Plane Spanning Trees in Edge-Colored Simple Drawings of K n
Oswin Aichholzer, Michael Hoffmann, Johannes Obenaus, Rosna Paul, Daniel Perz, Nadja Seiferth, Birgit Vogtenhuber, Alexandra Weinberger
PPlane Spanning Trees in Edge-ColoredSimple Drawings of K n(cid:63) Oswin Aichholzer , Michael Hoffmann , Johannes Obenaus ,Rosna Paul , Daniel Perz , Nadja Seiferth , Birgit Vogtenhuber , andAlexandra Weinberger Institute of Software Technology, Graz University of Technology, Graz, Austria { oaich,ropaul,daperz,bvogt,aweinber } @ist.tugraz.at Department of Computer Science, ETH Z¨urich, Switzerland [email protected] Institut f¨ur Informatik, Freie Universit¨at Berlin, Germany { johannes.obenaus,nadja.seiferth } @fu-berlin.de Abstract.
K´arolyi, Pach, and T´oth proved that every 2-edge-coloredstraight-line drawing of the complete graph contains a monochromaticplane spanning tree. It is open if this statement generalizes to otherclasses of drawings, specifically, to simple drawings of the complete graph.These are drawings where edges are represented by Jordan arcs, any twoof which intersect at most once. We present two partial results towardssuch a generalization. First, we show that the statement holds for cylin-drical simple drawings. (In a cylindrical drawing, all vertices are placedon two concentric circles and no edge crosses either circle.) Second, we in-troduce a relaxation of the problem in which the graph is k -edge-colored,and the target structure must be hypochromatic , that is, avoid (at least)one color class. In this setting, we show that every (cid:100) ( n + 5) / (cid:101) -edge-colored monotone simple drawing of K n contains a hypochromatic planespanning tree. (In a monotone drawing, every edge is represented as an x -monotone curve.) Keywords:
Simple drawing · Cylindrical drawing · Monotone drawing · Plane subdrawing. A simple drawing of a graph represents vertices by pairwise distinct points (inthe Euclidean plane) and edges by Jordan arcs connecting their endpoints such (cid:63) We are particularly grateful to Irene Parada for bringing this problem to our atten-tion. We also thank the organizers of the 4 th DACH Workshop on Arrangements, thattook place in February 2020 in Malchow and was funded by Deutsche Forschungs-gemeinschaft (DFG), the Austrian Science Fund (FWF) and the Swiss NationalScience Foundation (SNSF). M. H. is supported by SNSF Project 200021E-171681.R. P. and A. W. are supported by FWF grant W1230. J. O. is supported by ERCStG 757609. N. S. is supported by DFG Project MU3501/3-1. D. P. and B. V. aresupported by FWF Project I 3340-N35. a r X i v : . [ c s . C G ] A ug Aichholzer, Hoffmann, Obenaus, Paul, Perz, Vogtenhuber, and Weinberger that (1) no (relative interior of an) edge passes through a vertex and (2) everypair of edges intersect at most once, either in a common endpoint or in theirrelative interior, forming a proper crossing. Simple drawings (also called gooddrawings [7] or simple topological graphs [12]) have been well studied, amongstothers, in the context of crossing minimization (see e.g. [15]), as it is knownthat every crossing-minimal drawing of a graph is simple. Also every straight-line drawing is simple. Further well-known classes of simple drawings relevantfor this work are pseudolinear drawings , where every edge can be extended to abi-infinite Jordan arc such that every pair of them intersects exactly once; cylin-drical simple drawings , where all vertices are placed on two concentric circles, noedge crosses either circle, and edges between two vertices on the outer (inner)circle lie completely outside (inside) that circle; , where allvertices lie on a line and no edge crosses that line; and monotone simple drawings ,where all edges are x -monotone curves. Unless explicitly mentioned otherwise,all considered drawings are simple, and the term simple is mostly omitted.In this paper we are concerned with finding plane substructures in simpledrawings. Specifically, we study the existence of plane spanning trees in edge-colored simple drawings of the complete graph K n . A k -edge-coloring of a graphis a map from its edge set to a set of k colors. A subgraph H of a k -edge-coloredgraph G is hypochromatic if the edges of H use at most k − H avoids at least one of the k color classes. If all edges of H have the same color,then H is monochromatic . We are inspired by the following conjecture. Conjecture 1.
Every 2-edge-colored simple drawing of K n contains a monochro-matic plane spanning tree.K´arolyi, Pach, and T´oth [10] proved the statement for straight-line drawings,where the 2-edge-coloring can also be interpreted as a Ramsey-type setting,where one color corresponds to the edges of the graph and the other color to theedges of its complement. Such an interpretation is less natural in the topologicalsetting, where the edges are not implicitly defined by placing the vertices.Unfortunately, a proof of Conjecture 1 seems elusive. However, we show thatit holds for specific classes of simple drawings, such as 2-page book drawings,pseudolinear drawings, and cylindrical drawings. The result for 2-page bookdrawings can be shown straightforwardly. The statement for pseudolinear draw-ings follows from generalizing the proof for straight-line drawings by K´arolyi,Pach, and T´oth [10] to this setting. Proposition 1.
Every -edge-colored -page book drawing of K n contains aplane monochromatic spanning tree. Proposition 2.
Every 2-edge-colored pseudolinear drawing of K n contains aplane monochromatic spanning tree. See Appendix A for proofs of those statements. Note that the coloring need not be proper nor have any other special properties.lane Spanning Trees in Edge-Colored Simple Drawings of K n The result for cylindrical drawings is more involved; it forms our first maincontribution.
Theorem 1.
Every -edge-colored cylindrical simple drawing of K n contains amonochromatic plane spanning tree. In light of the apparent challenge in attacking Conjecture 1, we also considerthe following generalized formulation, which uses more colors.
Conjecture 2.
For k ≥
2, every k -edge-colored simple drawing of K n contains ahypochromatic plane spanning tree.Note that both conjectures are in fact equivalent: On the one hand, Con-jecture 2 implies Conjecture 1 by setting k = 2. On the other hand, assumingConjecture 2 holds for some k , it also holds for every larger k (cid:48) because we cansimply merge color classes until we are down to k colors. Avoiding any one ofthe resulting color classes also avoids at least one of the original color classes.Our second result is the following statement about monotone drawings. Theorem 2.
Every (cid:100) ( n + 5) / (cid:101) -edge-colored monotone simple drawing of K n contains a hypochromatic plane spanning tree. Finally, note that some assumptions concerning the drawing are necessary toobtain any result on the existence of plane substructures. Without any restric-tion, every pair of edges may cross. The class of simple drawings is formed bytwo restrictions: forbid adjacent edges to cross and forbid independent edges tocross more than once. Both restrictions are necessary in the statement of Con-jecture 1. If adjacent edges may cross, then one can construct drawings whereevery pair of adjacent edges crosses (e.g., in the neighborhood of the commonvertex), implying that no plane substructure can have a vertex of degree morethan one. And for star-simple drawings, where adjacent edges do not cross butindependent edges may cross more than once, already K admits 2-edge-coloredstar-simple drawings without any monochromatic plane spanning tree; see Fig. 1.Fig. 1: Star-simple drawings of K without monochromatic plane spanning tree. Aichholzer, Hoffmann, Obenaus, Paul, Perz, Vogtenhuber, and Weinberger
Related Work.
The problem of finding plane subdrawings in a given drawinghas gained some attention over the past decades. We mention only a few resultsfrom the vast literature on plane substructures. In 1988, Rafla [13] conjecturedthat every simple drawing of K n contains a plane Hamiltonian cycle. By nowthe conjecture is known to be true for n ≤ Lemma 1.
Let D be a k -edge-colored simple drawing of K n , for k ≥ . If oneof the color classes is not spanning, then D contains a hypochromatic planespanning tree. This section is devoted to Theorem 1, which states that every 2-edge-coloredcylindrical drawing of K n contains a monochromatic plane spanning tree. Wegive a detailed outline of the proof. The full proof can be found in Appendix B.For easier readability, we introduce some names for the different elementsof a cylindrical drawing (cf. Fig. 2). We call the vertices on the inner (outer)circle inner (outer) vertices . Similarly, we call edges connecting two inner (outer)vertices inner (outer) edges ; the remaining edges are called side edges . The edgesbetween consecutive vertices on the inner (outer) circle are called cycle edges and the union of all inner (outer) cycle edges are called inner (outer) cycle . Thedefinition of cylindrical drawings implies that all cycle edges are uncrossed. The rotation of a vertex v is the circular ordering of all edges incident to v . In thisordering, the cycle edges separate the inner (outer) edges from the side edges.Hence, the rotation of v induces a linear order on the side edges incident to v . Proof (sketch).
Our proof consists of two steps. In Step 1, we restrict consider-ations to drawings fulfilling two properties, for which we compute a monochro-matic plane spanning subgraph using a multi-stage sweep algorithm. In Step 2,we show how to handle drawings that do not fulfill all properties from Step 1.
Step 1.
Let D be a 2-edge-colored cylindrical drawing that fulfills the followingproperties:(P1) D has inner and outer vertices, and(P2) D ’s inner and outer cycle are both monochromatic, but of different color. lane Spanning Trees in Edge-Colored Simple Drawings of K n v w Fig. 2: Sketch of a cylindrical drawing. Inner edges are drawn blue, outer edgesred, and side edges black. vw is the first side edge in the clockwise rotation of v .Assume without loss of generality that the inner cycle of D is blue and hencethe outer cycle is red. We will refer to them as the blue and red cycle and to thevertices on them as blue and red vertices, respectively.We use the following algorithm to compute a (bichromatic) subdrawing H of D consisting of some side edges of D and their endpoints (cf. Fig. 3). Phase 0.
Initially, let H be empty. Choose an arbitrary inner vertex as initial rotation vertex v cur , set the rotation direction to clockwise, and set the firstside edge of v cur in the rotation direction as initial current edge e cur . Phase 1.
We repeat the following process while e cur is a side edge and while H is still missing vertices from the cycle of D not containing v cur : Add e cur to H ;If e cur does not have the same color as v cur , set v cur to be the other endpointof e cur and reverse the rotation direction (clockwise ↔ counterclockwise);In any case, set e cur to be the next edge incident to v cur after e cur in the(possibly changed) rotation direction. Phase 2. If H contains all vertices of D from the cycle not containing v cur :Return H . Phase 3.
Otherwise: Set H prev = H , reset H to be empty, reverse the rotationdirection, set e cur to be the first side edge of v cur in the new rotation direction,and restart with Phase 1.The following invariants hold for the algorithm (see Appendix B for a proof):(J1) At any time, the union of H and the two cycles of D forms a plane drawing.(J2) Any blue (red) vertex in H is incident to a red (blue) edge in H , exceptfor the current rotation vertex.(J3) Assume that Phase 1 is performed more than once and let V ( H ) be the setof vertices of H . Then for any i ≥
2, after round i of Phase 1, either V ( H )is a strict superset of V ( H prev ) or H contains all vertices from the cyclenot containing v cur , the current rotation vertex (or both conditions hold).Using those invariants, we can now complete Step 1: By (J3), the algorithmterminates. And by (J1) and (J2), at least one of the color classes of the unionof H and the two cycles of D is a monochromatic plane spanning graph for D . Aichholzer, Hoffmann, Obenaus, Paul, Perz, Vogtenhuber, and Weinberger v v vv (cid:48) v (cid:48) v (cid:48) w w w Fig. 3: The first steps of the algorithm. The black arc at vertex v indicates that vv (cid:48) is the first side edge of v in clockwise order (the initial rotation direction). Step 2.
Now assume that D violates at least one of the properties (P1) and (P2).If it violates (P1), then D is isomorphic to a 2-page book drawing and hencecontains a monochromatic plane spanning tree (see Proposition 1).If D does not fulfill (P2), then we remove vertices whose cycle edges areof different color until we reach a subdrawing D (cid:48) where both cycles are mono-chromatic, find a plane monochromatic spanning tree on D (cid:48) by either Step 1 orLemma 1, and then extend it to a monochromatic spanning tree on D . (cid:117)(cid:116) In this section, we prove the existence of hypochromatic plane spanning trees in k -edge-colored monotone drawings of K n , for k linear in n . Lemma 2.
Conjecture 1 holds for any simple drawing of K n with n ≤ vertices. For n ≤ n = 5 , . . . , of simple drawings of K n [1] and checked for allpossible 2-edge colorings that there exists a monochromatic plane spanning tree.Computations for n = 8 are currently out of reach, as there are 5,370,725 weakisomorphism classes of simple drawings [1] and more than 10 possible coloringsfor each of them. Proof (of Theorem 2).
Let d ≥ k = (cid:100) ( n + d − /d (cid:101) = (cid:100) ( n − /d (cid:101) + 1. The argument works for any d so that Conjecture 1holds for all monotone drawings on up to d + 1 vertices.Consider a k -edge-colored monotone drawing D of K n , and let v , v , . . . , v n − denote the sequence of vertices in increasing x -order. We partition the ver-tices into k − G , . . . , G k − of size at most d + 1 by setting G i =( v di , v di +1 , . . . , v di + d ). (The last group may have less than d + 1 vertices.) Ob-serve that G i ∩ G i +1 = { v d ( i +1) } . Two simple drawings of K n are weakly isomorphic iff they have the same crossingedge pairs.lane Spanning Trees in Edge-Colored Simple Drawings of K n We proceed in two phases. In both phases we consider each group separately.At the end of the first phase, we choose which color to remove. At the end of thesecond phase, we have an induced plane spanning tree T i for G i that avoids thechosen color, for each i ∈ { , . . . k − } . As D is monotone, the union (cid:83) k − i =0 T i forms a hypochromatic plane spanning tree in D .In the first phase, we consider each group G i , and check whether it has amonochromatic plane spanning tree in some color c . If so, we put c in a set S of colors to keep. If not, then by Conjecture 1 (which we assume to hold for G i ,as G i has at most d + 1 vertices) we can remove any single color and still find amonochromatic plane spanning tree in G i . (If c is the color to be removed, thenconsider the bicoloring where all colors other than c are merged into a singlesecond color.) As | S | ≤ k −
1, we can choose a color not in S to be removed atthe end of the first phase.In the second phase, for each group G i we either select a monochromaticplane spanning tree (if it exists), or find a plane spanning tree that avoids thechosen color.To obtain the statement of Theorem 2, we use the result of Lemma 2. (cid:117)(cid:116) Besides resolving the conjectures in full generality, it would be interesting toprove them for other specific classes of drawings (e.g., monotone). A useful stepin this direction would be to expand the range of k for which Conjecture 2 holds. References
1. ´Abrego, B., Aichholzer, O., Fern´andez-Merchant, S., Hackl, T., Pammer, J., Pilz,A., Ramos, P., Salazar, G., Vogtenhuber, B.: All good drawings of small com-plete graphs. In: Abstracts 31 st European Workshop on Computational Geometry(EuroCG’15). pp. 57–60 (2015)2. Aichholzer, O., Garc´ıa, A., Parada, I., Vogtenhuber, B., Weinberger, A.: Simpledrawings of K m,n contain shooting stars. In: Abstracts 35 th European Workshopon Computational Geometry (EuroCG’20). pp. 36:1–36:7 (2020)3. Bernhart, F., Kainen, P.C.: The book thickness of a graph. Journal of Combi-natorial Theory, Series B (3), 320–331 (1979). https://doi.org/10.1016/0095-8956(79)90021-24. Biniaz, A., Garc´ıa, A.: Partitions of complete geometric graphsinto plane trees. Computational Geometry , 101653 (2020).https://doi.org/10.1016/j.comgeo.2020.1016535. Bose, P., Hurtado, F., Rivera-Campo, E., Wood, D.R.: Partitions of complete ge-ometric graphs into plane trees. Computational Geometry (2), 116–125 (2006).https://doi.org/10.1016/j.comgeo.2005.08.0066. Brualdi, R.A., Hollingsworth, S.: Multicolored trees in complete graphs.Journal of Combinatorial Theory, Series B (2), 310–313 (1996).https://doi.org/10.1006/jctb.1996.00717. Erd˝os, P., Guy, R.: Crossing number problems. The American MathematicalMonthly , 52–58 (1973) Aichholzer, Hoffmann, Obenaus, Paul, Perz, Vogtenhuber, and Weinberger8. Erd˝os, P., Neˇsetril, J., R¨odl, V.: Some problems related to partitions of edges of agraph. Graphs and other combinatorial topics, Teubner, Leipzig (1983)9. Goodman, J.E.: Proof of a conjecture of Burr, Gr¨unbaum, and Sloane. DiscreteMathematics (1), 27–35 (1980). https://doi.org/10.1016/0012-365X(80)90096-510. K´arolyi, G., Pach, J., T´oth, G.: Ramsey-type results for geometricgraphs, I. Discrete & Computational Geometry , 247–255 (1997).https://doi.org/10.1007/PL0000931711. Keller, C., Perles, M.A., Rivera-Campo, E., Urrutia-Galicia, V.: Blockers for non-crossing spanning trees in complete geometric graphs. In: Thirty Essays on Geo-metric Graph Theory, pp. 383–397. Springer (2013). https://doi.org/10.1007/978-1-4614-0110-0 2012. Kynˇcl, J.: Enumeration of simple complete topological graphs. European Journalof Combinatorics , 1676–1685 (2009). https://doi.org/10.1016/j.ejc.2009.03.00513. Rafla, N.H.: The good drawings D n of the complete graph K n . Ph.D. thesis, McGillUniversity, Montreal (1988)14. Rivera-Campo, E., Urrutia-Galicia, V.: A sufficient condition for the existence ofplane spanning trees on geometric graphs. Computational Geometry (1), 1–6(2013). https://doi.org/10.1016/j.comgeo.2012.02.00615. Schaefer, M.: The graph crossing number and its variants: A sur-vey. Electronic Journal of Combinatorics, Dynamic Survey (4) (2020).https://doi.org/10.37236/2713 A Preliminary Results
Observation 1.
Let D be a 2-edge-colored simple drawing of K n and v be avertex incident to (monochromatically) uncrossed edges of both color classes. If D \ { v } contains a monochromatic plane spanning tree, then so does D . Using this observation, it is not hard to see that any 2-page book drawingcontains a monochromatic plane spanning tree.
Proposition 1.
Every -edge-colored -page book drawing of K n contains aplane monochromatic spanning tree.Proof. Let D be a 2-edge-colored 2-page book drawing of K n (with colors red andblue). As long as D contains vertices that are incident to at least one (monochro-matically) uncrossed blue edge and at least one (monochromatically) uncrossedred edge, iteratively remove these vertices to obtain a subdrawing D (cid:48) . Clearly, D (cid:48) remains a 2-page book drawing.Label the vertices in D (cid:48) with v , v , .., v j from left to right. By the propertiesof 2-page book drawings, the edges v i v i +1 between consecutive vertices are un-crossed. Since no vertex is adjacent to two differently colored, uncrossed edges,the path v . . . v j is monochromatic and thus forms a plane monochromatic span-ning tree for D (cid:48) .Finally, using Observation 1, re-add the previously removed vertices in inverseorder to obtain a plane monochromatic spanning tree of D . (cid:117)(cid:116) Proposition 2.
Every 2-edge-colored pseudolinear drawing of K n contains aplane monochromatic spanning tree. lane Spanning Trees in Edge-Colored Simple Drawings of K n The proof for the straight line case in [10] uses two concepts: The existence ofa convex hull and the monotonicity of all edges. We will observe that pseudolineardrawings fulfill both and then follow the lines of the straight-line proof.
Proof.
The proof goes by induction on the number n of vertices. As inductionbase let n = 2. Then there is a plane monochromatic spanning tree consisting ofthe only edge in the drawing. So assume that any pseudolinear drawing of K n − contains a plane monochromatic spanning tree.For the induction step we consider a 2-colored pseudolinear drawing of K n and call it D . We consider first the case that there exists a vertex v that isincident to an uncrossed red and an uncrossed blue edge. Then the subdrawing D \ { v } contains a plane monochromatic spanning tree by our induction hypoth-esis. Thus, the drawing D contains a plane monochromatic spanning tree byObservation 1.So let us assume that D does not contain any vertex that is incident totwo differently colored crossing-free edges. To prove that D contains a planemonochromatic spanning tree in this case, we will use the following well knownfact (whose proof we include for the sake of self-containment). Claim.
The outermost edges of any pseudolinear drawing of K n form an un-crossed cycle.Proof of Claim: Assume, for contradiction, there are edges e = u v , e = u v that lie partly on the boundary and cross each other. Let H be the subdraw-ing induced by { u , u , v , v } . Let e be extended by pseudoline (cid:96) and e be ex-tended by (cid:96) . The pseudolines (cid:96) and (cid:96) are intersected by all edges of H \{ e , e } in one of the vertices of H . Thus, they cannot have a crossing point with anyedges of H \ { e , e } in the interior of that edge. Thus, every edge other than e and e has to stay completely on one side of each of (cid:96) and (cid:96) , respectively. Theedges together form a cycle that completely encloses e and e ; see Figure 4.This means in particular that neither e nor e can lie (partly) on the boundaryof the drawing. (cid:4) Since D does not contain any vertices that are incident to crossing-free edgescolored in different colors, it follows that the boundary cycle of D is monochro-matic. Assume without loss of generality that the boundary cycle of D is red. Ifall vertices lie on the boundary, the boundary edges form a plane red spanningtree and we are done. Otherwise there exists at least one interior vertex. Sinceby [9], every pseudoline arrangement is isomorphic to a pseudoline arrangementin which every pseudoline is x -monotone, we can assume that our pseudolineardrawing is x -monotone. This implies that there are at least two more uncrossededges: One uncrossed edge is incident to the leftmost vertex and the leftmostvertex that is not on the boundary; another uncrossed edge is incident to therightmost vertex and the rightmost vertex that is not on the boundary. Bothedges have to be red, because D does not contain any vertices that are adjacentto two differently colored uncrossed edges. e e u v u v (cid:96) (cid:96) Fig. 4: The edges e and e cross and are extended by the (black, dashed) pseu-dolines (cid:96) and (cid:96) . The (blue) edges that are in the same K are forced by (cid:96) and (cid:96) to stay on one side of the crossing.By the assumption that our pseudolinear drawing is x -monotone we can labelthe vertices x , x , ..., x n in x -monotone order. By our induction hypothesis,the subdrawings induced by x , x , ..., x i and by x i , x i +1 , ...., x n contain planemonochromatic spanning trees for any i ∈ { , ..., n − } . Let T li be the planemonochromatic spanning tree of the subdrawing induced by x , x , ..., x i and T ri the plane monochromatic spanning tree subdrawing induced by x i , x i +1 ..., x n .If both of them have the same color, then T li (cid:83) T ri forms a plane monochromaticspanning tree for the whole drawing. So assume that they have different colors.We know from the color of the first and the last edge that T l and T rn − arered. Thus there has to be an i for which T li is red and T ri +1 is red as well. If theedge x i x i +1 is red, we can use it to connect the two spanning trees. If the edge isblue, it is not part of the boundary cycle. We can use the boundary edge aboveor the boundary edge below x i x i +1 to connect the two spanning trees. (cid:117)(cid:116) Lemma 1.
Let D be a k -edge-colored simple drawing of K n , for k ≥ . If oneof the color classes is not spanning, then D contains a hypochromatic planespanning tree.Proof. Assume, without loss of generality, that the edges of the red color classcontain no spanning tree (not even a crossing one). Then the subdrawing inducedby the red edges has at least two different components. Let A be the vertex setof one of those components and let B be the vertices that are not in A . Thereare no red edges between A and B . This means that the subdrawing inducedby the remaining edges contains a complete bipartite graph with sides of thepartition A and B . Every complete bipartite graph contains a plane spanningtree [2]. Thus D contains a plane hypochromatic spanning tree (consisting ofonly non-red edges). (cid:117)(cid:116) lane Spanning Trees in Edge-Colored Simple Drawings of K n B Full proof of Theorem 1
Theorem 1.
Every -edge-colored cylindrical simple drawing of K n contains amonochromatic plane spanning tree.Proof. Our proof consists of two steps. In Step 1, we restrict considerations todrawings fulfilling two properties, for which we compute a monochromatic planespanning subdrawing using a multi-stage sweep algorithm. In Step 2, we showhow to handle drawings that do not fulfill all properties from Step 1.
Step 1.
Let D be a 2-edge-colored cylindrical drawing that fulfills the followingproperties:(P1) D has inner and outer vertices, and(P2) D ’s inner and outer cycle are both monochromatic, but of different color.To simplify the description, we assume without loss of generality that theinner cycle of D is blue and hence the outer cycle is red. We will refer to themas the blue and red cycle and to the vertices on them as blue and red vertices,respectively. We remark that, if there are less than three vertices on a cycle, thenthe cycle is in fact not a cycle in the graph-theoretic meaning, as it has at mostone edge. Moreover, if a cycle has only one vertex and hence does not have anyedges, we can assume it to be of any color.We use the following algorithm to compute a (possibly bichromatic) sub-drawing H of D consisting of a subset of side edges of D and their endpoints(cf. Fig. 5). Phase 0.
Initially, let H be empty. Choose an arbitrary inner vertex as initial rotation vertex v cur , set the rotation direction to clockwise, and set the firstside edge of v cur in the rotation direction as initial current edge e cur . Phase 1.
We repeat the following process while e cur is a side edge and while H is still missing vertices from the cycle of D not containing v cur : Add e cur to H ;If e cur does not have the same color as v cur , set v cur to be the other endpointof e cur and reverse the rotation direction (clockwise ↔ counterclockwise);In any case, set e cur to be the next edge incident to v cur after e cur in the(possibly changed) rotation direction. Phase 2. If H contains all vertices of D from the cycle not containing v cur :Return H . Phase 3.
Otherwise: Set H prev = H , reset H to be empty, reverse the rotationdirection, set e cur to be the first side edge of v cur in the new rotation direction,and restart with Phase 1.Intuitively speaking, this algorithm sweeps back and forth in a zig-zag manner(see Fig. 5 for an illustration). We remark that Phase 1 adds at least one edgeto H , namely e cur = vv (cid:48) as set in Phase 0. Moreover, the active subdrawing H constructed in Phase 1 of the algorithm consists of a main path (also called backbone path ) of alternating red and blue edges corresponding to the switches v v vv (cid:48) v (cid:48) v (cid:48) w w w Fig. 5: The first steps of our algorithm. The black arc at vertex v indicates that vv (cid:48) is the first side edge of v in clockwise order (the first rotation direction).between the two cycles, i.e., each vertex along the backbone path has been arotation vertex. Additionally, each vertex of this backbone path may have anarbitrary number of monochromatic leaves attached. This graph structure iscalled caterpillar .An illustration of Phase 1 in reverse direction can be found in Fig. 8. As wewill see later, at least some edges causing a switch of cycles will differ from theprevious backbone edges. However, the first edges in the reverse process (untilthe first switch) are the same as the last edges of the previous iteration.Of course, the graph H (returned in Phase 2) is not the plane monochromaticspanning tree we are looking for. But we claim that either the red cycle togetherwith the red edges of H or the blue cycle together with the blue edges of H forms a plane monochromatic spanning subdrawing of D .To prove this and thereby the correctness of our algorithm, we need thefollowing invariants concerning the active subdrawing H .(J1) At any time, the union of H and the two cycles of D forms a plane drawing.(J2) Any blue (red) vertex in H is incident to a red (blue) edge in H , exceptfor the current rotation vertex.(J3) Assume that Phase 1 is performed more than once and let V ( H ) be the setof vertices of H . Then for any i ≥
2, after round i of Phase 1, either V ( H )is a strict superset of V ( H prev ) or H contains all vertices from the cyclenot containing v cur , the current rotation vertex (or both conditions hold).Before showing that the invariants (J1) – (J3) indeed hold, we first showhow to obtain a plane monochromatic spanning subdrawing from the outputof our algorithm under the assumption that (J1) – (J3) are true. Invariant (J3)guarantees the termination of our algorithm. Further, by (J1) and (J2), it followsthat the union of the result H of the algorithm and the two cycles contains amonochromatic plane spanning subdrawing. Indeed, let H be the output of ouralgorithm. Then V ( H ) contains all vertices of the cycle that does not containthe last rotation vertex. Assume first that this cycle is blue. As by (J2), all bluevertices are incident to a red edge in H , the red cycle together with the red edges lane Spanning Trees in Edge-Colored Simple Drawings of K n in H forms a spanning subdrawing in D , which, by (J1), is plane. Analogously,if the cycle not containing the last rotation vertex is red, then V ( H ) contains allvertices of the red cycle, each of which is incident to a blue edge by (J2). Hence,the blue cycle and the blue edges in H form a plane spanning subdrawing of D . Proving the invariants.
Invariants (J1) and (J2) follow quite straightforwardlyfrom the construction, whereas (J3) is more involved. Recall that in the rotationof any vertex v , all side edges incident to v appear consecutively. Moreover, westate the following observation, which will be useful for proving (J1) and (J3). Observation 2.
In the rotation of any vertex v , the order of edges to the verticesof each circle is the same as the order along that circle. In particular, if v , . . . , v k are all vertices on the circle not containing v in circular order, then there existsa ≤ j ≤ k such that v j , v j +1 , . . . v k , v , . . . v j − appear in that order in therotation around v . (J1). Observation 2 together with the fact that incident edges must not inter-sect and we stop as soon as we reach an edge incident to v or v (cid:48) implies (J1)(remember, v and v (cid:48) are the incident vertices of the very first edge of Phase 1). (J2). All leaves that are attached to the backbone path fulfill (J2) by construc-tion. Concerning, the vertices on the backbone path, we only switch cycles whenreaching an edge of color different than the current rotation vertex. Hence, allbut the last rotation vertex fulfill (J2). (J3).
Let i ≥ i of Phase 1. Let H be the active sub-drawing at the end of this current iteration and H prev the one at the end of theprevious iteration. Let v be the first and let x be the last rotation vertex of theprevious iteration, i.e., x is the first rotation vertex of the current iteration. Let z denote the last rotation vertex of the current iteration.Then, we need to show that H covers all vertices from the cycle not con-taining z or V ( H ) is a strict superset of V ( H prev ). To this end, we consider thefollowing cases depending on the relative position of v and x . Case 1: v and x lie on the same cycle.
In this case (the first and the last rotation vertex lie on the same cycle) weargue that our algorithm in fact covered all vertices from the other cycle,i.e., is already finished before triggering a new iteration.Without loss of generality, let v and x be blue vertices and assume there isa red vertex y that has not been covered by H prev . By Observation 2, thisvertex must lie “behind” the already considered vertices on the red cycle. If v is equal to x , i.e., we considered only a single rotation vertex, we coveredall red vertices. Otherwise, the edges vy and xy must intersect (see Fig. 6),which is not possible in a simple drawing. v xy Fig. 6: The black arcs around v and x indicate that there are no edges incidentto x (resp. v ) in this direction. This forces the black edges yv and yx to intersect,which is forbidden in a simple drawing. Case 2: v and x lie on different cycles.
This is the more interesting case, that indeed triggers a new iteration of ouralgorithm in the reverse direction.Assume, without loss of generality, that v is a blue vertex and x is red (i.e.,the previous iteration started on the blue cycle and the current iteration onthe red cycle). The argument of Case 1 of course also applies to the currentiteration and hence, we can safely assume z to be a blue vertex.Remember that every blue vertex u ∈ V ( H prev ) (of the previous iteration)is incident to a red edge uu r ∈ E ( H prev ) (due to (J2)). Then, the followingobservation turns out to be very helpful. Claim.
When rotating around a blue vertex u ∈ V ( H prev ) in the current iteration at latest we switch cycles with the edge uu r , i.e., it is not possibleto “skip” this red edge of the previous iteration.Proof of Claim: Assume that this is not true and let u be the first bluerotation vertex violating this property, i.e., u is incident (in H ) to a redvertex after u r . Let ru be the blue backbone edge (in H ) that led from r to u . In particular, r lies behind u r . This obviously also implies that r is notequal to x . So, let br be the red backbone edge (in H ) that led from b to r .In particular, the algorithm considered b before u . Moreover, since the edges uu r and br intersect, r must be behind b r (the neighbor of b incident to b ’sred backbone edge in the previous iteration). Hence, b must have skipped itsred edge bb r from the previous iteration (see Fig. 7). This is a contradictionto u being the first such blue vertex. (cid:4) To summarize, z is a blue vertex and by the above claim z cannot be in V ( H prev ) (except if the stopping condition of covering all vertices from theother cycle was reached earlier).Hence, it remains to show that all red vertices of H prev are also in H . If thiswas not the case, then in particular v (cid:48) is not in H (Observation 2) and theedge zv (cid:48) would intersect the edge vv (cid:48) (see Figure 8). Again, a contradictionto the drawing being simple. lane Spanning Trees in Edge-Colored Simple Drawings of K n v . . . . . .x v (cid:48) u u r b b r r Fig. 7: If the blue rotation vertex u is incident to some edge behind u r in theiteration from x to v (the dashed edge), then r , the neighbor of the blue backboneedge, is also behind u r . Hence, b must also be a blue rotation vertex that skippedits red edge bb r . v xw xwvv (cid:48) zv (cid:48) Fig. 8: On the left, the algorithm started from v and got stuck in x . On theright, the next iteration (in reverse direction) is illustrated. If we get stuck atthe edge zw (rotating around z ), there is no way to connect z and v (cid:48) withoutcrossing vv (cid:48) . Step 2.
Now let D be a 2-edge-colored cylindrical drawing that does not fulfillat least one of the properties (P1) and (P2).If it does not fulfill (P1), the inner or outer cycle is empty, which impliesthat D is isomorphic to a 2-page book drawing and hence contains a monochro-matic plane spanning tree (see Proposition 1).So assume that D fulfills (P1) but does not fulfill (P2). If at least one ofthe cycles of D is bichromatic (contains red and blue edges), then we iterativelyremove a vertex whose incident cycle edges are of different color until we obtaina subdrawing D (cid:48) of D in which both cycles are monochromatic. Clearly, D (cid:48) is acylindrical drawing, since removing a vertex cannot break any of the propertiesof a cylindrical drawing (all vertices still lie on the inner or outer circle, neithercircle is crossed, and all edges between two vertices on the inner (outer) circlestill lie completely inside (outside) that circle).If the two cycles of D (cid:48) are of different color, D (cid:48) fulfills the properties (P1)and (P2) and hence contains a plane monochromatic spanning tree by Step 1. If, on the other hand, the two cycles in D (cid:48) have the same color, then the unionof them plus one side edge of that color gives a monochromatic plane spanningsubdrawing for D (cid:48) , or, if such an edge does not exist, the according color classis not spanning and hence D (cid:48) contains a monochromatic plane spanning treeby Lemma 1. Finally, as cycle edges are always uncrossed, we can extend theobtained spanning tree for D (cid:48) to one for D by re-adding the removed vertices ininverse order by Observation 1.by re-adding the removed vertices ininverse order by Observation 1.