Small-Size Algorithms for Quaternion Discrete Fourier Transform

Abstract

The quaternion discrete Fourier transform (QDFT) is a powerful tool in modern digital signal processing, even though until recently this transformation seemed exotic. In recent years, quite a lot of publications have appeared devoted to effective ways to calculate this transformation. In particular, in one of our previous publications, we presented an economical algorithm for calculating one-dimensional QDFT and showed that this algorithm has the lowest computational complexity among all known algorithms of this type. This generalized algorithm is suitable for computing the QDFT of any sequence in which the number of elements is a power of two. However, as it turned out, there are additional possibilities that make it possible to further reduce the computational complexity of the developed algorithm for each specific N. In this article, we provide some examples of the synthesis of such algorithms for short-length input sequences (samples of signals). In particular, algorithms for N ∊ {2, 3, 4, 5, 6, 7, 8} are presented. A parallel implementation of the proposed algorithm allows for saving more than half of the number of required multipliers in each case compared with the parallel implementation of the naive methods of calculation.

1. Introduction

Discrete Fourier transform (DFT) is the most popular tool used for data analysis and processing in various fields of science and engineering [1]. In the most general case, the discrete Fourier transform is defined in the complex-valued domain [2,3,4,5,6,7]. This means that both the input data (signal samples) and the output data (DFT coefficients) are complex-valued. The sample values of the system of basic functions are also complex-valued. The rapid development in the area of signal and image processing has led to the introduction of data representation and processing in the hypercomplex domain. Today, hypercomplex numbers, especially quaternions, are used in various fields of science and technology, such as digital signal and image processing [8], deep learning and neural networks [9], computer graphics [10], machine vision and robotics [11], space–time block coding [12], time series compression [13], and many others [14]. There was a need to introduce the concept of quaternion DFT (QDFT). This is how the quaternion discrete Fourier transform was defined [15,16,17,18,19,20]. Regardless of the area in which the DFT was defined, its computation time has always been a stepping stone on the path to the development of high-speed DSP algorithms. This is why, over time, many efficient algorithms for calculating complex-valued DFT have emerged. They are called fast Fourier transform (FFT) algorithms [5,7]. In those years, no one had heard anything about hypercomplex DFT. But today, when hypercomplex signal processing is becoming increasingly popular, there is a need to define typical DSP operations in the hypercomplex domain. And these operations were defined [21,22,23,24,25,26,27,28]. One-dimensional and two-dimensional hypercomplex DFTs were also defined. The quaternion-valued DFT has become the most popular. Quaternion-valued DFT had the same problem as complex-valued DFT—its computational complexity was very significant. This is not surprising: if multiplying complex numbers requires 4 multiplications and 2 additions of real numbers, then multiplying two quaternions requires 16 multiplications and 12 additions of real numbers. The developers intended to search for effective algorithms for calculating the quaternion DFT with a reduced number of algorithmic operations. As a result, fast algorithms for calculating quaternion-valued DFTs appeared [26,27]. One of the latest works describes a universal algorithm for calculating one-dimensional QDFT, which, as it seems to us, has the least computational complexity among those known [28]. This algorithm allows one to calculate the QDFT of a sequence of any length N. However, it turned out that taking into account the structural properties of the base QDFT matrices for specific lengths N of input sequences makes it possible to find even better solutions. In this article, we propose a few algorithms for short lengths of N. In particular, fast QDFT algorithms for N ∊ {2, 3, 4, 5, 6, 7, 8} are presented.

2. Preliminary Remarks

The one-dimensional left-side quaternion discrete Fourier transform may be defined using the following expression:

$$y_m={\sum }_{n=1}^{N-1}\mathrm{e}\mathrm{x}\mathrm{p}(-\mu {\displaystyle \frac{2\pi }{N}}mn)x_n,\hspace{1em}m=0,1,...,N-1,$$

(1)

where

• $x_n=x_n^{\left(r\right)}+x_n^{\left(i\right)}i+x_n^{\left(j\right)}j+x_n^{\left(k\right)}k$ n-th quaternion-valued sample of discrete-time signal with N samples, $x_n∊H$, $x_n^{\left(r\right)},x_n^{\left(i\right)},x_n^{\left(j\right)},x_n^{\left(k\right)}∊R$;
• $y_m=y_m^{\left(r\right)}+y_m^{\left(i\right)}i+y_m^{\left(j\right)}j+y_m^{\left(k\right)}k$, is m-th quaternion-valued QDFT coefficient, $y_m∊H$, $y_m^{\left(r\right)},y_m^{\left(i\right)},y_m^{\left(j\right)},y_m^{\left(k\right)}∊R$, $\mu ={\mu }^{\left(i\right)}i+{\mu }^{\left(j\right)}j+{\mu }^{\left(k\right)}k$ is a unit pure quaternion, $\mu ∊H$, ${\mu }^{\left(i\right)},{\mu }^{\left(j\right)},{\mu }^{\left(k\right)}∊R$.

Most often, it is assumed that ${\mu }^{\left(i\right)}={\mu }^{\left(j\right)}={\mu }^{\left(k\right)}=1/\sqrt{3}$, and we will assume so as well.

In matrix notation, expression (1) can be rewritten as follows:

$${\mathsf{y}}_N={\mathsf{E}}_N{\mathsf{x}}_N,$$

(2)

where

• ${\mathsf{x}}_N=[x_0,x_1,...,x_{N-1}{]}^{{\rm T}}$ is a vector of size N containing quaternion-valued signal samples,
• ${\mathsf{y}}_N=[y_0,y_1,...,,y_{N-1}{]}^{{\rm T}}$ is a vector of size N containing the QDFT coefficients,
• ${\mathsf{E}}_N$ is a matrix whose entries $w^{mn}$ represent the values of quaternion-valued discrete exponential functions, and $w^{mn}=\mathrm{e}\mathrm{x}\mathrm{p}(-\mu {\displaystyle \frac{2\pi mn}{N}})$, $\mathrm{e}\mathrm{x}\mathrm{p}\left(-\mu {\displaystyle \frac{2\pi mn}{N}}\right)=\mathrm{c}\mathrm{o}\mathrm{s}{\displaystyle \frac{2\pi mn}{N}}-\mu \mathrm{s}\mathrm{i}\mathrm{n}{\displaystyle \frac{2\pi mn}{N}}$. It is easy to see that these values are quaternion-valued constants. Therefore, if the order of the matrix ${\mathsf{E}}_N$ is known in advance, then its entries can be calculated in advance.

It is the matrix representation of the left-side QDFT for which we will need to synthesize fast left-side QDFT algorithms. By analyzing the structures of matrices and finding successful tricks for their decomposition and factorization, it is possible to develop QDFT algorithms with reduced computational complexity. Let us show how we do it.

3. Small Size 1D-QDFT Algorithms

3.1. Algorithm 1, N = 2

Let ${\mathsf{x}}_2=[x_0,x_1{]}^{{\rm T}}$ and ${\mathsf{y}}_2=[y_0,y_1{]}^{{\rm T}}$ be two-entry input and output vectors, and $x_n∊H$, $y_m∊H$, $m,n=\mathrm{0,1}$.

Then, the algorithm for calculating the two-point QDFT can be represented as the following matrix–vector product:

$${\mathsf{y}}_{2\times 1}={\mathsf{E}}_2{\mathsf{x}}_{2\times 1},$$

(3)

where

• ${\mathsf{E}}_2=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]={\mathsf{H}}_2$ is an order two QDFT matrix.

It should be noted that exclusively in this case, the matrix ${\mathsf{E}}_2$coincides with the second-order Hadamard matrix ${\mathsf{H}}_2$. The operation of multiplying the second-order Hadamard matrix by a vector of length 2 is the basic operation of most fast algorithms. In some papers, it is called a butterfly [29].

Figure 1 shows a data-flow diagram of the proposed algorithm for implementing two-point QDFT.

In all our articles, we try to use the same notation and unify our drawings. Therefore, here, and in the rest of the article, data in all graphs flow from left to right by default. Solid lines represent data transfer operations (data paths). Dashed lines are used when the sign of the operand being passed is reversed. Circles denote left-side multiplications by a quaternion-valued twiddle factor inscribed within them. Points where two or more lines join represent summation operations. Rectangles marked with the letter ${\mathsf{H}}_2$indicate butterfly operations. Finally, we deliberately use regular lines without arrows in order to not to create clutter at the points where they converge.

3.2. Algorithm 2, N = 3

Let ${\mathsf{x}}_3=[x_0,x_1,x_2{]}^{{\rm T}}$ and ${\mathsf{y}}_3=[y_0,y_1,y_2{]}^{{\rm T}}$ be three-entry input and output vectors, and $x_n∊H,$ $y_m∊H$, $m,n=\mathrm{0,1},2$.

The 3-point QDFT can be represented as

$${\mathsf{y}}_3={\mathsf{E}}_3{\mathsf{x}}_3,$$

(4)

where

• ${\mathsf{E}}_3=\left[\begin{array}{ccc}1& 1& 1\\ 1& a_3& a_3^{*}\\ 1& a_3^{*}& a_3\end{array}\right]$ is an order three QDFT matrix, $a_3=-0.5-0.5i-0.5j-0.5k$, and symbol (*) denotes the conjugate of a quaternion number.

The calculation of the QDFT under expression (4) is redundant. To calculate this expression, 4 quaternion-valued multiplications and 6 quaternion-valued additions are required. We will show how one can reduce the number of operations required when calculating a three-point QDFT.

Let us perform the following decomposition:

$${\mathsf{E}}_3=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 1& 0& 0\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& a_3& a_3^{*}\\ 0& a_3^{*}& a_3\end{array}\right].$$

Taking advantage of the structure of the matrices obtained as a result of decomposition, the three-point QDFT algorithm can be written as follows:

$${\mathsf{y}}_3={\mathsf{A}}_{3\times 4}{\mathsf{W}}_4{\mathsf{D}}_4{\mathsf{W}}_4{\mathsf{A}}_{4\times 3}{\mathsf{x}}_3,$$

(5)

where

• ${\mathsf{A}}_{4\times 3}=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$, ${\mathsf{W}}_4={\mathsf{I}}_2\oplus {\mathsf{H}}_2=\left[\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}\end{array}\right]$, ${\mathsf{A}}_{3\times 4}=\left[\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 1& 0\\ 1& 0& 0& 1\end{array}\right]$,
• ${\mathsf{D}}_4=diag(\mathrm{1,1},s_0^{\left(3\right)},s_1^{\left(3\right)})$,
• $s_0^{\left(3\right)}=1/2(a_3+a_3^{\mathrm{*}})=-1/2$, $s_1^{\left(3\right)}=1/2(a_3-a_3^{\mathrm{*}})=(-i-j-k)/2$,
• ${\mathsf{H}}_2$ is (as we have already noted) the second-order Hadamard matrix, ${\mathsf{I}}_N$ is the order $N$ identity matrix, and “$\otimes$”, “$\oplus$” mean the tensor product and direct sum of two matrices, respectively [30].

Thus, we described an algorithm that requires 8 quaternion additions and only 2 trivial quaternion multiplications.

Figure 2 shows a data-flow diagram of the proposed algorithm for the implementation of 3-point QDFT. To improve the readability of the figures, in Figure 2 (and in the other figures), we decided not to write superscripts for the entries $s_0^{\left(3\right)}$ and $s_1^{\left(3\right)}$ of the diagonal matrix ${\mathsf{D}}_4$.

3.3. Algorithm 3, N = 4

Let ${\mathsf{x}}_4=[x_0,x_1,x_2,x_3{]}^{{\rm T}}$ and ${\mathsf{y}}_4=[y_0,y_1,y_2,y_3{]}^{{\rm T}}$ be four-entry input and output vectors, and $x_n∊H,$ $y_m∊H$, $m,n=\mathrm{0,1},\mathrm{2,3}$.

The 4-point QDFT can be represented as

$${\mathsf{y}}_4={\mathsf{E}}_4{\mathsf{x}}_4,$$

(6)

where

• ${\mathsf{E}}_4=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& a_4& -1& -a_4\\ 1& -1& 1& -1\\ 1& -a_4& -1& a_4\end{array}\right]$, $a_4=-0.5774i-0.5774j-0.5774k$.

To calculate expression (6), 4 quaternion-valued multiplications and 12 quaternion-valued additions are required. We will show how one can reduce the number of operations required when calculating a four-point QDFT.

Let us reorder the rows of the matrix ${\mathsf{E}}_4$ using the following permutation:

$${\pi }_4=\left(\begin{array}{cccc}0& 1& 2& 3\\ 0& 2& 1& 3\end{array}\right).$$

As a result, we obtain the following matrix:

$${\tilde{\mathsf{E}}}_4=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& 1& -1\\ 1& a_4& -1& -a_4\\ 1& -a_4& -1& a_4\end{array}\right]=\left[\begin{array}{cc}{\mathsf{A}}_2^{\left(4\right)}& {\mathsf{A}}_2^{\left(4\right)}\\ {\mathsf{B}}_2^{\left(4\right)}& -{\mathsf{B}}_2^{\left(4\right)}\end{array}\right],$$

where

$${\mathsf{A}}_2^{\left(4\right)}={\mathsf{H}}_2,{\mathsf{B}}_2^{\left(4\right)}=\left[\begin{array}{cc}1& a_4\\ 1& {-a}_4\end{array}\right]={\mathsf{H}}_2\left[\begin{array}{cc}1& 0\\ 0& a_4\end{array}\right].$$

Taking into account the structural properties of the resulting matrix, the four-point QDFT algorithm can be written in the following form:

$${\mathsf{y}}_4={\mathsf{E}}_4{\mathsf{x}}_4={\mathsf{P}}_4{\tilde{\mathsf{E}}}_4{\mathsf{x}}_4={\mathsf{P}}_4{\mathsf{W}}_4^{\left(2\right)}{\tilde{\mathsf{D}}}_4{\mathsf{W}}_4^{\left(1\right)}{\mathsf{x}}_4,$$

(7)

where

• ${\tilde{\mathsf{D}}}_4=diag(\mathrm{1,1},1,s_0^{\left(4\right)})$, $s_0^{\left(4\right)}=a_4$,

$${\mathsf{W}}_4^{\left(1\right)}={\mathsf{H}}_2\otimes {\mathsf{I}}_2=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& -1& 0\\ 0& 1& 0& -1\end{array}\right],{\mathsf{W}}_4^{\left(2\right)}={\mathsf{I}}_2\otimes {\mathsf{H}}_2=\left[\begin{array}{cccc}1& 1& 0& 0\\ 1& -1& 0& 0\\ 0& 0& 1& 1\\ 0& 0& 1& -1\end{array}\right],$$

$${\mathsf{P}}_4=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right].$$

Figure 3 shows a data-flow diagram of the proposed algorithm for implementing four-point QDFT. One can see that the algorithm requires performing only one multiplication and only eight additions of quaternion-valued operands.

3.4. Algorithm 4, N = 5

Let ${\mathsf{x}}_5=[x_0,x_1,x_2,x_3,x_4{]}^{{\rm T}}$ and ${\mathsf{y}}_5=[y_0,y_1,y_2,y_3,y_4{]}^{{\rm T}}$ be five-entry input and output vectors, and $x_n∊H$, $y_m∊H$, $m,n=\mathrm{0,1},\mathrm{2,3},4$.

The 5-point QDFT can be represented as

$${\mathsf{y}}_5={\mathsf{E}}_5{\mathsf{x}}_5,$$

(8)

where

$${\mathsf{E}}_5=\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& a_5& b_5& b_5^{*}& a_5^{*}\\ 1& b_5& a_5^{*}& a_5& b_5^{*}\\ 1& b_5^{*}& a_5& a_5^{*}& b_5\\ 1& a_5^{*}& b_5^{*}& b_5& a_5\end{array}\right],$$

$$a_5=0.30902-0.54909i-0.54909j-0.54909k,$$

$$a_5^{*}=0.30902+0.54909i+0.54909j+0.54909k,$$

$$b_5=-0.80902-0.33936i-0.33936j-0.33936k,$$

$$b_5^{*}=-0.80902+0.33936i+0.33936j+0.33936k.$$

Let us now change the order of the rows and columns of matrix ${\mathsf{E}}_5$ using the following permutation:

$${\pi }_5=\left(\begin{array}{ccccc}0& 1& 2& 3& 4\\ 0& 1& 2& 4& 3\end{array}\right).$$

Then, we obtain the following matrix:

$${\breve{\mathsf{E}}}_5=\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& a_5& b_5& a_5^{*}& b_5^{*}\\ 1& b_5& a_5^{*}& b_5^{*}& a_5\\ 1& a_5^{*}& b_5^{*}& a_5& b_5\\ 1& b_5^{*}& a_5& b_5& a_5^{*}\end{array}\right].$$

Let us perform the following decomposition:

$${\breve{\mathsf{E}}}_5=\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& a_5& b_5& a_5^{*}& b_5^{*}\\ 1& b_5& a_5^{*}& b_5^{*}& a_5\\ 1& a_5^{*}& b_5^{*}& a_5& b_5\\ 1& b_5^{*}& a_5& b_5& a_5^{*}\end{array}\right]=\underset{{\breve{\mathsf{E}}}_5^{\left(1\right)}}{\underbrace{\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 0& 0& 0& 0\\ 1& 0& 0& 0& 0\\ 1& 0& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right]}}+\underset{{\breve{\mathsf{E}}}_5^{\left(2\right)}}{\underbrace{\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& a_5& b_5& a_5^{*}& b_5^{*}\\ 0& b_5& a_5^{*}& b_5^{*}& a_5\\ 0& a_5^{*}& b_5^{*}& a_5& b_5\\ 0& b_5^{*}& a_5& b_5& a_5^{*}\end{array}\right]}},$$

$${\breve{\mathsf{E}}}_5^{\left(2\right)}=\left[\begin{array}{cc}0& \begin{array}{cccc}0& 0& 0& 0\end{array}\\ \begin{array}{c}0\\ 0\\ 0\\ 0\end{array}& {\mathsf{A}}_4^{\left(5\right)}\end{array}\right],{\mathsf{A}}_4^{\left(5\right)}=\left[\begin{array}{cc}{\mathsf{A}}_2^{\left(5\right)}& {\mathsf{B}}_2^{\left(5\right)}\\ {\mathsf{B}}_2^{\left(5\right)}& {\mathsf{A}}_2^{\left(5\right)}\end{array}\right],{\mathsf{A}}_2^{\left(5\right)}=\left[\begin{array}{cc}{a}_5& b_5\\ b_5& a_5^{*}\end{array}\right],{\mathsf{B}}_2^{\left(5\right)}=\left[\begin{array}{cc}{a}_5^{*}& b_5^{*}\\ b_5^{*}& a_5\end{array}\right].$$

Due to the special structure of the ${\mathsf{A}}_4^{\left(5\right)}$ matrix, it can be written as follows [31]:

$${\mathsf{A}}_4^{\left(5\right)}=\left({\mathsf{H}}_2\otimes {\mathsf{I}}_2\right)\left[{\displaystyle \frac{1}{2}}\left({\mathsf{A}}_2^{\left(5\right)}+{\mathsf{B}}_2^{\left(5\right)}\right)\oplus {\displaystyle \frac{1}{2}}\left({\mathsf{A}}_2^{\left(5\right)}-{\mathsf{B}}_2^{\left(5\right)}\right)\right]\left({\mathsf{H}}_2\otimes {\mathsf{I}}_2\right),$$

where the matrices

$${\displaystyle \frac{1}{2}}({\mathsf{A}}_2^{\left(5\right)}+{\mathsf{B}}_2^{\left(5\right)})={\displaystyle \frac{1}{2}}\left[\begin{array}{cc}{a}_5+a_5^{*}& b_5+b_5^{*}\\ b_5+b_5^{*}& a_5+a_5^{*}\end{array}\right],\mathrm{and}{\displaystyle \frac{1}{2}}\left({\mathsf{A}}_2^{\left(5\right)}-{\mathsf{B}}_2^{\left(5\right)}\right)={\displaystyle \frac{1}{2}}\left[\begin{array}{cc}{a}_5-a_5^{*}& b_5-b_5^{*}\\ b_5-b_5^{*}& a_5^{*}-a_5\end{array}\right],$$

can be written in the form

$${\displaystyle \frac{1}{2}}\left({\mathsf{A}}_2^{\left(5\right)}+{\mathsf{B}}_2^{\left(5\right)}\right)={\mathsf{H}}_{2}diag\left(s_0^{\left(5\right)},s_1^{\left(5\right)}\right){\mathsf{H}}_2,{\displaystyle \frac{1}{2}}\left({\mathsf{A}}_2^{\left(5\right)}-{\mathsf{B}}_2^{\left(5\right)}\right)={\mathsf{T}}_{2\times 3}diag\left(s_2^{\left(5\right)},s_3^{\left(5\right)},s_4^{\left(5\right)}\right){\mathsf{T}}_{3\times 2},$$

$${\mathsf{T}}_{2\times 3}=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\end{array}\right],{\mathsf{T}}_{3\times 2}=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right],$$

$$s_0^{\left(5\right)}=(a_5+a_5^{*}+b_5+b_5^{*})/4=-1/4,s_1^{\left(5\right)}=(a_5+a_5^{*}-b_5-b_5^{*})/4=0.55902$$

$$s_2^{\left(5\right)}=(a_5-a_5^{*}-b_5+b_5^{*})/2=-0.20973i-0.20973j-0.20973k,$$

$$s_3^{\left(5\right)}=(-a_5+a_5^{*}-b_5+b_5^{*})/2=0.88845i+0.88845j+0.88845k,$$

$$s_4^{\left(5\right)}={(b}_5-b_5^{*})/2=-0.33936i-0.33936j-0.33936.$$

Taking into account the structural properties of the resulting matrix, the five-point QDFT algorithm may be written in the following form:

$${\mathsf{y}}_5={\mathsf{E}}_5{\mathsf{x}}_5={\mathsf{P}}_5{\breve{\mathsf{E}}}_5{\mathsf{x}}_5{=\mathsf{P}}_5{\mathsf{A}}_{5\times 6}{\mathsf{W}}_6^{\left(1\right)}{\mathsf{W}}_{6\times 7}^{\left(3\right)}{\mathsf{D}}_7{\mathsf{W}}_{7\times 6}^{\left(2\right)}{\mathsf{W}}_6^{\left(1\right)}{\mathsf{A}}_{6\times 5}{\mathsf{P}}_5{\mathsf{x}}_5$$

(9)

where

$${\mathsf{P}}_5={\mathsf{I}}_3\oplus {\mathsf{J}}_2=\left[\begin{array}{cc}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}& {\mathbf{0}}_{3\times 2}\\ {\mathbf{0}}_{2\times 3}& \begin{array}{cc}0& 1\\ 1& 0\end{array}\end{array}\right],{\mathsf{J}}_2=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],$$

$${\mathsf{A}}_{6\times 5}=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 1& 1& 1& 1& 1\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right],$$

$${\mathsf{W}}_6^{\left(1\right)}={\mathsf{I}}_2\oplus \left({\mathsf{H}}_2\otimes {\mathsf{I}}_2\right)=\left[\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& {\mathbf{0}}_{2\times 4}\\ {\mathbf{0}}_{4\times 2}& \begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& -1& 0\\ 0& 1& 0& -1\end{array}\end{array}\right],$$

$${\mathsf{W}}_{7\times 6}^{\left(2\right)}={\mathsf{I}}_2\oplus {\mathsf{H}}_2\oplus {\mathsf{T}}_{3\times 2}=\left[\begin{array}{ccc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& {\mathbf{0}}_2& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_{3\times 2}& {\mathbf{0}}_{3\times 2}& \begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\end{array}\right],$$

$${\mathsf{D}}_7=diag\left(\mathrm{1,1},s_0^{\left(5\right)},s_1^{\left(5\right)},s_2^{\left(5\right)},s_3^{\left(5\right)},s_4^{\left(5\right)}\right),$$

$${\mathsf{W}}_{6\times 7}^{\left(3\right)}={\mathsf{I}}_2\oplus {\mathsf{H}}_2\oplus {\mathsf{T}}_{2\times 3}=\left[\begin{array}{ccc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& {\mathbf{0}}_2& {\mathbf{0}}_{2\times 3}\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}& {\mathbf{0}}_{2\times 3}\\ {\mathbf{0}}_2& {\mathbf{0}}_2& \begin{array}{ccc}1& 0& 1\\ 0& 1& 1\end{array}\end{array}\right],{\mathsf{A}}_{5\times 6}=\left[\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ 1& 0& 1& 0& 0& 0\\ 1& 0& 0& 1& 0& 0\\ 1& 0& 0& 0& 1& 0\\ 1& 0& 0& 0& 0& 1\end{array}\right].$$

Figure 4 shows a data-flow diagram of the proposed algorithm for the five-point QDFT with reduced computation complexity.

3.5. Algorithm 5, N = 6

Let ${\mathsf{x}}_6=[x_0,x_1,x_2,x_3,x_4,x_5{]}^{{\rm T}}$ and ${\mathsf{y}}_6=[y_0,y_1,y_2,y_3,y_4,y_5{]}^{{\rm T}}$ be six-entry input and output vectors, and $x_n∊H,y_m∊H$, $m,n=\mathrm{0,1},\mathrm{2,3},\mathrm{4,5}$.

The 6-point QDFT can be represented as

$${\mathsf{y}}_6={\mathsf{E}}_6{\mathsf{x}}_6,$$

(10)

where

$${\mathsf{E}}_6=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& a_6& {-a}_6^{*}& -1& -a_6& a_6^{*}\\ 1& {-a}_6^{*}& -a_6& 1& {-a}_6^{*}& -a_6\\ 1& -1& 1& -1& 1& -1\\ 1& -a_6& {-a}_6^{*}& 1& -a_6& {-a}_6^{*}\\ 1& a_6^{*}& -a_6& -1& {-a}_6^{*}& a_6\end{array}\right],$$

$$a_6=0.5-0.5i-0.5j-0.5k,a_6^{*}=0.5+0.5i+0.5j+0.5k.$$

Let us now reorder both rows and columns of the matrix ${\mathsf{E}}_6$ using the following permutation:

$${\pi }_6=\left(\begin{array}{cccccc}0& 1& 2& 3& 4& 5\\ 0& 4& 2& 3& 1& 5\end{array}\right).$$

Then, we obtain the following matrix:

$${\breve{\mathsf{E}}}_6=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& -a_6& {-a}_6^{*}& 1& -a_6& a_6^{*}\\ 1& {-a}_6^{*}& -a_6& 1& {-a}_6^{*}& -a_6\\ 1& 1& 1& -1& -1& -1\\ 1& -a_6& {-a}_6^{*}& -1& a_6& a_6^{*}\\ 1& {-a}_6^{*}& -a_6& -1& a_6^{*}& a_6\end{array}\right]=\left[\begin{array}{cc}{\mathsf{A}}_3^{\left(6\right)}& {\mathsf{A}}_3^{\left(6\right)}\\ {\mathsf{A}}_3^{\left(6\right)}& -{\mathsf{A}}_3^{\left(6\right)}\end{array}\right],{\mathsf{A}}_3^{\left(6\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 1& -a_6& {-a}_6^{*}\\ 1& {-a}_6^{*}& -a_6\end{array}\right].$$

We see that matrix ${\breve{\mathsf{E}}}_6$ has a specific structure that allows us to factorize this matrix as follows [31]:

$${\breve{\mathsf{E}}}_6=({\mathsf{A}}_3^{\left(6\right)}\oplus {\mathsf{A}}_3^{\left(6\right)})({\mathsf{H}}_2\otimes {\mathsf{I}}_3).$$

(11)

Matrix ${\mathsf{A}}_3^{\left(6\right)}$ has a structure similar to that which has already been analyzed above. Therefore, it can be decomposed in the same way:

$${\mathsf{A}}_3^{\left(6\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 1& 0& 0\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& -a_6& -a_6^{*}\\ 0& -a_6^{*}& -a_6\end{array}\right].$$

By applying to matrix ${\mathsf{A}}_3^{\left(6\right)}$ techniques similar to those we used in the case of matrix ${\mathsf{E}}_3$and taking into account factorization (11), we can synthesize an algorithm for calculating a six-point QDFT with reduced computational complexity:

$${\mathit{y}}_6={\mathsf{P}}_6^{\left(3\right)}{\mathsf{A}}_{6\times 8}{\mathsf{W}}_8^{\left(2\right)}{\mathsf{D}}_8{\mathsf{W}}_8^{\left(2\right)}{\mathsf{A}}_{8\times 6}{{\mathsf{P}}_6^{\left(2\right)}\mathsf{W}}_6^{\left(1\right)}{\mathsf{P}}_6^{\left(1\right)}{\mathsf{x}}_6,$$

(12)

where

$${\mathsf{P}}_6^{\left(1\right)}=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right],{\mathsf{W}}_6^{\left(1\right)}={\mathsf{H}}_2\otimes {\mathsf{I}}_3=\left[\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 1\\ 1& 0& 0& -1& 0& 0\\ 0& 1& 0& 0& -1& 0\\ 0& 0& 1& 0& 0& -1\end{array}\right],$$

$${\mathsf{P}}_6^{\left(2\right)}=\left[\begin{array}{cc}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}& {\mathbf{0}}_3\\ {\mathbf{0}}_3& \begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\end{array}\right],{\mathsf{A}}_{8\times 6}=\left[\begin{array}{cc}\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}& {\mathbf{0}}_{4\times 3}\\ {\mathbf{0}}_{4\times 3}& \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 1& 1\\ 0& 0& 1\end{array}\end{array}\right],$$

$${\mathsf{W}}_8^{\left(2\right)}=\left({\mathsf{I}}_2\oplus {\mathsf{H}}_2\right)\oplus \left({\mathsf{H}}_2\oplus {\mathsf{I}}_2\right)=\left[\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}\end{array}& {\mathbf{0}}_4\\ {\mathbf{0}}_4& \begin{array}{cc}\begin{array}{cc}1& 1\\ 1& -1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 0\\ 0& 1\end{array}\end{array}\end{array}\right],$$

$${\mathsf{D}}_8=diag\left(\mathrm{1,1},s_0^{\left(6\right)},s_1^{\left(6\right)},s_2^{\left(6\right)},s_3^{\left(6\right)},\mathrm{1,1}\right),$$

$$s_0^{\left(6\right)}=-{\displaystyle \frac{1}{2}}(a_6^{*}+a_6)=-{\displaystyle \frac{1}{2}},s_1^{\left(6\right)}=-{\displaystyle \frac{1}{2}}(a_6^{*}-a_6)={\displaystyle \frac{1}{2}}(i+j+k),s_2^{\left(6\right)}=s_0^{\left(6\right)},s_3^{\left(6\right)}=s_1^{\left(6\right)},$$

$${\mathsf{A}}_{6\times 8}=\left[\begin{array}{cc}\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 1& 0\\ 1& 0& 0& 1\end{array}& {\mathbf{0}}_{3\times 4}\\ {\mathbf{0}}_{3\times 4}& \begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 1\\ 0& 0& 1& 0\end{array}\end{array}\right],{\mathsf{P}}_6^{\left(3\right)}=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\end{array}\right].$$

Figure 5 shows a data-flow diagram of the proposed algorithm for the six-point QDFT with reduced computation complexity.

3.6. Algorithm 6, N = 7

Let ${\mathsf{x}}_7=[x_0,x_1,x_3,x_4,x_5,x_6{]}^{{\rm T}}$ and ${\mathsf{y}}_7=[y_0,y_1,y_3,y_4,y_5,y_6{]}^{{\rm T}}$ be seven-entry input and output vectors, and $x_n∊H,y_m∊H$, $m,n=\mathrm{0,1},\mathrm{2,3},\mathrm{4,5},6$.

The 7-point QDFT can be represented as

$${\mathsf{y}}_7={\mathsf{E}}_7{\mathsf{x}}_7,$$

(13)

where

$${\mathsf{E}}_7=\left[\begin{array}{ccccccc}1& 1& 1& 1& 1& 1& 1\\ 1& a_7& b_7& c_7& c_7^{*}& b_7^{*}& a_7^{*}\\ 1& b_7& c_7^{*}& a_7^{*}& a_7& c_7& b_7^{*}\\ 1& c_7& a_7^{*}& b_7& b_7^{*}& a_7& c_7^{*}\\ 1& c_7^{*}& a_7& b_7^{*}& b_7& a_7^{*}& c_7\\ 1& b_7^{*}& c_7& a_7& a_7^{*}& c_7^{*}& b_7\\ 1& a_7^{*}& b_7^{*}& c_7^{*}& c_7& b_7& a_7\end{array}\right],$$

$$a_7=0.62349-0.45139i-0.45139j-0.45139k,$$

$$b_7=-0.22252-0.56287i-0.56287j-0.56287k,$$

$$c_7=-0.90097-0.2505i-0.2505j-0.2505k.$$

Let us now change the order of the columns and rows of matrix ${\mathsf{E}}_7$ using the following permutation:

$${\pi }_7=\left(\begin{array}{ccccccc}0& 1& 2& 3& 4& 5& 6\\ 0& 1& 2& 3& 6& 5& 4\end{array}\right).$$

We obtain the following matrix:

$${\breve{\mathsf{E}}}_7=\left[\begin{array}{ccccccc}1& 1& 1& 1& 1& 1& 1\\ 1& a_7& b_7& c_7& a_7^{*}& b_7^{*}& c_7^{*}\\ 1& b_7& c_7^{*}& a_7^{*}& b_7^{*}& c_7& a_7\\ 1& c_7& a_7^{*}& b_7& c_7^{*}& a_7& b_7^{*}\\ 1& a_7^{*}& b_7^{*}& c_7^{*}& a_7& b_7& c_7\\ 1& b_7^{*}& c_7& a_7& b_7& c_7^{*}& a_7^{*}\\ 1& c_7^{*}& a_7& b_7^{*}& c_7& a_7^{*}& b_7\end{array}\right].$$

Let us perform the following decomposition:

$${\breve{\mathsf{E}}}_7=\left[\begin{array}{ccccccc}1& 1& 1& 1& 1& 1& 1\\ 1& a_7& b_7& c_7& a_7^{*}& b_7^{*}& c_7^{*}\\ 1& b_7& c_7^{*}& a_7^{*}& b_7^{*}& c_7& a_7\\ 1& c_7& a_7^{*}& b_7& c_7^{*}& a_7& b_7^{*}\\ 1& a_7^{*}& b_7^{*}& c_7^{*}& a_7& b_7& c_7\\ 1& b_7^{*}& c_7& a_7& b_7& c_7^{*}& a_7^{*}\\ 1& c_7^{*}& a_7& b_7^{*}& c_7& a_7^{*}& b_7\end{array}\right]=\underset{{\breve{\mathsf{E}}}_7^{\left(1\right)}}{\underbrace{\left[\begin{array}{ccccccc}1& 1& 1& 1& 1& 1& 1\\ 1& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0\end{array}\right]}}+\underset{{\breve{\mathsf{E}}}_7^{\left(2\right)}}{\underbrace{\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& a_7& b_7& c_7& a_7^{*}& b_7^{*}& c_7^{*}\\ 0& b_7& c_7^{*}& a_7^{*}& b_7^{*}& c_7& a_7\\ 0& c_7& a_7^{*}& b_7& c_7^{*}& a_7& b_7^{*}\\ 0& a_7^{*}& b_7^{*}& c_7^{*}& a_7& b_7& c_7\\ 0& b_7^{*}& c_7& a_7& b_7& c_7^{*}& a_7^{*}\\ 0& c_7^{*}& a_7& b_7^{*}& c_7& a_7^{*}& b_7\end{array}\right]}}$$

$${\breve{\mathsf{E}}}_7^{\left(2\right)}=\left[\begin{array}{cc}0& \begin{array}{cccccc}0& 0& 0& 0& 0& 0\end{array}\\ \begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}& {\mathsf{A}}_6^{\left(7\right)}\end{array}\right],{\mathsf{A}}_6^{\left(7\right)}=\left[\begin{array}{cccccc}{a}_7& b_7& c_7& a_7^{*}& b_7^{*}& c_7^{*}\\ b_7& c_7^{*}& a_7^{*}& b_7^{*}& c_7& a_7\\ c_7& a_7^{*}& b_7& c_7^{*}& a_7& b_7^{*}\\ a_7^{*}& b_7^{*}& c_7^{*}& a_7& b_7& c_7\\ b_7^{*}& c_7& a_7& b_7& c_7^{*}& a_7^{*}\\ c_7^{*}& a_7& b_7^{*}& c_7& a_7^{*}& b_7\end{array}\right],$$

$${\mathsf{A}}_6^{\left(7\right)}=\left[\begin{array}{cc}{\mathsf{A}}_3^{\left(7\right)}& {\mathsf{B}}_3^{\left(7\right)}\\ {\mathsf{B}}_3^{\left(7\right)}& {\mathsf{A}}_3^{\left(7\right)}\end{array}\right],{\mathsf{A}}_3^{\left(7\right)}=\left[\begin{array}{ccc}{a}_7& b_7& c_7\\ b_7& c_7^{*}& a_7^{*}\\ c_7& a_7^{*}& b_7\end{array}\right],{\mathsf{B}}_3^{\left(7\right)}=\left[\begin{array}{ccc}{a}_7^{*}& b_7^{*}& c_7^{*}\\ b_7^{*}& c_7& a_7\\ c_7^{*}& a_7& b_7^{*}\end{array}\right].$$

We see that matrix ${\mathsf{A}}_6^{\left(7\right)}$ has a specific structure that allows us to factorize this matrix as follows [31]:

$${\mathsf{A}}_6^{\left(7\right)}=\left({\mathsf{H}}_2\otimes {\mathsf{I}}_3\right)\left[{\displaystyle \frac{1}{2}}({\mathsf{A}}_3^{\left(7\right)}+{\mathsf{B}}_3^{\left(7\right)})\oplus {\displaystyle \frac{1}{2}}({\mathsf{A}}_3^{\left(7\right)}-{\mathsf{B}}_3^{\left(7\right)})\right]({\mathsf{H}}_2\otimes {\mathsf{I}}_3),$$

(14)

$${\displaystyle \frac{1}{2}}\left({\mathsf{A}}_3^{\left(7\right)}+{\mathsf{B}}_3^{\left(7\right)}\right)={\displaystyle \frac{1}{2}}\left[\begin{array}{ccc}{a}_7+a_7^{*}& b_7+b_7^{*}& c_7+c_7^{*}\\ b_7+b_7^{*}& c_7^{*}+c_7& a_7^{*}+a_7\\ c_7+c_7^{*}& a_7^{*}+a_7& b_7+b_7^{*}\end{array}\right]={\mathsf{C}}_3,$$

$${\displaystyle \frac{1}{2}}\left({\mathsf{A}}_3^{\left(7\right)}-{\mathsf{B}}_3^{\left(7\right)}\right)={\displaystyle \frac{1}{2}}\left[\begin{array}{ccc}{a}_7-a_7^{*}& b_7-b_7^{*}& c_7-c_7^{*}\\ b_7-b_7^{*}& c_7^{*}-c_7& a_7^{*}-a_7\\ c_7-c_7^{*}& a_7^{*}-a_7& b_7-b_7^{*}\end{array}\right]={\mathsf{D}}_3,$$

It is easy to see that matrix ${\mathsf{С}}_3$ is left-circulant. For it, there is a factorization leading to a reduction in the number of multiplications when calculating the matrix–vector product. We can write

$${\mathsf{С}}_3={\mathsf{\Lambda }}_3^{\left(2\right)}{\mathsf{\Lambda }}_{3\times 4}{\mathsf{D}}_4^{\left(1\right)}{\mathsf{\Lambda }}_{4\times 3}{\mathsf{\Lambda }}_3^{\left(1\right)},$$

(15)

where

$${\mathsf{\Lambda }}_3^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& -1\\ 0& 1& -1\end{array}\right],{\mathsf{\Lambda }}_{4\times 3}=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],{\mathsf{\Lambda }}_{3\times 4}=\left[\begin{array}{cccc}0& 1& 0& 0\\ -1& 0& 1& 0\\ -1& 0& 0& 1\end{array}\right],{\mathsf{\Lambda }}_3^{\left(2\right)}=\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 1& -1& -1\end{array}\right],$$

$${\mathsf{D}}_4^{\left(1\right)}=diag\left(s_0^{\left(7\right)},s_1^{\left(7\right)},s_2^{\left(7\right)},s_3^{\left(7\right)}\right),$$

$$s_0^{\left(7\right)}={\displaystyle \frac{1}{6}}\left[\left(a_7+a_7^{*}\right)+\left(c_7+c_7^{*}\right)-2\left(b_7+b_7^{*}\right)\right]=0.055854,$$

$$s_1^{\left(7\right)}={\displaystyle \frac{1}{6}}\left[\left(a_7+a_7^{*}\right)+\left(c_7+c_7^{*}\right)+\left(b_7+b_7^{*}\right)\right]=-0.16667,$$

$$s_2^{\left(7\right)}={\displaystyle \frac{1}{2}}[\left(a_7+a_7^{*}\right)-\left(b_7+b_7^{*}\right)=0.84601,$$

$$s_3^{\left(7\right)}={\displaystyle \frac{1}{2}}\left[\left(c_7+c_7^{*}\right)-\left(b_7+b_7^{*}\right)\right]=-0.67845.$$

Let us now multiply by (−1) the last row and last column of the matrix ${\mathsf{D}}_3$. Then, we obtain a left-circulant matrix ${\breve{\mathsf{D}}}_3$ of the form

$${\breve{\mathsf{D}}}_3={\displaystyle \frac{1}{2}}\left[\begin{array}{ccc}{a}_7-a_7^{*}& b_7-b_7^{*}& c_7^{*}-c_7\\ b_7-b_7^{*}& c_7^{*}-c_7& a_7-a_7^{*}\\ c_7^{*}-c_7& a_7-a_7^{*}& b_7-b_7^{*}\end{array}\right].$$

This matrix can also be successfully factorized (due to the symmetry of the data-flow diagram, a slightly different factorization method was chosen in this case than for the ${\mathsf{С}}_3$ matrix):

$${\breve{\mathsf{D}}}_3={\mathsf{\Lambda }}_3^{\left(2\right)}{\mathsf{V}}_{3\times 4}{\mathsf{D}}_4^{\left(2\right)}{\mathsf{V}}_{4\times 3}{\mathsf{\Lambda }}_3^{\left(1\right)},$$

(16)

where

$${\mathsf{V}}_{4\times 3}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 1& 1\end{array}\right],{\mathsf{V}}_{3\times 4}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& -1\\ 0& 0& 1& -1\end{array}\right],$$

$${\mathsf{D}}_4^{\left(2\right)}=diag\left(s_4^{\left(7\right)},s_5^{\left(7\right)},s_6^{\left(7\right)},s_7^{\left(7\right)}\right),$$

$$s_4^{\left(7\right)}={\displaystyle \frac{1}{6}}\left[\left(a_7-a_7^{*}\right)+\left(c_7^{*}-c_7\right)+\left(b_7-b_7^{*}\right)\right]=-0.25459i-0.25459j-0.25459k,$$

$$s_5^{\left(7\right)}={\displaystyle \frac{1}{2}}\left[\left(a_7-a_7^{*}\right)-\left(b_7-b_7^{*}\right)\right]=0.11148i+0.11148j+0.11148k,$$

$$s_6^{\left(7\right)}={\displaystyle \frac{1}{2}}\left[\left(c_7^{*}-c_7\right)-\left(b_7-b_7^{*}\right)\right]=0.81338i+0.81338j+0.81338k,$$

$$s_7^{\left(7\right)}={\displaystyle \frac{1}{6}}\left[\left(a_7-a_7^{*}\right)+\left(c_7^{*}-c_7\right)-2\left(b_7-b_7^{*}\right)\right]=0.30829i+0.30829j+0.30829k,$$

and the remaining matrices are the same as in (15).

Since

$${\mathsf{D}}_3={\mathsf{B}}_3^{\left(1\right)}{\breve{\mathsf{D}}}_3{\mathsf{B}}_3^{\left(1\right)},$$

(17)

where

$${\mathsf{B}}_3^{\left(1\right)}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right],$$

we can write

$${\mathsf{D}}_3={\mathsf{V}}_3^{\left(2\right)}{\mathsf{V}}_{3\times 4}{\mathsf{D}}_4^{\left(2\right)}{\mathsf{V}}_{4\times 3}{\mathsf{V}}_3^{\left(1\right)}$$

(18)

where

$${\mathsf{V}}_3^{\left(1\right)}={\mathsf{\Lambda }}_3^{\left(1\right)}{\mathsf{B}}_3^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& -1\\ 1& 0& 1\\ 0& 1& 1\end{array}\right],{\mathsf{V}}_3^{\left(2\right)}={{\mathsf{B}}_3^{\left(1\right)}\mathsf{\Lambda }}_3^{\left(2\right)}=\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ -1& 1& 1\end{array}\right],$$

and the remaining matrices are the same as in (16).

Taking into account expressions (15), (18) and substituting them into (14), we can write the final seven-point QDFT algorithm with reduced computational complexity:

$${\mathsf{y}}_7={\mathsf{P}}_7{\mathsf{A}}_{7\times 8}{\mathsf{W}}_8^{\left(1\right)}{\mathsf{W}}_8^{\left(5\right)}{{\mathsf{W}}_{8\times 10}^{\left(4\right)}\mathsf{D}}_{10}{\mathsf{W}}_{10\times 8}^{\left(3\right)}{\mathsf{W}}_8^{\left(2\right)}{\mathsf{W}}_8^{\left(1\right)}{\mathsf{A}}_{8\times 7}{\mathsf{P}}_7{\mathsf{x}}_7,$$

(19)

where

$${\mathsf{P}}_7={\mathsf{I}}_4\oplus {\mathsf{J}}_3=\left[\begin{array}{cc}\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}& {\mathbf{0}}_{4\times 3}\\ {\mathbf{0}}_{3\times 4}& \begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\end{array}\right],{\mathsf{J}}_3=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right],$$

$${\mathsf{A}}_{8\times 7}=\left[\begin{array}{ccccccc}1& 1& 1& 1& 1& 1& 1\\ 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1\\ 1& 0& 0& 0& 0& 0& 0\end{array}\right],$$

$${\mathsf{W}}_8^{\left(1\right)}={\mathsf{I}}_1\oplus \left({\mathsf{H}}_2\otimes {\mathsf{I}}_3\right)\oplus {\mathsf{I}}_1=\left[\begin{array}{ccc}1& {\mathbf{0}}_{1\times 6}& 0\\ {\mathbf{0}}_{6\times 1}& \begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 1\\ 1& 0& 0& -1& 0& 0\\ 0& 1& 0& 0& -1& 0\\ 0& 0& 1& 0& 0& -1\end{array}& {\mathbf{0}}_{6\times 1}\\ 0& {\mathbf{0}}_{1\times 6}& 1\end{array}\right],$$

$${\mathsf{W}}_8^{\left(2\right)}={\mathsf{I}}_1\oplus {\mathsf{\Lambda }}_3^{\left(1\right)}\oplus {\mathsf{V}}_3^{\left(1\right)}\oplus {\mathsf{I}}_1=\left[\begin{array}{cccc}1& {\mathbf{0}}_{1\times 3}& {\mathbf{0}}_{1\times 3}& 0\\ {\mathbf{0}}_{3\times 1}& \begin{array}{ccc}1& 1& 1\\ 1& 0& -1\\ 0& 1& -1\end{array}& {\mathbf{0}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}& {\mathbf{0}}_3& \begin{array}{ccc}1& 1& -1\\ 1& 0& 1\\ 0& 1& 1\end{array}& {\mathbf{0}}_{3\times 1}\\ 0& {\mathbf{0}}_{1\times 3}& {\mathbf{0}}_{1\times 3}& 1\end{array}\right],$$

$${\mathsf{W}}_{10\times 8}^{\left(3\right)}={\mathsf{I}}_1\oplus {\mathsf{\Lambda }}_{4\times 3}\oplus {\mathsf{V}}_{4\times 3}\oplus {\mathsf{I}}_1=\left[\begin{array}{cccc}1& {\mathbf{0}}_{1\times 3}& {\mathbf{0}}_{1\times 3}& 0\\ {\mathbf{0}}_{4\times 1}& \begin{array}{ccc}0& 1& 1\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}& {\mathbf{0}}_{4\times 3}& {\mathbf{0}}_{4\times 1}\\ {\mathbf{0}}_{4\times 1}& {\mathbf{0}}_{4\times 3}& \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 1& 1\end{array}& {\mathbf{0}}_{4\times 1}\\ 0& {\mathbf{0}}_{1\times 3}& {\mathbf{0}}_{1\times 3}& 1\end{array}\right],$$

$${\mathsf{D}}_{10}=diag\left(1,s_0^{\left(7\right)},s_1^{\left(7\right)},s_2^{\left(7\right)},s_3^{\left(7\right)},s_4^{\left(7\right)},s_5^{\left(7\right)},s_6^{\left(7\right)},s_7^{\left(7\right)},1\right),$$

$${\mathsf{W}}_{8\times 10}^{\left(4\right)}={\mathsf{I}}_1\oplus {\mathsf{\Lambda }}_{3\times 4}{\oplus \mathsf{V}}_{3\times 4}\oplus {\mathsf{I}}_1=\left[\begin{array}{cccc}1& {\mathbf{0}}_{1\times 4}& {\mathbf{0}}_{1\times 4}& 0\\ {\mathbf{0}}_{3\times 1}& \begin{array}{cccc}0& 1& 0& 0\\ -1& 0& 1& 0\\ -1& 0& 0& 1\end{array}& {\mathbf{0}}_{3\times 4}& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}& {\mathbf{0}}_{3\times 4}& \begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& -1\\ 0& 0& 1& -1\end{array}& {\mathbf{0}}_{3\times 1}\\ 0& {\mathbf{0}}_{1\times 4}& {\mathbf{0}}_{1\times 4}& 1\end{array}\right],$$

$${\mathsf{W}}_8^{\left(5\right)}={\mathsf{I}}_1\oplus {\mathsf{\Lambda }}_3^{\left(2\right)}\oplus {\mathsf{V}}_3^{\left(2\right)}\oplus {\mathsf{I}}_1=\left[\begin{array}{cccc}1& {\mathbf{0}}_{1\times 3}& {\mathbf{0}}_{1\times 3}& 0\\ {\mathbf{0}}_{3\times 1}& \begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 1& -1& -1\end{array}& {\mathbf{0}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}& {\mathbf{0}}_3& \begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ -1& 1& 1\end{array}& {\mathbf{0}}_{3\times 1}\\ 0& {\mathbf{0}}_{1\times 3}& {\mathbf{0}}_{1\times 3}& 1\end{array}\right],$$

$${\mathsf{A}}_{7\times 8}=\left[\begin{array}{cccccccc}1& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 1\\ 0& 0& 1& 0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0& 0& 1\\ 0& 0& 0& 0& 0& 1& 0& 1\\ 0& 0& 0& 0& 0& 0& 1& 1\end{array}\right].$$

Figure 6 shows a data-flow diagram of the proposed algorithm for the seven-point QDFT with reduced computation complexity.

3.7. Algorithm 7, N = 8

Let ${\mathsf{x}}_8=[x_0,x_1,x_3,x_4,x_5,x_6,x_7{]}^{{\rm T}}$ and ${\mathsf{y}}_8=[y_0,y_1,y_3,y_4,y_5,y_6,y_7{]}^{{\rm T}}$ be the eight-entry input and output vectors, and $x_n∊H,y_m∊H$, $m,n=\mathrm{0,1},\mathrm{2,3},\mathrm{4,5},\mathrm{6,7}$.

The 8-point QDFT can be represented as

$${\mathsf{y}}_8={\mathsf{E}}_8{\mathsf{x}}_8,$$

(20)

where

$${\mathsf{E}}_8=\left[\begin{array}{cccccccc}1& 1& 1& 1& 1& 1& 1& 1\\ 1& a_8& b_8& -a_8^{*}& -1& -a_8& -b_8& a_8^{*}\\ 1& b_8& -1& -b_8& 1& b_8& -1& -b_8\\ 1& -a_8^{*}& {-b}_8& a_8& -1& a_8^{*}& b_8& -a_8\\ 1& -1& 1& -1& 1& -1& 1& -1\\ 1& -a_8& b_8& a_8^{*}& -1& a_8& -b_8& -a_8^{*}\\ 1& -b_8& -1& b_8& 1& -b_8& -1& b_8\\ 1& a_8^{*}& -b_8& -a_8& -1& -a_8^{*}& b_8& a_8\end{array}\right],$$

and

$$a_8=0.70711-0.40825i-0.40825j-0.40825k,$$

$$b_8=-0.57735i-0.57735j-0.57735k.$$

Let us now change the order of the rows of matrix ${\mathsf{E}}_8$ using the following permutation:

$${\pi }_8=\left(\begin{array}{cccccccc}0& 1& 2& 3& 4& 5& 6& 7\\ 0& 4& 2& 5& 1& 6& 3& 7\end{array}\right).$$

We obtain the following matrix:

$${\breve{\mathsf{E}}}_8=\left[\begin{array}{cccccccc}1& 1& 1& 1& 1& 1& 1& 1\\ 1& -1& 1& -1& 1& -1& 1& -1\\ 1& b_8& -1& -b_8& 1& b_8& -1& {-b}_8\\ 1& -b_8& -1& b_8& 1& -b_8& -1& b_8\\ 1& a_8& b_8& -a_8^{*}& -1& -a_8& -b_8& a_8^{*}\\ 1& -a_8^{*}& -b_8& a_8& -1& a_8^{*}& b_8& -a_8\\ 1& -a_8& b_8& a_8^{*}& -1& a_8& -b_8& -a_8^{*}\\ 1& a_8^{*}& -b_8& -a_8& -1& -a_8^{*}& b_8& a_8\end{array}\right]=\left[\begin{array}{cc}{\mathsf{A}}_4^{\left(8\right)}& {\mathsf{A}}_4^{\left(8\right)}\\ {\mathsf{B}}_4^{\left(8\right)}& -{\mathsf{B}}_4^{\left(8\right)}\end{array}\right],$$

where

$${\mathsf{A}}_4^{\left(8\right)}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& 1& -1\\ 1& b_8& -1& -b_8\\ 1& -b_8& -1& b_8\end{array}\right],{\mathsf{B}}_4^{\left(8\right)}=\left[\begin{array}{cccc}1& a_8& b_8& -a_8^{*}\\ 1& -a_8^{*}& -b_8& a_8\\ 1& -a_8& b_8& a_8^{*}\\ 1& a_8^{*}& -b_8& -a_8\end{array}\right].$$

The matrix ${\breve{\mathsf{E}}}_8$ of such a structure can be written as

$${\breve{\mathsf{E}}}_8=({\mathsf{A}}_4^{\left(8\right)}\oplus {\mathsf{B}}_4^{\left(8\right)})\left({\mathsf{H}}_2{\otimes \mathsf{I}}_4\right).$$

(21)

The matric ${\mathsf{A}}_4^{\left(8\right)}$ has analogous structure so

$${\mathsf{A}}_4^{\left(8\right)}=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& 1& -1\\ 1& b_8& -1& -b_8\\ 1& -b_8& -1& b_8\end{array}\right]=\left[\begin{array}{cc}{\mathsf{A}}_2^{\left(8\right)}& {\mathsf{A}}_2^{\left(8\right)}\\ {\mathsf{B}}_2^{\left(8\right)}& -{\mathsf{B}}_2^{\left(8\right)}\end{array}\right]=\left({\mathsf{A}}_2^{\left(8\right)}\oplus {\mathsf{B}}_2^{\left(8\right)}\right)\left({\mathsf{H}}_2{\otimes \mathsf{I}}_2\right),$$

(22)

where

$${\mathsf{A}}_2^{\left(8\right)}=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]={\mathsf{H}}_2,{\mathsf{B}}_2^{\left(8\right)}=\left[\begin{array}{cc}1& b_8\\ 1& -b_8\end{array}\right]={\mathsf{H}}_2\left[\begin{array}{cc}1& 0\\ 0& b_8\end{array}\right].$$

Let us reorder the columns of the matrix ${\mathsf{B}}_4^{\left(8\right)}$ using the ${\pi }_4$ permutation. We obtain

$${\breve{\mathsf{B}}}_4^{\left(8\right)}=\left[\begin{array}{cccc}1& b_8& a_8& -a_8^{*}\\ 1& -b_8& -a_8^{*}& a_8\\ 1& b_8& -a_8& a_8^{*}\\ 1& -b_8& a_8^{*}& -a_8\end{array}\right]=\left[\begin{array}{cc}{\mathsf{B}}_2^{\left(8\right)}& {\mathsf{C}}_2^{\left(8\right)}\\ {\mathsf{B}}_2^{\left(8\right)}& -{\mathsf{C}}_2^{\left(8\right)}\end{array}\right]=({\mathsf{H}}_2{\otimes \mathsf{I}}_2)({\mathsf{B}}_2^{\left(8\right)}\oplus {\mathsf{C}}_2^{\left(8\right)}),$$

(23)

where

$${\mathsf{C}}_2^{\left(8\right)}=\left[\begin{array}{cc}{a}_8& -a_8^{*}\\ -a_8^{*}& a_8\end{array}\right]={\mathsf{H}}_2\left[\begin{array}{cc}{\displaystyle \frac{1}{2}}(a_8-a_8^{*})& 0\\ 0& {\displaystyle \frac{1}{2}}(a_8+a_8^{*})\end{array}\right]{\mathsf{H}}_2.$$

Taking into account the obtained factorization schemes, the final computational procedure can be written as follows:

$${\mathsf{y}}_8={\mathsf{P}}_8^{\left(2\right)}{\mathsf{W}}_8^{\left(4\right)}{\mathsf{W}}_8^{\left(3\right)}{\mathsf{D}}_8{\mathsf{W}}_8^{\left(2\right)}{\mathsf{P}}_8^{\left(1\right)}{\mathsf{W}}_8^{\left(1\right)}{\mathsf{x}}_8,$$

(24)

where

$${\mathsf{W}}_8^{\left(1\right)}={\mathsf{H}}_2\otimes {\mathsf{I}}_4=\left[\begin{array}{cccccccc}1& 0& 0& 0& 1& 0& 0& 0\\ 0& 1& 0& 0& 0& 1& 0& 0\\ 0& 0& 1& 0& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 0& 0& -1& 0& 0& 0\\ 0& 1& 0& 0& 0& -1& 0& 0\\ 0& 0& 1& 0& 0& 0& -1& 0\\ 0& 0& 0& 1& 0& 0& 0& -1\end{array}\right],$$

$${\mathsf{P}}_8^{\left(1\right)}={\mathsf{I}}_4\oplus {\mathsf{P}}_4,{\mathsf{P}}_4=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right],$$

$${\mathsf{W}}_8^{\left(2\right)}={(\mathsf{I}}_2\otimes {\mathsf{H}}_2)\oplus ({\mathsf{I}}_2\oplus {\mathsf{H}}_2)=\left[\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& \begin{array}{cc}1& 0\\ 0& 1\end{array}\\ \begin{array}{cc}1& 0\\ 0& 1\end{array}& \begin{array}{cc}-1& 0\\ 0& -1\end{array}\end{array}& {\mathbf{0}}_4\\ {\mathbf{0}}_4& \begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}\end{array}\end{array}\right],$$

$${\mathsf{W}}_8^{\left(3\right)}={\mathsf{I}}_4\otimes {\mathsf{H}}_2=\left[\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}1& 1\\ 1& -1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}\end{array}& {\mathbf{0}}_4\\ {\mathbf{0}}_4& \begin{array}{cc}\begin{array}{cc}1& 1\\ 1& -1\end{array}& {\mathbf{0}}_2\\ {\mathbf{0}}_2& \begin{array}{cc}1& 1\\ 1& -1\end{array}\end{array}\end{array}\right],$$

$${\mathsf{W}}_8^{\left(4\right)}={\mathsf{I}}_4\oplus \left({\mathsf{H}}_2\otimes {\mathsf{I}}_2\right)=\left[\begin{array}{cc}\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}& {\mathbf{0}}_4\\ {\mathbf{0}}_4& \begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& -1& 0\\ 0& 1& 0& -1\end{array}\end{array}\right],$$

$${\mathsf{P}}_8^{\left(2\right)}=\left[\begin{array}{cccccccc}1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right],$$

$${\mathsf{D}}_8=diag\left(\mathrm{1,1},1,s_0^{\left(8\right)},1,s_1^{\left(8\right)},s_2^{\left(8\right)},s_3^{\left(8\right)}\right),$$

$$s_0^{\left(8\right)}=s_1^{\left(8\right)}=b_8=-0.57735i-0.57735j-0.57735k,$$

$$s_2^{\left(8\right)}={\displaystyle \frac{1}{2}}\left(a_8-a_8^{*}\right)=-0.40825i-0.40825j-0.40825k,$$

$$s_3^{\left(8\right)}={\displaystyle \frac{1}{2}}(a_8+a_8^{*})=0.70711.$$

Figure 7 shows a data-flow diagram of the proposed algorithm for the eight-point QDFT with reduced computation complexity.

4. Discussion of Computation Complexity

Since the input data lengths are quite small and the diagrams representing the data flow during the computation are relatively simple, it is not difficult to estimate the computational complexity of the synthesized algorithms.

Table 1. The number of quaternion-valued arithmetic operations necessary to calculate small-size QDFT.

Size	Naive Methods		Proposed Algorithms
N	Number of multiplications	Number of additions	Number of multiplications	Number of additions
2	—	2	—	2
3	4	6	2	8
4	4	12	1	8
5	16	20	5	23
6	16	30	4	22
7	36	42	8	46
8	32	56	4	26

and

Table 2. The number of real-valued arithmetic operations necessary to calculate small-size QDFT.

Size	Naive Methods		Proposed Algorithms
N	Number of multiplications	Number of additions	Number of multiplications	Number of additions	Number of right or left shifts
2	—	8	—	8	—
3	64	72	—	40	8
4	64	96	4	40	—
5	256	272	12	116	4
6	256	312	—	104	16
7	576	600	32	216	—
8	512	608	16	128	—

show the number of quaternion and real-valued arithmetic operations for computing small-size QDFTs. One can see that the implementation of the obtained solutions requires fewer multiplications than the implementation based on naive methods for computing QDFTs. Reducing the number of multiplications is especially important when designing specialized fully parallel VLSI processors. Minimizing the number of required multipliers allows reducing the dissipated power and reduces the cost of implementing the entire system. This is because the hardware multiplier is more complex than the adder. It also takes up much more space on the chip than the adder. It is known that the complexity of implementing a hardware multiplier grows quadratically with the operand size, while the hardware complexity of a binary adder grows only linearly [32]. Therefore, reducing the number of multipliers is of great importance in the hardware implementation of the developed algorithms, even if this leads to a small increase in the number of adders.

5. Conclusions

In this paper, we studied the possibilities of reducing the computational complexity of calculating small-sized left-handed QDFTs. We also proposed new algorithms for implementing these transforms for N = 2, 3, 4, 5, 6, 7, and 8. The implementation of the synthesized algorithms takes less time than the implementation based on naive, non-optimized solutions. It is impossible not to notice that accounting for the specific properties of the structure of the basic transform matrices for specific N allows us to obtain better solutions than those provided by the method described in [28]. The use of these algorithms allows us to reduce the computational complexity of these transforms, thereby reducing the complexity of their hardware implementation. From Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6 and Figure 7, we can conclude that the proposed algorithms have a pronounced parallel modular structure. This allows us to simplify the mapping of the algorithms into the ASIC structure and unify its implementation in FPGAs. Therefore, acceleration of calculations when implementing these algorithms can also be achieved by parallelizing the computational processes. In the case of hardware implementation of QDFT, the minimization of the number of built-in binary multipliers turns out to be even more significant than in the case of implementation of conventional calculations. This is primarily because the implementation of just one quaternion multiplier requires the use of as many as 16 conventional binary multipliers. In one of our latest papers, a quaternion multiplier structure was proposed that uses only eight conventional binary multipliers [33]. This multiplier uses the well-known fast Walsh–Hadamard transform algorithm to implement the quaternion multiplication operation. Thus, the design of specialized hardware QDFT processors should be considered, taking into account the new solutions obtained. In future works, we plan to consider these issues and design modules for implementing small-size QDFTs using the Verilog language. These issues require a more in-depth discussion and are currently beyond the scope of this paper.