 1
generate_rhs(discretization_points, f, u0, u1, rhs.data());
solve_system(executor_string, discretization_points, row_ptrs.data(),
col_idxs.data(), values.data(), rhs.data(), u.data(), 1e-12);
print_solution(discretization_points, 0, 1, u.data());
std::cout << "The average relative error is "
<< calculate_error(discretization_points, u.data(), correct_u) /
discretization_points
<< std::endl;
}
Results
This is the expected output:
0
0.00010798
0.000863838
0.00291545
0.0069107
0.0134975
0.0233236
0.037037
0.0552856
0.0787172
0.10798
0.143721
0.186589
0.237231
0.296296
0.364431
0.442285
0.530504
0.629738
0.740633
0.863838
1
The average relative error is 1.87318e-15
Comments about programming and debugging
The plain program
#include <ginkgo/ginkgo.hpp>
#include <iostream>
#include <map>
#include <string>
#include <vector>
void generate_stencil_matrix(int discretization_points, int *row_ptrs,
int *col_idxs, double *values)
{
int pos = 0;
const double coefs[] = {-1, 2, -1};
row_ptrs[0] = pos;
for (int i = 0; i < discretization_points; ++i) {
for (auto ofs : {-1, 0, 1}) {
if (0 <= i + ofs && i + ofs < discretization_points) {
values[pos] = coefs[ofs + 1];
col_idxs[pos] = i + ofs;
++pos;
}
}
row_ptrs[i + 1] = pos;
}
}
template <typename Closure>
void generate_rhs(int discretization_points, Closure f, double u0, double u1,
double *rhs)
{
const auto h = 1.0 / (discretization_points + 1);
for (int i = 0; i < discretization_points; ++i) {
const auto xi = (i + 1) * h;
rhs[i] = -f(xi) * h * h;
}
rhs[0] += u0;
rhs[discretization_points - 1] += u1;
}
void print_solution(int discretization_points, double u0, double u1,
const double *u)
{
std::cout << u0 << '\n';
for (int i = 0; i < discretization_points; ++i) {
std::cout << u[i] << '\n';
}
std::cout << u1 << std::endl;
}
template <typename Closure>
double calculate_error(int discretization_points, const double *u,
Closure correct_u)
{
const auto h = 1.0 / (discretization_points + 1);
auto error = 0.0;
for (int i = 0; i < discretization_points; ++i) {
using std::abs;
const auto xi = (i + 1) * h;
error +=
abs(u[i] - correct_u(xi)) /
abs(correct_u(xi));
}
return error;
}
void solve_system(const std::string &executor_string,
unsigned int discretization_points, int *row_ptrs,
int *col_idxs, double *values, double *rhs, double *u,
double accuracy)
{
using vec = gko::matrix::Dense<double>;
using mtx = gko::matrix::Csr<double, int>;
using cg = gko::solver::Cg<double>;
const auto &dp = discretization_points;
std::map<std::string, std::shared_ptr<gko::Executor>> exec_map{
{"omp", omp},
{"reference", gko::ReferenceExecutor::create()}};
const auto exec = exec_map.at(executor_string);
const auto app_exec = exec_map["omp"];
val_array::view(app_exec, 3 * dp - 2, values),
idx_array::view(app_exec, 3 * dp - 2, col_idxs),
idx_array::view(app_exec, dp + 1, row_ptrs));
val_array::view(app_exec, dp, rhs), 1);
val_array::view(app_exec, dp, u), 1);
auto solver_gen =
cg::build()
.with_criteria(
gko::stop::Iteration::build().with_max_iters(dp).on(exec),
.with_reduction_factor(accuracy)
.on(exec))
.with_preconditioner(bj::build().on(exec))
.on(exec);
auto solver = solver_gen->generate(
gko::give(matrix));
}
int main(int argc, char *argv[])
{
if (argc < 2) {
std::cerr << "Usage: " << argv[0] << " DISCRETIZATION_POINTS [executor]"
<< std::endl;
std::exit(-1);
}
const int discretization_points = argc >= 2 ? std::atoi(argv[1]) : 100;
const auto executor_string = argc >= 3 ? argv[2] : "reference";
auto correct_u = [](double x) { return x * x * x; };
auto f = [](double x) { return 6 * x; };
auto u0 = correct_u(0);
auto u1 = correct_u(1);
std::vector<int> row_ptrs(discretization_points + 1);
std::vector<int> col_idxs(3 * discretization_points - 2);
std::vector<double> values(3 * discretization_points - 2);
std::vector<double> rhs(discretization_points);
std::vector<double> u(discretization_points, 0.0);
generate_stencil_matrix(discretization_points, row_ptrs.data(),
col_idxs.data(), values.data());
generate_rhs(discretization_points, f, u0, u1, rhs.data());
solve_system(executor_string, discretization_points, row_ptrs.data(),
col_idxs.data(), values.data(), rhs.data(), u.data(), 1e-12);
print_solution(discretization_points, 0, 1, u.data());
std::cout << "The average relative error is "
<< calculate_error(discretization_points, u.data(), correct_u) /
discretization_points
<< std::endl;
}