15-451 11/30/04
* Multiplying polynomials
* Fast Fourier Transform (FFT) Handout: mini 5
===========================================================================
Today we are going to talk about the Fast Fourier Transform, a widely
used algorithm in areas like signal processing, speech understanding,
digital audio, radar.
We'll develop it in trying to solve the problem of "how to multiply
quickly" we talked about on the first day of class. This is not how
it was invented historically (and it obscures the connection to the
usual kind of Fourier Transform), but I think it's most natural from the
perspective of Algorithms.
Warning: even though the algorithm in the end won't be that complicated,
this will be one of the most difficult lectures of the class!
===================================================================
At the start of class we talked about the problem of multiplying big
integers. Let's look at a simpler version of the problem,
which is the problem of multiplying polynomials. It sounds more
complicated, but really it's just the problem of multiplying integers
without the carries.
MULTIPLYING POLYNOMIALS
=======================
E.g., multiply (x^2 + 2x + 3)(2x^2 + 5) =
2x^2 + 0x + 5
1x^2 + 2x + 3
-------------
6 0 15
4 0 10
2 0 5
----------------------------
2x^4 + 4x^3 + 11x^2 +10x+15
MODEL: view each individual small multiplication as a unit cost operation.
More generally, given A(x) = a_0 + a_1 x + ... + a_{n-1}x^{n-1},
B(x) = b_0 + b_1 x + ... + b_{n-1}x^{n-1}
Our goal is to compute the polynomial C(x) = A(x)*B(x).
c_i = a_0*b_i + a_1*b_{i-1} + ... + a_i*b_0.
If we think of A and B as vectors, then the C-vector is called the
"convolution" of A and B.
- Straightforward computation is O(n^2) time. Karatuba is n^{1.58..}
- we'll use FFTs to do in O(n log n) time. This is then used in
Schonhage-Strassen integer multiplication algorithm that multiplies
two n-bit integers in O(n log n loglog n) time. We're only going to do
polynomial multiplication.
High Level Idea of Algorithm
============================
Let m = 2n-1. [so degree of C is less than m]
1. Pick m points x_0, x_1, ..., x_{m-1} according to a secret formula.
2. Evaluate A at each of the points: A(x_0),..., A(x_{m-1}).
3. Same for B.
4. Now compute C(x_0),..., C(x_{m-1}), where C is A(x)*B(x)
5. Interpolate to get the coefficients of C.
This approach is based on the fact that a polynomial of degree < m is
uniquely specified by its value on m points. It seems patently crazy
since it looks like steps 2 and 3 should take O(n^2) time just in
themselves. However, the FFT will allow us to quickly move from
"coefficient representation" of polynomial to the "value on m points"
representation, and back, for our special set of m points. (Doesn't
work for *arbitrary* m points. The special points will turn out to be
roots of unity).
The reason we like this is that multiplying is easy in the "value on m
points" representation. We just do: C(x_i) = A(x_i)*B(x_i). So, only
O(m) time for step 4.
Let's focus on forward direction first. In that case, we've reduced
our problem to the following:
GOAL: Given a polynomial A of degree < m, evaluate A at m points of our
choosing in total time O(m log m). We can assume m is a power
of 2 (if it's not we just pad with zeroes).
The FFT:
=======
Let's first develop it through an example. Say m=8 so we have a polynomial
A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + a_6x^6 + a_7x^7.
And we want to evaluate at eight points of our choosing. Here is an
idea. Let A_even be a degree < m/2 polynomial defined by the even
coefficients, and A_odd be a degree < m/2 poly defined by the odd
coeffs. I.e.,
A_even(x) = a_0 + a_2 x + a_4 x^2 + a_6 x^3
A_odd(x) = a_1 + a_3 x + a_5 x^2 + a_7 x^3.
How can we write A(x) in terms of A_even and A_odd?
A(x) = A_even(x^2) + x A_odd(x^2).
What's nice is that the effort spent computing A(x) will give us A(-x)
almost for free. So, let's say our special set of m points will have
the property:
The 2nd half of points are the negative of the 1st half (*)
E.g., {1,2,3,4,-1,-2,-3,-4}.
Now, things look good: Let T(m) = time to evaluate a degree-m
polynomial at our special set of m points. We're doing this by
evaluating two degree-m/2 polynomials at m/2 points each (the
squares), and then doing O(m) work to combine the results. This is
great because the recurrence T(m) = 2T(m/2) + O(m) solves to O(m log m).
But, we're deluding ourselves by saying "just do it recursively".
Why is that? The problem is that recursively, our special points (now
{1, 4, 9, 16}) have to satisfy property (*). E.g., they should really
look like {1, 4, -1, -4}. BUT THESE ARE SQUARES!! How to fix? Just
use complex numbers! E.g., if these are the squares, what do the
original points look like?
{1, 2, i, 2i, -1, -2, -i, -2i}
so then their squares are: 1, 4, -1, -4
and their squares are: 1, 16
But, at the next level we again need property (*). So, we want to
have {1, -1} there. This means we want the level before that to be
{1, i, -1, -i}, which is the same as {1, i, i^2, i^3}. So, for the
original level, let w = sqrt(i) = 0.707 + 0.707i, and then
our original set of points will be:
1, w, w^2, w^3, w^4 (= -1), w^5 (= -w), w^6 (= -w^2), w^7 (= -w^3)
so that the squares are: 1, i, i^2 (= -1), i^3 (= -i)
and *their* squares are: 1, -1
and *their* squares are: 1
The "w" we are using is called the "primitive eighth root of unity"
(since w^8 = 1 and w^k != 1 for 0 < k < 8).
In general, the mth primitive root of unity is the vector
w = cos(2*pi/m) + i*sin(2*pi/m)
Alternatively, we can use MODULAR ARITHMETIC!
============================================
E.g., 2 is a primitive 8th root of unity mod 17.
{2^0,2^1,2^2,...,2^7} = {1,2,4,8 16,15,13, 9}
= {1,2,4,8,-1,-2,-4,-8}.
Then when you square them, you get {1,4,-1,-4}, etc.
This is nice because we don't need to deal with messy floating-points.
THE FFT ALGORITHM
=================
Here is the general algorithm in pseudo-C:
Let A be array of length m, w be primitive mth root of unity.
Goal: produce DFT F(A): evaluation of A at 1, w, w^2,...,w^{m-1}.
FFT(A, m, w)
{
if (m==1) return vector (a_0)
else {
A_even = (a_0, a_2, ..., a_{m-2})
A_odd = (a_1, a_3, ..., a_{m-1})
F_even = FFT(A_even, m/2, w^2) //w^2 is a primitive m/2-th root of unity
F_odd = FFT(A_odd, m/2, w^2)
F = new vector of length m
x = 1
for (j=0; j < m/2; ++j) {
F[j] = F_even[j] + x*F_odd[j]
F[j+m/2] = F_even[j] - x*F_odd[j]
x = x * w
}
return F
}
THE INVERSE OF THE FFT
======================
Remember, we started all this by saying that we were going to multiply
two polynomials A and B by evaluating each at a special set of m
points (which we can now do in time O(m log m)), then multiply the
values point-wise to get C evalauated at all these points (in O(m)
time) but then we need to interpolate back to get the coefficients.
In other words, we're doing F^{-1}(F(A) \cdot F(B)).
So, we need to compute F^{-1}. Here's how.
First, we can look at the forward computation (computing A(x) at 1, w,
w^2, ..., w^{m-1}) as an implicit matrix-vector product:
+---------------------------------------+ +-----+ +-------+
| 1 1 1 1 ... 1 | | a_0 | | A(1) |
| 1 w w^2 w^3 ... w^{m-1} | | a_1 | | A(w) |
| 1 w^2 w^4 w^6 ... w^{2(m-1)} | | a_2 | == | A(w^2)|
| 1 w^3 w^6 w^9 ... w^{3(m-1)} | | a_3 | | A(w^3)|
| ..... ..... | | ... | | ... |
| 1 w^{-1} w^{-2} w^{-3} ... w | |a_m-1| | ... |
+---------------------------------------+ +-----+ +-------+
(Note w^{m-1} = w^{-1} since w^m = 1)
(We're doing this "implicitly" in the sense that we don't even have
time to write down the matrix.)
To do the inverse transform, what we want is to multiply by the
inverse of the F matrix. As it turns out, this inverse looks very
much like F itself. In particular, notice that w^{-1} is also a
principal mth root of unity. Let's define \bar{F} to be the fourier
transform matrix using w^{-1} instead of w. Then,
Claim: F^{-1} = (1/m) * \bar{F}. I.e., 1/m * \bar{F} * F = identity.
Proof: What is the i,j entry of \bar{F} * F? It is:
1 + w^{j-i} + w^{2j-2i} + w^{3j-3i} + ... + w^{(m-1)j - (m-1)i}
If i=j, then these are all = 1, so the sum is m. Then when we
divide by m we get 1.
If i!=j, then the claim is these all cancel out and we get zero.
Maybe easier to see if we let z = w^{j-i}, so then the sum is:
1 + z + z^2 + z^3 + z^4 + ... + z^{m-1}.
Then can see these cancel by picture. For instance, try z = w, z = w^2.
Or can use the formula for summations: (1 - z^m)/(1-z) = 0/(1-z) = 0.
So, the final algorithm is:
Let F_A = FFT(A, m, w) // time O(n log n)
Let F_B = FFT(B, m, w) // time O(n log n)
Let F_C = F_A . F_B [ "." is dot-product] // time O(n)
Output C = 1/m * FFT(F_C, m, w^{-1}). // time O(n log n)
NOTE: If you're an EE or Physics person, what we're calling the "Fourier
Transform" is what you would usually call the "inverse Fourier
transform" and vice-versa.