Strong Matching of Points with Geometric Shapes

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1 Strong Matching of Points with Geometric Shaes Ahmad Biniaz Anil Maheshwari Michiel Smid School of Comuter Science, Carleton University, Ottawa, Canada December 9, 05 In memory of Ferran Hurtado. Abstract Let P be a set of n oints in general osition in the lane. Given a convex geometric shae S, a geometric grah G S (P ) on P is defined to have an edge between two oints if and only if there exists a homothet of S having the two oints on its boundary and whose interior is emty of oints of P. A matching in G S (P ) is said to be strong, if the homothets of S reresenting the edges of the matching are airwise disjoint, i.e., they do not share any oint in the lane. We consider the roblem of comuting a strong matching in G S (P ), where S is a diametral disk, an equilateral triangle, or a square. We resent an algorithm that comutes a strong matching in G S (P ); if S is a diametral-disk, then it comutes a strong matching of size at least n 7, and if S is an equilateral-triangle, then it comutes a strong matching of size at least n 9. If S can be a downward or an uward equilateral-triangle, we comute a strong matching of size at least n in G S(P ). When S is an axis-aligned square, we comute a strong matching of size at least n in G S(P ), that imroves the revious lower bound of n 5. Introduction Let S be a comact and convex set in the lane that contains the origin in its interior. A homothet of S is obtained by scaling S with resect to the origin by some factor µ 0, followed by a translation to a oint b in the lane: b + µs = {b + µa : a S}. For a oint set P in the lane, we define G S (P ) as the geometric grah on P that has a straight-line edge between two oints and q if and only if there exists a homothet of S having and q on its boundary and whose interior does not contain any oint of P. If P is in general osition, i.e., no four oints of P lie on the boundary of any homothet of S, then G S (P ) is lane [9]. Hereafter, we assume that P is a set of n oints in the lane that is in general osition with resect to S (see Definition for a formal definition). If S is a disk whose center is the origin, then G (P ) is the Delaunay triangulation of P. If S is an equilateral triangle whose barycenter is the origin, then G (P ) is the triangular-distance Delaunay grah of P, which has been introduced by Chew [0]. A matching in a grah G is a set of edges that do not share any vertices. A maximum matching is a matching of maximum cardinality. A erfect matching is a matching that matches all the vertices of G. Let M be a matching in G S (P ). The matching M is referred to as a matching of oints with shae S, e.g., a matching in G (P ) is a matching of oints with disks. Research suorted by NSERC.

2 Let S M be a set of homothets of S reresenting the edges of M. The matching M is called a strong matching if there exists a set S M whose elements are airwise disjoint, i.e., the objects in S M do not share any oint in the lane. Otherwise, M is a weak matching. See Figure. To be consistent with the definition of the matching in the grah theory, we use the term matching to refer to a weak matching. Given a oint set P in the lane and a shae S, the (strong) matching roblem is to comute a (strong) matching of maximum cardinality in G S (P ). (a) (b) (c) Figure : Point set P and (a) a erfect weak matching in G (P ), (b) a erfect strong matching in G (P ), and (c) a erfect strong matching in G (P ). Let denote a closed disk whose center is the origin. Let denote a closed axis-aligned square whose center is the origin. Let denote a closed downward equilateral triangle whose barycenter is the origin and whose lowest vertex is on the negative y-axis. For two oints and q, the closed disk that has the line segment q as its diameter is called the diametral-disk between and q. Let denote a diametral-disk between two oints. Let P be a set of oints in the lane. G (P ) is the grah that has an edge between two oints, q P if there exists a homothet of that has and q on its boundary and does not contain any oint of P in its interior. Similarly, G (P ) is the grah that has an edge between two oints, q P if there exists a homothet of that has and q on its boundary and does not contain any oint of P in its interior. G (P ) is the grah that has an edge between two oints, q P if the diametral-disk between and q and does not contain any oint of P in its interior. G (P ) is the grah that has an edge between two oints, q P if there exists a homothet of that has and q on its boundary and does not contain any oint of P in its interior. If we consider an uward triangle, then G (P ) is defined similarly. The grah G (P ) is defined as the union of G (P ) and G (P ). Definition. Given a oint set P and a shae S {,,, }, we say that P is in general osition with resect to S if S = : no four oints of P lie on the boundary of any homothet of. S = : no four oints of P lie on the boundary of any between any two oints of P. S = : the line assing through any two oints of P does not make angles 0, 60, or 0 with the horizontal. This imlies that no four oints of P are on the boundary of any homothet of. S = : (i) no two oints in P have the same x-coordinate or the same y-coordinate, and (ii) no four oints of P lie on the boundary of any homothet of. In this aer we consider the strong matching roblem of oints in general osition in the lane with resect to a given shae. Let P be a set of oints in the lane that is in general

3 Table : Lower bounds on the size of weak and strong matchings in G S (P ). S weak matching reference strong matching reference n n [] 8 [] n n [7] 7 Theorem n n [] 9 Theorem or n n [] Theorem 5 n [, ] n 5 [, ] n Theorem osition with resect to S {,,, }. If S =, then G (P ) is the Delaunay triangulation of P, DT (P ). If S =, then G (P ) is the L -Delaunay grah of P. If S =, then G (P ) is the Gabriel grah of P, GG(P ). If S =, then G (P ) is the half-theta six grah of P, Θ 6(P ) [8], that is in turn the triangular-distance Delaunay grah of P, which was introduced by Chew [0]. Moreover, G (P ) is the theta six grah of P, Θ 6 (P ) [8].. Previous Work Let P be a set of n oints in the lane that is in general osition with resect to a given shae S {,,, }. The roblem of comuting a maximum (strong) matching in G S (P ) is one of the fundamental roblems in comutational geometry and grah theory [,,, 5, 7, 6, ]. Dillencourt [] and Ábrego et al. [] considered the roblem of matching oints with disks. Dillencourt [] roved that G (P ) contains a erfect matching. Ábrego et al. [] roved that G (P ) has a strong matching of size at least (n )/8. They also showed that for arbitrarily large n, there exists a set P of n oints in the lane such that G (P ) does not contain a strong matching of size more than 6 7 n. As for diametral disks, Biniaz et al. [7] roved that G (P ) has a matching of size at least (n )/, and that this bound is tight. The roblem of matching of oints with equilateral triangles has been considered by Babu et al. []. They roved that G (P ) has a matching of size at least (n )/, and that this bound is tight. Since G (P ) is a subgrah of G (P ), the lower bound of (n )/ on the size of a maximum matching in G (P ) holds also for G (P ). The roblem of strong matching of oints with axis-aligned rectangles is trivial. An obvious algorithm is to reeatedly match the two leftmost oints. The roblem of matching oints with axis-aligned squares was considered by Ábrego et al. [, ]. They roved that G (P ) has a erfect matching and a strong matching of size at least n/5. Further, they showed that there exists a set P of n oints in the lane with arbitrarily large n, such that G (P ) does not contain a strong matching of size more than 5 n. Table summarizes the results. Bereg et al. [5] concentrated on matching oints of P with axis-aligned rectangles and squares, where P is not necessarily in general osition. They roved that any set of n oints in the lane has a strong rectangle matching of size at least n, and that such a matching can be comuted in O(n log n) time. As for squares, they resented a Θ(n log n)-time algorithm that decides whether a given matching has a weak square realization, and an O(n log n)-time algorithm for the strong square matching realization. They also roved that it is NP-hard to decide whether a given oint set has a erfect strong square matching.. Our results In this aer we consider the roblem of comuting a strong matching in G S (P ), where S {,, }. In Section, we rovide some observations and rove necessary lemmas. Given a

4 oint set P in general osition with resect to a given shae S, in Section, we resent an algorithm that comutes a strong matching in G S (P ). In Section, we rove that if S is a diametral disk, then the algorithm of Section comutes a strong matching of size at least (n )/7 in G (P ). In Section 5, we rove that if S is an equilateral triangle, then the algorithm of Section comutes a strong matching of size at least (n )/9 in G (P ). In Section 6, we comute a strong matching of size at least (n )/ in G (P ); this imroves the revious lower bound of n/5. In Section 7, we comute a strong matching of size at least (n )/ in G (P ). A summary of the results is given in Table. In Section 8 we discuss a ossible way to further imrove uon the result obtained for diametral-disks in Section. Concluding remarks and oen roblems are given in Section 9. Preliminaries Let S {, }, and let S and S be two homothets of S. We say that S is smaller than S if the area of S is smaller than the area of S. For two oints, q P, let S(, q) be a smallest homothet of S having and q on its boundary. If S is a diametral-disk or a downward equilateral-triangle, then we denote S(, q) by D(, q) or t(, q), resectively. If S is a diametraldisk, then D(, q) is uniquely defined by and q. If S is an equilateral-triangle, then S has the shrinkability roerty: if there exists a homothet S of S that contains two oints and q, then there exists a homothet S of S such that S S, and and q are on the boundary of S. Moreover, we can shrink S further, such that each side of S contains either or q. Then, t(, q) is uniquely defined by and q. Thus, we have the following observation: Observation. For two oints, q P, D(, q) is uniquely defined by and q, and it has the line segment q as a diameter. t(, q) is uniquely defined by and q, and it has one of and q on a corner and the other oint is on the side oosite to that corner. r q t(, q) t(q, r) D(, r) D(q, r) q t(, r) r D(, q) Figure : Illustration of Observation. Given a shae S {, }, we define an order on the homothets of S. Let S and S be two homothets of S. We say that S S if the area of S is less than the area of S. Similarly, S S if the area of S is less than or equal to the area of S. We denote the homothet with the larger area by max{s, S }. As illustrated in Figure, if S(, q) contains a oint r, then both S(, r) and S(q, r) have smaller area than S(, q). Thus, we have the following observation: Observation. If S(, q) contains a oint r in its interior, then max{s(, r), S(q, r)} S(, q). Given a oint set P in general osition with resect to a given shae S {, }, let K S (P ) be a comlete edge-weighted geometric grah on P. For each edge e = (, q) in K S (P ), we define

5 S(e) to be the shae S(, q), i.e., a smallest homothet of S having and q on its boundary. We say that S(e) reresents e, and vice versa. Furthermore, let the weight w(e) (res. w(, q)) of e be equal to the area of S(e). Thus, w(, q) < w(r, s) if and only if S(, q) S(r, s). Note that G S (P ) is a subgrah of K S (P ), and has an edge (, q) if and only if S(, q) does not contain any oint of P \ {, q}. Lemma. Let P be a set of n oints in the lane that is in general osition with resect to a given shae S {, }. Then, any minimum sanning tree of K S (P ) is a subgrah of G S (P ). Proof. The roof is by contradiction. Assume there exists an edge e = (, q) in a minimum sanning tree T of K S (P ) such that e / G S (P ). Since (, q) is not an edge in G S (P ), S(, q) contains a oint r P \ {, q}. By Observation, max{s(, r), S(q, r)} S(, q). Thus, w(, r) < w(, q) and w(q, r) < w(, q). By relacing the edge (, q) in T with either (, r) or (q, r), we obtain a sanning tree in K S (P ) that is shorter than T. This contradicts the minimality of T. Lemma. Let G be an edge-weighted grah with edge set E and edge-weight function w : E R +. For any cycle C in G, if the maximum-weight edge in C is unique, then that edge is not in any minimum sanning tree of G. Proof. The roof is by contradiction. Let e = (u, v) be the unique maximum-weight edge in a cycle C in G such that e is in a minimum sanning tree T of G. Let T u and T v be the two trees obtained by removing e from T. Let e = (x, y) be an edge in C that connects a vertex x T u to a vertex y T v. By assumtion, w(e ) < w(e). Thus, by relacing e with e in T, we obtain a tree T = T u T v {(x, y)} in G such that w(t ) < w(t ). This contradicts the minimality of T. Recall that t(, q) is the smallest homothet of that has and q on its boundary. Similarly, let t (, q) denote the smallest uward equilateral-triangle having and q on its boundary. Note that t (, q) is uniquely defined by and q, and it has one of and q on a corner and the other oint is on the side oosite to that corner. In addition the area of t (, q) is equal to the area of t(, q). Note that G (P ) is the triangular-distance Delaunay grah TD-DG(P ), that is in turn a half theta-six grah Θ 6(P ) [8]. A half theta-six grah on P, and equivalently G (P ), can be constructed in the following way. For each oint in P, let l be the horizontal line through. Define l γ as the line obtained by rotating l by γ degrees in counterclockwise direction around. Thus, l 0 = l. Consider the three lines l, 0 l 60, and l 0, which artition the lane into six disjoint cones with aex. Let C,..., C 6 be the cones in counter-clockwise order around as shown in Figure. C, C, C 5 will be referred to as odd cones, and C, C, C 6 will be referred to as even cones. For each even cone C, i connect to the nearest oint q in C. i The distance between and q, is defined as the Euclidean distance between C C Figure : G (P ). l 0 q C 5 C t(, q) l 60 C C 6 l 0 The construction of and the orthogonal rojection of q onto the bisector of C i. See Figure. In other words, the nearest oint to in C i is a oint q in C i that minimizes the area of t(, q). The resulting grah is the half theta-six grah, which is defined by even cones [8]. Moreover, the resulting 5

6 grah is G (P ) that is defined with resect to the homothets of. By considering the odd cones, G (P ) is obtained. By considering the odd cones and the even cones, G (P ) that is equal to Θ 6 (P ) is obtained. Note that G (P ) is the union of G (P ) and G (P ). Let X(, q) be the regular hexagon centered at that has q on its boundary, and its sides are arallel to l, 0 l 60, and l 0. Then, we have the following observation: Observation. If X(, q) contains a oint r in its interior, then t(, r) t(, q). Strong Matching in G S (P ) Given a oint set P in general osition with resect to a given shae S {, }, in this section we resent an algorithm that comutes a strong matching in G S (P ). Recall that K S (P ) is the comlete edge-weighted grah on P with the weight of each edge e is equal to the area of S(e), where S(e) is a smallest homothet of S reresenting e. Let T be a minimum sanning tree of K S (P ). By Lemma, T is a subgrah of G S (P ). For each edge e T we denote by T (e + ) the set of all edges in T whose weight is at least w(e). Moreover, we define the influence set of e as the set of all edges in T (e + ) whose reresenting shaes overla with S(e), i.e., Inf(e) = {e : e T (e + ), S(e ) S(e) }. Note that Inf(e) is not emty, as e Inf(e). Consequently, we define the influence number of T to be the maximum size of a set among the influence sets of edges in T, i.e., Inf(T ) = max{ Inf(e) : e T }. Algorithm receives P and S as inut and comutes a strong matching in P with resect to S as follows. The algorithm starts by comuting G S (P ), where the weight of each edge is equal to the area of its reresenting shae. Then it comutes a minimum sanning tree T of G S (P ). Then it initializes a forest F by T, and a matching M by an emty set. Afterwards, as long as F is not emty, the algorithm adds the smallest edge e in F to M, and removes the influence set of e from F. Finally, it returns M. Algorithm StrongMatching(P, S) : comute G S (P ) : T MST(G S (P )) : F T : M 5: while F do 6: e smallest edge in F 7: M M {e} 8: F F Inf(e) 9: return M Theorem. Given a set P of n oints in the lane and a shae S {, }, Algorithm comutes a strong matching of size at least n Inf(T ) in G S(P ), where T is a minimum sanning tree of G S (P ). Proof. Let M be the matching returned by Algorithm. First we show that M is a strong matching. If M contains one edge, then trivially, M is a strong matching. Consider any two edges e and e in M. Without loss of generality assume that e is considered before e in the 6

7 while loo. At the time e is added to M, the algorithm removes the edges in Inf(e ) from F, i.e., all the edges whose reresenting shaes intersect S(e ). Since e remains in F after the removal of Inf(e ), we know that e / Inf(e ). This imlies that S(e ) S(e ) =, and hence M is a strong matching. In each iteration of the while loo we select e as the smallest edge in F, where F is a subgrah of T. Then, all edges in F have weight at least w(e). Thus, F T (e + ); that imlies that the set of edges in F whose reresenting shaes intersect S(e) is a subset of Inf(e). Therefore, in each iteration of the while loo, out of at most Inf(e) many edges of T, we add one edge to M. Since Inf(e) Inf(T ) and T has n edges, we conclude that M n Inf(T ). Remark Let T be the minimum sanning tree comuted by Algorithm. Let e = (u, v) be an edge in T. Recall that T (e + ) contains all the edges of T whose weight is at least w(e). We define the degree of e as deg(e) =deg(u)+deg(v), where deg(u) and deg(v) are the numbers of edges incident to u and v in T (e + ), resectively. Note that all the edges incident to u or v in T (e + ) are in the influence set of e. Thus, Inf(e) deg(e), and consequently Inf(T ) deg(e). Strong Matching in G (P ) In this section we consider the case where S is a diametral-disk. Recall that G (P ) is an edge-weighted geometric grah, where the weight of an edge (, q) is equal to the area of D(, q). G (P ) is equal to the Gabriel grah, GG(P ). We rove that G (P ), and consequently GG(P ), has a strong diametral-disk matching of size at least n 7. We run Algorithm on G (P ) to comute a matching M. By Theorem, M is a strong matching of size at least n Inf(T ), where T is a minimum sanning tree in G (P ). By Lemma, T is a minimum sanning tree of the comlete grah K (P ). Observe that T is a Euclidean minimum sanning tree for P as well. In order to rove the desired lower bound, we show that Inf(T ) 7. Since Inf(T ) is the maximum size of a set among the influence sets of edges in T, it suffices to show that for every edge e in T, the influence set of e contains at most 7 edges. Lemma. Let T be a minimum sanning tree of G (P ), and let e be any edge in T. Then, Inf(e) 7. We will rove this lemma in the rest of this section. Recall that, for each two oints, q P, D(, q) is the closed diametral-disk with diameter q. Let D denote the set of diametral-disks reresenting the edges in T. Since T is a subgrah of G (P ), we have the following observation: Observation. Each disk in D does not contain any oint of P in its interior. Lemma. For each air D i and D j of disks in D, D i does not contain the center of D j. Proof. Let (a i, b i ) and (a j, b j ) be the edges of T that corresond to D i and D j, resectively. Let c i and c j be the centers of D i and D j, resectively. Let C i and C j be the circles reresenting the boundaries of D i and D j, resectively. Without loss of generality assume that C j is the bigger circle, i.e., a i b i < a j b j. By contradiction, suose that C j contains the center c i of C i. Let x and y denote the intersections of C i and C j. Let x i (res. x j ) be the intersection of C i (res. C j ) with the line through y and c i (res. c j ). Similarly, let y i (res. y j ) be the intersection of C i (res. C j ) with the line through x and c i (res. c j ). As illustrated in Figure, the arcs x i x, ŷ i y, x j x, and ŷ j y are the otential ositions for the oints a i, b i, a j, and b j, resectively. First we will show that the line segment x i x j asses through x and a i a j x i x j. The angles x i xy and x j xy are right angles, thus the line segment x i x j 7

8 a j x i a i x x j C i c i c j C j y i b i y b j y j Figure : Illustration of Lemma : C i and C j intersect, and C j contains the center of C i. goes through x. Since x i x < π (res. x j x < π), for any oint a i x i x, a i x x i x (res. a j x j x, a j x x j x ). Therefore, a i a j a i x + xa j x i x + xx j = x i x j. Consider triangle x i x j y, which is artitioned by segment c i x j into t = x i x j c i and t = c i x j y. It is easy to see that x i c i in t is equal to c i y in t, and the segment c i x j is shared by t and t. Since c i is inside C j and ŷx j = π, the angle yc i x j is greater than π. Thus, x ic i x j in t is smaller than π (and hence smaller than yc ix j in t ). That is, x i x j in t is smaller than x j y in t. Therefore, a i a j x i x j < x j y = a j b j. By symmetry b i b j < a j b j. Therefore max{ a i a j, b i b j } < max{ a i b i, a j b j }. Therefore, the cycle a i, a j, b j, b i, a i contradicts Lemma, that is, not both (a i, b i ) and (a j, b j ) can be edges of T. Let e = (u, v) be an edge in T. Without loss of generality, we suose that D(u, v) has radius and is centered at the origin o = (0, 0) such that u = (, 0) and v = (, 0). For any oint in the lane, let denote the distance of from o. Let D(e + ) be the disks in D reresenting the edges of T (e + ). Recall that T (e + ) contains the edges of T whose weight is at least w(e), where w(e) is equal to the area of D(u, v). Since the area of any circle is directly related to its radius, we have the following observation: Observation 5. The disks in D(e + ) have radius at least. Let C(x, r) (res. D(x, r)) be the circle (res. closed disk) of radius r centered at oint x in the lane. Let I(e + ) = {D,..., D k } be the set of disks in D(e + ) \ {D(u, v)} intersecting D(u, v). We show that I(e + ) contains at most sixteen disks, i.e., k 6. For i {,..., k}, let c i denote the center of the disk D i. In addition, let c i be the intersection oint between C(o, ) and the ray that starts in o and asses through c i. Let the oint i be c i, if c i <, and c i, otherwise. See Figure 5. Finally, let P = {o, u, v,,..., k }. Observation 6. Let c j be the center of a disk D j in I(e + ), where c j. Then, D( j, ) D(c j, c j ) D j. See Figure 5. 8

9 c i c k k c k α i j c j c j - u o v D j D(u, v) C(o, ) Figure 5: Proof of Lemma 5; i = c i, j = c j, and k = c k. Lemma 5. The distance between any air of oints in P is at least. Proof. Let x and y be two oints in P. We are going to rove that xy. We distinguish between the following three cases. x, y {o, u, v}. In this case the claim is trivial. x {o, u, v}, y {,..., k }. If y =, then y is on C(o, ), and hence xy. If y <, then y is the center of a disk D i in I(e + ). By Observation, D i does not contain u and v, and by Lemma, D i does not contain o. Since D i has radius at least, we conclude that xy. x, y {,..., k }. Without loss of generality assume x = i and y = j, where i < j k. We differentiate between three subcases: i < and j <. In this case i and j are the centers of D i and D j, resectively. By Lemma and Observation 5, we conclude that i j. i < and j =. By Observation 6 the disk D( j, ) is contained in the disk D j. By Lemma, i is not in the interior of D j, and consequently, it is not in the interior of D( j, ). Therefore, i j. i = and j =. Recall that c i and c j are the centers of D i and D j, resectively, and that c i and c j. Without loss of generality, assume that c i c j. For the sake of contradiction assume that i j <. Then, for the angle α = c i oc j we have sin(α/) <. Then, cos(α) > sin (α/) = 7 8. By the law of cosines in the triangle c i oc j, we have c i c j < c i + c j 8 c i c j. () By Observation 6, the disk D(c j, c j ) is contained in D j ; see Figure 5. By Lemma, c i is not in the interior of D j, and consequently, c i is not in the interior of D(c j, c j ). Thus, c i c j c j. In combination with Inequality (), this gives ( ) c j 8 c i < c i. () 9

10 In combination with the assumtion that c i c j, Inequality () gives c i c i + < 0. To satisfy this inequality, we should have c i <, contradicting the fact that c i. This comletes the roof. By Lemma 5, the oints in P have mutual distance. Moreover, the oints in P lie in D(o, ). Bateman and Erdős [] roved that it is imossible to have 0 oints in a closed disk of radius such that one of the oints is at the center and all of the mutual distances are at least. Therefore, P contains at most 9 oints, including o, u, and v. This imlies that k 6, and hence I(e + ) contains at most sixteen edges. This comletes the roof of Lemma. Theorem. Algorithm comutes a strong matching of size at least n 7 in G (P ). 5 Strong Matching in G (P ) In this section we consider the case where S is a downward equilateral triangle whose barycenter is the origin and one of its vertices is on the negative y-axis. In this section we assume that P is in general osition, i.e., for each oint P, there is no oint of P \ {} on l, 0 l 60, and. In combination with Observation, this imlies that for two oints, q P, no oint of P \ {, q} is on the boundary of t(, q) (res. t (, q)). Recall that t(, q) is the smallest homothet of having one of and q on a corner and the other oint on the side oosite to that corner. We rove that G (P ), and consequently Θ 6(P ), has a strong triangle matching l 0 of size at least n 9. We run Algorithm on G (P ) to comute a matching M. Recall that G (P ) is an edgeweighted grah where the weight of each edge (, q) is equal to the area of t(, q). By Theorem, M is a strong matching of size at least n Inf(T ), where T is a minimum sanning tree in G (P ). In order to rove the desired lower bound, we show that Inf(T ) 9. Since Inf(T ) is the maximum size of a set among the influence sets of edges in T, it suffices to show that for every edge e in T, the influence set of e has at most nine edges. Lemma 6. Let T be a minimum sanning tree of G (P ), and let e be any edge in T. Then, Inf(e) 9. v s s v v t s s s v v s v t (s ) t (a) (b) (c) t t t (v ) Figure 6: (a) Labeling the vertices and the sides of a downward triangle. vertices and the sides of an uward triangle. (c) Two intersecting triangles. (b) Labeling the We will rove this lemma in the rest of this section. We label the vertices and the sides of a downward equilateral-triangle, t, and an uward equilateral-triangle, t, as deicted in 0

11 Figures 6(a) and 6(b). We refer to a vertex v i and a side s i of a triangle t by t(v i ) and t(s i ), resectively. Recall that F is a subgrah of the minimum sanning tree T in G (P ). In each iteration of the while loo in Algorithm, let T denote the set of triangles reresenting the edges in F. By Lemma and the general osition assumtion we have Observation 7. Let t(, q) be a triangle in T. P \ {, q} in its interior or on its boundary. Then t(, q) does not contain any oint of Consider two intersecting triangles t (, q ) and t (, q ) in T. By Observation, each side of t contains either or q, and each side of t contains either or q. Thus, by Observation 7, we argue that no side of t is comletely in the interior of t, and vice versa. Therefore, either exactly one vertex (corner) of t is in the interior of t, or exactly one vertex of t is in the interior of t. Without loss of generality assume that a corner of t is in the interior of t, as shown in Figure 6(c). In this case we say that t intersects t through the vertex t (v ), or symmetrically, t intersects t through the side t (s ). The following two lemmas have been roved by Biniaz et al. [6] (see Figure 7(a)): Lemma 7 (Biniaz et al. [6]). Let t be a downward triangle that intersects a downward triangle t through t (s ), and let a horizontal line l intersect both t and t. Let and q be two oints on t (s ) and t (s ), resectively, that are above t (s ). Let and q be two oints on t (s ) and t (s ), resectively, that are above l. Then, max{t(, ), t(q, q )} max{t, t }. Lemma 8 (Biniaz et al. [6]). For every four triangles t, t, t, t T, t t t t =. As a consequence of Lemma 7, we have the following corollary (see Figure 7(a)): Corollary. Let t, t, t be three triangles in T. Then t, t, and t cannot make a chain configuration such that t intersects t through t (s ), and t intersects both t and t through t (s ) and t (s ). t t q t t l t q t (s ) q t (s ) t (a) (b) Figure 7: (a) Illustration of Lemma 7. (b) Illustration of Lemma 9. For the following lemma refer to Figure 7(b). Lemma 9. Let t be a downward triangle that intersects a downward triangle t through t (v ). Let be a oint on t (s ) and to the left of t (s ), and let q be a oint on t (s ) and to the right of t (s ). Then, t(, q) max{t, t }. Proof. Let t (s ) be the art of the line segment t (s ) that is to the left of t (s ), and let t (s ) be the art of the line segment t (s ) that is to the right of t (s ). Without loss of generality assume that t (s ) is larger than t (s ). Let t be an uward triangle having t (s ) as its left

12 side. Then, t t, which imlies that t max{t, t }. Since t has both and q on its boundary, the area of the downward triangle t(, q) is smaller than the area of t. Therefore, t(, q) t ; which comletes the roof. Because of symmetry, the statement of Lemma 9 holds even if is above t (s ) and q is on t (s ). Consider the six cones with aex at, as shown in Figure. Lemma 0. Let T be a minimum sanning tree in G (P ). Then, in T, every oint is adjacent to at most one oint in each cone C, i where i 6. Proof. If i is even, then by the construction of G (P ), which is given in Section, is adjacent to at most one oint in C i. So, assume that i is odd. For the sake of contradiction, assume that in T, the oint is adjacent to two oints q and r in the same cone C i. Then, t(, q) has q on a corner, and t(, r) has r on a corner. Without loss of generality, assume that t(, r) t(, q). Then, the hexagon X(q, ) has r in its interior. Thus, t(q, r) t(, q). Then the cycle r,, q, r contradicts Lemma. Therefore, is adjacent to at most one oint in each of the six cones. In Algorithm, in each iteration of the while loo, let T (e + ) be the set of triangles reresenting the edges of F. Recall that e is the smallest edge in F, and hence, t(e) is a smallest triangle in T (e + ). Let e = (, q) and let I(e + ) be the set of triangles in T (e + ) (excluding t(e)) that intersect t(e). We show that I(e + ) contains at most eight triangles. We artition the triangles in I(e + ) into I I such that every triangle τ I shares only or q with t = t(e) = t(, q), i.e., I = {τ : τ I(e + ), τ t {, q}}, and every triangle τ I intersects t either through a side or through a corner that is neither nor q. By Observation, for each triangle t(, q), one of C and q is a corner of t(, q) and the other one is on the side oosite to that corner. Without loss of generality, C assume that is on the corner t(v ), and hence, q is on t(s ) the side t(s ). See Figure 8. Note that the other cases, t(, q) where is on t(v ) or on t(v ), are similar. Let τ I t(s ) reresents an edge e in T. Since the intersection of t q C 6 with any triangle in I is either or q, τ has either or t(s ) q on its boundary. In combination with Observation 7, Cq t(s ) this imlies that, either or q is an endoint of e. As illustrated in Figure 8, the other endoint of e can be either in C, C, C, 6 or in Cq, because otherwise Figure 8: Illustration of the triangles τ t {, q}. By Lemma 0, has at most one neighbor in I. in each of C, C, C, 6 and q has at most one neighbor in Cq. Therefore, I contains at most four triangles. We are going to show that I also contains at most four triangles. The oint q divides t(s ) into two arts. Let t(s ) and t(s ) be the arts of t(s ) that are below and above q, resectively; see Figure 8. The triangles in I intersect t either through t(s ) t(s ) or through t(s ) t(s ); the two sets are shown by red and blue olylines in Figure 8. We show that at most two triangles in I intersect t through each of t(s ) t(s ) or t(s ) t(s ). Because of symmetry, we only rove this for t(s ) t(s ). When a triangle t intersects t through both t(s ) and t(s ), we say t intersects t through t(v ). In the next lemma, we rove that at most one triangle in I intersects t through each of t(s ), t(s ). Again, because of symmetry, we only rove this for t(s ). Lemma. At most one triangle in I intersects t through t(s ). Proof. The roof is by contradiction. Assume that two triangles t (, q ) and t (, q ) in I intersect t through t(s ). Without loss of generality, assume that i is on t i (s ) and q i is on

13 t q t (v ) t q q t q t (v ) t (v ) t t t t (v ) q t (v ) = q (b) (a) Figure 9: Illustration of Lemma : (a) t (v ) t. (b) t (v ) / t and t (v ) / t. t i (s ) for i =,. Recall that t t and t t. If t (v ) is in the interior of t (as shown in Figure 9(a)) or t (v ) is in the interior of t, then we get a contradiction to Corollary. Thus, we assume that t (v ) / t and t (v ) / t. Without loss of generality, assume that t (s ) is above t (s ); see Figure 9(b). By Lemma 9, we have t(, ) max{t, t } t. If q is in X(, q), then by Observation, t(, q ) t. Then, the cycle,, q, contradicts Lemma. Thus, assume that q / X(, q,). In this case t (s ) is to the left of t (s ), because otherwise q lies in t which contradicts Observation 7. Since both t and t are larger than t, t intersects t through t (s ), and hence t (v ) is in the interior of t. This imlies that q = t (v ). In addition, is on the art of t (s ) that lies in the interior of X(, q). By Observation and Lemma 9, we have t(, ) t and t(q, q ) max{t, t }, resectively. Thus, the cycle,, q, q,, contradicts Lemma. Lemma. At most two triangles in I intersect t through t(v ). Proof. For the sake of contradiction assume three triangles t, t, t I intersect t through t(v ). This imlies that t(v ) belongs to four triangles t, t, t, t, which contradicts Lemma 8. Lemma. If two triangles in I intersect t through t(v ), then no other triangle in I intersects t through t(s ) or through t(s ). Proof. The roof is by contradiction. Assume that two triangles t (, q ) and t (, q ) in I intersect t through t(v ), and a triangle t (, q ) in I intersects t through t(s ) or t(s ). Let i be the inut oint that lies on t i (s ) for i =,,. By Lemma, t cannot intersect both t(s ) and t(s ). Thus, t intersects t either through t(s ) or through t(s ). We rove the former case; the roof for the latter case is similar. Assume that t intersects t through t(s ). By Lemma 9, t(, ) t. See Figure 0. In addition, both t (s ) and t (s ) are to the left of t (s ), because otherwise q lies in t t X(, q). If q t t, we get a contradiction to Observation 7. If q X(, q) then by Observation, we have t(, q ) t, and hence, the cycle,, q, contradicts Lemma. Without loss of generality, assume that t (s ) is above t (s ); see Figure 0. If t (v ) t or t (v ) t, then we get a contradiction to Corollary. Thus, assume that t (v ) / t and t (v ) / t. This imlies that either (i) t (s ) is to the right of t (s ) or (ii) t (s ) is to the left of t (s ). We show that both cases lead to a contradiction. In case (i), lies in the interior of X(, q,), and then by Observation, we have t(, ) t; see Figure 0(a). In addition, Lemma 9 imlies that t(, q ) max{t, t } t. Thus, the cycle,, q,, contradicts Lemma.

14 q q t t q t t t c q t q q (a) (b) q q t q t t t q q t t q q τ q (c) (d) Figure 0: Illustration for the roof of Lemma : (a) is to the right of t (s ), (b) q C 5 t(v ), (c) q C 6 t(v ), and (d) q C t(v ). Now consider case (ii) where t (s ) is above t (s ) and t (s ) is to the left of t (s ). If is to the right of t, then as in case (i), the cycle,, q,, contradicts Lemma. Thus, assume that is to the left of t, as shown in Figure 0(b). By Lemma 9, we have t(q, ) max{t, t } t. Each side of t contains either or q, while is on the art of t (s ) that is to the left of t, thus, q is on t (s ). Consider the six cones around t(v ); see Figure 0(b). We have three cases: (a) q C 5 t(v ), (b) q C 6 t(v ) or (c) q C t(v ). In case (a), which is shown in Figure 0(b), by Lemma 7, we have max{t(, ), t(q, q )} max{t, t }. Thus, the cycle,, q, q, contradicts Lemma. In Case (b), which is shown in Figure 0(c), we have t(q, q ) t, because if we ma t to a downward triangle τ of area equal to the area of t that has τ(v ) on t(v ), then τ contains both q and q. Therefore, the cycle,, q, q,, q, contradicts Lemma. In Case (c), which is shown in Figure 0(d), by Observation, t(, q ) t, and then, the cycle, q,, q, contradicts Lemma. Lemma. If three triangles intersect t through t(s ), t(v ) and t(s ), then at least one of the three triangles is not in I. Proof. The roof is by contradiction. Assume that three triangles t (, q ), t (, q ), t (, q ) in I intersect t through t(s ), t(v ), t(s ), resectively. Let i be the oint that lies on t i (s ) for i =,,. See Figure (a). By Lemma 9, we have t(, ) t and t(q, ) t. If q is in the interior of X(, q), then by Observation, t(, q ) t, and hence, the cycle,, q, contradicts Lemma. If q is in X(q, ), then by Observation, t(q, q ) t, and hence, the cycle q, q,, q contradicts Lemma ; see Figure (b). Thus, assume that q / X(, q) and q / X(q, ). Let

15 q q t t t t t (s ) t (s ) q t q q t q (a) (b) Figure : Illustration for the roof of Lemma : (a) t (s ), and (b) t (s ). t (s ) and t (s ) be the arts of t (s ) that are to the right of t(s ) and to the left of t(s ), resectively. Consider the oint that lies on t (s ). If t (s ), then X(, q) and by Observation, t(, ) t. In addition, Lemma 9 imlies that t(, q ) t. Thus, the cycle,, q,, contradicts Lemma ; see Figure (a). If t (s ), then X(q, ) and by Observation, t(q, ) t. In addition, Lemma 9 imlies that t(, q ) t. Thus, the cycle q,, q,, q contradicts Lemma ; see Figure (b). Putting Lemmas,,, and together, imlies that at most two triangles in I intersect t through t(s ) t(s ), and consequently, at most two triangles in I intersect t through t(s ) t(s ). Thus, I contains at most four triangles. Recall that I contains at most four triangles. Then, I(e + ) contains at most eight triangles. Therefore, the influence set of e contains at most 9 edges (including e itself). This comletes the roof of Lemma 6. Theorem. Algorithm comutes a strong matching of size at least n 9 in G (P ). +ɛ +ɛ +ɛ +ɛ q +ɛ +ɛ +ɛ Figure : Four triangles in I (in red) and four triangles in I (in blue) intersect with t(, q). The bound obtained by Lemma 6 is tight. Figure shows a configuration of 0 oints in general osition such that the influence set of a minimal edge is 9. In Figure, t = t(, q) reresents a smallest edge of weight ; the minimum sanning tree is shown in bold-green line segments. The weight of all edges the area of the triangles reresenting these edges is at least. The red triangles are in I and share either or q with t. The blue triangles are in I and intersect t through t(s ) t(s ) or through t(s ) t(s ); as shown in Figure, two of them share only the oints t(v ) and t(v ). 5

16 6 Strong Matching in G (P ) In this section we consider the roblem of comuting a strong matching in G (P ), where is an axis-aligned square whose center is the origin. We assume that P is in general osition, i.e., (i) no two oints have the same x-coordinate or the same y-coordinate, and (ii) no four oints are on the boundary of any homothet of. Recall that G (P ) is equal to the L -Delaunay grah on P. Ábrego et al. [, ] roved that G (P ) has a strong matching of size at least n/5. Using a similar aroach as Ábrego et al. [, ], we rove that G (P ) has a strong matching of size at least n. Theorem. Let P be a set of n oints in general osition in the lane. Let S be an axis-arallel square that contains P. Then, it is ossible to find a strong matching of size at least n for G (P ) such that for each edge e in this matching, the square corresonding to e is in S. Proof. The roof is by induction. Assume that any oint set of size n n in an axisarallel square S has a strong matching of size at least n in S. If n is 0 or, then there is no matching in S, and if n {,,, 5}, then by shrinking S, it is ossible to find a strongly matched air. Now suose that n 6, and n = m + r, where r {0,,, }. If r {0,, }, then n = (n ), and by induction we are done. So we may assume that that n = m +, for some m. We rove that there are n = m + disjoint squares in S, each of them matching a air of oints in P. To this end we artition S into four equal area squares S, S, S, S that contain n, n, n, n oints, resectively; see Figure (a). Let n i = m i + r i for i, where r i {0,,, }. Let R be the multiset {r, r, r, r }. By induction, in S S S S, we have a strong matching of size at least n n n n A = () Claim : A m. Proof. By Equation (), we have A = ni i= i= n i Since A and m are integers, we argue that A m. = n = m + = m. If A > m, then we are done. Assume that A = m; in fact, by the induction hyothesis we have a strong matching of size at least m for P. In order to comlete the roof, we have to get one more strongly matched air. Let R be the multiset {r, r, r, r }. Claim : If A = m, then either (i) R = {,,, } or (ii) R = {0, 0,, }. Proof. Let α = r + r + r + r, where 0 r i. Then n = (m + m + m + m ) + α. Since n = m+, α = k+, for some 0 k. Thus, n = m+, where m = m +m +m +m +k. By induction, in S i, we get a matching of size at least (m i+r i ) = m i + r i. Hence, in S S S S, we get a matching of size at least r A = m + m + m + m + + r + r + r Since A = m and m = m + m + m + m + k, we have r r r r k = () 6.

17 Note that 0 k. We go through some case analysis: (i) k = 0, (ii) k =, (iii) k =. In case (i), we have α = k+ = r +r +r +r =. In order to have k equal to 0 in Equation (), no element in R can be greater than ; this haens only if two elements in R are equal to 0 and the other two elements are equal to. In case (ii), we have α = r + r + r + r = 6. In order to have k equal to in Equation (), at most one element in R should be greater than ; this haens only if three elements in R are equal to and the remaining element is equal to (note that all elements in R are less than ). In case (iii), we have α = r + r + r + r = 0. In order to have k equal to in Equation (), at most two elements in R should be greater than ; which is not ossible. In both cases of Claim we show how to augment a strong matching of size m by one more air such that the resulting matching is strong and has size m +. We define S -x as the smallest axis-arallel square contained in S and anchored at the toleft corner of S, that contains all the oints in S excet x oints. If S contains less than x oints, then the area of S -x is zero. We also define S +x as the smallest axis-arallel square that contains S and anchored at the to-left corner of S, that has all the oints in S lus x other oints of P. See Figure (a). Similarly we define the squares S -x, S +x, S -x, S +x, and S -x, S +x that are anchored at the to-right corner of S, the bottom-left corner of S, and the bottom-right corner of S, resectively. Case (i): R = {,,, }. In this case, we have m = m + m + m + m +. Without loss of generality, assume that r = and r = r = r =. Consider the squares S -, S -, S -, and S -. Note that the area of some of these squares but not all may be equal to zero. See Figure (b). By induction, we get matchings of sizes at least m +, m, m, and m, in S -, S -, S -, and S -, resectively. Now consider the largest square among S -, S -, S -, and S -. Because of symmetry, we have only three cases: (i) S - is the largest, (ii) S - is the largest, and (iii) S - is the largest. S - S - S - S - S - S l l S + S + l S + S S S - S - S - S - l (a) (b) (c) l l Figure : (a) Slit S into four equal area squares. (b) S - is larger than S -, S -, and S -. (c) S - is larger than S -, S -, and S -. S - is the largest square. Consider the lines l and l that contain the bottom side and right side of S -, resectively; see the dashed lines in Figure (b). Note that l and l and their mirrored versions l and l do not intersect any of S-, S -, and S -. If any oint of S is to the right of l, then by induction, we get a matching of size at least (m + ) + (m + ) + m + m in S - S + S - S -. Note that S + is searated from S - by l and from S - by l (since we assume that S- is the largest of the four squares). 7

18 Otherwise, by induction, we get a matching of size at least (m + ) + m + (m + ) + m in S - S - S + S -, which, again, is a disjoint union. In both cases we get a matching of size at least m + in S. S - is the largest square. Consider the lines l and l that contain the bottom side and left side of S -, resectively; the dashed lines in Figure (c). Note that l and l do not intersect any of S -, S -, and S -. If any oint of S is below l, then by induction, we get a matching of size at least (m + ) + m + m + (m + ) in S - S - S - S +. Otherwise, by induction, we get a matching of size at least (m + ) + m + m + m in S + S - S - S - ; see Figure (c). In all cases we get a matching of size at least m + in S. S - is the largest square. Consider the lines l and l that contain the to side and left side of S -, resectively. If any oint of S is above l, then by induction, we get a matching of size at least (m +)+(m +)+m +m in S - S + S - S -. Otherwise, by induction, we get a matching of size at least (m + ) + m + (m + ) + m in S - S - S + S -. In all cases we get a matching of size at least m + in S. Case (ii): R = {0, 0,, }. In this case, we have m = m + m + m + m. Due to symmetry, only the following two cases may arise: r = r = and r = r = 0. Consider the squares S -, S -, S -, and S -. By induction, we, resectively.. Because of symmetry, we is the largest. In case (a) we get one. In case (b) we get one more matched air get matchings of sizes at least m, m, m, and m, in S -, S -, S -, and S - Now consider the largest square among S -, S -, S -, and S - have only two cases: (a) S - is the largest, (b) S - more matched air either in S + or in S + either in S + or in S +. r = r = and r = r = 0. Consider the squares S -, S -, S -, and S -. By induction, we, resectively.. Because of symmetry, we is the largest. In case (a) we get one. In case (b) we get one more matched air get matchings of sizes at least m, m, m, and m, in S -, S -, S -, and S - Now consider the largest square among S -, S -, S -, and S - have only two cases: (a) S - is the largest, (b) S - more matched air either in S + or in S + either in S + or in S +. 7 Strong Matching in G (P ) In this section we consider the roblem of comuting a strong matching in G (P ). Recall that G (P ) is the union of G (P ) and G (P ), and is equal to the grah Θ 6 (P ). We assume that P is in general osition, i.e., for each oint P, there is no oint of P \ {} on l, 0 l 60, and. A matching M in G (P ) is a strong matching if for each edge e in M there is a homothet of or a homothet of reresenting e such that these homothets are airwise disjoint. See Figure (b). Using a similar aroach as in Section 6, we rove the following theorem: l 0 Theorem 5. Let P be a set of n oints in general osition in the lane. Let S be an uward or a downward equilateral-triangle that contains P. Then, it is ossible to find a strong matching of size at least n for G (P ) such that for each edge e in this matching, the triangle corresonding to e is in S. 8

19 Proof. The roof is by induction. Assume that any oint set of size n n in a triangle S has a strong matching of size at least n in S. Without loss of generality, assume that S is an uward equilateral-triangle. If n is 0 or, then there is no matching in S, and if n {,,, 5}, then by shrinking S, it is ossible to find a strongly matched air; the statement of the theorem holds. Now suose that n 6, and n = m + r, where r {0,,, }. If r {0,, }, then n = (n ), and by induction we are done. So we may assume that n = m +, for some m. We rove that there are n = m + disjoint equilateral-triangles (uward or downward) in S, each of them matching a air of oints in P. To this end we artition S into four equal area equilateral triangles S, S, S, S containing n, n, n, n oints, resectively; see Figure (a). Let n i = m i + r i, where r i {0,,, }. By induction, in S S S S, we have a strong matching of size at least n n n n A = In the roof of Theorem, we have shown the following two claims: Claim : A m. Claim : If A = m, then either (i) R = {,,, } or (ii) R = {0, 0,, }. If A > m, then we are done. Assume that A = m; in fact, by the induction hyothesis we have an strong matching of size at least m in S. By Claim we have two cases. In both cases of Claim we show how to augment a strong matching of size m by one more air such that the resulting matching is strong and has size m +. We show how to find one more strongly matched air in each case of Claim. We define S -x as the smallest uward equilateral-triangle contained in S and anchored at the to corner of S, that contains all the oints in S excet x oints. If S contains less than x oints, then the area of S -x is zero. We also define S +x as the smallest uward equilateraltriangle that contains S and anchored at the to corner of S, that has all the oints in S lus x other oints of P. Similarly we define uward triangles S -x and S +x that are anchored at the left corner of S. Moreover, we define uward triangles S -x and S +x that are anchored at the right corner of S. We define downward triangles S -x l, S -x r, S -x b that are anchored at the to-left corner, to-right corner, and bottom corner of S, resectively. See Figure (a). Case (i): R = {,,, }. In this case, we have m = m + m + m + m +. Because of symmetry, we have two cases: (i) r =, (ii) r j = for some j {,, }. r =. In this case n = m +. We differentiate between two cases: the case that all the elements of the multiset {m, m, m } are equal to zero, and the case that some of them are greater than zero. All elements of {m, m, m } are equal zero. In this case, we have m = m +. Consider the triangles S + and Sr. - See Figure (a). Note that S + and S - r are disjoint, S + contains two oints, and S - r contains m + oints. By induction, we get a matched air in S + and a matching of size at least m + in Sr. - Thus, in total, we get a matching of size at least + (m + ) = m + in S. 9