Mathematical Properties of Pyramid-Transform-Based Resolution Conversion and Its Applications

Abstract

In this paper, we aim to clarify the statistical and geometric properties of linear resolution conversion for registration between different resolutions observed using the same modality. The pyramid transform is achieved by smoothing and downsampling. The dual operation of the pyramid transform is achieved by linear smoothing after upsampling. The rational-order pyramid transform is decomposed into upsampling for smoothing and the conventional integer-order pyramid transform. By controlling the ratio between upsampling for smoothing and downsampling in the pyramid transform, the rational-order pyramid transform is computed. The tensor expression of the multiway pyramid transform implies that the transform yields orthogonal base systems for any ratio of the rational pyramid transform. The numerical evaluation of the transform shows that the rational-order pyramid transform preserves the normalised distribution of greyscale in images.

10 Appendix

1.1 10.1 Alternative Proof of Theorem 1

The relation

$$\begin{aligned}{} & {} \int _{\textbf{R}^n}\int _{\textbf{R}^n} f(\varvec{y})w(\varvec{x}-\varvec{y})g(\varvec{x})\textrm{d}\varvec{y}\textrm{d}\varvec{x} \end{aligned}$$

$$\begin{aligned}{} & {} \qquad =\int _{\textbf{R}^n}\int _{\textbf{R}^n} f(\varvec{x})w(\varvec{x}-\varvec{y})g(\varvec{y})\textrm{d}\varvec{x}\textrm{d}\varvec{y} \end{aligned}$$

(174)

is satisfied if $w(-\varvec{x})=w(\varvec{x})$.

We have the relation

$$\begin{aligned} (R_\sigma f,g)= & {} \int _{\textbf{R}^n} \left( \int _{\textbf{R}^n}w_\sigma (\varvec{y})f(\sigma \varvec{x}-\varvec{y})\textrm{d}\varvec{y}\right) g(\varvec{x}) \textrm{d}\varvec{x}\nonumber \\= & {} \int _{\textbf{R}^n} \left( \int _{\textbf{R}^n}w_\sigma (\sigma \varvec{x}-\varvec{y})f(\varvec{y})\textrm{d}\varvec{y}\right) g(\varvec{x}) \textrm{d}\varvec{x}\nonumber \\= & {} \int _{\textbf{R}^n} f(\varvec{x})\left( \int _{\textbf{R}^n} w_\sigma (\varvec{x}-\varvec{y})\sigma ^ng\left( \frac{\varvec{y}}{\sigma }\right) d\varvec{y}\right) \textrm{d}\varvec{x}\nonumber \\= & {} \int _{\textbf{R}^n} f(\varvec{x})\left( \sigma ^n\int _{\textbf{R}^n} w_\sigma (\varvec{u}) g\left( \frac{\varvec{x}-\varvec{u}}{\sigma }\right) \textrm{d}\varvec{u}\right) \textrm{d}\varvec{x}\nonumber \\= & {} (f, E_\sigma g). \end{aligned}$$

(175)

1.2 10.2 Discrete Scale Space Transform and Binomial Coefficients

The coefficients of the binomial polynomial

$$\begin{aligned}{} & {} (1+x)^{2n}=c_{2n}(0)+c_{2n}(1)x\nonumber \\{} & {} \qquad +c_{2n}(2)x^2+\cdots +c_{2n}(2n)x^{2n} \end{aligned}$$

(176)

are

$$\begin{aligned} c_{2n}(k)= \left( \begin{array}{c} 2n\\ k \end{array}\right) =\frac{(2n!)}{k!(2n-k)!} \end{aligned}$$

(177)

for $k=0,1,\ldots ,2n$. The discrete function

$$\begin{aligned} p_{2n}(k)=\frac{1}{2^{2n}}c_{2n}(k+n)=\frac{1}{2^{2n}} \left( \begin{array}{c} 2n\\ n+k \end{array}\right) , \end{aligned}$$

(178)

for $k=-n,-n+1,\ldots , n$, satisfies the conditions $p_{2n}(k)\ge 0$ for $|k|\le n$, $p_{2n}(k)=0$ for $|k|>n$, $p_{2n}(k)=p_{2n}(-k)$, $\max p_{2n}(k)=p_{2n}(0)$ and $\sum _{k=-n}^n p_{2n}(k)=1$.

Since the equality

$$\begin{aligned} \left( \begin{array}{c} 2n\\ n+k \end{array} \right) = \left( \begin{array}{c} 2n\\ n-k \end{array} \right) \end{aligned}$$

(179)

is satisfied for $k=0,1,2,\ldots , n$, the discrete function

$$\begin{aligned} q_{2n}(k)=\frac{1}{2^{2n}}\left( \begin{array}{c} 2n\\ n-|k| \end{array} \right) \end{aligned}$$

(180)

implies the relations

$$\begin{aligned} q_{2n}(k)=p_{2n}(k)=\frac{1}{2^{2n}}c_{2n}(k+n) \end{aligned}$$

(181)

for $k=0,1,2,\ldots , n$ and $q_{2n}(k)=q_{2n}(-k)$ for $k=0,-1,-2,\ldots ,-n$.

Since

$$\begin{aligned} (1+x)^{2m}(1+x)^{2n}=(1+x)^{2(m+n)}=\sum _{k=0}^{2(m+n)}c_{2(m+n)}(k)x^k \nonumber \\ \end{aligned}$$

(182)

and

$$\begin{aligned} q_{2(m+n)}(k)=p_{2(m+n)}(k)= \frac{1}{2^{2(m+n)}}c_{2(m+n)}(k), \end{aligned}$$

(183)

the relation

$$\begin{aligned} \sum _{l=-\infty }^\infty q_{2m}(l)q_{2n}(k-l)= q_{2(m+n)}(k) \end{aligned}$$

(184)

is satisfied. Therefore, the kernel function $\{q_{2n}(k)\}_{k=-n}^n$ satisfies the semi-group property with respect to the parameter n.

1.3 10.3 Scaling Linear Scale-Space Transform

For a triplet of non-negative functions $f(\varvec{x})$, $g(\varvec{x})$ and $h(\varvec{x})$, the equality

$$\begin{aligned}{} & {} \int _{\textbf{R}^n}\int _{\textbf{R}^n}g(\varvec{x})h(\varvec{y}-\varvec{x})f(\varvec{y})\textrm{d}\varvec{y}\textrm{d}\varvec{x}\nonumber \\= & {} \int _{\textbf{R}^n}\int _{\textbf{R}^n}g(\varvec{x})h(\varvec{x}-\varvec{y})f(\varvec{y})\textrm{d}\varvec{x}\textrm{d}\varvec{y}\nonumber \\= & {} \int _{\textbf{R}^n}\int _{\textbf{R}^n}h(\varvec{x}-\varvec{y})g(\varvec{x})f(\varvec{y})\textrm{d}\varvec{x}\textrm{d}\varvec{y} \end{aligned}$$

(185)

is satisfied assuming $h(-\varvec{x})=h(\varvec{x})$. Let

$$\begin{aligned} G_\tau (x)=\frac{1}{\sqrt{2\pi \tau }}\exp \left( -\frac{x^2}{2\tau }\right) . \end{aligned}$$

(186)

Setting

$$\begin{aligned} h(y)=\int _{-\infty }^\infty G_\tau (x-y)f(\varvec{x})\textrm{d}y, \end{aligned}$$

(187)

the quadratic form satisfies the relation

$$\begin{aligned}{} & {} \int _{-\infty }^\infty \int _{-\infty }^\infty f(x)G_\tau (y-x) s g\left( \frac{y}{s}\right) \textrm{d}y\textrm{d}x\nonumber \\= & {} \int _{-\infty }^\infty h(y)s g\left( \frac{y}{s}\right) \textrm{d}y\nonumber \\= & {} \int _{-\infty }^\infty h(sy) g(y)\textrm{d}y. \end{aligned}$$

(188)

Therefore, the dual transform of the scaling linear scale-space transform is

$$\begin{aligned} f(x)=G_\tau ^{s*} g(x) =s\int _{-\infty }^\infty G_\tau (y)f\left( \frac{x-y}{s}\right) \textrm{d}y. \end{aligned}$$

(189)

The decomposition of the scaling linear scale-space transform

$$\begin{aligned} G_\tau ^\sigma f(x)=\int _{-\infty }^\infty G_\tau (s y-x)f(x)\textrm{d}y \end{aligned}$$

(190)

as $G_\tau ^s=S_s G_\tau$ implies the inequality $| G_\tau ^\sigma f|_2^2\le \frac{1}{\sigma ^n}|f|_2^2$ since $\int _{-\infty }^\infty G_s^\tau (x)dx=1$.

1.4 10.4 Numerical Methods

1.4.1 10.4.1 Vectorisation of Discrete Functions

Using the two-dimensional discrete function $\{f_{i,j}\}_{i=1,j=1}^{n,n}$ constructed from the sample $f_{i,j}=f(x_i,y_j)$ of the function f(x, y), we the image matrix

$$\begin{aligned} \varvec{F}=\left( \begin{array}{cccc} f_{1,1}&{}f_{1,2}&{}\ldots &{}f_{1,n}\\ f_{2,1}&{}f_{2,2}&{}\ldots &{}f_{2,n}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ f_{n,1}&{}f_{n,2}&{}\ldots &{}f_{n,n} \end{array}\right) . \end{aligned}$$

(191)

Next, we construct the vector

$$\begin{aligned} \varvec{f}=\varvec{e}_1\otimes \varvec{F}\varvec{e}_1+\varvec{e}_2\otimes \varvec{F}\varvec{e}_2, \end{aligned}$$

(192)

where $\varvec{e}_1=(1,0)^\top$ and $\varvec{e}_2=(0,1)^\top$, for numerical computation of the partial differential equation.

From the discrete vector field $\{u_{1,i,j}, u_{2,i,j}\}_{i=1,j=1}^{n,n}$ for $u_{1,i,j}=u_1(x_i,y_j)$ and $u_{2,i,j}=u_2(x_i,y_j)$ of the vector-valued function $\varvec{u}(x,y)=(u_1(x,y),u_2(x,y))^\top$, we construct the vector

$$\begin{aligned} \varvec{u}=\left( \begin{array}{c} \varvec{u}_1\\ \varvec{u}_2 \end{array}\right) =\varvec{e}_1\otimes \varvec{u}_2+\varvec{e}_2\otimes \varvec{u}_2, \end{aligned}$$

(193)

where, for $i=1,2$,

$$\begin{aligned}{} & {} \varvec{u}_1=\varvec{e}_1\otimes \varvec{U}_1\varvec{e}_1+\varvec{e}_2\otimes \varvec{U}_2\varvec{e}_2, \nonumber \\{} & {} \varvec{u}_2=\varvec{e}_1\otimes \varvec{U}_2\varvec{e}_1+\varvec{e}_2\otimes \varvec{U}_2\varvec{e}_2 \end{aligned}$$

(194)

and

$$\begin{aligned} \varvec{U}_1= & {} \left( \begin{array}{cccc} u_{1,1,1}&{}u_{1,1,2}&{}\ldots &{}u_{1,1,n}\\ u_{1,2,1}&{}u_{1,2,2}&{}\ldots &{}u_{1,2,n}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ u_{1,n,1}&{}u_{1,n,2}&{}\ldots &{}u_{1,n,n} \end{array}\right) , \end{aligned}$$

(195)

$$\begin{aligned} \varvec{U}_2= & {} \left( \begin{array}{cccc} u_{2,1,1}&{}u_{2,1,2}&{}\ldots &{}u_{2,1,n}\\ u_{2,2,1}&{}u_{2,2,2}&{}\ldots &{}u_{2,2,n}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ u_{2,n,1}&{}u_{2,n,2}&{}\ldots &{}u_{2,n,n} \end{array}\right) . \end{aligned}$$

(196)

1.4.2 10.4.2 Discretisation of Linear Diffusion Equation

The hierarchical images in the linear scale space are computed by solving the three-dimensional diffusion equation [99, 127]

$$\begin{aligned} \frac{\partial f}{\partial \tau }=\frac{1}{2}\left( \frac{\partial ^2 f}{\partial x^2} + \frac{\partial ^2 f}{\partial y^2} + \frac{\partial ^2 f}{\partial z^2}\right) \end{aligned}$$

(197)

with the Neumann condition. The numerical computation of the diffusion equation is achieved by semi-implicit discretisation [99, 126, 127]

$$\begin{aligned} \frac{\varvec{f}^{(m+1)}-\varvec{f}^{(m)}}{\tau } =\frac{1}{2}\varvec{K}_n^{(3)}\varvec{f}^{(m+1)}, \end{aligned}$$

(198)

where

$$\begin{aligned} \varvec{K}_n^{(3)}= & {} \frac{1}{4}(\varvec{D}_n\oplus \varvec{D}_n\oplus \varvec{D}_n)\nonumber \\= & {} \frac{1}{4}\varvec{D}_n\otimes \varvec{I}_n\otimes \varvec{I}_n\nonumber \\{} & {} \hspace{1cm}+\frac{1}{4}\varvec{I}_n\otimes \varvec{D}_n\otimes \varvec{I}_n\nonumber \\{} & {} \hspace{2cm}+\frac{1}{4}\varvec{I}_n\otimes \varvec{I}_n\otimes \varvec{D}_n, \end{aligned}$$

(199)

for $n\times n$ matrices.

1.4.3 10.4.3 Variational Image Registration

The solutions of variational problems for image registration are the solutions for $t=\infty$ of the system of diffusion equations

$$\begin{aligned} \frac{\partial \varvec{v}}{\partial t}= & {} \Delta \varvec{v} -\frac{1}{\lambda }(f(\varvec{y}+\varvec{R}_\sigma \varvec{u})\nonumber \\{} & {} -R_\sigma g(\varvec{y}-\varvec{v})) \nabla _{\varvec{v}} R_\sigma g(\varvec{y}-\varvec{v}), \end{aligned}$$

(200)

$$\begin{aligned} \frac{\partial \varvec{u}}{\partial t}= & {} \Delta \varvec{u} -\frac{1}{\lambda }(E_\sigma f(\varvec{x}+\varvec{E}_\sigma \varvec{v})\nonumber \\{} & {} -g(\varvec{x}-\varvec{u}))\nabla _{\varvec{u}} g(\varvec{x}-\varvec{u}), \end{aligned}$$

(201)

with the boundary conditions

$$\begin{aligned} \frac{\partial }{\partial \varvec{m}}\varvec{v}=0,\, \, \, \frac{\partial }{\partial \varvec{n}}\varvec{u}=0. \end{aligned}$$

(202)

Semi-implicit discretisation of Eqs. (200) and (201) yields the system of iterations

$$\begin{aligned}{} & {} \left( \begin{array}{cc} 1,&{}0\\ 0,&{}1 \end{array}\right) \otimes (\varvec{I}-\tau \varvec{K}_n)\varvec{v}^{(m+1)}\nonumber \\{} & {} =\varvec{v}^{(m)}\nonumber \\{} & {} - \frac{\tau }{\lambda } (f(\varvec{y}+\varvec{R}_{q/p} \varvec{u}^{(m)})\nonumber \\{} & {} -R_{q/p} g(\varvec{y}-\varvec{v}^{(m)})) \nabla _{\varvec{v}} R_{q/p} g(\varvec{y}-\varvec{v}^{(m)}), \end{aligned}$$

(203)

$$\begin{aligned}{} & {} \left( \begin{array}{cc} 1,&{}0\\ 0,&{}1 \end{array}\right) \otimes (\varvec{I}-\tau \varvec{K}_{np/q})\varvec{u}^{(m+1)}\nonumber \\{} & {} =\varvec{u}^{(m)}\nonumber \\{} & {} -\frac{\tau }{\lambda }(E_{q/p} f(\varvec{x}+\varvec{E}_{q/p} \varvec{v}^{(m)})\nonumber \\{} & {} -g(\varvec{x}-\varvec{u}^{(m)})) \nabla _{\varvec{u}} g(\varvec{x}-\varvec{u}^{(m)}). \end{aligned}$$

(204)

For the Neumann boundary condition, the Laplacian matrix is expressed as

$$\begin{aligned} \varvec{K}_k =\frac{1}{4}\varvec{D}_k\otimes \varvec{I}_k+\frac{1}{4}\varvec{I}_k\otimes \varvec{D}_k, \end{aligned}$$

(205)

for $k\times k$ matrices.

1.4.4 10.4.4 Resolution Interpolation

As the minimiser of Eq. (172), we have the system of differential equations

$$\begin{aligned} \Delta h&=\frac{1}{\mu } \left\{ (h(\varvec{x}+\varvec{u})-f_{q/p}(\varvec{x}) )\right. \nonumber \\&\hspace{1cm}\left. +(h(\varvec{x}+\varvec{v})-g^{r/s}(\varvec{x})) \right\} , \end{aligned}$$

(206)

$$\begin{aligned} \Delta \varvec{u}&= \frac{1}{\lambda }\left\{ (h(\varvec{x}+\varvec{u})-f_{q/p}(\varvec{x}) )\nabla _{\varvec{u}} h(\varvec{x}+\varvec{u}) \right\} , \end{aligned}$$

(207)

$$\begin{aligned} \Delta \varvec{v}&= \frac{1}{\lambda }\left\{ (h(\varvec{x}+\varvec{v})-g^{r/s}(\varvec{x}) )\nabla _{\varvec{v}}h(\varvec{x}+\varvec{v}) \right\} . \end{aligned}$$

(208)

Then, the semi-implicit discretisation of the system of the diffusion equations

$$\begin{aligned} \frac{\partial h}{\partial t}= & {} \Delta h-a(\varvec{x}), \end{aligned}$$

(209)

$$\begin{aligned} \frac{\partial \varvec{u}}{\partial t}= & {} \Delta \varvec{u}-\varvec{b}(\varvec{x}), \end{aligned}$$

(210)

$$\begin{aligned} \frac{\partial \varvec{v}}{\partial t}= & {} \Delta \varvec{v}-\varvec{c}(\varvec{x}), \end{aligned}$$

(211)

for

$$\begin{aligned} a(\varvec{x})= & {} \frac{1}{\mu } \left\{ (h(\varvec{x}+\varvec{u})-f_{q/p}(\varvec{x}) )\right. \nonumber \\{} & {} \left. +(h(\varvec{x}+\varvec{v})-g^{r/s}(\varvec{x})) \right\} , \end{aligned}$$

(212)

$$\begin{aligned} \varvec{b}(\varvec{x})= & {} \frac{1}{\lambda }\left\{ (h(\varvec{x}+\varvec{u})-f_{q/p}(\varvec{x}) )\nabla _{\varvec{u}} h(\varvec{x}+\varvec{u}) \right\} , \end{aligned}$$

(213)

$$\begin{aligned} \varvec{c}(\varvec{x})= & {} \frac{1}{\lambda }\left\{ (h(\varvec{x}+\varvec{v})-g^{r/s}(\varvec{x}) )\nabla _{\varvec{v}}h(\varvec{x}+\varvec{v}) \right\} \end{aligned}$$

(214)

yields the system of iterations

$$\begin{aligned} \frac{h^{(m+1)}-h^{(m)}}{\tau }= & {} \varvec{K}_nh^{(m+1)}-\frac{1}{\mu }a^{(m)}, \end{aligned}$$

(215)

$$\begin{aligned} \frac{\varvec{u}^{(m+1)}-\varvec{u}^{(m)}}{\tau }= & {} \left( \begin{array}{cc} 1,&{}0\\ 0,&{}1 \end{array}\right) \otimes \varvec{K}_n\varvec{u}^{(m+1)} -\frac{1}{\lambda }\varvec{b}^{(m)},\end{aligned}$$

(216)

$$\begin{aligned} \frac{\varvec{v}^{(m+1)}-\varvec{v}^{(m)}}{\tau }= & {} \left( \begin{array}{cc} 1,&{}0\\ 0,&{}1 \end{array}\right) \otimes \varvec{K}_n\varvec{v}^{(m+1)} -\frac{1}{\lambda }\varvec{c}^{(m)}, \end{aligned}$$

(217)

where

$$\begin{aligned} a^{(m)}= & {} (h(\varvec{x}+\varvec{u}^{(m)})^{(m)}-f_{q/p}(\varvec{x}) )\nonumber \\{} & {} +(h(\varvec{x}+\varvec{v}^{(m)})^{(m)}-g^{s/r}(\varvec{x})), \end{aligned}$$

(218)

$$\begin{aligned} \varvec{b}^{(m)}= & {} (h(\varvec{x}+\varvec{u}^{(m)})^{(m)}\nonumber \\{} & {} -f_{q/p}(\varvec{x}) )^{(m)} \nabla _{\varvec{u}} h(\varvec{x}+\varvec{u}^{(m)})^{(m)}, \end{aligned}$$

(219)

$$\begin{aligned} \varvec{c}^{(m)}= & {} (h(\varvec{x}+\varvec{v}^{(m)})^{(m)}\nonumber \\{} & {} -g_{s/r}(\varvec{x})) \nabla _{\varvec{v}}h(\varvec{x}+\varvec{v}^{(m)})^{(m)}. \end{aligned}$$

(220)