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Home/ Questions/Q 8578087
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T20:18:26+00:00 2026-06-11T20:18:26+00:00

I’ve written a simple serial 1D convolution function (below). I’m also experimenting with GPU

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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.

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.

How can I do a correct 1D convolution implementation without using any if/else statements?

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;
}

A couple of quick notes:

  • 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.
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T20:18:27+00:00Added an answer on June 11, 2026 at 8:18 pm

    Why don’t you do something like this?

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

    Sorry If I didn’t understand the question.

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