使用自定义内核或 CUBLAS 对矢量张量积进行 CUDA 优化
CUDA optimization for a vector tensor product using a custom kernel or CUBLAS
我有两个向量 a
和 b
。每个向量包含一个3d点的坐标x
, y
, z
vector3f
.
struct Vector3f
{
float x;
float y;
float z;
}
向量 a
的大小为 n = 5000
点,向量 b
的大小为 m = 4000
。我需要像图片右侧那样在它们之间做一个张量向量积。结果向量的长度应为 5000 * 4000
,并包含浮点数,结果存储在 c
。
__global__ void tensor3dProdcutClassic(const int n, const int m, const Vector3f *a, const Vector3f *b, float *c) {
int i = blockIdx.x * blockDim.x + threadIdx.x;
// int j = blockIdy.y * blockDim.y + threadIdx.y;
//check if the idx is out of range
if (i < n) {
for (int j = 0; j < m; j++) {
int idx = j + m * i;
c[idx] = a[i].x * b[j].x + a[i].y * b[j].y + a[i].z * b[j].z;
}
}
}
dim3 blockSize(32, 1, 1);
dim3 gridSize((n + blockSize.x - 1) / blockSize.x, 1, 1);
tensor3dProdcutClassic<<<gridSize, blockSize>>>(n, m, x, y, out);
我在 Volta arch 上的执行时间很长。
我的问题是如何优化内核以减少时间,这主要是因为内核中的 for 循环。我在这里知道所有全局读写都没有合并。
您可以让内核同时通过 a
和 b
,像这样:
__global__ void tensor3dProdcutClassic(const int n, const int m, const Vector3f *a, const Vector3f *b, float *c)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
int j = blockIdy.y * blockDim.y + threadIdx.y;
if (i < n && j < m)
{
int idx = j + m * i;
c[idx] = a[i].x * b[j].x + a[i].y * b[j].y + a[i].z * b[j].z;
}
}
dim3 blockSize(32, 32);
dim3 gridSize((int)ceil(n / 32.0), (int)ceil(m / 32.0));
tensor3dProdcutClassic<<<gridSize, blockSize>>>(n, m, x, y, out);
更新
我尝试修改代码以使用带和不带共享内存的单个数组,不带共享内存的代码总是快 3 或 4 倍。
共享内存:
#define BLOCK_SIZE 32
void tensor3dProdcut(const int n, const int m, const float* a, const float* b, float* c)
{
float* d_a;
size_t size = (uint64_t)n * 3 * sizeof(float);
cudaMalloc(&d_a, size);
cudaMemcpy(d_a, a, size, cudaMemcpyHostToDevice);
float* d_b;
size = (uint64_t)m * 3 * sizeof(float);
cudaMalloc(&d_b, size);
cudaMemcpy(d_b, b, size, cudaMemcpyHostToDevice);
float* d_c;
size = (uint64_t)n * m * sizeof(float);
cudaMalloc(&d_c, size);
dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE);
dim3 dimGrid((int)ceil((double)n / BLOCK_SIZE), (int)ceil((double)m / BLOCK_SIZE));
tensor3dProdcutKernel<<<dimGrid, dimBlock>>>(d_a, d_b, d_c, n, m);
cudaMemcpy(c, d_c, size, cudaMemcpyDeviceToHost);
cudaFree(d_a);
cudaFree(d_b);
cudaFree(d_c);
}
__global__ void tensor3dProdcutKernel(float* a, float* b, float* c, int n, int m)
{
int i, blockRow, blockCol, row, col;
float Cvalue;
blockRow = blockIdx.x;
blockCol = blockIdx.y;
row = threadIdx.x;
col = threadIdx.y;
if (blockRow * BLOCK_SIZE + row >= n || blockCol * BLOCK_SIZE + col >= m)
return;
__shared__ double as[BLOCK_SIZE][3];
__shared__ double bs[BLOCK_SIZE][3];
for (i = 0; i < 3; i++)
{
as[row][i] = a[(BLOCK_SIZE * blockRow + row) * 3 + i];
bs[col][i] = b[(BLOCK_SIZE * blockCol + col) * 3 + i];
}
__syncthreads();
Cvalue = 0;
for (i = 0; i < 3; i++)
Cvalue += as[row][i] * bs[col][i];
c[(BLOCK_SIZE * blockRow + row) * m + BLOCK_SIZE * blockCol + col] = Cvalue;
}
没有共享内存:
__global__ void tensor3dProdcutKernel(float* a, float* b, float* c, int n, int m)
{
int i, blockRow, blockCol, row, col;
float Cvalue;
blockRow = blockIdx.x;
blockCol = blockIdx.y;
row = threadIdx.x;
col = threadIdx.y;
if (blockRow * BLOCK_SIZE + row >= n || blockCol * BLOCK_SIZE + col >= m)
return;
Cvalue = 0;
for (i = 0; i < 3; i++)
Cvalue += a[(BLOCK_SIZE * blockRow + row) * 3 + i] * b[(BLOCK_SIZE * blockCol + col) * 3 + i];
c[(BLOCK_SIZE * blockRow + row) * m + BLOCK_SIZE * blockCol + col] = Cvalue;
}
我有两个向量 a
和 b
。每个向量包含一个3d点的坐标x
, y
, z
vector3f
.
struct Vector3f
{
float x;
float y;
float z;
}
向量 a
的大小为 n = 5000
点,向量 b
的大小为 m = 4000
。我需要像图片右侧那样在它们之间做一个张量向量积。结果向量的长度应为 5000 * 4000
,并包含浮点数,结果存储在 c
。
__global__ void tensor3dProdcutClassic(const int n, const int m, const Vector3f *a, const Vector3f *b, float *c) {
int i = blockIdx.x * blockDim.x + threadIdx.x;
// int j = blockIdy.y * blockDim.y + threadIdx.y;
//check if the idx is out of range
if (i < n) {
for (int j = 0; j < m; j++) {
int idx = j + m * i;
c[idx] = a[i].x * b[j].x + a[i].y * b[j].y + a[i].z * b[j].z;
}
}
}
dim3 blockSize(32, 1, 1);
dim3 gridSize((n + blockSize.x - 1) / blockSize.x, 1, 1);
tensor3dProdcutClassic<<<gridSize, blockSize>>>(n, m, x, y, out);
我在 Volta arch 上的执行时间很长。
我的问题是如何优化内核以减少时间,这主要是因为内核中的 for 循环。我在这里知道所有全局读写都没有合并。
您可以让内核同时通过 a
和 b
,像这样:
__global__ void tensor3dProdcutClassic(const int n, const int m, const Vector3f *a, const Vector3f *b, float *c)
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
int j = blockIdy.y * blockDim.y + threadIdx.y;
if (i < n && j < m)
{
int idx = j + m * i;
c[idx] = a[i].x * b[j].x + a[i].y * b[j].y + a[i].z * b[j].z;
}
}
dim3 blockSize(32, 32);
dim3 gridSize((int)ceil(n / 32.0), (int)ceil(m / 32.0));
tensor3dProdcutClassic<<<gridSize, blockSize>>>(n, m, x, y, out);
更新
我尝试修改代码以使用带和不带共享内存的单个数组,不带共享内存的代码总是快 3 或 4 倍。
共享内存:
#define BLOCK_SIZE 32
void tensor3dProdcut(const int n, const int m, const float* a, const float* b, float* c)
{
float* d_a;
size_t size = (uint64_t)n * 3 * sizeof(float);
cudaMalloc(&d_a, size);
cudaMemcpy(d_a, a, size, cudaMemcpyHostToDevice);
float* d_b;
size = (uint64_t)m * 3 * sizeof(float);
cudaMalloc(&d_b, size);
cudaMemcpy(d_b, b, size, cudaMemcpyHostToDevice);
float* d_c;
size = (uint64_t)n * m * sizeof(float);
cudaMalloc(&d_c, size);
dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE);
dim3 dimGrid((int)ceil((double)n / BLOCK_SIZE), (int)ceil((double)m / BLOCK_SIZE));
tensor3dProdcutKernel<<<dimGrid, dimBlock>>>(d_a, d_b, d_c, n, m);
cudaMemcpy(c, d_c, size, cudaMemcpyDeviceToHost);
cudaFree(d_a);
cudaFree(d_b);
cudaFree(d_c);
}
__global__ void tensor3dProdcutKernel(float* a, float* b, float* c, int n, int m)
{
int i, blockRow, blockCol, row, col;
float Cvalue;
blockRow = blockIdx.x;
blockCol = blockIdx.y;
row = threadIdx.x;
col = threadIdx.y;
if (blockRow * BLOCK_SIZE + row >= n || blockCol * BLOCK_SIZE + col >= m)
return;
__shared__ double as[BLOCK_SIZE][3];
__shared__ double bs[BLOCK_SIZE][3];
for (i = 0; i < 3; i++)
{
as[row][i] = a[(BLOCK_SIZE * blockRow + row) * 3 + i];
bs[col][i] = b[(BLOCK_SIZE * blockCol + col) * 3 + i];
}
__syncthreads();
Cvalue = 0;
for (i = 0; i < 3; i++)
Cvalue += as[row][i] * bs[col][i];
c[(BLOCK_SIZE * blockRow + row) * m + BLOCK_SIZE * blockCol + col] = Cvalue;
}
没有共享内存:
__global__ void tensor3dProdcutKernel(float* a, float* b, float* c, int n, int m)
{
int i, blockRow, blockCol, row, col;
float Cvalue;
blockRow = blockIdx.x;
blockCol = blockIdx.y;
row = threadIdx.x;
col = threadIdx.y;
if (blockRow * BLOCK_SIZE + row >= n || blockCol * BLOCK_SIZE + col >= m)
return;
Cvalue = 0;
for (i = 0; i < 3; i++)
Cvalue += a[(BLOCK_SIZE * blockRow + row) * 3 + i] * b[(BLOCK_SIZE * blockCol + col) * 3 + i];
c[(BLOCK_SIZE * blockRow + row) * m + BLOCK_SIZE * blockCol + col] = Cvalue;
}