On the Structure of Higher Order Voronoi Cells

Transcription

1 On the Structure of Higher Order Voronoi Cells Juan Enrique Martínez-Legaz, Vera Roshchina, Maxim Todorov November 7, 018 arxiv: v1 [math.mg] 6 Nov 018 Abstract The classic Voronoi cells can be generalized to a higher-order version by considering the cells of points for which a given k-element subset of the set of sites consists of the k closest sites. We study the structure of the k-order Voronoi cells and illustrate our theoretical findings with a case study of two-dimensional higher-order Voronoi cells for four points. 1 Introduction The classic Voronoi cells partition the Euclidean space into polyhedral regions that consist of points nearest to one of the sites from a given finite set. We consider higher order or multipoint) Voronoi cells that correspond to the subsets of points nearest to k several sites see an illustration in Fig. 1). AE AC DE CE E A EF AB C D CD F B BD DF BC BF Figure 1: The classic and order two Voronoi diagrams on six sites shown in dashed lines and shaded regions, respectively). Notice that some two-point combinations generate empty cells. Departament d Economia i d Història Econòmica, Universitat Autònoma de Barcelona, and BGS- Math, Spain School of Mathematics and Statistics, UNSW Sydney, Australia Department of Physics and Mathematics, UDLAP, Puebla, Mexico. On leave from Institute of Mathematics and Informatics, BAS, Sofia, Bulgaria 1

2 To our best knowledge, the earliest mention of k-point Voronoi cells appears in [13], where a tessellation of the plane by such cells was called the Voronoi diagram of order k; that paper also provides bounds on the number of nonempty cells in a plane and complexity estimates for the construction of such diagrams; in [11] the complexity of constructing the higher-order diagrams for line segments was studied. The multipoint or k-order Voronoi diagrams discussed in this paper are one possible way to generalize the classic construction. Some notable generalizations are the cells of more general sets [5,11], which are particularly useful in computer graphics; the use of non- Euclidean metrics [1, 6, 9] and the abstract cells that are defined via manifold partitions of the space rather than distance relations [7]. Much of the recent work mentioned above is focussed on the algorthmic complexity of constructing planar Voronoi diagrams of various types. In this paper we rather focus on the structure of multipoint Voronoi cells, and in particulat obtain constructive characterizations of cells with nonempty interior, also of bounded and empty cells. We use these results for a case study of multipoint cells defined on at most four sites. We prove that perhaps counterintuitively some convex polygons, including triangles and cyclic quadrilaterals, can not be such cells, and provide explicit algorithms for the construction of sites for a given cell in other cases. Finally, we would like to mention a wealth of emerging application of higher order Voronoi cells, predominantly driven by the recent advancements in big data and mobile sensor technology. For instance, in [10] such cells are utilized in a numerical technique for smoothing point clouds from experimental data; in [1] k-order cells are used for detecting and rectifying coverage problems in wireless sensor networks; in [] the k-order diagrams are used to analyse coalitions in the US supreme court voting decisions. A well-known application of the higher order Voronoi cells is in a k-nearest neighbor problem in spatial networks [8], however, the practical implementations are limited due to the complexity of higher order diagrams and the lack of readily available software. Our work is organized as follows. In Section we study the structure of higher order Voronoi cells, and in particular prove the conditions for the cell to be bounded and nonempty. In Section 3 we study the special case of higher order cells on no more than four sites. We will refer to the higher order cells as multipoint cells, to highlight the discrete nature of our construction. High order Voronoi cells in R n Let T R n be finite. For a nonempty proper subset S T we define the multipoint Voronoi cell as the set of points that are not farther from each point of S than from each point of T \ S, { } V T S) := x R n : max distx, s) min distx, t), s S t T \S where distx, y) is the Euclidean distance function. When S is a singleton, i.e. S = {s}, the set V T S) is a classic Voronoi cell. We abuse the notation slightly and write V T s) := V T {s}). It is evident that V T S) = s S V {s} T \S) s). 1) It is not difficult to observe that each multipoint Voronoi cell is a convex polyhedron, i.e. the intersection of finitely many closed halfspaces, since each cell is defined by finitely

3 many linear inequalities. Explicitly, we have the following representation. Proposition 1. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. Then V T S) is the intersection of S T S ) closed halfspaces: { V T S) = x R n : t s, x 1 t s )}. ) s S t T \S Proof. Observe that, from the definition, { V T S) = x R n : max = s S = s S s S distx, s) min t T \S } distx, t) } { x R n : distx, s) min distx, t) t T \S t T \S {x R n : distx, s) distx, t)}. Explicitly for the Euclidean distance function we have distx, s) distx, t) x x, s + s x x, t + t, from where the desired representation follows. As a consequence of a well known necessary and sufficient condition for the inconsistency of an arbitrary system of linear inequalities [3, Theorem 4.4i)], from ) we obtain the following characterization of empty Voronoi cells. Theorem. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. Then V T S) = iff ) { ) } 0n t s cone 1 t s, s S, t T \ S. 3) We use the characterization in Theorem to obtain two well known statements about the classic Voronoi cells. Corollary 3. For a finite subset T of R n we have Moreover iff t is an extreme point vertex) of conv T. V T s) s T. V T T \ {t}) Proof. Suppose that there exists s T such that V T s) =. Then 3) holds for S = {s}, and therefore there exist λ t 0 for t T \ {s} such that λ t t s) = 0 n and λ t t s ) = 1, t T \{s} 3 t T \{s}

4 whereby λ 0 := t T \{s} λ t > 0 and s = t T \{s} λ t λ 0 t. Since the the square of the Euclidean norm. is a strictly convex function, we have t T \{s} λ t t + 1 = r T \{s} = t T \{s} λ r s = λ 0 s < λ 0 λ t t, t T \{s} λ t λ 0 t which is a contradiction. Now, suppose that V T T \ {t}) =. Denote S := T \ {t}. From Theorem, λ s t s) = 0 n and λ s t s ) = 1, s S s S whereby λ s > 0 and t = s S s S λ s r S λ s. r Hence, t is not an extreme point. If t is not an extreme point, then there exist λ s 0, s S with s S λ s = 1 and t = s S λ ss. Since. is a strictly convex function, t < s S λ s s ; therefore, setting µ s := λ s for s S, we have µ r S λr r t s 0, s S µ s t s) = 0 n and s S µ s t s ) = 1, showing that 3) holds, which, in view of Theorem, proves the corollary. The following result generalizes the if statement in the last part of Corollary 3. Corollary 4. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. If conv S) T \ S), then V T S) =. Proof. Taking t conv S) T \ S), since t is not an extreme point of conv S, by Corollary 3 we have V T S) V S {t} S) =. In fact we can prove a more general geometric statement which yields the preceding corollary. Theorem 5. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. Then V T S) iff there exists an open Euclidean ball B S such that B T \ S) =. 4

5 Proof. First assume that V T S). Then there exists c V T S). Let R := max distc, s), s S and let B be the open Euclidean ball of radius R centered at c. It is evident that B T \ S) =, otherwise we would have distc, t) < distc, s) for some t T \ S and s S, hence, c / V T S), which contradicts our choice of c. Now assume that there exists some open Euclidean ball B such that B T \ S) =. The centre of the ball B is contained in V T S), hence, V T S). We give an explicit example of an empty cell with S = and T = 3. Example 6. Let s 1 = 1, 0), s = 1, 0), t = 0, 0). It is not difficult to observe that ) becomes { V T S) = x 1, x ) : x 1 1, x 1 1 } =. This configuration is shown in Fig.. Figure : Minimal configuration for an empty cell Notice that we can likewise construct an empty cell with T > S by making sure that T \ S) conv S using Corollary 4). Theorem 7. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. Then V T S) is bounded iff cone {t s, s S, t T \ S} = R n. Proof. It suffices to observe that from the linear representation ) we obtain that the first moment cone of V T S) is cone {t s, s S, t T \ S}. Remark 8. It follows from Theorem 7 that if n and T 3 all nonempty cells are unbounded. We can strengthen the result in the preceding remark as follows. Theorem 9. Let T be a finite subset of R n. If then for any S T the cell V T S) is bounded. T < n + 1, 4) 5

6 Proof. The proof is based on the observation that a nonempty bounded polyhedron in R n must be defined by at least n + 1 inequalities. Let p := T, k := S. Then the number of inequalities that feature in the representation ) is φk) = kp k). Observe that φ attains its maximum at p/ for even p and at p 1)/ for odd p. Hence for even p and for odd p kp k) p kp k) p 1 p p ) = p 4 = p 4 p p 1 ) = p 1 4, p = 4 hence, ensuring 4) yields at most n inequalities that define each cell, and so all cells are unbounded. Proposition 10. Let S T R n, with S = 3 and T = 4. Then V T S) is either empty or unbounded. Proof. By Corollary 4 we have V T S) if and only if t conv S here T \ S = {t}). If t / conv S, then t can be separated from S, and by Theorem 7 the cell has to be unbounded. The following statement will be useful later for a discussion on planar quadrilateral cells. Proposition 11. Let S T R, with S = and T = 4. Then V T S) is bounded iff s 1, s ) t 1, t ). Proof. The configuration of the points of T in Fig. 3 means that if we take the line trough, Figure 3: Configuration for a bounded cell s 1 and s, then t 1 and t belong to the two opposite open halfspaces defined by this line. The same holds true if we interchange s 1 and s with t 1 and t. Obviously, the first moment cone M is equal to cone {t i s j, i, j = 1, }. Let us consider that the configuration of the points of T is like in Fig. 3. Then, we are going to prove that M = R, which, by Theorem 7, implies that V T {S} is bounded. What we are actually going to prove is the equivalent assertion that the polar cone M reduces to {0 }. To this aim, let p M and assume, w.l.o.g., that s 1, s ) t 1, t ) = {0 }. Then there exist λ, µ > 0 such that s = λs 1 and t = µt 1. Since p, t 1 s 1 0 and p, t 1 + λs 1 = p, t 1 s 0, we have p, t 1 0. This inequality combined with p, µt 1 s 1 = p, t s 1 0 yields p, s 1 0; hence, in view of p, µt 1 + λs 1 = p, t s 0, it turns out that p, t 1 = 0 = p, s 1. Since 6

7 s 1 and t 1 are linearly independent, we conclude that p = 0, as was to be proved. Second, let the Voronoi cell V T {s 1, s } be bounded. Then, by Theorem 7, the first moment cone cone {t i s j, i, j = 1, } is the whole of R. This clearly implies that t 1 and t are not on a common halfplane of the two determined by the straight line determined by s 1 and s, and the same assertion holds true when we interchange t 1 and t with s 1 and s s. This rules out the possibility that conv {s 1, s, t 1, t } be a segment, a triangle, or a quadrilateral having s 1 and s as adjacent vertices. Therefore s 1 and s are opposite vertices of the quadrilateral conv {s 1, s, t 1, t }, which clearly implies that s 1, s ) t 1, t ). The proof is completed. Theorem 1. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. Then int V T S) iff { ) } t s 0 n+1 / conv t s, s S, t T \ S. Proof. The proof comes from the well known characterization of the Slater condition for a linear system of inequalities [4, Theorem 3.1]. Example 13. Consider a system T of four points in the plane, T = {0, 0), 1, 1), 1, 0), 0, 1)}, S = {0, 0), 1, 1)}. This is illustrated in Fig. 4. Using Theorems, 7 and 1, it is easy to check that V T S) Figure 4: A singleton cell the intersection of the two shaded regions). is nonempty and bounded, but int V T S) =. Indeed, V T S) = { 1, 1 )}. The next statement is a specific characterization for a three-point system, which we will use in what follows. Proposition 14. Let T, T R n be such that T = T = 3 and T and T differ by exactly one point i.e. T T = ). Let s T \ T and s T \ T. If V T s) V T s ), then all points in the set T T belong to the same straight line. 7

8 Proof. For notational convenience we will prove the result for T = {t 1, t, 0} and T = {t 1, t, s}, where the points t 1, t, s are all nonzero and pairwise distinct. Let { F := V T 0) = x R n : t j, x 1 } t j, j = 1,, F := V T s) = { x R n : t j s, x 1 tj s ) }, j = 1,. The two inequalities defining F are consequence relations of the inequalities defining F. Therefore there exist λ ij 0, i, j = 1, such that t i s = λ ij t j, i = 1, and 5) j=1 t i s λ ij t j, i = 1,. j=1 Hence s = 1 λ ii ) t i λ 3 i t 3 i, i = 1, and 6) s 1 λ ii ) t i λ 3 i t 3 i, i = 1,. 7) We can subtract the two representations 6) of s to obtain 1 λ 11 + λ 1 ) t λ 1 + λ ) t = 0 n. If t j, j = 1, are linearly independent, we have 1 = λ 11 λ 1 and 1 = λ λ 1. Together with 7) this yields 0 s 1 λ 11 ) t 1 λ 1 t = λ 1 t 1 λ 1 t 0, which contradicts the condition s 0. Together with 5) this finishes the proof. Therefore t j, j = 1, are linearly dependent. Remark 15. This proposition means that it is impossible to enlarge a Voronoi cell of a single point in a three-point affinely independent system by moving this point. Proposition 16. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. If S, then V T S) = s S [ VT \{s} S \ {s}) V S s) ]. 8) Proof. Denote Observe that for any s S we have A s := V T \{s} S \ {s}) V S s). A s = {x R n : distx, s) distx, s) distx, t) s S, t T \ S}. 8

9 It is evident that A s V T S) for every s S, hence, A s V T S). To prove the reverse inclusion, assume that x V T S). Let s be a closest point to x in S. It is evident that x V S s). At the same time, it is not difficult to observe that V T S) V T \{ s} S \ { s}). Hence x V T \{ s} S \ { s}) V S s) = A s, and therefore V T S) A s. s S Proposition 17. Let T be a finite subset of R n, and let S be a nonempty proper subset of T. If there exist s 1, s S and t 1, t T \ S such that the inequalities s S s 1 x t 1 x and s x t x 9) define the same halfspace, then these inequalities are nonessential for V T S), i.e. they can be dropped from the system ). Proof. It is evident from the representation ) that these inequalities can be written as t 1 s 1, x 1 t1 s 1 ), t s, x 1 t s ). 10) If 9) define the same halfspace, then it is evident from 10) that there exists α > 0 such that t s = αt 1 s 1 ) and 1 t s ) = α 1 t1 s 1 ). 11) Every point x V T S) also satisfies the system t 1 s, x 1 t1 s ), t s 1, x 1 Adding these inequalities together we obtain t 1 s + t s 1 1 and substituting 11) we have a consequence of 1) 1 + α) t 1 s 1, x 1 + α) 1 t1 s + t s 1 ), t1 s 1 ), t s 1 ). 1) which defines the same halfspace as 10). Assume that the halfspace defined by 10) is different to one of the halfspaces defined by 1). Without loss of generality the first inequality in 1) defines the same halfspace as the first) inequality in 10). Then we have for some β > 0 t 1 s 1 = αt s ) = βt 1 s ), from where it is easy to see that t 1, t, s 1, s are collinear. Theorem 18. Let T R n be a finite set, let S := {s 1, s } T be a two-point set, and let H := {x R n : x s 1 = x s } = {x R n : s 1 s, x = 1 s 1 s )}. If int V T S), then H ri F = for every facet F of V T S). 9

10 Proof. If there is x H rif for some facet F of V T S), then without loss of generality for some t 0 T we have t 0 s 1, x = 1 t 0 s 1 ). 13) Notice that then t 0 s, x = 1 t 0 s ). 14) Otherwise we must have and subtracting 13) from 15) we obtain t 0 s, x < 1 t 0 s ), 15) s 1 s, x < 1 s 1 s ), which contradicts our choice of x on the bisector of [s 1, s ]. Since int V T S), then there is exactly one halfspace defined by the inequalities ) such that every point of rif belongs to the boundary hyperplane of this halfspace, and to the interior of the remaining ones. Hence the equalities 13) and 14) define the same hyperplane. Substituting a solution t 0 + s 1 )/ of 13) into 14) we have after elementary algebraic manipulation t 0 s, s 1 s = 0, meaning that s 1 s must be orthogonal to t 0 s. At the same time from the coincidence of 13) into 14) the points t 0, s 1, s must be collinear. This yields s 1 = s, which is impossible. 3 Case study 3.1 Singletons A singleton zero-dimensional) cell {c} can be obtained by placing the pairs of points s 1, s ) and t 1, t ) in the opposite corners of a square centred at c. This was already demonstrated in Example 13. Note that this is a minimal representation, since we need at least three inequalities to obtain a bounded cell cf. Theorem 9), and hence T 4. In fact the square can be replaced by a rectangle or a general cyclic quadrilateral a quadrilateral is cyclic if the sum of opposite angles equals π). Proposition 19. Let S T R, with S = and T = 4. The following statements are equivalent: a) V T S) is nonempty and at most one-dimensional. b) The points of T are the vertices of a cyclic quadrilateral, with s 1 and s located opposite to each other across a diagonal). c) V T S) is a singleton. 10

11 Proof. a) b). Without loss of generality assume that 0 V T S) while int V T S) =. By Theorem 1 we have 0 = λ ij t j s i ), 0 = λ ij t j s i ), 16) i,j {1,} i,j {1,} where λ ij are convex combination coefficients. Since 0 V T S), we have from ) that Without loss of generality assume that t j s i i, j {1, }. s s 1 t 1 t. If s 1 < t 1, then the second equality in 16) is impossible. Hence, s 1 = t 1. If s 1 > s, then λ 1 = λ = 0 and s 1 = λ 11 t 1 + λ 1 t, so s 1 [t 1, t ] and s 1 = λ 11 t 1 + λ 1 t, which holds only when t 1 = t = s 1 by the strict convexity of the square norm. This is impossible. Likewise, when t 1 < t we have λ 1 = λ = 0, then t 1 [s 1, s ], which by Corollary 4 yields V T S) =, again a contradiction. We proved that s = s 1 t 1 = t, and our sites are the vertices of a cyclic quadrilateral. It is now easy to observe that s 1, s are located opposite to each other because by the first equality in 16) [s 1, s ] [t 1, t ]. b) c) Since the points of T lie on some circle, without loss of generality we may assume that the centre of the circle is the origin, and t 1 = t = s 1 = s. In this case the right-hand side of system ) is zero, and we have for every point x V T S) t j s i, x 0 i, j {1, }. 17) It is evident that x = 0 is a solution of this system, hence, V T S). On the other hand, from 17) it folloes that V T S) is a cone. From these facts, using Proposition 11 we immediately deduce that V T S) = {0}. The implication c) a) is obvious. Note that for the case S = and T = 3 it is impossible to have a nonempty bounded cell due to Theorem 7: the conic hull of two vectors is always a proper subset of R. This means that we do not need to consider this configuration when discussing the subsequent cases of bounded polygons. Furthermore, in the case S = 3 and T = 4 it is impossible to have a bounded cell, as was shown in Proposition 10. Since we have determined that we can not have a nonempty bounded cell for T = S +1, the only possibility to have a singleton cell is for S = and T = 4. Furthermore, we can focus on the latter case when studying other bounded cells. 3. One-dimensional cells It follows from the preceding discussion that it is impossible to obtain line segments as multipoint Voronoi cells in our setting. Corollary 0. Let S T R, with S = and T = 4. Then V T S) is not onedimensional. 11

12 Proof. Follows directly from Proposition 19. It follows from Corollary 0 that it is impossible to have a one-dimensional cell for T = 4, S =, so both rays and lines are impossible in this configuration. Now consider the case T S = 1. By Corollary 4 we must have for {t} = T \ S that t / conv S. By the separation theorem this yields the existence of some d, d = 1 such that t s, d < 0 s S, which yields the existence of a sufficiently small ball B ε d) centred at d such that t s, y < 0 s S, y B ε d). Then for any x 0 V T S) and any y B ε d) we have t s, x 0 + y = t s, x 0 + t s, y < t s, x 0. It is hence clear from the representation in Proposition 1 that x 0 + B ε y) V T S), which contradicts the fact that V T S) is one-dimensional. Thus we have proved the following statement. Proposition 1. Let S T R n, with T = S +1. Then V T S) is not one-dimensional. 3.3 Triangles A somewhat surprising result is that a two-point cell cannot be a triangle. As discussed previously, we only need to prove this for the case S = and T = 4. Proposition. Let S T R, with S = and T = 4. Then V T S) is not a triangle. Proof. Suppose that V T S) is a triangle. Denote T = {s 1, s, t 1, t }, S = {s 1, s }. The cell V T S) is the solution set of the linear system of inequalities c ij, x α ij i, j = 1, ), 18) with c ij := t i s j and α ij := 1 ti s j ). Notice that ) ) ) c c11 c1 = + + α α 11 α 1 c1 α 1 ). 19) Without loss of generality, we will assume that 0 int V T S), that is, α ij > 0 i, j = 1, ). Since 18) defines a triangle, one of its inequalities, say the one corresponding to i = = j is redundant, so that the triangle is actually the solution set of the system consisting of the other three. Hence the vectors c 11, c 1 and c 1, being exterior normals to the sides, are such that for some β 1, β 1 > 0 one has c 11 + β 1 c 1 + β 1 c 1 = 0. 0) We claim that c 11, c 1 and c 1, are affinely independent, that is, they are not contained in a common straight line. Indeed, suppose that one has p, c 11 = γ, p, c 1 = γ, p, c 1 = γ for some p R and some γ R. Then, multiplying 0) with p we get 1 + β 1 + β 1 ) γ = 0, that is, γ = 0, which implies that c 11, c 1 and c 1 have the same 1

13 direction and hence s 1, s, t 1 and t belong to a common straight line, which is impossible because V T S) is supposed to be a triangle. This proves the claim that c 11, c 1 and c 1 are affinely independent. Since the inequality corresponding to i = = j is redundant, by Farkas lemma there exist λ ij 0 i, j = 1, ) such that c = λ 11 c 11 + λ 1 c 1 + λ 1 c 1 and α λ 11 α 11 + λ 1 α 1 + λ 1 α 1. 1) Consider the vertex x of V T S) defined by c 1, x = α 1 and c 1, x = α 1. From 19) we obtain that α = α 11 + α 1 + α 1 c 11, x + c 1, x + c 1, x = c 11 + c 1 + c 1, x = c, x α ; hence, by 1), we deduce that c, x = α λ 11 α 11 + λ 1 α 1 + λ 1 α 1 λ 11 c 11, x + λ 1 c 1, x + λ 1 c 1, x = λ 11 c 11 + λ 1 c 1 + λ 1 c 1, x = c, x. Therefore λ 11 α 11 + λ 1 α 1 + λ 1 α 1 = α, that is, ) ) c c11 c1 = λ α 11 + λ α 1 11 α 1 ) + λ 1 c1 α 1 ) c11 c1 Comparing this equality with 19) and taking into account that the vectors, ) α 11 α 1 c1 and are linearly independent since c 11, c 1 and c 1 are affinely independent), we α 1 deduce that λ 11 = 1, a contradiction. 3.4 Bounded quadrilaterals We next show that any non-cyclic bounded quadrilateral is a Voronoi cell of points with T = 4. We also prove that it is impossible for a cyclic quadrilateral to be a Voronoi cell of points when T = 4). Proposition 3. Let F R be a non-cyclic bounded quadrilateral. Then there exist S T R, with S = and T = 4, such that V T S) = F. Proof. We construct the sets S and T explicitly, hence proving the proposition. First, looking at Fig. 5, we would like to mention that α > 0, β > 0 and w.l.o.g. π > α + β. The angles CBD = β and CAD = α. Symmetrically to AB we draw the lines which take angles α at the point A and β at B. In this way, we get the points t 1 and t. We take s 1 and s as symmetric points of t 1, with respect to the lines CA and CB, respectively. The only thing that we have to prove is that t s 1 AD and t s BD. B is a circumcenter of the circumcircle passing trough the points t 1, t and s. Therefore, t 1 s t = π β, which looking at the quadrilateral C s D B implies that t s BD. A is a circumcenter of the circumcircle passing trough the points t 1, t and s 1. Therefore, t 1 s 1 t = π α, which looking at the quadrilateral C 1 s 1 D 1 A implies t s 1 AD. Observe that the algorithm does not work if π α + β. However it is not difficult to observe that for a non-cyclic quadrilateral it is always possible to choose the corners of the qualdrilateral to ensure α + β < π. We have the following negative result. 13 ). )

14 Figure 5: Bounded quadrilateral. Proposition 4. Let S T R, with S = and T = 4. Then V T S) is not a cyclic quadrilateral. Proof. Assume that the Voronoi cell V T S) of some set S = {s 1, s }, with T = {s 1, s, t 1, t }, is a cyclic quadrilateral. Then each side of this quadrilateral is defined by the bisector between s i and t j for i, j {1, }. First, we will show that any two sides defined by the bisectors of disjoint pairs, say, {s 1, t 1 } and {s, t }, cannot be adjoint. Assume the contrary: then without loss of generality the intersection u of the two bisectors is a vertex of V T S). We have using the representation ) t 1 s, u 1 t 1 s ), t s 1, u 1 t s 1 ), ) and since u is the intersection of the two bisectors, t 1 s 1, u = 1 t 1 s 1 ), t s, u = 1 t s ). 3) Adding the two equalities in 3) and rearranging, we obtain t 1 s, u + t s 1, u = 1 t 1 s ) + 1 t s 1 ). Together with ) this yields equalities in ), and hence the four lines that define the sides of the quadrilateral must intersect at u. This is impossible, hence, the assumption is wrong. Now, let us consider the quadrilateral with vertices s 1, s, t 1 and t. Looking at Fig. 6, it is easy to see that the angles at t 1 and C are equal, and so they are the angles at t and A. This means that this quadrilateral is cyclic too, that is, s 1, s, t 1 and t lie on a circumference. It follows from Proposition 19 that the Voronoi cell V T S) is a singleton, which contradicts our assumption. 14

15 3.5 Halfspaces Figure 6: An illustration to the proof of Proposition 4. A halfspace cell can be obtained by putting the two points of S on a line perpendicular to the boundary line of the halfspace making sure that S is in the interior of the halfspace. An additional point t is placed on the same line on the opposite side of the hyperplane at the same distance from the hyperplane as the distance to the hyperplane from the farthest point in S see Fig. 7). We hence conclude that a halfspace can be constructed Figure 7: The intersection of the two halfspaces is V T s 1, s ). using S = and T = 3. Observe that it is also possible to do the same construction for S {, 3} and T = 4. We prove this explicitly in the next statement. Proposition 5. Let F R n be a halfspace. Then for any two integers τ > σ 1 there exist S T R n, with S = σ and T = τ, such that V T S) = F. Proof. Note that any halfspace F can be represented as F = {x R x, d γ} for some d, d = 1 and γ R. Choose any x 0 such that x 0, d = d, and let S = {s 0,..., s m }, T = {t 0,..., t p } S, where m = σ 1, p = τ σ 1, s 0 := x 0 d, t 0 := x 0 + d, s i = x 0 α i d, 0 < α i < 1 i {1,..., m}, t j := x 0 + β j d, β j > 1 j {1,..., p}, and the constants α i and β j are all different to make sure that the sites do not coincide). We will next show that V T S) = {x R x, d γ}. 15

16 We have from the representation in Proposition 1 V T S) = { x : t j s i, x 1 tj s i )}, i {0,...,m} j {1,...,p} and so each inequality becomes for i {0,..., m}, j {0,..., p} β j + α i ) d, x 1 β j α i )β j + α i ) + β j + α i ) x 0, d ), where α 0 = β 0 = 1. Dividing by the common factor β j + α i > 0, we have hence, V T S) = d, x d, x 0 β j α i, { x : d, x γ 1 } minβ j α i ) = {x : d, x γ}, i,j precisely the halfspace that we were aiming for. 3.6 Intersections of parallel halfspaces We consider a set F R n represented by the inequalities α d, x β, with d = 1 and α < β. One can easily check that F = V T S) for T := {s 1, s, t 1, t } and S := {s 1, s } with, for instance, s 1 := 3α+β d, s 4 := α+β d, t 1 := 3α β d and t := 7β 3α. We have shown 4 the following result. Proposition 6. Let F R n be a nonempty intersection of two parallel halfspaces with opposite normals. Then there exist S T R n, with S = and T = 4, such that V T S) = F. Note that it is impossible to produce a strip with T S = 1: indeed, it is clear from the fact that all constraints are defined by the parallel lines that all vectors s t, s S, t T should be collinear, lying on some line orthogonal to the inequalities. Now, if for the unique t T \ S we have t conv S, then the cell is empty by Corollary 4. However if t / conv S, then the cell has to be a halfspace. Hence the only possibility is T = 4, S =. 3.7 Wedges Now, let us have an angle F := {x R n : c i x α i, c i 0 n, i = 1, }. Let us take an arbitrary point from t int F := {x R n : c i x α i, c i 0 n, i = 1, } and construct the two symmetric points s 1 and s with respect to both hyperplanes defining F. If T := {s 1, s, t} then VT s 1, s ) = F T t) = F. More generally, if V is the intersection of two non-parallel halfplanes, V = {x R x, v 1 a, v 1, x, v 1 a, v }, where v 1, v, a R, and {v 1, v } is a linearly independent system, we can choose any point t in the set V = {x R x, v 1 > a, v 1, x, v 1 > a, v }, 16

17 Figure 8: Construction of the set of sites for a wedge and let s 1 and s be the two reflections of t with respect to the lines x, v i = a, v i, i = 1, see Fig. 8). It is possible to add one more site s 3 to S in such a way that the halfspace s 3 x t x includes the original wedge. Proposition 7. Let F := {x R n x, v i b i i = 1, )}, with v 1, v R n linearly independent unit vectors and b 1, b R. Then there exist S T R n, with S = and T = 4, such that V T S) = F. Proof. Without loss of generality, we assume that b i = 0 i = 1, ). Set β := v 1, v, i := 3 i for i = 1,, and define s i := v i v i i = 1, ). Let t i be the point symmetric to s i with respect to the hyperplane defined by x, v i = 0, that is, t i := v i β) v i i = 1, ), and define S := {s 1, s } and T := {s 1, s, t 1, t }. We have T = 4, since β > 1. Given that t i = 5 + 4β = s i i = 1, ), one has V T S) = {x R n x, t i s i 0, x, t i s i 0 i = 1, )}. 4) Hence, to prove that V T S) = F, it will suffice to show that, for i = 1,, the inequality x, t i s i 0 is equivalent to x, v i 0 and that the remaining two inequalities in 4) are redundant. The first assertion is obvious, since t i s i = + β) v i and β > 1. Let us now prove that the inequalities x, t i s i 0 are redundant, that is, that they are consequences of the system x, v i 0 i = 1, ). But this is also immediate, since t i s i = v i β) v i and β > 1. 17

18 3.8 Unbounded polygons with three sides Our next construction is for an unbounded polygon like the one in Fig. 9, where the two unbounded sides are non-parallel. Figure 9: Unbounded polygon with three sides Proposition 8. Let F be an unbounded polygon with non-parallel sides. Then there exist S T R, with S = and T = 4, such that V T S) = F. Proof. We construct the sets S and T explicitly, hence proving the proposition. In the notation of Fig. 9, we would like to mention that 0 < α < π, 0 < β < π, α + β > π and β α. If we consider the triangle ABC,the angle at C is α + β π. Now, we draw a line trough the point A, which takes an angle α + β π with the line BA. On this line we consider arbitrary point t 1, sufficiently close to A. Let us take a symmetry points of t 1 with respect to AB s 1 and AC s, respectively. We shall prove that the line trough the points s 1, s is perpendicular to BC. For this purpose, we would like to mention that the point A is circumcenter of the circumcircle passing trough the points s 1, s and t 1. This means that the angle s 1 s t 1 = α +β π. If we consider the quadrilateral formed by the lines BC, AC, s 1 s and t 1 s we get the desired fact. At the end we construct the site t 1 as a symmetric point of s 1 with respect to the line BC. Proposition 9. Let S T R, with S = and T = 4. Then V T S) is not an unbounded polygon with parallel sides and just two vertices. Proof. Similarly to the proof of Proposition, suppose that V T S) is an unbounded polygon with parallel sides and just two vertices, denote S = {s 1, s } and T = {s 1, s, t 1, t }, and suppose further, w.l.o.g., that the redundant inequality in the linear system t i s j, x 1 ti s j ) i, j = 1, ) 5) which has V T S) as its solution set) is the one corresponding to i = = j. Then two of the three vectors t 1 s 1, t 1 s and t s 1 are exterior normals to the parallel sides, and hence they make an angle of π. Those two vectors cannot be t 1 s 1 and t 1 s, because with such a configuration it is clear that V T S) would be empty. Let us consider the two remaining cases. We start with the case when the two vectors are t 1 s 1 and t s 1. Since in this case s does not belong to the straight line determined by s 1, t 1 and t, there is a circumference containing s, t 1 and t, and we may assume, w.l.o.g., that 18

19 the center of this circumference is 0. Let r be the radius of this circunference. We have t 1 = t = t = r > s 1, this inequality following from the fact that s 1 belongs to the relative interior of the segment with endpoints t 1 and t. System 5) can be rewritten as t 1 s 1, x 1 r s 1 ), t 1 s, x 0, t s 1, x 1 r s 1 ), t s, x 0, and recall that the latter inequality is redundant. Since t 1 s 1 and t s are linearly independent, there exists x R such that t 1 s, x 0 and t s, x > 0. Then, for sufficiently small ɛ > 0, the point ɛx satisfies the first three inequalities as the first and the third right hand sides are strictly positive) but not the fourth one, which is a contradiction with the redundancy of the latter. It only remains to consider the case when t 1 s and t s 1 are the vectors that make an angle of π. In this case, s 1 and s are opposite vertices of the quadrilateral determined by s 1, s, t 1 and t. But this contradicts the unboundedness of V T S), in view of Proposition 11. We note here that it is possible to have an unbounded polygon with three sides for the case S = 3 and T = 4 if and only if the unbounded sides are non-parallel. Indeed, in this case we can first build a wedge that defines the two unbounded sides, and then add an extra site to define the extra inequality. For the case of unbounded parallel sides, it is clear that the point t should at the same time lie outside of each of these parallel sides, which is impossible. 3.9 Unbounded polygons with four sides In the next proposition we shall consider the case of an unbounded quadrilateral with two parallel sides, shown in Fig. 10. Figure 10: Unbounded polygon with parallel sides 19

20 Proposition 30. Let F be an unbounded quadrilateral with two parallel sides. Then there exist S T R, with S = and T = 4, such that V T S) = F. Proof. We construct the sets S and T explicitly, hence proving the proposition. In the notation of Fig. 10, we have l 1 l, 0 < α < π and the point A belongs to the interior of the parallel band. First, we move, parallel to the line l 1, the point A to an arbitrary point M. Through M we draw a line perpendicular to l 1. Take a point S 1, on the same line, to the same distance to l as M to l 1. Now, we consider the symmetrical points of S 1, T 1 with respect to l and T with respect to l 1. In this way M is a midpoint of the segment [T 1 T ]. Next we move parallel the line trough the points T 1 and T till the configuration in Fig. 10, i.e., for the new points t 1, t, s 1 and m, t 1 At = π α and mat = π α. To find the last site, we take s as a symmetric point of t with respect to the line trough the points A and B. We shall prove that the line trough the points t 1, s is perpendicular to AC. For this purpose, we would like to mention that the point A is circumcenter of the circumcircle passing trough the points t 1, t and s. This means that the angle t s t 1 = π α. If we consider the quadrilateral formed by the lines BA, AC, t 1 s and t s we get the desired fact. At the end, after finding the four sites, we would like to mention that this construction is impossible if α = π. We next consider the case of an arbitrary unbounded quadrilateral with nonparallel sides, presented in Fig. 11. Figure 11: Configuration for a general unbounded quadrilateral Proposition 31. Let F be an unbounded quadrilateral with no sides parallel. Then there exist S T R, with S = and T = 4, such that V T S) = F. Proof. We construct the sets S and T explicitly, hence proving the proposition. In the notation of Fig. 10, First, looking at Fig. 11, we would like to mention that π > α > β > 0. For the quadrilateral CAD, we prolong the lines passing trough the points C and D to obtain the point B. The angles CBD = β and CAD = α. Symmetrically 0

21 to AB we draw the lines which take angles π α at the point A and β at B. In this way, we get the points t 1 and t. We take s 1 and s as symmetric points of t 1, with respect to the lines CA and CB, respectively. The only thing that we have to prove is that t s 1 AD and t s BD. B is a circumcenter of the circumcircle passing trough the points t 1, t and s. Therefore, t 1 s t = π β, which looking at the quadrilateral C s D B implies that t s BD. A is a circumcenter of the circumcircle passing trough the points t 1, t and s 1. Therefore, t 1 s 1 t = π α, which looking at the quadrilateral C 1 s 1 D 1 A implies t s 1 AD. 4 Acknowledgments The first author was supported by the MINECO of Spain, Grant MTM C--P, and the Severo Ochoa Programme for Centres of Excellence in R&D [SEV ]. He is affiliated to MOVE Markets, Organizations and Votes in Economics). He thanks The School of Mathematics and Statistics of UNSW Sydney for sponsoring a visit to Sidney to complete this work. The second author is grateful to the Australian Research Council for continuous financial support via grants DE and DP , which in particular sponsored a trip to Barcelona that initiated this collaboration. The third author was partially supported by MINECO of Spain and ERDF of EU, Grant MTM C-1-P, and Sistema Nacional de Investigadores, Mexico. References [1] A. Gemsa, D. T. Lee, C.-H. Liu, and D. Wagner. Higher order city Voronoi diagrams. In Algorithm theory SWAT 01, volume 7357 of Lecture Notes in Comput. Sci., pages Springer, Heidelberg, 01. [] N. Giansiracusa and C. Ricciardi. Spatial analysis of u.s. supreme court 5-to-4 decisions, 018. [3] M. A. Goberna and M. A. López. Linear semi-infinite optimization, volume of Wiley Series in Mathematical Methods in Practice. John Wiley & Sons, Ltd., Chichester, [4] M. A. Goberna, M. A. Lopez, and M. Todorov. Stability theory for linear inequality systems. SIAM J. Matrix Anal. Appl., 174): , [5] M. A. Goberna, J. E. Martínez-Legaz, and V. N. Vera de Serio. The Voronoi inverse mapping. Linear Algebra Appl., 504:48 71, 016. [6] H. Kaplan, W. Mulzer, L. Roditty, P. Seiferth, and M. Sharir. Dynamic planar Voronoi diagrams for general distance functions and their algorithmic applications. In Proceedings of the Twenty-Eighth Annual ACM-SIAM Symposium on Discrete Algorithms, pages SIAM, Philadelphia, PA, 017. [7] R. Klein. Abstract Voronoĭ diagrams and their applications extended abstract). In Computational geometry and its applications Würzburg, 1988), volume 333 of Lecture Notes in Comput. Sci., pages Springer, New York,

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