Discrete and Continuous Models for Partitioning Problems

Abstract

Recently, variational relaxation techniques for approximating solutions of partitioning problems on continuous image domains have received considerable attention, since they introduce significantly less artifacts than established graph cut-based techniques. This work is concerned with the sources of such artifacts. We discuss the importance of differentiating between artifacts caused by discretization and those caused by relaxation and provide supporting numerical examples. Moreover, we consider in depth the consequences of a recent theoretical result concerning the optimality of solutions obtained using a particular relaxation method. Since the employed regularizer is quite tight, the considered relaxation generally involves a large computational cost. We propose a method to significantly reduce these costs in a fully automatic way for a large class of metrics including tree metrics, thus generalizing a method recently proposed by Strekalovskiy and Cremers (IEEE conference on computer vision and pattern recognition, pp. 1905–1911, 2011).

Appendix

Proof

(Proposition 3) Let $v \in \mathcal D _{\mathrm{loc}}^m$ and $i \ne j$. Denote a shortest path in $\mathcal G ^m$ from $i$ to $j$ by $(k_1=i,k_2,\ldots ,k_{n-1},k_n=j)$ for some $n \geqslant 2$. Then, due to the triangle inequality of the norm,

$$\begin{aligned} \Vert v^i - v^j\Vert _2 \leqslant \Vert v^{k_1} - v^{k_2} \Vert _2 + \ldots + \Vert v^{k_{n-1}} - v^{k_n}\Vert _2. \end{aligned}$$

(124)

Since $v \in \mathcal D _{\mathrm{loc}}^m$, the terms on the right can be bounded,

$$\begin{aligned} \Vert v^i - v^j\Vert _2 \leqslant m(k_1,k_2) + \ldots +m(k_{n-1},k_n). \end{aligned}$$

(125)

By definition, the sum equals $\bar{m}(i,j)$ and therefore by assumption $\bar{m^{\prime }}(i,j)$, thus

$$\begin{aligned} \Vert v^i - v^j\Vert _2 \leqslant \bar{m}(i,j)=\bar{m^{\prime }}(i,j) \leqslant m^{\prime }(i,j). \end{aligned}$$

(126)

Consequently $v \in \mathcal D _{\mathrm{loc}}^{m^{\prime }}$, and, since $v$ was arbitrary, $\mathcal D _{\mathrm{loc}}^m \subseteq \mathcal D _{\mathrm{loc}}^{m^{\prime }}$. By symmetry the reverse inclusion holds as well, and therefore equality. $\square$

Proof

(Proposition 4) Since $d$ is a metric and thus satisfies the triangle inequality, (114) is equivalent to

$$\begin{aligned} d(i,k) + d(k,j) = d(i,j). \end{aligned}$$

(127)

Since $d$ is the shortest-path metric of $m$, Eq. (127) states that there exists a shortest path from $i$ to $j$ in $\mathcal G ^m$ passing through $k$. This path consists of at least two edges which both have positive weight by the definition of $\mathcal M$. Therefore the path cannot contain the edge $(i,j)$, since $m(i,j) \geqslant d(i,j)$ which would contradict (127).

This means that the path is also contained in $\mathcal G ^{m^{\prime }}$. By construction of $m^{\prime }$ it must then also be a shortest path from $i$ to $j$ in $m^{\prime }$, which shows that $\bar{m^{\prime }}(i,j) = \bar{m}(i,j)$. Since $m^{\prime }(i^{\prime },j^{\prime }) = m(i^{\prime },j^{\prime })$ for all $(i^{\prime },j^{\prime }) \ne (i,j)$, this immediately implies $\bar{m^{\prime }} = \bar{m}$ and therefore the assertion. $\square$

Proof

(Proposition 5) Define

$$\begin{aligned} m(i,j):=\left\{ \begin{array}{ll} +\infty , &{} \exists k\ne i,j : d(i,j) \geqslant d(i,k) + d(k,j)\\ d(i,j), &{} \text{ otherwise }. \end{array}\right. \end{aligned}$$

An iterative application of Proposition 4 shows that $\bar{m} = d$. Proposition 3 then states that their associated dual constraint sets coincide,

$$\begin{aligned} \mathcal D _{\mathrm{loc}}^m = \mathcal D _{\mathrm{loc}}^d. \end{aligned}$$

(128)

Since $\bigcap _{(i,j) \in \mathcal R } \mathcal D _{i j} = \mathcal D _{\mathrm{loc}}^m$ by construction, this concludes the proof. $\square$