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  • 稀疏矩阵 part 3

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

    ● MAT 乘法

     1 int dotCPU(const MAT *a, const MAT *x, MAT *y)
     2 {
     3     checkNULL(a); checkNULL(x); checkNULL(y);
     4     if (a->col != x->row)
     5     {
     6         printf("dotMATCPU dimension mismatch!
    ");
     7         return 1;
     8     }
     9     
    10     y->row = a->row;
    11     y->col = x->col;
    12     for (int i = 0; i < a->row; i++)
    13     {
    14         format sum = 0;
    15         for (int j = 0; j < a->col; j++)        
    16             sum += a->data[i * a->col + j] * x->data[j];                
    17         y->data[i] = sum;        
    18     }
    19     COUNT_MAT(y);
    20     return 0;
    21 }

    ● CSR 乘法

     1 int dotCPU(const CSR *a, const MAT *x, MAT *y)
     2 {
     3     checkNULL(a); checkNULL(x); checkNULL(y);
     4     if (a->col != x->row)
     5     {
     6         printf("dotCSRCPU dimension mismatch!
    ");
     7         return 1;
     8     }
     9     
    10     y->row = a->row;
    11     y->col = x->col;
    12     for (int i = 0; i < a->row; i++)                            // i 遍历 ptr,j 遍历行内数据,A 中为 0 的元素不参加乘法
    13     {
    14         format sum = 0;
    15         for (int j = a->ptr[i]; j < a->ptr[i + 1]; j++)
    16             sum += a->data[j] * x->data[a->index[j]];
    17         y->data[i] = sum;
    18     }
    19     COUNT_MAT(y);
    20     return 0;
    21 }

    ● ELL 乘法

     1 int dotCPU(const ELL *a, const MAT *x, MAT *y)      // CPU ELL乘法
     2 {
     3     checkNULL(a); checkNULL(x); checkNULL(y);
     4     if (a->colOrigin != x->row)
     5     {
     6         printf("dotELLCPU dimension mismatch!
    ");
     7         return 1;
     8     }
     9 
    10     y->row = a->col;
    11     y->col = x->col;
    12     for (int i = 0; i<a->col; i++)
    13     {
    14         format sum = 0;
    15         for (int j = 0; j < a->row; j++)
    16         {
    17             int temp = a->index[j * a->col + i];
    18             if (temp < 0)                                   // 跳过无效元素
    19                 continue;
    20             sum += a->data[j * a->col + i] * x->data[temp];
    21         }
    22         y->data[i] = sum;
    23     }
    24     COUNT_MAT(y);
    25     return 0;
    26 }

    ● COO 乘法

     1 int dotCPU(const COO *a, const MAT *x, MAT *y)
     2 {
     3     checkNULL(a); checkNULL(x); checkNULL(y);
     4     if (a->col != x->row)
     5     {
     6         printf("dotCOOCPU null!
    ");
     7         return 1;
     8     }
     9 
    10     y->row = a->row;
    11     y->col = x->col;
    12     for (int i = 0; i<a->count; i++)
    13         y->data[a->rowIndex[i]] += a->data[i] * x->data[a->colIndex[i]];
    14     COUNT_MAT(y);
    15     return 0;
    16 }

    ● DIA 乘法

     1 int dotCPU(const DIA *a, const MAT *x, MAT *y)
     2 {
     3     checkNULL(a); checkNULL(x); checkNULL(y);
     4     if (a->colOrigin != x->row)
     5     {
     6         printf("dotDIACPU null!
    ");
     7         return 1;
     8     }    
     9     y->row = a->row;
    10     y->col = x->col;
    11     int * inverseIndex = (int *)malloc(sizeof(int) * a->col);
    12     for (int i = 0, j = 0; i < a->row + a->col - 1; i++)
    13     {
    14         if (a->index[i] == 1)
    15         {
    16             inverseIndex[j] = i;
    17             j++;
    18         }
    19     }
    20     for (int i = 0; i < a->row; i++)
    21     {
    22         format sum = 0;
    23         for (int j = 0; j < a->col; j++)
    24         {
    25             if (i < a->row - 1 - inverseIndex[j] || i > inverseIndex[a->col - 1] - inverseIndex[j])
    26                 continue;
    27             sum += a->data[i * a->col + j] * x->data[i + inverseIndex[j] - a->row + 1];
    28         }
    29         y->data[i] = sum;
    30     }
    31     COUNT_MAT(y);
    32     free(inverseIndex);
    33     return 0;
    34 }
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  • 原文地址:https://www.cnblogs.com/cuancuancuanhao/p/10428493.html
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