Parallel/Distributed FFT

Parallel FFT

Suppose we are given a polynomial $P(X) = \sum_{i=0}^{2^{2n} - 1} c_i X^i$. The Discrete Fourier Transform (DFT) of this polynomial is defined as:

$$\hat{c}_j = \sum_{i=0}^{2^{2n} - 1} c_i \cdot \omega^{ij}, \quad \text{where } 0 \le j < 2^{2n}$$

Here, $\omega$ is a primitive $2^{2n}$-th root of unity.

We can decompose this equation using index splitting as follows (where $0 \le j_0, j_1 < 2^n$):

$$\begin{align*} \hat{c}_{j_1 \cdot 2^n + j_0} &= \sum_{i_0 = 0}^{2^n - 1} \sum_{i_1 = 0}^{2^n - 1} c_{i_1 \cdot 2^n + i_0} \cdot \omega^{(i_1 \cdot 2^n + i_0)(j_1 \cdot 2^n + j_0)} \\ &= \underbrace{ \left(\sum_{i_0 = 0}^{2^n - 1} \underbrace{\left( \omega^{i_0 j_0} \underbrace{ \left( \sum_{i_1 = 0}^{2^n - 1} c_{i_1 \cdot 2^n + i_0} \cdot (\omega^{2^n})^{(j_1 \cdot 2^n + j_0)i_1} \right)}_{\text{Step 1}} \right)}_{\text{Step 2}} \cdot (\omega^{2^n})^{i_0 j_1}\right)}_{\text{Step 3}} \end{align*}$$

Example ($n=2$)

Let $P(X) = c_0 + c_1 X + \cdots + c_{15} X^{15}$, and suppose $\omega^{16} = 1$. Define the following input vectors:

$$\begin{aligned} \mathbf{c}^{(0)} &= (c_0, c_4, c_8, c_{12}) \\ \mathbf{c}^{(1)} &= (c_1, c_5, c_9, c_{13}) \\ \mathbf{c}^{(2)} &= (c_2, c_6, c_{10}, c_{14}) \\ \mathbf{c}^{(3)} &= (c_3, c_7, c_{11}, c_{15}) \\ \end{aligned}$$

Step 1: Perform FFT on each $\mathbf{c}^{(i_0)}$

$$\begin{aligned} \widehat{\mathbf{c}^{(0)}} &= (\widehat{c^{(0)}}_0, \widehat{c^{(0)}}_1, \widehat{c^{(0)}}_2, \widehat{c^{(0)}}_3) \\ \widehat{\mathbf{c}^{(1)}} &= (\widehat{c^{(1)}}_0, \widehat{c^{(1)}}_1, \widehat{c^{(1)}}_2, \widehat{c^{(1)}}_3) \\ \widehat{\mathbf{c}^{(2)}} &= (\widehat{c^{(2)}}_0, \widehat{c^{(2)}}_1, \widehat{c^{(2)}}_2, \widehat{c^{(2)}}_3) \\ \widehat{\mathbf{c}^{(3)}} &= (\widehat{c^{(3)}}_0, \widehat{c^{(3)}}_1, \widehat{c^{(3)}}_2, \widehat{c^{(3)}}_3) \end{aligned}$$

Each transform is defined by:

$$\widehat{c^{(i_0)}}_{j_0} = \sum_{i_1 = 0}^3 c_{4 \cdot i_1 + i_0} \cdot (\omega^4)^{(4j_1 + j_0)i_1}$$

Step 2: Construct $\mathbf{z}^{[j_0]}$

$$\begin{aligned} \mathbf{z}^{[0]} &= (\widehat{c^{(0)}}_0, \widehat{c^{(1)}}_0, \widehat{c^{(2)}}_0, \widehat{c^{(3)}}_0) \\ \mathbf{z}^{[1]} &= (\widehat{c^{(0)}}_1, \omega^1 \widehat{c^{(1)}}_1, \omega^2 \widehat{c^{(2)}}_1, \omega^3 \widehat{c^{(3)}}_1) \\ \mathbf{z}^{[2]} &= (\widehat{c^{(0)}}_2, \omega^2 \widehat{c^{(1)}}_2, \omega^4 \widehat{c^{(2)}}_2, \omega^6 \widehat{c^{(3)}}_2) \\ \mathbf{z}^{[3]} &= (\widehat{c^{(0)}}_3, \omega^3 \widehat{c^{(1)}}_3, \omega^6 \widehat{c^{(2)}}_3, \omega^9 \widehat{c^{(3)}}_3) \end{aligned}$$

Each $\mathbf{z}^{[j_0]}$ can be defined generally as:

$$\mathbf{z}^{[j_0]} := \left( \widehat{c^{(0)}}_{j_0}, \omega^{j_0} \widehat{c^{(1)}}_{j_0}, \omega^{2j_0} \widehat{c^{(2)}}_{j_0}, \omega^{3j_0} \widehat{c^{(3)}}_{j_0} \right)$$

Step 3: Final FFT on each $\mathbf{z}^{[j_0]}$

$$\begin{aligned} \widehat{\mathbf{z}^{[0]}} &= (\hat{c}_0, \hat{c}_4, \hat{c}_8, \hat{c}_{12}) \\ \widehat{\mathbf{z}^{[1]}} &= (\hat{c}_1, \hat{c}_5, \hat{c}_9, \hat{c}_{13}) \\ \widehat{\mathbf{z}^{[2]}} &= (\hat{c}_2, \hat{c}_6, \hat{c}_{10}, \hat{c}_{14}) \\ \widehat{\mathbf{z}^{[3]}} &= (\hat{c}_3, \hat{c}_7, \hat{c}_{11}, \hat{c}_{15}) \end{aligned}$$

Each transform is again a standard FFT:

$$\widehat{z^{[j_0]}}_{j_1} = \sum_{i_0 = 0}^3 {z^{[j_0]}}_{i_0} \cdot (\omega^4)^{i_0 j_1}$$

Verification Examples

• $\widehat{z^{[1]}}_2 = \hat{c}_9$ $(j_1 = 2, j_0 = 1)$

$$\begin{align*} \widehat{z^{[1]}}_2 &= {z^{[1]}}_0 + {z^{[1]}}_1 \omega^{8} + {z^{[1]}}_2 \omega^{16} + {z^{[1]}}_3 \omega^{24} \\ &= \widehat{c^{(0)}}_1 + \widehat{c^{(1)}}_1 \omega^{9} + \widehat{c^{(2)}}_1 \omega^{18} + \widehat{c^{(3)}}_1 \omega^{27} \\ &= \sum_{i_1=0}^3 c_{4i_1} (\omega^4)^{9i_1} + \sum_{i_1=0}^3 c_{4i_1 + 1} (\omega^4)^{9i_1}\omega^9 + \sum_{i_1=0}^3 c_{4i_1 + 2} (\omega^4)^{9i_1}\omega^{18} + \sum_{i_1=0}^3 c_{4i_1 + 3} (\omega^4)^{9i_1}\omega^{27} \\ &= \hat{c}_9 \end{align*}$$

• $\widehat{z^{[2]}}_1 = \hat{c}_6$ $(j_1 = 1, j_0 = 2)$

$$\begin{align*} \widehat{z^{[2]}}_1 &= {z^{[2]}}_0 + {z^{[2]}}_1 \omega^{4} + {z^{[2]}}_2 \omega^{8} + {z^{[2]}}_3 \omega^{12} \\ &= \widehat{c^{(0)}}_2 + \widehat{c^{(1)}}_2 \omega^{6} + \widehat{c^{(2)}}_2 \omega^{12} + \widehat{c^{(3)}}_2 \omega^{18} \\ &= \sum_{i_1=0}^3 c_{4i_1} (\omega^4)^{6i_1} + \sum_{i_1=0}^3 c_{4i_1 + 1} (\omega^4)^{6i_1} \omega^6 + \sum_{i_1=0}^3 c_{4i_1 + 2} (\omega^4)^{6i_1} \omega^{12} + \sum_{i_1=0}^3 c_{4i_1 + 3} (\omega^4)^{6i_1} \omega^{18} \\ &= \hat{c}_6 \end{align*}$$

The inverse FFT is structurally similar, but we do not cover it in this article.

Distributed FFT

The Parallel FFT algorithm described above can be implemented as a Distributed FFT by mapping it onto the MapReduce model. The diagram above illustrates how the same example from the parallel FFT is executed within the MapReduce paradigm.

• Map phase: Each executor performs the first-stage FFTs over $\mathbf{c}^{(i_0)}$

• Shuffle (transpose + twist): Results are reorganized into $\mathbf{z}^{[j_0]}$

• Reduce phase: Second-stage FFTs are computed over each $\mathbf{z}^{[j_0]}$

This structure enables FFT computations over extremely large input sizes to be parallelized and scaled effectively in distributed systems like Spark or Hadoop.

Reference

• https://www.csd.uwo.ca/~mmorenom/SNC-11-file-for-ACM/p54-Sze.pdf

Written by ryan Kim from A41