Efficiently Testing Digital Convexity and Recognizing Digital Convex Polygons

Abstract

A set $S \subset \mathbb {Z}^2$ of integer points is digital convex if ${{\,\mathrm{conv}\,}}(S) \cap \mathbb {Z}^2 = S$, where ${{\,\mathrm{conv}\,}}(S)$ denotes the convex hull of S. In this paper, we consider the following two problems. The first one is to test whether a given set S of n lattice points is digital convex. If the answer to the first problem is positive, then the second problem is to find a polygon $P\subset \mathbb {Z}^2$ with minimum number of edges and whose intersection with the lattice $P\cap \mathbb {Z}^2$ is exactly S. We provide linear-time algorithms for these two problems. The algorithm is based on the well-known quickhull algorithm. The time to solve both problems is $O(n + h \log r) = O(n + n^{1/3} \log r)$, where $h = \min (|{{\,\mathrm{conv}\,}}(S)|, n^{1/3})$ and r is the diameter of S.

1 Introduction

Digital geometry is the field of mathematics that studies the geometry of points with integer coordinates, also known as lattice points [1]. Although the subsets of $\mathbb {Z}^d$ are not convex in the usual meaning of the term, a simple notion of convexity is induced by the convexity of $\mathbb {R}^d$ [2]. A set of lattice points $S \subset \mathbb {Z}^d$ is digital convex if ${{\,\mathrm{conv}\,}}(S) \cap \mathbb {Z}^d = S$, where ${{\,\mathrm{conv}\,}}(S)$ denotes the convex hull of S in $\mathbb {R}^d$. In other words, S is digital convex if it is the intersection of a convex subset of $\mathbb {R}^d$ with the lattice $\mathbb {Z}^d$ (Fig. 1). Digital convex lattice sets are then directly related to the lattice polytopes investigated in geometry of numbers since the works of Minkowski [3]. Digital convexity is preserved by homeomorphisms of $\mathbb {Z}^d$.

Let us remark that a digital convex lattice set S is not necessarily connected while the convex sets of $\mathbb {R}^d$ are arc-connected or simply connected. In $\mathbb {Z}^2$ and $\mathbb {Z}^3$, the lack of connectivity has led to the introduction of some alternative definitions of digital convexity that we will not consider [4,5,6,7,8].

Herein, we consider the following two problems in the plane.

1.1 Testing Convexity

The first problem is to determine whether a given finite lattice set S is convex.

Problem TestConvexity

Input: Set $S \subset \mathbb {Z}^2$ of n lattice points given by their coordinates.

Output: Determine whether S is digital convex.

The input of TestConvexity is an unstructured finite lattice set (without repeating elements). Related work considered more structured data, in which S is assumed to be connected. The contour of a connected set S of lattice points is the ordered list of the points of S having a grid neighbor (a lattice point whose Chebyshev distance to the point is one) not belonging to S. When S is connected, it is possible to represent S by its contour, either directly as in [9] or encoded by its Freeman chains code [10]. The algorithms presented in [9, 10] test digital convexity in linear time on the respective input representations.

Our work, however, does not make any assumption on S being connected, or any particular ordering of the input. In this setting, a naive approach to test the digital convexity is:

• 1. Compute the convex hull ${{\,\mathrm{conv}\,}}(S)$ of the n lattice points of S.

• 2. Compute the number $n'$ of lattice points inside the convex hull of S.

• 3. If $n=n'$, then S is convex. Otherwise, it is not.

Step 1 consists in computing the convex hull of n points. The field of computational geometry provides a plethora of algorithms for computing the convex hull of a finite set $S\subset \mathbb {R}^2$ of n points [11]. The fastest algorithms take $O(n \log n)$ time [12], which matches the lower bound in the algebraic decision tree model of computation [13]. If we also take into consideration the output size h, i.e., the number of vertices of the convex hull, then the fastest algorithms take $O(n \log h)$ time [14, 15].

Fig. 1

Digital convexity. The first set is digital convex, while the second set is not because of the red lattice points that are inside the convex hull of the set but not in the set itself (Color figure online)

Step 2 consists in computing the number of lattice points inside a convex polygon (represented by its vertices), which is a well studied problem. In dimension 2, it can be solved using Pick’s formula [16]. In higher dimension, the question has been widely investigated in the framework of the geometry of numbers, from Ehrhart theory [17] to Barvinok’s algorithm [18]. The currently best-known algorithms have a complexity of $O(n^{O(d)})$ for fixed dimension d [19]. Overall, the time complexity of this naive approach is at least that of the computation of the convex hull.

1.2 Digital Convex Polygon Minimization

In the case where the set S is convex, the second problem is to determine a convex polygon P having as few edges as possible and whose intersection $P\cap \mathbb {Z}^d$ with the lattice is exactly S (Fig. 2).

Problem Minimization

Input: Set $S \subset \mathbb {Z}^2$ of n lattice points given by their coordinates.

Output: Find a convex polygon P with minimum number q of edges verifying $P\cap \mathbb {Z}^2 =S$.

Fig. 2

Recognition. The input of the Minimization problem is a finite lattice set. The question is to find a convex polygon P with the smallest number of edges $P\cap \mathbb {Z}^2=S$ and whose intersection with $\mathbb {Z}^2$ is exactly S

The problem Minimization has been mentioned in a survey of open questions in Digital geometry [20]. Contrary to what is claimed, a minimal decomposition of the boundary of the set in digital straight segments (Min-DSS) as in [21] does not yield a solution. We disprove the reduction of Minimization to Min-DSS with the counter-example provided in Fig. 3. A weaker form of the problem, assuming 8-connectivity of the input set, has been studied in [22].

Fig. 3

Min-DSS fails to solve Minimization. According to [20], the problem Minimization can be solved by a technique decomposing the contour of a 4-connected shape in a minimal number of digital segments. The above example disproves it. The min-DSS decomposition of the boundary of the lattice set is done with only 3 digital segments while the set cannot be separated from its complement in $\mathbb {Z}^2$ by a triangle. It can be seen by considering the four exterior colored points in c. No pair of colored points can be separated from the lattice set by a line, which proves that the problem Minimization admits only a solution with at least $q=4$ edges and not 3 as in Min-DSS (Color figure online)

The problem Minimization is a fundamental problem in geometry, related for instance to combinatorial optimization. In this field, solutions are often characterized by an exponential number of linear constraints, and the reduction of the number of linear inequalities characterizing them is a major concern. It is related to the question that we address but in our framework the set of integer points is explicitly given through the list of the coordinates of the points. Even with the assumption that the lattice set is given, the state of the art about Minimization is restricted to a few results. The problem Minimization is decidable in dimensions $d=2$ and 3. In arbitrary dimensions, the problem is only known to be decidable if S is a non-hollow convex polytope (non-hollow means that there is some lattice points in the interior of its convex hull) [23,24,25]. These questions of decidability have been investigated without focus on the efficiency of the algorithms. Providing algorithms of low complexity for solving Minimization remained a fully open question, which we solve in this paper in dimension 2.

1.3 Our Results

In Sect. 2, we consider the problem of testing the digital convexity of a given lattice set S. We recall the linear time solution already presented in the conference version [26]. Our main result is an algorithm to solve TestConvexity in $O(n + h \log r) = O(n + n^{1/3} \log r)$ time, where h is the number of edges of the convex hull and r is the diameter of S. Furthermore, when the set S is digital convex, the algorithm returns the convex hull of S.

In Sect. 3, we consider the problem Minimization. We present the first linear-time algorithm to recognize a digital convex polygon. This algorithm uses the convex hull of S computed in TestConvexity for digital convex sets and then solves Minimization in $O(h \log r)$ time.

For a given number of edges h, the number of points n and the diameter r are not bounded (consider for instance large triangles). Similarly, for a given n, the diameter r is not bounded (consider the pair of points (0, 0) and (1, r)), however the number of vertices or edges h is bounded by $O(n^{1/3})$ [27]. At last, given the diameter r, the number of points n is clearly at most $O(r^2)$ and h is at most $O(r^{2/3})$ [28]. Expressed only with n and r, our algorithm for solving TestConvexity takes $O(n + n^{1/3} \log r)$ time while a post-processing step in $O(n^{1/3} \log r)$ time allows us to solve Minimization.

2 Digital Convexity

The purpose of this section is to provide an algorithm to test the convexity of a finite lattice $S \subset \mathbb {Z}^2$ in linear time in n. To achieve this goal, we first show that the convex hull of a digital convex set S can be computed in linear time using the well-known quickhull algorithm [29].

Quickhull is one of the many early algorithms to compute the convex hull in dimension 2. Its worst-case time is $O(n^2)$. However, for some inputs and variations of the algorithm, the average time complexity is reduced to $O(n \log n)$ or O(n) [11, 30].

The quickhull algorithm starts by initializing a convex polygon in the following manner. First it computes the top-most and bottom-most points of the set. Let $\ell$ be the line defined by these two points. Then, the algorithm computes the farthest point from $\ell$, on each side of $\ell$. The (at most) four points we computed describe a convex polygon that we call a partial hull, which is a subset of the vertices of the convex hull of S. All points contained in the interior of the partial hull are discarded from S. Furthermore, horizontal lines and lines parallel to the top-most to bottom-most line passing through these points define an outlying bounding box containing the convex hull (Fig. 4).

Fig. 4

Quickhull initialization. Points inside the partial hull (light brown) are discarded. The remaining points are potentially part of the hull (Color figure online)

After the initial step, the algorithm adds vertices one by one to the partial hull until it obtains the entire convex hull. For each edge of the partial hull, we apply the following steps. Let v denote the edge’s outwards normal vector. The algorithm searches for the extreme point in direction v. If this point distance from the edge is 0, then the edge is part of the convex hull. Otherwise, we add to the convex hull the farthest point found, discarding the points that are inside the new partial hull. Throughout this paper, we call a step of the quickhull algorithm the computation of the farthest point of every edge for a given partial hull. When adding new vertices to the partial hull, the region inside the partial hull expands. Points inside that expansion are discarded by quickhull and herein we call this region discarded region. The points not belonging to the partial hull are preserved, and are the elements of the preserved region (Fig. 5).

We show that quickhull steps take linear time for any digital convex lattice set and that, in this case, at each step half of the remaining input points are discarded. Therefore, the total running time remains linear, as in standard decimation algorithms (see for example [31]). In Sect. 2.2, we explain how to use this algorithm to test the digital convexity of any lattice set in linear time in n.

Theorem 1

If the input is a digital convex set of n points, then quickhull has O(n) time and space complexities.

2.1 Proof of Theorem 1

We prove Theorem 1 as follows.

Proof

During quickhull algorithm, we discard from S the points that become useless for the next computation and add some of them as vertices of the partial hull. The algorithm discards all the points that are in the interior or on the boundary of the current partial hull. The theorem is a consequence of the following two propositions, which we prove next: (i) At each step, the running time is linear in the number of points remaining in S. (ii) At least half of the remaining points are discarded at each iteration. We start by proving proposition (ii).

Consider one step of the algorithm. Let ab be the edge defining the step. When a was added to the hull, it was as the farthest point in a given direction. Hence, there is no point beyond the line orthogonal to this direction going through a (Fig. 5b). The same holds for b. Let c be the intersection point of these two lines going through a and b. We know that any remaining point of S is in the interior of such a triangle $\triangle abc$ to which it is allocated. We proceed with the remaining points of S in $\triangle abc$ as follows. We are looking for the point that is the farthest from the supporting line of ab in the triangle $\triangle abc$ (Fig. 6). Three cases might occur. If the triangle $\triangle abc$ does not contain any remaining point, then ab is an edge of the partial hull and we stop the computation for this edge in the following steps. If there is a unique remaining point of the triangle $\triangle abc$ which is the farthest from the line ab, then we denote it d. If there are multiple points which are farthest from ab in the interior of the triangle $\triangle abc$, we denote the two extreme points of S on this segment d and $d'$.

Fig. 5

Quickhull regions. The preserved region (region in which we look for the next vertex to be added to the partial hull) is a triangle. This stays true when adding new vertices to the hull (as shown here in the bottom right corner). The partial hull (whose interior is shown in light brown) grows at each vertex insertion to the partial hull. The points in or on the boundary of the new region of the partial hull are discarded (Color figure online)

Let us consider the case where the point d is the unique farthest point from the line ab. Let e and f be the intersections between the line parallel to ab going through d, and respectively, ac and bc. The point d is the unique remaining point in the triangle $\triangle cef$. Adding d to the partial hull creates two other edges to be further processed: One is ad and the other is bd. Then, we insert the vertex d in the partial hull and remove from S all the points which are neither in the interior of the triangles $\triangle ade$ nor $\triangle bdf$. The points of S in the interior of the triangle $\triangle abc$ that we do not discard are allocated either to $\triangle ade$ or to $\triangle bdf$ according to their positions.

We denote, respectively, $c_1$ and $c_2$ the midpoints of ad and bd. All the lattice points in the interior of the triangles $\triangle ade$ and $\triangle dbf$ have different symmetric lattice points toward $c_1$ and $c_2$ in the interior the triangle $\triangle ade$. Since S is digital convex, those lattice points are in S, they also are discarded due to their positions. (Fig. 6a). In other words, at this step, for each remaining points of S, one point of S is discarded. It proves (ii). This proposition also holds in the case where there are two extreme points d and $d'$ from S on the line ef. In this case, we insert the two vertices d and $d'$ in the partial hull. We discard from S all the points of the triangle $\triangle abc$ which are not in the interiors of the triangles $\triangle ade$ and $\triangle d'bf$. As previously, any of the remaining points has a different symmetric point which is discarded (Fig. 6b). It proves (ii) in this case. In both cases, our initial assumption is preserved: All the remaining points are in the interior of the triangle to which they are allocated. At last, we can easily provide an initialization of the partial hull and of the set of remaining points satisfying this condition.

Fig. 6

Symmetrical regions. At each step, we discard from S all the points of the triangle $\triangle abc$ which are not in the interior of $\triangle ade$ or of $\triangle dbf$ ($\triangle d'bf$ in b). By considering the symmetries through $c_1$ and $c_2$, any of these remaining points has a symmetric lattice point in the interior of $\triangle abd$ which is discarded

For proving (i), the computation of the farthest point from the line ab among the remaining points of S in the triangle $\triangle abc$ takes linear time. For all points in the triangle, we test if they are in the interior of either the triangles $\triangle ade$ or $\triangle dbf$ (or $\triangle d'bf$ in the second case). We allocate them to their containing triangle or discard them. The operation takes a constant time per point. In the second case, where we have two extreme points d and $d'$, these two points are also computed in linear time; This proves (i). Consequently, the number of operations is proportional to $n\sum \nolimits _{i=0}^\infty (\frac{1}{2})^i = 2n$ and quickhull takes linear time for digital convex sets. $\square$

2.2 Testing Digital Convexity

By running quickhull on any given set S, and stopping the computation if any step of the algorithm discards less than half of the remaining points, we ensure both that the running time is linear, and that if S is digital convex, quickhull finishes and returns the convex hull of S. If the computation finishes for S, we still need to test its digital convexity. To do so, we use the previously computed convex hull and compute $|{{\,\mathrm{conv}\,}}(S) \cap \mathbb {Z}^2|$ using Pick’s formula [16]. The set S is digital convex if $|{{\,\mathrm{conv}\,}}(S) \cap \mathbb {Z}^2| = |S|$. Hence, the resulting Algorithm 1.

Theorem 2

Algorithm 1 tests digital convexity of S in $O(n + h \log r) = O(n + n^{1/3} \log r)$ time, where $h = \min (|{{\,\mathrm{conv}\,}}(S)|, n^{1/3})$ and r is the diameter of S.

Proof

As Algorithm 1 runs quickhull, but stops as soon as less than half the remaining points have been removed, the running time of the quickhull part is bounded by the series $n\sum \nolimits _{i=0}^\infty (\frac{1}{2})^i = 2n$, and is hence linear. Thanks to Theorem 1 we know that the computation of quickhull will be completed for any digital convex set. Computing $|{{\,\mathrm{conv}\,}}(S) \cap \mathbb {Z}^2|$ using Pick’s formula requires the computation of both the area of ${{\,\mathrm{conv}\,}}(S)$ in O(h) and the number of lattice points on its boundary, which requires the computation of a greatest common divisor. Hence, this takes $O(h \log r)$ time where h is the number of edges of ${{\,\mathrm{conv}\,}}(S)$ and r is the diameter of S. As S is digital convex if and only if $|S| = |{{\,\mathrm{conv}\,}}(S) \cap \mathbb {Z}^2|$, Algorithm 1 effectively tests digital convexity in $O(n + h \log r)$ time. $\square$

3 Minimum Digital Convex Polygon

In this section, we consider a fundamental question of pattern recognition: the recognition of digital convex polygons, namely problem Minimization from Sect. 1.2. In this problem, we are given a set $S \subset \mathbb {Z}^2$ of n points and the goal is to find a convex polygon P with minimum number of edges such that $P \cap \mathbb {Z}^2 = S$. Notice that the vertices of P are not necessarily lattice points. We prove the following theorem:

Theorem 3

Given a finite lattice set $S\subset \mathbb {Z}^2$ of n points, the Algorithm 3 solves the problem Minimization. The running time is $O(n + h \log r) = O(n + n^{1/3} \log r)$, where $h = \min (|{{\,\mathrm{conv}\,}}(S)|, n^{1/3})$ and r is the diameter of S.

3.1 Strategy

The problem Minimization can be rephrased as the following polygonal separation problem with the set $IN =S$ and its complement $OUT =\mathbb {Z}^2 \setminus S$.

Problem: Polygonal Separation

Input: A set $IN \subset \mathbb {Z}^2$ of inliers and a set $OUT \subset \mathbb {Z}^2$ of outliers.

Output: A convex polygon $P \subset \mathbb {R}^2$ with as few edges as possible and such that all points of $IN$ and none of $OUT$ are inside P.

Polygonal separability has been widely investigated in the literature. An optimal algorithm for PolygonalSeparation that takes $O((| IN |+| OUT |) \log (| IN |+| OUT |))$ time is presented in [32]. However, it cannot be applied to Minimization since the set of outliers $OUT =\mathbb {Z}^2 \setminus S$ is not finite.

The strategy to solve Minimization is as follows: We start by testing the digital convexity of S in linear time using Theorem 2. If S is not digital convex, then there is no solution. Otherwise, the algorithm quickhull computes the convex hull of S in linear time and we can proceed to the second step.

The second step of the algorithm is to reduce the set of outliers $OUT =\mathbb {Z}^2 \setminus S$ to a finite subset $OUT ' \subseteq OUT$ of O(n) points. In fact, we do not explicitly compute $OUT '$. Instead, we compute an implicit description of $OUT '$ of size O(h) in $O(h \log r)$ time, where h is the number of edges of ${{\,\mathrm{conv}\,}}(S)$.

The third step is to separate $OUT '$ from S using the smallest number of edges. We could use the polygonal separability algorithm from [32], but that would lead to a running time of $O(n \log r + n \log n) = O(n \log r)$. Instead, we provide an algorithm that takes benefit of the lattice structure to achieve a running time of $O(h \log r)$ after the convex hull computation and digital convexity tests of the first step, that takes $O(n + h \log r)$ time.

As the first step, i.e., testing the digital convexity, is already addressed in the previous section, we present the second and third steps of the algorithm in the two following sections.

3.2 Reduction

In this section, we assume that the set S is digital convex and show how to reduce the set of outliers $OUT =\mathbb {Z}^2 \setminus S$ to a finite set of O(n) points. To do this, we use the notion of jewels introduced in [23, 33] for testing digital circularity and recognizing digital polyhedra. We say that a point $p \in \mathbb {Z}^2 \setminus S$ is a jewel of S if ${{\,\mathrm{conv}\,}}(S \cup p) \cap \mathbb {Z}^2 = S \cup p$ (Fig. 7). The set of all the jewels of S is denoted ${{\,\mathrm{Jewel}\,}}(S)$ and it has the property that a convex set separates S from $\mathbb {Z}^2 \setminus S$ if and only if it separates separates S from ${{\,\mathrm{Jewel}\,}}(S)$ [23]. Hence, the infinite set of the outliers of our problem of separability can be reduced from $OUT =\mathbb {Z}^2 \setminus S$ to $OUT '={{\,\mathrm{Jewel}\,}}(S)$.

Fig. 7

Jewel’s hull. In black, the set S, its convex hull is in dark red. The point a is not a jewel because of the red point, any convex polygon that includes both S and a also includes the red point. The point b is a jewel because its union $S\cap \{ b\}$ with S is still convex. In other words, the convex hull of the union $S\cap \{ b\}$ does not contain any other lattice points (Color figure online)

The number of jewels is infinite if and only if S is the intersection of a line segment and $\mathbb {Z}^2$ [23]. In this case, it is easy to see that the set S forms a digital triangle. A simple way to establish bounds on the number of jewels has been discovered by French high school students during the national contest TFJM2017. They presented the following structure of the set of jewels: The jewels of the lattice set S are the lattice points that lie on the edges of a polygon J surrounding the convex hull of S. This surrounding polygon $J \supset {{\,\mathrm{conv}\,}}(S)$ is the arithmetic dilation of ${{\,\mathrm{conv}\,}}(S)$ obtained by moving the support lines of the edges of the ${{\,\mathrm{conv}\,}}(S)$ to the next Diophantine lines toward the exterior (Fig. 8). We define J as the jewel hull of S (Fig. 8) and define it more formally as follows.

Fig. 8

Jewels.In black, the set S, its convex hull is in dark red. The halfplanes $H'_i$ are delimited by the dashed lines, and form the jewel hull that surrounds the convex hull of S. The jewel hull has three properties: Its edges are parallel to the ones of the convex hull of S, there are no point between the convex hull and the jewel hull and all the jewels (drawn in red are) on its boundary (Color figure online)

Given S, let $E =\{e_1, e_2,\ldots e_h\}$ be the edges of ${{\,\mathrm{conv}\,}}(S)$. For each i, let $HP_i: a_ix + b_iy + c_i \le 0$ ($a_i$ and $b_i$ co-prime integers) be the closed supporting halfplane associated with $e_i$ such that $S \subset HP_i$. Notice that ${{\,\mathrm{conv}\,}}(S) = \bigcap _i HP_i$. Consider the open halfplanes $HP'_i: a_ix + b_iy + c_i < 1$. Notice that there is no integer point in $HP'_i \setminus HP_i$. The jewel hull of S is the closure of the intersection of the half-planes $HP'_i$ (Fig. 8).

The jewel hull J of S has three main properties. (i) By construction, its edges are parallel to the edges of ${{\,\mathrm{conv}\,}}(S)$. (ii) It is easy to prove that there is no integer point between ${{\,\mathrm{conv}\,}}(S)$ and the jewel hull J. Finally, (iii) the jewels of S are a subset of J. This last property is in fact a corollary of the first Lemma of [34] which is reformulated in the next lemma.

Lemma 1

For any three lattice points $p_1,p_2,p_3$ such that $p_1,p_2$ lie on the line $ax + by + c = 0$ (coefficients a and b are coprime) and $p_3$ does not, we have that the triangle $p_1p_2p_3$ either contains a lattice point on the line $ax + by + c + 1 = 0$ or on the line $ax + by + c - 1 = 0$.

Proof

Up to a lattice preserving affine isomorphism, we can assume $p_1=(0,0)$ and $p_2=(0,u)$ while the images of the two lines are $x = -1$ and $x = 1$. We assume $p_3$ lies on the right of $p_1p_2$ (the other case is identical by symmetry). Hence, there exists three integers u, v, w with $u,v>0$ such that $p_1=(0,0)$, $p_2=(0,u)$, and $p_3=(v,w)$ and we want to prove that the triangle $p_1p_2p_3$ contains an integer point on the line $x=1$. The lower and upper points of the triangle in the line $x=1$ are the two intersection points of $x=1$ and each of the two segments $p_1p_3$ and $p_2p_3$. Their coordinates are, respectively, $(1,\frac{w}{v})$ and $(1,u+\frac{w-u}{v})$. Then, the intersection of the line $x=1$ and the triangle $p_1p_2p_3$ contains an integer point if and only if the interval $[\frac{w}{v},\frac{uv+w-u}{v}]$ contains an integer namely if the interval $[w , w+u(v-1)]$ contains a multiple of v, which is trivially true since there is necessarily a multiple of v in any interval $[w,w+v[$ and then in $[w,w+v-1]\subset [w , w+u(v-1)]$ as $u \ge 1$. $\square$

Fig. 9

Turn algorithm. We start from a chosen starting jewel $p_{i_0}$ and search for its last successor $p_{j_0}$ that can be separated from S simultaneously with $p_{i_0}$ by a single line. We then take the successor of $p_{j_0}$ as new starting jewel $p_{i_1}$ and search for the last successor $p_{j_1}$ of $p_{i_1}$ that can be separated with $p_{i_1}$... We repeat the process until reaching the predecessor of $p_{i_0}$

The area of the jewel hull of S is finite unless all the points of S are colinear. This case is easy to detect, and it is easy to see that in this case there exists a triangle with vertices in $\mathbb {R}^2$ that separates S from $\mathbb {Z}^2 \setminus S$.

The jewel hull consists of the intersection of a set of h halfplanes. Computing the vertices of the intersection of halfplanes is the dual [11, Chapter 8] of the computation of the convex hull of a given points set. In the general case, computing the intersection of h halfplanes takes $O(h \log h)$ time [11, Chapter 4]. However, since we already have the h halfplanes sorted by slope, we can use Graham Scan [11, Chapter 1] to compute the jewel hull in O(h) time. Notice that not all h halfplanes appear on the boundary of the jewel hull, which is the dual of the fact that some points may be in the interior of the convex hull.

3.3 Jewel Separation

The jewel separation is the final step to solve the

Minimization problem. The jewel hull J has been computed and the problem is the polygonal separation of $IN =S$ and the jewel set $OUT '=Jewel (S)$. The previous step does not provide the set of jewels but the ordered list of edges of the jewel hull J as a sequence of linear equalities $\ell _i: a_i x + b_i y + c_i =1$ with coprime integers $a_i$ and $b_i$. An initial lattice point $d_i$ of each given Diophantine straight lines $\ell _i$ can be computed with the extended Euclid algorithm in $O ( \log r )$ time. We can go from this first point to the other integer points of the line $\ell _i$ through translations of vectors $k\overrightarrow{(-b_i , a_i)}$ where $k\in \mathbb {Z}$. Nevertheless, J is a rational polytope. Its vertices are the intersection point of consecutive Diophantine lines $\ell _i$ but they are not necessarily integer points. It is even possible that some edges of the jewel hull do not contain any integer point. By computing the vertices of each edge $e_i$ we can count all the jewels on $\ell _i$ and obtain a generating formula for them in O(1) time and space for each edge. The jewels on $\ell _i$ are: $\bigcup _k d_i + k(-b_i , a_i)$. The computation of an integer point $d_i$ per line $\ell _i$ for each one of our at most h Diophantine lines takes $O(h \log r )$. The computation of the vertices of J takes O(h) time, and hence the computation of the formulas generating the jewels takes $O(h \log r)$ time and O(h) space.

The jewels are determined in counterclockwise order according to their order of appearance in the jewel hull. Their cyclic index i goes from 0 to $|jewel (S)|-1$. Furthermore, any pair of indices i, j with $i<j$ defines two intervals of indices, the interval $I_{i\rightarrow j}$ containing the indices of the successors of i until j and the interval $I_{j\rightarrow i}$ containing the indices of the successors of j until i. We introduce now the precise meaning of separation. We say that a real line $\ell$separates some jewels from S if S lies entirely on one side of $\ell$ while the jewels lie strictly on the other side. The fact that all jewels lie on the boundary of a convex polygon leads to the following simple lemma:

Lemma 2

If $\ell$ is a line separating the jewels of indices i and j from S, then the line $\ell$ separates S from either the jewels with indices in $I_{i\rightarrow j}$ or the jewels with indices in $I_{j\rightarrow i}$.

A naive approach to solve the polygonal separation problem of the sorted set of jewels from S is the following: Choose a starting jewel of index $i_0$. Search for the index $j_0$ such that the jewels with indices in the interval $I_{i_0 \rightarrow j_0}$ can be separated from S and $|I_{i_0 \rightarrow j_0}|$ is maximized. The method used to compute $j_0$ in constant time using our representation of the jewels will be detailed later. We then define $i_1$ as the successor of $j_0$ and repeat the process: search for $j_1$ such that $I_{i_1 \rightarrow j_1}$ can be separated from S and the number of jewels in the interval is maximized. We repeat until we find an interval $I_{i_k \rightarrow j_k}$ which contains the predecessor of $i_0$. The number of lines of the solution is the number $k+1$ of intervals considered. This algorithm is illustrated Fig. 9. We call this greedy algorithm the turn routine since the strategy is to turn around the set S from a starting jewel $p_{i_0}$.

The difficulty of this approach is that different choices of the starting point $p_{i_0}$ may lead to different numbers of separating lines (actually, they may differ by at most 1 line). The strategy to find the minimum number of separating lines is to test several starting jewels. Dynamic programming approaches might be used to find an optimal solution as in [21], but in the framework of our minimization problem in the lattice, we are able to obtain a major simplification.

The strategy to simplify the problem is the following. There are two families of jewels: The ones which chosen as starting jewel in the turn routine provide a minimal number of lines, their indices are denoted $I_{ OPT }$, and the ones that provide a non optimal number of lines. Notice that if the index $i_0$ is in $I_{ OPT }$, then all the indices $i_k$ computed during the turn routine are also in $I_{ OPT }$ since it can be easily seen that they provide also optimal solutions. In the general case of polygonal separability, a large set of starting points has to be investigated until finding one leading to an optimal solution but in the framework of the separation of $IN =S$ and $OUT '={{\,\mathrm{Jewel}\,}}(S)$, we can provide a subset of at most 4 jewels containing at least one in $I_{ OPT }$. It means that testing these four jewels as starting points of the turn routine is enough to find the optimal solution. The properties of the set $I_{ OPT }$ are presented in the next two lemmas.

The first lemma states that there is no line that simultaneously separates two jewels of a line $\ell _i$ and two jewels of $\ell _{i+1}$.

Lemma 3

Let $\ell _1$ and $\ell _2$ be two jewel lines. (i) If $\ell _1 \cap \ell _2 \notin \mathbb {Z}^2$ then there is no line that separates two jewels of $\ell _1$ and two jewels of $\ell _2$. (ii) If $\ell _1 \cap \ell _2 \in \mathbb {Z}^2$ then there is no line that separates three jewels of $\ell _1$ and three jewels of $\ell _2$.

Proof

(i) Let order the jewels on $\ell _1$: $J_1 = \{p_{1-1},p_{1-2},\ldots \}$ according to their distance to $\ell _2$, and order the jewels on $\ell _2$: $J_2 = \{p_{2-1},p_{2-2},\ldots \}$ according to their distance to $\ell _1$. Assume that there is a line l such that l separates two jewels of $\ell _1$ and two jewels of $\ell _2$ from ${{\,\mathrm{conv}\,}}(S)$. Then, l separates $p_{1-1},p_{1-2},p_{2-1}$ and $p_{2-2}$ from ${{\,\mathrm{conv}\,}}(S)$. Hence, the triangle $\triangle p_{1-1}p_{1-2}p_{2-2}$ lies inside the jewel hull and outside of ${{\,\mathrm{conv}\,}}(S)$ (Fig. 10a). As the triangle $\triangle p_{1-1}p_{1-2}p_{2-1}$ is not degenerated we have $Area(\triangle p_{1-1}p_{1-2}p_{2-1}) \ge \frac{1}{2}$. Hence, the inequality $Area(\triangle p_{1-1}p_{1-2}p_{2-2}) > Area(\triangle p_{1-1}p_{1-2}p_{2-1})$ leads to $Area(\triangle p_{1-1}p_{1-2}p_{2-2}) > \frac{1}{2}$. Using Pick’s theorem we can conclude that $\triangle p_{1-1}p_{1-2}p_{2-2}$ contains at least four lattice points. However, since $p_{1-1}p_{1-2}$ are two consecutive lattice points of $\ell _1$, this means that there is a lattice point strictly inside the jewel hull and outside ${{\,\mathrm{conv}\,}}(S)$, which is impossible. Hence, l does not exist. The proof of (ii) is the same, we just have to consider $p_{1-0} = p_{2-0} = \ell _1 \cap \ell _2$. $\square$

Fig. 10

Jewel separation. a If a single line separates both $p_{1-2}$ and $p_{2-2}$, then the triangle $\triangle p_{1-1}p_{1-2}p_{2-2}$ is larger than $\triangle p_{1-1}p_{1-2}p_{2-1}$ and hence must contain a fourth lattice point, which is impossible. b No jewels lie between $e_{i+1}$ and $\ell _{i+1}$ hence it is impossible to separate simultaneously jewels from $\ell _i$ and jewels from $\ell _j$

We complete Lemma 3 with a lemma about the separation of jewels which are not in consecutive lines $\ell _i$ and $\ell _{i+1}$.

Lemma 4

If $\ell _i$ and $\ell _j$ are two non-consecutive jewels lines: $j \ge i+2$, then there is no line that separates any jewel that belongs only to $\ell _i$ and any jewel that belongs only to $\ell _j$.

Proof

Consider $\ell _{i+1}$ and its associated edge on ${{\,\mathrm{conv}\,}}(S)$: $e_{i+1}$. By construction, there is no lattice points between $\ell _{i+1}$ and $e_{i+1}$ (Fig. 10b). Assume that there is a line l that separates jewels of both $\ell _i$ and $\ell _j$ As all the jewels belonging only to $\ell _i$ and all the jewels belonging only to $\ell _j$ lies on the same side $s_j$ of $e_{i+1}$ as S, l has to be in $s_j$ to separate jewels of $\ell _i$, then has to leave $s_j$ in order to not intersect ${{\,\mathrm{conv}\,}}(S)$, and finally has to go back in $s_j$ to separate jewels of $\ell _{i+1}$. Hence, l intersects $e_{i+1}$ twice which is impossible. $\square$

We now explain how to use Lemmas 3 and 4 to determine at most four jewels such that at least one of them leads to an optimal solution with the turn routine. In other words, we provide four indices with the guarantee that at least one of them is in $I_{ OPT }$. For convenience, the successor of the index s is now simply denoted $s+1$ and so on with the successor of the successor denoted $s+2$. In the same manner, we also use $s-1,s-2,\ldots$ to denote the predecessors of s. When looking for a jewel in $I_{ OPT }$, several cases might occur:

• 1. The jewel hull J has an edge $e_i$ which does not contain any integer point. If we denote s the index of the first jewel after this edge, then $I_{ OPT }$ contains s. It is a corollary of Lemma 4. Considering an optimal solution, the vertex of index s cannot be included in the interval $I_{i_r\rightarrow j_r}$ containing $s-1$ because the interval would contain jewels of the lines $\ell _{i-1}$ and $\ell _{i+1}$ which is excluded by Lemma 4. Hence, the index s is a starting index, namely an index of the form $i_r$ of the considered optimal solution. As the indices $i_r$ of the intervals $I_{i_r\rightarrow j_r}$ computed from an optimal starting index $i_0$ are also optimal, s is included in $I_{ OPT }$.

• 2. The jewel hull J has an edge $e_i$ with only one jewel s, hence $I_{ OPT }$ contains either s or $s+1$. Considering an optimal solution, it follows from Lemma 4 that $s-1$ and $s+1$ cannot be in an interval of the form $I_{i_r\rightarrow j_r}$ since they are on distant lines $\ell _{i-1}$ and $\ell _{i+1}$. Hence, there exist either an index $i_r$ equal to s or to $s+1$. It proves that one of these two indices s or $s+1$ is in $I_{ OPT }$.

• 3. The jewel hull has an edge with only two jewels. Their indices are s and $s+1$. Considering an optimal solution, according to Lemma 4 the indices $s-1$ and $s+2$ cannot be in the same interval $I_{i_r\rightarrow j_r}$ because they belong to the distant lines $\ell _{i-1}$ and $\ell _{i+1}$. Hence, there is at least a beginning of interval in s, $s+1$ or $s+2$. One of these three indices s, $s+1$, $s+2$ is in $I_{ OPT }$

• 4. The edges of the jewel hull all contain at least three jewels. We choose any edge $e_i$ and denote s, $s+1$, $s+2$ the indices of its three firsts jewels. According to Lemma 3, the indices $s+2$ and $s-2$ cannot be in the same interval $I_{i_r\rightarrow j_r}$. Hence, there is at least a beginning of interval in $s-1$, s, $s+1$ or $s+2$. One of these four indices $s-1$, s, $s+1$, $s+2$ is in $I_{ OPT }$.

In any case, we can determine a set of at most four starting jewels with the guarantee that the turn algorithm provides an optimal solution for at least one of them. We now explain how, in the turn algorithm 2, for a given jewel $p_i$ we compute its last successor $p_j$ that can be separated alongside him with a single line. Let $p_i$ be on the jewel line $\ell _i$, and let $v_i$ be the end vertex of the edge of the convex hull parallel to $\ell _i$. Consider the line $p_i v_i$. S lies on one side of $p_i v_i$, all the jewels that lies strictly on the other side can be separated alongside $p_i$ (Fig. 9). It is clear that all jewels located on $\ell _i$ can be separated with $p_i$, and using Lemma 4 we know that the jewels located on $\ell _{i+2}$ cannot. Hence, all we have to do is determine the last jewel of $\ell _{i+1}$ that lies on the correct side of $p_i v_i$. This is easily done by computing the intersection point q of $p_i v_i$ and $\ell _{i+1}$ and expressing q as $d_{i+1} + \lambda (-b_{i+1} , a_{i+1})$ (We remind that the jewels on $\ell _{i+1}$ are expressed as: $\bigcup _k d_i + k(-b_i , a_i)$). From there a separating line can be computed by rotating slightly $p_i v_i$ around any points in between $p_i$ and $v_i$.

The time complexity of the turn algorithm 2 is hence $O(h) = O(n^{1/3})$. This follows from the fact that h is an upper bound to the number of edges of the solution of the Minimization problem and $h=O(n^{1/3})$. Starting from any jewel, the algorithm computes a polygon that has at most one edge more than the optimal solution and each edge is computed in O(1) time.

As the jewel hull is computed $O(h \log r)$ time, the set of O(1) starting jewels can be computed in constant time, and the turn algorithm 2 runs in O(h) time. The minimization algorithm, once provided with the convex hull of S runs in $O(h \log r)$ time, which proves Theorem 3.

4 Perspectives

We showed that the convex hull of a digital convex set in dimension 2 can be computed in linear time, and we can determine the minimum digital convex polygon in the same complexity. Can the convex hull of digital convex sets be computed in linear time in dimension 3, or more generally, what is the complexity of convex hull computation of a digital convex set in any fixed dimension? We note that the number of faces of any digital convex set in d dimensions is $O(V^{(d-1)/(d+1)})$, where V is the volume of the polytope [35, 36]. Therefore, the lower bound of $\Omega (n^{\left\lfloor (d-1)/2\right\rfloor })$ for the complexity of the convex hull of arbitrary polytopes does not hold for digital convex sets. The decidability of the polygon minimization problem has been proven in dimension 3 [25], but no polynomial-time algorithm have been presented yet. Even the decidability of the problem remains an open problem for dimensions higher than 3.