Golden section transform

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An example of 2-Level DCT-Low Golden Section Transform(DCT-LGST).

The golden section transform was proposed and used in digital image multiresolution analysis in 2007 by Jun Li.[1] The ratio of 1 level "Low" part length to "High" part length is the approximation of the golden ratio $\varphi$ (Type L) or $\frac{1}{\varphi}$ (Type H),

where

$\varphi = \frac{1+\sqrt{5}}{2} = 1.618\ldots ;$
$\frac{1}{\varphi}= \frac{-1+\sqrt{5}}{2} = 0.618\ldots$

DCT-golden section transform (DCT-GST)

It is well known that the Fibonacci sequence is:

$0,\;1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;89,\;144,\;233,\;377,\;610,\;987\ldots\;$ Template:OEIS
$F_0 = 0,\; F_1= 1,F_n = F_{n-1} + F_{n-2},$ make a golden section tree of the Fibonacci number 8:
The golden section tree of 8
Fibonacci number DCT-LGST Reverse order Symmetrical form DCT-HGST Reverse order symmetrical form
1 1 1 1
2 2 2 2 2 2 2
3 12 21 21 3 3 3
5 212 212 212 23 32 32
8 12212 21221 21212 323 323 323
13 21212212 21221212 21212212 23323 32332 32323
21 1221221212212 2122121221221 2121221221212 32323323 32332323 32323323

DCT-low golden section transform (DCT-LGST)

Take a sequence $(p_0,p_1,p_2,p_3,p_4,p_5,p_6,p_7)$ of 8 elements for example, we get the form of 8 is $12212$, 1-level DCT-LGST of the 8 elements is:

$p_0,\;\frac{p_1 + p_2}{2},\;\frac{p_3 + p_4}{2},\;p_5,\;\frac{p_6 + p_7}{2},\;\frac{p_1 - p_2}{2},\;\frac{p_3 - p_4}{2},\;\frac{p_6 - p_7}{2}$

the "low" part of the result is:

$p_0,\;\frac{p_1 + p_2}{2},\;\frac{p_3 + p_4}{2},\;p_5,\;\frac{p_6 + p_7}{2}$

according to the form $212$ of 5, 2-level DCT-LGST of the 8 elements is:

$\frac{p_0 + \frac{p_1 + p_2}{2}}{2},\;\frac{p_3 + p_4}{2},\;\frac{p_5 + \frac{p_6 + p_7}{2}}{2},\;\frac{p_0 - \frac{p_1 + p_2}{2}}{2},\;\frac{p_5 - \frac{p_6 + p_7}{2}}{2},\;\frac{p_1 - p_2}{2},\;\frac{p_3 - p_4}{2},\;\frac{p_6 - p_7}{2}$
An example of 1-Level DCT-Low Golden Section Transform(DCT-LGST).

Let $L_{N}$ be the $N\times N$ non-normalized DCT-LGST matrix, where $N=F_k, (k\ge3,k \in \mathbb{Z})$

when $N=2$, 1-level non-normalized DCT-LGST matrix is:

$L_{2}=\frac{1}{2}\begin{bmatrix} 1 & 1\\1 & -1\\ \end{bmatrix}$

when $N=3$, 2-level non-normalized DCT-LGST matrix and the inverse matrix is:

$L_{3}=\frac{1}{4}\begin{bmatrix} 2 & 1 & 1\\2 & -1 & -1\\0 & 2 & -2\\ \end{bmatrix}$; $L_{3}^{-1}=\begin{bmatrix} 1 & 1 & 0\\1 & -1 & 1\\1 & -1 & -1\\ \end{bmatrix}$

when $N=5$, 3-level non-normalized DCT-LGST matrix and the inverse matrix is:

$L_{5}=\frac{1}{8}\begin{bmatrix} 2 & 2 & 2 & 1 & 1\\2 & 2 & -2 & -1 & -1 \\0 & 0 & 4 & -2 & -2\\4 & -4 & 0 & 0 & 0 \\0 & 0 & 0 & 4 & -4\\ \end{bmatrix}$; $L_{5}^{-1}=\begin{bmatrix} 1 & 1 & 0 & 1 & 0\\1 & 1 & 0 & -1 & 0 \\1 & -1 & 1 & 0 & 0\\1 & -1 & -1 & 0 & 1 \\1 & -1 & -1 & 0 & -1\\ \end{bmatrix}$

when $N=8$, 4-level non-normalized DCT-LGST matrix and the inverse matrix is:

$L_{8}=\frac{1}{16}\begin{bmatrix} 4 & 2 & 2 & 2 & 2 & 2 & 1 & 1\\ 4 & 2 & 2 & -2 & -2 & -2 & -1 & -1\\ 0 & 0 & 0 & 4 & 4 & -4 & -2 & -2\\ 8 & -4 & -4 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 8 & -4 & -4\\ 0 & 8 & -8 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 8 & -8 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 8 & -8 \end{bmatrix}$;$L_{8}^{-1}=\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & -1 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & -1 & 0 & -1 & 0 & 0\\ 1 & -1 & 1 & 0 & 0 & 0 & 1 & 0\\ 1 & -1 & 1 & 0 & 0 & 0 & -1 & 0\\ 1 & -1 & -1 & 0 & 1 & 0 & 0 & 0\\ 1 & -1 & -1 & 0 & -1 & 0 & 0 & 1\\ 1 & -1 & -1 & 0 & -1 & 0 & 0 & -1 \end{bmatrix}$

when $N=8$, 4-level non-normalized reverse-order DCT-LGST matrix and the inverse matrix is:

$L_8=\frac{1}{16}\begin{bmatrix} 1 & 1 & 2 & 2 & 2 & 2 & 2 & 4\\ 1 & 1 & 2 & 2 & 2 & -2 & -2 & -4\\ 2 & 2 & 4 & -4 & -4 & 0 & 0 & 0\\ 4 & 4 & -8 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 4 & 4 & -8\\ 8 & -8 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 8 & -8 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 8 & -8 & 0 \end{bmatrix}$;$L_{8}^{-1}=\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 1 & 1 & 0 & -1 & 0 & 0\\ 1 & 1 & 1 & -1 & 0 & 0 & 0 & 0\\ 1 & 1 & -1 & 0 & 0 & 0 & 1 & 0\\ 1 & 1 & -1 & 0 & 0 & 0 & -1 & 0\\ 1 & -1 & 0 & 0 & 1 & 0 & 0 & 1\\ 1 & -1 & 0 & 0 & 1 & 0 & 0 & -1\\ 1 & -1 & 0 & 0 & -1 & 0 & 0 & 0 \end{bmatrix}$

also, when $N=8$, 4-level non-normalized symmetrical form DCT-LGST matrix and the inverse matrix is:

$L_{8}=\frac{1}{16}\begin{bmatrix} 1 & 1 & 2 & 2 & 2 & 4 & 2 & 2\\ 1 & 1 & 2 & 2 & 2 & -4 & -2 & -2\\ 2 & 2 & 4 & -4 & -4 & 0 & 0 & 0\\ 4 & 4 & -8 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 8 & -4 & -4\\ 8 & -8 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 8 & -8 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 8 & -8 \end{bmatrix}$;$L_{8}^{-1}=\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 1 & 1 & 0 & -1 & 0 & 0\\ 1 & 1 & 1 & -1 & 0 & 0 & 0 & 0\\ 1 & 1 & -1 & 0 & 0 & 0 & 1 & 0\\ 1 & 1 & -1 & 0 & 0 & 0 & -1 & 0\\ 1 & -1 & 0 & 0 & 1 & 0 & 0 & 0\\ 1 & -1 & 0 & 0 & -1 & 0 & 0 & 1\\ 1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 \end{bmatrix}$

DCT-high golden section transform (DCT-HGST)

An example of 1-Level DCT-High Golden Section Transform(DCT-HGST).

Similar to DCT-LGST, 1-level DCT-HGST of the 8 elements by the form $323$ is:

$\frac{p_0 + p_1+p_2}{3},\;\frac{p_3 + p_4}{2},\;\frac{p_5+p_6 + p_7}{3},\; \frac{\sqrt{6}}{6}p_0 - \frac{\sqrt{6}}{6}p_2,\;\frac{\sqrt{2}}{6}p_0 -\frac{\sqrt{2}}{3}p_1 +\frac{\sqrt{2}}{6}p_2,\;\frac{p_3 - p_4}{2} ,\;$

$\frac{\sqrt{6}}{6}p_5 - \frac{\sqrt{6}}{6}p_7 ,\;\frac{\sqrt{2}}{6}p_5 -\frac{\sqrt{2}}{3}p_6 +\frac{\sqrt{2}}{6}p_7$

DCT-HGST can be also defined by a $N\times N$ matrix $H_{N}$, where $N=F_k, (k\ge3,k \in \mathbb{Z})$, it is known that the $3\times 3$ non-normalized DCT matrix is:

$W=\frac{1}{3} \begin{bmatrix} 1 & 1& 1 \\ \frac{\sqrt{6}}{2} &0&-\frac{\sqrt{6}}{2} \\ \frac{\sqrt{2}}{2} &-\sqrt{2}& \frac{\sqrt{2}}{2} \end{bmatrix}$

when $N=8$, 2-level non-normalized DCT-HGST matrix and the inverse matrix is:

$H_{8}=\frac{1}{18}\begin{bmatrix} 2 & 2 & 2 & 3 & 3 & 2 & 2 & 2\\ \sqrt{6}&\sqrt{6}&\sqrt{6}& 0 & 0 & -\sqrt{6}&-\sqrt{6}&-\sqrt{6}\\ \sqrt{2}&\sqrt{2}&\sqrt{2}& -3\sqrt{2}& -3\sqrt{2}& \sqrt{2}&\sqrt{2}&\sqrt{2}\\ 3\sqrt{6} & 0 & -3\sqrt{6} & 0 & 0 & 0 & 0 & 0\\ 3\sqrt{2}&-6\sqrt{2}&3\sqrt{2}& 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 9 & -9 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 3\sqrt{6} & 0 & -3\sqrt{6}\\ 0 & 0 & 0 & 0 & 0 & 3\sqrt{2}&-6\sqrt{2}&3\sqrt{2}\\ \end{bmatrix};$

$H_{8}^{-1}=\begin{bmatrix} 1 & \frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 & 0\\ 1 & \frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & 0 & -\sqrt{2} & 0 & 0 & 0\\ 1 & \frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 & 0\\ 1 & 0 & -\sqrt{2} & 0 & 0 & 1 & 0 & 0\\ 1 & 0 & -\sqrt{2} & 0 & 0 & -1 & 0 & 0\\ 1 & -\frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 & 0 & \frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2}\\ 1 & -\frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 & 0 & 0 & -\sqrt{2}\\ 1 & -\frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} & 0 & 0 & 0 & -\frac{\sqrt{6}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix}$

DFT-golden section transform (DFT-GST)

Similarly, DFT-golden section transform can be defined as well, the $2\times 2$ DFT matrix is identical with $2\times 2$ DCT matrix, the $3\times 3$ non-normalized DFT matrix is:

$W=\frac{1}{3} \begin{bmatrix} 1 & 1& 1 \\ 1 &-\frac{1}{2} - \frac{\sqrt{3}}{2}i&-\frac{1}{2} + \frac{\sqrt{3}}{2}i \\ 1 &-\frac{1}{2} + \frac{\sqrt{3}}{2}i& -\frac{1}{2} - \frac{\sqrt{3}}{2}i \end{bmatrix}$

The lifting scheme of reverse-order DCT-LGST

Let $F_0 = 0,\; F_1= 1,$ let $\{X_u\}\,\!$ be the original sequence, let $\{S_v\}\,\!$ be a sequence as the "Low" part of reverse-order DCT-LGST, let $\{D_w\}\,\!$ be a sequence as the "High" part of reverse-order DCT-LGST, we have

$\{X_u\}=X_0, X_1, X_2, \ldots, X_u;\;$ $\{S_v\}=S_0, S_1, S_2, \ldots, S_v;\;$ $\{D_w\}=D_0, D_1, D_2, \ldots, D_w\;$

where $u=F_k-1,\;v=F_{k-1}-1,\;w=F_{k-2}-1\;(k\ge3,k \in \mathbb{Z})$

The lifting scheme of normalized reverse-order DCT-LGST is:

$\begin{cases}S_{M_n}^{0} = X_{L_n-1},\; D_{n-1}^{0} = X_{L_n}\\ D_{n-1}^{1} = S_{M_n}^{0}-D_{n-1}^{0},S_{M_n}^{1} = S_{M_n}^{0}-0.5D_{n-1}^{1}\\S_{M_n} = \sqrt{2}S_{M_n}^{1}, \;S_{L_m} = X_{H_m}, \;D_{n-1}=D_{n-1}^{1}/\sqrt{2}\end{cases}$

and the reconstruction algorithm is:

$\begin{cases}S_{M_n}^1 = S_{M_n}/\sqrt{2},\;D_{n-1}^1 =\sqrt{2}D_{n-1} \\S_{M_n}^0 = S_{M_n}^1+0.5D_{n-1}^1,\;D_{n-1}^0 = S_{M_n}^0-D_{n-1}^1 \\X_{L_n-1}= S_{M_n}^0,\;X_{H_m}=S_{L_m},\;X_{L_n}=D_{n-1}^0 \end{cases}$

where

$\{L_n\}=1, 4, 6, 9, 12, 14, 17, 19, \ldots\;$ Template:OEIS
$\{M_n\}=0, 2, 3, 5, 7, 8, 10, 11, \ldots\;$ Template:OEIS
$\{H_n\}=2, 7, 10, 15, 20, 23, 28, 31, \ldots\;$ Template:OEIS
$L_n = \left\lfloor \frac{3+\sqrt{5}}{2}n\right\rfloor-1 , M_n = \left\lfloor \frac{1+\sqrt{5}}{2}n\right\rfloor-1 , H_n = 2\left\lfloor \frac{1+\sqrt{5}}{2}n\right\rfloor+n-1$

where $m=1, 2, 3, 4, \ldots\;F_{k-3};\;n=1, 2, 3, 4, \ldots\;F_{k-2}\;(m,n \in \mathbb{N}^*)$