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稀疏矩阵 part 4

时间:2019-02-24 22:57:18      阅读:171      评论:0      收藏:0      [点我收藏+]

? 各种稀疏矩阵数据结构下 y(n,1) = A(n,m) * x(m,1) 的实现,GPU版本

● MAT 乘法

 1 __global__ void dotGPU(const MAT *a, const MAT *x, MAT *y)
 2 {
 3     int id = blockIdx.x * blockDim.x + threadIdx.x;
 4     if (id < a->row)
 5     {
 6         format sum = 0;
 7         for (int i = 0; i < a->col; i++)
 8             sum += a->data[id * a->col + i] * x->data[i];
 9         y->data[id] = sum;
10     }
11     if (id == 0)
12     {
13         y->row = a->row;
14         y->col = x->col;
15         COUNT_MAT(y);
16     }
17     return;
18 }

 

● CSR 乘法

 1 __global__ void dotGPU(const CSR *a, const MAT *x, MAT *y)
 2 {
 3     int id = blockIdx.x * blockDim.x + threadIdx.x;
 4     if (id < a->row)
 5     {
 6         format sum = 0;
 7         for (int j = a->ptr[id]; j < a->ptr[id + 1]; j++)
 8             sum += a->data[j] * x->data[a->index[j]];
 9         y->data[id] = sum;
10     }
11     if (id == 0)
12     {
13         y->row = a->row;
14         y->col = x->col;
15         COUNT_MAT(y);
16     }
17     return;
18 }

 

● ELL 乘法

 1 __global__ void dotGPU(const ELL *a, const MAT *x, MAT *y)
 2 {
 3     int id = blockIdx.x * blockDim.x + threadIdx.x;
 4     if (id < a->col)
 5     {
 6         format sum = 0;
 7         for (int j = 0; j < a->row; j++)            
 8             sum += a->data[id + j * a->col] * (a->index[id + j * a->col] < 0 ? 0 : x->data[a->index[id + j * a->col]]);
 9         y->data[id] = sum;
10     }
11     if (id == 0)
12     {
13         y->row = a->col;
14         y->col = x->col;
15         COUNT_MAT(y);
16     }
17     return;
18 }

 

● COO 乘法

 1 __global__ void dotGPU(const ELL *a, const MAT *x, MAT *y)// GPU ELL乘法
 2 {
 3     int id = blockIdx.x * blockDim.x + threadIdx.x;
 4     if (id < a->col)
 5     {
 6         format sum = 0;
 7         for (int j = 0; j < a->row; j++)            
 8             sum += a->data[id + j * a->col] * (a->index[id + j * a->col] < 0 ? 0 : x->data[a->index[id + j * a->col]]);
 9         y->data[id] = sum;
10     }
11     if (id == 0)
12     {
13         y->row = a->col;
14         y->col = x->col;
15         COUNT_MAT(y);
16     }
17     return;
18 }

 

● DIA 乘法,留坑

 

稀疏矩阵 part 4

原文:https://www.cnblogs.com/cuancuancuanhao/p/10428510.html

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