1D没有if-else语句的卷积（非FFT）？

问题描述

我写了一个简单的串行1D卷积函数（如下）。我也在试验GPU卷积实现。这主要是为了我自己的好奇心;我试图学习在各种非FFT实现策略之间的性能折衷。

I've written a simple serial 1D convolution function (below). I'm also experimenting with GPU convolution implementations. This is mostly for my own curiosity; I'm trying to learn the performance tradeoffs among various non-FFT implementation strategies.

避免分支对我的GPU卷积实验很重要，因为分支在Nvidia GPU上很贵。我的一个朋友提到有一种方法来实现下面的代码，而不用 if / else 语句，但他不记得它的工作原理。

Avoiding branching will be important for my GPU convolution experiments, since branching is expensive on Nvidia GPUs. One of my friends mentioned that there's a way to implement the code below without if/else statements, but he couldn't remember how it works.

如何在不使用任何 if / else 语句

这是我在C ++中的基本1D序列码：

Here's my basic 1D serial code in C++:

vector<int> myConv1d(vector<int> vec, vector<int> kernel)
{
    int paddedLength = vec.size() + kernel.size() - 1;
    vector<int> convolved(paddedLength); //zeros
    reverse(kernel.begin(), kernel.end()); //flip the kernel (if we don't flip it, then we have correlation instead of convolution)
    for(int outputIdx=0; outputIdx<paddedLength; outputIdx++) //index into 'convolved' vector
    {
        int vecIdx = outputIdx - kernel.size() + 1; //aligns with leftmost element of kernel
        for(int kernelIdx=0; kernelIdx<kernel.size(); kernelIdx++)
        {
            if( (vecIdx+kernelIdx) >= 0  &&  (vecIdx+kernelIdx) < vec.size() ) //TODO: FIND A WAY TO REMOVE THIS
            {
                convolved[outputIdx] += kernel[kernelIdx]*vec[vecIdx+kernelIdx];
            }
        }
    }
    return convolved;
}

几个快速笔记：

• 我发现 ，但我不太明白避免条件语句的策略。


• 也写了一个2D卷积实现，我希望将这个SO post的结果也应用到2D版本。


• 这不是家庭作业。

• I did find some related posts, but I didn't quite understand the strategy avoiding conditional statements.
• I've also written a 2D convolution implementation, and I'm hoping to apply the results of this SO post to the 2D version as well.
• This is NOT homework. It's marginally related to one of our research projects, but it's mostly for the sake of learning.

推荐答案

int lowerBound = std::max( 0, -vecIdx );
int upperBound = std::min( kernel.size(), vec.size() - vecIdx );
for( int kernelIdx = lowerBound; kernelIdx < upperBound; kernelIdx++ )

不明白这个问题。

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