On Compatible Matchings
Oswin Aichholzer, Alan Arroyo, Zuzana Masárová, Irene Parada, Daniel Perz, Alexander Pilz, Josef Tkadlec, Birgit Vogtenhuber
OOn Compatible Matchings (cid:63)
Oswin Aichholzer , Alan Arroyo , Zuzana Masárová , Irene Parada ,Daniel Perz , Alexander Pilz , Josef Tkadlec , and Birgit Vogtenhuber Institute of Software Technology, Graz University of Technology, Austria {oaich,daperz,apilz,bvogt}@ist.tugraz.at IST Austria {alanmarcelo.arroyoguevara,zuzana.masarova,josef.tkadlec}@ist.ac.at TU Eindhoven, The Netherlands [email protected]
Abstract.
A matching is compatible to two or more labeled point setsof size n with labels { , . . . , n } if its straight-line drawing on each ofthese point sets is crossing-free. We study the maximum number of edgesin a matching compatible to two or more labeled point sets in generalposition in the plane. We show that for any two labeled convex setsof n points there exists a compatible matching with (cid:98)√ n (cid:99) edges. Moregenerally, for any (cid:96) labeled point sets we construct compatible matchingsof size Ω ( n /(cid:96) ) . As a corresponding upper bound, we use probabilisticarguments to show that for any (cid:96) given sets of n points there existsa labeling of each set such that the largest compatible matching has O ( n / ( (cid:96) +1) ) edges. Finally, we show that Θ (log n ) copies of any set of n points are necessary and sufficient for the existence of a labeling suchthat any compatible matching consists only of a single edge. Keywords: compatible graphs · crossing-free matchings · geometric graphs
For plane drawings of geometric graphs, the term compatible is used in two ratherdifferent interpretations. In the first variant, two plane drawings of geometricgraphs are embedded on the same set P of points. They are called compatible(to each other with respect to P ) if their union is plane (see e.g. [3,19]). Notethat this is different to simultaneous planar graph embedding, as it is requirednot only that the two graphs are plane, but also that their union is crossing-free.In the second setting, which is the one that we will consider in this work, one planar graph G is drawn straight-line on two or more labeled point sets (with (cid:63) A.A. funded by the Marie Skłodowska-Curie grant agreement No 754411. Z.M. par-tially funded by Wittgenstein Prize, Austrian Science Fund (FWF), grant no. Z342-N31. I.P., D.P., and B.V. partially supported by FWF within the collaborativeDACH project
Arrangements and Drawings as FWF project I 3340-N35. A.P. sup-ported by a Schrödinger fellowship of the FWF: J-3847-N35. J.T. partially supportedby ERC Start grant no. (279307: Graph Games), FWF grant no. P23499-N23 andS11407-N23 (RiSE). a r X i v : . [ c s . C G ] J a n he same label set). We say that G is compatible to the point sets if the drawingof G is plane for each of them (where each vertex of G is mapped to a uniquelabel and thereby identified with a unique point of each point set). Note thatthe labelings of the point sets can be predefined or part of the solution. As anexample, we mention the compatible triangulation conjecture [4]: For any two sets P and P with the same number of points and the samenumber of extreme points, there is a labeling of the two sets such that there existsa triangulation which is compatible to both sets, P and P . Motivation and related work.
The study of the type of compatibility con-sidered in this work (the second type from above) is motivated by applicationsin morphing [7,16,17], 2D shape animation [12], or cartography [22].Compatible triangulations were first introduced by Saalfeld [22] for labeledpoint sets, who studied the construction of compatible triangulations usingSteiner points (compatible triangulations do not always exist for pairs of labeledpoint sets). Aronov et al. [9] and Babikov et al. [11] showed that O ( n ) Steinerpoints are always sufficient, while Pach et al. [21] showed that Ω ( n ) Steinerpoints are sometimes necessary. The compatible triangulation conjecture statesthat – in contrast to labeled point sets – two unlabeled point sets (in generalposition and with the same number of extreme points) can always be compatiblytriangulated without using Steiner points. To date, the conjecture has only beenproven for point sets with at most three interior points [4]. Krasser [20] showedthat more than two point sets cannot always be compatibly triangulated.Concerning compatible paths, Hui and Schaefer [18] showed that it is NP-hard to decide whether two labeled point sets admit a compatible spanningpath. Arseneva et al. [10] presented efficient algorithms for finding a monotonecompatible spanning path, or a compatible spanning path inside simple polygons(if they exist). Czyzowicz et al. [15] showed that any two labeled point sets admita compatible path of length at least √ n and also presented an O ( n log n ) algorithm to find such a path for two convex point sets. In a similar direction,results from Czabarka and Wang [14] imply a lower bound of ( √ n − / onthe length of the longest cycle compatible to two convex point sets.In this paper we will focus on compatible matchings. To the best of ourknowledge, previous results on (geometric) matchings study only compatibilityof the first type, that is, where two matchings are embedded on the same pointset. A well-studied question in this setting is whether any two perfect matchingscan be transformed into each other by a sequence of steps such that at everystep the intermediate graph is a perfect matching and the union of any twoconsecutive matchings is plane. Aichholzer at al. [3] proved that such a sequenceof at most O (log n ) steps always exists. Questions of whether any matching of agiven point set can be transformed into any other and how many steps it takes(that is, the connectivity of and the distance in the so-called reconfigurationgraph of matchings, as well as its other properties) have been investigated alsofor matchings on bicolored point sets and for edge-disjoint compatible matchings,see for example [5,8,19]. (a)
14 23 14 32 13 24 (b)
Fig. 1: (a) There is no perfect matching compatible to the two labeled sets.(b) Any possible pair of matching edges crosses in exactly one of the three sets.
Our results.
We study the second type of compatibility for matchings on twoor more point sets. This is a setting for which no previous comprehensive theoryappears to exist. Let us mention that throughout this paper we denote unlabeledpoint sets with P and labeled point sets with P . We start by considering convexpoint sets: given two unlabeled convex point sets P , P , both with n points,we study the largest guaranteed size ccm( n ) of a compatible matching across allpairs of labelings of P and P . More formally, ccm( n ) is the minimum over allpairs of labelings of the maximum compatible matching size for the accordinglylabeled n -point set pairs. The largest compatible matching for two labeled pointsets is not necessarily perfect, see Fig. 1(a). In Section 2 we present upper andlower bounds on ccm( n ) . In particular, for any n that is a multiple of 10, weconstruct two labeled convex sets P , P of n points each, for which the largestcompatible matching has n/ edges. Using probabilistic arguments, we obtainan upper bound of ccm( n ) = O ( n / ) . For the lower bound, we show that forany pair of labeled convex point sets P , P there exists a compatible matchingconsisting of (cid:98)√ n (cid:99) edges. This implies that ccm( n ) = Ω ( √ n ) .We further extend our study to consider (cid:96) point sets in general position in-stead of just two point sets in convex position. Given (cid:96) unlabeled sets P , . . . , P (cid:96) of n points in general position, we denote by cm( n ; P , . . . , P (cid:96) ) the largest guaran-teed size of a compatible matching across all (cid:96) -tuples of labelings of P , . . . , P (cid:96) .We remark that the size n of the point sets is included in the notation onlyfor the sake of clarity (since our bounds depend on n ). In Section 3 we givebounds on cm( n ; P , . . . , P (cid:96) ) for any sets P , . . . , P (cid:96) of n points in general po-sition. Building on the ideas of the proofs for two convex sets, we show that cm( n ; P , . . . , P (cid:96) ) = O ( n / ( (cid:96) +1) ) and that cm( n ; P , . . . , P (cid:96) ) = Ω ( n /(cid:96) ) .Finally, we investigate the question of how many labeled copies of a givenunlabeled point set are needed so that the largest compatible matching consistsof a single edge. Already for four points in convex position, three different setsare needed (and sufficient, see Fig. 1(b)). In Section 4 we prove that for anygiven set of n ≥ points in general position, Θ (log n ) copies of it are necessaryand sufficient for the existence of labelings forcing that the largest compatiblematching consists of a single edge.For brevity, a plane matching that consists of k edges is called a k -matching . Two convex sets
Throughout this section we consider two labeled convex point sets P , P con-sisting of n points each. Without loss of generality we assume that P is labeled (1 , , . . . , n ) in clockwise order and that P is labeled ( π (1) , π (2) , . . . , π ( n )) inclockwise order for some permutation π : [ n ] → [ n ] .In the following, we present lower and upper bounds on the largest guaranteedsize ccm( n ) of a compatible matching of any two such sets. Starting with lowerbounds, we present four pairwise incomparable results (Theorem 1), each ofthem giving rise to a polynomial-time algorithm for constructing a compatible k -matching with k = Ω ( √ n ) edges. The results are ordered by the size of theobtained compatible k -matching, where the last one gives the best lower boundfor ccm( n ) , while the three other results yield compatible matchings of specialstructure. The second result can be generalized to any number of (not necessarilyconvex) sets (Theorem 3). We remark that [15] implies a lower bound of √ n/ ,which is weaker than the fourth result.Before stating the theorem, we introduce the notion of a shape of a matchingon a convex point set which, informally stated, captures “how the matchinglooks”. Consider a labeled point set P and a plane matching M on P . Let P M ⊆P be the points of P that are incident to an edge of M . The shape of M is thecombinatorial embedding of the union of M and the boundary of the convexhull of P M . Further, M is called non-nested if its shape is a cycle, that is, alledges of M lie on the boundary of the convex hull of P M . Note that the shapeof M also determines the number of its edges (even though some or all of theedges might be “hidden” in the boundary of the convex hull of P M ). We say thattwo matchings have the same shape , if their shapes are identical, possibly up toa reflection. Theorem 1 (Lower bound for two convex sets).
For any two labeledconvex sets P , P of n points each, it holds that: (i) If n ≥ (2 k − + 2 thenfor any shape of a k -matching there exists a compatible k -matching having thatshape in both P and P . (ii) If n ≥ k + 2 k − then any maximal compatiblematching consists of at least k edges. (iii) If n ≥ k + k then there exists acompatible k -matching that is non-nested in both P and P . (iv) If n ≥ k + k then there exists a compatible k -matching.Proof. (i) By the circular Erdős-Szekeres Theorem [14], the permutation π containsa monotone subsequence σ having length k . The sequence S = { x i | i ∈ σ } of points whose labels belong to σ has the same cyclic order in both sets P , P (possibly once clockwise and once counter-clockwise), hence anyplane matching on S in P is also plane in P and has the same shape. (ii) Suppose we have already found a compatible matching M consisting of m ≤ k − edges. This leaves at least n − m ≥ k +1 points yet unmatched. The combinatorial embedding fixes the cyclic order of incident edges for each vertex. he unmatched points are split by the m matching edges into at most m + 1 ≤ k subsets, both in P and in P . Since there are at most k different ways to choose one such subset from P and one from P , thereexist two yet unmatched points x , y that lie in the same subset in P andin the same subset in P . Hence xy can be added to the matching M . (iii) This claim is equivalent to Problem 5 given at IMO 2017. For complete-ness we sketch a proof (see Fig. 2): split the perimeter of P into k con-tiguous blocks B , . . . , B k consisting of k + 1 points each (that is, block B consists of points labeled π (1) , . . . , π ( k + 1) and so on). We aim to drawone matching edge per block. We process points x i in order i = 1 , . . . , n in which they appear in P . Once some block, say B (cid:63) , contains two pro-cessed points, say x u and x v , we draw edge x u x v , discard other alreadyprocessed points and discard other points in B (cid:63) . In this way, any time wedraw an edge in some block, we discard at most one point from each otherblock. Since each block initially contains k + 1 points, we will eventuallydraw one edge in each block. The produced matching contains one edgeper block, hence it is non-nested in P . Since points x i are processed inorder i = 1 , . . . , n , the matching is also non-nested in P . (iii) (iv) P P
11 1111 111 11 1 (iv)
Fig. 2: Theorem 1, Claim (iii): Illustration with n = 12 points and k = 3 blocks(grey). After drawing an edge we switch the color of processed points (red togreen to blue). Claim (iv): The permutation matrix Π and two -balls (yellow).A 2-ball centered at [5 , would intersect a 2-ball at [7 , , so drawing the edgebetween points labeled 7, 9 forces us to discard at most 2 other points (6 and 8). (iv) The idea is to find two points x i , x j that are close to each other in thecyclic order in both P and P . Then draw the edge x i x j , omit all pointson the shorter arcs of x i x j in both P and P , and proceed recursively.Consider the permutation matrix Π given by π , that is, an n × n matrixsuch that Π i,j = 1 if π ( i ) = j and 0 otherwise. Given an integer r > and a cell Π i,j containing a digit 1, the r -ball centered at Π i,j is a set B ( Π i,j , r ) = { Π u,v : | i − u | + | j − v | ≤ r } of cells whose L -distance from , Problem C4. i,j is at most r , where all indices are considered cyclically modulo n (see Fig. 2). Note that an r -ball contains r + 2 r + 1 cells.Now suppose n and r satisfy n ≤ r + 2 r and consider r -balls centered atall n cells containing a digit 1. The balls in total cover n · (2 r +2 r +1) > n cells, hence some two r -balls intersect and their centers Π i,π ( i ) , Π j,π ( j ) have L -distance at most r . This means that the shorter arcs between pointslabeled π ( i ) and π ( j ) contain, together in both point sets P and P , atmost r − other points. Drawing an edge π ( i ) π ( j ) and removing these r − other points leaves convex sets in both P and P whose convexhulls do not intersect the matched edge π ( i ) π ( j ) .The rest is induction. The claim holds for k ∈ { , } . Suppose that k = 2 r is even and that n = k + k = 2 r + 2 r . By the above argument, find a“short” edge x i x j and remove up to r − other points. This leaves n − r ( < r + 2 r ) points, so find another edge x u x v and remove up to r − other points. This leaves r − r = 2( r − + 2( r − points and theinduction applies. Last, note that the above shows that having r pointsimplies a (2 r − -matching. Since r = (cid:100) (2 r − + 2 r − (cid:101) , the case of k = 2 r − odd and n = (cid:100) (2 r − + 2 r − (cid:101) is also settled. (cid:117)(cid:116) For the remainder of this section, we consider upper bounds on the size ofcompatible matchings for pairs of convex point sets.We first describe an explicit construction of two labeled point sets P id and P π ,where n is a multiple of 10, the set P id is labeled (1 , , . . . , n ) in clockwise order,and the set P π is labeled ( π (1) , π (2) , . . . , π ( n )) in clockwise order, by defining aspecific permutation π : [ n ] → [ n ] . We will show that any compatible matchingof P id and P π misses at least n/ of the points.Our building block for π is the permutation (2 , , , , of five elements. Forlabeling the n = 5 k points of P π (with k ≥ even) we use the permutation π =(2 , , , , , , , , , , . . . , k − , k − , k − , k − , k − that yields k blocks of 5 points each in both P and P (see Fig. 3). P id P π ... ... Fig. 3: The two labeled point sets P id and P π for the permutation π . Proposition 1 (Constructive upper bound for two convex sets).
Thelargest compatible matching of the two labeled n -point sets P id and P π definedabove contains n edges.Proof. We show that any compatible matching misses at least one point withineach of the k blocks. This gives n unmatched points and thus at most n edgesin any compatible matching.e classify the edges of any compatible matching M into two types: thosethat connect two points in one block (that is, edges with both labels in { i +1 , i +2 , i +3 , i +4 , i +5 } for some ≤ i ≤ k − ; we call them short edges) and allother edges, which connect two points from different blocks (we call them long edges). To show that M misses at least one point of each block B , we distinguishtwo cases: Case 1: B contains at least one short edge. We show that there is always at least one unmatched point in B . W.l.o.g., let B be the block with labels , . . . , . For each of the possible (cid:0) (cid:1) = 10 shortedges, it is straightforward to see that if we include it, then we inevitablyobtain a point of B that can not be matched in either P or P (see Fig. 3).For example, (i) edge (1 , induces point to be unmatchable by P ; (ii)edge (1 , induces point to be unmatchable due to P ; (iii) edge (1 , forces edge (2 , by P , which makes point unmatchable due to P ; (iv)edge (1 , forces one of the points , and to be unmatched due to P ;and so on. Case 2:
All five points in B are matched by a long edge. We argue that, under the assumption that all five points in B are matched(by a long edge), all those five edges in fact must go to the same block B (cid:48) ,which we then show to be impossible. Consider a pair of numbers a , b thatlie in the same block whose relative position within that block is different in P and in P (for example, 1 and 2 but not 1 and 3). Suppose b is matchedto b (cid:48) . Then a has to be matched to a point on the same side of the line bb (cid:48) as a , in both P and P . This is impossible unless a is matched to a point inthe same block as b (cid:48) that moreover happens to lie on the correct side of b (cid:48) inboth P and P . Hence for any such pair a , b , the points a and b are matchedto the same block. It remains to notice that (2 , , (1 , , (4 , , (3 , are alleligible ( a, b ) pairs, hence all five edges go to the same block B (cid:48) . However,there is only one non-crossing perfect matching of B and B (cid:48) in P and weeasily check that it is not compatible with P .To see that the bound is tight, note that within each block of P π we canmatch the first two points and the next two points. This yields a compatiblematching of P id and P π with k = n edges consisting only of short edges. (cid:117)(cid:116) The above construction yields an upper bound of ccm( n ) ≤ (cid:100) n (cid:101) . However,this bound is not tight. We next show in a probabilistic way that there exists apermutation π : [ n ] → [ n ] for which the largest compatible matching consists of k = O ( n / ) edges. In Section 3, we will extend this approach to any number ofpoint sets, not necessarily in convex position (Theorem 4). Theorem 2 (Probabilistic upper bound for two convex sets).
Fix n andlet k = 4 n / . Then two convex sets P , P of n points each can be labeled suchthat the largest compatible matching consists of fewer than k edges.roof. Let P be P with labeling (1 , , . . . , n ) in clockwise order and let P notyet be labeled. The idea for this proof is that for large n there are more ways tolabel P than there are ways to draw a compatible k -matching.For any k ≤ n , let f ( k ) be the number of plane k -matchings of P i , i ∈ { , } (that is, matchings leaving n − k points unmatched). As there are (cid:0) n k (cid:1) ways toselect the k points to be matched and the number of plane perfect matchingon those points is k +1 (cid:0) kk (cid:1) (the k -th Catalan number), we obtain f ( k ) = (cid:0) n k (cid:1) · k +1 (cid:0) kk (cid:1) ≤ n !( n − k )! · k ! · k ! .Given two plane k -matchings, one of P and one of P , there are exactly g ( k ) = ( n − k )! · k ! · k labelings of P for which those two matchings constitutea compatible k -matching: there are ( n − k )! ways to label the unmatched pointsof P , k ! ways to pair up the matching edges and k ways to label their endpoints.Therefore, ( f ( k )) g ( k ) is an upper bound for the number of labelings π of P such that there is a compatible k -matching for P and P ( P with labeling π ).On the other hand, there are n ! labelings of P in total.Our goal is to show that ( f ( k )) · g ( k ) < n ! . If we succeed, then there exists alabeling π of P such that there is no compatible k -matching for P and P ( P with labeling π ). Canceling some of the factorials and using standard bounds ( n/ e) n < n ! < n n on the remaining ones (where e denotes Euler’s number), weobtain ( f ( k )) · g ( k ) n ! ≤ n ! · k ( n − k )! · ( k !) ≤ n k · k ( k/ e) k = (cid:18) n k (cid:19) k . For k ≥ n / , the above expression is less than one (we have < ), whichcompletes the proof. (cid:117)(cid:116) In this section we generalize our results in two ways, by considering point setsin general position and more than two sets. We again start with lower bounds.Theorem 3, which is a generalization of the second result of Theorem 1, impliesthat for any (cid:96) -tuple of point sets P , . . . , P (cid:96) we have cm( n ; P , . . . , P (cid:96) ) = Ω ( n /(cid:96) ) . Theorem 3 (Lower bound for multiple sets).
Let P , P , . . . , P (cid:96) be labeledsets of n points each. If n ≥ k (cid:96) + 2 k − , then any maximal compatible matchingconsists of at least k edges.Proof. We extend the idea from the proof of Theorem 1, part (ii): suppose wehave already found a compatible matching M consisting of m ≤ k − edges.This leaves at least k (cid:96) + 2 k − − m ≥ k (cid:96) + 2 k − − k −
1) = k (cid:96) + 1 points yetunmatched. Imagine the (cid:96) point sets live in (cid:96) different planes. We process the m matching edges one by one. When an edge is processed, we extend it alongits line in both directions until it hits another matching edge or an extension ofa previously processed edge (in all (cid:96) planes). In this way, the m lines partitioneach plane into m + 1 ≤ k convex regions. By simple counting ( k (cid:96) + 1 > k (cid:96) ),here exist two yet unmatched points x , y that lie in the same region in each ofthe (cid:96) planes. Hence xy can be added to the matching M . (cid:117)(cid:116) Regarding upper bounds, the following theorem implies that for any fixed (cid:96) and any (cid:96) -tuple of point sets P , . . . , P (cid:96) , we have cm( n ; P , . . . , P (cid:96) ) = O ( n / ( (cid:96) +1) ) . Theorem 4 (Probabilistic upper bound for multiple sets).
Fix n and (cid:96) and let k = 125 · n / ( (cid:96) +1) . Then any (cid:96) sets P , . . . , P (cid:96) of n points each, where each P i is in general position, can be labeled such that the largest compatible matchingconsists of fewer than k edges.Proof. We extend the idea from the proof of Theorem 2. Let P be P withlabeling (1 , , . . . , n ) in clockwise order and suppose that the remaining (cid:96) − sets are not yet labeled. Let k ≤ n and let f i ( k ) , ≤ i ≤ (cid:96) , be the number of k -matchings of P i . Sharir, Sheffer and Welzl showed in [24] that the number ofplane perfect matchings of any set P of k points in general position is at most · (3 / ( k/ · tr ( P ) , where tr ( P ) denotes the number of triangulations of P .Sharir and Sheffer also showed in [23] that the number of triangulations of P isat most k . This implies that there are also at most · . k different perfectmatchings of P . By counting this upper bound for every possible k -point subsetof P i , we obtain f i ( k ) ≤ (cid:0) n k (cid:1) · · . k . Next, consider an (cid:96) -tuple ( M , . . . , M (cid:96) ) of matchings on P , . . . , P (cid:96) , respec-tively, consisting of k edges each. Notice that any such (cid:96) -tuple forms a compatiblematching for g ( k ) = (cid:0) ( n − k )! · k ! · k (cid:1) (cid:96) − combinations of labelings for P , . . . , P (cid:96) .On the other hand, there are ( n !) (cid:96) − such combinations of labelings for P , . . . , P (cid:96) in total. It suffices to show that (cid:32) (cid:96) (cid:89) i =1 f i ( k ) (cid:33) · g ( k ) < ( n !) (cid:96) − for guaranteeing the existence of a combination of labelings for P , . . . , P (cid:96) suchthat there is no compatible k -matching for the resulting labeled sets P , . . . , P (cid:96) .As before, we expand the binomials into factorials, cancel some of them anduse standard bounds ( n/e ) n < n ! < n n on the remaining ones (where e againdenotes Euler’s number) to obtain (cid:16)(cid:81) (cid:96)i =1 f i ( k ) (cid:17) · g ( k )( n !) (cid:96) − = n ! · (8 · . k ) (cid:96) ( k !) (cid:96) − (2 k ) (cid:96) − ( n − k )!((2 k )!) (cid:96) ≤ (cid:96) n k · (66 . (cid:96) ) k ( k (cid:96) − ) k ( e (cid:96) ) k ((2 k ) (cid:96) ) k ≤ (cid:96) (cid:20) (16 . · e ) (cid:96) · n k (cid:96) +1 (cid:21) k . hen k = 125 · n / ( (cid:96) +1) , then also k ≥ holds. Further, the expression insidethe brackets is less than . (cid:96) (we have . · e < . ). Since < . ≤ . k , this completes the proof. (cid:117)(cid:116) We remark that the upper bound of . k for the number of plane perfectmatchings of any set of k points in the plane in general position is by far nottight. Actually, Sharir and Welzl [25] showed that this number can be boundedby O (10 . k ) . However, for the above proof, we require an explicit upper boundthat holds for any value of k ≥ and hence we did not use this result. In this section we consider the following question: Given an unlabeled point set P with n points, is there an integer (cid:96) such that there exist (cid:96) labelings of P forwhich every compatible matching has at most one edge? If (cid:96) exists, we denote as force( n ; P ) the minimum number (cid:96) of copies of P such that cm( n ; P, . . . , P ) = 1 (where P appears force( n ; P ) times). Otherwise, we set force( n ; P ) = ∞ . Inother words, we are asking for the existence (and minimal number) of labelingsof the set P so that any pair of labeled edges crosses for at least one labeling. Weremark that, again, the size n of the point sets is included in the notation onlyfor the sake of clarity. Note that force( n ; P ) = ∞ if and only if the straight-linedrawing of K n on P contains no crossing. Hence force( n ; P ) is finite for any set P of n ≥ points.We first focus on upper bounds and on the case when P is in convex position.We denote by cforce( n ) the minimum number of copies of a convex set with n points that need to be labeled so that the largest compatible matching consistsof only a single edge.Let b ( n ) = (cid:100) log n (cid:101) , which is the number of bits that are needed to representthe labels 1 to n . We construct a family of b ( n ) labeled convex n -point sets suchthat all pairs of edges cross in at least one set. First consider three labeled convexpoint sets, which are obtained by partitioning the set of labels into four blocks A , B , C , D , and combining those blocks in different orders and orientations asdepicted in Fig. 4. The order within a block is arbitrary, but identical for allthree sets (up to reflection; those block orientations are indicated by arrows). Lemma 1.
Consider three convex point sets P , P , and P that are labeled asin Fig. 4 for some partition A , B , C , and D of their label set. Then any pair ofindependent edges, where none of them has both labels in one of the blocks A , B , C , and D , forms a crossing in at least one of P , P , and P .Proof. We consider three cases, depending on the number x of blocks containingendpoints of the edges e and f . Case x = 2 : Then e and f are spanned by the same two subsets. As any pairof blocks shows up in the same and in inverse orientation in at least one Two edges are independent if they do not share an endpoint.
CBD (a)
A DCB (b)
A BDC (c)
Fig. 4: Three labeled point sets obtained from different orders and orientationsof four blocks A , B , C , and D .of the three drawings, this guarantees a crossing. For example, let the twosubsets be A and B . They have the same orientation in Fig. 4(a) and 4(b),but inverse orientation in Fig. 4(c). An analogous property holds for theremaining five combinations. Case x = 3 : W.l.o.g. e and f have their starting point in the same subset, butthe endpoints in different subsets. There are 12 possible configurations ofone common and two disjoint subsets, and it is straightforward to check thateach situation shows up in both possible orientations w.r.t. the common set.For example, let the common set be A , and the other sets C and D . Theorientation of A is the same in all three drawings, but the order of C and D is inverted in Fig. 4(a) and 4(c). If the common set is B and the two othersets are again C and D , then order of the three sets is the same in all threedrawings, but the orientation of the common set B is inverted in Fig. 4(a)and 4(c). Thus, in both cases a crossing is guaranteed. Case x = 4 : In this case the orientation of the subsets is not relevant. There areonly three possible combinations of such edges ( A − B with C − D , A − C with B − D , and A − D with B − C ) and the three drawings cover one caseeach. (cid:117)(cid:116) We next identify a small number of 4-partitions of the label set { , , . . . , n } such that each edge pair fulfills the condition of Lemma 1 in at least one ofthe partitions (when the four subsets form blocks). This yields the followingconstructive upper bound for cforce( n ) . Proposition 2 (Constructive upper bound on cforce( n ) ). For any n ≥ and for b ( n ) = (cid:100) log n (cid:101) , we can define (cid:0) b ( n )2 (cid:1) labeled convex sets of n pointssuch that the largest matching compatible to all of them consists of a single edge.Proof. Given a convex set of n points, we construct (cid:0) b ( n )2 (cid:1) i, j , ≤ i (cid:54) = j < b ( n ) , of the labels, partitionthe label set { , , . . . , n } so that A contains all labels where those two bits arezero, B those where the bits are zero-one, C those with one-zero, and finally D the ones with both one. This gives (cid:0) b ( n )2 (cid:1) different partitions.ow consider two arbitrary edges e and f . Then there is a bit position inwhich the two endpoints of e have different values, and the same is true for f .Let i and j , respectively, be those positions. If this would give i = j , then choose j arbitrarily but not equal to i . By Lemma 1, the edges e and f cross in one ofthe three labeled point sets for the partition generated for i and j . (cid:117)(cid:116) The upper bound O (log n ) of cforce( n ) from Proposition 2 is constructivebut it is not asymptotically tight. Next we present a probabilistic argumentwhich shows that we actually have force( n ; P ) = O (log n ) for any point set P of n ≥ points. Lemma 2 (Probabilistic upper bound on force( n ; P ) ). Given a set P of n ≥ points in general position, there exists a constant c P ≥ / such that force( n ; P ) ≤ log c P (3 (cid:0) n (cid:1) ) = O (log n ) .Proof. Fix P and let α P ∈ (0 , be the proportion of 4-tuples of points in P that are in convex position. Note that since any 5-tuple of points contains atleast one 4-tuple in convex position, we have α P ≥ / (here we use n ≥ ).There are r = 3 (cid:0) n (cid:1) pairs of non-incident edges. Fix one of them, say ac and bd . Note that when P is labeled uniformly at random, the edges ac , bd intersectwith constant probability α P / : indeed, the edges intersect if their 4 endpointsform a convex quadrilateral and the points a , b , c , d lie on its perimeter in twoout of the six possible cyclic orders.Now set c P = (1 − α P / − ≥ / and consider (cid:96) > log c P ( r ) copies of P labeled independently and uniformly at random. We say that a pair of edges is bad if it is non-crossing in all (cid:96) point sets. Then any one fixed pair of edges isbad with probability ρ = (1 − α P / (cid:96) < /r . By linearity of expectation, theexpected number of bad pairs of edges is r · ρ < . Therefore there exists alabeling of the (cid:96) point sets for which no pair of edges is bad. In such a labeling,the largest compatible matching consists of a single edge. (cid:117)(cid:116) We remark that for a fixed set P , one can often obtain a better bound onparameter α P used in the proof and thus a better bound on c P , which then givesa constant factor improvement on force( n ; P ) . Specifically, for sufficiently large n , finding the maximum constant α is a topic of high relevance in connectionwith the rectilinear crossing number of the complete graph, see [2] for a nicesurvey of this area. The currently best known bounds are . < α < . [1,6]. Moreover, when P is in convex position we have α P = 1 andthus the above proof implies cforce( n ) ≤ log / (3 (cid:0) n (cid:1) ) . On the other hand, noneof these observations leads to an asymptotic improvement on the upper boundof force( n ; P ) or cforce( n ) . In the following we show that any such asymptoticimprovement is in fact impossible. Lemma 3 (Lower bound on force( n ; P ) ). Fix k ≥ and let P be any setof n = 2 k + 3 points in general position. Then force( n ; P ) ≥ k + 2 = Ω (log n ) .Proof. We use a similar argument as the one used in Theorem 3. Denote by P , . . . , P k +1 any k + 1 labeled copies of P . Take an arbitrary edge ab on theonvex hull of P k +1 . The line containing ab divides each of P , . . . , P k into twoparts (one possibly empty). Since there are n − k +1 > k unmatched points,there exist two points, say x , y , that lie in the same part, for each i = 1 , . . . , k .Thus the two edges xy and ab form a compatible 2-matching implying that force( n ; P ) ≥ k + 2 . (cid:117)(cid:116) Combining upper and lower bounds for force( n ; P ) from Lemmata 2 and 3,we obtain the following conclusion: Theorem 5.
For every set P of n ≥ points in general position, it holds that force( n ; P ) = Θ (log n ) . Besides the presented results there are several related directions of research.Below we list some of them, together with open questions emerging from theprevious sections.1. A natural open problem is the computational complexity of finding compat-ible matchings of a certain size or even deciding their existence: How fastcan we decide if two (or more) given (general or convex) labeled point setshave a perfect compatible matching (or a compatible matching of size k )?2. Can we close the gaps between the lower and upper bounds, Ω ( √ n ) and O ( n / ) respectively, for the size of the largest compatible matchings for twolabeled point sets (general, convex, or of any other particular order type)?3. Can we improve the constructions to match their according probabilisticbounds?4. Consider the following game version: two players alternately add an edgewhich must not intersect any previously added edge. The last player who isable to add such an edge wins. It is not hard to see that for a single set ofpoints in convex position, this is the well-known game Dawson’s Kayles, seee.g. [13]. This game can be perfectly solved using the nimber theory devel-oped by Sprague-Grundy, see also [13] for a nice introduction to the area.An interesting generalization of Dawson’s Kayles occurs when the playersuse two (or more) labeled (convex) point sets and add compatible edges.5. What are tight bounds for the size of compatible paths and compatible cycles(see [14,15])? In ongoing work we have obtained first results in this direction. References
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