如何在Halide中进行正反FFT

Question

我目前正在尝试先做一个正向fft，然后再做一个反向fft，但是它似乎不起作用。我使用的快速傅立叶变换是在fft.cpp (Halide/apps/ FFT )中找到的。我目前的目标只是尝试保存16x16平铺的图像。该16x16瓦片应该是正向的，然后是16x16瓦片的反向fft。我的问题是，由于某种原因，我的输出缓冲区的值是9000。下面是我的代码：

//A program to make an fft of an image both ways (r2c, c2r)
//Plan of action:
//1.)Load in image into buffer using load_image (uint8)
//2.)Then cast it to a float
//3.)Then convert float buffer to a function
//4.)Then set fft2d settings
//5.)Then call real to complex
//6.)Then call complex to real
//7.)Then realize it to an output buffer
//8.)Then save the image

#include <stdio.h>
#include "Halide.h"
#include "fft.h"
#include "halide_image_io.h"

using namespace Halide;
using namespace Halide::Tools;
using namespace std;


template <typename Type1, typename Type2>
void compare(Halide::Buffer<Type1> org, Halide::Buffer<Type2> other);

Var x{"x"}, y{"y"}, c{"c"};
Func real_result;
ComplexFunc complex_result("Complex_Result");
int colour_channel_to_fft = 1; //or 1 , or 2
int tileSize = 16;

int main(){
    Halide::Buffer<uint8_t> unsignedIntTempBuffer = load_image("rgb.png");

    //2.) Then cast it to a float
    Func uint8_tToFloat;
    uint8_tToFloat(x,y,c) = Halide::cast<float>(unsignedIntTempBuffer(x,y,c));

    Halide::Buffer<float> input;
    input = uint8_tToFloat.realize(unsignedIntTempBuffer.width(),unsignedIntTempBuffer.height(),unsignedIntTempBuffer.channels()); //Input becomes a float buffer

    //3.)Then convert float buffer to a greysacle function
    Func in;
    in(x,y) = input(x,y,colour_channel_to_fft); //Third parameter states which RGB grey scale to use

    Halide::Buffer<float> temp;
    temp = in.realize(input.width(), input.height());

    //4.)Then set fft2d settings - the current setting are defaulted
    Fft2dDesc desc;
    desc.gain = 1.0f;
    desc.vector_width = 0;
    desc.parallel = false;

    //5.)Then call real to complex
    complex_result = fft2d_r2c(in, tileSize, tileSize,get_jit_target_from_environment(), desc);    //Max dimension size of 767

    //Load the complex result into the complexBuffer
    Halide::Buffer<float> complexBuffer;
    complexBuffer = complex_result.realize();

    ComplexFunc cmplxIn;
    cmplxIn(x, y) = ComplexExpr(re(complexBuffer(x, y)), im(complexBuffer(x, y))); //IN GENERATOR THEY USE CHANNEL 1 & 0? Not possible due to us only using one channel for real input
    //6.)Then call complex to real
    real_result = fft2d_c2r(cmplxIn,tileSize,tileSize,get_jit_target_from_environment(),desc);
    Halide::Buffer<float>output;
    output = real_result.realize(); // as output(x,y,c) = re(complex_result(x,y)); doesn't work (seg fault)

    Func floatToUInt8;
    floatToUInt8(x,y,c) = Halide::cast<uint8_t>(output(x,y));

    Halide::Buffer<uint8_t> finalOutput = floatToUInt8.realize(tileSize, tileSize, input.channels());//, input.channels());
    save_image(finalOutput, "forwardThenReverseFFT.png");
    cout << "Success" << endl;
    //Func -> Buffer must use a realize
}

template <typename Type1, typename Type2>
void compare(Halide::Buffer<Type1> org, Halide::Buffer<Type2> other){
    string channel = "";
    if (colour_channel_to_fft == 0) channel = "Red";
    else if (colour_channel_to_fft == 1) channel = "Green";
    else if (colour_channel_to_fft == 2) channel = "Blue";
    else cout<< "You have chosen an incorrect channel";
    std::cout << "Original: " << std::endl << channel << " channel value at (0,0) = " << org(3,3) << std::endl;
    std::cout << "FFTd: " << std::endl << channel << " channel value at (0,0) = " << other(0,0) << std::endl << std::endl;
}

保存的图像是：https://i.stack.imgur.com/9Rqtm.png，它似乎与任何通道上的原始图像没有相关性。

你知道我做错了什么吗？

Answer 1

SleuthEye的发展方向是正确的。我认为问题的一部分是fft2d_r2c的结果是一个元值函数(参见https://halide-lang.org/tutorials/tutorial_lesson_13_tuples.html)。ComplexExpr/ComplexFunc是围绕Tuple和元值Func的包装器，令人有点惊讶的是，它甚至会编译/运行来将这一点的实现分配给Halide::Buffer。

另一个问题是实现需要一个大小。(您可能不需要这样做，因为FFT对输出的边界有要求。)对于FFT来说，这有点棘手，因为复杂的域只有部分定义。对于16x16FFT，复数域是16x9，其余的域可以通过利用FFT的共轭对称特性来计算。(https://github.com/halide/Halide/blob/b465318a583ff58114ac63ecd8125ca7e648ae35/apps/fft/fft.h#L55-L56)。

我猜想这就是你所需要的：

//Load the complex result into the complexBuffer
Halide::Realization complexBuffer = complex_result.realize(16, 9);
Halide::Buffer<float> realPart = complexBuffer[0];
Halide::Buffer<float> imagPart = complexBuffer[1];

// Now construct the complex part for `fft2d_c2r` from these two buffers.

我认为，如果你试图认识到(在复杂的域中去往/离开Halide )，事情将会比必要的更困难。通常，如果您使用fft2d_r2c，做一些工作，然后使用fft2d_c2r，所有这些都在一个管道中，Halide的边界推断将意味着您不需要太担心DFT域的奇数边界。