On Maximum-Sum Matchings of Points

Huemer et al. (Discrete Mathematics, 2019) proved that for any two point sets $R$ and $B$ with $|R|=|B|$, the perfect matching that matches points of $R$ with points of $B$, and maximizes the total \emph{squared} Euclidean distance of the matched pairs, verifies that all the disks induced by the matching have a common point. Each pair of matched points $p\in R$ and $q\in B$ induces the disk of smallest diameter that covers $p$ and $q$. Following this research line, in this paper we consider the perfect matching that maximizes the total Euclidean distance. First, we prove that this new matching for $R$ and $B$ does not always ensure the common intersection property of the disks. Second, we extend the study of this new matching for sets of $2n$ uncolored points in the plane, where a matching is just a partition of the points into $n$ pairs. As the main result, we prove that in this case all disks of the matching do have a common point. This implies a big improvement on a conjecture of Andy Fingerhut in 1995, about a maximum matching of $2n$ points in the plane.


Introduction
Let R and B be two disjoint point sets in the plane with |R| = |B| = n, n ≥ 2. The points in R are red, and those in B are blue. A matching of R ∪ B is a partition of R ∪ B into n pairs such that each pair consists of a red and a blue point. A point p ∈ R and a point q ∈ B are matched if and only if the (unordered) pair (p, q) is in the matching. For every p, q ∈ R 2 , we use pq to denote the segment connecting p and q, and p − q to denote its length, which is the Euclidean norm of the vector p − q. Let D pq denote the disk with diameter equal to p − q , that is centered at the midpoint p+q 2 of the segment pq. For any matching M, we use D M to denote the set of the disks associated with the matching, that is, D M = {D pq | (p, q) ∈ M}.
Huemer et al. [14] proved that if M is any matching that maximizes the total squared Euclidean distance of the matched points, i.e., it maximizes (p,q)∈M p − q 2 , then all disks of D M have a point in common.
In this paper, we will consider the max-sum matching E, as the matching that maximizes the total Euclidean distance of the matched points. For any matching M, let cost(M) denote the sum (p,q)∈M p − q . Thus, E is such that cost(E) is maximum among all matchings. In Section 2, we prove that every pair of disks in E have a common point, but it cannot be guaranteed that all disks have a common point.
In this paper, we will also consider max-sum matchings of sets of 2n uncolored points in the plane, where a matching is just a partition of the points into n pairs. Then, we study the following problem: Fingerhut [12] conjectured that given a set P of 2n uncolored points in the plane and a max-sum matching {(a i , b i ), i = 1, . . . , n} of P , there exists a point o of the plane, not necessarily a point of P , such that The statement of equation (1) is equivalent to stating that the intersection E 1 ∩ E 2 ∩ · · · ∩ E n is not empty, where E i is the region bounded by the ellipse with foci a i and b i , and semimajor axis length (1/ √ 3) · a i − b i , for all i ∈ {1, . . . , n} [12]. Then, by Helly's Theorem, it is sufficient to prove equation (1) for n = 3. As noted by Andy Fingerhut, the factor 2/ √ 3 is the minimum possible. It is enough to consider an equilateral triangle, where at each vertex two points are located. The max-sum matching of the six points is made of pairs of vertex-opposed points, and the regions defined by the three ellipses using the factor 2/ √ 3 have exactly one point in common.
Andy Fingerhut was also interested in a small constant close to 2/ √ 3, and Eppstein [12] proved that the result holds with 2.5 instead of 2/ √ 3. The proof is simple and does not require Helly's theorem. Let o be the midpoint of the shortest edge in the matching. Namely, for all i ∈ {1, . . . , n}, In Section 3, we first show that the constant in this inequality can be improved to √ 5 ≈ 2.236 by using the same point o, but refined arguments. Second, as the main result of this paper, we improve even further this constant. Precisely, we prove that for any point set P of 2n uncolored points in the plane and a max-sum matching E = {(a i , b i ), i = 1, . . . , n} of P , all disks in D E have a common intersection. This directly implies that any point o in the common intersection satisfies is the Euclidean distance from o to conv(P i ), and diam(P ) is the diameter of P . Our result can improve the bound to d(o, conv(P i )) ≤ 1 2 diam(P i ) ≤ 1 2 diam(P ) for d = 2 and n = 2k (using the max-sum matching for P and taking o in the intersection of the disks). For d = 2 and arbitrary values of k, another bound of d(o, conv(P i )) ≤ 1 √ 3 diam(P ) is implied by Jung's theorem [15] (taking as o the center of the smallest disk enclosing P ).

Related problems
As described by Huemer et al. [14], this class of problems is well studied in discrete and computational geometry, starting from the classic result that n red points and n blue points can always be perfectly matched with n pairwise non-crossing segments, where each segment connects a red point with a blue point [16]. The study has been continued in plenty of directions, for both the monochromatic and bichromatic versions, by using pairwise disjoint objects inducing the matching: segments [4,11], rectangles [1,2,6,7,9], and more general geometric objects [5].
Our results, as that of [14], are in the direction opposite to that of above mentioned known results on matching points: The goal is that all matching objects (i.e., the disks in D M ) have a common intersection, whereas in previous work it is required that all matching objects are pairwise disjoint.
The problem of stabbing a finite set of pairwise intersecting disks with four or five points [8,10,13], is also related with this paper, since we deal with finite sets of pairwise intersecting disks.
Particularly, the conjecture posted by Fingerhut [12] relates to an abstract version of a problem in designing communication networks. The given 2n points represent nodes that should be connected by a network. The cost cost(E) of a max-sum matching E is a lower bound on the cost of any feasible network, and the point o is a place where one would like to place the center of a "star" network, in which all given points are connected directly to the center of the star.

Red-blue matchings
Let R and B be two disjoint point sets defined as above, where |R| = |B| = n, n ≥ 2. In this section, we prove that the perfect matching E of R and B that maximizes the total Euclidean distance cost(E) does not always ensure the common intersection property of the disks in D E . Nevertheless, although the common intersection is not always possible, all disks must be pairwise intersecting, as proved in the next proposition.

Proposition 1. Every pair of disks in D E have a non-empty intersection.
Proof. Let (a, a ) and (b, b ) be two different pairs of E, with a, b ∈ R and a , b ∈ B. Since E is a maximum matching, we must have

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On Maximum-Sum Matchings of Points of their radii. The following equations ensure this condition: Hence, we conclude that D aa ∩ D bb = ∅ for every pair of disks D aa and D bb of D E .

Theorem 2.
There exist point sets R ∪ B, with R ∩ B = ∅ and |R| = |B| = 3, such that, for any max-sum matching E of R and B, the intersection of the disks of D E is the empty set. Figure 1).
Note that and we need to ensure that

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That is, the matching {(a, a ), (b, b ), (c, c )} has larger total Euclidean distance than the matchings Furthermore, since Hence, any ε > 0 satisfying (4) (and then also (3) To see that, it is straightforward to show (based on the fact that c / ∈ D aa and c / ∈ D bb ) that all points of D aa ∩ D cc have negative x-coordinates, and all points of D bb ∩ D cc have positive x-coordinates.
Let |R| = |B| = n. If n = 2, then the disks of D E always intersect by Proposition 1. If n = 3, then the disks of D E may not intersect by Theorem 2. One could expect that the disks of D E always intersect if n is sufficiently large. Next, we answer this question negatively. Proof. We construct a set R of n red points, and a set B of n blue points, as follows. First, take six points a, b, c ∈ R and a , b , c ∈ B as in Theorem 2. Second, we add n − 3 red points and n − 3 blue points in the ε-neighborhood of point c, as explained below (see Figure 2, a zoomed-in view of Figure 1), where ε > 0 is a sufficiently small number that will be specified later.
Add n − 3 blue points, denoted a 1 , a 2 , . . . , a n−3 , on the segment b a . Note that they are within distance ε from c since c − b = c − a = ε. Add n − 3 red points, denoted a 1 , a 2 , . . . , a n−3 , on the horizontal line through c (perpendicular to cc ), and within distance ε from c (see Furthermore, the members of the remaining n − 2 pairs of M 2 are within distance ε from c. Thus, cost(M 2 ) ≤ √ 10 + 2 + ε + 2(n − 2)ε. We choose ε > 0 such that or what is the same,

On Fingerhut's Conjecture
In this section, we first show that the constant in inequality (2) can be improved to √ 5 ≈ 2.236. We start by proving the next technical lemma: Lemma 4. Let p and q be two points of the plane, and let r pq be the radius of disk D pq . Let D be a second disk with center o and radius r ≤ r pq such that D ∩ D pq = ∅. Then, S. Bereg et al.

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Proof. Let o pq denote the center of D pq , and note that o − o pq ≤ r + r pq ≤ 2 · r pq = p − q , because D ∩ D pq = ∅. Then, we have that: which implies the lemma.
Lemma 4 and Proposition 1 imply the following theorem.
Note that the bound of √ 5 is tight if the point o is always the midpoint of the shortest segment. Namely, consider 2 red points and 2 blue points, as vertices of a square, such that diagonal-opposed vertices have the same color.
We now strengthen Theorem 5. Precisely, we prove that for any point set P of 2n uncolored points in the plane and a max-sum matching M = {(a i , b i ), i = 1, . . . , n} of P , all disks in D M have a common intersection. We will use Helly's theorem, that is, we will prove the claim for n = 3.
We note that P might contain different points with the same coordinates. Furthermore, the common intersection might be a singleton. Namely, consider six points a, b, c, a , b , and c ,  where a,  Let p, q, and r be three points of the plane. We denote by ∆pqr the triangle with vertices p, q, and r; by (p, q) the straight line through p and q oriented from p to q; by τ (p, q) the ray with apex p that goes through q; by pq the segment pq oriented from p to q; and by C pq the circle bounding D pq . Furthermore, given a fourth point s, we say that pq points to rs if q is in the interior of ∆prs ∩ D rs . See Figure 3 (left), where segment cd points to ab.
The first observation that we state is that if M is a max-sum matching, then the disks of D M intersect pairwise, as a consequence of Proposition 1. Second, every pair of segments of M either cross or one oriented segment points to the other one, as formally proved in the following lemma. Third, the four vertices of any two segments cannot be in convex position, because the matching would not be max-sum by the triangle inequality.
} is a max-sum matching, the equation can be extended to is smaller than the radius a − m = a − b /2 of D ab , whose center is m. These observations imply the result.
As mentioned above, we will elaborate a proof for n = 3, that is, matchings of three segments. Since every pair of segments of a max-sum matching cross, or one of them points to the other one, we distinguish ten cases of relative position (or order-type) of the three segments, as shown in Figure 4, enumerated from (A) to (J). In the rest of this section, we will first show with direct proofs that in each of the cases from (A) to (G) the three disks have a common point. After that, we will use a proof by contradiction for each of the cases from (H) to (J).
The general idea of these latter proofs by contradiction is the following: Since the disks pairwise intersect, but they do not have a common intersection, we can extend one of the segments (by moving one of its vertices) such that the new three disks have a singleton common intersection. In this new setting, the three new segments must fall again in one of the cases from (H) to (J). Then, we show that the new segments must correspond to a max-sum matching given that the original ones do, but contradictorily such a new matching is not max-sum. ∈ D bb and b / ∈ D cc , then the triangle ∆c bu is such that the interior angles at c and b, respectively, are both acute. Hence, the altitude h from vertex u goes through the interior of ∆c bu, and let v ∈ c b be the other vertex of h. We have v ∈ D bb ∩ D cc , by Thales' theorem. Furthermore, since c , b ∈ D aa , we have that segment Cases (E), (F), and (G): In each of these cases, the same oriented segment points to each of the other two ones: Say, segment aa points to both bb and cc . Then, we have that a ∈ D bb ∩ D cc , by Lemma 6. Hence, a ∈ D aa ∩ D bb ∩ D cc . This completes the proof.
We now prove several technical lemmas, which will be used for the cases from (H) to (J). The following lemma guarantee that if we extend one segment by moving one of the points, then the resulting segments correspond to a max-sum matching of the resulting point set. The rest of the lemmas impose monotone properties, and situations in which  {(a, a ), (b, b ), (c, c )} is not a max-sum matching of point set {a, b, c, a , b , c }, all of them associated with cases from (H) to (J).
Hence, the lemma follows.

Lemma 9.
Let p, p , q, and q be four points such that pp points to qq , and q is to the right of (p, p ). Let z be a point to the left of both (p, p ) and (q, q ) such that: (i) q is to the left of (z, p); (ii) vectors p − z and p − z are orthogonal, and vectors q − z and q − z are also orthogonal. Refer to Figure 6a. Then, we have that Proof. Rearranging terms in equation (5), we need to prove that Conditions (i) and (ii), the fact that pp points to qq , and the location of z, imply that q is to the right of (z, p ) if and only if segments pq and qz have a common point. Suppose that q is to the right of (z, p ) (see Figure 6b), case where segments pq and qz have a common point. Then, p, q, q , and z are the vertices of a convex quadrilateral with non-empty interior and diagonals pq and zq. Hence, Equation (6) holds by the triangle inequality.
Suppose now that q is not to the right of (z, p ) (see Figure 6c), then segments pq and qz do not intersect. Let z be the reflection of z about the center of segment qq , also the center of D qq . Then, we have q − z = q − z and q − q = z − z , by condition (ii) and Thales' theorem. The fact that p belongs to the interior of D qq given that pp points to qq , implies that z must be to the right of line (p, z). Then, since pq and qz do not intersect, we have that z belongs to triangle ∆pz q . This implies Proof. We divide the proof into two cases: p − p ≥ q − p ; and p − p < q − p . In both cases, let α be the arc of the hyperbola with foci p and q that goes through p .
In the first case (see Figure 8a), let z be the intersection point of C pp and pq . Note that p − p ≥ q − p implies that the region H(α, q ) is convex. Furthermore, line (p , z ) is perpendicular to the line (p, q ) through the foci of α, and this implies that z belongs to H(α, q ), and then z belongs to the convex intersection H( (p , z ), q ) ∩ H(α, q ). Given that z belongs to the interior of α ∪ H(α, q ), p is on the boundary of α ∪ H(α, q ), and that q ∈ zp , q = p , we also have that q belongs to H(α, q ). This last fact, equivalent to q / ∈ α ∪ H(α, p), implies q − q − p − q < q − p − p − p , by Proposition 13, hence the result.
Consider now the second case, p − p < q − p (see Figure 8b). Let β be the bisector of the interior angle at p of triangle ∆op q . By well-known properties of hyperbolas, β is the tangent of α at p . Furthermore, p − p < q − p implies that β separates α from vertex q . Let z be the intersection point between oq and β. We use the following claim to show that o − z < o − p : If A, B, and C are the vertices of a triangle, and point E belongs to side AB, such that line (C, E) is the bisector of the interior angle at C, then C − B > B − E . The claim follows from the fact that in any triangle, precisely in ∆BCE, larger sides correspond to larger opposed interior angles (see Figure 8c). Applying the claim to ∆op q , we have o − z < o − p , which implies that β separates point z and arc α. Furthermore, β separates point q and α because q ∈ zp \ {p }. This last fact shows that q is to left of α in the direction from q to p (i.e. q / ∈ α ∪ H(α, p)), implying q − q − p − q < q − p − p − p , by Proposition 13. The results thus follows. a, b, c, a , b , c , and  Proof. We can apply Lemma 9 for b, b , c, c , and z (where b and c play the role of p and q, respectively); and for c, c , a, a , and z (where c and a play the role of p and q, respectively). Hence, we obtain the following two inequalities:

Lemma 15. Let
Let o denote the midpoint of segment aa , also the center of D aa . Given that z − c and z − c are orthogonal, and cc points to aa , we have that c is to the left of line (o, z). Note that if c is to the right of, or in, (o, z), then c would not belong to the interior of D aa . Similarly, since z − b and z − b are orthogonal, and b b points to aa , we have that b is to the right of line (o, z). Then, since bb points to cc , we have that rays τ (b, b ) and τ (o, z) must intersect. If point b belongs to triangle ∆aa z (see Figure 9b), then the facts that z − b and z − b are orthogonal, and τ (b, b ) ∩ τ (o, z) = ∅ imply that segments bz and ab have a common point.
Adding the three inequalities above, we obtain again inequality (7), implying the result.
In the contrary case, b does not belong to ∆aa z, (see Figure 9a), let us assume by contradiction that the matching {(a, a ), (b, b ), (c, c )} is max-sum. Then, the matching {(a, a ), (b, b )} is also max-sum. Let w and w be the intersection points of (b, b ) with τ (o, z) and za , respectively. Note that bb ⊂ ww , and by Lemma 8, {(a, a ), (w, w )} is a max-sum matching of {a, a , w, w }. But we can apply Lemma 14, where a and w play the role of p and q, respectively, to have that {(a, a ), (w, w )} is not max-sum, which implies the result. We are now ready to give the proofs to the missing cases, those from (H) to (J).

Lemma 16. If the segments of a max-sum matching of six points fall in one of the cases from (H) to (J), then the three disks of the matching have a common intersection.
Proof. Suppose by contradiction that the three disks, denoted D 1 , D 2 , and D 3 , intersect pairwise, but without a common intersection (see Figure 10). Let u 1,2 , u 2,3 , and u 3,1 be the vertices of the pairwise disjoint lenses D 1 ∩ D 2 , D 2 ∩ D 3 , and D 3 ∩ D 1 , respectively, located in the triangle with vertices the centers of D 1 , D 2 , and D 3 , respectively.
The idea is to use Lemma 8, in combination with Lemmas 11, 12, and 15, such that the point z of these lemmas is among u 1,2 , u 2,3 , and u 3,1 . To this end, we need to guarantee that point z is not an extreme point of some segment of the matching. This is done in the next paragraph.
Note that two vertices among u 1,2 , u 2,3 , and u 3,1 cannot be the extreme points of a same segment of the matching. Furthermore, if each of the three vertices is an extreme point of some segment of the matching, then at least one pair of disjoint segments violates Lemma 6. That is, the extreme point of one segment, in the interior of the convex hull of the four involved points, is not in the interior of the disk corresponding to the other segment. Hence, we can assume that at least one vertex among u 1,2 , u 2,3 , and u 3,1 is not an extreme point of a segment of the matching: say vertex u 1,2 . This implies that we can extend the segment of disk D 3 by moving one of its extreme points such that the new three matching disks have a singleton common intersection at u 1,2 . Let z = u 1,2 , where z is distinct from all the new six points.
Let the new six points be denoted as a, b, c, a , b , and c , in such a way that the new segments are precisely aa , bb , and cc , and for each u ∈ {a, b, c} point z is to the left of line (u, u ). By Lemma 8, {(a, a ), (b, b ), (c, c )} is a max-sum matching of {a, b, c, a , b , c }. It is important to note the following: If the original segments are in Case (H), then by extending one segment we can stay in Case (H), or go to Case (I). If the original segments are in Case (I), then by extending one segment we can stay in Case (I), or go to Case (C) with a non-singleton common intersection of the three disks. Otherwise, if the original segments are in Case (J), then by extending one segment we can stay in Case (J), or go to Case (B) or (D) with a non-singleton common intersection of the three disks. Hence, since the common intersection of the new three disks D aa , D bb , and D cc is singleton, we can