Topological Properties of the First Non-Local Digitally Well-Composed Interpolation on n-D Cubical Grids

Abstract

In discrete topology, we like digitally well-composed (shortly DWC) interpolations because they remove pinches in cubical images. Usual well-composed interpolations are local and sometimes self-dual (they treat in a same way dark and bright components in the image). In our case, we are particularly interested in n-D self-dual DWC interpolations to obtain a purely self-dual tree of shapes. However, it has been proved that we cannot have an n-D interpolation which is at the same time local, self-dual and well-composed. By removing the locality constraint, we have obtained an n-D interpolation with many properties in practice: it is self-dual, DWC, and in-between (this last property means that it preserves the contours). Since we did not publish the proofs of these results before, we propose to provide in a first time the proofs of the two last properties here (DWCness and in-betweeness) and a sketch of the proof of self-duality (the complete proof of self-duality requires more material and will come later). Some theoretical and practical results are given.

Appendix

1.1 Proof of the First Intrinsic Property of the FPA

Lemma 4

Let $U : {\mathcal {D}}' \subseteq ({\mathbb {Z}}/2)^n\leadsto {\mathbb {R}}$ be an n-D interval-valued map, and let $u^{\flat }= {{\mathfrak {F}}}{{\mathfrak {P}}}(U) : {\mathcal {D}}' \rightarrow {\mathbb {R}}$ be the function resulting from the front propagation algorithm applied on U. Now, let $a,m \in {\mathcal {D}}'$ be 2n-neighbors in $({\mathbb {Z}}/2)^n$ such that $U(a) \subseteq U(m)$. Then $u^{\flat }(m) < u^{\flat }(a)$ implies that $u^{\flat }(a) = \lfloor U \rfloor (a)$ and $u^{\flat }(m) > u^{\flat }(a)$ implies that $u^{\flat }(a) = \lceil U \rceil (a)$.

Proof

Let us begin with the case $t(a) < t(m)$, that is, a has been enqueued before m. Three cases are possible.

The first subcase corresponds to $\ell (a) > \lceil U \rceil (a)$. Then $u^{\flat }(a) = \lceil U \rceil (a)$, $Q[u^{\flat }(a)] \supseteq \{a\}$ at $t = t(a)$, and the current level $\ell$ remains greater than or equal to $u^{\flat }(a)$ until a has been processed, because no jump of non-empty queue level is allowed. Since m is enqueued after a (by hypothesis) and at the latest during the processing of a (because a and m are 2n-neighbors), $\ell (m) \ge u^{\flat }(a)$. Since $\lceil U \rceil (m) \ge \lceil U \rceil (a) \ge u^{\flat }(a)$, we obtain finally the relation $u^{\flat }(m) \ge u^{\flat }(a)$ (Case 1.1).

The second subcase corresponds to $\ell (a) \in U(a)$. In this subcase, $u^{\flat }(a) = \ell (a)$, $Q[u^{\flat }(a)] \supseteq \{a\}$ at time $t = t(a)$, and the current level $\ell$ stays at the value $u^{\flat }(a)$ until a is processed (at least). Since a and m are 2n-neighbors, and since m is enqueued after a, m is enqueued after t(a) and at the latest while a is processed. This way, $\ell (m) = u^{\flat }(a)$ and then $u^{\flat }(m) = u^{\flat }(a)$ since $U(a) \subseteq U(m)$ (Case 1.2).

The third subcase corresponds to $\ell (a) < \lfloor U \rfloor (a)$. We reason by symmetry and we obtain that $u^{\flat }(a) = \lfloor U \rfloor (a)$ and $u^{\flat }(m) \le u^{\flat }(a)$ (Case 1.3).

Let us follow with the case $t(a) > t(m)$. Then five subcases are possible.

If $\ell (m) > \lceil U \rceil (m)$, then $u^{\flat }(m) = \lceil U \rceil (m)$, $Q[u^{\flat }(m)] \supseteq \{m\}$ at $t = t(m)$, and the current level $\ell$ remains greater than or equal to $u^{\flat }(m)$ until m has been processed, because no jump of non-empty queue level is allowed. Since a is enqueued after m (by hypothesis) and at the latest during the processing of m (because a and m are 2n-neighbors), $\ell (a) \ge u^{\flat }(m)$. Then two subcases are possible:

• either $\lceil U \rceil (m) >\lceil U \rceil (a)$ and then:

  $$\begin{aligned} u^{\flat }(a) = \lceil U \rceil (a) < u^{\flat }(m), \ \ \text {(Case 2.1.a)} \end{aligned}$$

• or $\lceil U \rceil (m) = \lceil U \rceil (a)$ and then:

  $$\begin{aligned} u^{\flat }(a) = \lceil U \rceil (a) = u^{\flat }(m). \ \ \text {(Case 2.1.b)} \end{aligned}$$

If $\ell (m) \in ]\lceil U \rceil (a), \lceil U \rceil (m)]$, assuming that $\lceil U \rceil (a) < \lceil U \rceil (m)$, $u^{\flat }(m) = \ell (m)$, $Q[u^{\flat }(m)] \supseteq \{m\}$ at $t = t(m)$, and the current level $\ell$ stays at the value $u^{\flat }(m)$ until m is processed (at least). Since a and m are 2n-neighbors, and since a is enqueued after m, a is enqueued after t(m) and at the latest while m is processed. This way, $\ell (a) = u^{\flat }(m)$, and then $u^{\flat }(a) = \lceil U \rceil (a) < u^{\flat }(m)$ (Case 2.2).

If $\ell (m) \in U(a)$, $u^{\flat }(m) = \ell (m)$ (since we have $U(a) \subseteq U(m)$) and $Q[u^{\flat }(m)] \supseteq \{m\}$ at $t = t(m)$. Then the current level $\ell$ stays at the value $u^{\flat }(m)$ until m is processed (at least). Since a and m are 2n-neighbors, and since a is enqueued after m, a is enqueued after t(m) and at the latest while m is processed. This way, $\ell (a) = u^{\flat }(a)$ and then $u^{\flat }(a) = u^{\flat }(m)$ (Case 2.3).

If $\ell (m) \in [\lfloor U \rfloor (m), \lfloor U \rfloor (a)]$ (assuming that $\lfloor U \rfloor (m) < \lfloor U \rfloor (a)$), we reason by symmetry and we obtain that $u^{\flat }(a) = \lfloor U \rfloor (a) > u^{\flat }(m)$ (Case 2.4).

If $\ell (m) < \lfloor U \rfloor (m)$, we reason again by symmetry and we obtain that:

• either $\lfloor U \rfloor (m) < \lfloor U \rfloor (a)$ and:

  $$\begin{aligned} u^{\flat }(a) = \lfloor U \rfloor (a) > u^{\flat }(m), \ \ \text {(Case 2.5.a)} \end{aligned}$$

• or $\lfloor U \rfloor (m) = \lfloor U \rfloor (a)$ and:

  $$\begin{aligned} u^{\flat }(a) = \lfloor U \rfloor (a) = u^{\flat }(m).\ \ \text {(Case 2.5.b)} \end{aligned}$$

Let us summarize the different cases:

Case	Relation 1	Relation 2	Relation 3
(1.1)	$t(a) < t(m)$	$u^{\flat }(a) = \lceil U \rceil (a)$	$u^{\flat }(m) \ge u^{\flat }(a)$
(1.2)	$t(a) < t(m)$	$u^{\flat }(a) \in U(a)$	$u^{\flat }(m) = u^{\flat }(a)$
(1.3)	$t(a) < t(m)$	$u^{\flat }(a) = \lfloor U \rfloor (a)$	$u^{\flat }(m) \le u^{\flat }(a)$
(2.1.a)	$t(m) < t(a)$	$u^{\flat }(a) = \lceil U \rceil (a)$	$u^{\flat }(m) > u^{\flat }(a)$
(2.1.b)	$t(m) < t(a)$	$u^{\flat }(a) = \lceil U \rceil (a)$	$u^{\flat }(m) = u^{\flat }(a)$
(2.2)	$t(m) < t(a)$	$u^{\flat }(a) = \lceil U \rceil (a)$	$u^{\flat }(m) > u^{\flat }(a)$
(2.3)	$t(m) < t(a)$	$u^{\flat }(a) \in U(a)$	$u^{\flat }(m) = u^{\flat }(a)$
(2.4)	$t(m) < t(a)$	$u^{\flat }(a) = \lfloor U \rfloor (a)$	$u^{\flat }(m) < u^{\flat }(a)$
(2.5.a)	$t(m) < t(a)$	$u^{\flat }(a) = \lfloor U \rfloor (a)$	$u^{\flat }(m) < u^{\flat }(a)$
(2.5.b)	$t(m) < t(a)$	$u^{\flat }(a) = \lfloor U \rfloor (a)$	$u^{\flat }(m) = u^{\flat }(a)$

We obtain finally that $u^{\flat }(a) < u^{\flat }(m)$ implies that we are in Cases 1.1, 2.1.a, or 2.2 and then $u^{\flat }(a) = \lceil U \rceil (a)$, and that $u^{\flat }(a) > u^{\flat }(m)$ implies that we are in Cases 1.3, 2.4 or 2.5.a, and then $u^{\flat }(a) = \lfloor U \rfloor (a)$. This concludes the proof. $\square$

1.2 Proof of the Secund Intrinsic Property of the FPA

Lemma 5

Let $U : {\mathcal {D}}' \subseteq ({\mathbb {Z}}/2)^n\leadsto {\mathbb {R}}$ be an n-D interval-valued map, and let $u^{\flat }= {{\mathfrak {F}}}{{\mathfrak {P}}}(U) : {\mathcal {D}}' \rightarrow {\mathbb {R}}$ be the function resulting from the front propagation algorithm applied on U. Now, let r be a point of ${\mathcal {D}}'$. We can observe the two following implications:

$$\begin{aligned} \left\{ \begin{array}{ll} u^{\flat }(r) < \lceil U \rceil (r) \Rightarrow \ell (r) \le u^{\flat }(r) &{} (1)\\ u^{\flat }(r) > \lfloor U \rfloor (r) \Rightarrow \ell (r) \ge u^{\flat }(r) &{} (2) \end{array}\right. \end{aligned}$$

Proof

By a case-by-case study, we can establish a correlation between $\ell (r)$ and $u^{\flat }(r)$ for any given point $r \in {\mathcal {D}}'$. The possible cases are $\ell (r) < \lfloor U \rfloor (r)$ (1), $\ell (r) \in U(r)$ (2), and $\ell (r) > \lceil U \rceil (r)$ (3):

• 1. we obtain that $\ell (r) < u^{\flat }(r)$ because $u^{\flat }(r) \in U(r)$, and at the same time, $u^{\flat }(r)$ is equal to $\lfloor U \rfloor (r)$ because it is the nearest value to $\ell (r)$ in U(r);

• 2. we obtain that $u^{\flat }(r) = \ell (r)$ because the nearest value to $\ell (r)$ in U(r) is $\ell (r)$ itself, and at the same time we obtain simply the initial property $u^{\flat }(r) \in U(r)$ (no additional assumption is possible);

• 3. we obtain that $\ell (r) > u^{\flat }(r)$ because $u^{\flat }(r) \in U(r)$, and at the same time $u^{\flat }(r) = \lceil U \rceil (r)$ because this is the nearest value to $\ell (r)$ into U(r).

Finally, we obtain this table:

Case	Relation 1	Relation 2
(1) : $\ell (r) < \lfloor U \rfloor (r)$	$\ell (r) < u^{\flat }(r)$	$u^{\flat }(r) = \lfloor U \rfloor (r)$
(2) : $\ell (r) \in U(r)$	$\ell (r) = u^{\flat }(r)$	$u^{\flat }(r) \in U(r)$
(3) : $\ell (r) > \lceil U \rceil (r)$	$\ell (r) > u^{\flat }(r)$	$u^{\flat }(r) = \lceil U \rceil (r)$

Then we can observe that if $u^{\flat }(r) < \lceil U \rceil (r)$, that is, if $u^{\flat }(r) \ne \lceil U \rceil (r)$, we are then either in the case (1) or in the case (2) and then we obtain that $\ell (r) \le u^{\flat }(r)$.

Conversely, if $u^{\flat }(r) > \lfloor U \rfloor (r)$, that is, if $u^{\flat }(r) \ne \lfloor U \rfloor (r)$, we are then either in the case (2) or in the case (3) and then we obtain that $\ell (r) \ge u^{\flat }(r)$. $\square$

Lemma 6

Let $U : {\mathcal {D}}' \subseteq ({\mathbb {Z}}/2)^n\leadsto {\mathbb {R}}$ be an n-D interval-valued map, and let $u^{\flat }= {{\mathfrak {F}}}{{\mathfrak {P}}}(U) : {\mathcal {D}}' \rightarrow {\mathbb {R}}$ be the gray-level function resulting from the front propagation algorithm applied on U. Let $p,q \in {\mathcal {D}}'$ be two 2n-neighbors in $({\mathbb {Z}}/2)^n$ and $\lambda \in {\mathbb {R}}$. Then, it is impossible to get the following set of properties together:

$$\begin{aligned} \left\{ \begin{array}{llll} u^{\flat }(p) &{}\, \le &{} \, \lambda , &{}\,({\mathcal {H}}1) \\ \lceil U \rceil (p) &{}\,>&{} \, \lambda , &{}\,({\mathcal {H}}2) \\ u^{\flat }(q) &{}\, >&{} \, \lambda , &{}\,({\mathcal {H}}3) \\ \lfloor U \rfloor (q) &{}\, \le &{} \, \lambda . &{}\,({\mathcal {H}}4)\\ \end{array}\right. \end{aligned}$$

Fig. 29

The 4 possible scenarios when only two 2n-neighbors p and q in ${\mathcal {D}}'$ are considered

Proof

Now, let p, q be two 2n-neighbors in ${\mathcal {D}}'$ and let us assume that there exists a value $\lambda \in {\mathbb {R}}$ verifying $({\mathcal {H}}1)$, $({\mathcal {H}}2)$, $({\mathcal {H}}3)$ and $({\mathcal {H}}4)$.

We can observe easily thanks to $({\mathcal {H}}1)$ and $({\mathcal {H}}2)$ that $u^{\flat }(p) < \lceil U \rceil (p)$ and then using Lemma 5, we obtain:

$$\begin{aligned} \ell (p) \le u^{\flat }(p)\ \ \ ({\mathcal {H}}5). \end{aligned}$$

In addition, thanks to $({\mathcal {H}}3)$ and $({\mathcal {H}}4)$, we obtain $u^{\flat }(q) > \lfloor U \rfloor (q)$ and using Lemma 5, this results in:

$$\begin{aligned} \ell (q) \ge u^{\flat }(q) \ \ \ ({\mathcal {H}}6). \end{aligned}$$

Taking into consideration the two 2n-neighbors p and q, we have 4 possible scenarios as depicted in Fig. 29:

• 1. either p is enqueued before q, then two subcases are possible:

  • (a) either q is enqueued when p is the current position,

  • (b) or q is enqueued before p is the current position.

• 2. either q is enqueued before p, then two subcases are possible:

  • (a) either p is enqueued when q is the current position,

  • (b) or p is enqueued before q is the current position.

Let us notice that since p and q are 2n-neighbors, q cannot be enqueued after p is the current position, and similarly p cannot been enqueued after q is the current position (all the 2n-neighbors of the current position will have been enqueued when it has been processed).

Now let us show that whatever the scenario we choose, we always obtain a contradiction.

(1.a): p is enqueued before q, and then q is enqueued when p is the current position. It means that $\ell (q) = u^{\flat }(p)$. However, we have seen that $u^{\flat }(p) \le \lambda$ by $({\mathcal {H}}1)$, and that $\ell (q) \ge u^{\flat }(q) > \lambda$ by $({\mathcal {H}}6)$ and $({\mathcal {H}}3)$. This leads to a contradiction.

(1.b): p is enqueued before q, and q is enqueued before the current position is set at p. This way, since the current level $\ell$ at t(p) is equal to $\ell (p) \le u^{\flat }(p)$, it is equal to $\ell (q) \le u^{\flat }(p)$ at t(q) (no jump of the non-empty queue level $Q[u^{\flat }(p)]$ is allowed by the algorithm). This means by $({\mathcal {H}}1)$ that $\ell (q) \le \lambda$. However, by $({\mathcal {H}}6)$ and $({\mathcal {H}}3)$, $\ell (q) > \lambda$. This leads to a contradiction.

(2.a) is the symmetrical case of (1.a) and (2.b) is the one of (1.b) and then they lead also to contradictions.

The conclusion is that whatever the scenario (and one of these scenarios happens during the computation of the interpolation), the combination of hypotheses $({\mathcal {H}}1)$, $({\mathcal {H}}2)$, $({\mathcal {H}}3)$ and $({\mathcal {H}}4)$ leads to a contradiction. These hypotheses are then incompatible. $\square$

1.3 The Proof that Adding a Constant Border Preserves DWCNess

Proposition 7

Let us denote by $\delta$ the dilation operator and by $\text {se}$ the structuring element defined such that

$$\begin{aligned} \text {se}:= \left\{ p \in \left( \frac{{\mathbb {Z}}}{2}\right) ^n\; ; \; ||p||_{\infty } \le 1/2\right\} . \end{aligned}$$

Let $U_0 : {\mathcal {D}}\subset \left( \frac{{\mathbb {Z}}}{2}\right) ^n\leadsto {\mathbb {Z}}$ be a DWC interval-valued map defined on a bounded hyperrectangle ${\mathcal {D}}$ in $\left( \frac{{\mathbb {Z}}}{2}\right) ^n$. Now, let $U_1 : {\mathcal {D}}' \leadsto {\mathbb {Z}}$ be another interval-valued map defined on a bounded hyperrectangle ${\mathcal {D}}' = \delta ({\mathcal {D}}, \text {se})$, such that $U_1|_{{\mathcal {D}}} = U_0$ and for any $p \in {\mathcal {D}}' {\setminus } {\mathcal {D}}$, $U'(p) = \{c\}$ (where c in a given constant in ${\mathbb {R}}$). Then, $U_1$ is a DWC interval-valued map.

Fig. 30

Two possible configurations when dilating the domain ${\mathcal {D}}_{k-1}$ into ${\mathcal {D}}_k$ with our structuring elements

Proof

First let us introduce some notations. Let $(\text {se}^k)_{k \in \llbracket 1,2n \rrbracket }$ be a sequence of structuring elements defined s.t. $\forall k \in \llbracket 1,2n \rrbracket$:

$$\begin{aligned} \text {se}^k = \left\{ \mathbf{0}, \frac{1}{2} \ (-1)^k \ e^{\lfloor \frac{(k+1)}{2}\rfloor }\right\} , \end{aligned}$$

and let $({\mathcal {D}}_k)_{k \in \llbracket 0,2n \rrbracket }$ be a sequence of domains s.t. ${\mathcal {D}}_0 = {\mathcal {D}}$ and s.t., $\forall k \in \llbracket 1,2n \rrbracket$:

$$\begin{aligned} {\mathcal {D}}_k = \delta ({\mathcal {D}}_{k-1},\text {se}^k). \end{aligned}$$

In this manner, ${\mathcal {D}}_{2n} = \delta ({\mathcal {D}},\text {se}) = {\mathcal {D}}'$.

We want to show that $U_1$ is digitally well-composed on ${\mathcal {D}}'$, and for that we are going to show by an induction process that, $\forall k \in \llbracket 0,2n \rrbracket$, $U_1\big |_{{\mathcal {D}}_k}$ is digitally well-composed.

Initialization ($k = 0$): $U_1\big |_{{\mathcal {D}}_0} = U_0$ which is DWC by hypothesis.

Heredity ($k \in \llbracket 1,2n \rrbracket$): assuming that the image $U_1\big |_{{\mathcal {D}}_{k-1}}$ is DWC, let us show that $U_1\big |_{{\mathcal {D}}_k}$ is DWC too. Two cases are then possible.

• either k is odd, then $\text {se}^k = \{\mathbf{0}, - \ \frac{1}{2} \ e^{\frac{k+1}{2}}\}$, and then we obtain the configuration depicted in Fig. 30 (Case 1),

• or k is even, then $\text {se}^k = \{\mathbf{0}, \frac{1}{2} \ e^{\frac{k}{2}}\}$, and then we obtain the configuration depicted in Fig. 30 (Case 2).

Let us now denote $\varDelta {\mathcal {D}}$ the set equal to ${\mathcal {D}}_k {\setminus } {\mathcal {D}}_{k-1}$. Let us remark that this set is also an hyperrectangle. In addition, let us denote by $u^+_k$, $u^-_k$, $u^+_{k-1}$ and $u^-_{k-1}$ the images $\lceil U_1 \rceil \big |_{{\mathcal {D}}_k}$, $\lfloor U_1 \rfloor \big |_{{\mathcal {D}}_k}$, $\lceil U_1 \rceil \big |_{{\mathcal {D}}_{k-1}}$, $\lfloor U_1 \rfloor \big |_{{\mathcal {D}}_{k-1}}$, respectively. We can say that $U_1\big |_{{\mathcal {D}}_k}$ is DWC iff $\forall S \in {\mathcal {B}}({\mathcal {D}}_k,\left( \frac{{\mathbb {Z}}}{2}\right) ^n)$ s.t. $\dim (S) \ge 2$, $\forall p,p' \in S$ s.t. $p' = \text {antag}_S(p)$, we have the following relations:

$$\begin{aligned} \left\{ \begin{array}{lll} &{}\text {intvl}(u^+_k(p),u^+_k(p')) \\ &{}\cap \ \text {Span}\{u^+_k(q) \; ; \; q \in S {\setminus } \{p,p'\}\} \ne \emptyset , &{} \hbox {(A)}\\ &{}\text {intvl}(u^-_k(p),u^-_k(p')) \\ &{}\cap \ \text {Span}\{u^-_k(q) \; ; \; q \in S {\setminus } \{p,p'\}\} \ne \emptyset . &{} \hbox {(B)}\\ \end{array}\right. \end{aligned}$$

So let S be such a block of ${\mathcal {D}}_k$ into $\left( \frac{{\mathbb {Z}}}{2}\right) ^n$, and let $p^{\min }$ and $p^{\max }$ be two elements of S such that, for any $i \in \llbracket 1,n \rrbracket$,

$$\begin{aligned} \left\{ \begin{array}{l} p^{\min }_i = \min \{p_i \; ; \; p \in S\},\\ p^{\max }_i = \max \{p_i \; ; \; p \in S\}.\\ \end{array}\right. \end{aligned}$$

In this manner, $p^{\min }$ and $p^{\max }$ are antagonists in S. Then, 4 cases are possible:

• 1. $p^{\min }$ and $p^{\max }$ belong to ${\mathcal {D}}_{k-1}$, then $S \subseteq {\mathcal {D}}_{k-1}$, and in this way, $\forall p \in S$, $u^+_k(p) = u^+_{k-1}(p)$ and $u^-_k(p) = u^-_{k-1}(p)$, which implies that the intersections in (A) and (B) are non-empty since $u^+_{k-1}$ and $u^-_{k-1}$ are DWC and $\dim (S) \ge 2$,

• 2. or $p^{\min }$ and $p^{\max }$ belong to $\varDelta {\mathcal {D}}$, then $S \subseteq \varDelta {\mathcal {D}}$, and then $\forall p \in S$, $u^+_k(p) = u^-_{k}(p) = c$, which means that (A) and (B) are true since $\dim (S) \ge 2$,

• 3. or $p^{\min }\in {\mathcal {D}}_{k-1}$ and $p^{\max }\in \varDelta {\mathcal {D}}$. Then we are in the second case in Fig. 30. In other words,

  $$\begin{aligned} \left\{ \begin{array}{l} S \cap {\mathcal {D}}_{k-1} = \{p \in S \; ; \; p_{k/2} = p^{\min }_{k/2}\},\\ S \cap \varDelta {\mathcal {D}}= \{p \in S \; ; \; p_{k/2} = p^{\max }_{k/2}\},\\ \end{array}\right. \end{aligned}$$

  which means that S can decomposed into two blocks of dimension $\dim (S) - 1 \ge 1$, the first being $S \cap {\mathcal {D}}_{k-1}$ and the second being $S^* := S \cap \varDelta {\mathcal {D}}$. Since $S^*$ verifies that $\forall p \in S^*$, $u^+_k(p) = u^-_k(p) = c$ and that $\dim (S^*) \ge 1$, there exist two points $p,q \in S^*$ which are not antagonists into S and such that $u^+_k(p) = u^+_k(q)$ and $u^-_k(p) = u^-_k(q)$, then (A) and (B) are both true,

• 4. or $p^{\max }\in {\mathcal {D}}_{k-1}$ and $p^{\min }\in \varDelta {\mathcal {D}}$. Then we are in the first case in Fig. 30. A dual reasoning leads to the fact that (A) and (B) are both true.

We can then conclude by induction that $U_1$ is DWC. $\square$

1.4 Proof that digital well-composedness Implies Equivalent Connectivities

Let us recall the definition of well-composedness based on the equivalence of connectivities[8] (EWCness).

Definition 21

Let X be a digital set in ${\mathbb {Z}}^n$. X is said to be EWC or well-composed based on the equivalence of its connectivities if the two following conditions hold:

• any of its 2n-components is also one of its $(3^n-1)$-components and vice versa,

• any 2n-component of $X^c$ is also a $(3^n-1)$-component of $X^c$ and vice versa.

We can underline that this definition is clearly self-dual, and since the connectivity does not matter for this class of sets, we will sometimes say that their connectivities (and the ones of their complement in ${\mathbb {Z}}^n$) are equivalent. In addition, this definition is the “natural” extension of the one of Latecki [26] for 2D sets.

Now that we have the definition of EWCness for sets, we can define EWCness for gray-level images.

Definition 22

A gray-level image $u : {\mathcal {D}}\subseteq {\mathbb {Z}}^n \rightarrow {\mathbb {Z}}$ is said well-composed based on the equivalence of connectivities (EWC) if all its threshold sets are well-composed based on the equivalence of connectivities.

The definitions of EWCness for sets and images are extended naturally to $({\mathbb {Z}}/2)^n$: a subset X of $({\mathbb {Z}}/2)^n$ is said EWC if the connected components of X and of $({\mathbb {Z}}/2)^n{\setminus } X$ do not depend on the chosen connectivity, and a gray-level image $u : {\mathcal {D}}\subseteq ({\mathbb {Z}}/2)^n\rightarrow {\mathbb {Z}}$ is said EWC if all its threshold sets are EWC.

Let us recall that EWCness is a global property, since it is based on connected components, and that DWCness is based on local properties, that is, there is no critical configurations. That shows that the link between DWCness and EWCness is not so obvious. Before proving that DWCness implies EWCness in any (finite) dimension n, $n \ge 2$, let us announce some lemmas.

Lemma 7

Let $p,p' \in {\mathbb {Z}}^n$ be two points in a digitally well-composed set $X \subset {\mathbb {Z}}^n$. If p and $p'$ are $(3^n-1)$-connected into X, they are also 2n-connected into X.

Proof

Let $p,p'$ be two points in $X \subset {\mathbb {Z}}^n$ which is digitally well-composed. Assuming that p and $p'$ are $(3^n-1)$-connected into X, there exists a $(3^n-1)$-path

$$\begin{aligned} \pi = (q^0 = p, q^1, \dots ,q^{k-1},q^k = p') \end{aligned}$$

of length $k \ge 0$ joining them into X. For any $i \in \llbracket 0,k-1 \rrbracket$, $q^i$ and $q^{i+1}$ are $(3^n-1)$-adjacent, and then antagonists in a block $S(q^i,q^{i+1})$. Since X is digitally well-composed and $q^i$ and $q^{i+1}$ belong to X, by Theorem 1, there exists a 2n-path joining $q^i$ and $q^{i+1}$ into $X \cap S(q^i,q^{i+1})$. Then, p and $p'$ are 2n-connected into X. $\square$

Fig. 31

DWCness implies EWCness

Theorem 5

Let $X \subset {\mathbb {Z}}^n$ be a digitally well-composed set. Then, X is well-composed based on the equivalence of connectivities(EWC). In other words, we have:

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {C}}}_{2n}(X) = {{\mathcal {C}}}{{\mathcal {C}}}_{3^n-1}(X), \end{aligned}$$

and

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {C}}}_{2n}(X^c) = {{\mathcal {C}}}{{\mathcal {C}}}_{3^n-1}(X^c). \end{aligned}$$

Proof

Let assume that $X \subset {\mathbb {Z}}^n$ is DWC. By Lemma 7, each 2n-component of X is also a $(3^n-1)$-component of X (see details in Fig. 31a), and each $(3^n-1)$-component of X is also a 2n-component of X (see details in Fig. 31b). $\square$

Fig. 32

EWCness does not imply DWCness in n-D ($n \ge 3$)

Recall that the converse of Theorem 5 is not true in 3D (see Fig. 32): a 3D subset of ${\mathbb {Z}}^n$ can be EWC without being DWC, since the $(3^n-1)$-components and the 2n-components of this set are equal, but it contains a 2D critical configuration at the top and then is not DWC (the reasoning holds for any dimension $n \ge 3$).

Corollary 1

Let $u : {\mathcal {D}}\rightarrow {\mathbb {Z}}$ be a gray-level image. Then, when u is DWC, u is EWC.

Proof

This follows directly from Theorem 5. $\square$