Orders for Simplifying Partial Partitions

Abstract

This paper is the third in a series devoted to orders on partial partitions in the framework of image analysis; the first two (with a view to image filtering and segmentation) identified 4 basic operations on blocks involved in such orders: merging, apportioning, creating and inflating blocks. Here, we consider orders where growing a partial partition decreases its support and diminishes the number of blocks. This can be done by a combination of merging (or apportioning) blocks with removing or deflating blocks (the opposite of creating or inflating blocks). We also introduce related operations on blocks: partial apportioning and partial merging. The new orders that we obtain can be useful in relation to skeletonization, image simplification, for processing segmentation markers or to describe the evolution of object boundaries in hierarchies. There are also possible applications in geographic information processing.

Appendix: More About Single Overlap

In [15] we indicated that the singularity relation satisfies the following:

We have the counterpart of it for single overlap:

(17)

In particular, single overlap is preserved by restriction: for any $A \in \mathcal {P}(E)$, $\pi _1 \vdash \pi _2 \;\Rightarrow \;\pi _1 \cap \mathcal {P}(A) \vdash \pi _2 \cap \mathcal {P}(A)$. Now the following complements Proposition 7:

Proposition 17

For any $\pi _1,\pi _2 \in \varPi ^*(E)$,

$$\begin{aligned} \begin{array}{lrrl} \forall \, \pi _0 \in \varPi ^*(E), \quad &{} \pi _0 \le \pi _1 \vdash \pi _2 &{}\implies &{} \pi _0 \vdash \pi _2 , \\ \forall \, A \in \mathcal {P}(E), \quad &{} \pi _1 \vdash \pi _2 &{}\implies &{} \pi _1 \vdash \pi _2 \wedge \mathbf {1}_A , \\ \forall \, A \supseteq \mathsf {supp}(\pi _1), \quad &{} \pi _1 \vdash \pi _2 &{}\iff &{} \pi _1 \vdash \pi _2 \wedge \mathbf {1}_A . \end{array} \end{aligned}$$

In particular, single overlap is preserved by truncation: for any $A \in \mathcal {P}(E)$, $\pi _1 \vdash \pi _2 \;\Rightarrow \;\pi _1 \wedge \mathbf {1}_A \vdash \pi _2 \wedge \mathbf {1}_A$.

Proof

Let $\pi _1 \vdash \pi _2$, that is (by Proposition 7), $\pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)} \le \pi _2$. Then for $\pi _0 \in \varPi ^*(E)$ such that $\pi _0 \le \pi _1$, we get $\pi _0 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)} \le \pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)} \le \pi _2$, that is, $\pi _0 \vdash \pi _2$.

For $A \in \mathcal {P}(E)$ we have $\mathsf {supp}(\pi _2 \wedge \mathbf {1}_A) = \mathsf {supp}(\pi _2) \cap A$, so $\mathbf {1}_{\mathsf {supp}(\pi _2 \wedge \mathbf {1}_A)} = \mathbf {1}_{\mathsf {supp}(\pi _2) \cap A} = \mathbf {1}_{\mathsf {supp}(\pi _2)} \wedge \mathbf {1}_A$, hence

$$\begin{aligned} \pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2 \wedge \mathbf {1}_A)} = (\pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)}) \wedge \mathbf {1}_A \le \pi _2 \wedge \mathbf {1}_A , \end{aligned}$$

thus $\pi _1 \vdash \pi _2 \wedge \mathbf {1}_A$. Combining both results with $\pi _0 = \pi _1 \wedge \mathbf {1}_A$, we get $\pi _1 \vdash \pi _2 \;\Rightarrow \;\pi _1 \wedge \mathbf {1}_A \vdash \pi _2 \;\Rightarrow \;\pi _1 \wedge \mathbf {1}_A \vdash \pi _2 \wedge \mathbf {1}_A$: single overlap is preserved by truncation.

Suppose finally that $A \supseteq \mathsf {supp}(\pi _1)$; then $\pi _1 \le \mathbf {1}_A$; now let $\pi _1 \vdash \pi _2 \wedge \mathbf {1}_A$, that is, $\pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2 \wedge \mathbf {1}_A)} \le \pi _2 \wedge \mathbf {1}_A$; we saw above that $\mathbf {1}_{\mathsf {supp}(\pi _2 \wedge \mathbf {1}_A)} = \mathbf {1}_{\mathsf {supp}(\pi _2)} \wedge \mathbf {1}_A$; hence we get

$$\begin{aligned} \pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)}= & {} (\pi _1 \wedge \mathbf {1}_A) \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)} = \pi _1 \wedge (\mathbf {1}_{\mathsf {supp}(\pi _2)} \wedge \mathbf {1}_A) \\= & {} \pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2 \wedge \mathbf {1}_A)} \le \pi _2 \wedge \mathbf {1}_A \le \pi _2 , \end{aligned}$$

therefore $\pi _1 \vdash \pi _2$. $\square$

As a counterpart of single overlap, we could consider the relation given by “every block of $\pi _1$ overlaps (at least) one block of $\pi _2$”, which is equivalent both to $\pi _1 \wedge \mathbf {1}_{\mathsf {supp}(\pi _2)} \Subset \pi _1$ and to $\pi _1 \wedge \pi _2 \Subset \pi _1$. However, it seems that this relation is too general to produce a new partial order relation by intersecting it with other relations.