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Combination of Piecewise-Geodesic Curves for Interactive Image Segmentation

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Curves and Surfaces (Curves and Surfaces 2014)

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Abstract

Boundary-based interactive image segmentation methods aim to build a closed contour, very often using paths linking a set of user-provided landmark points, ordered along the contour of the object of interest. Among these methods, the geodesically-linked active contour (GLAC) model generates a piecewise-geodesic curve, by linking each pair of successive landmark points by a geodesic curve. As an important shortcoming, the geodesically linked active contour model in its initial formulation does not guarantee the curve to be simple. It may have multiple points, creating self-tangencies and self-intersections, which is inconsistent with respect to the purpose of segmentation. To overcome this issue, we study some properties of non-simple closed curves and introduce a novel energy term to quantity the amount of non-simplicity. We propose to extract a relevant contour from a set of possible paths, such that the resulting structure fits the image data and is simple. We develop a local search method to choose the best combination among possible paths, integrating the novel energy term.

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Notes

  1. 1.

    See detailed proof in appendix B.

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Appendices

Appendix

A Self-Overlap Term

Let \(\mathcal {C}\) be a regular curve parameterized over [0, L]. Let \(\phi \) be a \(C^1\) function defined over \([0,L]^2\) representing the distance between two positions on the curve:

$$\begin{aligned} \phi _\mathcal {C}(u,v) = \left\| \mathcal {C}(u)-\mathcal {C}(v) \right\| ^p \end{aligned}$$

where p is an arbitrary positive real exponent. The length of the zero level set of \(\phi _\mathcal {C}\),

$$\begin{aligned} \left| \mathcal {Z}_\mathcal {C}\right| = \int _0^L \int _0^L \delta (\phi (u,v))\left\| \nabla \phi (u,v) \right\| d u d v, \end{aligned}$$
(9)

quantifies the self-overlap of \(\mathcal {C}\).

Proposition: If \(\mathcal {C}\) is simple, i.e. without self-intersection and self-tangency, then \(\left| \mathcal {Z}_\mathcal {C}\right| = L\sqrt{2}\)

Proof: As a preliminary calculation, let us express the gradient of \(\phi \) (partial derivatives are written using the indexed notation):

$$\begin{aligned} \begin{array}{rl} \nabla \phi (u,v) = &{} \left[ \phi _u(u,v)~~\phi _v(u,v) \right] ^T \\ = &{} p \left\| \mathcal {C}(u)-\mathcal {C}(v) \right\| ^{p-2} \left[ \begin{aligned} \mathcal {C}'(u) \cdot (\mathcal {C}(u)-\mathcal {C}(v)) \\ -\mathcal {C}'(v) \cdot (\mathcal {C}(u)-\mathcal {C}(v)) \end{aligned} \right] \end{array} \end{aligned}$$

If \(\mathcal {C}\) is regular and simple, varying with respect to u in range [0, L], \(\phi (u,v)\) is nowhere zero except when \(u=v\). Hence, for a fixed v, we have:

$$\begin{aligned} \delta (\phi (u,v)) = \frac{\delta (u-v)}{\left| \phi _u(v,v) \right| } \end{aligned}$$
(10)

Integrating (10) into (9) and applying the definition of measure \(\delta \):

$$\begin{aligned} \begin{array}{rl} \left| \mathcal {Z}_\mathcal {C}\right| = &{} \int _0^L \int _0^L \delta (\phi (u,v))\left\| \nabla \phi (u,v) \right\| d u d v \\ = &{} \int _0^L \int _0^L \frac{\delta (u-v)}{\left| \phi _u(v,v) \right| } \left\| \nabla \phi (u,v) \right\| d u d v \\ = &{} \int _0^L \frac{\left\| \nabla \phi (v,v) \right\| }{\left| \phi _u(v,v) \right| } d v \end{array} \end{aligned}$$

Trivially, \(\phi (v,v)=0\). However expanding the gradient gives:

$$\begin{aligned} \begin{array}{rl} \left| \mathcal {Z}_\mathcal {C}\right| = &{} \int _0^L \frac{p \left\| \mathcal {C}(v)-\mathcal {C}(v) \right\| ^{p-2} \sqrt{2(\mathcal {C}'(v)\cdot (\mathcal {C}(v)-\mathcal {C}(v)))^2}}{ p \left\| \mathcal {C}(v)-\mathcal {C}(v) \right\| ^{p-2} \left| \mathcal {C}'(v)\cdot (\mathcal {C}(v)-\mathcal {C}(v))\right| } d v \\ = &{} \int _0^L \sqrt{2}~ d v \\ = &{} L\sqrt{2} \end{array} \end{aligned}$$

B Exteriority Term

Let \(\mathcal {C}\) be a piecewise-smooth regular curve parameterized over [0, 1],

$$\begin{aligned} \mathcal {C}: u \longmapsto \mathcal {C}(u) = \left[ x(u)~y(u)\right] ^T\!. \end{aligned}$$

If it is simple and positively oriented such that normal vector \({\mathcal {C}}'^\perp \) points inward, its inner area may be expressed using Green’s theorem:

$$\begin{aligned} \left| {{\varOmega _{ in }}}(\mathcal {C}) \right| = \frac{1}{2} \int _0^1 \mathcal {C}^\perp (u) \cdot {\mathcal {C}}'(u)~ d u = \frac{1}{2} \int _0^1 x(u){y}'(u) - {x}'(u)y(u)~ d u \end{aligned}$$

When one calculates the previous expression on a non-simple closed curve, one gets the signed area, in which positively and negatively oriented connected components have positive and negative contributions, respectively.

Fig. 7.
figure 7

The exteriority of an open curve is measured as the signed area of the multiple connected region that it forms with the line segment joining its two endpoints

Proposition: The signed area formed by an open curve \(\mathcal {C}\) over [0, 1] and the line segment from \(\mathcal {C}(1)\) returning to \(\mathcal {C}(0)\) (see Fig. 7), which we use to as the exteriority measure in the paper, may be expressed as:

$$\begin{aligned} \mathcal {X}[\mathcal {C}] = \frac{1}{2} \int _0^1 \mathcal {C}^\perp \cdot {\mathcal {C}}' d u + \frac{1}{2}\mathcal {C}^\perp (1) \cdot \mathcal {C}(0) \end{aligned}$$

Proof: Let S be the parametrization of the line segment joining \(\mathcal {C}(1)\) and \(\mathcal {C}(0)\), over [0, 1]:

$$\begin{aligned} S(u) = (1-u)\mathcal {C}(1)+u\mathcal {C}(0) \end{aligned}$$

The signed area is then obtained by applying Green’s theorem on a piecewise basis:

$$\begin{aligned} \begin{array}{rcl} \mathcal {X}[\mathcal {C}] &{}=&{} \frac{1}{2} \int _0^1 \mathcal {C}^\perp \cdot {\mathcal {C}}' d u + \frac{1}{2} \int _0^1 S^\perp \cdot {S}' d u \\ &{}=&{} \frac{1}{2} \int _0^1 \mathcal {C}^\perp \cdot {\mathcal {C}}' d u + \frac{1}{2} \int _0^1 ((1-u)\mathcal {C}^\perp (1)+u\mathcal {C}^\perp (0))\cdot (\mathcal {C}(0)-\mathcal {C}(1)) d u \\ &{}=&{} \frac{1}{2} \int _0^1 \mathcal {C}^\perp \cdot {\mathcal {C}}' d u + \frac{1}{2} \int _0^1 (1-u)\mathcal {C}^\perp (1) \cdot \mathcal {C}(0) +u\mathcal {C}^\perp (1)\cdot \mathcal {C}(0) d u \\ &{}=&{} \frac{1}{2} \int _0^1 \mathcal {C}^\perp \cdot {\mathcal {C}}' d u + \frac{1}{2} \mathcal {C}^\perp (1) \cdot \mathcal {C}(0) \end{array} \end{aligned}$$

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Mille, J., Bougleux, S., Cohen, L.D. (2015). Combination of Piecewise-Geodesic Curves for Interactive Image Segmentation. In: Boissonnat, JD., et al. Curves and Surfaces. Curves and Surfaces 2014. Lecture Notes in Computer Science(), vol 9213. Springer, Cham. https://doi.org/10.1007/978-3-319-22804-4_25

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