Packing Short Plane Spanning Graphs in Complete Geometric Graphs
Oswin Aichholzer, Thomas Hackl, Matias Korman, Alexander Pilz, Günter Rote, André van Renssen, Marcel Roeloffzen, Birgit Vogtenhuber
PPacking Plane Spanning Graphs with ShortEdges in Complete Geometric Graphs ∗ Oswin Aichholzer † Thomas Hackl † Matias Korman ‡ Alexander Pilz † G¨unter Rote § Andr´e van Renssen ¶ Marcel Roeloffzen (cid:107)
Birgit Vogtenhuber † April 5, 2019
Abstract
Given a set of points in the plane, we want to establish a connectedspanning graph between these points, called connection network, thatconsists of several disjoint layers. Motivated by sensor networks, our goalis that each layer is connected, spanning, and plane. No edge in thisconnection network is too long in comparison to the length needed toobtain a spanning tree.We consider two different approaches. First we show an almost optimalcentralized approach to extract two layers. Then we consider a distributedmodel in which each point can compute its adjacencies using only infor-mation about vertices at most a predefined distance away. We show aconstant factor approximation with respect to the length of the longestedge in the graphs. In both cases the obtained layers are plane.
Given a set S of n points in the plane and an integer k , we are interested infinding k edge-disjoint non-crossing spanning graphs H , H , . . . , H k on S suchthat the length BE ( H ∪ H ∪ · · · ∪ H k ) of the bottleneck edge (the longest edgewhich is used) is as short as possible. Each subgraph H i is referred to as a ∗ A preliminary version of this paper was presented in the 27th International Symposiumon Algorithms and Computation (ISAAC 2016), Sydney, Australia [2]. † Institute of Software Technology, Graz University of Technology, Austria. { oaich|thackl|apilz|bvogt } @ist.tugraz.at ‡ Department of Computer Science, Tufts University, Medford, MA, USA. [email protected] § Institut f¨ur Informatik, Freie Universit¨at Berlin, Germany. [email protected] ¶ University of Sydney, Sydney, Australia. [email protected] (cid:107)
Department of Mathematics and Computer Science, TU Eindhoven, The Netherlands. [email protected] a r X i v : . [ c s . C G ] A p r ayer of the complete graph G on the n points. We require each layer to benon-crossing, but edges from different layers are allowed to cross each other. For k = 1, the minimum spanning tree MST ( S ) solves the problem: its longest edge BE ( MST ( S )) is a lower bound on the bottleneck edge of any spanning subgraph,and it is non-crossing. For larger k , we take BE ( MST ( S )) as the yardstick andmeasure the solution quality in terms of BE ( MST ( S )) and k .Although we find the problem to be of its own (theoretical) interest, thisparticular variation comes motivated from the field of sensor networks. In sensornetworks, the energy consumed in transmission drastically grows as the distancebetween the two points increases [5, 6]. Since we want to avoid high energyconsumption, it is desirable to apply the bottleneck criterion in order to minimizethe maximum length of the whole network.Once the network is built, we want to send messages through it without havingto store the network explicitly at each node. One of the most commonly usedmethods for doing so is called face routing [7], which uses only local informationand guarantees delivery as long as the underlying network is plane. In fact,most local routing algorithms can only route on plane graphs. Extending thesealgorithms for non-plane graphs is a long-standing open problem in the field.In this paper, we provide a different way to avoid this obstacle. Rather thanone plane graph, we construct several disjoint plane spanning graphs. If we splitall the messages among the different layers we can potentially spread the loadamong a larger number of edges. Previous Work.
This problem falls into the family of graph packing problems, where we are given a graph G = ( V, E ) and a family F of sub-graphs of G . The aim is to pack as many pairwise edge-disjoint subgraphs H = ( V, E ) , H = ( V, E ) , . . . as possible into G .A related problem is the decomposition of G . In this case, we also look fordisjoint subgraphs but require that (cid:83) i E i = E . For example, there are knowncharacterizations of when we can decompose the complete graph of n points intopaths [10] (for n even) and stars [9] (for n odd). Dor and Tarsi [4] showed thatto determine whether we can decompose a graph G into subgraphs isomorphicto a given graph H is NP-complete. Concerning graph packing, Aichholzer etal. [1] showed that we can pack Ω( √ n ) edge-disjoint plane spanning trees inthe complete graph on any set S of n points. This bound has been improvedto (cid:98) n/ (cid:99) by Biniaz and Garc´ıa [3]. Note that a trivial (cid:98) n/ (cid:99) upper bound followsfrom the number of edges in the complete graph. Thus, the latter result is closeto optimal.In our case, the graph G is the complete graph on a given point set S , and F consists of all plane spanning graphs of G . In addition to proving results fora large (fixed) number of layers, we are interested in minimizing a geometricconstraint (Euclidean length of the longest edge among the selected graphs of F ).To the best of our knowledge, this is the first packing problem of such type.2 esults. Recall that both the point set S and the integer k are givenand that we aim to find k edge-disjoint connected plane spanning graphs H , H , . . . , H k on S such that the length BE ( H ∪ H ∪· · ·∪ H k ) of the bottleneckedge (the longest edge that is used) is minimized.We give two different approaches to solve the problem. In Section 2 wegive a construction for two spanning trees, i.e., k = 2. This construction iscentralized in a classic model that assumes that the positions of all points areknown and computed in a single place. Our construction creates two trees andguarantees that all edges (except possibly one) have length at most 2 BE ( MST ( S )).The remaining edge has length at most 3 BE ( MST ( S )). We complement thisconstruction with a matching worst-case lower bound that shows that for twospanning trees this is the shortest length the longest edge in the graphs can have.In Section 3 we use a different approach to construct k edge-disjoint connectedplane spanning graphs (not necessarily trees). The construction works for any k ≤ n/
12 in an almost local fashion, i.e., using only information about verticesat most a certain maximum distance away. The only global information thatis needed is β : BE ( MST ( S )) or some upper bound on it. Each point of S cancompute its adjacencies by only looking at nearby points, namely, those atdistance O ( kβ ).A simple adversary argument shows that it is impossible to construct spanningnetworks locally without knowing BE ( MST ( S )) (or an upper bound). The lowerbound of Section 2 shows that a neighborhood of radius Ω( k BE ( MST ( S ))) maybe needed for the network, so we conclude that our construction is asymptoticallyoptimal in terms of the neighborhood.For simplicity, throughout the paper we make the usual general positionassumption that no three points are collinear. Without this assumption, it mightbe impossible to obtain more than a single plane layer (for example, when allpoints lie on a line). Note however, that if collinear and partially overlappingedges are considered as non-crossing, our algorithms do not require the point setto be in general position. In this section we present a centralized algorithm to construct two layers. Westart with some properties on the minimum spanning tree of a set of points.
Lemma 1. If | uw | > max {| uv | , | vw |} for three points u, v, w ∈ S , the edge uw does not belong to MST ( S ) .Proof. This is a special case of the more general well-known statement thatthe longest edges of any cycle in a graph, if it is unique, does not belong toits minimum spanning tree: The greedy algorithm would first pick all otheredges of the cycle unless their endpoints are already connected. Thus, when thealgorithm looks at the longest edge, its endpoints are already connected, andthe edge is not included in the minimum spanning tree.3 emma 2.
Let S be a finite set of points in the plane and let uv and vw be twoedges of MST ( S ) . Then the triangle uvw does not contain any other point of S .Proof. Suppose for the sake of contradiction that the triangle uvw containsa point p ∈ S . Then the sum of the angles vpu and vpw is at least π ; seeFigure 2 (a). Hence, one of these angles, say, vpu is least π/
2. But then vu is the longest edge in the triangle vpu , and by Lemma 1, vu cannot belong to MST ( S ), a contradiction. v v i v i − v i +1 u W i u v wp (a) (b) Figure 1: (a) Proof of Lemma 2. (b) Proof of Lemma 3: The neighbors of v i cannot lie outside the wedge W i defined by its two siblings in MST ( S ). Lemma 3.
Let S be a finite set of points in the plane. Let v ∈ S be a pointwith k ≥ neighbors v , . . . , v k − in MST ( S ) in counterclockwise order. Thenfor every triple ( v i − , v i , v i +1 ) ( indices modulo k ) , the neighbors of v i in MST ( S ) are inside the wedge W i that is bounded by the rays vv i − and vv i +1 and containsthe edge vv i .Proof. Assume for the sake of contradiction that v i has a neighbor u in MST ( S )that does not lie in W i ; see Figure 2 (b). Then the edge v i u intersects one ofthe boundary rays of W i , say, vv i +1 . As MST ( S ) is plane, the edge vv i +1 doesnot intersect the edge v i u . Hence, the point v i +1 lies in the triangle vv i u . As vv i and v i u are in MST ( S ), this contradicts Lemma 2.We denote by MST ( S ) the square of MST ( S ), the graph connecting all pairsof points of S that are at distance at most 2 in MST ( S ). We call the edges of MST ( S ) short edges and all remaining edges of MST ( S ) long edges. For everylong edge uw , the points u and w have a unique common neighbor v in MST ( S ),which we call the witness of uw . We define the wedge of uw to be the area thatis bounded by the rays vu and vw and contains the segment uw .We now characterize edge crossings in MST ( S ); see Figure 2. Lemma 4.
Let S be a finite set of points in the plane. Two edges e and f of MST ( S ) cross if and only if one of the following two conditions is fulfilled: uvw f uvweW W W (cid:48) ee (cid:48) (a) f uvw e (b) xy ? zg (c) Figure 2: The different cases of Lemma 4. Short edges are solid and long edgesare dashed. (a) Condition 1, showing two options e and e (cid:48) , (b) Condition 2, and(c) the contradiction used in the proof.
1. At least one of e and f is a long edge with witness v and wedge W , andthe other edge has v as an endpoint and lies inside W .2. Both e and f are long edges with the same witness v , their wedges intersect,but none is contained in the other.Proof. Clearly, if both e and f are short, they cannot cross. Without loss ofgenerality assume that f = uw is a long edge with witness v and wedge W .If e is incident to v , then it must lie in W in order to cross f , and we satisfyCondition 1.In the remaining case, e = xz with x, z ∈ S \ { u, v, w } . By Lemma 2, x and z cannot lie in the triangle uvw ; hence, e must cross one of the MST edges uv or vw in addition to the edge f = uw . It follows that e cannot be short, and ithas some witness y and some wedge W (cid:48) . We distinguish three possibilities for y :(i) If y = v , we satisfy Condition 2: W (cid:48) is not contained in W because e crosses uv or vw , and by swapping the roles of e with f , we conclude that W isnot contained in W (cid:48) . The wedges W and W (cid:48) must overlap because otherwise e and f could not intersect.(ii) If y = u or y = w , we can swap the roles of e and f , thus satisfyingCondition 1.(iii) We are left with the case that all six points u, v, w, x, y, z are distinct.Let g = uv or g = vw be the edge that is intersected by e . By Lemma 2, thetriangle xyz is empty; thus, g must intersect a second edge xy or yz of thistriangle, in addition to e = xz . This is a contradiction, since the edges g , xy ,and yz are edges of the MST .It is easy to see that the two conditions are sufficient for a crossing: In bothsituations of Condition 1 and Condition 2 (Figure 2), if there were no crossingbetween e and f , an endpoint of one edge would be contained in the trianglespanned by the other edge and its witness, contradicting Lemma 2. With the above observations we can proceed to show a construction that almostworks for two layers: a single edge will be part of both layers, while all other5dges occur in at most one tree. To this end we consider the minimum spanningtree
MST ( S ) to be rooted at an arbitrary leaf r . For any v ∈ S , we define its level (cid:96) ( v ) as its distance to r in MST ( S ). That is, (cid:96) ( v ) = 0 if and only if v = r .Likewise, (cid:96) ( v ) = 1 if and only if v is adjacent to r etc.For any v ∈ S \ { r } , we define its parent p ( v ) as the first vertex traversedin the unique shortest path from v to r in MST ( S ). Similarly, we define its grandparent g ( v ) as g ( v ) = p ( p ( v )) if (cid:96) ( v ) ≥ g ( v ) = r otherwise (i.e., g ( v ) = p ( v ) = r if (cid:96) ( v ) = 1). Each vertex q for which v = p ( q ) is called a child of v . r p p p p p p Figure 3: Example of Construction 1.
MST ( S ) is drawn in solid black, and thered and blue graphs in dashed and dot dashed, respectively. Note that the onlycommon edge between the red and blue trees is the one from the root to its onlyneighbor in MST ( S ). Construction 1.
Let S be a finite set of points in the plane and let MST ( S ) berooted at an arbitrary leaf r ∈ S . We construct two graphs R = G ( S, E R ) and B = G ( S, E B ) as follows: For any vertex v o ∈ S whose level is odd, we add theedge v o p ( v o ) to E R and the edge v o g ( v o ) to E B . For any vertex v e ∈ S \ { r } whose level is even, we add the edge v e g ( v e ) to E R and the edge v e p ( v e ) to E B . For simplicity we say that the edges of R = G ( S, E R ) are colored red andthe edges of B = G ( S, E B ) are colored blue. An edge in both graphs is calledred-blue. See Figure 3 for a sketch of the construction. Theorem 1.
Let MST(S) be rooted at r . The two graphs R = G ( S, E R ) and B = G ( S, E B ) from Construction 1 fulfill the following properties:1. Both R and B are plane spanning trees.2. max { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )) .3. E R ∩ E B = { rs } , with r = p ( s ) , that is, | E R ∩ E B | = 1 .Proof. Recall from Construction 1 that r is a leaf of MST ( S ). Hence r has aunique neighbor s in MST ( S ) and we have r = p ( s ) = g ( s ) and (cid:96) ( s ) = 1. Let S o ⊂ S \{ s } be all v o ∈ S whose level (cid:96) ( v o ) is odd. Likewise, let S e ⊂ S \{ r } beall v e ∈ S whose level (cid:96) ( v e ) is even. By construction, E R contains all the edgesfrom odd-leveled nodes to their parents, those from even-leveled nodes to theirgrandparents and rs . More formally, E R = (cid:91) v o ∈ S o { v o p ( v o ) } ∪ (cid:91) v e ∈ S e { v e g ( v e ) } ∪ { rs } . E B contains edges from odd-leveled nodes to their grandparents, thosefrom even-leveled nodes to their parents and rs , that is E B = (cid:91) v o ∈ S o { v o g ( v o ) } ∪ (cid:91) v e ∈ S e { v e p ( v e ) } ∪ { rs } . Thus, the edge rs is the only shared edge between the sets E R and E B , as statedin Property 3 (we call this unique edge the double-edge ).As E R and E B are subsets of the edge set of MST ( S ), the vertices of everyedge in E R and E B have link distance at most 2 in MST ( S ), and the bound onmax { BE ( R ) , BE ( B ) } stated in Property 2 follows.Further, both R and B are spanning trees, that is, connected and cycle-freegraphs, as each vertex except r is connected either to its parent or grandparentin MST ( S ). To prove Property 1, it remains to show that both trees are plane.Assume for the sake of contradiction that an edge f is crossed by an edge e of the same color. Recall that all edges of E R and E B are edges of MST ( S )whose endpoints have different levels. By Lemma 4, at least one of { e, f } hasto be a long edge. Without loss of generality let f = uw be a long edge andlet v be the witness of f with (cid:96) ( u ) = (cid:96) ( v ) − (cid:96) ( w ) −
2. First note that v cannot be an endpoint of e : since the level of v has different parity than theone of u and w , then v must a leaf in this tree. Moreover, its only neighbormust be u and thus the edges uv and f = uw cannot cross (and in particular v cannot be an endpoint of e as claimed). We further claim that v cannot bethe witness of e . Any edge that has v as its witness is an edge from a child of v to u and therefore cannot cross f = uw . As e is neither incident to v nor has v as a witness, e crossing f contradicts Lemma 4. This proves Property 1 andconcludes the proof.The properties of our construction imply a first result stated in the followingcorollary. Corollary 2.
For any set S of n points in the plane, there are two planespanning trees R = G ( S, E R ) and B = G ( S, E B ) such that | E R ∩ E B | = 1 and max { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )) . Although the construction might seem to generalize to more layers by usingedges of
MST k ( S ), this is not the case. Already for k = 3, we can show thatthe trees may be non-plane. Take the example of Figure 4, where the full edgesdenote the minimum spanning tree. However, if a is chosen as the root of the tree,the edge ad will be crossed by the edge from e to either its parent, grandparentor great-grandparent. In this example the problem can be remedied by choosinga different root. But now consider placing a horizontally mirrored copy of thisconstruction to the left so that a and its mirrored version are connected by anedge. Regardless of which root is chosen, in one of the two subtrees the node a or its mirrored equivalent is the root of the respective subtree. Hence, any rootwill create a crossing. 7 b c de Figure 4: Example graph where choosing a as root creates a crossing when wegeneralize the above construction to three trees. rsN − N + rs rsT − T + S − S + Figure 5: Illustration of the various definitions used in Section 2.2. Grey trianglesillustrate further subtrees, one shows interior vertices purely for illustrativepurposes.
Construction 1 is almost valid in the sense that only one edge was shared betweenboth trees. In the following we modify this construction to avoid the sharededge.Let N − ⊂ ( S \ { r } ) be the set of neighbors v − ∈ ( S \ { r } ) of s in MST ( S )such that the ordered triangle rsv − is oriented clockwise. Let N + ⊂ ( S \ { r } ) bethe set of neighbors v + ∈ ( S \ { r } ) of s in MST ( S ) such that the ordered triangle rsv + is oriented counterclockwise. Let T − be the subtree of MST ( S ) that isconnected to s via the vertices in N − and let T + be the subtree of MST ( S )that is connected to s via the vertices in N + . Let S − ⊂ S consist of r andthe set of vertices from T − and let S + ⊂ S consist of r and the set of verticesfrom T + . Observe that S − ∩ S + = { r, s } (see Figure 5). Let E R and E B besets of red and blue edges as defined in the Construction 1. Then let E − R ⊂ E R ( E − B ⊂ E B ) be the subset of edges that have at least one endpoint in S − \ { r, s } E + R ⊂ E R ( E + B ⊂ E B ) be the subset of edges that have at least oneendpoint in S + \ { r, s } . Note that by this definition E R = E − R ∪ E + R ∪ { rs } and E B = E − B ∪ E + B ∪ { rs } . With this we define the subgraphs R − = G ( S − , E − R ), R + = G ( S + , E + R ), B − = G ( S − , E − B ), and B + = G ( S + , E + B ) and prove a usefulnon-crossing property between these graphs. Lemma 5.
For any set S of n points in the plane, let R = G ( S, E R ) and B = G ( S, E B ) be the graphs from Construction 1. Then no edge in E − R crossesan edge in E + B and no edge in E + R crosses an edge in E − B .Proof. Consider any edge e ∈ E − R that is not incident to r . By Lemma 4, suchan edge e can be crossed only by an edge incident to at least one vertex of S − \ { r, s } . Hence, e does not cross any edge of E + B .Assume for the sake of contradiction that there is an edge f ∈ E + B incidentto r that crosses an edge e ∈ E − R . By construction, e = rz is a long edge of MST ( S ) with witness s and wedge W . By Lemma 4, f has to be incident to s ,since s cannot be the witness of any blue edges by construction. If f is a shortedge, then f is not in W by our definition of S − and S + , which contradictsLemma 4. Hence, let f = sc be a long edge of MST ( S ) with witness b . FollowingLemma 4, the witness b must be s , which contradicts the fact that s cannot be awitness of any blue edge. This concludes the proof that no edge in E − R is crossedby an edge in E + B . Symmetric arguments prove that no edge in E + R is crossed byan edge in E − B .With this observation we can now prove that the two spanning trees (rootedat an arbitrary leaf r ) from Construction 1 actually exist in 4 different colorcombination variants. Lemma 6.
Let S be a set of n points in the plane. Let R = G ( S, E R ) and B = G ( S, E B ) be the graphs from Construction 1 and let R − = G ( S − , E − R ) , R + = G ( S + , E + R ) , B − = G ( S − , E − B ) , and B + = G ( S + , E + B ) be subgraphs asdefined above. Then R and B can be recolored to any of the four versions below,where each version fulfills the properties of Theorem 1.(1) R = G ( S, E R ) and B = G ( S, E B ) (the original coloring )(2) R = G ( S, E B ) and B = G ( S, E R ) (the inverted coloring )(3) R = G ( S, E − B ∪ E + R ∪ { rs } ) and B = G ( S, E − R ∪ E + B ∪ { rs } ) (the − sideinverted coloring )(4) R = G ( S, E − R ∪ E + B ∪ { rs } ) and B = G ( S, E − B ∪ E + R ∪ { rs } ) (the + sideinverted coloring )Proof. The statement is trivially true for recolorings (1) and (2). It is easy toobserve that this corresponds to a simple recoloring. Hence, Properties 2 and 3of Theorem 1 are also obviously true. By Lemma 5, both R and B are plane forthe recolorings (3) and (4) and thus fulfill Property 1 of Theorem 1 as well.9ith these tools we can now show how to construct two disjoint spanningtrees. For technical reasons we use two different constructions based on theexistence of a vertex v in the minimum spanning tree where no two consecutiveadjacent edges span an angle larger than π . Theorem 3.
Let S be a set of n points in the plane, and v a vertex of S .Assume that there is a minimum spanning tree MST ( S ) such that the anglebetween any two consecutive adjacent edges of v in MST ( S ) is smaller than π .Then there exist two plane spanning trees R = G ( S, E R ) and B = G ( S, E B ) suchthat E R ∩ E B = ∅ and max { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )) .Proof. We build the two spanning trees by using the vertex v to decompose theminimum spanning tree into trees where v is a leaf. For each of these subtreeswe apply Construction 1 and possibly recolor them in one of the variants fromLemma 6.Let S v = { v , . . . , v k } be the set of vertices incident to v in MST ( S ), labeledin counterclockwise order as they appear around v . Observe that k ≥ S v contains no points of S except v . We start by constructing twoplane spanning trees of S v ∪ { v } . The red spanning tree R v = G ( S v ∪ { v } , E vR )contains all edges incident to v except vv , plus the edge v v , which lies onthe convex hull boundary of S v . The blue spanning tree B v = G ( S v ∪ { v } , E vB )contains all edges on the convex hull boundary of S v except v v , plus the edge vv . Observe that R v and B v are plane spanning trees, E vR ∩ E vB = ∅ , andmax { BE ( R v ) , BE ( B v ) } ≤ BE ( MST ( S v ∪ { v } )).Next consider a vertex v i of S v and let M i be the maximal subtree of MST ( S )that is connected to v by v i . Let S i ⊂ S be the vertex set of M i . Note that M i = MST ( S i ) and that v is a leaf in M i . Thus, we can use Construction 1 toget spanning trees R i = G ( S i , E iR ) and B i = G ( S i , E iB ), all rooted at v . Thegraphs R i and B i fulfill the three properties of Theorem 1 and the only edgeshared between R i and B i is vv i .Considering Lemma 4 and the fact that for i (cid:54) = j the edges of E iR ∪ E iB haveno point, except for the root v , in common with E jR ∪ E jB , it is easy to seethat no edge of E iR ∪ E iB crosses any edge of E jR ∪ E jB . In order to join thegraphs to two plane spanning trees on S , we adapt them slightly, while keepingthe properties of Theorem 1. We first state how we combine the edge sets of thedifferent plane spanning trees to get R = G ( S, E R ) and B = G ( S, E B ) and thenreason why the claim in the theorem is true for this construction. E R = E vR ∪ ( E R \{ vv } ) ∪ . . . ∪ ( E kR \{ vv k } ) ∪ ( E − R ∪ E +1 B ) ∪ ( E − B ∪ E +2 R ) E B = E vB ∪ ( E B \{ vv } ) ∪ . . . ∪ ( E kB \{ vv k } ) ∪ ( E − B ∪ E +1 R ) ∪ ( E − R ∪ E +2 B )First we add the construction for S v ∪ { v } to both edge sets. This is the baseto which all other trees will be attached. Then the graphs from the subtrees M i for 1 ≤ i ≤ k are added to this base. The edges vv i are already used in R v or B v , so we add the edges vv i neither from E iR nor E iB . As both v and v i are connected to both colors (both spanning trees), the construction stays10onnected. As we did not add any additional edges the construction obviouslystays cycle-free and the edge length bound is maintained.It remains to argue the planarity of the resulting graphs. By Lemma 4, edgesof E iR or E iB that cross any edge of E vR and E vB have to be incident to v . ByLemma 3, only the edges e − i = v i − v i and e + i = v i v i +1 (indices modulo k ) arecrossed by edges of E iR \{ vv i } and E iB \{ vv i } .Using the original coloring (see Lemma 6) for R i and B i only red edges (edgesof E iR \{ vv i } ) cross e − i and e + i . For any 3 ≤ i ≤ k , e − i and e + i are blue, i.e., e − i , e + i ∈ E vB .For i = 1, the edge e + i is red. In this case, we use the + side inverted coloring(see Lemma 6) for R i and B i (and exclude the edge vv i ): E iR = E − iR ∪ E + iB and E iB = E − iB ∪ E + iR . Using this coloring, all shown properties remain valid (seeLemma 6), all edges from R i and B i that cross the blue edge e − i remain red,and all edges from R i and B i that cross the red edge e + i are now blue.In a similar manner we fix the case of i = 2, where the edge e − i is red. Weuse the − side inverted coloring (see Lemma 6) for R i and B i (and exclude theedge vv i ): E iR = E − iB ∪ E + iR and E iB = E − iR ∪ E + iB . Again, all shown propertiesremain valid (see Lemma 6). All edges from R i and B i that cross the red edge e − i are now blue, and all edges from R i and B i that cross the blue edge e + i remain red.Hence, with this slightly adapted construction (and coloring), R = G ( S, E R )and B = G ( S, E B ) are plane spanning trees that solely use edges of MST ( S )and have no edge in common.In the remaining case, for every vertex in an MST ( S ) there are two consecutiveadjacent edges that span an angle larger than π . In such an MST ( S ), everyvertex has degree at most three, since the angle between adjacent edges is atleast π/ Theorem 4.
Consider a set S of n ≥ points in the plane for which ev-ery vertex in the minimum spanning tree MST ( S ) has two consecutive ad-jacent edges spanning an angle larger than π . Then there exist two planespanning trees R = G ( S, E R ) and B = G ( S, E B ) such that E R ∩ E B = ∅ and max { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )) . In addition, at most one edge of E R ∪ E B is longer than BE ( MST ( S )) .Proof. As before in Theorem 3 we will use our construction scheme for treesrooted at a leaf for the majority of the points and use a small local constructionthat avoids double edges. In this case, instead of removing a single vertex v todecompose the tree we use a set of four vertices as follows. We start at a leaf of MST ( S ) to generate a connected graph P with four vertices that is a subgraphof MST ( S ). Then we show how to construct R and B for S for the differentcases of P in combination with the remainder of MST ( S ). The construction of P = { v , v , v , v } : Let v be a leaf of MST ( S ).For the construction of P , we root MST ( S ) at v . We call the number of verticesin the (sub)tree of which that vertex is a root of (including itself) the weight of11 vertex. Hence, the weight of v is n . Further, the angle between two successiveincident edges at a vertex of MST ( S ) that is larger than π is called the big angle . (a) (b) (c)(d) (e) v v v v v v (f) v v v v v v v v v v v v v v v v v v Figure 6: Case 1 for P and the connections to the rest of MST ( S ). The graytriangles indicate possible subtrees of MST ( S ) and where and how they mightbe connected. Dotted edges are from MST ( P ). Note that the subtree with root v can be on either side of the supporting line of v v and even on both sides asindicated in the figure.Let v be the unique child of v in MST ( S ) (i.e., v = p ( v )). To define v and v we use a case distinction. Consider the set C of the children of v thatare not spanning the big angle with v at v . (Note that v may or may not bespanning the big angle at v and that C contains 0, 1, or 2 vertices).1. If v has only a single child (i.e., C is empty), or if C contains a vertexthat is not a leaf in MST ( S ), we choose it (or one of them) as v . Weassume w.l.o.g. that v is the successor of v in clockwise order around v .Further, we choose v as a child of v such that v and v are consecutivearound v and do not span the big angle at v . If v has two children, andthis is true for both, then we choose v such that it is the successor of v incounterclockwise order around v . See Figure 6 (a)–(f) for the six differentvariations of this case, i.e., all possible distributions of the positions ofthe big angles. Explicit definitions for these cases can be found below.W.l.o.g., we require the subtree at v to be nonempty in cases (d)–(f) andthe subtree of v to be nonempty in cases (c) and (f). For all cases weassume without loss of generality that the angle v v v is clockwise lessthan π .(1a) The edge v v is adjacent to the big angle at v and the angle v v v is clockwise less than π .(1b) The edge v v is adjacent to the big angle at v , the angle v v v isclockwise greater than π , and the edge v v is adjacent to the bigangle at v . 121c) The edge v v is adjacent to the big angle at v , the angle v v v isclockwise greater than π , and the edge v v is not adjacent to thebig angle at v .(1d) The edge v v is not adjacent to the big angle at v and the angle v v v is clockwise less than π .(1e) The edge v v is not adjacent to the big angle at v , the angle v v v is clockwise greater than π , and the edge v v is adjacent to the big angle at v .(1f) The edge v v is not adjacent to the big angle at v , the angle v v v is clockwise greater than π , and the edge v v is not adjacent tothe big angle at v .2. C contains at least one vertex and all vertices in C are leaves in MST ( S ).Note that this implies that v has degree exactly three in M ST ( S ), as thedegree of any vertex with a big angle in a minimum spanning tree is atmost three and thus C would be empty if v had degree at most two. Wechoose a vertex of C as v (assuming w.l.o.g. that v is the successor of v in clockwise order around v ), and choose the other child of v as v . Notethat if n ≥ v cannot be a leaf in MST ( S ) and hence v spans a bigangle with v at v . Therefore, taking the location of the big angle at v into account, there are two possibilities for the counterclockwise order ofincident edges around v . See Figure 7 (a) and (b).(2a) The angle v v v is clockwise less than π (the edge v v is adjacent to the big angle at v .)(2b) The angle v v v is clockwise greater than π (the edge v v is notadjacent to the big angle at v ). Here, both v and v must be leavesimplying that n = 4 and we do not have any subtrees. (a) (b) v v v v v v v v Figure 7: Case 2 for the selection of P and the connections to the rest of MST ( S ).Vertices drawn with squares are leaves of MST ( S ). The gray triangles indicatepossible subtrees of MST ( S ) and where they might be connected. Dotted edgesare from MST ( P ). Note that the subtree with root v can be on either sideof the supporting line of v v and even on both sides as indicated in the figure.Further note that case (b) can appear only if S contains exactly 4 vertices. The construction of R and B : First we show how to construct thetrees R P = G ( { v , v , v , v } , E P R ) and B P = G ( { v , v , v , v } , E P B ). The13ertices of P can either be in convex position or form a triangle with oneinterior point, with v interior for the cases shown in Figure 6 (b) and (e),and v interior for the cases shown in Figure 6 (e) and (f); there are no othernon-convex versions: otherwise either the path v , v , v , v could not be in MST ( S ), or one of the vertices of P could not be incident to a big angle (recallLemma 2). As in any non-convex case the complete graph on { v , v , v , v } is crossing-free, any construction of R P and B P for the convex cases is alsovalid for the non-convex cases. Let us thus assume the points of P to be inconvex position. For Cases 1a and 1d the vertices { v , v , v , v } must appearin this order on their convex hull and we set R P = { v v , v v , v v } and B P = { v v , v v , v v } , as illustrated in Figure 8 (a). For Cases 1b, 1c, 1e, and1f the vertices appear in the order { v , v , v , v } on their convex hull. (Theorder { v , v , v , v } violates the fact that the clockwise angle v v v is not largeand the order { v , v , v , v } has edges v v and v v crossing, which are bothedges of MST ( S ); no other orderings exist when accounting for symmetry.) Forthese cases we set R P = { v v , v v , v v } and B P = { v v , v v , v v } , as inFigure 8 (b). All edges except the edge v v (in E P B ) are from
MST ( S ) andhave endpoints with different levels in MST ( S ) rooted at v . In contrast, v v isan edge of MST ( S ), which could be crossed by other edges of the construction.We will later discuss how to handle this.For Case 2 the placement must be convex as v , v and v are adjacent to v and the clockwise v v v and v v v are convex for Cases 2a and 2b, respectively.Since the ordering around v is fixed (modulo symmetry) by the case definition,the vertices appear in the order { v , v , v , v } for Case 2a and { v , v , v , v } forCase 2b. For Case 2a we set R P = { v v , v v , v v } and B P = { v v , v v , v v } and for Case 2b we set R P = { v v , v v , v v } and B P = { v v , v v , v v } asillustrated in Figure 8 (c) and (d). For both cases, all edges are from MST ( S ). (a) (b) v v v v v v v v (d)(c) v v v v v v v v Figure 8: The different colorings for P : (a) for the cases from Figure 6 (a)and (d), (b) for the cases from Figure 6 (b), (c), (e), and (f), (c) for the casefrom Figure 7 (a), and (d) for the case from Figure 7 (b). The bold edge in (a)and (b) is an edge of MST ( S ), that is, an edge with link distance 3 in MST ( S ).With R P and B P as a base, we now create red and blue trees for all remainingsubtrees of M ST ( S ) and “attach” them to the base. For Case 1 we define threepossible subtrees. Let T (cid:48) = G ( S (cid:48) , E (cid:48) ) be the subtree (i.e., connected component)of MST ( S ) that contains v when removing v (and its incident edges) from MST ( S ). Likewise, let T (cid:48) = G ( S (cid:48) , E (cid:48) ) be the subtree of MST ( S ) that contains v when removing v and v from MST ( S ), and let T (cid:48) = G ( S (cid:48) , E (cid:48) ) be the subtree of MST ( S ) that contains v when removing v and v from MST ( S ). For Case 2(a)14here is one possible subtree T (cid:48) = G ( S (cid:48) , E (cid:48) ), which is the subtree of MST ( S )that contains v when removing v from MST ( S ). (Case 2(b) appears only if n = 4 and hence the construction is already completed.) The subtrees T (cid:48) , T (cid:48) ,and T (cid:48) are shown as (pairs of) gray triangles in Figure 6 and 7. To connectthese subtrees to the bases R P and B P , we create corresponding trees T , T ,and T depending on the different shown cases, then apply Construction 1 tothem, and possibly recolor them using Lemma 6.We first consider the different subtrees for Case 1. In essence, for each treewe pick a neighbor from { v , v , v , v } to add to T (cid:48) , T (cid:48) , T (cid:48) , which we then use asroot for Construction 1. When there is a choice we pick a root that is adjacentto the outgoing edge from v i into the subtree T (cid:48) i as defined more precisely below. T : For all cases, we consider the subtree T = G ( S , E ) of MST ( S ), with S = S (cid:48) ∪ { v } , E = E (cid:48) ∪ { v v } . We root T at r = v (observe, v is a leaf in T with unique child s = v ) and apply Construction 1 to get R = G ( S , E R )and B = G ( S , E B ), with the double-edge rs removed from both E R and E B . T : For the cases depicted in Figure 6 (a), (b), (d), and (e), we define T = G ( S , E ) of MST ( S ), such that S = S (cid:48) ∪ { v } , E = E (cid:48) ∪ { v v } .We root T at r = v (observe, v is a leaf in T with unique child v ) andapply Construction 1 to get R = G ( S , E R ) and B = G ( S , E B ), with thedouble-edge rs removed from both E R and E B .In the cases shown in Figure 6 (c) and (f), we define T = G ( S , E ) of MST ( S ), such that S = S (cid:48) ∪ { v } , E = E (cid:48) ∪ { v v } . We root T at r = v (observe, v is a leaf in T with unique child v ) and apply Construction 1 to get R = G ( S , E R ) and B = G ( S , E B ), with the double-edge rs removed fromboth E R and E B . T : For the cases depicted in Figure 6 (a)–(c), let T = G ( S , E ) be asubtree of MST ( S ), such that S = S (cid:48) ∪ { v } , E = E (cid:48) ∪ { v v } . We root T at r = v (observe, v is a leaf in T with unique child v ) and apply Construction 1to get R = G ( S , E R ) and B = G ( S , E B ), with the double-edge rs removedfrom both E R and E B .For the cases depicted in Figure 6 (d)–(f), T = G ( S , E ) is the subtree of MST ( S ), such that S = S (cid:48) ∪ { v } , E = E (cid:48) ∪ { v v } . We root T at r = v (observe, v is a leaf in T with unique child v ) and apply Construction 1 to get R = G ( S , E R ) and B = G ( S , E B ), with the double-edge rs removed fromboth E R and E B .It is easy to see that the edge sets E P R , E R , E R , E R , E P B , E B , E B ,and E B are all individually edge disjoint. In the following, we describe howthese edge sets are combined to form the two plane spanning trees R and B inthe different cases; see Figure 9 for the convex versions and Figure 10 for thenon-convex versions of the corresponding cases illustrated in Figure 6. For thenon-convex cases only points v and v can be in the interior as v or v being inthe interior would violate Lemma 2. Furthermore, in some of the cases (a)–(f),further restrictions apply as listed below.(a) Neither v nor v can be the middle point as both clockwise angles v v v and v v v have angle less than π .15b) Only v can be in the center (the subtree of v is nonempty and hence v would not be incident to a big angle).(c) Neither v nor v can be in the center (both subtrees are nonempty).(d) Similar to (a).(e) Both v and v may be the middle point.(f) Only v can be in the center (the subtree of v is non-empty). (a) (b) (c)(d) (e) (f) v v v v v v v v v v v v v v v v v v v v v v v v Figure 9: The different plane spanning trees R and B for case 1 when P is inconvex position. The bold blue edges v v are edges of MST ( S ), i.e., edges withlink distance 3 in MST ( S ), that might still be crossed. Note that v v can onlybe crossed in cases (a) and (b), as in the other cases this edge is surrounded by“uncrossable” MST ( S )-edges. v v v v (e) v v v v (e) v v v (f) v v v v v (b) Figure 10: The different plane spanning trees R and B for case 1 when P is notin convex position, with v or v in the interior. The case numberings are thesame as the ones in Figure 6.First we add the red and blue trees for T . By construction, only edges of E R connect to v (only crossing edges of E P B ) and the edges of E B do notcross any edge outside T . For the cases (a), (b), (d), and (e) we use the invertedcoloring (see Lemma 6) for the two trees of T . For the remaining cases (c)16nd (f) we use the original coloring (see Lemma 6) for the two trees of T . Foradding the red and blue trees for T we use the original coloring for the cases(a-c), and the inverted coloring for the cases (d-f). For the sake of simplicity,we exchange the names of the according sets E iR and E iB whenever we use theinverted coloring for a T i , i ∈ { , } .Since the edge sets E R , E R , and E R form spanning trees connecting thenodes of T , T , and T and since E P R connects their roots and v with a spanningtree, it follows that R = G ( S, E
P R ∪ E R ∪ E R ∪ E R ) is a spanning tree for S . The same argument applies to show that B = G ( S, E
P B ∪ E B ∪ E B ∪ E B )is a spanning tree. All edges in E R , E R , E R , E B , E B , and E B are from MST ( S ) and all edges of E P R and E P B are from
MST ( S ) ∪ { v v } , it followsthat max { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )), with only the edge v v , which mayoccur in E P B being possibly larger than 2 BE ( MST ( S )). What remains is toshow that both R and B are non-crossing.The MST ( S )-edge v v from E P B is also the only edge that could cause acrossing, see Lemma 4 and Theorem 1. Hence, R is a plane spanning tree. If v v is not crossed by any other edge of E B then also B is a plane spanning treeand we are done. Otherwise, note first that by Lemma 2, the triangles v v v and v v v cannot contain any points of S . Then observe that for the cases inFigure 9 (b), (c), (e) and (f) any edge crossing v v that does not have v , v , v or v as an endpoint must cross an MST -edge between v , v , v and v . Thisimplies that any MST -edge that crosses v v must have v , v , v or v as itswitness by Lemma 4. By construction, v , v , v and v are not a witness to anyblue edge between vertices in S \ { v , v , v , v } . The edges in E P B are incidentto v or v so they also cannot cross v v .For cases from Figure 9 (a) and (d), as well as all cases from Figure 10,observe first that v v , v v , and v v cannot be crossed, again due to Lemma 4and the fact that v , v , v and v cannot be a witness to any long edge in thegrey subtrees by construction. So any edge that crosses v v must have anendpoint in the interior of the convex hull of P or connect directly to v or v .The latter however cannot happen: The only points connecting with blue edgesto v or v are direct neighbors of these vertices, which reside in the large-angledwedge v v v or v v v , respectively. (Here, the large-angled wedge v v v isthe wedge spanned by the a ray from v to v and a ray from v to v so thatthe wedge has an opening angle larger than π . The large-angled wedge v v v isdefined analogously.) Hence, if v v is crossed by some blue edge, there mustbe a nonempty set X ⊂ S \ P that resides in the interior of the convex hullof P . In the cases depicted in Figure 9 (a) and (d), X lies in the triangle ∆spanned by v , v , and the intersection of v v and v v . In the cases depictedin Figure 10, the X lies in the triangle ∆ spanned by v , v , and the vertex of P in the interior of the convex hull of P . Further, removing the edge v v from B splits B into two connected components that are each plane trees, where v is in one and v is in the other component. Now consider the convex hull of X ∪ { v , v } , and the path along the boundary of that convex hull between v and v that contains at least one vertex of X . This path contains exactly oneedge e that connects the two components of B . Due to the construction of B R , e can neither be part of R (as the two endpoints of e must reside in twodifferent subtrees of v , v or v ) nor cross any edge of B (as the only possiblyintersected segment of the convex hull boundary of X was v v ). Further, thelength of e must be less than 3 BE ( MST ( P )), as e lies inside the triangle ∆, andas all sides of ∆ are bounded from above by 3 BE ( MST ( P )). Hence, as v v wasthe only edge that could be crossed by others, the replacement of v v by e in B results in two edge-disjoint plane spanning trees R and B with maximum edgelength less than 3 BE ( MST ( P )).As for Case 2 (b), S consists only of four vertices and hence Figure 8 (d)already shows all of the two trees R and B , it remains to consider the subtreefor Case 2 (a). v v v v Figure 11: The plane spanning trees R and B for Case 2. T : Consider the subtree T = G ( S , E ) of MST ( S ), with S = S (cid:48) ∪ { v } , E = E (cid:48) ∪ { v v } . We root T at r = v (observe, v is a leaf in T withunique child s = v ) and apply Construction 1 to obtain edge sets E (cid:48) R and E (cid:48) B ,since we will add connectivity between r = v and s = v using E P R and E P B we remove the edge rs to obtain R = G ( S , E R ) and B = G ( S , E B ), with E R = E (cid:48) R \ { rs } and E B = E (cid:48) B \ { rs } . We use the inverted coloring as definedin Lemma 6 for the two trees of T , implying that the edges connecting to v and crossing red edges of E P R are all blue. Hence the graphs R = G ( S, E
P R ∪ E R ) and B = G ( S, E
P B ∪ E B ) are plane spanning trees, E R ∩ E B = ∅ , andmax { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )).This concludes the proof. Corollary 5.
For any set S of n ≥ points in the plane, there are two planespanning trees R = G ( S, E R ) and B = G ( S, E B ) such that E R ∩ E B = ∅ and max { BE ( R ) , BE ( B ) } ≤ BE ( MST ( S )) . We now show that the above construction is worst-case optimal.
Theorem 6.
For any n > and k > there is a set S of n points such thatfor any k disjoint spanning trees, at least one has a bottleneck edge larger than ( k + 1) BE ( MST ( S )) .Proof. A counterexample simply consists of n points equally distributed on aline segment. The points can be slightly perturbed to obtain general position(similar to Figure 3). In this problem instance there are kn − ( k ( k + 1) /
2) edgeswhose distance is strictly less than ( k + 1) BE ( MST ( S )) = k + 1. However, weneed kn − k edges for k disjoint trees and thus it is impossible to construct thatmany trees with sufficiently short edges.18 Distributed Approach
The previous construction relies heavily on the minimum spanning tree of S . Itis well known that this tree cannot be constructed locally, thus we are implicitlyassuming that the network is constructed by a single processor that knows thelocation of all other vertices. In ad-hoc networks, it is often desirable that eachvertex can compute its adjacencies using only local information, i.e., using onlyinformation about vertices at most a certain maximum distance away.In the following, we provide an alternative construction. Although, for anyfixed k , the length of the edges is increased by a constant factor of 12 √ k (seeTheorem 9 below for details), it has the benefit that it can be constructed locallyand that it can be extended to compute k layers, for k ≤ n/
12. The only globalproperty that is needed is a value β that should be at least BE ( MST ( S )). Wealso note that these plane disjoint graphs are not necessarily trees, as large cyclescannot be detected locally.Before we describe our approach, we report the result of Biniaz and Garc´ıa [3]that every point set of at least 3 k points contains k layers. Since the details ofthis construction are important for our construction, we add a proof sketch. Theorem 7 ([3]) . Every finite point set that consists of at least k pointscontains k layers.Proof. First, recall that for every set of n points, there is a center point c suchthat every line through c splits the point set into two parts that each containat least (cid:98) n/ (cid:99) points, see e.g. Chapter 1 in [8] (note that c need not be one ofthe initial n points). For ease of explanation, we assume that every line through c contains at most one point. Number the points v , v , . . . , v n − in clockwisecircular order around c . We split the plane into three angular regions by the threerays originating from c and passing through v , v (cid:98) n (cid:99) , and v (cid:98) n (cid:99) ; see Figure 12.Since every line through the center contains at least n/ π and thus thethree angular regions are convex. We declare v to be the representative of theangular region between the rays through v and v (cid:98) n (cid:99) and connect the vertices v , ..., v (cid:98) n (cid:99) in this region to v . Similarly, we assign v (cid:98) n (cid:99) to be the representativeof angle between the rays center through v (cid:98) n (cid:99) and v (cid:98) n (cid:99) and connect vertices v (cid:98) n (cid:99) +1 , ..., v (cid:98) n (cid:99) to v (cid:98) n (cid:99) . Finally, we connect vertices v (cid:98) n (cid:99) +1 , ..., v n − to v (cid:98) n (cid:99) .This results in a non-crossing spanning tree.For the second tree, we rotate the construction and we use v , v (cid:98) n (cid:99) +1 , and v (cid:98) n (cid:99) +1 to define the three regions, and so on.While this construction provides a simple method of constructing the k layers, it does not give any guarantee on the length of the longest edge in thisconstruction. To give such a guarantee, we combine it with a bucketing approach:we partition the point set using a grid (whose size will depend on k and β ), solvethe problem in each box with sufficiently many points independently, and thencombine the subproblems to obtain a global solution (see Figure 13).19 v v (cid:98) n (cid:99) v (cid:98) n (cid:99) v (cid:98) n (cid:99) v (cid:98) n (cid:99) cc (a) (b) Figure 12: Extracting one layer: (a) The three sectors defined by v , v (cid:98) n (cid:99) , and v (cid:98) n (cid:99) . (b) Connecting the points to the representative of their sector. The rededges connect the representatives.We place a grid with cells of height and width 6 kβ and classify the pointsaccording to which grid cell contains them (if a point lies exactly on a separatingline, we assign it an arbitrary adjacent cell). We say that a grid cell is a densebox if it contains at least 3 k points of S . Similarly, it is a sparse box if itcontains points of S but is not dense. Two boxes are adjacent if they share someboundary or vertex. Hence, each box has 8 neighbors. This is also referred toas the 8-neighbor topology. We observe that dense and sparse boxes satisfy thefollowing properties. Lemma 7.
Given two non-adjacent boxes B and B (cid:48) , the points in B and B (cid:48) cannot be connected by edges of length at most β using only points from sparseboxes.Proof. Suppose the contrary and let B and B (cid:48) be two boxes such that there isa path that uses edges of length at most β between a point in B and a pointin B (cid:48) visiting only points in sparse boxes. This path crosses the sides of acertain number of boxes in a given order; let σ be the sequence of these sides,after repeatedly removing adjacent duplicates. Observe first that horizontaland vertical sides alternate in σ , as otherwise the path would traverse the cellwidth of 6 kβ using at most 3 k − β . Since B and B (cid:48) are non-adjacent, w.l.o.g., there is a vertical side s that hastwo adjacent horizontal sides in σ with different y -coordinates. Hence, betweenthe two horizontal sides, the corresponding part of the path has length at least6 kβ , and may use only the points in the two boxes adjacent to s . But sinceany sparse box contains at most 3 k − β , that part of the path can havelength at most (6 k − β , a contradiction. Corollary 8.
Dense boxes are connected by the 8-neighbor topology.
Lemma 8.
Any finite set S of at least · (3 k − points with β ≥ BE ( MST ( S )) contains at least one dense box. a) (b) Figure 13: The distributed approach: a grid is placed over the point set anddifferent representatives construct different graphs ((a) and (b)). The red andblack edges form the tree in each dense cell, blue edges connect the dense cells,and orange edges connect the vertices in sparse cells.
Proof.
Assume S and β induce only sparse boxes. This implies that the pointsare distributed over at least five boxes, and thus, there is a pair of boxes that isnon-adjacent. Using Lemma 7, this means that these boxes cannot be connectedusing edges of length at most BE ( MST ( S )), a contradiction. Lemma 9.
In any finite set S of at least · (3 k −
1) + 1 points with β ≥ BE ( MST ( S )) , all sparse boxes are adjacent to a dense box.Proof. This follows from Lemma 7, since any sparse box that is not adjacent toa dense box cannot be connected to any dense box using edges of length at most β ≥ BE ( MST ( S )).Next, we assign all points to dense boxes. In order to do this, let c B be thecenter of a dense box B . Note that c B is not necessarily the center point of thepoints in this box. We consider the Voronoi diagram of the centers of all denseboxes and assign a point p to B if p lies in the Voronoi cell of c B . Let S B be theset of points of S that are associated with a dense box B . We note that eachdense box B gets assigned at least all points in its own box, since in the case ofadjacent dense boxes, the boundary of the Voronoi cell coincides with the sharedboundary of these boxes (see Figure 14).Furthermore, we can compute the points assigned to each box locally. ByLemma 9 all sparse boxes are adjacent to a dense box, and hence for any point p in a sparse box B its distance to its nearest center is at most 3 (cid:96)/ √
2, where (cid:96) = 6 kβ . It follows that only the centers of cells of neighbors and neighbors ofneighbors need to be considered. 21igure 14: The Voronoi cells of the centers of the dense boxes. Lemma 10.
For any two dense boxes B and B (cid:48) , we have that the convex hullsof S B and S B (cid:48) are disjoint.Proof. We observe that the convex hull of S B is contained in the Voronoi cellof c B . Hence, since the Voronoi cells of different dense boxes are disjoint, theconvex hulls of the points assigned to them are also disjoint.For each dense box B , we apply Theorem 7 on the points inside the densebox to compute k disjoint layers of S B . Next, we connect all sparse points in S B to the representative of the sector that contains them in each layer. Since allpoints in the same sector connect to the same representative and the sectors ofthe same layer do not overlap, we obtain a plane graph for each layer within theconvex hull of each S B .Hence, we obtain k pairwise disjoint layers such that in each layer the pointsassociated to each dense box are connected. Moreover, since the created edgesstay within the convex hull of each subproblem and by Lemma 10 those hullsare disjoint, each layer is plane. Thus, to assure that each layer is connected, wemust connect the construction between dense boxes.We connect adjacent dense boxes in a tree-like manner using the followingrules: • Connect every dense box to any dense box below it. • Always connect every dense box to any dense box to the left of it. • If neither the box below nor the one to the left of it is dense, connect thebox to the dense box diagonally below and to the left of it. • If neither the box above nor the one to the left of it is dense, connect thebox to the dense box diagonally above and to the left of it.To connect two dense boxes, we find and connect two representatives p and q (one from each dense box) such that p lies in the sector of q and q lies in thesector of p ; see Figure 15 (a). Lemma 11.
For any layer and any two adjacent dense boxes B and B (cid:48) , thereare two representatives p and q in B and B (cid:48) , respectively, such that p lies in thesector of q and q lies in the sector of p . q c c (cid:48) (cid:96) (cid:96) (cid:96) r (cid:48) (cid:96) (cid:48) W W W W (cid:48) (a) (b) Figure 15: Connecting two dense boxes by means of p and q . The half-circlesin (a) indicate which sector each representative covers. The red edges connectthe dense boxes internally and the blue edge connects the two dense cells. (b)illustrates the sectors involved in connecting two neighboring dense boxes. Proof.
Consider two boxes B and B (cid:48) with center points (of their respective pointsets) c and c (cid:48) . Now let W and W (cid:48) with representatives r and r (cid:48) denote thesectors containing c (cid:48) and c , respectively; see Figure 15. The other sectors W and W of B with representatives r and r are ordered clockwise. We use (cid:96) i to denote the ray from c containing r i . If r ∈ W (cid:48) and r (cid:48) ∈ W we are done.So assume that r (cid:48) (cid:54)∈ W , the case when r (cid:54)∈ W (cid:48) (or when both r (cid:54)∈ W (cid:48) and r (cid:48) (cid:54)∈ W ) is symmetric. It follows that r (cid:48) is in sector W if the line segment c (cid:48) r (cid:48) intersects (cid:96) or sector W if the segment intersects (cid:96) and (cid:96) . Assume that r (cid:48) isin sector W (again the argument is symmetric when r (cid:48) is in sector W ). Now r can be positioned on (cid:96) between c and the intersection point with c (cid:48) r (cid:48) or behindthis intersection point when viewed from c . In the former case r (cid:48) is in W and r is in W (cid:48) and we are done. In the latter case the segments cr and c (cid:48) r (cid:48) cross.Since c, r ∈ B and c (cid:48) , r (cid:48) ∈ B (cid:48) this crossing would imply that B and B (cid:48) are notdisjoint, a contradiction.Now that we have completed the description of the construction, we showthat each layer of the resulting graph is plane and connected, and that the lengthof the longest edge is bounded. Lemma 12.
Each layer is plane.Proof.
Since dense boxes are internally plane and the addition of edges to thesparse points do not violate planarity, it suffices to show that the edges betweendense boxes cannot cross any previously inserted edges and that these edgescannot intersect other edges used to connect dense boxes.We first show that the edge used to connect boxes B and B (cid:48) is contained inthe union of the Voronoi cells of these two boxes. If B and B (cid:48) are horizontally orvertically adjacent, the connecting edge stays in the union of the two dense boxes,23hich is contained in their Voronoi cells. If B and B (cid:48) are diagonally adjacent,we connect them only if their shared horizontal and vertical neighbors are notdense. This implies that at least the two triangles defined by the sides of B and B (cid:48) that are adjacent to their contact point are part of the union of the Voronoicells of these boxes. Hence, the edge used to connect B and B (cid:48) cannot intersectthe Voronoi cell of any other dense box. Since all points of a dense box in asector connect to the same representative and these edges lie entirely inside thesector, the edge connecting two adjacent boxes can intersect only at one of thetwo representatives, but does not cross them. Therefore, an edge connecting twoadjacent dense boxes by connecting the corresponding representatives cannotcross any previously inserted edge.Next, we show that edges connecting two pairs of dense boxes cannot cross.Since any edge connecting two dense boxes stays within the union of the Voronoicells of B and B (cid:48) , the only way for two edges to intersect is if they connect to thesame box B and intersect in the Voronoi cell of B . If the connecting edges liein the same sector of B , they connect to the same representative and thus theycannot cross. If they lie in different sectors of B , the edges lie entirely insidetheir respective sectors. Since these sectors are disjoint, this implies that theedges cannot intersect. Lemma 13.
Each layer is connected.Proof.
Since the sectors of the representatives of the dense boxes cover the plane,each point in a sparse box is connected to a representative of the dense box it isassigned to. Hence, showing that the dense boxes are connected completes theproof.By Corollary 8, the dense boxes are connected using the 8-neighbor topology.This implies that there is a path between any pair of dense boxes where every stepis one to a horizontally, vertically, or diagonally adjacent box. Since we alwaysconnect horizontally or vertically adjacent boxes and we connect diagonallyadjacent boxes when they share no horizontal and vertical dense neighbor, thelayer is connected after adding edges as described in the proof of Lemma 11.
Lemma 14.
The distance between a representative in a dense box B and anypoint connecting to it is at most √ kβ .Proof. Since the representatives of B are connected only to points from denseand sparse boxes adjacent B , the distance between a representative and a pointconnected to it is at most the length of the diagonal of the 2 × B as one of its boxes. Since a box has width 6 kβ , this diagonal has length2 √ · kβ = 12 √ kβ . Theorem 9.
For all finite point sets with at least k −
1) + 1 points, we can ex-tract k plane layers with the longest edge having length at most √ k BE ( MST ( S )) . Conclusions
We presented two algorithms for constructing k edge-disjoint non-crossing planespanning graphs on a given point set such that the length of the bottleneckedge is minimized. The first algorithm uses global properties in order to keepall edges as small as possible. We also give matching worst-case lower bounds,making the algorithm tight. The main drawback is that this method can only beused to construct two layers, and it is unlikely that a similar approach can workfor more. Our second algorithm works for a large number of layers (up to n/ √ BE ( MST ( S )) implied by thismethod is far larger than the 3 BE ( MST ( S )) of the first approach.So far, there is no centralized method to construct more than two trees.Finding such a method and comparing it to the distributed method presentedhere is an interesting direction of future research. Another direction would be tolower the length of the longest edge in the distributed construction, though froma purely worst-case theoretical point of view this is likely to require a differentapproach from the one used in this paper. Acknowledgments
This research was initiated during the 10th European Research Week on Geo-metric Graphs (GGWeek 2013), Illgau, Switzerland. We would like to thank allparticipants for fruitful discussions. O.A., A.P., and B.V. were partially sup-ported by the ESF EUROCORES programme EuroGIGA - ComPoSe, AustrianScience Fund (FWF): I 648-N18. T.H. was supported by the Austrian ScienceFund (FWF): P23629-N18 ‘Combinatorial Problems on Geometric Graphs’. M.K.was supported in part by the ELC project (MEXT KAKENHI No.17K12635)and NSF award CCF-1423615.. A.P. is supported by an Erwin Schr¨odingerfellowship, Austrian Science Fund (FWF): J-3847-N35. A.v.R. and M.R. weresupported by JST ERATO Grant Number JPMJER1201, Japan.