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cuda(2) 矩阵乘法优化过程

Created on 2013-8-5
URL : http://blog.sina.com.cn/s/blog_a502f1a30101mjch.html
@author: zhxfl
转载请说明出处

  1 #include <stdio.h>
  2 #include <time.h>
  3 #include <cuda_runtime.h>
  4 __global__ void matrixMulCUDA(int *A,int *B,int * C, 
  5     dim3 dimsA,dim3 dimsB, dim3 dimsC)
  6 {
  7     int i = blockIdx.x;
  8     int j = threadIdx.x;
  9 
 10     for(int k = 0; k < dimsA.y; k++)
 11     {
 12         C[i * dimsC.y + j] += A[i * dimsA.y + k] * B[k * dimsB.y + j];
 13         //printf("id = %d %d %d A = %d B = %d C = %d 
", i,j,k, A[i * dimsA.y + k],
 14         //     B[k * dimsB.y + j],
 15         //     C[i * dimsC.y + j]);
 16     }
 17 }
 18 
 19 int* matrixMultiplyByGpu(int *h_A, int n1,int m1,int *h_B,int n2,int m2)
 20 {
 21     int *d_A, *d_B, *d_C;
 22     int *h_C;
 23 
 24     dim3 dimsA(n1,m1);
 25     dim3 dimsB(n2,m2);
 26     dim3 dimsC(n1,m2);
 27 
 28     int mem_size_A = dimsA.x * dimsA.y * sizeof(int);
 29     int mem_size_B = dimsB.x * dimsB.y * sizeof(int);
 30     int mem_size_C = dimsC.x * dimsC.y * sizeof(int);
 31 
 32     cudaMalloc((void**)&d_A, mem_size_A);
 33     cudaMalloc((void**)&d_B, mem_size_B);
 34     cudaMalloc((void**)&d_C, mem_size_C);
 35 
 36     cudaMemcpy(d_A, h_A, mem_size_A, cudaMemcpyHostToDevice);
 37     cudaMemcpy(d_B, h_B, mem_size_B, cudaMemcpyHostToDevice);
 38 
 39     h_C = (int*)malloc(sizeof(int)*mem_size_C);
 40     for(int i = 0; i<dimsC.x * dimsC.y;i++)h_C[i] = 0;
 41     cudaMemcpy(d_C, h_C, mem_size_C, cudaMemcpyHostToDevice);
 42     dim3 grid(dimsC.x,dimsC.y);
 43     matrixMulCUDA<<<dimsC.x,dimsC.y>>>(d_A,d_B,d_C,dimsA,dimsB,dimsC);
 44     cudaMemcpy(h_C, d_C, mem_size_C, cudaMemcpyDeviceToHost);
 45     cudaFree(d_A);
 46     cudaFree(d_B);
 47     cudaFree(d_C);
 48     return h_C;
 49 }
 50 
 51 int* matrixMultiplyByCpu(int *h_A, int n1,int m1,int *h_B,int n2,int m2)
 52 {
 53     int *h_C = new int [n1 * m2];
 54     for(int i = 0; i < n1 * m2; i++)h_C[i] = 0;
 55 
 56     for(int i = 0; i < n1; i ++)
 57     {
 58         for(int j = 0; j < m2; j++)
 59         {
 60             for(int k = 0; k < m1; k++)
 61             {
 62                 //h_C[i][j] = h_A[i][k] * h_B[k][j];
 63                 h_C[i * m2 + j] += h_A[i * m1 + k] * h_B[k * m2 + j];
 64             }
 65         }
 66     }
 67     return h_C;
 68 }
 69 
 70 void outPutMatrix(char c,int *g, int n,int m)
 71 {
 72     return;
 73     printf("matrix %c [%3d %3d]
", c, n, m);
 74     for(int i = 0; i < n * m;i++)
 75     {
 76         printf("%5d ", g[i]);
 77         if((i + 1) % m == 0)printf("
");
 78     }
 79 }
 80 
 81 const int base = 1000;
 82 const int large = 2;
 83 int main()
 84 {
 85     int n1 = base;
 86     int m1 = base + 1;
 87     int n2 = m1;
 88     int m2 = base;
 89     int *g1 = new int[n1 * m1];
 90     int *g2 = new int[n2 * m2];
 91     for(int i = 0; i < n1 * m1;i++)g1[i] = rand() % large;
 92     for(int i = 0; i < n2 * m2;i++)g2[i] = rand() % large;
 93     outPutMatrix('A',g1,n1,m1);
 94     outPutMatrix('B',g2,n2,m2);
 95     int *gg1,*gg2;
 96     
 97     clock_t start, finish;
 98 
 99     start = clock();
100     gg1 = matrixMultiplyByGpu(g1,n1,m1,g2,n2,m2);
101     finish = clock();
102     printf("GPU time = %f
",(double)(finish - start) / CLOCKS_PER_SEC);
103     
104     start = clock();
105     gg2 = matrixMultiplyByCpu(g1,n1,m1,g2,n2,m2);
106     finish = clock();
107     printf("CPU time = %f
",(double)(finish - start) / CLOCKS_PER_SEC);
108 
109     printf("check---");
110     for(int i = 0; i< n1*m2;i++)
111     {
112         if(gg1[i] != gg2[i])
113         {
114             printf("wrong ans
");
115             break;
116         }
117     }
118     outPutMatrix('1',gg1,n1,m2);
119     outPutMatrix('2',gg2,n1,m2);
120 }

版本一

版本一分析：

n 约等于 maxThreadsPerBlock

这里我们的矩阵空间复杂度大概是o(n^2),两个这样矩阵的乘法复杂度大概是0(n^3)，这里使用GPU优化的方案是开启n个block，每个block有n个thread。这样我们的并发量就是n^2,也就是计算复杂度大概是0(n)。

版本一测试：

n 约等于 maxThreadsPerBlock

这里请注意，你的base + 1 < min(maxThreadsPerBlock,maxGridSize[0])，不然将超过cuda的最大计算量，会导致你的计算结果错误。

根据我的机子的情况 n = 1000，运行时间如下，可以看出计算时间大概是13.87倍

  1 #include <stdio.h>
  2 #include <time.h>
  3 #include <cuda_runtime.h>
  4 __global__ void matrixMulCUDA(float *A,float *B,float * C, 
  5     dim3 dimsA,dim3 dimsB, dim3 dimsC)
  6 {
  7     int i = blockIdx.x;
  8     int j = threadIdx.x;
  9 
 10     for(int k = 0; k < dimsA.y; k++)
 11     {
 12         C[i * dimsC.y + j] += A[i * dimsA.y + k] * B[k * dimsB.y + j];
 13         //printf("id = %d %d %d A = %d B = %d C = %d 
", i,j,k, A[i * dimsA.y + k],
 14         //     B[k * dimsB.y + j],
 15         //     C[i * dimsC.y + j]);
 16     }
 17 }
 18 
 19 float* matrixMultiplyByGpu(float *h_A, int n1,int m1,float *h_B,int n2,int m2)
 20 {
 21     float *d_A, *d_B, *d_C;
 22     float *h_C;
 23 
 24     dim3 dimsA(n1,m1);
 25     dim3 dimsB(n2,m2);
 26     dim3 dimsC(n1,m2);
 27 
 28     int mem_size_A = dimsA.x * dimsA.y * sizeof(float);
 29     int mem_size_B = dimsB.x * dimsB.y * sizeof(float);
 30     int mem_size_C = dimsC.x * dimsC.y * sizeof(float);
 31 
 32     cudaMalloc((void**)&d_A, mem_size_A);
 33     cudaMalloc((void**)&d_B, mem_size_B);
 34     cudaMalloc((void**)&d_C, mem_size_C);
 35 
 36     cudaMemcpy(d_A, h_A, mem_size_A, cudaMemcpyHostToDevice);
 37     cudaMemcpy(d_B, h_B, mem_size_B, cudaMemcpyHostToDevice);
 38 
 39     h_C = (float*)malloc(sizeof(float)*mem_size_C);
 40     for(int i = 0; i<dimsC.x * dimsC.y;i++)h_C[i] = 0;
 41     cudaMemcpy(d_C, h_C, mem_size_C, cudaMemcpyHostToDevice);
 42     dim3 grid(dimsC.x,dimsC.y);
 43     matrixMulCUDA<<<dimsC.x,dimsC.y>>>(d_A,d_B,d_C,dimsA,dimsB,dimsC);
 44     cudaMemcpy(h_C, d_C, mem_size_C, cudaMemcpyDeviceToHost);
 45     cudaFree(d_A);
 46     cudaFree(d_B);
 47     cudaFree(d_C);
 48     return h_C;
 49 }
 50 
 51 float* matrixMultiplyByCpu(float *h_A, int n1,int m1,float *h_B,int n2,int m2)
 52 {
 53     float *h_C = new float [n1 * m2];
 54     for(int i = 0; i < n1 * m2; i++)h_C[i] = 0;
 55 
 56     for(int i = 0; i < n1; i ++)
 57     {
 58         for(int j = 0; j < m2; j++)
 59         {
 60             for(int k = 0; k < m1; k++)
 61             {
 62                 //h_C[i][j] = h_A[i][k] * h_B[k][j];
 63                 h_C[i * m2 + j] += h_A[i * m1 + k] * h_B[k * m2 + j];
 64             }
 65         }
 66     }
 67     return h_C;
 68 }
 69 
 70 void outPutMatrix(char c,float *g, int n,int m)
 71 {
 72     return;
 73     printf("matrix %c [%3d %3d]
", c, n, m);
 74     for(int i = 0; i < n * m;i++)
 75     {
 76         printf("%5f ", g[i]);
 77         if((i + 1) % m == 0)printf("
");
 78     }
 79 }
 80 
 81 const int base = 1000;
 82 const int large = 2;
 83 int main()
 84 {
 85     int n1 = base;
 86     int m1 = base + 1;
 87     int n2 = m1;
 88     int m2 = base;
 89     float *g1 = new float[n1 * m1];
 90     float *g2 = new float[n2 * m2];
 91     for(int i = 0; i < n1 * m1;i++)g1[i] = rand() % large + 1.0f / 3.0f;
 92     for(int i = 0; i < n2 * m2;i++)g2[i] = rand() % large + 1.0f / 3.0f;
 93     outPutMatrix('A',g1,n1,m1);
 94     outPutMatrix('B',g2,n2,m2);
 95     float *gg1,*gg2;
 96     
 97     clock_t start, finish;
 98 
 99     start = clock();
100     gg1 = matrixMultiplyByGpu(g1,n1,m1,g2,n2,m2);
101     finish = clock();
102     printf("GPU time = %f
",(double)(finish - start) / CLOCKS_PER_SEC);
103     
104     start = clock();
105     gg2 = matrixMultiplyByCpu(g1,n1,m1,g2,n2,m2);
106     finish = clock();
107     printf("CPU time = %f
",(double)(finish - start) / CLOCKS_PER_SEC);
108 
109     printf("check---");
110     for(int i = 0; i< n1*m2;i++)
111     {
112         if(fabs(gg1[i] - gg2[i]) > 0.01)
113         {
114             printf("%f
 %f
wrong ans
",gg1[i],gg2[i]);
115             break;
116         }
117     }
118     outPutMatrix('1',gg1,n1,m2);
119     outPutMatrix('2',gg2,n1,m2);
120 }

版本二

版本一分析：

在版本一的基础上改成float运算

版本一测试：

结果如下，没有太大区别，本来预期是GPU的浮点计算能力会比CPU好很多的，但这里看来，并没有很明显的区别。

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原文地址：https://www.cnblogs.com/zhxfl/p/3238277.html