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网站开发有哪些竞赛,云优化seo软件,专业的企业智能建站价格便宜,龙岩市永定区疫情最新消息背景：网上很多关于矩阵乘的编程优化思路，本着看理论分析万遍，不如实际代码写一遍的想法，大概过一下优化思路。矩阵乘的定义如下，约定矩阵的形状及存储方式为: A[M, K], B[K, N], C[M, N]。 CPU篇朴素实现方法按照…

背景：网上很多关于矩阵乘的编程优化思路，本着看理论分析万遍，不如实际代码写一遍的想法，大概过一下优化思路。

矩阵乘的定义如下，约定矩阵的形状及存储方式为: A[M, K], B[K, N], C[M, N]。

$C_{i,j}=\sum_{k=0}^{n}A_{ik}\times B_{kj}$

CPU篇

朴素实现方法

按照常规的思路，实现矩阵乘时如下的3层for循环。

#define OFFSET(row, col, ld) ((row) * (ld) + (col))
void cpuSgemm(float *a, float *b, float *c, const int M, const int N, const int K) 
{for (int m = 0; m < M; m++) {for (int n = 0; n < N; n++) {float psum = 0.0;for (int k = 0; k < K; k++) {psum += a[OFFSET(m, k, K)] * b[OFFSET(k, n, N)];}c[OFFSET(m, n, N)] = psum;}}
}

数据访存连续的优化

矩阵B的存储默认为N方向连续，所以可以将上面的第2，3层循环互换顺序，这样B的取数就不会跨行了，而是连续取数，达到访问连续的效果。

void cpuSgemm_1(float *a, float *b, float *c, const int M, const int N, const int K) 
{for (int m = 0; m < M; m++) {for (int k = 0; k < K; k++) {for (int n = 0; n < N; n++){c[OFFSET(m, n, N)] += a[OFFSET(m, k, K)] * b[OFFSET(k, n, N)];}           }}
}

数据重排/数据复用的优化

上面将M，N，K的for循环调整为M，K，N的循环顺序，导致我们K方向累加不能缓存了，增加了多次访问C矩阵的开销，所以我们不放先直接将B矩阵转置处理，然后再按照原始的M，N，K的for循环来处理。

void cpuSgemm_2(float *a, float *b, float *c, const int M, const int N, const int K) 
{float* b1=(float*) malloc(sizeof(float)*K*N);for(int i=0; i<K; i++){for (int j=0; j<N; j++){b1[OFFSET(j,i,K)]= b[OFFSET(i,j,N)];}}for (int m = 0; m < M; m++) {for (int n = 0; n < N; n++) {float psum = 0.0;for (int k = 0; k < K; k++) {psum += a[OFFSET(m, k, K)] * b1[OFFSET(n, k, K)];}c[OFFSET(m, n, N)] = psum;}}
}

性能表现

如下是测试CPU环境下这几种方法的时间情况，其中M=N=512, K =256。可以发现经过优化后的代码在时间上是逐步减少的。

CPU的优化思路还有其他的，比如循环展开，intrinsic函数，基于cache的矩阵切分等，注意本文并没有都实现出来。

cpuSgemm, Time measured: 416889 microseconds.
cpuSgemm_1, Time measured: 405259 microseconds.
cpuSgemm_2, Time measured: 238786 microseconds.

GPU篇

grid线程循环矩阵乘法

输出矩阵C有M*N个点，每个点是K个数的乘积和，所以可以定义每个线程计算K个点的乘积和，即grid线程循环矩阵乘法。

__global__ void matrix_multiply_gpu_0(float*a, float*b, float*c, int M, int N, int K)
{int tidx =threadIdx.x;int bidx = blockIdx.x;int idx = bidx * blockDim.x +tidx;int row = idx/N;int col = idx%N;if(row<M && col < N){float tmp =0.0;for(int k=0; k<K; k++){tmp+=a[row*K+k] * b[k*N+col];}c[row*N+col] = tmp;}
}

block线程循环矩阵乘法

grid内线程循环的矩阵乘法有如下缺憾：一个block内线程可能需要计算C矩阵不同行的矩阵元素，block内thread对相应的A矩阵访存不一致，导致无法广播和额外的访存开销，导致执行时间增加。

针对这个问题，可以做如下改进：每个block计算C矩阵的一行，block内的thread以固定跳步步长blockDim.x的方法循环计算C矩阵的一行，每一行启动一个block，共计M个block。

__global__ void matrix_multiply_gpu_1(float*a, float*b, float*c, int M, int N, int K)
{int tidx =threadIdx.x;int bidx = blockIdx.x;float tmp;for(;bidx<M; bidx += gridDim.x){for(;tidx<N; tidx+=blockDim.x ){tmp=0.0;for(int k=0; k<K; k++){tmp+=a[bidx*K +k] * b[k*N+tidx];}c[bidx*N+tidx] = tmp;}              }
}

行共享存储矩阵乘法

共享存储与L1 Cache同级，其访存延迟较全局存储小一个量级。用共享存储代替全局存储是GPU最重要的优化手段之一。采用共享存储优化的关键是数据复用，数据复用次数越多，共享存储优化可获得的收益也越高。

在block循环乘法中，1个block内所有thread都会用到A矩阵的一行，此时与B矩阵每一列相乘，A矩阵中该行复用了N次。故可以考虑将A矩阵的一行读入shared memory，运算时候从shared memory读取相应的数据。

注意代码中TILE_WIDTH>=K。

#define TILE_WIDTH 256
__global__ void matrix_multiply_gpu_2(float*a, float*b, float*c, int M, int N, const int K)
{__shared__ float data[TILE_WIDTH];int tid = threadIdx.x;int row = blockIdx.x;int i,j;for(i=tid; i<K; i+=blockDim.x){data[i]=a[row*K +i];}__syncthreads();float tmp;for(j=tid; j<N; j+=blockDim.x){tmp=0.0;for(int k=0; k<K; k++){tmp += data[k]*b[k*N+j];}c[row*N+j] = tmp;}
}

分块共享存储矩阵乘法

根据上面共享存储的理解，我们很自然的想到把B矩阵也考虑数据复用，所以可以同时把A，B矩阵都分成棋盘似的小尺寸的数据块，从全局内存读取到共享内存，这样可以有效降低数据访问时间，充分复用矩阵乘的局部数据。

#define TILE_SIZE 32
__global__ void matrix_multiply_gpu_3(float*a, float*b, float*c, int M, int N, const int K)
{__shared__ float matA[TILE_SIZE][TILE_SIZE];__shared__ float matB[TILE_SIZE][TILE_SIZE];int bx = blockIdx.x;int by = blockIdx.y;int tx = threadIdx.x;int ty = threadIdx.y;int Col = bx * TILE_SIZE + tx;int Row = by * TILE_SIZE + ty;float Pervalue = 0.0;for(int i = 0;i < K / TILE_SIZE;i++)  {matA[ty][tx] = a[Row * K + (i * TILE_SIZE + tx)];matB[ty][tx] = b[Col + (i * TILE_SIZE + ty) * N];__syncthreads();for(int k = 0;k < TILE_SIZE;k++) Pervalue += matA[ty][k] * matB[k][tx];__syncthreads();}c[Row * N + Col] = Pervalue;}

性能表现

利用nvprof工具，统计各个核函数的执行时间如下，可以发现每一步优化思路都能直观的带来的性能提升。

完整代码:

GitHub - Briwisdom/study_CUDA_examples: some demos for study CUDA program.

#include <iostream>
#include <chrono>using namespace std;#define OFFSET(row, col, ld) ((row) * (ld) + (col))void initDate(float *arr,int Len, bool randFlag=true)
{if (randFlag){for (int i = 0; i < Len; i++) {arr[i] = rand()/1000000;}}else{float value =0.0;for (int i = 0; i < Len; i++) {arr[i] = value;}}  
}void compare_result(float *x, float *y, int n, char *name)
{int cnt=0;for (int i=0; i<n; i++){if (x[i]!=y[i]){cnt++;printf("x= %f, y= %f\n", x[i],y[i]);}}printf("%s, ", name);if(cnt ==0)printf("result matched.\n");elseprintf("something error! result not match number = %d int total number: %d .\n", cnt, n);}void cpuSgemm(float *a, float *b, float *c, const int M, const int N, const int K) 
{for (int m = 0; m < M; m++) {for (int n = 0; n < N; n++) {float psum = 0.0;for (int k = 0; k < K; k++) {psum += a[OFFSET(m, k, K)] * b[OFFSET(k, n, N)];}c[OFFSET(m, n, N)] = psum;}}
}void cpuSgemm_1(float *a, float *b, float *c, const int M, const int N, const int K) 
{for (int m = 0; m < M; m++) {for (int k = 0; k < K; k++) {for (int n = 0; n < N; n++){c[OFFSET(m, n, N)] += a[OFFSET(m, k, K)] * b[OFFSET(k, n, N)];}           }}
}void cpuSgemm_2(float *a, float *b, float *c, const int M, const int N, const int K) 
{float* b1=(float*) malloc(sizeof(float)*K*N);for(int i=0; i<K; i++){for (int j=0; j<N; j++){b1[OFFSET(j,i,K)]= b[OFFSET(i,j,N)];}}for (int m = 0; m < M; m++) {for (int n = 0; n < N; n++) {float psum = 0.0;for (int k = 0; k < K; k++) {psum += a[OFFSET(m, k, K)] * b1[OFFSET(n, k, K)];}c[OFFSET(m, n, N)] = psum;}}
}void operation(void (*func)(float*,float*, float*, int, int, int), float *a, float *b, float *c, const int M, const int N, const int K, int repeat, char* name)
{auto begin0 = std::chrono::high_resolution_clock::now();for(int i=0; i<repeat; i++){(*func)(a,b,c, M, N, K);}auto end0 = std::chrono::high_resolution_clock::now();auto elapsed0 = std::chrono::duration_cast<std::chrono::microseconds>(end0 - begin0);printf("%s, Time measured: %d microseconds.\n", name, int(elapsed0.count()/repeat));
}__global__ void matrix_multiply_gpu_0(float*a, float*b, float*c, int M, int N, int K)
{int tidx =threadIdx.x;int bidx = blockIdx.x;int idx = bidx * blockDim.x +tidx;int row = idx/N;int col = idx%N;if(row<M && col < N){float tmp =0.0;for(int k=0; k<K; k++){tmp+=a[row*K+k] * b[k*N+col];}c[row*N+col] = tmp;}
}__global__ void matrix_multiply_gpu_1(float*a, float*b, float*c, int M, int N, int K)
{int tidx =threadIdx.x;int bidx = blockIdx.x;float tmp;for(;bidx<M; bidx += gridDim.x){for(;tidx<N; tidx+=blockDim.x ){tmp=0.0;for(int k=0; k<K; k++){tmp+=a[bidx*K +k] * b[k*N+tidx];}c[bidx*N+tidx] = tmp;}              }
}#define TILE_WIDTH 256
__global__ void matrix_multiply_gpu_2(float*a, float*b, float*c, int M, int N, const int K)
{__shared__ float data[TILE_WIDTH];int tid = threadIdx.x;int row = blockIdx.x;int i,j;for(i=tid; i<K; i+=blockDim.x){data[i]=a[row*K +i];}__syncthreads();float tmp;for(j=tid; j<N; j+=blockDim.x){tmp=0.0;for(int k=0; k<K; k++){tmp += data[k]*b[k*N+j];}c[row*N+j] = tmp;}
}#define TILE_SIZE 32
__global__ void matrix_multiply_gpu_3(float*a, float*b, float*c, int M, int N, const int K)
{__shared__ float matA[TILE_SIZE][TILE_SIZE];__shared__ float matB[TILE_SIZE][TILE_SIZE];int bx = blockIdx.x;int by = blockIdx.y;int tx = threadIdx.x;int ty = threadIdx.y;int Col = bx * TILE_SIZE + tx;int Row = by * TILE_SIZE + ty;float Pervalue = 0.0;for(int i = 0;i < K / TILE_SIZE;i++)  {matA[ty][tx] = a[Row * K + (i * TILE_SIZE + tx)];matB[ty][tx] = b[Col + (i * TILE_SIZE + ty) * N];__syncthreads();for(int k = 0;k < TILE_SIZE;k++) Pervalue += matA[ty][k] * matB[k][tx];__syncthreads();}c[Row * N + Col] = Pervalue;}int main()
{int M=512;int N=512;int K=256;float *a = (float*) malloc(M*K * sizeof(float));float *b = (float*) malloc(N*K * sizeof(float));float *c = (float*) malloc(M*N * sizeof(float));float *c1 = (float*) malloc(M*N * sizeof(float));float *c2 = (float*) malloc(M*N * sizeof(float));float *c_gpu_0 = (float*) malloc(M*N * sizeof(float));float *c_gpu_1 = (float*) malloc(M*N * sizeof(float));float *c_gpu_2 = (float*) malloc(M*N * sizeof(float));float *c_gpu_3 = (float*) malloc(M*N * sizeof(float));initDate(a,M*K);initDate(b,N*K);initDate(c, M*N, false);initDate(c1, M*N, false);initDate(c2, M*N, false);initDate(c_gpu_0, M*N, false);initDate(c_gpu_1, M*N, false);initDate(c_gpu_2, M*N, false);initDate(c_gpu_3, M*N, false);//ensure result is right.cpuSgemm(a,b,c,M,N,K);cpuSgemm_1(a,b,c1,M,N,K);cpuSgemm_2(a,b,c2,M,N,K); compare_result(c, c1, M*N,"sgemm1");compare_result(c, c2,  M*N,"sgemm2");//test the prerformance.int repeat =10;operation(cpuSgemm,a,b,c,M,N,K,repeat,"cpuSgemm");operation(cpuSgemm_1,a,b,c1,M,N,K,repeat,"cpuSgemm_1");operation(cpuSgemm_2,a,b,c2,M,N,K,repeat,"cpuSgemm_2");float* d_a, *d_b, *d_c0, *d_c1, *d_c2, *d_c3;cudaMalloc((void**) &d_a, sizeof(float)*(M*K));cudaMalloc((void**) &d_b, sizeof(float)*(N*K));cudaMalloc((void**) &d_c0, sizeof(float)*(M*N));cudaMalloc((void**) &d_c1, sizeof(float)*(M*N));cudaMalloc((void**) &d_c2, sizeof(float)*(M*N));cudaMalloc((void**) &d_c3, sizeof(float)*(M*N));cudaMemcpy(d_a, a, sizeof(float)*M*K, cudaMemcpyHostToDevice);cudaMemcpy(d_b, b, sizeof(float)*N*K, cudaMemcpyHostToDevice);int threadnum=64;int blocks =(M*N+threadnum-1)/threadnum;cudaMemcpy(d_c0, c_gpu_0, sizeof(float)*M*N, cudaMemcpyHostToDevice);matrix_multiply_gpu_0<<<blocks, threadnum>>>(d_a, d_b, d_c0, M, N, K);cudaMemcpy(c_gpu_0, d_c0, sizeof(float)*M*N, cudaMemcpyDeviceToHost);compare_result(c, c_gpu_0,  M*N,"gpu_0");cudaFree(d_c0);cudaMemcpy(d_c1, c_gpu_1, sizeof(float)*M*N, cudaMemcpyHostToDevice);matrix_multiply_gpu_1<<<M, threadnum>>>(d_a, d_b, d_c1, M, N, K);cudaMemcpy(c_gpu_1, d_c1, sizeof(float)*M*N, cudaMemcpyDeviceToHost);compare_result(c, c_gpu_1,  M*N,"gpu_1");cudaFree(d_c1);cudaMemcpy(d_c2, c_gpu_2, sizeof(float)*M*N, cudaMemcpyHostToDevice);matrix_multiply_gpu_2<<<M, threadnum>>>(d_a, d_b, d_c2, M, N, K);cudaMemcpy(c_gpu_2, d_c2, sizeof(float)*M*N, cudaMemcpyDeviceToHost);compare_result(c, c_gpu_2,  M*N,"gpu_2");cudaFree(d_c2);threadnum=32;dim3 gridSize(M / threadnum,N / threadnum);dim3 blockSize(threadnum,threadnum);cudaMemcpy(d_c3, c_gpu_3, sizeof(float)*M*N, cudaMemcpyHostToDevice);matrix_multiply_gpu_3<<<gridSize, blockSize>>>(d_a, d_b, d_c3, M, N, K);cudaMemcpy(c_gpu_3, d_c3, sizeof(float)*M*N, cudaMemcpyDeviceToHost);compare_result(c, c_gpu_3,  M*N,"gpu_3");cudaFree(d_c3);free(a);free(b);free(c);free(c1);free(c2);free(c_gpu_0);free(c_gpu_1);free(c_gpu_2);free(c_gpu_3);cudaFree(d_a);cudaFree(d_b);}