Radix 3,5 FFT in C++

Question

I'm trying to implement A Radix-5,Radix-3 FFT in C++, I already managed to write a Radix-2 but I have some type of bug when it comes to Radix 3 or 5, let's say I do an FFT of 3 samples, that would show the correct results, however if I do FFT of 9 which is 3 * 3, it doesn't show the correct results.

I originally took the code from Python, it works there and I'm trying to simply "copy" it to C++, Here is the original Python code:

def fft(x):
    """
    radix-2,3,5 FFT algorithm
    """
    N = len(x)
    if N <= 1:
        return x
    elif N % 2 == 0:
        # For multiples of 2 this formula works
        even = fft(x[0::2])
        odd =  fft(x[1::2])
        T = [np.exp(-2j*np.pi*k/N)*odd[k] for k in range(N//2)]
        return [even[k] + T[k] for k in range(N//2)] + \
               [even[k] - T[k] for k in range(N//2)]
    elif N % 3 == 0:
        # Optional, implementing factor 3 decimation
        p0 = fft(x[0::3])
        p1 = fft(x[1::3])
        p2 = fft(x[2::3])
        # This will construct the output output without the simplifications
        # you can do explorint symmetry
        
        for k in range(N):
            return [p0[k % (N//3)] +
                    p1[k % (N//3)] * np.exp(-2j*np.pi*k/N) + 
                    p2[k % (N//3)] * np.exp(-4j*np.pi*k/N)]
    elif N % 5 == 0:
        #factor 5 decimation
        p0 = fft(x[0::5])
        p1 = fft(x[1::5])
        p2 = fft(x[2::5])
        p3 = fft(x[3::5])
        p4 = fft(x[4::5])

        return [p0[k % (N//5)] +
                p1[k % (N//5)] * np.exp(-2j*np.pi*k/N) + 
                p2[k % (N//5)] * np.exp(-4j*np.pi*k/N) + 
                p3[k % (N//5)] * np.exp(-6j*np.pi*k/N) +
                p4[k % (N//5)] * np.exp(-8j*np.pi*k/N)
               for k in range(N)]

x = [1,1.00071,1.00135,1.00193,1.00245,1.0029,1.00329,1.00361,1.00387]
assert(np.allclose(fft(x), np.fft.fft(x)))

And here is my code in C++ (fft.hpp):

#define _USE_MATH_DEFINES
#pragma once
#include <cmath>
#include <vector>
#include <complex>


using std::vector;
using std::complex;

vector<complex<float>> slicing(vector<complex<float>> vec, unsigned int X, unsigned int Y, unsigned int stride)
{
    // To store the sliced vector
    vector<complex<float>> result;

    // Copy vector using copy function()
    int i = X;
    while (result.size() < Y)
    {
        result.push_back(vec[i]);
        i = i + stride;
    }
    // Return the final sliced vector
    return result;
}

void fft(vector<complex<float>>& x)
{
    // Check if it is splitted enough

    const size_t N = x.size();
    if (N <= 1)
        return;

    else if (N % 2 == 0)
    {
        //Radix-2
        vector<complex<float>> even = slicing(x, 0, N / 2, 2); //split the inputs in even / odd indices subarrays
        vector<complex<float>>  odd = slicing(x, 1, N / 2, 2);

        // conquer
        fft(even);
        fft(odd);

        // combine
        for (size_t k = 0; k < N / 2; ++k)
        {
            complex<float> t = std::polar<float>(1.0, -2 * M_PI * k / N) * odd[k];
            x[k] = even[k] + t;
            x[k + N / 2] = even[k] - t;
        }
    }
    else if (N % 3 == 0)
    {
        //Radix-3
        //factor 3 decimation
        vector<complex<float>> p0 = slicing(x, 0, N / 3, 3);
        vector<complex<float>> p1 = slicing(x, 1, N / 3, 3);
        vector<complex<float>> p2 = slicing(x, 2, N / 3, 3);

        fft(p0);
        fft(p1);
        fft(p2);

        for (int i = 0; i < N; i++)
        {
            complex<float> temp = p0[i % (int)N / 3];
            temp += (p1[i % (int)N / 3] * std::polar<float>(1.0, -2 * M_PI * i / N));
            temp += (p2[i % (int)N / 3] * std::polar<float>(1.0, -4 * M_PI * i / N));
            x[i] = temp;
        }
    }
    else if (N % 5 == 0)
    {
        //Radix-5
        //factor 5 decimation
        vector<complex<float>> p0 = slicing(x, 0, N / 5, 5);
        vector<complex<float>> p1 = slicing(x, 1, N / 5, 5);
        vector<complex<float>> p2 = slicing(x, 2, N / 5, 5);
        vector<complex<float>> p3 = slicing(x, 3, N / 5, 5);
        vector<complex<float>> p4 = slicing(x, 4, N / 5, 5);

        fft(p0);
        fft(p1);
        fft(p2);
        fft(p3);
        fft(p4);
        for (int i = 0; i < N; i++)
        {
            complex<float> temp = p0[i % (int)N / 5];
            temp += (p1[i % (int)N / 5] * std::polar<float>(1.0, -2 * M_PI * i / N));
            temp += (p2[i % (int)N / 5] * std::polar<float>(1.0, -4 * M_PI * i / N));
            temp += (p3[i % (int)N / 5] * std::polar<float>(1.0, -6 * M_PI * i / N));
            temp += (p4[i % (int)N / 5] * std::polar<float>(1.0, -8 * M_PI * i / N));
            x[i] = temp;
        }
    }
}

And main.cpp:

#define _USE_MATH_DEFINES
#include <stdio.h>
#include <iostream>
#include "fft.hpp"


typedef vector<complex<float>> complexSignal;

int main()
{
    complexSignal abit;
    int N = 9;
    abit.push_back({1,0});
    abit.push_back({1.00071 ,0 });
    abit.push_back({1.00135 ,0 });
    abit.push_back({1.00193 ,0 });
    abit.push_back({1.00245 ,0 });
    abit.push_back({1.0029 ,0 });
    abit.push_back({1.00329 ,0 });
    abit.push_back({1.00361 ,0 });
    abit.push_back({1.00387 ,0 });
    //abit.push_back({ 1.0029 ,0 });
    std::cout << "Before:" << std::endl;
    for (int i = 0; i < N; i++)
    {
        //abit.push_back(data[0][i]);
        std::cout << abit[i] << std::endl;
    }
    std::cout << "After:" << std::endl;
    fft(abit);
    for (int i = 0; i < N; i++)
    {
        std::cout << abit[i] << std::endl;
    }
    return 0;
}

I'm getting the following output from CPP:

(9.02011,0)
(5.83089,-4.89513)
(0.700632,-3.98993)
(-0.000289979,0.000502368)
(-0.00218513,0.000362784)
(-0.00179241,0.00139188)
(-0.000289979,-0.000502368)
(0.000175771,-0.00354373)
(-0.003268,-0.00558837)

While I should get:

(9.020109999999999+0j)
(-0.0032675770104925446+0.005588577982060319j)
(-0.0023772289746976797+0.0024179090499282354j)
(-0.0022250000000012538+0.0011691342951078987j)
(-0.002185194014811494+0.00036271471530890747j)
(-0.0021851940148113033-0.00036271471530980844j)
(-0.0022249999999994774-0.0011691342951105632j)
(-0.002377228974696629-0.0024179090499291786j)
(-0.00326757701049002-0.005588577982061138j)

As you can see only the first result is correct.

I didn't do a full review but : Replace your floats with doubles and this part especialy looks suspect, an integer division in the middle of a floating point one in std::polar which is i / N try : static_cast(i)/static_cast(N), also replace -2 with -2.0, -4 with -4.0 etc that also changes the underlying type from int to double. — Pepijn Kramer, Oct 16 '21 at 14:52
What do you mean by underlying type from int to double? @PepijnKramer — yarin Cohen, Oct 16 '21 at 15:13
Roughly : C++ treats whole numbers as ints. And numbers with a . in it as floating points. A more complete overview and explanation can be found here : https://www.learncpp.com/cpp-tutorial/literals/ — Pepijn Kramer, Oct 16 '21 at 15:19
I see, always a good thing to learn new things, however I'm still looking for an answer as all of this did not fix it — yarin Cohen, Oct 16 '21 at 15:28
Have you debugged and are all your slices exactly as you want them? It is always good to test small bits of your code seperately. Here is how I would write the slice code : https://onlinegdb.com/R021-X2zf (don't worry about the template it will accept your std::vector> too — Pepijn Kramer, Oct 16 '21 at 16:12

Cris Luengo · Accepted Answer · 2021-10-16T15:51:29.777

0

In Python it says k % (N//3), in C++ it says i % (int)N / 3. These are not the same! See operator precedence in C++. You need to use parenthesis to execute the operations in the right order: i % ((int)N / 3).

Pepijn Kramer’s comment about changing -2 * M_PI * i / N into -2.0 * M_PI * static_cast<double>(i)/static_cast<double>(N) is good practice, in my opinion, but shouldn’t make a difference in this case because of the left-to-right evaluation and the automatic promotion of integers to floats in a float context. Still, ensuring all your constants are floats where needed doesn’t cost much effort and can prevent difficult to spot bugs.

edited Oct 16 '21 at 15:51

answered Oct 16 '21 at 15:01

Cris Luengo

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What is equivalent to `k % (N//3)` in C++? – yarin Cohen Oct 16 '21 at 15:17
1

@Cris Luengo Fair enough, years of experience have taught me to be precise, this by the way also includes adding extra ( ) even if operator precedence is clear. – Pepijn Kramer Oct 16 '21 at 15:22
@yarinCohen I’ve made the answer more explicit. It matters if you do the division or the modulo operation first. – Cris Luengo Oct 16 '21 at 15:52
@PepijnKramer I agree with you. Explicit and precise. I have a hard time remembering operator precedence, so tend to use tons of parenthesis. And I explicitly cast everything, have my compiler warn on implicit casts, it catches bugs some times. – Cris Luengo Oct 16 '21 at 15:54

Radix 3,5 FFT in C++

1 Answers1