Fast Walsh–Hadamard transform

In computational mathematics, the Hadamard ordered fast Walsh–Hadamard transform (FWHT_h) is an efficient algorithm to compute the Walsh–Hadamard transform (WHT). A naive implementation of the WHT of order ${\displaystyle n=2^{m}}$ would have a computational complexity of O(${\displaystyle n^{2}}$). The FWHT_h requires only ${\displaystyle n\log n}$ additions or subtractions.

The FWHT_h is a divide-and-conquer algorithm that recursively breaks down a WHT of size ${\displaystyle n}$ into two smaller WHTs of size ${\displaystyle n/2}$. ^[1] This implementation follows the recursive definition of the ${\displaystyle 2^{m}\times 2^{m}}$ Hadamard matrix ${\displaystyle H_{m}}$:

${\displaystyle H_{m}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}H_{m-1}&H_{m-1}\\H_{m-1}&-H_{m-1}\end{pmatrix}}.}$

The ${\displaystyle 1/{\sqrt {2}}}$ normalization factors for each stage may be grouped together or even omitted.

The sequency-ordered, also known as Walsh-ordered, fast Walsh–Hadamard transform, FWHT_w, is obtained by computing the FWHT_h as above, and then rearranging the outputs.

A simple fast nonrecursive implementation of the Walsh–Hadamard transform follows from decomposition of the Hadamard transform matrix as ${\displaystyle H_{m}=A^{m}}$, where A is m-th root of ${\displaystyle H_{m}}$. ^[2]

import math
def fwht(a) -> None:
    """In-place Fast Walsh–Hadamard Transform of array a."""
    assert math.log2(len(a)).is_integer(), "length of a is a power of 2"
    h = 1
    while h < len(a):
        # perform FWHT
        for i in range(0, len(a), h * 2):
            for j in range(i, i + h):
                x = a[j]
                y = a[j + h]
                a[j] = x + y
                a[j + h] = x - y
        # normalize and increment
        a /= math.sqrt(2)
        h *= 2

• Fast Fourier transform

• Charles Constantine Gumas, A century old, the fast Hadamard transform proves useful in digital communications

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