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Show Source |    | About   «  24.2. The Transformation Concept   ::   Contents   ::   25.1. Introduction to Probabilistic Algorithms  »

24.3. The Fast Fourier Transform

24.3.1. The Fast Fourier Transform

In this module we continue the discussion on how to speed up the multiplication of larg polyonmials. Recall that we can get the result of multiplying two polynomials by the process of evaluating both at a sufficient number of points, doing pair-wise multiplication on the evaluation values, and then using interpolation to construct the solution polynomial from the resulting values. Doing this at arbitrary points on the polynomials is no faster than brute-force multiplication, since both evaluation and interpolation of \(n\) points will normally take \(\Theta(n^2)\) time. But in this module we show that we can find a way to take advantage of symmetry to speed up the process.

Evaluating any polynomial at 0 is easy, since only the constant term is non-zero. Evaluating at either 1 or -1 is relatively easy, because we don't need to actually multiply the \(x\) values. If we evaluate the polynomial at both 1 and -1, we can share a lot of the work between the two evaluations. Consider again multiplying polynomials \(A = x^2 + 1\) and \(B = 2x^2 -x + 1\). Since the end result is a degree-4 polynomial, we would need five points to nail down polynomial \(AB\). Fortunately, we can speed processing for any pair of values \(c\) and \(-c\). This seems to indicate some promising ways to speed up the process of evaluating polynomials. But, evaluating two points in roughly the same time as evaluating one point only speeds the process by a constant factor. Is there some way to generalized these observations to speed things up further? And even if we do find a way to evaluate many points quickly, we will also need to interpolate the five values to get the coefficients of \(AB\) back.

So we see that we could multiply two polynomials in less than \(\Theta(n^2)\) operations if a fast way could be found to do evaluation/interpolation of \(2n + 1\) points. Before considering further how this might be done, first observe again the relationship between evaluating a polynomial at values \(c\) and \(-c\). In general, we can write \(P_a(x) = E_a(x) + O_a(x)\) where \(E_a\) is the even powers and \(O_a\) is the odd powers. That is,

\[P_a(x) = \sum_{i=0}^{n/2-1} a_{2i} x^{2i} + \sum_{i=0}^{n/2-1} a_{2i+1} x^{2i+1} = E_a(x) + O_a(x)\]
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The key to fast polynomial multiplication is finding the right points to use for evaluation/interpolation to make the process efficient. In particular, we want to take advantage of symmetries, such as the one we see for evaluating \(x\) and \(-x\). But this symmetry is only enough to let us do selected pairs of points in half the time, so only a constant factor in savings. We need to find even more symmetries between points if we want to do more than cut the work in half. We have to find symmetries not just between pairs of values, but also further symmetries between pairs of pairs, and then pairs of pairs of pairs, and so on.

Recall that a complex number \(z\) has a real component and an imaginary component. We can consider the position of \(z\) on a number line if we use the \(y\) dimension for the imaginary component. Now, we will define a \(z\) to be a primitive nth root of unity if

  1. \(z^n = 1\) and
  2. \(z^k \neq 1\) for \(0 < k < n\).

\(z^0, z^1, ..., z^{n-1}\) are called the nth roots of unity. For example, when \(n=4\), then \(z = i\) because \(i^4 = 1\). The following identities will also be useful to us: \(e^{i\pi} = -1\), and \(z^j = e^{2\pi ij/n} = -1^{2j/n}\). The significance is that we can find as many points on a unit circle as we would need (see Figure 24.3.1). But these points are special in that they will allow us to do just the right computation necessary to get the needed symmetries to speed up the overall process of evaluating many points at once.

Figure 24.3.1: Examples of the 4th, 5th, and 8th roots of unity.

Now we want to turn these ideas into an actual, detailed algorithm. This process will be easier to both understand and implement if we assume that the number of coefficients is a power of two, so we will assume that this is the case. (We can always fill out the polynomials to be the proper size by adding zero-valued coefficients.)

Define an \(n \times n\) matrix \(A_{z}\) with row \(i\) and column \(j\) as

\[A_{z}[i,j] = (z^{ij}).\]

The idea is that there is a row for each root (row \(i\) for \(z^i\)) while each column corresponds to the power of the exponent of the \(x\) value in the polynomial. For example, when \(n = 4\) we have \(z = i\). Thus, the \(A_{z}\) array appears as follows.

Let \(a = [a_0, a_1, ..., a_{n-1}]^T\) be a vector that stores the coefficients for the polynomial being evaluated. We can then do the calculations to evaluate the polynomial at the \(n\) th roots of unity by multiplying the \(A_{z}\) matrix by the coefficient vector. The resulting vector \(F_{z}\) is called the Discrete Fourier Transform (DFT) for the polynomial. (Note that we also use the name \(b\) for \(F_z\), just to make the subscripting notation easier to read in our descriptions.)

\[F_{z} = b = A_{z}a.\] \[b_i = \sum_{k=0}^{n-1} a_kz^{ik}.\]
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We still have two problems. We need to be able to multiply this matrix and the vector faster than just by performing a standard matrix-vector multiplication, otherwise the cost is still \(n^2\) multiplies to do the evaluation. Even if we can multiply the matrix and vector cheaply, we still need to be able to reverse the process. That is, after transforming the two input polynomials by evaluating them, and then pair-wise multiplying the evaluated points, we must interpolate those points to get the resulting polynomial back that corresponds to multiplying the original input polynomials.

Let's get the second problem out of the way first. It turns out that the interpolation step is nearly identical to the evaluation step.

\[F_{z}^{-1} = A_{z}^{-1}b' = a'.\]

We just need to find \(A_{z}^{-1}\). This turns out to be simple to compute, and is defined as follows.

\[A_{z}^{-1} = \frac{1}{n}A_{1/z}.\]

In other words, interpolation (the inverse transformation) requires the same computation as evaluation, except that we substitute \(1/z\) for \(z\) (and multiply by \(1/n\) at the end). So, if we can do one of these steps fast, we can also do the other step fast.

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If you examine the example \(A_z\) matrix for \(n=8\), you should see that there are symmetries within the matrix. For example, the top half is identical to the bottom half with suitable sign changes on some rows and columns. Likewise for the left and right halves. An efficient divide and conquer algorithm exists to perform both the evaluation and the interpolation in \(\Theta(n \log n)\) time. This is called the Fast Fourier Transform. It is a recursive function that decomposes the matrix multiplications, taking advantage of the symmetries made available by doing evaluation at the \(n\) th roots of unity.

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Todo

type:AV

Practice fft algorithm. Maybe only practice the final for loop since the rest of the algorithm is recursivly dividing the polynomial and performing fft on the smaller polynomials.

Thus, the full process for multiplying polynomials \(A\) and \(B\) using the Fourier transform is as follows.

  1. Represent an \(n-1\) -degree polynomial as \(2n-1\) coefficients:

    \[[a_0, a_1, ..., a_{n-1}, 0, ..., 0]\]
  2. Perform Fourier transform on the representations for \(A\) and \(B\)

  3. Pairwise multiply the results to get \(2n-1\) values.

  4. Perform the inverse Fourier transform to get the \(2n-1\) degree polynomial \(AB\).

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