Analysis of Amoeba Active Contours

Abstract

Subject of this paper is the theoretical analysis of structure-adaptive median filter algorithms that approximate curvature-based PDEs for image filtering and segmentation. These so-called morphological amoeba filters are based on a concept introduced by Lerallut et al. They achieve similar results as the well-known geodesic active contour and self-snakes PDEs. In the present work, the PDE approximated by amoeba active contours is derived for a general geometric situation and general amoeba metric. This PDE is structurally similar but not identical to the geodesic active contour equation. It reproduces the previous PDE approximation results for amoeba median filters as special cases. Furthermore, modifications of the basic amoeba active contour algorithm are analysed that are related to the morphological force terms frequently used with geodesic active contours. Experiments demonstrate the basic behaviour of amoeba active contours and its similarity to geodesic active contours.

Appendix: Details of Proofs

1.1 Proof of Corollary 2

For the $L^1$ amoeba norm, one has $\nu (s)=1+|s|$, thus $\nu '(s)=\mathrm {sgn}\,s$. Inserting these into (5) yields

$$\begin{aligned} J_1(s,\alpha )&= \int \limits _{\alpha -\pi /2}^{\alpha +\pi /2}\!\! \frac{\sin ^2\vartheta \,\,\mathrm {sgn}\,\!\cos \vartheta }{(1+s\,|\cos \vartheta |)^4}~\mathrm {d}\vartheta \nonumber \\&=\int \limits _{\alpha -\pi /2}^{\pi /2}\!\! \frac{\sin ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta -\!\!\!\!\int \limits _{\pi /2}^{\alpha +\pi /2}\!\! \frac{\sin ^2\vartheta }{(1-s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&=\int \limits _{\alpha -\pi /2}^{\pi /2}\!\! \frac{\sin ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta -\!\!\!\!\int \limits _{\pi /2-\alpha }^{\pi /2}\!\! \frac{\sin ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&=\int \limits _{-\pi /2+\alpha }^{\pi /2-\alpha }\!\! \frac{\sin ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \end{aligned}$$

(64)

where we have assumed without loss of generality $\alpha \in [0,\pi ]$. Evaluating the indefinite integrals

$$\begin{aligned}&\int \frac{\sin ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&\quad = \left\{ \begin{array}{ll} \frac{1}{2\,(s^2-1)^{5/2}} \ln \frac{\sqrt{s+1}+\sqrt{s-1}\tan \frac{\vartheta }{2}}{\sqrt{s+1}-\sqrt{s-1}\tan \frac{\vartheta }{2}}\;, &{} s>1\;,\\ \frac{1}{(1-s^2)^{5/2}} \arctan \left( \sqrt{\frac{1-s}{1+s}}\,\tan \frac{\vartheta }{2}\right) \;, &{} s<1 \end{array}\right. \nonumber \\&\qquad -\frac{\left\{ \begin{array}{@{}r@{}} \sin \vartheta \bigl ((2\,s^3+s)\cos ^2\vartheta +3(s^2+1)\cos \vartheta \\ -2\,s^3+5\,s\bigr ) \end{array}\right. }{6(s^2-1)^2(1+s\cos \vartheta )^2} \end{aligned}$$

(65)

and

$$\begin{aligned}&\int \frac{\sin ^2\vartheta }{(1+\cos \vartheta )^4}~\mathrm {d}\vartheta =\frac{\sin ^3\vartheta \,(4+\cos \vartheta )}{15\,(1+\cos \vartheta )^4} \end{aligned}$$

(66)

at the integration boundaries $\pm (\pi /2-\alpha )$ and inserting $\tan \left( \frac{\pi }{4}-\frac{\alpha }{2}\right) =\frac{\cos \alpha }{1+\sin \alpha }$ yields (27), (29) and (31). The proof for $\alpha \in [-\pi ,0]$ is analogous but the integral is split for the $\mathrm {sgn}\,\cos \vartheta$ factor at $-\pi /2$, finally leading to integration boundaries $\pm (\pi /2+\alpha )$. As a consequence, all instances of $\sin \alpha$ are replaced with $-\sin \alpha$, which is subsumed by the use of $|\sin \alpha |$ in (27), (29) and (31). Analogously, one has for $\alpha \in [0,\pi ]$

$$\begin{aligned} J_3(s,\alpha )&= \int \limits _{\alpha -\pi /2}^{\alpha +\pi /2} \frac{\cos ^2\vartheta \,\mathrm {sgn}\,\cos \vartheta }{(1+s\,|\cos \vartheta |)^4}~\mathrm {d}\vartheta \nonumber \\&=\int \limits _{-\pi /2+\alpha }^{\pi /2-\alpha } \frac{\cos ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \end{aligned}$$

(67)

which is evaluated via the indefinite integrals

$$\begin{aligned}&\int \frac{\cos ^2\vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&\quad = \left\{ \begin{array}{ll} \frac{-4\,s^2-1}{2\,(s^2-1)^{7/2}} \ln \frac{\sqrt{s+1}+\sqrt{s-1}\tan \frac{\vartheta }{2}}{\sqrt{s+1}-\sqrt{s-1}\tan \frac{\vartheta }{2}}\;, &{} s>1\;,\\ \frac{4\,s^2+1}{(1-s^2)^{7/2}} \arctan \left( \sqrt{\frac{1-s}{1+s}}\,\tan \frac{\vartheta }{2}\right) \;, &{} s<1 \end{array}\right. \nonumber \\&\qquad +\frac{\left\{ \begin{array}{@{}r@{}} \sin \vartheta \bigl ((6\,s^5+10\,s^3-s)\cos ^2\vartheta \\ +3(2\,s^4+9\,s^2-1)\cos \vartheta \\ +(2\,s^3+13\,s)\bigr ) \end{array}\right. }{6(s^2-1)^2(1+s\cos \vartheta )^2}\;, \end{aligned}$$

(68)

$$\begin{aligned}&\int \frac{\cos ^2\vartheta }{(1+\cos \vartheta )^4}~\mathrm {d}\vartheta =\nonumber \\&\quad \frac{2\,\sin \vartheta \,(13\cos ^3\vartheta +52\cos ^2\vartheta +32\cos \vartheta +8)}{105\,(1+\cos \vartheta )^4} \end{aligned}$$

(69)

to obtain (28), (30) and (32). As before, the inclusion of the case $\alpha \in [-\pi ,0]$ implies the use of $|\sin \alpha |$ in all three equations. Finally, one has for $J_2$ and $\alpha \in [0,\pi ]$

$$\begin{aligned} J_2(s,\alpha )&= \int \limits _{\alpha -\pi /2}^{\alpha +\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta \,\,\mathrm {sgn}\,\!\cos \vartheta }{(1+s\,|\cos \vartheta |)^4}~\mathrm {d}\vartheta \nonumber \\&=\int \limits _{\alpha -\pi /2}^{\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta -\!\!\int \limits _{\pi /2}^{\alpha +\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1-s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&=\int \limits _{\alpha -\pi /2}^{\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta -\!\!\int \limits _{-\pi /2}^{\alpha -\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&=\underbrace{\int \limits _{-\pi /2}^{\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta }_{{}=0} -~2\!\!\!\!\int \limits _{-\pi /2}^{\alpha -\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \nonumber \\&=~2\!\!\!\!\int \limits _{\pi /2-\alpha }^{\pi /2}\!\! \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta \end{aligned}$$

(70)

and the indefinite integral

$$\begin{aligned}&\int \frac{\sin \vartheta \,\cos \vartheta }{(1+s\cos \vartheta )^4}~\mathrm {d}\vartheta =\frac{3\,s\cos \vartheta +1}{6\,s^2\,(1+s\,\cos \vartheta )^3} \end{aligned}$$

(71)

from which (26) is obtained in a straightforward way. As before, the case $\alpha \in [-\pi ,0]$ is subsumed by inserting modulus bars around $\sin \alpha$.

1.2 Relation between $\tilde{J}_1$ and $\tilde{J}_3$

To complete the proof of Corollary 3, we show that $h(s)=s\,g'(s)$. We notice first that

$$\begin{aligned}&\frac{~\mathrm {d}}{~\mathrm {d}\vartheta }\left( \frac{\sin \vartheta }{\nu (\beta \,s\cos \vartheta )^3}\right) \nonumber \\&\quad = \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3} +3\,\beta \,s\,\frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4}\sin ^2\vartheta \end{aligned}$$

(72)

where the last summand is essentially the integrand of $\tilde{J}_1(\beta \,s)$. By integration it follows that

$$\begin{aligned} 3\,\beta \,s\,\tilde{J}_1(\beta \,s)&= \underbrace{\left[ \frac{\sin \vartheta }{\nu (\beta \,s\cos \vartheta )^3}\right] ^{\vartheta =+\pi /2}_{\vartheta =-\pi /2}}_{{}=2}\nonumber \\ {}&\quad \!\!-\!\! \int \limits _{-\pi /2}^{+\pi /2}\!\! \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}~\mathrm {d}\vartheta \;. \end{aligned}$$

(73)

Substituting this into (34) yields

$$\begin{aligned} g(s)&=\frac{1}{2}\int \limits _{-\pi /2}^{+\pi /2} \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}~\mathrm {d}\vartheta \end{aligned}$$

(74)

from which one easily calculates

$$\begin{aligned} s\,g'(s)&= \frac{s}{2}\,\int \limits _{-\pi /2}^{+\pi /2} \frac{-3\,\beta \,\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^4} \nu '(\beta \,s\cos \vartheta )\cos \vartheta ~\mathrm {d}\vartheta \nonumber \\&=-\frac{3}{2}\,\beta \,s\,\tilde{J}_3(\beta \,s) = h(s)\;. \end{aligned}$$

(75)

1.3 Equivalence of Corollary 3 to the Result from [27]

In [27] it was shown that iterated amoeba median filtering approximates the PDE (4) as in Corollary 3 with the edge-stopping function $g$ given by

$$\begin{aligned} g(s)&= \frac{3\,I_1(\beta \,s)}{\beta ^2\,s^2\,\psi \left( \frac{1}{\beta \,s} \right) ^3}\;, \end{aligned}$$

(76)

$$\begin{aligned} I_1(\beta \,s)&= \int \limits _0^1\xi ^2\,\sqrt{ \left( \psi ^{-1}\left( \frac{1}{\xi }\psi \left( \frac{1}{\beta \,s}\right) \right) \right) ^2-\frac{1}{\beta ^2\,s^2}\,}~\mathrm {d}\xi \;, \end{aligned}$$

(77)

where the function $\psi$ is related to $\nu$ via

$$\begin{aligned} \psi (q) =q\,\nu \left( \frac{1}{q}\right) \;, \end{aligned}$$

(78)

and $\psi ^{-1}$ denotes the inverse function of $\psi$. Substituting

$$\begin{aligned} \xi&= \frac{\psi \left( \frac{1}{\beta \,s}\right) }{\psi \left( \frac{1}{\beta \,s\cos \vartheta }\right) }\;, \end{aligned}$$

(79)

$$\begin{aligned} ~\mathrm {d}\xi&= -\frac{\psi \left( \frac{1}{\beta \,s}\right) }{\beta \,s}\, \frac{\psi '\left( \frac{1}{\beta \,s\cos \vartheta }\right) }{\psi \left( \frac{1}{\beta \,s\cos \vartheta }\right) ^2}\, \frac{\sin \vartheta }{\cos ^2\vartheta }~\mathrm {d}\vartheta \end{aligned}$$

(80)

into $I_1$ yields

$$\begin{aligned}&I_1(\beta \,s)= -\int \limits _{\pi /2}^0 \frac{\psi \left( \frac{1}{\beta \,s}\right) ^2}{\psi \left( \frac{1}{\beta \,s\cos \vartheta }\right) ^2} {}\times \nonumber \\&\qquad \times \sqrt{\left( \psi ^{-1}\left( \frac{\psi \left( \frac{1}{\beta \,s\cos \vartheta }\right) }{\psi \left( \frac{1}{\beta \,s}\right) }\, \psi \left( \frac{1}{\beta \,s}\right) \right) \right) -\frac{1}{\beta ^2\,s^2}\,} \times {}\nonumber \\&\qquad \times \frac{-\psi \left( \frac{1}{\beta \,s}\right) }{\beta \,s}\, \frac{\psi '\left( \frac{1}{\beta \,s\cos \vartheta }\right) }{\psi \left( \frac{1}{\beta \,s\cos \vartheta }\right) ^2}\, \frac{\sin \vartheta }{\cos ^2\vartheta }~\mathrm {d}\vartheta \nonumber \\&\quad =-\frac{\psi \left( \frac{1}{\beta \,s}\right) ^3}{\beta ^2\,s^2} \int \limits _0^{\pi /2}\frac{\psi '\left( \frac{1}{\beta \,s\cos \vartheta }\right) }{\psi \left( \frac{1}{\beta \,s\cos \vartheta }\right) ^4}\, \frac{\sin ^2\vartheta }{\cos ^3\vartheta }~\mathrm {d}\vartheta \end{aligned}$$

(81)

where the inverse function has been cancelled due to $\psi ^{-1}\circ \psi \equiv \mathrm {id}$. Inserting this into (76) and rewriting $\psi$ into $\nu$ via (78) and

$$\begin{aligned} \psi '(q) = \nu \left( \frac{1}{q}\right) -\frac{1}{q}\,\nu '\left( \frac{1}{q}\right) \end{aligned}$$

(82)

gives

$$\begin{aligned} g(s)&=\frac{3}{\beta ^4s^4}\! \int \limits _0^{\pi /2} \frac{\nu (\beta \,s\cos \vartheta )-\beta \,s\cos \vartheta \, \nu '(\beta \,s\cos \vartheta )}{\frac{1}{\beta ^4s^4\cos ^4\vartheta }\,\nu (\beta \,s\cos \vartheta )^4}\, \frac{\sin ^2\vartheta }{\cos ^3\vartheta }~\mathrm {d}\vartheta \nonumber \\&=3\int \limits _0^{\pi /2} \frac{\sin ^2\vartheta \cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3} ~\mathrm {d}\vartheta \nonumber \\&\quad -3\,\beta \,s \int \limits _0^{\pi /2}\frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4}\sin ^2\vartheta \cos ^2\vartheta ~\mathrm {d}\vartheta \;. \end{aligned}$$

(83)

Integration by parts using (72) gives for the first summand

$$\begin{aligned}&\int \limits _0^{\pi /2} \frac{\sin ^2\vartheta \cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3} ~\mathrm {d}\vartheta = \int \limits _0^{\pi /2} \frac{\sin \vartheta \cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}\, \frac{\sin (2\vartheta )}{2}~\mathrm {d}\vartheta \nonumber \\&\quad = \left[ \frac{\sin \vartheta }{\nu (\beta \,s\cos \vartheta )^3}\, \left( -\frac{1}{4}\cos (2\vartheta )\right) \right] _{\vartheta =0}^{\vartheta =\pi /2}\nonumber \\&\qquad +\frac{1}{4}\int \limits _0^{\pi /2} \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}\,\cos (2\vartheta ) ~\mathrm {d}\vartheta \nonumber \\&\qquad +\frac{3}{4}\,\beta \,s\int \limits _0^{\pi /2} \frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4} \sin ^2\vartheta \cos (2\vartheta )~\mathrm {d}\vartheta \nonumber \\&\quad =\frac{1}{4}+\frac{1}{4}\int \limits _0^{\pi /2} \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}~\mathrm {d}\vartheta -\frac{1}{2}\int \limits _0^{\pi /2} \frac{\cos \vartheta \sin ^2\vartheta }{\nu (\beta \,s\cos \vartheta )^3} ~\mathrm {d}\vartheta \nonumber \\&\qquad +\frac{3}{4}\,\beta \,s\int \limits _0^{\pi /2} \frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4} \sin ^2\vartheta \cos (2\vartheta )~\mathrm {d}\vartheta \end{aligned}$$

(84)

and after reordering of terms and division by $3/2$

$$\begin{aligned}&\int \limits _0^{\pi /2}\frac{\sin ^2\vartheta \cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3} ~\mathrm {d}\vartheta = \frac{1}{6} +\frac{1}{6}\int \limits _0^{\pi /2} \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}~\mathrm {d}\vartheta \nonumber \\&\qquad +\frac{1}{2}\,\beta \,s\int \limits _0^{\pi /2} \frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4} \sin ^2\vartheta \cos (2\vartheta )~\mathrm {d}\vartheta \;. \end{aligned}$$

(85)

Making once more use of (72), we calculate

$$\begin{aligned}&\int \limits _0^{\pi /2} \frac{\cos \vartheta }{\nu (\beta \,s\cos \vartheta )^3}~\mathrm {d}\vartheta = \underbrace{\left[ \frac{\sin \vartheta }{\nu (\beta \,s\cos \vartheta )^3}\right] _{\vartheta =0}^{\vartheta =\pi /2}}_{{}=1}\nonumber \\&\qquad -3\,\beta \,s\int \limits _0^{\pi /2}\frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4} \sin ^2\vartheta ~\mathrm {d}\vartheta \;. \end{aligned}$$

(86)

Substituting (85) and (86) into (83) eventually leads to

$$\begin{aligned} g(s)&= 1-3\,\beta \,s\int \limits _0^{\pi /2} \frac{\nu '(\beta \,s\cos \vartheta )}{\nu (\beta \,s\cos \vartheta )^4}\sin ^2\vartheta (\sin ^2\vartheta +\cos ^2\vartheta ) ~\mathrm {d}\vartheta \nonumber \\&=1-\frac{3}{2}\,\beta \,s\,\tilde{J}_1(\beta \,s) \end{aligned}$$

(87)

in accordance with the representation from Corollary 3. This completes the proof.