0


从头开始进行CUDA编程:线程间协作的常见技术

在前一篇文章中,我们介绍了如何使用 GPU 运行的并行算法。这些并行任务是那些完全相互独立的任务,这点与我们一般认识的编程方式有很大的不同,虽然我们可以从并行中受益,但是这种奇葩的并行运行方式对于我们来说肯定感到非常的复杂。所以在本篇文章的Numba代码中,我们将介绍一些允许线程在计算中协作的常见技术。

首先还是载入相应的包

 from time import perf_counter
 
 import numpy as np
 
 import numba
 from numba import cuda
 
 print(np.__version__)
 print(numba.__version__)
 
 cuda.detect()
 
 # 1.21.6
 # 0.55.2
 
 # Found 1 CUDA devices
 # id 0             b'Tesla T4'                              [SUPPORTED]
 #                       Compute Capability: 7.5
 #                            PCI Device ID: 4
 #                               PCI Bus ID: 0
 #                                     UUID: GPU-bcc6196e-f26e-afdc-1db3-6eba6ff160f0
 #                                 Watchdog: Disabled
 #              FP32/FP64 Performance Ratio: 32
 # Summary:
 # 1/1 devices are supported
 # True

不要忘记,我们这里是CUDA编程,所以NV的GPU是必须的,比如可以去colab或者Kaggle白嫖。

线程间的协作

简单的并行归约算法

我们将从一个非常简单的问题开始本节:对数组的所有元素求和。这个算法非常简单。如果不使用NumPy,我们可以这样实现它:

 def sum_cpu(array):
     s = 0.0
     for i in range(array.size):
         s += array[i]
     return s

这看起来不是很 Pythonic。但它能够让我们了解它正在跟踪数组中的所有元素。如果 s 的结果依赖于数组的每个元素,我们如何并行化这个算法呢?首先,我们需要重写算法以允许并行化, 如果有无法并行化的部分则应该允许线程相互通信。

到目前为止,我们还没有学会如何让线程相互通信……事实上,我们之前说过不同块中的线程不通信。我们可以考虑只启动一个块,但是我们上次也说了,在大多数 GPU 中块只能有 1024 个线程!

如何克服这一点?如果将数组拆分为 1024 个块(或适当数量的threads_per_block)并分别对每个块求和呢?然后最后,我们可以将每个块的总和的结果相加。下图显示了一个非常简单的 2 块拆分示例。

上图就是对数组元素求和的“分而治之”方法。

如何在 GPU 上做到这一点呢?首先需要将数组拆分为块。每个数组块将只对应一个具有固定数量的线程的CUDA块。在每个块中,每个线程可以对多个数组元素求和。然后将这些每个线程的值求和,这里就需要线程进行通信,我们将在下一个示例中讨论如何通信。

由于我们正在对块进行并行化,因此内核的输出应该被设置为一个块。为了完成Reduce,我们将其复制到 CPU 并在那里完成工作。

 threads_per_block = 1024  # Why not!
 blocks_per_grid = 32 * 80  # Use 32 * multiple of streaming multiprocessors
 
 # Example 2.1: Naive reduction
 @cuda.jit
 def reduce_naive(array, partial_reduction):
     i_start = cuda.grid(1)
     threads_per_grid = cuda.blockDim.x * cuda.gridDim.x
     s_thread = 0.0
     for i_arr in range(i_start, array.size, threads_per_grid):
         s_thread += array[i_arr]
 
     # We need to create a special *shared* array which will be able to be read
     # from and written to by every thread in the block. Each block will have its
     # own shared array. See the warning below!
     s_block = cuda.shared.array((threads_per_block,), numba.float32)
     
     # We now store the local temporary sum of a single the thread into the
     # shared array. Since the shared array is sized
     #     threads_per_block == blockDim.x
     # (1024 in this example), we should index it with `threadIdx.x`.
     tid = cuda.threadIdx.x
     s_block[tid] = s_thread
     
     # The next line synchronizes the threads in a block. It ensures that after
     # that line, all values have been written to `s_block`.
     cuda.syncthreads()
 
     # Finally, we need to sum the values from all threads to yield a single
     # value per block. We only need one thread for this.
     if tid == 0:
         # We store the sum of the elements of the shared array in its first
         # coordinate
         for i in range(1, threads_per_block):
             s_block[0] += s_block[i]
         # Move this partial sum to the output. Only one thread is writing here.
         partial_reduction[cuda.blockIdx.x] = s_block[0]

这里需要注意的是必须共享数组,并且让每个数组块变得“小”

这里的“小”:确切大小取决于 GPU 的计算能力,通常在 48 KB 和 163 KB 之间。请参阅此表中的“每个线程块的最大共享内存量”项。(https://docs.nvidia.com/cuda/cuda-c-programming-guide/index.html#features-and-technical-specifications__technical-specifications-per-compute-capability)

在编译时有一个已知的大小(这就是我们调整共享数组threads_per_block而不是blockDim.x的原因)。我们总是可以为任何大小的共享数组定义一个工厂函数……但要注意这些内核的编译时间。

这里的数组需要为 Numba 类型指定的 dtype,而不是 Numpy 类型(这个没有为什么!)。

 N = 1_000_000_000
 a = np.arange(N, dtype=np.float32)
 a /= a.sum() # a will have sum = 1 (to float32 precision)
 
 s_cpu = a.sum()
 
 # Highly-optimized NumPy CPU code
 timing_cpu = np.empty(21)
 for i in range(timing_cpu.size):
     tic = perf_counter()
     a.sum()
     toc = perf_counter()
     timing_cpu[i] = toc - tic
 timing_cpu *= 1e3  # convert to ms
 
 print(f"Elapsed time CPU: {timing_cpu.mean():.0f} ± {timing_cpu.std():.0f} ms")
 # Elapsed time CPU: 354 ± 24 ms
 
 dev_a = cuda.to_device(a)
 dev_partial_reduction = cuda.device_array((blocks_per_grid,), dtype=a.dtype)
 
 reduce_naive[blocks_per_grid, threads_per_block](dev_a, dev_partial_reduction)
 s = dev_partial_reduction.copy_to_host().sum()  # Final reduction in CPU
 
 np.isclose(s, s_cpu)  # Ensure we have the right number
 # True
 
 timing_naive = np.empty(21)
 for i in range(timing_naive.size):
     tic = perf_counter()
     reduce_naive[blocks_per_grid, threads_per_block](dev_a, dev_partial_reduction)
     s = dev_partial_reduction.copy_to_host().sum()
     cuda.synchronize()
     toc = perf_counter()
     assert np.isclose(s, s_cpu)    
     timing_naive[i] = toc - tic
 timing_naive *= 1e3  # convert to ms
 
 print(f"Elapsed time naive: {timing_naive.mean():.0f} ± {timing_naive.std():.0f} ms")
 # Elapsed time naive: 30 ± 12 ms

在谷歌Colab上测试,得到了10倍的加速。

题外话:上面这个方法之所以说是简单的规约算法,是因为这个算法最简单,也最容易实现。我们在大数据中常见的Map-Reduce算法就是这个算法。虽然实现简单,但是他容易理解,所以十分常见,当然他慢也是出名的,有兴趣的大家可以去研究研究。

一种更好的并行归约算法

上面的算法最 “朴素”的,所以有很多技巧可以加快这种代码的速度(请参阅 CUDA 演示文稿中的 Optimizing Parallel Reduction 以获得基准测试)。

在介绍更好的方法之前,让我们回顾一下内核函数的的最后一点:

 if tid == 0:  # Single thread taking care of business
     for i in range(1, threads_per_block):
         s_block[0] += s_block[i]
     partial_reduction[cuda.blockIdx.x] = s_block[0]

我们并行化了几乎所有的操作,但是在内核的最后,让一个线程负责对共享数组 s_block 的所有 threads_per_block 元素求和。为什么不能把这个总和也并行化呢?

听起来不错对吧,下图显示了如何在 threads_per_block 大小为 16 的情况下实现这一点。我们从 8 个线程开始工作,第一个将对 s_block[0] 和 s_block[8] 中的值求和。第二个求和s_block[1]和s_block[9]中的值,直到最后一个线程将s_block[7]和s_block[15]的值相加。

在下一步中,只有前 4 个线程需要工作。第一个线程将对 s_block[0] 和 s_block[4] 求和;第二个,s_block[1] 和 s_block[5];第三个,s_block[2] 和 s_block[6];第四个也是最后一个,s_block[3] 和 s_block[7]。

第三步,只需要 2 个线程来处理 s_block 的前 4 个元素。

第四步也是最后一步将使用一个线程对 2 个元素求和。

由于工作已在线程之间分配,因此它是并行化的。这不是每个线程的平等划分,但它是一种改进。在计算上,这个算法是 O(log2(threads_per_block)),而上面“朴素”算法是 O(threads_per_block)。比如在我们这个示例中是 1024 次操作,用于 了两个算法差距有10倍

最后还有一个细节。在每一步,我们都需要确保所有线程都已写入共享数组。所以我们必须调用 cuda.syncthreads()。

 # Example 2.2: Better reduction
 @cuda.jit
 def reduce_better(array, partial_reduction):
     i_start = cuda.grid(1)
     threads_per_grid = cuda.blockDim.x * cuda.gridDim.x
     s_thread = 0.0
     for i_arr in range(i_start, array.size, threads_per_grid):
         s_thread += array[i_arr]
 
     # We need to create a special *shared* array which will be able to be read
     # from and written to by every thread in the block. Each block will have its
     # own shared array. See the warning below!
     s_block = cuda.shared.array((threads_per_block,), numba.float32)
     
     # We now store the local temporary sum of the thread into the shared array.
     # Since the shared array is sized threads_per_block == blockDim.x,
     # we should index it with `threadIdx.x`.
     tid = cuda.threadIdx.x
     s_block[tid] = s_thread
     
     # The next line synchronizes the threads in a block. It ensures that after
     # that line, all values have been written to `s_block`.
     cuda.syncthreads()
 
     i = cuda.blockDim.x // 2
     while (i > 0):
         if (tid < i):
             s_block[tid] += s_block[tid + i]
         cuda.syncthreads()
         i //= 2
 
     if tid == 0:
         partial_reduction[cuda.blockIdx.x] = s_block[0]
 
 
 reduce_better[blocks_per_grid, threads_per_block](dev_a, dev_partial_reduction)
 s = dev_partial_reduction.copy_to_host().sum()  # Final reduction in CPU
 
 np.isclose(s, s_cpu)
 # True
 
 timing_naive = np.empty(21)
 for i in range(timing_naive.size):
     tic = perf_counter()
     reduce_better[blocks_per_grid, threads_per_block](dev_a, dev_partial_reduction)
     s = dev_partial_reduction.copy_to_host().sum()
     cuda.synchronize()
     toc = perf_counter()
     assert np.isclose(s, s_cpu)    
     timing_naive[i] = toc - tic
 timing_naive *= 1e3  # convert to ms
 
 print(f"Elapsed time better: {timing_naive.mean():.0f} ± {timing_naive.std():.0f} ms")
 # Elapsed time better: 22 ± 1 ms

可以看到,这比原始方法快25%。

重要说明:你可能很想将同步线程移动到 if 块内,因为在每一步之后,超过当前线程数一半的内核将不会被使用。但是这样做会使调用同步线程的 CUDA 线程停止并等待所有其他线程,而所有其他线程将继续运行。因此停止的线程将永远等待永远不会停止同步的线程。如果您同步线程,请确保在所有线程中调用 cuda.syncthreads()。

 i = cuda.blockDim.x // 2
 while (i > 0):
     if (tid < i):
         s_block[tid] += s_block[tid + i]
         #cuda.syncthreads()  这个是错的
     cuda.syncthreads()  # 这个是对的
     i //= 2

Numba 自动归约

其实归约算法并不简单,所以Numba 提供了一个方便的 cuda.reduce 装饰器,可以将二进制函数转换为归约。所以上面冗长而复杂的算法可以替换为:

 @cuda.reduce
 def reduce_numba(a, b):
     return a + b
 
 # Compile and check
 s = reduce_numba(dev_a)
 
 np.isclose(s, s_cpu)
 # True
 
 # Time
 timing_numba = np.empty(21)
 for i in range(timing_numba.size):
     tic = perf_counter()
     s = reduce_numba(dev_a)
     toc = perf_counter()
     assert np.isclose(s, s_cpu)    
     timing_numba[i] = toc - tic
 timing_numba *= 1e3  # convert to ms
 
 print(f"Elapsed time better: {timing_numba.mean():.0f} ± {timing_numba.std():.0f} ms")
 # Elapsed time better: 45 ± 0 ms

上面的运行结果我们可以看到手写代码通常要快得多(至少 2 倍),但 Numba 给我们提供的方法却非常容易使用。这对我们来说是格好事,因为终于有我们自己实现的Python方法比官方的要快了😂

这里还有一点要注意就是默认情况下,要减少复制因为这会强制同步。为避免这种情况可以使用设备上数组作为输出调用归约:

 dev_s = cuda.device_array((1,), dtype=s)
 
 reduce_numba(dev_a, res=dev_s)
 
 s = dev_s.copy_to_host()[0]
 np.isclose(s, s_cpu)
 # True

二维规约示例

并行约简技术是非常伟大的,如何将其扩展到更高的维度?虽然我们总是可以使用一个展开的数组(array2 .ravel())调用,但了解如何手动约简多维数组是很重要的。

在下面这个例子中,将结合刚才所学的知识来计算二维数组。

 threads_per_block_2d = (16, 16)  #  256 threads total
 blocks_per_grid_2d = (64, 64)
 
 # Total number of threads in a 2D block (has to be an int)
 shared_array_len = int(np.prod(threads_per_block_2d))
 
 # Example 2.4: 2D reduction with 1D shared array
 @cuda.jit
 def reduce2d(array2d, partial_reduction2d):
     ix, iy = cuda.grid(2)
     threads_per_grid_x, threads_per_grid_y = cuda.gridsize(2)
 
     s_thread = 0.0
     for i0 in range(iy, array2d.shape[0], threads_per_grid_x):
         for i1 in range(ix, array2d.shape[1], threads_per_grid_y):
             s_thread += array2d[i0, i1]
 
     # Allocate shared array
     s_block = cuda.shared.array(shared_array_len, numba.float32)
 
     # Index the threads linearly: each tid identifies a unique thread in the
     # 2D grid.
     tid = cuda.threadIdx.x + cuda.blockDim.x * cuda.threadIdx.y
     s_block[tid] = s_thread
 
     cuda.syncthreads()
 
     # We can use the same smart reduction algorithm by remembering that
     #     shared_array_len == blockDim.x * cuda.blockDim.y
     # So we just need to start our indexing accordingly.
     i = (cuda.blockDim.x * cuda.blockDim.y) // 2
     while (i != 0):
         if (tid < i):
             s_block[tid] += s_block[tid + i]
         cuda.syncthreads()
         i //= 2
     
     # Store reduction in a 2D array the same size as the 2D blocks
     if tid == 0:
         partial_reduction2d[cuda.blockIdx.x, cuda.blockIdx.y] = s_block[0]
 
 
 N_2D = (20_000, 20_000)
 a_2d = np.arange(np.prod(N_2D), dtype=np.float32).reshape(N_2D)
 a_2d /= a_2d.sum() # a_2d will have sum = 1 (to float32 precision)
 
 s_2d_cpu = a_2d.sum()
 
 dev_a_2d = cuda.to_device(a_2d)
 dev_partial_reduction_2d = cuda.device_array(blocks_per_grid_2d, dtype=a.dtype)
 
 reduce2d[blocks_per_grid_2d, threads_per_block_2d](dev_a_2d, dev_partial_reduction_2d)
 s_2d = dev_partial_reduction_2d.copy_to_host().sum()  # Final reduction in CPU
 
 np.isclose(s_2d, s_2d_cpu)  # Ensure we have the right number
 # True
 
 timing_2d = np.empty(21)
 for i in range(timing_2d.size):
     tic = perf_counter()
     reduce2d[blocks_per_grid_2d, threads_per_block_2d](dev_a_2d, dev_partial_reduction_2d)
     s_2d = dev_partial_reduction_2d.copy_to_host().sum()
     cuda.synchronize()
     toc = perf_counter()
     assert np.isclose(s_2d, s_2d_cpu)    
     timing_2d[i] = toc - tic
 timing_2d *= 1e3  # convert to ms
 
 print(f"Elapsed time better: {timing_2d.mean():.0f} ± {timing_2d.std():.0f} ms")
 # Elapsed time better: 15 ± 4 ms

设备函数

到目前为止,我们只讨论了内核函数,它是启动线程的特殊GPU函数。内核通常依赖于较小的函数,这些函数在GPU中定义,只能访问GPU数组。这些被称为设备函数(Device functions)。与内核函数不同的是,它们可以返回值。

我们将展示一个跨不同内核使用设备函数的示例。该示例还将展示在使用共享数组时同步线程的重要性。

在CUDA的新版本中,内核可以启动其他内核。这被称为动态并行,但是Numba 的CUDA API还不支持。

我们将在固定大小的数组中创建波纹图案。首先需要声明将使用的线程数,因为这是共享数组所需要的。

 threads_16 = 16
 
 import math
 
 @cuda.jit(device=True, inline=True)  # inlining can speed up execution
 def amplitude(ix, iy):
     return (1 + math.sin(2 * math.pi * (ix - 64) / 256)) * (
         1 + math.sin(2 * math.pi * (iy - 64) / 256)
     )
 
 # Example 2.5a: 2D Shared Array
 @cuda.jit
 def blobs_2d(array2d):
     ix, iy = cuda.grid(2)
     tix, tiy = cuda.threadIdx.x, cuda.threadIdx.y
 
     shared = cuda.shared.array((threads_16, threads_16), numba.float32)
     shared[tiy, tix] = amplitude(iy, ix)
     cuda.syncthreads()
 
     array2d[iy, ix] = shared[15 - tiy, 15 - tix]
 
 # Example 2.5b: 2D Shared Array without synchronize
 @cuda.jit
 def blobs_2d_wrong(array2d):
     ix, iy = cuda.grid(2)
     tix, tiy = cuda.threadIdx.x, cuda.threadIdx.y
 
     shared = cuda.shared.array((threads_16, threads_16), numba.float32)
     shared[tiy, tix] = amplitude(iy, ix)
 
     # When we don't sync threads, we may have not written to shared
     # yet, or even have overwritten it by the time we write to array2d
     array2d[iy, ix] = shared[15 - tiy, 15 - tix]
 
 
 N_img = 1024
 blocks = (N_img // threads_16, N_img // threads_16)
 threads = (threads_16, threads_16)
 
 dev_image = cuda.device_array((N_img, N_img), dtype=np.float32)
 dev_image_wrong = cuda.device_array((N_img, N_img), dtype=np.float32)
 
 blobs_2d[blocks, threads](dev_image)
 blobs_2d_wrong[blocks, threads](dev_image_wrong)
 
 image = dev_image.copy_to_host()
 image_wrong = dev_image_wrong.copy_to_host()
 
 import matplotlib.pyplot as plt
 
 fig, (ax1, ax2) = plt.subplots(1, 2)
 ax1.imshow(image.T, cmap="nipy_spectral")
 ax2.imshow(image_wrong.T, cmap="nipy_spectral")
 for ax in (ax1, ax2):
     ax.set_xticks([])
     ax.set_yticks([])
     ax.set_xticklabels([])
     ax.set_yticklabels([])

左:同步(正确)内核的结果。正确:来自不同步(不正确)内核的结果。

总结

本文介绍了如何开发需要规约模式来处理1D和2D数组的内核函数。在这个过程中,我们学习了如何利用共享数组和设备函数。如果你对文本感兴趣,请看源代码:

https://colab.research.google.com/drive/1GkGLDexnYUnl2ilmwNxAlWAH6Eo5ZK2f?usp=sharing

作者:Carlos Costa, Ph.D.

标签: 深度学习 CUDA GPU

记录[+]

21 天前 已修改

“从头开始进行CUDA编程:线程间协作的常见技术”的评论:

还没有评论