Combinatorics of the Gauss Digitization Under Translation in 2D

Étienne Baudrier; Loïc Mazo

doi:10.1007/s10851-018-0846-5

Journal of Mathematical Imaging and Vision

February 2019, Volume 61, Issue 2, pp 224–236 | Cite as

Combinatorics of the Gauss Digitization Under Translation in 2D

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Étienne Baudrier
Loïc Mazo

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First Online: 08 September 2018

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Abstract

The action of a translation on a continuous object before its digitization generates several digital objects. This paper focuses on the combinatorics of the generated digital objects up to integer translations. In the general case, a worst-case upper bound is exhibited and proved to be reached on an example. Another upper bound is also proposed by making a link between the number of the digital objects and the boundary curve, through its self-intersections on the torus. An upper bound, quadratic in digital perimeter, is then derived in the convex case and eventually an asymptotic upper bound, quadratic in the grid resolution, is exhibited in the polygonal case. A few significant examples finish the paper.

Keywords

Discrete geometry Gauss digitization Translation Combinatorics

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Supplementary material

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Convex Sets

The proof of Lemma 5 relies on the two following lemmas about convex sets that seem obvious at the first glance. Nevertheless, since we did not find any result related to these lemmas in the literature, we provide our own justifications of the two statements.

Lemma 3

(Chords of convex sets) Let [a, b] be a chord of the boundary \(\varGamma \) of a closed convex set S. If \([a,b]\nsubseteq \varGamma \), then \((ab)\cap S=[a,b]\) and \((ab)\cap \varGamma =\{a,b\}\).

Proof

Since \([a,b]\nsubseteq \varGamma \), the line (ab) does not support S at any point (the notion of supporting line of a convex set is exposed for instance in [2]). So, there exists two supporting lines of S at a and b that cross the line (ab). Then, \((ab)\cap S\) is included in [a, b]. Let \(c\in [a,b]\cap \varGamma \), \(c\ne a\). Applying the first part of the proof to the chord [a, c], we derive that \((ac)\cap S\subseteq [a,c]\) and, since \(b\in (ac)\cap S\), we conclude that \(b=c\). \(\square \)

Lemma 4

(Cuts of convex sets) Let a, b be two points of the boundary \(\varGamma \) of a closed convex set S. If \([a,b]\nsubseteq \varGamma \), then the open curve segments of \(\varGamma \), \(\sigma _1\), \(\sigma _2\), whose extremities are a and b are included in distinct open half-planes bounded by the line (ab).

Proof

Let \(H^-\) and \(H^+\) be the two open half-planes bounded by (ab). Since \([a,b]\nsubseteq \varGamma \), from Lemma 3, \((ab)\cap \varGamma =\{a,b\}\). Thereby, by connectivity, either \(\sigma _1\subset H^-\) or \(\sigma _1\subset H^+\) and \(\sigma _2\subset H^-\) or \(\sigma _2\subset H^+\). Suppose for instance that \(\sigma _1\subset H^-\) and \(\sigma _2\subset H^-\). Then, S, which is the connected subset of \(\mathbb R^2\) bounded by \(\sigma _1\cup \sigma _2\cup \{a,b\}\) is included in \(H^-\cup (ab)\) and, since \((ab)\cap S=[a,b]\) from Lemma 3, \([a,b]\subset \varGamma \): contradiction. \(\square \)

Lemma 5

(Intersection of two segments of a convex curve) Let \(\varGamma \) be a Jordan curve whose interior is convex. Let \(\varGamma _1\) and \(\varGamma _2\) two disjoint closed segments of the curve \(\varGamma \) and \(\tau \) a vector of \(\mathbb R^2\). Then the intersection of \(\varGamma _1 \) and \(\tau +\varGamma _2\) is composed of none, one, two points or a line segment.

Proof

Let p, q be two distinct points in \(\varGamma _1 \cap (\tau +\varGamma _2)\) if such a pair exists. We denote by \( \varSigma _1 \) the open segment of \( \varGamma _1 \) between p and q. Alike, \( \varSigma _2 \) is the open segment of \( \varGamma _2 \) between \(-\tau + p \) and \(-\tau + q \). We set \(\overline{\varSigma }_1=\varSigma _1\cup \{p,q\}\) and \(\overline{\varSigma }_2=\varSigma _2\cup \{p,q\}\). Firstly, we prove that \(\varSigma _1\cup (\tau +\varSigma _2)\) is a straight line segment whenever it contains more than two points. First case: \(\varSigma _1\cup (\tau +\varSigma _2)\subseteq (pq)\). Then, since \(\overline{\varSigma }_1\) and \(\overline{\varSigma }_2\) are connected and \(\varGamma \) is simple, \(\overline{\varSigma }_1=\overline{\varSigma }_2=[p,q]\). Second case: \(\exists x\in \big (\varSigma _1\cup (\tau +\varSigma _2)\big ){\setminus }(pq)\). For instance, we assume \(x\in \varSigma _1{\setminus }(pq)\). By Lemma 4, \(\varSigma _1=\varGamma \cap H_1\), where \(H_1\) is the open half-plane bounded by the line (pq) and containing x, and \(-\tau +p\) is in \(\mathbb R^2{\setminus } H_1\). Then, it can easily be seen that \(-\tau +H_1\) is the open half-plane bounded by the line joining \(-\tau +p\) and \(-\tau +q\) and including p. Thanks to Lemma 4, we derive that \(\varSigma _2\) does not intersect \(-\tau +H_1\). Thus, \(\tau +\varSigma _2\) does not intersect \(H_1\). In particular, \((\tau +\varSigma _2)\cap \varSigma _1=\emptyset \). This achieves the first part of the proof. Now, let r be a point in \(\varGamma _1 \cap (\tau +\varGamma _2)\) which is not in \(\varSigma _1\) (if such a point exists). For instance, p belongs to the segment of \(\varGamma _1\) between q and r. Then, the first part of the proof, applied to the points q and r, implies that \(\varGamma _1 \cap (\tau +\varGamma _2)\) includes the segments [q, r]. We straightforwardly concludes that either the intersection of \(\varGamma _1 \) and \(\tau +\varGamma _2\) is composed of at most two points or it is a line segment. \(\square \)

Examples and Counterexamples

Building Examples Without Proper Congruent Digitizations in the Image of the Dual

Let u and v be two vectors in \([0,1)^2\) such that the sets \(u+S\) and \(v+S\) have distinct but congruent digitizations. Then, there exists an integer vector w, \(w\ne 0\), such that \((u+S)\cap \mathbb Z^2=w+((v+S)\cap \mathbb Z^2) = (w+v+S)\cap \mathbb Z^2\). Let p be a point in the digitization core. Then, \(p\in (u+S)\cap \mathbb Z^2\) and \(p\in (v+S)\cap \mathbb Z^2\). Therefore, \(w+p\in (u+S)\cap \mathbb Z^2\) and \(-w+p\in (v+S)\cap \mathbb Z^2\), which can be rewritten as \(p\in ((-w+u)+S)\cap \mathbb Z^2\) and \(p\in ((w+v)+S)\cap \mathbb Z^2\). Then, at least one of the vectors \(w+v\) or \(-w+u\) has one of its coordinates which is negative. We derive that if there exists a point in the digitization core which is maximal in S for both coordinates then there is no proper congruent digitizations in the dual. An example with such a maximal point in the digitization core is provided in Fig. 10.

Building Toric Partitions in One-to-one Correspondence with the Power Set of the Toggling Boundary

In this section, we exhibit a way to modify the boundary of the set S in order to ensure that any subset of the toggling boundary is represented in the dual. To do so, we move along \(\mathcal B\), ordered in the same way as in Definition 1. Then, a new boundary is built thanks to the approximations of the Hilbert filling curve: the segment of \(\varGamma \) intersecting the n-th cell of \(\mathcal B\) is replaced by a n-th approximation of the Hilbert filling curve \(H_n\) (extended at its extremities to ensure the continuity of the boundary): see Fig. 10. We consider the family of binary partitions \(\mathcal {P}_n\) of the unit square that comes with the curves \(H_n\). We claim that each curve \(H_{n+1}\) crosses each cell of the partition \(\bigwedge _{i=1}^n\mathcal P_i\) so that the size of the final torus partition is \(2^N\) where N is the cardinal of the toggling boundary. To justify our claim, we divide the unit square in a family of \(2^n\times 2^n\) small squares \((K_{i,j}^n)_{1\le i,j\le 2^n}\) (\(n\ge 0\)) whose sizes are \(\frac{1}{2^n}\times \frac{1}{2^n}\). It can be seen that on the one hand, the Hilbert curve \(H_n\) passes through the center of each of the squares \(K_{i,j}^n\) and, on the other hand, does not intersect any of the interior of the squares \(K_{i,j}^{n+1}\) (\(H_0\) is just the center of the unit square). Thereby, the partition \(\bigwedge _{i=1}^n\mathcal {P}_i\) is coarser than the partition \(\{K_{i,j}^{n+1}\mid {1\le i,j<n}\}\) (the boundaries of the squares \(K_{i,j}^k\) are assigned to the cells so as to coincide with \(\mathcal {P}_n\)). Since \(H_{n+1}\) passes through the center of each of the squares \(K_{i,j}^{n+1}\), it passes in each cell of \(\bigwedge _{i=1}^n\mathcal {P}_i\) which gives the claim.

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Étienne Baudrier
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Loïc Mazo
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1.ICube, University of Strasbourg, CNRSIllkirchFrance

Combinatorics of the Gauss Digitization Under Translation in 2D

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