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Homography Estimation from Ellipse Correspondences Based on the Common Self-polar Triangle

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Abstract

This paper presents new implementation algorithms for estimating the homography from ellipse correspondences based on the common self-polar triangle. Firstly, we propose an analytical solution with a fourfold ambiguity to homography based on converting two ellipse correspondences to three common pole correspondences. Secondly, after exploring the position information of the common poles, we propose the analytical algorithms for estimating the homography from only two ellipse correspondences. We also propose an analytical linear algorithm for estimating the homography by using the common pole correspondences with the known projective scale factors when given three or more ellipse correspondences. Unlike the previous methods, our algorithms are very easy to implement and furthermore may usually provide a unique solution (at most two solutions). Experimental results in synthetic data and real images show the accuracy advantage and the usefulness of our proposed algorithms.

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Notes

  1. This property can be used to match the corresponding ellipses between the two views.

  2. Note that since the determinant of \({\widetilde{\mathbf{C }}}_i\) is one, if \({\varvec{\xi }}_j\) lies inside \({\widetilde{\mathbf{C }}}_i\) then it must have \({\varvec{\xi }}_j^\mathrm{T}{\widetilde{\mathbf{C }}}_{i}{\varvec{\xi }}_j>0\) else if \({\varvec{\xi }}_j\) lies outside \({\widetilde{\mathbf{C }}}_i\) then it must have \({\varvec{\xi }}_j^\mathrm{T}{\widetilde{\mathbf{C }}}_{i}{\varvec{\xi }}_j<0\).

  3. http://www.ee.oulu.fi/~jkannala/bmvc.html.

  4. http://cmp.felk.cvut.cz/~chum/code/ell2h.html.

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Acknowledgements

The author thanks the anonymous referees for valuable suggestions.

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  1. Department of Mathematics, Northeastern University, Shenyang, 110819, People’s Republic of China

    Yang Guo

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  1. Yang Guo
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A Proof of “The ellipses \({\widetilde{\mathbf{C }}}_1\), \({\widetilde{\mathbf{C }}}_2\) have at least a common self-polar triangle if and only if the matrix \({\widetilde{\mathbf{C }}}_2^{-1}{\widetilde{\mathbf{C }}}_1\) has three linearly independent eigenvectors”

A Proof of “The ellipses \({\widetilde{\mathbf{C }}}_1\), \({\widetilde{\mathbf{C }}}_2\) have at least a common self-polar triangle if and only if the matrix \({\widetilde{\mathbf{C }}}_2^{-1}{\widetilde{\mathbf{C }}}_1\) has three linearly independent eigenvectors”

First suppose that the triangle \({\varDelta }{\varvec{\xi }}_1{\varvec{\xi }}_2{\varvec{\xi }}_3\) is a common self-polar triangle for the ellipses \({\widetilde{\mathbf{C }}}_1\), \({\widetilde{\mathbf{C }}}_2\). Then according to

$$\begin{aligned}&{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=0, \quad {\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=0,\\&{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1=0, \quad {\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1=0, \end{aligned}$$

we have that \({\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1\) is equal (up to scale) \({\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1\), namely \({\varvec{\xi }}_1\) is an eigenvector of the matrix \({\widetilde{\mathbf{C }}}_2^{-1}{\widetilde{\mathbf{C }}}_1\). Similarly, so are \({\varvec{\xi }}_2\) and \({\varvec{\xi }}_3\). Consequently, the matrix \({\widetilde{\mathbf{C }}}_2^{-1}{\widetilde{\mathbf{C }}}_1\) must have three linearly independent eigenvectors.

Conversely, suppose that the matrix \({\widetilde{\mathbf{C }}}_2^{-1}{\widetilde{\mathbf{C }}}_1\) has three linearly independent eigenvectors \({\varvec{\xi }}_1\), \({\varvec{\xi }}_2\), \({\varvec{\xi }}_3\) satisfying

$$\begin{aligned} \begin{aligned} {\widetilde{\mathbf{C }}}_2^{-1}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_i=\lambda _i{\varvec{\xi }}_i,\quad i=1,2,3, \end{aligned} \end{aligned}$$

where \(\lambda _1\), \(\lambda _2\), \(\lambda _3\) are the three eigenvalues.

When \(\lambda _1\ne \lambda _2\ne \lambda _3\), according to \({\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=\lambda _1{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1\) and \({\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_2=\lambda _2{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_2\), we may get

$$\begin{aligned} \begin{aligned} {\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=\lambda _1{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1,\quad {\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_2=\lambda _2{\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_2. \end{aligned} \end{aligned}$$

On one hand, since \(\lambda _1\ne \lambda _2\), we have

$$\begin{aligned} \begin{aligned} {\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=0, \quad {\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1=0. \end{aligned} \end{aligned}$$

Similarly, we also may get

$$\begin{aligned} \begin{aligned}&{\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=0, \quad {\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1=0, \\&{\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_2=0, \quad {\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_2=0. \end{aligned} \end{aligned}$$

On the other hand, since \({\varvec{\xi }}_1\), \({\varvec{\xi }}_2\), \({\varvec{\xi }}_3\) are linearly independent, we also have

$$\begin{aligned} \begin{aligned}&{\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1\ne 0,\quad {\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1\ne 0,\\&{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_2\ne 0,\quad {\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_2\ne 0,\\&{\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_3\ne 0,\quad {\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_3\ne 0, \end{aligned} \end{aligned}$$

namely the triangle \({\varDelta }{\varvec{\xi }}_1{\varvec{\xi }}_2{\varvec{\xi }}_3\) is a common self-polar triangle for the ellipses \({\widetilde{\mathbf{C }}}_1\), \({\widetilde{\mathbf{C }}}_2\).

When \(\lambda _1=\lambda _2\ne \lambda _3\), according to the above discussion, we may immediately get

$$\begin{aligned} \begin{aligned}&{\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_3=0,\quad {\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_3=0,\\&{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_3=0,\quad {\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_3=0,\\&{\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_3\ne 0,\quad {\varvec{\xi }}_3^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_3\ne 0. \end{aligned} \end{aligned}$$

That means that the line \(\mathbf{l }\) passing through the points \({\varvec{\xi }}_1\) and \({\varvec{\xi }}_2\) is the polar line of \({\varvec{\xi }}_3\) with respect to \({\widetilde{\mathbf{C }}}_1\), \({\widetilde{\mathbf{C }}}_2\).

Additionally, since

$$\begin{aligned} \begin{aligned}&{\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=\lambda _1{\varvec{\xi }}_1^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1,\\&{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_1=\lambda _1{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_1,\\&{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_1{\varvec{\xi }}_2=\lambda _2{\varvec{\xi }}_2^\mathrm{T}{\widetilde{\mathbf{C }}}_2{\varvec{\xi }}_2,\\ \end{aligned} \end{aligned}$$

we may know

$$\begin{aligned} \begin{aligned} ({\varvec{\xi }}_1+t{\varvec{\xi }}_2)^\mathrm{T}{\widetilde{\mathbf{C }}}_1({\varvec{\xi }}_1+t{\varvec{\xi }}_2)=\lambda _1({\varvec{\xi }}_1+t{\varvec{\xi }}_2)^\mathrm{T}{\widetilde{\mathbf{C }}}_2({\varvec{\xi }}_1+t{\varvec{\xi }}_2), \end{aligned} \end{aligned}$$

where t is an unknown variable. That means that the common polar line \(\mathbf{l }\) of \({\varvec{\xi }}_3\) with respect to \({\widetilde{\mathbf{C }}}_1\) and \({\widetilde{\mathbf{C }}}_2\) passes through the two intersection points of \({\widetilde{\mathbf{C }}}_1\) and \({\widetilde{\mathbf{C }}}_2\). Further, according to the pole–polar relation, we may assert that \({\widetilde{\mathbf{C }}}_1\) is double contact with \({\widetilde{\mathbf{C }}}_2\). Consequently, the point \({\varvec{\xi }}_3\) and any two points of the line \(\mathbf{l }\) which are harmonic with respect to the two common tangent points of \({\widetilde{\mathbf{C }}}_1\) and \({\widetilde{\mathbf{C }}}_2\) may form a common self-polar triangle of \({\widetilde{\mathbf{C }}}_1\) and \({\widetilde{\mathbf{C }}}_2\) [21].

Last, when \(\lambda _1=\lambda _2=\lambda _3\), we may have that \({\widetilde{\mathbf{C }}}_1\) is equal (up to scale) to \({\widetilde{\mathbf{C }}}_2\), namely the two ellipses are coincident. Obviously, the ellipses \({\widetilde{\mathbf{C }}}_1\), \({\widetilde{\mathbf{C }}}_2\) have an infinity of common self-polar triangles. This completes the proof.

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Guo, Y. Homography Estimation from Ellipse Correspondences Based on the Common Self-polar Triangle. J Math Imaging Vis 62, 169–188 (2020). https://doi.org/10.1007/s10851-019-00928-6

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