How to Make n-D Plain Maps Defined on Discrete Surfaces Alexandrov-Well-Composed in a Self-Dual Way

Abstract

In 2013, Najman and Géraud proved that by working on a well-composed discrete representation of a gray-level image, we can compute what is called its tree of shapes, a hierarchical representation of the shapes in this image. This way, we can proceed to morphological filtering and to image segmentation. However, the authors did not provide such a representation for the non-cubical case. We propose in this paper a way to compute a well-composed representation of any gray-level image defined on a discrete surface, which is a more general framework than the usual cubical grid. Furthermore, the proposed representation is self-dual in the sense that it treats bright and dark components in the image the same way. This paper can be seen as an extension to gray-level images of the works of Daragon et al. on discrete surfaces.

Appendix

In this section, we prove some remarkable properties specific to the framework of this paper.

Lemma 2

Let X be a suborder of a poset Y of rank $n \ge 0$, with $X = \cup _{x \in X_n} \alpha _Y(x)$. Then, we have:

$$\begin{aligned} \alpha _Y(Y {\setminus } X) \sqcup \mathrm {Int} _Y(X) = Y. \end{aligned}$$

Proof

Let us first prove that the union is disjoint. Let us assume that there exists some $p \in \alpha _Y(Y {\setminus } X) \cap \mathrm {Int} (X)$. Since $p \in \alpha _Y(Y {\setminus } X)$, there exists some $z \in Y {\setminus } X$ such that $p \in \alpha _Y(z)$, that is, $\beta _Y(z) \subseteq \beta _Y(p)$. Besides, $Y {\setminus } X$ is open since X is closed, then $\beta _Y(z) \subseteq Y {\setminus } X$. However, $p \in \mathrm {Int} _Y(X)$ implies that $\beta _Y(p) \subseteq X$ and then:

$$\begin{aligned} \beta _Y(z) \subseteq \beta _Y(p) \subseteq X, \end{aligned}$$

which contradicts $\beta _Y(z) \subseteq Y {\setminus } X$.

Let us now prove that the union is equal to Y. The fact that $\alpha _Y(Y {\setminus } X) \sqcup \mathrm {Int} _Y(X) \subseteq Y$ is obvious. Now, let us prove the converse inclusion. Let h be a face of Y. Two cases are possible:

• either $\beta _Y(h) \subseteq X$, then $h \in \mathrm {Int} _Y(X)$,

• or $\beta _Y(h) \not \subseteq X$, then $\beta _Y(h) \cap (Y {\setminus } X) \ne \emptyset$, and then there exists some $p \in \beta _Y(h) \cap (Y {\setminus } X)$; that is, $h \in \alpha _Y(p)$ and $p \in Y {\setminus } X$. In other words, $h \in \alpha _Y(Y {\setminus } X)$.

The proof is done. $\square$

Proposition 8

Let X be a suborder of a poset Y of rank $n \ge 0$, with $X = \cup _{x \in X_n} \alpha _Y(x)$. Then the topological boundary:

$$\begin{aligned} \alpha _Y(X) {\setminus } \mathrm {Int} _Y(X) \end{aligned}$$

of X in Y is equal to the combinatorial boundary:

$$\begin{aligned} \alpha _Y(X) \cap \alpha _Y(Y {\setminus } X) \end{aligned}$$

of X in Y.

Proof

This proposition follows directly from Lemma 2. $\square$

Property 8

The set-valued map $U_{\mathrm {USC}}: Y \rightarrow \mathbb {I}_{\mathbb {R}}$ is upper semi-continuous.

Proof

The fact that $U_{\mathrm {USC}}$ is USC relies on the fact that for any $z \in Y$ and for any $z' \in \beta _Y(z)$:

$$\begin{aligned} U_{\mathrm {USC}} (z') \subseteq U_{\mathrm {USC}} (z). \end{aligned}$$

Indeed, let z be an element of Y and let $z'$ be an element of $\beta _Y(z)$:

• when $z \in \mathrm {bd} (X,Y)$:

  $$\begin{aligned} U_{\mathrm {USC}} (z) = \mathrm {Span} \{\mathfrak {M},\mathrm {Span} \{U(q) ; \; q \in \beta _Y(z) \cap X_n\}\}, \end{aligned}$$

  • when $z' \in \beta _Y(z) \cap \mathrm {bd} (X,Y)$, $U_{\mathrm {USC}} (z')$ is equal to:

    $$\begin{aligned} \mathrm {Span} \{\mathfrak {M},\mathrm {Span} \{U(q) ; \; q \in \beta _Y(z') \cap X_n\}\}, \end{aligned}$$

    which is included in $U_{\mathrm {USC}} (z)$ since $\beta _Y(z') \subseteq \beta _Y(z)$,

  • when $z' \in \beta _Y(z)$ such that $z' \not \in \mathrm {bd} (X,Y)$, either $z' \in X {\setminus } \mathrm {bd} (X,Y)$ (which is an open set), which implies $\beta _Y(z') \subseteq X$ and $U_{\mathrm {USC}} (z') = U(z') \subseteq U_{\mathrm {USC}} (z)$, or $z' \in Y {\setminus } X$ (which is an open set since it is equal to $Y {\setminus } \mathrm {bd} (X,Y)$), which implies $\beta _Y(z') \subseteq Y {\setminus } X$ and $U_{\mathrm {USC}} (z') = \{\mathfrak {M} \} = U(z') \subseteq U_{\mathrm {USC}} (z)$,

• when $z \in X {\setminus } \mathrm {bd} (X,Y)$, then $\beta _Y(z) \subseteq X$, which means that $z' \in \beta _Y(z)$ belongs to X, and then:

  $$\begin{aligned} U_{\mathrm {USC}} (z') = U(z') \subseteq U(z), \end{aligned}$$

• when $z \in Y {\setminus } X$, then $\beta _Y(z) \subseteq Y {\setminus } X$, then for any $z' \in \beta _Y(z)$, $U_{\mathrm {USC}} (z') = \{\mathfrak {M} \} = U_{\mathrm {USC}} (z)$.

This concludes the proof. $\square$