CUDA编程(九)
矩阵乘法
在之前我们一直围绕着一个非常简单的求立方和的小程序学习CUDA,从编写到优化,学习了很多,包括CUDA GPU的架构,如何评估程序,并行优化,内存优化,等等,把程序的运行时间从679680304个时钟周期(对于我的显卡是0.853S)最终优化到了133133个时钟周期(对于我的显卡是1.67e-4S),优化的效果还是非常明显的,前后总共加速了5015倍。
不过这个立方和的小程序实际上没有什么实用价值,之前也提到过了,CUDA广泛用于神经网络,计算机视觉这些领域,因为这些领域的算法往往可并行性极强,运算量大,非常适合使用GPU计算,说白了就是有大量的浮点数矩阵计算。
所以接下来我们就想办法用CUDA去并行一个常用的矩阵运算,矩阵加法没什么好说的,所以我们接下来去并行一下矩阵乘法~
矩阵乘法
为了简单起见,我们以方阵为例,矩阵的乘法大家应该都是比较熟悉的,比如两个方阵A,B
C = AB
for(i = 0; i < n; i++) { for(j = 0; j < n; j++) { C[i][j] = 0; for(k = 0; k < n; k++) { C[i][j] += A[i][k] * B[k][j]; } }}
计算的思路还是非常简单清晰的,那么我们如何把这个过程并行呢?
并行矩阵乘法
我们先试着写一下最简单的并行方式,之后再慢慢优化~
现在我们先考虑最核心的核函数,仿照不并行的程序,首先我们需要有AB和C三个浮点数矩阵,还要知道它们的大小,之后还需要计算时间,所以我们核函数的参数就非常明确了:
__global__ static void matMultCUDA(const float* a, const float* b, float* c, int n, clock_t* time){}
我们之前也说了,程序不可能一蹴而就,所以先用最简单的形式写好核函数:
__global__ static void matMultCUDA(const float* a, const float* b, float* c, int n, clock_t* time){ const int tid = threadIdx.x; const int bid = blockIdx.x; const int idx = bid * THREAD_NUM + tid; const int row = idx / n; const int column = idx % n; int i; clock_t start; if (tid == 0) time[bid] = clock(); if (row < n && column < n) { float t = 0; for (i = 0; i < n; i++) { t += a[row * n + i] * b[i * n + column]; } c[row * n + column] = t; } if (tid == 0) { time[bid + blocks_num] = clock(); }}
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注释也写得比较清楚了,我们一开始就是用最简单的形式来完成计算,优化之后再说。下面我们先让这个程序能跑起来。
编写程序
和第一个程序一样,我们先引入需要的库,定义thread数量,方阵的大小,block的数量需要根据矩阵的大小进行计算:
#include <stdio.h>#include <stdlib.h>#include <time.h>#include <cuda_runtime.h>#define THREAD_NUM 256#define MATRIX_SIZE 1000int blocks_num = (MATRIX_SIZE + THREAD_NUM - 1) / THREAD_NUM;
然后第一步还是要初始化CUDA,打印设备信息:
打印信息的方法:
//打印设备信息void printDeviceProp(const cudaDeviceProp &prop){ printf("Device Name : %s.\n", prop.name); printf("totalGlobalMem : %d.\n", prop.totalGlobalMem); printf("sharedMemPerBlock : %d.\n", prop.sharedMemPerBlock); printf("regsPerBlock : %d.\n", prop.regsPerBlock); printf("warpSize : %d.\n", prop.warpSize); printf("memPitch : %d.\n", prop.memPitch); printf("maxThreadsPerBlock : %d.\n", prop.maxThreadsPerBlock); printf("maxThreadsDim[0 - 2] : %d %d %d.\n", prop.maxThreadsDim[0], prop.maxThreadsDim[1], prop.maxThreadsDim[2]); printf("maxGridSize[0 - 2] : %d %d %d.\n", prop.maxGridSize[0], prop.maxGridSize[1], prop.maxGridSize[2]); printf("totalConstMem : %d.\n", prop.totalConstMem); printf("major.minor : %d.%d.\n", prop.major, prop.minor); printf("clockRate : %d.\n", prop.clockRate); printf("textureAlignment : %d.\n", prop.textureAlignment); printf("deviceOverlap : %d.\n", prop.deviceOverlap); printf("multiProcessorCount : %d.\n", prop.multiProcessorCount);}
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CUDA初始化的方法:
bool InitCUDA(){ int count; cudaGetDeviceCount(&count); if (count == 0) { fprintf(stderr, "There is no device.\n"); return false; } int i; for (i = 0; i < count; i++) { cudaDeviceProp prop; cudaGetDeviceProperties(&prop, i); printDeviceProp(prop); if (cudaGetDeviceProperties(&prop, i) == cudaSuccess) { if (prop.major >= 1) { break; } } } if (i == count) { fprintf(stderr, "There is no device supporting CUDA 1.x.\n"); return false; } cudaSetDevice(i); return true;}
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下一步要生成我们要计算的矩阵,上个立方和的程序是一个生成大量随机数的程序,这里要随机生成一个浮点数方阵,我们的矩阵采用i * n + j
的方式来表示,所以我们要传入方阵的尺寸。
随机生成矩阵的方法:
void matgen(float* a, int n) { int i, j; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { a[i * n + j] = (float)rand() / RAND_MAX + (float)rand() / (RAND_MAX * RAND_MAX); } }}
有了这三个方法,我们其他的工作直接在main中完成就好了:
int main(){ if (!InitCUDA()) return 0; float *a, *b, *c, *d; int n = MATRIX_SIZE; a = (float*)malloc(sizeof(float)* n * n); b = (float*)malloc(sizeof(float)* n * n); c = (float*)malloc(sizeof(float)* n * n); d = (float*)malloc(sizeof(float)* n * n); srand(0); matgen(a, n); matgen(b, n); float *cuda_a, *cuda_b, *cuda_c; clock_t* time; cudaMalloc((void**)&cuda_a, sizeof(float)* n * n); cudaMalloc((void**)&cuda_b, sizeof(float)* n * n); cudaMalloc((void**)&cuda_c, sizeof(float)* n * n); cudaMalloc((void**)&time, sizeof(clock_t)* blocks_num * 2); cudaMemcpy(cuda_a, a, sizeof(float)* n * n, cudaMemcpyHostToDevice); cudaMemcpy(cuda_b, b, sizeof(float)* n * n, cudaMemcpyHostToDevice); matMultCUDA << < blocks_num, THREAD_NUM, 0 >> >(cuda_a , cuda_b , cuda_c , n , time); clock_t time_use[blocks_num * 2]; cudaMemcpy(c, cuda_c, sizeof(float)* n * n, cudaMemcpyDeviceToHost); cudaMemcpy(&time_use, time, sizeof(clock_t)* blocks_num * 2, cudaMemcpyDeviceToHost); cudaFree(cuda_a); cudaFree(cuda_b); cudaFree(cuda_c); cudaFree(time); clock_t min_start, max_end; min_start = time_use[0]; max_end = time_use[blocks_num]; for (int i = 1; i < blocks_num; i++) { if (min_start > time_use[i]) min_start = time_use[i]; if (max_end < time_use[i + blocks_num]) max_end = time_use[i + blocks_num]; } clock_t final_time = max_end - min_start; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { double t = 0; for (int k = 0; k < n; k++) { t += a[i * n + k] * b[k * n + j]; } d[i * n + j] = t; } } float max_err = 0; float average_err = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (d[i * n + j] != 0) { float err = fabs((c[i * n + j] - d[i * n + j]) / d[i * n + j]); if (max_err < err) max_err = err; average_err += err; } } } printf("Max error: %g Average error: %g\n",max_err, average_err / (n * n)); printf("gputime: %d\n", final_time);return 0;}
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在GPU上计算完成之后,我们又从CPU上计算了一次,注意这里使用的是double,用来提高精度,然后通过与GPU的结果进行做差比较,计算了精度上的差距(最大相对误差和平均相对误差)。
下面是完整程序:
#include <stdio.h>#include <stdlib.h>#include <time.h>#include <cuda_runtime.h>#define THREAD_NUM 256#define MATRIX_SIZE 1000const int blocks_num = MATRIX_SIZE*(MATRIX_SIZE + THREAD_NUM - 1) / THREAD_NUM;void printDeviceProp(const cudaDeviceProp &prop){printf("Device Name : %s.\n", prop.name);printf("totalGlobalMem : %d.\n", prop.totalGlobalMem);printf("sharedMemPerBlock : %d.\n", prop.sharedMemPerBlock);printf("regsPerBlock : %d.\n", prop.regsPerBlock);printf("warpSize : %d.\n", prop.warpSize);printf("memPitch : %d.\n", prop.memPitch);printf("maxThreadsPerBlock : %d.\n", prop.maxThreadsPerBlock);printf("maxThreadsDim[0 - 2] : %d %d %d.\n", prop.maxThreadsDim[0], prop.maxThreadsDim[1], prop.maxThreadsDim[2]);printf("maxGridSize[0 - 2] : %d %d %d.\n", prop.maxGridSize[0], prop.maxGridSize[1], prop.maxGridSize[2]);printf("totalConstMem : %d.\n", prop.totalConstMem);printf("major.minor : %d.%d.\n", prop.major, prop.minor);printf("clockRate : %d.\n", prop.clockRate);printf("textureAlignment : %d.\n", prop.textureAlignment);printf("deviceOverlap : %d.\n", prop.deviceOverlap);printf("multiProcessorCount : %d.\n", prop.multiProcessorCount);}bool InitCUDA(){ int count; cudaGetDeviceCount(&count); if (count == 0) { fprintf(stderr, "There is no device.\n"); return false; } int i; for (i = 0; i < count; i++) { cudaDeviceProp prop; cudaGetDeviceProperties(&prop, i); printDeviceProp(prop); if (cudaGetDeviceProperties(&prop, i) == cudaSuccess) { if (prop.major >= 1) { break; } } } if (i == count) { fprintf(stderr, "There is no device supporting CUDA 1.x.\n"); return false; } cudaSetDevice(i); return true;}void matgen(float* a, int n){ int i, j; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { a[i * n + j] = (float)rand() / RAND_MAX + (float)rand() / (RAND_MAX * RAND_MAX); } }}__global__ static void matMultCUDA(const float* a, const float* b, float* c, int n, clock_t* time){ const int tid = threadIdx.x; const int bid = blockIdx.x; const int idx = bid * THREAD_NUM + tid; const int row = idx / n; const int column = idx % n; int i; clock_t start; if (tid == 0) time[bid] = clock(); if (row < n && column < n) { float t = 0; for (i = 0; i < n; i++) { t += a[row * n + i] * b[i * n + column]; } c[row * n + column] = t; } if (tid == 0) { time[bid + blocks_num] = clock(); }}int main(){ if (!InitCUDA()) return 0; float *a, *b, *c, *d; int n = MATRIX_SIZE; a = (float*)malloc(sizeof(float)* n * n); b = (float*)malloc(sizeof(float)* n * n); c = (float*)malloc(sizeof(float)* n * n); d = (float*)malloc(sizeof(float)* n * n); srand(0); matgen(a, n); matgen(b, n); float *cuda_a, *cuda_b, *cuda_c; clock_t* time; cudaMalloc((void**)&cuda_a, sizeof(float)* n * n); cudaMalloc((void**)&cuda_b, sizeof(float)* n * n); cudaMalloc((void**)&cuda_c, sizeof(float)* n * n); cudaMalloc((void**)&time, sizeof(clock_t)* blocks_num * 2); cudaMemcpy(cuda_a, a, sizeof(float)* n * n, cudaMemcpyHostToDevice); cudaMemcpy(cuda_b, b, sizeof(float)* n * n, cudaMemcpyHostToDevice); matMultCUDA << < blocks_num, THREAD_NUM, 0 >> >(cuda_a , cuda_b , cuda_c , n , time); clock_t time_use[blocks_num * 2]; cudaMemcpy(c, cuda_c, sizeof(float)* n * n, cudaMemcpyDeviceToHost); cudaMemcpy(&time_use, time, sizeof(clock_t)* blocks_num * 2, cudaMemcpyDeviceToHost); cudaFree(cuda_a); cudaFree(cuda_b); cudaFree(cuda_c); cudaFree(time); clock_t min_start, max_end; min_start = time_use[0]; max_end = time_use[blocks_num]; for (int i = 1; i < blocks_num; i++) { if (min_start > time_use[i]) min_start = time_use[i]; if (max_end < time_use[i + blocks_num]) max_end = time_use[i + blocks_num]; } clock_t final_time = max_end - min_start; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { double t = 0; for (int k = 0; k < n; k++) { t += a[i * n + k] * b[k * n + j]; } d[i * n + j] = t; } } float max_err = 0; float average_err = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (d[i * n + j] != 0) { float err = fabs((c[i * n + j] - d[i * n + j]) / d[i * n + j]); if (max_err < err) max_err = err; average_err += err; } } } printf("Max error: %g Average error: %g\n",max_err, average_err / (n * n)); printf("gputime: %d\n", final_time);return 0;}
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运行结果:
这里我们看到,非常明显的,执行效率相当的低下,用了189967999个周期,大概是0.23秒,这是非常差的一个结果了。
同时精度也非常差,最大相对误差偏高,理想上应该要低于 1e-6。
计算结果的误差偏高的原因是,在 CPU 上进行计算时,我们使用 double(即 64 bits 浮点数)来累进计算过程,而在 GPU 上则只能用 float(32 bits 浮点数)。在累加大量数字的时候,由于累加结果很快会变大,因此后面的数字很容易被舍去过多的位数。
不过我们已经算是完成了程序的初级版本,精度和速度的问题我们慢慢优化。
总结:
这篇博客我们用CUDA完成了矩阵乘法,问题也比较简单,基于上一个立方和程序的经验,完成这个程序也不算太难,但是当然会存在很多问题,毕竟我们还没有开始优化,不过除了速度问题,GPU浮点数运算的精度也成了一个大问题,这些我们后面再一步步解决~
希望我的博客能帮助到大家~
参考资料:《深入浅出谈CUDA》