Hexagonal fast Fourier transform

The linear transforms ${\displaystyle g_{a}}$ and ${\displaystyle f_{a}}$ are similar to the rectangular Fourier kernel where a linear transform is applied along each dimension of the 2-D rectangular data.[1] Note that each of these equations, discussed above, is a combination of four rectangular arrays that serve as precursors to the HDFT. Two, out of those four rectangular ${\displaystyle g_{a}}$ terms contribute to the sub-array of HFFT. Now, by switching the binary coordinate, we have four different forms of equations. In,[7] the three out of those four expressions have been evaluated using what the author called "non-standard transforms (NSTs)" (shown below) while one expression is computed using any correct and applicable FFT algorithm.

${\displaystyle g_{a}(0,r,d)=\sum _{c}x(a,r,c)\exp \left(-j2\pi {\frac {(c)(d)}{m}}\right)}$

${\displaystyle g_{a}(1,r,d)=\sum _{c}x(a,r,c)\exp \left(-j2\pi {\frac {(c)(2d+1)}{2m}}\right)}$

${\displaystyle f_{a}(0,s,d)=\sum _{r}g_{a}(a,r,d)\exp \left(-j2\pi {\frac {(r)(2s)}{n}}\right)}$

${\displaystyle f_{a}(1,s,d)=\sum _{r}g_{a}(a,r,d)\exp \left(-j2\pi {\frac {(r)(2s+1)}{n}}\right)}$

Looking at the second expression, ${\displaystyle g_{a}(1,r,d)}$, we see that it is nothing more than a standard discrete Fourier transform (DFT) with a constant offset along the rows of rectangular sub-arrays of a hexagonally-sampled image ${\displaystyle x(a,r,c)}$.[1] This expression is nothing more than a circular rotation of the DFT. Note that the shift must happen in the integer number of samples for the property to hold. This way, the function ${\displaystyle g_{a}}$ can be computed using the standard DFT, in same number of operations, without introducing an NST.

If we have a look at 0-array ${\displaystyle f_{a}}$, the expression will always be symmetric about half its spatial period. Because of this, it is enough to compute only half of it. We find that this expression is the standard DFT of the columns of ${\displaystyle g_{a}}$, which is decimated by a factor of 2 and then is duplicated to span the space of r for the identical second period of the complex exponential.[1] Mathematically,

${\displaystyle {\begin{aligned}X_{\text{even}}[k]&=\sum _{n=0}^{N-1}x[n]e^{-{\tfrac {2j\pi }{N}}2kn}\\[5pt]&=\sum _{n=0}^{{\tfrac {N}{2}}-1}x[n]e^{-{\tfrac {2j\pi }{N/2}}kn}+\sum _{n={\tfrac {N}{2}}}^{N-1}x[n]e^{-{\tfrac {2j\pi }{N/2}}kn}\\[5pt]&=\sum _{n=0}^{{\tfrac {N}{2}}-1}x[n]e^{-{\tfrac {2j\pi }{N/2}}kn}+\sum _{n=0}^{{\tfrac {N}{2}}-1}x\left[n+{\tfrac {N}{2}}\right]e^{-{\tfrac {2j\pi }{N/2}}kn}\\[5pt]&=\sum _{n=0}^{{\tfrac {N}{2}}-1}\left(x[n]+x\left[n+{\tfrac {N}{2}}\right]\right)e^{-{\tfrac {2j\pi }{N/2}}kn}\end{aligned}}}$

The expression for the 1-array ${\displaystyle f_{a}}$ is equivalent to the 0-array expression with a shift of one sample. Hence, the 1-array expression can be expressed as columns of the DFT of ${\displaystyle g_{a}}$ decimated by a factor of two, starting with second sample providing a constant offset needed by 1-array, and then doubled in space to span the range of s. Thus, the method developed by James B. Birdsong and Nicholas I. Rummelt in [1] is able to successfully compute the HFFT using the standard FFT routines unlike the previous work in.[7]