在 cusp 中,有一个乘法来计算 spmv(稀疏矩阵向量乘法),它需要一个 reduce 和一个 combine:
template <typename LinearOperator,
typename MatrixOrVector1,
typename MatrixOrVector2,
typename UnaryFunction,
typename BinaryFunction1,
typename BinaryFunction2>
void multiply(const LinearOperator& A,
const MatrixOrVector1& B,
MatrixOrVector2& C,
UnaryFunction initialize,
BinaryFunction1 combine,
BinaryFunction2 reduce);
从界面看来,对于任何矩阵/向量乘法,自定义合并和归约都应该是可能的。我认为 cusp 支持使用在推力/功能.h 中定义的其他组合和减少函数,除了乘法和加法来计算 spmv。例如,我可以使用thrust::plus 来代替原来的combine 函数(即乘法)吗?而且我猜,这个缩放的 spmv 也支持 coo、csr、dia、hyb 格式的稀疏矩阵。
但是,当我在矩阵 A 为 coo 格式的 a.cu 中测试以下示例时,我得到了错误的答案。它使用加号运算符进行组合。我用 cmd 编译它:nvcc a.cu -o a
to 。
#include <cusp/csr_matrix.h>
#include <cusp/monitor.h>
#include <cusp/multiply.h>
#include <cusp/print.h>
#include <cusp/krylov/cg.h>
int main(void)
{
// COO format in host memory
int host_I[13] = {0,0,1,1,2,2,2,3,3,3,4,5,5}; // COO row indices
int host_J[13] = {0,1,1,2,2,4,6,3,4,5,5,5,6}; // COO column indices
int host_V[13] = {1,1,1,1,1,1,1,1,1,1,1,1,1};
// x and y arrays in host memory
int host_x[7] = {1,1,1,1,1,1,1};
int host_y[6] = {0,0,0,0,0,0};
// allocate device memory for COO format
int * device_I;
cudaMalloc(&device_I, 13 * sizeof(int));
int * device_J;
cudaMalloc(&device_J, 13 * sizeof(int));
int * device_V;
cudaMalloc(&device_V, 13 * sizeof(int));
// allocate device memory for x and y arrays
int * device_x;
cudaMalloc(&device_x, 7 * sizeof(int));
int * device_y;
cudaMalloc(&device_y, 6 * sizeof(int));
// copy raw data from host to device
cudaMemcpy(device_I, host_I, 13 * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(device_J, host_J, 13 * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(device_V, host_V, 13 * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(device_x, host_x, 7 * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(device_y, host_y, 6 * sizeof(int), cudaMemcpyHostToDevice);
// matrices and vectors now reside on the device
// *NOTE* raw pointers must be wrapped with thrust::device_ptr!
thrust::device_ptr<int> wrapped_device_I(device_I);
thrust::device_ptr<int> wrapped_device_J(device_J);
thrust::device_ptr<int> wrapped_device_V(device_V);
thrust::device_ptr<int> wrapped_device_x(device_x);
thrust::device_ptr<int> wrapped_device_y(device_y);
// use array1d_view to wrap the individual arrays
typedef typename cusp::array1d_view< thrust::device_ptr<int> > DeviceIndexArrayView;
typedef typename cusp::array1d_view< thrust::device_ptr<int> > DeviceValueArrayView;
DeviceIndexArrayView row_indices (wrapped_device_I, wrapped_device_I + 13);
DeviceIndexArrayView column_indices(wrapped_device_J, wrapped_device_J + 13);
DeviceValueArrayView values (wrapped_device_V, wrapped_device_V + 13);
DeviceValueArrayView x (wrapped_device_x, wrapped_device_x + 7);
DeviceValueArrayView y (wrapped_device_y, wrapped_device_y + 6);
// combine the three array1d_views into a coo_matrix_view
typedef cusp::coo_matrix_view<DeviceIndexArrayView,
DeviceIndexArrayView,
DeviceValueArrayView> DeviceView;
// construct a coo_matrix_view from the array1d_views
DeviceView A(6, 7, 13, row_indices, column_indices, values);
std::cout << "\ndevice coo_matrix_view" << std::endl;
cusp::print(A);
cusp::constant_functor<int> initialize;
thrust::plus<int> combine;
thrust::plus<int> reduce;
cusp::multiply(A , x , y , initialize, combine, reduce);
std::cout << "\nx array" << std::endl;
cusp::print(x);
std::cout << "\n y array, y = A * x" << std::endl;
cusp::print(y);
cudaMemcpy(host_y, device_y, 6 * sizeof(int), cudaMemcpyDeviceToHost);
// free device arrays
cudaFree(device_I);
cudaFree(device_J);
cudaFree(device_V);
cudaFree(device_x);
cudaFree(device_y);
return 0;
}
我得到了以下答案。
device coo_matrix_view
sparse matrix <6, 7> with 13 entries
0 0 (1)
0 1 (1)
1 1 (1)
1 2 (1)
2 2 (1)
2 4 (1)
2 6 (1)
3 3 (1)
3 4 (1)
3 5 (1)
4 5 (1)
5 5 (1)
5 6 (1)
x array
array1d <7>
(1)
(1)
(1)
(1)
(1)
(1)
(1)
y array, y = A * x
array1d <6>
(4)
(4)
(6)
(6)
(2)
(631)
我得到的向量 y 很奇怪,我认为正确的答案 y 应该是:
[9,
9,
10,
10,
8,
9]
所以我不确定combine和reduce的这种替换是否可以适应其他稀疏矩阵格式,比如coo。或者我上面写的代码调用乘法可能是不正确的。你能给我一些帮助吗?任何信息都会有所帮助。
谢谢!