Fast Fourier Transform: Scaling Multi-Point Evaluation

Trigger warning: specialized mathematical topic

Special thanks to Karl Floersch for feedback

Background

Ok fine, Fourier transforms also have really important applications in signal processing, quantum mechanics, and other areas, and help make significant parts of the global economy happen. But come on, elephants are cooler.

The kind of Fourier transform we'll be talking about in this post is a similar algorithm, except instead of being a continuous Fourier transform over real or complex numbers, it's a discrete Fourier transform over finite fields (see the "A Modular Math Interlude" section  for a refresher on what finite fields are).

Instead of talking about converting between "frequency domain" and "time domain", here we'll talk about two different operations: multi-point polynomial evaluation (evaluating a degree <N polynomial at N different points) and its inverse, polynomial interpolation (given the evaluations of a degree <N polynomial at N different points, recovering the polynomial). For example, if we are operating in the prime field with modulus 5, then the polynomial y=x²+3 (for convenience we can write the coefficients in increasing order: [3,0,1]) evaluated at the points [0,1,2] gives the values [3,4,2] (not [3,4,7] because we're operating in a finite field where the numbers wrap around at 5), and we can actually take the evaluations [3,4,2] and the coordinates they were evaluated at ([0,1,2]) to recover the original polynomial [3,0,1].

There are algorithms for both multi-point evaluation and interpolation that can do either operation in O(N²) time. Multi-point evaluation is simple: just separately evaluate the polynomial at each point. Here's python code for doing that:

def eval_poly_at(self, poly, x, modulus):
    y = 0
    power_of_x = 1
    for coefficient in poly:
        y += power_of_x * coefficient
        power_of_x *= x
    return y % modulus

The algorithm runs a loop going through every coefficient and does one thing for each coefficient, so it runs in O(N) time. Multi-point evaluation involves doing this evaluation at N different points, so the total run time is O(N²).

Lagrange interpolation is more complicated (search for "Lagrange interpolation"  for a more detailed explanation). The key building block of the basic strategy is that for any domain D and point x, we can construct a polynomial that returns 1 for x and 0 for any value in D other than x. For example, if D=[1,2,3,4] and x=1, the polynomial is:

You can mentally plug in 1, 2, 3 and 4 to the above expression and verify that it returns 1 for x=1 and 0 in the other three cases.

We can recover the polynomial that gives any desired set of outputs on the given domain by multiplying and adding these polynomials. If we call the above polynomial P1, and the equivalent ones for x=2, x=3, x=4, P2, P3 and P4, then the polynomial that returns [3,1,4,1] on the domain [1,2,3,4] is simply 3⋅P1+P2+4⋅P3+P4. Computing the Pi polynomials takes O(N²) time (you first construct the polynomial that returns to 0 on the entire domain, which takes O(N²) time, then separately divide it by (x−xi) for each xi), and computing the linear combination takes another O(N²) time, so it's O(N²) runtime total.

Fast Fourier Transforms

There is a price you have to pay for using this much faster algorithm, which is that you cannot choose any arbitrary field and any arbitrary domain. Whereas with Lagrange interpolation, you could choose whatever x coordinates and y coordinates you wanted, and whatever field you wanted (you could even do it over plain old real numbers), and you could get a polynomial that passes through them., with an FFT, you have to use a finite field, and the domain must be a multiplicative subgroup of the field (that is, a list of powers of some "generator" value).

For example, you could use the finite field of integers modulo 337, and for the domain use [1,85,148,111,336,252,189,226] (that's the powers of 85 in the field, eg. 85^3 % 337=111; it stops at 226 because the next power of 85 cycles back to 1). Futhermore, the multiplicative subgroup must have size 2n (there's ways to make it work for numbers of the form 2^m⋅3^n and possibly slightly higher prime powers but then it gets much more complicated and inefficient). The finite field of intergers modulo 59, for example, would not work, because there are only multiplicative subgroups of order 2, 29 and 58; 2 is too small to be interesting, and the factor 29 is far too large to be FFT-friendly. The symmetry that comes from multiplicative groups of size 2^n lets us create a recursive algorithm that quite cleverly calculate the results we need from a much smaller amount of work.

To understand the algorithm and why it has a low runtime, it's important to understand the general concept of recursion. A recursive algorithm is an algorithm that has two cases: a "base case" where the input to the algorithm is small enough that you can give the output directly, and the "recursive case" where the required computation consists of some "glue computation" plus one or more uses of the same algorithm to smaller inputs. For example, you might have seen recursive algorithms being used for sorting lists. If you have a list (eg. [1,8,7,4,5,6,3,2,9]), then you can sort it using the following procedure:

• If the input has one element, then it's already "sorted", so you can just return the input.

• If the input has more than one element, then separately sort the first half of the list and the second half of the list, and then merge the two sorted sub-lists (call them A and B) as follows. Maintain two counters, apos and bpos, both starting at zero, and maintain an output list, which starts empty. Until either apos or bpos is at the end of the corresponding list, check if A[apos] or B[bpos] is smaller. Whichever is smaller, add that value to the end of the output list, and increase that counter by 1. Once this is done, add the rest of whatever list has not been fully processed to the end of the output list, and return the output list.

Note that the "glue" in the second procedure has runtime O(N): if each of the two sub-lists has N elements, then you need to run through every item in each list once, so it's O(N) computation total. So the algorithm as a whole works by taking a problem of size N, and breaking it up into two problems of size N2, plus O(N) of "glue" execution.

There is a theorem called the  that lets us compute the total runtime of algorithms like this. It has many sub-cases, but in the case where you break up an execution of size N into k sub-cases of size N/k with O(N) glue (as is the case here), the result is that the execution takes time O(N⋅log(N)).

An FFT works in the same way. We take a problem of size N, break it up into two problems of size N2, and do O(N) glue work to combine the smaller solutions into a bigger solution, so we get O(N⋅log(N)) runtime total - much faster than O(N2). Here is how we do it. I'll describe first how to use an FFT for multi-point evaluation (ie. for some domain D and polynomial P, calculate P(x) for every x in D), and it turns out that you can use the same algorithm for interpolation with a minor tweak.

Suppose that we have an FFT where the given domain is the powers of x in some field, where x^2^k=1 (eg. in the case we introduced above, the domain is the powers of 85 modulo 337, and 85^2^3=1). We have some polynomial, eg. y=6x7+2x6+9x5+5x4+x3+4x2+x+3 (we'll write it as p=[3,1,4,1,5,9,2,6]). We want to evaluate this polynomial at each point in the domain, ie. at each of the eight powers of 85. Here is what we do.

First, we break up the polynomial into two parts, which we'll call evens and odds: evens=[3,4,5,2] and odds [1,1,9,6] (or evens=2x3+5x2+4x+3 and odds=6x3+9x2+x+1; yes, this is just taking the even-degree coefficients and the odd-degree coefficients). Now, we note a mathematical observation: p(x)=evens(x2)+x⋅odds(x2) and p(−x)=evens(x2)−x⋅odds(x2) (think about this for yourself and make sure you understand it before going further).

The "glue" is relatively easy (and O(N) in runtime): we receive the evaluations of evens and odds as size-N/2 lists, so we simply do p[i]=evens_result[i]+domain[i]⋅odds_result[i] and p[N2+i]=evens_result[i]−domain[i]⋅odds_result[i] for each index i.

def fft(vals, modulus, domain):
    if len(vals) == 1:
        return vals
    L = fft(vals[::2], modulus, domain[::2])
    R = fft(vals[1::2], modulus, domain[::2])
    o = [0 for i in vals]
    for i, (x, y) in enumerate(zip(L, R)):
        y_times_root = y*domain[i]
        o[i] = (x+y_times_root) % modulus
        o[i+len(L)] = (x-y_times_root) % modulus
    return o

>>> fft([3,1,4,1,5,9,2,6], 337, [1, 85, 148, 111, 336, 252, 189, 226])
[31, 70, 109, 74, 334, 181, 232, 4]

And we can check the result; evaluating the polynomial at the position 85, for example, actually does give the result 70. Note that this only works if the domain is "correct"; it needs to be of the form [x^i % modulus for i in range(n)] where x^n=1.

def inverse_fft(vals, modulus, domain):
    vals = fft(vals, modulus, domain)
    return [x * modular_inverse(len(vals), modulus) % modulus for x in [vals[0]] + vals[1:][::-1]]

>>> domain = [1, 85, 148, 111, 336, 252, 189, 226]
>>> def modular_inverse(x, n): return pow(x, n - 2, n)
>>> values = fft([3,1,4,1,5,9,2,6], 337, domain)
>>> values
[31, 70, 109, 74, 334, 181, 232, 4]
>>> inverse_fft(values, 337, domain)
[3, 1, 4, 1, 5, 9, 2, 6]

Now, what can we use this for? Here's one fun use case: we can use FFTs to multiply numbers very quickly. Suppose we wanted to multiply 1253 by 1895. Here is what we would do. First, we would convert the problem into one that turns out to be slightly easier: multiply the polynomials [3,5,2,1] by [5,9,8,1] (that's just the digits of the two numbers in increasing order), and then convert the answer back into a number by doing a single pass to carry over tens digits.

We can multiply polynomials with FFTs quickly, because it turns out that if you convert a polynomial into evaluation form (ie. f(x) for every x in some domain D), then you can multiply two polynomials simply by multiplying their evaluations. So what we'll do is take the polynomials representing our two numbers in coefficient form, use FFTs to convert them to evaluation form, multiply them pointwise, and convert back:

>>> p1 = [3,5,2,1,0,0,0,0]
>>> p2 = [5,9,8,1,0,0,0,0]
>>> x1 = fft(p1, 337, domain)
>>> x1
[11, 161, 256, 10, 336, 100, 83, 78]
>>> x2 = fft(p2, 337, domain)
>>> x2
[23, 43, 170, 242, 3, 313, 161, 96]
>>> x3 = [(v1 * v2) % 337 for v1, v2 in zip(x1, x2)]
>>> x3
[253, 183, 47, 61, 334, 296, 220, 74]
>>> inverse_fft(x3, 337, domain)
[15, 52, 79, 66, 30, 10, 1, 0]

This requires three FFTs (each O(N⋅log(N)) time) and one pointwise multiplication (O(N) time), so it takes O(N⋅log(N)) time altogether (technically a little bit more than O(N⋅log(N)), because for very big numbers you would need replace 337 with a bigger modulus and that would make multiplication harder, but close enough). This is much faster than schoolbook multiplication, which takes O(N2) time:

     3  5  2  1
   ------------
5 | 15 25 10  5
9 |    27 45 18  9
8 |       24 40 16  8
1 |           3  5  2  1
   ---------------------
    15 52 79 66 30 10  1

So now we just take the result, and carry the tens digits over (this is a "walk through the list once and do one thing at each point" algorithm so it takes O(N) time):

[15, 52, 79, 66, 30, 10, 1, 0]
[ 5, 53, 79, 66, 30, 10, 1, 0]
[ 5,  3, 84, 66, 30, 10, 1, 0]
[ 5,  3,  4, 74, 30, 10, 1, 0]
[ 5,  3,  4,  4, 37, 10, 1, 0]
[ 5,  3,  4,  4,  7, 13, 1, 0]
[ 5,  3,  4,  4,  7,  3, 2, 0]

And if we read the digits from top to bottom, we get 2374435. Let's check the answer....

>>> 1253 * 1895
2374435

Yay! It worked. In practice, on such small inputs, the difference between O(N⋅log(N)) and O(N^2) isn't that large, so schoolbook multiplication is faster than this FFT-based multiplication process just because the algorithm is simpler, but on large inputs it makes a really big difference.

FFTs and binary fields

Prime fields are not the only kind of finite field out there. Another kind of finite field (really a special case of the more general concept of an extension field, which are kind of like the finite-field equivalent of complex numbers) are binary fields. In an binary field, each element is expressed as a polynomial where all of the entries are 0 or 1, eg. x^3+x+1.

Adding polynomials is done modulo 2, and subtraction is the same as addition (as −1=1mod2). We select some irreducible polynomial as a modulus (eg. x^4+x+1; x^4+1 would not work because x^4+1 can be factored into (x2+1)⋅(x2+1) so it's not "irreducible"); multiplication is done modulo that modulus. For example, in the binary field mod x^4+x+1, multiplying x2+1 by x3+1 would give x5+x3+x2+1 if you just do the multiplication, but x^5+x^3+x^2+1=(x^4+x+1)⋅x+(x^3+x+1), so the result is the remainder x^3+x+1.

We can express this example as a multiplication table. First multiply [1,0,0,1] (ie. x^3+1) by [1,0,1] (ie. x^2+1):

    1 0 0 1
   --------
1 | 1 0 0 1
0 |   0 0 0 0
1 |     1 0 0 1
   ------------
    1 0 1 1 0 1

The multiplication result contains an x^5 term so we can subtract (x4+x+1)⋅x:

   1 0 1 1 0 1
  -   1 1 0 0 1    [(x⁴ + x + 1) shifted right by one to reflect being multipled by x]
   ------------
    1 1 0 1 0 0 

And we get the result, [1,1,0,1] (or x3+x+1).

Addition and multiplication tables for the binary field mod x^4+x+1. Field elements are expressed as integers converted from binary (eg. x^3+x^2→1100→12)

Binary fields are interesting for two reasons. First of all, if you want to erasure-code binary data, then binary fields are really convenient because N bytes of data can be directly encoded as a binary field element, and any binary field elements that you generate by performing computations on it will also be N bytes long. You cannot do this with prime fields because prime fields' size is not exactly a power of two; for example, you could encode every 2 bytes as a number from 0...65536 in the prime field modulo 65537 (which is prime), but if you do an FFT on these values, then the output could contain 65536, which cannot be expressed in two bytes.

Second, the fact that addition and subtraction become the same operation, and 1+1=0, create some "structure" which leads to some very interesting consequences. One particularly interesting, and useful, oddity of binary fields is the "" theorem: (x+y)^2=x^2+y^2 (and the same for exponents 4,8,16... basically any power of two).

But if you want to use binary fields for erasure coding, and do so efficiently, then you need to be able to do Fast Fourier transforms over binary fields. But then there is a problem: in a binary field, there are no (nontrivial) multiplicative groups of order 2^n. This is because the multiplicative groups are all order 2^n-1. For example, in the binary field with modulus x^4+x+1, if you start calculating successive powers of x+1, you cycle back to 1 after 15 steps - not 16. The reason is that the total number of elements in the field is 16, but one of them is zero, and you're never going to reach zero by multiplying any nonzero value by itself in a field, so the powers of x+1 cycle through every element but zero, so the cycle length is 15, not 16. So what do we do?

The reason we needed the domain to have the "structure" of a multiplicative group with 2n elements before is that we needed to reduce the size of the domain by a factor of two by squaring each number in it: the domain [1,85,148,111,336,252,189,226] gets reduced to [1,148,336,189] because 1 is the square of both 1 and 336, 148 is the square of both 85 and 252, and so forth.

But what if in a binary field there's a different way to halve the size of a domain? It turns out that there is: given a domain containing 2^k values, including zero (technically the domain must be a ), we can construct a half-sized new domain D′ by taking x⋅(x+k) for x in D using some specific k in D. Because the original domain is a subspace, since k is in the domain, any x in the domain has a corresponding x+k also in the domain, and the function f(x)=x⋅(x+k) returns the same value for x and x+k so we get the same kind of two-to-one correspondence that squaring gives us.

So now, how do we do an FFT on top of this? We'll use the same trick, converting a problem with an N-sized polynomial and N-sized domain into two problems each with an N/2-sized polynomial and N/2-sized domain, but this time using different equations. We'll convert a polynomial p into two polynomials evens and odds such that p(x)=evens(x⋅(k−x))+x⋅odds(x⋅(k−x)). Note that for the evens and odds that we find, it will also be true that p(x+k)=evens(x⋅(k−x))+(x+k)⋅odds(x⋅(k−x)). So we can then recursively do an FFT to evens and odds on the reduced domain [x⋅(k−x) for x in D], and then we use these two formulas to get the answers for two "halves" of the domain, one offset by k from the other.

Converting p into evens and odds as described above turns out to itself be nontrivial. The "naive" algorithm for doing this is itself O(N^2), but it turns out that in a binary field, we can use the fact that (x2−kx)2=x4−k2⋅x2, and more generally (x2−kx)2i=x2i+1−k2i⋅x2i , to create yet another recursive algorithm to do this in O(N⋅log(N)) time.

And if you want to do an inverse FFT, to do interpolation, then you need to run the steps in the algorithm in reverse order. You can find the complete code for doing this here: , and a paper with details on more optimal algorithms here: 

So what do we get from all of this complexity? Well, we can try running the implementation, which features both a "naive" O(N^2) multi-point evaluation and the optimized FFT-based one, and time both. Here are my results:

>>> import binary_fft as b
>>> import time, random
>>> f = b.BinaryField(1033)
>>> poly = [random.randrange(1024) for i in range(1024)]
>>> a = time.time(); x1 = b._simple_ft(f, poly); time.time() - a
0.5752472877502441
>>> a = time.time(); x2 = b.fft(f, poly, list(range(1024))); time.time() - a
0.03820443153381348

And as the size of the polynomial gets larger, the naive implementation (_simple_ft) gets slower much more quickly than the FFT:

>>> f = b.BinaryField(2053)
>>> poly = [random.randrange(2048) for i in range(2048)]
>>> a = time.time(); x1 = b._simple_ft(f, poly); time.time() - a
2.22434
>>> a = time.time(); x2 = b.fft(f, poly, list(range(2048))); time.time() - a
0.07896280288696289

And voila, we have an efficient, scalable way to multi-point evaluate and interpolate polynomials. If we want to use FFTs to recover erasure-coded data where we are missing some pieces, then algorithms for this , though they are somewhat less efficient than just doing a single FFT. Enjoy!

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