# Haskell FFT 6: Implementing the Mixed-Radix FFT

[The code from this article is available as a Gist]

In this article, we’re going to implement the full “sorted prime factorisation mixed-radix decimation-in-time Cooley-Tukey FFT”. That’s a bit of a mouthful, and you might want to review some of the algebra in the last article to remind yourself of some of the details of how it works.

The approach we’re going to take to exploring the implementation of this mixed-radix FFT is to start with a couple of the smaller and simpler parts (prime factor decomposition, digit reversal ordering), then to look at the top-level driver function, and finally to look at the generalised Danielson-Lanczos step, which is the most complicated part of things. This code has an unavoidably large number of moving parts (which is probably why this algorithm is rarely covered in detail in textbooks!), so we’ll finish up by writing some QuickCheck properties to make sure that everything worksAn admission: it took me quite a while to get everything working correctly here. The powers-of-two FFT took about an hour to write an initial version and perhaps another hour to tidy things up. The mixed-radix FFT took several hours of off-line thinking time (mostly walking my dog), a couple of hours to put together an initial (non-working) version, then several more hours of debugging time..

Let’s start with a couple of type synonyms to make things more readable (typing `Vector (Complex Double)`

or `Vector (Vector (Complex Double))`

gets old quickly):

type CD = Complex Double type VCD = Vector CD type VVCD = Vector (Vector CD)

Calculating the prime factors for the decomposition of the input vector length is simple:

-- From Haskell wiki. primes :: [Int] primes = 2 : primes' where primes' = sieve [3, 5 ..] 9 primes' sieve (x:xs) q ps@ ~(p:t) | x < q = x : sieve xs q ps | True = sieve [x | x <- xs, rem x p /= 0] (P.head t^2) t -- Simple prime factorisation: small factors only; largest/last factor -- picked out as "special". factors :: Int -> (Int, Vector Int) factors n = let (lst, rest) = go n primes in (lst, fromList rest) where go cur pss@(p:ps) | cur == p = (p, []) | cur `mod` p == 0 = let (lst, rest) = go (cur `div` p) pss in (lst, p : rest) | otherwise = go cur ps

The only clever thing here is the efficient prime sieve used to generate an infinite list of primes: this is taken directly from the Haskell wiki page about primes. We treat the “last” prime factor specially, since it’s going to give us the size of the DFT calculations we’re going to do at the “bottom” of the recursive decomposition.

It’s probably worth saying a word or two about this special treatment of the last factor. Instead of treating the last factor specially, we could do one more decomposition step to end up with $N$ length-1 subvectors. The FFT of a length-1 vector is just the identityWhat else could it be? The DFT is a linear operation, and for a length-1 vector $v$, $(\mathcal{F}^{-1} \cdot \mathcal{F}) v = v$., and we could then compose those length-1 FFT results just as for the other levels of the decomposition. This is precisely what we did for the powers-of-two case: the `id`

in the definition of the `recomb`

function in the FFT calculation in article 3 is just this length-1 identity transform. In fact, if you work through the calculation, this lowest-level of the FFT decomposition is exactly equivalent to the naïve DFT that we’re going to use instead. The main motivation for treating this last factor specially and explicitly pulling out the simple DFT is that we may want to replace the “bottom level” DFT with a more sophisticated algorithm for prime-length vectors later on, and pulling it out here makes this more convenient.

A useful way to think of this prime factor decomposition is as a sort of “generalised place value” scheme for representing numbers. For example: if we want to represent 32 different values, we can do this exactly with five binary digits; if we want to represent 30 different values, they don’t quite “fit” into a binary scheme, but if we write numbers as $x_5 y_3 z_2$ where $x \in \{ 0, 1, 2, 3, 4 \}$, $y \in \{ 0, 1, 2 \}$, $z \in \{ 0, 1 \}$ and $z$ represents “units” in our place value system, $y$ represents “threes” and $x$ “sixes”, then we can write the numbers zero to thirty as:

$0_5 0_3 0_2$ | $0_5 0_3 1_2$ | $0_5 1_3 0_2$ | $0_5 1_3 1_2$ | $0_5 2_3 0_2$ | $0_5 2_3 1_2$ |

$1_5 0_3 0_2$ | $1_5 0_3 1_2$ | $1_5 1_3 0_2$ | $1_5 1_3 1_2$ | $1_5 2_3 0_2$ | $1_5 2_3 1_2$ |

$2_5 0_3 0_2$ | $2_5 0_3 1_2$ | $2_5 1_3 0_2$ | $2_5 1_3 1_2$ | $2_5 2_3 0_2$ | $2_5 2_3 1_2$ |

$3_5 0_3 0_2$ | $3_5 0_3 1_2$ | $3_5 1_3 0_2$ | $3_5 1_3 1_2$ | $3_5 2_3 0_2$ | $3_5 2_3 1_2$ |

$4_5 0_3 0_2$ | $4_5 0_3 1_2$ | $4_5 1_3 0_2$ | $4_5 1_3 1_2$ | $4_5 2_3 0_2$ | $4_5 2_3 1_2$ |

This prime factor digit approach gives us a kind of customised place value scheme for any value, and allows us to represent the generalisation of the even/odd splitting in the powers-of-two FFT as a removal of the lowest order digit in our place value representation–recall that the even/odd splitting resulted from taking the lowest order bit from the input data length each time we split the Fourier matrix using the Danielson-Lanczos lemma.

As for the powers-of-two case, where we needed to reorder our input data in bit-reversed order before doing the Danielson-Lanczos steps to get our FFT result, in the general case, we need a sort of “sorted prime factor decomposition digit reversal order”. That’s a bit long to type, so I’m just going to call it “digit reversed order” with the understanding that we’re only dealing with this sorted prime factor decomposition for the moment.

Here’s a simple example of this digit reversal ordering:

This case is particular simple because it represents an even/odd split (we’re splitting off a factor of two from the input data of length $N=6$), which we can see quite clearly from the pattern–even indexed elements and odd indexed elements are segregated by the rearrangement.

Although we could use a similar “place value” approach to reordering the data as we used for the $2^N$ FFT case, it turns out to be simpler to just compose appropriate permutations of indexes into the input vector. The bit-reversal approach gets a little confusing here because we need to keep track of the place values of each digit, and they change when we swap the digit order. The permutation code is a lot simpler to understand:

-- Generate digit reversal permutation using elementary "modulo" -- permutations: last digit is not permuted to match with using a -- simple DFT at the "bottom" of the overall algorithm. digrev :: Int -> (Int, Vector Int) -> Vector Int digrev n (lastf, fs) = foldl1' (%.%) $ map dupperm subperms where -- Sizes of the individual permutations that we need, one per -- factor. sizes = scanl div n fs -- Partial sub-permutations, one per factor. subperms = reverse $ zipWith perm sizes fs -- Duplicate a sub-permutation to fill the required vector length. dupperm p = let sublen = length p shift di = map (+(sublen * di)) p in concatMap shift $ enumFromN 0 (n `div` sublen) -- Generate a single "modulo" permutation. perm n fac = concatMap doone $ enumFromN 0 fac where n1 = n `div` fac doone i = generate n1 (\j -> j * fac + i) -- Composition of permutations. (%.%) :: Vector Int -> Vector Int -> Vector Int p1 %.% p2 = backpermute p2 p1

We construct basic “modulo” permutations, the generalisation to arbitrary factors of even/odd permutations, then compose them to form the final vector of indexes giving the digit reversed reordering of the input. The only extra wrinkle is that we need to duplicate sub-permutations for each lower recursive level of the Fourier matrix decomposition: if the input vector length is $30 = 2 \times 3 \times 5$, for instance, we split the original input vector into two vectors of length 15; we then split each of those two sub-vectors into three, so we need to duplicate the modulo-3 permutation to treat each of the sub-vectors in turn. This approach is relatively efficient, but it doesn’t really matter anyway since we eventually plan to only ever do this computation once at compile time. It doesn’t need to be optimised too much: it just needs to work!

Let’s step back a little now and look at the main driver function for the full mixed-radix FFT:

-- FFT and inverse FFT drivers. fft, ifft :: VCD -> VCD fft = fft' 1 1 ifft v = fft' (-1) (1.0 / (fromIntegral $ length v)) v -- Mixed-radix decimation-in-time Cooley-Tukey FFT. fft' :: Int -> Double -> VCD -> VCD fft' sign scale h = if n == 1 then h else if null fs then dft' sign scale h else map ((scale :+ 0) *) fullfft where -- Full vector length. n = length h -- Factorise input vector length. splitfs@(lastf, fs) = factors n -- Generate sizing information for Danielson-Lanczos step. wfacs = map (n `div`) $ scanl (*) 1 fs dlinfo = zip wfacs fs -- Compose all Danielson-Lanczos steps and "final" DFT. recomb = foldr (.) (mdft lastf) $ map (dl sign) dlinfo -- Apply Danielson-Lanczos steps and "final" DFT to digit reversal -- ordered input vector. fullfft = recomb $ backpermute h (digrev n splitfs) -- Multiple DFT for "bottom" of algorithm. mdft :: Int -> VCD -> VCD mdft factor h = concatMap (dft' sign 1) $ slicevecs factor h

This has a lot in common with the driver for the powers-of-two algorithm, but it’s complicated slightly by the need to keep track of the factors we’re using to decompose the input vector length. The key step is the definition of `recomb`

, where we compose the generalised Danielson-Lanczos steps for the Fourier matrix decomposition (represented by the function `dl`

, which we’ll get to in a moment), along with a “multiple” DFT to deal with the last level of the decomposition (“multiple” in the sense that it’s a size $M$ DFT repeated $N/M$ times to cover each of the decomposed sub-vectors in the input).

The generalised Danielson-Lanczos step itself looks like this:

-- Single Danielson-Lanczos step: process all duplicates and -- concatenate into a single vector. dl :: Int -> (Int, Int) -> VCD -> VCD dl sign (wfac, split) v = concatMap doone $ slicevecs wfac v where -- Overall vector length. n = length v -- Size of each diagonal sub-matrix. ns = wfac `div` split -- Root of unity for the size of diagonal matrix we need. w = omega $ sign * wfac -- Basic diagonal entries for a given column. ds c = map ((w^^) . (c *)) $ enumFromN 0 ns -- Twiddled diagonal entries in row r, column c (both -- zero-indexed), where each row and column if a wfac x wfac -- matrix. d r c = map (w^(ns*r*c) *) (ds c) -- Process one duplicate by processing all rows and concatenating -- the results into a single vector. doone v = concatMap single $ enumFromN 0 split where vs :: VVCD vs = slicevecs ns v -- Multiply a single block by its appropriate diagonal -- elements. mult :: Int -> Int -> VCD mult r c = zipWith (*) (d r c) (vs!c) -- Multiply all blocks by the corresponding diagonal -- elements in a single row. single :: Int -> VCD single r = foldl1' (zipWith (+)) $ map (mult r) $ enumFromN 0 split

This is kind of complicated, mostly because it involves keeping track of sub-matrices and sub-vectors all over the place, and it’s where the tricky “twiddle factor” calculation is done. To try to make things as clear as possible, the `d`

function is used to calculate the $D_j^{(i)}$ sub-matrices from the $I+D$ matrix explicitly as a function of the row and column of the sub-matrix in the full $I+D$ matrix. This corresponds exactly to how we laid this calculation out in the previous article (e.g. the calculations of the decompositions of $F_6$ and $F_9$ there). As for the “bottom level” DFT calculation, we need to duplicate the multiplication by the $I+D$ matrix across each of the sub-vectors we’re dealing with at the current level of the decomposition. For example, if we’re dealing with a vector of length 30, we split two ways at the first level, giving us two sub-vectors of length 15, *each of which* then has to be split three ways.

And that’s it. “That’s it?”, you say, in horror. “How do you know that this works?” There really is quite a lot going on in this algorithm. First, we need to trust that our prime factor decomposition of the Fourier matrix is correct. Hopefully, some experimentation with the “toy algebra system” code in the previous article is enough to gain some confidence on this point. We can build up the full decomposition of a Fourier matrix bit by bit, multiply them all together and compare to the full DFT matrix. And that does all work. But then we have prime factorisation, digit reversal ordering, the top-level driver and the generalised Danielson-Lanczos computation. There’s a lot that could go wrong.

Up-front thinking and algebra only gets us so far on this second point. In algorithms like this one, where there are multiple sizes of things floating around, there are different $\omega_N$ factors at different levels of the Fourier matrix decomposition, there are these tricksy twiddle factors, and so on, we really do need to do some testing. What are good tests in this case? First, we have the naïve DFT algorithm to serve as a benchmark. In all cases, the results from our FFT algorithm should be identical to those from the simple DFT, apart from minor differences due to the ordering of floating point operations. Second, the inverse FFT really should be the inverse of the forward FFT! The only difference between the forward and reverse versions of the algorithm is the overall scaling of the result and the sign of the exponent in the $\omega_N$ factors. Sign errors are notoriously easy to make, so this might be a good test–if we’ve made a sign error somewhere in the exponents, it’s unlikely that our inverse FFT will truly be the inverse of the forward FFT.

Both of these tests are stated in terms of properties that our algorithm should have for all input vectors. During the development phase of an algorithm like this, we might try our code with some simple “hand-made” input vectors, but if you only do that, it’s quite easy to convince yourself that your code works perfectly when in fact it *only* works for your simple test data. For an algorithm like this one, before we declare it to be working, we should turn to QuickCheck. Here are the definition of QuickCheck properties for the FFT vs. DFT (`prop_dft_vs_fft`

) and inverse FFT (`prop_ifft`

) tests, along with an `Arbitrary`

instance for `Vector (Complex Double)`

, which we need to use the tests:

-- Clean up number display. defuzz :: VCD -> VCD defuzz = map (\(r :+ i) -> df r :+ df i) where df x = if abs x < 1.0E-6 then 0 else x -- Check FFT against DFT. check :: VCD -> (Double, VCD) check v = let diff = defuzz $ zipWith (-) (fft v) (dft v) in (maximum $ map magnitude diff, diff) -- Check FFT-inverse FFT round-trip. icheck :: VCD -> (Double, VCD) icheck v = let diff = defuzz $ zipWith (-) v (ifft $ fft v) in (maximum $ map magnitude diff, diff) -- QuickCheck property for FFT vs. DFT testing. prop_dft_vs_fft (v :: VCD) = fst (check v) < 1.0E-6 -- QuickCheck property for inverse FFT round-trip testing. prop_ifft (v :: VCD) = maximum (map magnitude diff) < 1.0E-6 where diff = zipWith (-) v (ifft $ fft v) -- Non-zero length arbitrary vectors. instance Arbitrary VCD where arbitrary = fromList <$> listOf1 arbitrary

Once we have these things, we can do this in GHCi:

> :load FFT-v2.hs [1 of 1] Compiling Main ( FFT-v2.hs, interpreted ) Ok, modules loaded: Main. > import Test.QuickCheck > quickCheck prop_dft_vs_fft +++ OK, passed 100 tests. > quickCheck prop_ifft +++ OK, passed 100 tests.

Of course, random testing of this sort is no *proof* that our code is correct, but it provides strong confidence that things are working.

In the next article, we’ll take a bit of a breather to summarise where we’ve got to, and to lay out what we’re going to do next.