Fitting concentric elliptical shapes under general model

Abstract

Fitting concentric ellipses is a crucial yet challenging task in image processing, pattern recognition, and astronomy. To address this complexity, researchers have introduced simplified models by imposing geometric assumptions. These assumptions enable the linearization of the model through reparameterization, allowing for the extension of various fitting methods. However, these restrictive assumptions often fail to hold in real-world scenarios, limiting their practical applicability. In this work, we propose two novel estimators that relax these assumptions: the Least Squares method (LS) and the Gradient Algebraic Fit (GRAF). Since these methods are iterative, we provide numerical implementations and strategies for obtaining reliable initial guesses. Moreover, we employ perturbation theory to conduct a first-order analysis, deriving the leading terms of their Mean Squared Errors and their theoretical lower bounds. Our theoretical findings reveal that the GRAF is statistically efficient, while the LS method is not. We further validate our theoretical results and the performance of the proposed estimators through a series of numerical experiments on both real and synthetic data.

Appendices

Appendix A Overview of fitting ellipse

Although the methods proposed in this paper are for fitting concentric ellipses to data and must be handled differently than that of the single ellipse fitting problem, we devote this section to discuss briefly single ellipse case.

Fig. 7

Ellipse geometry

Figure 7 shows the geometric (natural) parameters of the ellipse and the orthogonal distance $d_i$ from the observation $(x_{i}, y_{i})$ to the true curve. If the observation is close to the curve then the algebraic distance is also close to zero, i.e.,

$$\begin{aligned} Ax_{j}^2 + 2Bx_{i}y_{i}+Cy_{i}^2+2Dx_{i}+2Ey_{i}+F\approx 0, \end{aligned}$$

(A1)

In terms of $\varvec{\mathcal {A}}=\bigl (A,B,C,D,E,F\bigr )^\top$ and $\varvec{\xi }_i=\bigl (x_{i}^2, 2x_{i}y_{i},y_{i}^2,2x_{i},2y_{i},1\bigr )^\top$, then $\varvec{\xi }_{i}^\top \varvec{\mathcal {A}}\approx 0$.

The simplest approach to estimate the parameters of $\varvec{\mathcal {A}}$ is the Least Squares (LS), which minimizes ${\mathcal {J}}(\varvec{\mathcal {A}})=\varvec{\mathcal {A}}^\top \varvec{M}\varvec{\mathcal {A}}$, where $\varvec{M}=\sum _{j=1}^{n}\varvec{\xi }_j\varvec{\xi }_j^\top$. The advantage of using this algebraic representation $\varvec{\xi }_j^\top \varvec{\mathcal {A}}$ is that it provides closed-form solutions for ${\mathcal {J}}$. The parametric space of the conic includes an ellipse, hyperbola, parabola, parallel lines, imaginary lines, and many other shapes, depending on the values of the parameters. For example, if $AC-B^2=0$, then (A1) will be the expression for a parabola, but if $AC-B^2 < 0$, the expression is that of a hyperbola. Only if $AC-B^2>0$ will the expression be that of an ellipse. Therefore, a constraint shall be imposed on the parametric space to remove any indeterminacy. To avoid this, a constraint, for example, $\Vert \varvec{\mathcal {A}}\Vert ^2=1$, is imposed. This is a constrained minimizing problem. Differentiating $\varvec{\mathcal {A}}^\top \varvec{M}\varvec{\mathcal {A}}-\lambda \left( \Vert \varvec{\mathcal {A}}\Vert ^2-1\right)$ with respect to $\varvec{\mathcal {A}}$ and setting the result equal to zero yields $\varvec{M}\hat{\varvec{\mathcal {A}}_{\textrm{L}}}=\lambda \hat{\varvec{\mathcal {A}}_{\textrm{L}}}$, which is an eigenvalue problem. The LS estimator suffers from heavy bias. Consequently, other methods under the algebraic approach have been proposed by imposing different constraints. These constraints are typically of the form $\varvec{\mathcal {A}}^\top \varvec{N}\varvec{\mathcal {A}}=1$. In this paper, we will use the widely known Taubin method to get an initial guess for our iterative methods. However, the Taubin method imposes a different constraint which depends on the data and it takes the form $\varvec{N}_\textrm{T}=4\sum _{j=1}^{n}\varvec{V}_i$, where

$$\begin{aligned} \varvec{V}_j=\begin{bmatrix} x_j^2 &{} x_jy_j &{} 0 &{} x_j &{} 0 &{} 0 \\ x_jy_j &{} x_j^2+y_j^2 &{} x_jy_j &{} y_j &{}0 &{} 0\\ 0 &{} x_jy_j &{} y_j^2 &{} 0 &{} 0 &{} 0\\ x_j &{} y_j &{} 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0&{} 0 &{} 0&{} 0 &{} 0 \\ 0 &{} 0&{} 0 &{} 0&{} 0 &{} 0 \end{bmatrix}_{6\times 6}. \end{aligned}$$

Therefore, the Taubin method solves $\varvec{M}\hat{\varvec{\mathcal {A}}_\textrm{T}}=\lambda \varvec{N}_\textrm{T}{\hat{\varvec{\mathcal {A}}}}_\textrm{T}$. Since $\varvec{M}$ and $\varvec{N}_\textrm{T}$ are positive semi-definite matrices, and $\lambda$ is non-negative too. Thus, the estimator $\hat{\varvec{\mathcal {A}}_T}$ is the generalized eigenvector corresponding to the smallest generalized eigenvalue of the generalized eigenvalue problem. The Taubin method is considered to be a good method for obtaining an estimate for the parameter vector ${\varvec{\mathcal {A}}}$.

Note that after $\varvec{\mathcal {A}}=\bigl (A,B,C,D,E,F\bigr )^\top$ is estimated, the geometric parameters $\varvec{\theta } = \bigl (c_x,c_y,a,b,\psi \bigr )^\top$ can be computed from the following relations:

$$\begin{aligned} \begin{aligned}&c_x = \tfrac{CD - BE}{B^2-AC} \\&c_y = \tfrac{AE - BD}{B^2-AC} \\&a = -\tfrac{\sqrt{2(AE^2+CD^2-BDE+(B^2-AC)F)(A+C+\sqrt{(A-C)^2+B^2})}}{B^2-AC} \\&b = -\tfrac{\sqrt{2(AE^2+CD^2-BDE+(B^2-AC)F)(A+C-\sqrt{(A-C)^2+B^2})}}{B^2-AC} \\&\psi = \arctan {\tfrac{C-A-\sqrt{(A-C)^2+B^2}}{B}}, \ \textrm{for } \,\,\,\ B \ne 0. \end{aligned} \end{aligned}$$

(A2)

Also, the conversion formulas for transforming the geometric parameters to algebraic parameters are

$$\begin{aligned} \begin{aligned} A&= \tfrac{\cos ^2(\psi )}{a^2} + \tfrac{\sin ^2(\psi )}{b^2},\\ B&= \cos (\psi )\sin (\psi )\bigl (\tfrac{1}{a^2} - \tfrac{1}{b^2}\bigr ) \\ C&=\tfrac{\sin ^2(\psi )}{a^2} + \tfrac{\cos ^2(\psi )}{b^2},\\ D&= -A c_{x}-B c_{y}\\ E&= -C c_{y} - B c_{x},\\ F&=A c_{x}^2+C c_{y}^2+2B c_{x} c_{y}-1. \end{aligned} \end{aligned}$$

(A3)

Appendix B Derivation of CCRB

To derive the CCRB of $\varvec{\tilde{\phi }}$ and $\tilde{\varvec{m}}$, we need to state our problem in the benchmark of this formulation. Then, with the aid of some elements of linear algebra, we will derive the CCRB for our parameters. First, we have here

$$\begin{aligned} \frac{\partial {\tilde{p}}_{ij} }{\partial \varvec{m}_{ij}^\top }&= 2\left( {\tilde{T}}_{ij}{\tilde{A}}_i{\tilde{C}}_i-{\tilde{T}}_{ij}'{\tilde{B}}_i{\tilde{S}}_i , {\tilde{T}}_{ij}{\tilde{A}}_i{\tilde{S}}_i+{\tilde{T}}_{ij}'{\tilde{B}}_i{\tilde{C}}_i \right) \\ \frac{\partial {\tilde{p}}_{ij} }{\partial {\tilde{x}}_c}&=2({\tilde{T}}_{ij}'{\tilde{B}}_i{\tilde{S}}_i - {\tilde{T}}_{ij}{\tilde{A}}_i{\tilde{C}}_i)\\ \frac{\partial {\tilde{p}}_{ij} }{\partial {\tilde{y}}_c}&=-2({\tilde{T}}_{ij}'{\tilde{B}}_i{\tilde{C}}_i + {\tilde{T}}_{ij}{\tilde{A}}_i{\tilde{S}}_i)\\ \frac{\partial {\tilde{p}}_{ij} }{\partial {\tilde{A}}_i}&={\tilde{T}}_{ij}^2\\ \frac{\partial {\tilde{p}}_{ij} }{\partial {\tilde{B}}_i}&={\tilde{T}}_{ij}^{'2}\\ \frac{\partial {\tilde{p}}_{ij} }{\partial {\tilde{\psi }}_i}&=2({\tilde{T}}_{ij}{\tilde{T}}_{ij}'{\tilde{A}}_i-{\tilde{T}}_{ij}{\tilde{T}}_{ij}'{\tilde{B}}_i). \end{aligned}$$

Therefore, $\varvec{\tilde{F}}$ can be written in the compact form $\varvec{\tilde{F}}_{n\times m } = \left( \varvec{\tilde{C}}_{n\times 2n}\, \,\, \varvec{\tilde{S}}_{n \times d} \right) ,$ where

$$\begin{aligned} \varvec{\tilde{C}}= \textrm{Diag}\left( \varvec{\tilde{C}}_1,\ldots ,\varvec{\tilde{C}}_k \right) _{n\times 2n}, \end{aligned}$$

(B4)

where $\varvec{\tilde{C}}_i= \textrm{Diag}\bigl (\varvec{\tilde{r}}_{11}^\top , \ldots , \varvec{\tilde{r}}_{k1}^\top \bigr )_{n_i\times 2n_i}$ with

$$\begin{aligned} \varvec{\tilde{r}}_{ij} = \nabla _{({\tilde{x}},{\tilde{y}})}{\tilde{p}}_{ij} = 2\bigl ({\tilde{T}}_{ij}{\tilde{A}}_i\tilde{C_i}-{\tilde{T}}_{ij}'{\tilde{B}}_i\tilde{S_i},\, {\tilde{T}}_{ij}{\tilde{A}}_i\tilde{S_i}+{\tilde{T}}_{ij}'{\tilde{B}}_i\tilde{C_i}\bigr ) ^\top . \end{aligned}$$

(B5)

Here, $\varvec{\tilde{C}}_{i}$ is an $n_i\times 2n_i$ matrix that has its $j{\mathrm{{th}}}$ row equal to $\bigl ( \tfrac{\partial {\tilde{p}}_{ij}}{\partial \tilde{\varvec{m}}_{11}^\top },\ldots ,\tfrac{\partial {\tilde{p}}_{ij}}{\partial \tilde{\varvec{m}}_{1n_1}^\top },\ldots ,\tfrac{\partial {\tilde{p}}_{ij}}{\partial \tilde{\varvec{m}}_{k1}^\top },\ldots ,\tfrac{\partial {\tilde{p}}_{ij}}{\partial \tilde{\varvec{m}}_{kn_k}^\top } \bigr )_{1\times 2n_i}$. The $n\times d$ matrix $\varvec{\tilde{S}}$ is defined as

$$\begin{aligned} \varvec{\tilde{S}}= \begin{bmatrix} \varvec{\tilde{V}}_1 &{}\varvec{\tilde{S}}_1 &{} &{} &{}\\ \varvec{\tilde{V}}_2 &{} &{} \varvec{\tilde{S}}_2 &{} &{}\\ \vdots &{} &{} &{} \ddots &{}\\ \varvec{\tilde{V}}_k &{} &{} &{} &{} \varvec{\tilde{S}}_k \end{bmatrix}_{n\times d}, \end{aligned}$$

(B6)

where the $j{\mathrm{{th}}}$ row of $\varvec{\tilde{V}}_{i}$ and $\varvec{\tilde{S}}_{i}$ are, respectively, equal to

$$\begin{aligned} \nabla _{({\tilde{x}}_c,{\tilde{y}}_c)}{\tilde{p}}_{ij}=-2\left( {\tilde{T}}_{ij}{\tilde{A}}_i\tilde{C_i}-{\tilde{T}}_{ij}'{\tilde{B}}_i\tilde{S_i},\, {\tilde{T}}_{ij}{\tilde{A}}_i\tilde{S_i}+{\tilde{T}}_{ij}'{\tilde{B}}_i\tilde{C_i}\right) , \end{aligned}$$

(B7)

and $\nabla _{({\tilde{A}}_i, {\tilde{B}}_i, {\tilde{\psi }}_i)}{\tilde{p}}_{ij}=\bigl ({\tilde{T}}_{ij}^2,,\,{\tilde{T}}_{ij}^{'2}\,,\, 2{\tilde{T}}_{ij}{\tilde{T}}_{ij}'({\tilde{A}}_i-{\tilde{B}}_i) \bigr )$. For this structure of $\varvec{\tilde{F}}$, we need to find $\varvec{\tilde{U}}$ such that $\varvec{\tilde{F}}\varvec{\tilde{U}}=\varvec{0}_{n\times (m-n)}$, where $n=\sum _{i=1}^kn_i$. If we define

$$\begin{aligned} \varvec{\tilde{U}}= \begin{bmatrix} \varvec{\tilde{C}}_{2n\times n}^\perp &{} \varvec{\tilde{V}}_{2n\times d} \\ \varvec{0}_{(2+3k)\times n} &{} \varvec{I}_{d} \end{bmatrix}_{m\times (m-n)}, \end{aligned}$$

and $\varvec{\tilde{C}}_i^\perp = \textrm{Diag}\bigl ((\varvec{\tilde{r}}_{i1}^\perp )^\top , \ldots , (\varvec{\tilde{r}}_{in_i}^\perp )^\top \bigr )_{{2n_i}\times n_i}$ with $\varvec{\tilde{r}}_{ij}^\perp = 2\bigl ({\tilde{T}}_{ij}{\tilde{A}}_i\tilde{S_i}+{\tilde{T}}_{ij}'{\tilde{B}}_i\tilde{C_i}\,,\, {\tilde{T}}_{ij}'{\tilde{B}}_i\tilde{S_i}-{\tilde{T}}_{ij}{\tilde{A}}_i\tilde{C_i}\bigr ) ^\top$, then it is easy to see that $\varvec{\tilde{C}}\varvec{\tilde{C}}^\perp =\varvec{0}_{n \times n}$. Thus, in order to show that $\varvec{\tilde{F}}\varvec{\tilde{U}}= \varvec{0}_{n\times (m-n)}$ we need to define $\varvec{\tilde{V}}$ such that $\varvec{\tilde{C}}\varvec{\tilde{V}}+\varvec{\tilde{S}}= \varvec{0}_{n\times d}$. If we define

$$\begin{aligned} \varvec{\tilde{V}}&=\begin{bmatrix} \varvec{I}_{2} &{} \varvec{\tilde{V}}_1 &{} &{} &{} \\ \vdots &{} &{} \varvec{\tilde{V}}_2 &{} &{} \\ \vdots &{} &{} &{} \ddots &{} \\ \varvec{I}_{2} &{} &{} &{} &{} \varvec{\tilde{V}}_k \end{bmatrix}_{2n\times d}, \end{aligned}$$

(B8)

where

$$\begin{aligned} \varvec{\tilde{V}}_i&= \begin{bmatrix} -{\tilde{T}}_{i1}^2\tilde{\varvec{u}}_{i1} &{} -{\tilde{T}}_{i1}^{'2}\tilde{\varvec{u}}_{i1} &{} -2{\tilde{T}}_{i1}{\tilde{T}}_{i1}'({\tilde{A}}_i-{\tilde{B}}_i)\tilde{\varvec{u}}_{i1} \\ \vdots &{}\vdots &{} \vdots \\ -{\tilde{T}}_{in_i}^2\tilde{\varvec{u}}_{in_i} &{} -{\tilde{T}}_{in_i}^{'2}\tilde{\varvec{u}}_{in_i} &{} -2{\tilde{T}}_{in_i}{\tilde{T}}_{in_i}'({\tilde{A}}_i-{\tilde{B}}_i)\tilde{\varvec{u}}_{in_i} \end{bmatrix}_{2n_i\times 3} ,\, \end{aligned}$$

and $\tilde{\varvec{u}}_{ij}=\tfrac{\varvec{\tilde{r}}_{ij}}{\Vert \varvec{\tilde{r}}_{ij}\Vert ^2}$, then $\varvec{\tilde{C}}\varvec{\tilde{V}}= -\varvec{\tilde{S}}$. Thus, this choice of $\varvec{\tilde{U}}$ satisfies $\varvec{\tilde{F}}\varvec{\tilde{U}}=\varvec{0}_{n \times (m-n)}$. Now, using the definition of $\varvec{J}$, we have

$$\begin{aligned} \varvec{\tilde{U}}^\top \varvec{J}\varvec{\tilde{U}}= \begin{bmatrix} (\varvec{\tilde{C}}^\perp )^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{C}}^\perp &{} (\varvec{\tilde{C}}^\perp )^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{V}}\\ \varvec{\tilde{V}}^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{C}}^\perp &{} \varvec{\tilde{V}}^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{V}}\end{bmatrix}. \end{aligned}$$

(B9)

Next, we apply the following lemma in order to obtain the CCRB of $\varvec{{\hat{\phi }}}$.

Lemma 2

(Mathai and Provost 1992) Let ${\varvec{A}}$ be a nonsingular matrix of size $(n+p)$ and ${\varvec{B}}={\varvec{A}}^{-1}$, where

$$\begin{aligned} {\varvec{A}}=\begin{bmatrix} \varvec{A_{11}} &{} \varvec{A_{12}}\\ \varvec{A_{21}} &{} \varvec{A_{22}} \end{bmatrix},\quad {\varvec{B}}=\begin{bmatrix} {\varvec{B}}_{11} &{} {\varvec{B}}_{12}\\ {\varvec{B}}_{21} &{}{\varvec{B}}_{22} \end{bmatrix}. \end{aligned}$$

If $\varvec{A_{22}}^{-1}$ exists, then

$$\begin{aligned} {\varvec{B}}_{11}&= {\varvec{A}}_{11}^{-1} + {\varvec{A}}_{11}^{-1}{\varvec{A}}_{12}\bigl ({\varvec{A}}_{22}-{\varvec{A}}_{21}{\varvec{A}}_{11}^{-1}{\varvec{A}}_{12}\bigr )^{-1}{\varvec{A}}_{21}{\varvec{A}}_{11}^{-1}\\ {\varvec{B}}_{22}&= \bigl ({\varvec{A}}_{22}-{\varvec{A}}_{21}{\varvec{A}}_{11}^{-1}{\varvec{A}}_{12}\bigr )^{-1}. \end{aligned}$$

(a) The CCRB of $\varvec{{\hat{\phi }}}$ We are mainly interested in finding the lower bound for $\varvec{{\hat{\phi }}}$. Because of the structure of $\varvec{\tilde{F}}$, the bottom right block of $(\varvec{\tilde{U}}^\top \varvec{J}\varvec{\tilde{U}})^{-1}$ is all we are interested in. To obtain this, we apply Lemma 2 to (B9). Accordingly, the bottom right block matrix $(\varvec{\tilde{U}}^\top \varvec{J}\varvec{\tilde{U}})^{-1}$ is

$$\begin{aligned} {\varvec{B}}_{22}&=\Bigl (\varvec{\tilde{V}}^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{V}}- \varvec{\tilde{V}}^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{C}}^\perp ((\varvec{\tilde{C}}^\perp )^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{C}}^\perp )^{-1}(\varvec{\tilde{C}}^\perp )^\top \varvec{\Sigma }_{2n}^{-1}\varvec{\tilde{V}}\Bigr )^{-1} \\&=\left( \varvec{\tilde{V}}^\top \varvec{\Sigma }_{2n}^{-1/2}\left( \varvec{I}_d-\varvec{\tilde{U}}(\varvec{\tilde{U}}^\top \varvec{\tilde{U}})^{-1}\varvec{\tilde{U}}^\top \right) \varvec{\Sigma }_{2n}^{-1/2}\varvec{\tilde{V}}\right) ^{-1}, \end{aligned}$$

where $\varvec{\tilde{U}}=\varvec{\Sigma }_{2n}^{-1/2}\varvec{\tilde{C}}^{\perp }$. It is worth mentioning here that $\varvec{I}_d-\varvec{\tilde{U}}\bigl (\varvec{\tilde{U}}^\top \varvec{\tilde{U}}\bigr )^{-1}\varvec{\tilde{U}}^\top$ is the projection matrix onto the orthogonal components of the column space of $\varvec{\tilde{U}}$. Since $\varvec{\tilde{C}}\varvec{\tilde{C}}^{\perp }={\varvec{0}}$, then $\varvec{\tilde{U}}^{\perp }=\varvec{\tilde{C}}\varvec{\Sigma }_{2n}^{1/2}$ is orthogonal to $\varvec{\tilde{U}}$. Then $\varvec{I}_d-\varvec{\tilde{U}}\bigl (\varvec{\tilde{U}}^\top \varvec{\tilde{U}}\bigr )^{-1}\varvec{\tilde{U}}^\top$ is the projection matrix onto the column space of $(\varvec{\tilde{U}}^{\perp })^\top$. This implies that

$$\begin{aligned} \varvec{I}_d-\varvec{\tilde{U}}\bigl (\varvec{\tilde{U}}^\top \varvec{\tilde{U}}\bigr )^{-1}\varvec{\tilde{U}}^\top =(\varvec{\tilde{U}}^{\perp })^\top \bigl ( (\varvec{\tilde{U}}^{\perp })(\varvec{\tilde{U}}^{\perp )^\top }\bigr )^{-1}\varvec{\tilde{U}}^{\perp }. \end{aligned}$$

Accordingly, substituting $\varvec{\tilde{U}}^{\perp }=\varvec{\tilde{C}}\varvec{\Sigma }_{2n}^{1/2}$ and using the fact $\varvec{\tilde{C}}\varvec{\tilde{V}}= -\varvec{\tilde{S}}$ in ${\varvec{B}}_{22}$ yields

$$\begin{aligned} {\varvec{B}}_{22}&=\left( \varvec{\tilde{V}}^\top \varvec{\Sigma }_{2n}^{-1/2} (\varvec{\tilde{U}}^{\perp })^\top \bigl ( (\varvec{\tilde{U}}^{\perp })(\varvec{\tilde{U}}^{\perp )^\top }\bigr )^{-1}\varvec{\tilde{U}}^{\perp }\varvec{\Sigma }_{2n}^{-1/2}\varvec{\tilde{V}}\right) ^{-1}\nonumber \\&=\left( \varvec{\tilde{S}}^\top \bigl (\varvec{\tilde{C}}\varvec{\Sigma }_{2n}\varvec{\tilde{C}}^\top \bigr )^{-1} \varvec{\tilde{S}}^\top \right) ^{-1}, \end{aligned}$$

(B10)

and as such,

$$\begin{aligned} \textrm{CCRB}(\varvec{{\hat{\phi }}}) = \nabla _{\varvec{\tilde{\phi }}}{\mathbb {E}}(\varvec{{\hat{\phi }}})\left( \varvec{\tilde{S}}^\top \bigl (\varvec{\tilde{C}}\varvec{\Sigma }_{2n}\varvec{\tilde{C}}^\top \bigr )^{-1}\varvec{\tilde{S}}\right) ^{-1}\nabla _{\varvec{\tilde{\phi }}}{\mathbb {E}}(\varvec{{\hat{\phi }}})^\top . \end{aligned}$$

(B11)