1 Introduction
A pioneering paper by Pach and Wenger [DBLP:journals/gc/PachW01] studied the problem of computing a planar drawing of a graph with the constraint that the mapping of the vertices to the points in the plane, that represent the vertices of , is given as part of the input. Pach and Wenger proved that, for any given mapping, a planar graph with vertices admits a planar drawing such that the curve complexity, i.e. the number of bends per edge, is . Furthermore, they proved that the bound on the curve complexity is (almost surely) tight as tends to infinity when has independent edges. This implies that the curve complexity of a planar drawing with vertices at fixed locations may be even for structurally very simple graphs such as paths or matchings, for which the number of independent edges is linear in .
These results have motivated the study of a relaxed version of the problem where the function that associates vertices of the graph to points of the plane is not a bijection. Namely, an instance of the colored point set embeddability problem receives as input an vertex planar graph such that every vertex is given one of distinct colors and a set of distinct points, such that, each point is given one of the distinct colors. The number of points of having a certain color is the same as the number of vertices of having color . The goal is to compute a planar drawing of with curve complexity independent of where every vertex of a specific color is represented by a point of the same color. When the colored point set embeddability problem coincides with the problem of computing a drawing with vertices at fixed locations and thus the lower bounds by Pach and Wenger hold. Therefore several papers have focused on small values of (typically ) to see whether better bounds on the curve complexity could be achieved in this scenario (see, e.g., [DBLP:journals/tcs/BadentGL08, DBLP:conf/walcom/GiacomoL10, DBLP:journals/ijfcs/GiacomoLT06, DBLP:journals/algorithmica/GiacomoLT10, DBLP:conf/gd/FratiGLLMN12]).
For , Kaufmann and Wiese [DBLP:journals/jgaa/KaufmannW02] proved that every planar graph admits a colored point set embedding onto any point set with curve complexity at most . For , outerplanar graphs always admit a colored point set embedding with curve complexity [DBLP:journals/jgaa/GiacomoDLMTW08]. However, for , there are connected colored planar graphs for which a colored point set embedding may require bends on edges [DBLP:journals/tcs/BadentGL08]. This result extends the lower bound of Pach and Wenger [DBLP:journals/gc/PachW01] to a much more relaxed set of constraints on the location of the vertices, but it does so by using 2connected graphs instead of just (not necessarily connected) planar graphs. For example, the problem of establishing tight bounds on the curve complexity of colored forests for small values of is a long standing open problem (see, e.g., [DBLP:journals/tcs/BadentGL08]). We explicitly address this gap in the literature and consider the colored point set embeddability problem for acyclic graphs and . Our main results are as follows.

In Section 3, we prove that a planar drawing of a forest of three stars and vertices may require edges with bends each, even if the mapping of the vertices to the points is defined by using a set of colors with . In contrast, a constant number of bends per edge can always be achieved if the number of stars is at most two (for any number of colors) or the number of colors is at most two (for any number of stars).

Since the above result implies that colored point set embeddings of colored caterpillars may have a nonconstant curve complexity, in Section 4 we study subfamilies of colored caterpillars for which constant curve complexity is possible. We prove that every colored path and every colored caterpillar whose leaves all have the same color admit a colored pointset emebdding with constant curve complexity onto any colored point set.

Finally, still in Section 4, we prove that any colored path such that the vertices of colors and precede all vertices of colors and when moving along has a colored point set embedding with at most five bends per edge onto any colored pointset.
Concerning the lower bound, it is worth mentioning that the argument by Pach and Wenger [DBLP:journals/gc/PachW01] does not apply to families of graphs where the number of independent edges is not a function of . Hence, our lower bound extends the one by Pach and Wenger about the curve complexity of planar drawings with vertices at fixed locations also to those graphs for which the number of independent edges does not grow with . Some proofs can be found in the appendix.
2 Preliminaries
Let be a graph. A coloring of is a partition of . The integers are called colors and is called a colored graph. For each vertex we denote by the color of .
Let be a set of distinct points in the plane. For any point , we denote by and the  and coordinates of , respectively. We denote by the convex hull of . Throughout the paper we always assume that the points of have different coordinates (if not we can rotate the plane so to achieve this condition). A coloring of is a partition of . A set of points with a coloring is called a colored point set. For each point , denotes the color of . A colored point set is compatible with a colored graph if for every . If is planar we say that has a topological pointset embedding on if there exists a planar drawing of such that: (i) every vertex is mapped to a distinct point of with , (ii) each edge of is drawn as simple Jordan arc. We say that has a colored pointset embedding on if there exists a planar drawing of such that: (i) every vertex is mapped to a distinct point of with , (ii) each edge of is drawn as a polyline . A point shared by any two consecutive segments of is called a bend of . The maximum number of bends along an edge is the curve complexity of the colored pointset embedding. A colored sequence is a sequence of (possibly repeated) colors , , , such that (). We say that is compatible with a colored graph if color occurs times in . Let be a colored point set. Let be the points of with . The colored sequence is called the colored sequence induced by , and is denoted as . A set of points is onesided convex if they are in convex position and the two points with minimum and maximum coordinate are consecutive along . In a colored onesided convex point set, the sequence of colors encountered clockwise along , starting from the point with minimum coordinate, coincides with .
A Hamiltonian cycle of a graph is a simple cycle that contains all vertices of . A graph that admits a Hamiltonian cycle is said to be Hamiltonian. A planar graph is subHamiltonian if either is Hamiltonian or can be augmented with dummy edges (but not with dummy vertices) to a graph that is Hamiltonian and planar. A subdivision of a graph is a graph obtained from by replacing each edge by a path with at least one edge. Internal vertices on such a path are called division vertices. Every planar graph has a subdivision that is subHamiltonian. Let be a planar graph and let be a subHamiltonian subdivision of . The graph is called a Hamiltonian augmentation of and will be denoted as . Let be the Hamiltonian cycle of a Hamiltonian augmentation of . Let be an edge of , let be the Hamiltonian path obtained by removing from , and let be the vertices of in the order they appear along . Finally, let be a colored sequence. is a colored Hamiltonian path consistent with if (). is a colored Hamiltonian cycle consistent with if there exists an edge such that is a colored Hamiltonian path consistent with . is called a colored Hamiltonian augmentation of consistent with . The following theorem has been proved in [DBLP:journals/jgaa/GiacomoDLMTW08] (see also [DBLP:journals/tcs/BadentGL08, DBLP:journals/algorithmica/GiacomoLT10]).
Theorem 2.1
[DBLP:journals/jgaa/GiacomoDLMTW08] Let be a colored planar graph and be a colored point set consistent with . If has a colored Hamiltonian augmentation consistent with and at most division vertices per edge then admits a colored pointset embedding on with at most bends per edge.
The next lemma can be easily derived from Theorem 2.1.
Lemma 1
Let be a colored graph, and be a colored onesided convex point set compatible with . If has a topological colored pointset embedding on such that each edge crosses at most times, then admits a colored pointset embedding on with at most bends per edge.
Let be a planar graph. A topological book embedding of is a planar drawing such that all vertices of are represented as points of a horizontal line , called the spine. Each of the halfplanes defined by is a page. Each edge of a topological book embedding is either in the top page, or completely in the bottom page, or it can be on both pages, in which case it crosses the spine. Each crossing between an edge and the spine is called a spine crossing. It is also assumed that in a topological book embedding every edge consists of one or more circular arcs, such that no two consecutive arcs are in the same page^{1}^{1}1The more general concept of page topological book embedding exists, where each arc can be drawn on one among different pages. For simplicity we use the term topological book embedding to mean page topological book embedding.. Let be a colored graph and let be a colored sequence compatible with . A topological book embedding of is consistent with if the sequence of vertex colors along the spine coincides with . Let be a colored point set compatible with a colored planar graph and let be the colored sequence induced by . The following lemma can be proved similarly to Lemma 1.
Lemma 2
If admits a topological book embedding consistent with and having at most spine crossing per edge, then admits a pointset embedding on with curve complexity at most .
3 Pointset Embeddings of Stars
In this section we establish that a colored pointset embedding of a forest of three stars may require bends along edges by exploiting a previous result about biconnected outerplanar graphs. We start by recalling the result in [DBLP:journals/jgaa/GiacomoDLMTW08]. An alternating point set is a colored onesided convex point set such that: (i) has points for each color , , and , and (ii) when going along the convex hull of in clockwise order, the sequence of colors encountered is . Each set of consecutive points colored is called a triplet.
A fan, denoted as , is a colored outerplanar graph with vertices () and defined as follows. consists of a simple cycle formed by vertices of color , followed (in the counterclockwise order) by vertices of color , followed by vertices of color . The vertex of color adjacent in the cycle to a vertex of color (indices taken modulo ) is denoted as . Also, in every vertex colored is adjacent to () and vertices form a cycle of . See, e.g. Fig. 1. The following theorem has been proved in [DBLP:journals/jgaa/GiacomoDLMTW08].
Theorem 3.1
[DBLP:journals/jgaa/GiacomoDLMTW08] Let be a positive integer and let be a fan for , and let be an alternating point set compatible with . In every colored pointset embedding of on there is one edge with more than bends.
The forest of stars that we use to establish our lower bound is called a sky and is denoted by . It consists of three stars , , such that: (i) each () has vertices (); (ii) all the vertices of each () have the same color .
Let be a pointset embedding of on . An uncrossed triplet of is a triplet of points of such that, when moving along in clockwise order, no edge of crosses between and and between and . A triplet is crossed times if the total number of times that is crossed by some edges between and and between and is . A leaf triplet of is a triplet of whose points represent leaves of . Analogously, a root triplet is a triplet of whose points represent the three roots of . The following lemma establishes the first relationship between the curve complexity of some special types of colored pointset embeddings of and those of a fan .
Lemma 3
Let be a sky, be an alternating point set compatible with , and be a colored topological pointset embedding of on . If has an uncrossed leaf triplet and each edge of crosses at most times, then the fan has a colored topological pointset embedding on such that each edge crosses at most times.
Proof
We show how to use to construct a topological pointset embedding of the fan on with at most crossings of per edge.
Let be an uncrossed leaf triplet. Every point of the triplet represents a leaf of a different star (because they have different color). Denote by the point of representing the root of () and denote by the edge connecting to . The idea is to connect the three points with a cycle that does not cross any existing edges. For each edge we draw two curves that from run very close to until they reach . The two curves are drawn on the same side of such that they are consecutive in the circular order of the edges around (see Fig. 2 for an illustration). These two curves do not intersect any existing edges and cross the same number of times as . The six drawn curves are now suitably connected to realize a cycle connecting . Depending on which side the various curves reach , the connections are different. However in all cases we can connect two curves to form a single edge by crossing at most two additional times and without violating planarity (see Fig. 2, 2, and 2). Thus, we have added to three edges connecting to (indices taken modulo ), each crossing at most times. Also, since the two curves that follow an edge are both drawn on the same side of , the cycle does not have any vertices inside. Notice that, depending on the case with respect to the connection of the curves, , , and appear along either in the clockwise or in the counterclockwise order. W.l.o.g. we assume that the clockwise order is , , and .
The obtained drawing is not yet a topological pointset embedding of because the cycle connecting all the vertices is missing. We first add edges connecting leaves of the same color. Let be the edges incident to in the circular order around (this is the counterclockwise order under our assumption that , , and are located in the clockwise order along ). We add an edge between the leaf of and the leaf of (for ) as follows. Starting from the leaf of , we draw a curve following the edge until we arrive very close to and then we follow until we reach the leaf of . The added edges do not cross any existing edges and cross a number of times equal to the number of times that crosses plus the number of times that crosses , so at most .
It remains to add the edges of connecting vertices of different colors. There are three such edges and they connect vertex () of to a vertex of color (indices taken modulo ). We add an edge connecting to a leaf of color as follows. Let be the edge incident to that follows in the clockwise order around (this is an edge connecting to a leaf of color ). Starting from we draw a curve following the edge until we arrive very close to and then we follow until we reach the leaf of . The constructed curve connects to a leaf of color and does not cross any existing edge. It crosses at most the number of times that crosses (that is ) plus the number of times that crosses (that is ). Thus the total number of crossing of is at most .∎
The next two lemmas explain how to obtain a colored topological book embedding that satisfies Lemma 3.
Lemma 4
Let be a sky, be an alternating point set compatible with , and be a colored topological pointset embedding of on with a root triplet. If has a leaf triplet that is crossed times () and each edge crosses at most times, then there exists a sky which is a subgraph of and an alternating point set which is a subset of such that: (i) ; (ii) there exists a colored topological pointset embedding of on such that each edge crosses at most times; (iii) is an uncrossed leaf triplet of .
Lemma 5
Let be a sky, be an alternating point set compatible with , and be a colored topological pointset embedding of on . If each edge of crosses at most times, then there exists a sky which is a subgraph of and an alternating point set which is a subset of such that: (i) ; (ii) there exists a colored topological pointset embedding of on such that each edge crosses at most times; (iii) has a root triplet.
Lemma 6
Let be a positive integer, be a sky for , and be an alternating point set compatible with . In every colored pointset embedding of on there exist at least edges with more than bends.
Proof (sketch)
Let , , be a sky for and let be an alternating point set compatible with . We prove by induction on that in every colored pointset embedding of on there exist edges with more than bends. Notice that for , we have .
Base case: : We have to prove that in any colored pointset embedding of on with , there exists one edge with more than bends. Suppose as a contradiction that there exists a colored pointset embedding of on with curve complexity . is also a colored topological pointset embedding of on such that each edge crosses at most times (each edge consists of at most segments). By Lemma 5 there exists a colored pointset embedding of a sky on an alternating point set such that: (i) ; (ii) each edge of crosses at most times; (iii) has a root triplet.
Since each edge of crosses at most times and there are edges in total, there are at most crossings of in total. The number of leaf triplets in is . It follows that there is at least one leaf triplet crossed at most times. By Lemma 4 there exists a colored pointset embedding of a sky on an alternating point set such that: (i) ; (ii) each edge of crosses at most times; (iii) is uncrossed. By Lemma 3, the fan has a colored topological pointset embedding on such that each edge crosses at most times and by Lemma 1 a colored point set embedding with curve complexity at most . On the other hand, since , we have that and by Theorem 3.1, in every colored pointset embedding of on at least one edge that has more than bends – a contradiction.
Inductive step: . We have to prove that in any colored pointset embedding of on with , there exist edges with more than bends.
We first prove that there exists at least one edge with more than bends. Suppose as a contradiction that there exists a colored pointset embedding of on with curve complexity . With the same reasoning as in the base case, there would exist a colored point set embedding with curve complexity at most of a fan , with . Since , we have that and by Theorem 3.1, in every colored pointset embedding of on at least one edge has more than bends – again a contradiction.
This proves that there is at least one edge crossed more than times. We now remove this edge and the whole triplet that contains the point representing the leaf of . We then arbitrarily remove triplets. The resulting drawing is a colored pointset embedding of on for . By induction, it contains edges each having more than bends. It follows that has edges each having more than bends. Since for we have , the statement follows. ∎
Theorem 3.2
For sufficiently large , there exists a colored forest consisting of three monochromatic stars with vertices and a colored point set in convex position compatible with such that any colored pointset embedding of on has edges having bends.
We conclude this section with some results deriving from Theorem 3.2 and/or related to it. Firstly, Theorem 3.2 extends the result of Theorem 3.1 since it implies that a colored point set embedding of may require edges with bends each. Moreover, the result of Theorem 3.2 implies an analogous result for a colored forest of at least three stars for every . In particular, when we have the following result that extends the one by Pach and Wenger [DBLP:journals/gc/PachW01].
Corollary 1
Let be a forest of three vertex stars. Every planar drawing of with vertices at fixed vertex locations has edges with bends each.
One may wonder whether the lower bound of Theorem 3.2 also holds when the number of colors or the number of stars is less than three. However, it is immediate to see that this is not the case, i.e., the following theorem holds.
Theorem 3.3
Let be a colored forest of stars and be a set of points compatible with . If then has a colored pointset embedding on with curve complexity at most .
Since a caterpillar can be regarded as a set of stars whose roots are connected in a path, the lower bound of Theorem 3.2 also holds for caterpillars. This answers an open problem in [DBLP:journals/tcs/BadentGL08] about the curve complexity of colored pointset embeddings of trees for . Note that curve complexity for colored outerplanar graphs has been proved in [DBLP:journals/jgaa/GiacomoDLMTW08].
Corollary 2
For sufficiently large , a colored pointset embedding of a colored caterpillar may require edges having bends.
4 Pointset Embeddings of Paths and Caterpillars
In the light of Corollary 2, one may ask whether there exist subclasses of colored caterpillars for which constant curve complexity can be guaranteed. In this section we first prove that this is the case for colored paths and then we extend the result to colored caterpillars whose leaves all have the same color.
Based on Lemma 2, we prove that a colored path has a topological book embedding consistent with and having a constant number of spine crossings. Namely, we first remove the vertices and points of one color from and , obtaining a colored path and a compatible colored point set . Next, we construct a topological book embedding of consistent with with at most two spine crossings per edge and with suitable properties. Then we use such properties to reinsert the third color and obtain a topological book embedding of consistent with .
and can be regarded as two binary strings of the same size where one color is represented by bit and the other one by bit . and are balanced if the number of ’s (’s, resp.) in equals the number of ’s (’s, resp.) in . and are a minimally balanced pair if there does not exist a prefix of and a corresponding prefix of that are balanced.
Lemma 7
Let and be a minimally balanced pair of length . Let denote the th bit of and denote the th bit of . Then , , and .
Let be a topological book embedding, be the spine of , and be a point of (possibly representing a vertex). We say that is visible from above (below) if the vertical ray with origin at and lying in the top (bottom) page does not intersect any edge of . We say that the segment is visible from above (below) if each point in the segment is visible from above (below). Let and be two vertices of that are consecutive along the spine , we say that segment is accessible if it contains a segment that is visible from below. A vertex of is hook visible if there exists a segment of the spine such that is visible from below and for any point of we can add an edge in the top page of connecting with without crossing any other edges of (see Fig. 3); is the access interval for vertex . If the access interval is to the right (left) of we say that is hook visible from the right (left).
Lemma 8
Let be a colored path and be a colored sequence compatible with . Path admits a topological book embedding consistent with and with the following properties: (a) Every edge of crosses the spine at least once and at most twice. (b) For any two vertices and that are consecutive along the spine of , segment is accessible from below. (c) Every spine crossing is visible from below. (d) The first vertex of is visible from above; the last vertex of is hook visible from the right; to the right of its access interval there is only one vertex and no spine crossing.
Proof
We prove the statement by induction on the length of (and of ). If the statement trivially holds. If we draw the unique edge of with one spine crossing immediately to the left of the leftmost vertex in the drawing (see Fig. 3). Also in this case the statement holds. Suppose that and that the statement holds for every . We distinguish between two cases.
Case 1: and are a minimally balanced pair. By Lemma 7 the first vertex of has the same color as the last element of , the last element of has the same color as the first element of and these two colors are different. This means that by removing the first and the last elements from both and , we obtain a new colored path of length and a new colored sequence compatible with . By induction, admits a topological book embedding consistent with and satisfying properties (a)–(d). To create a topological book embedding of consistent with , we add a point before all the points of , whose color is the same as the last vertex of , and a point after all points of , whose color is the same as the first vertex of . Vertex is mapped to and vertex is mapped to . We connect to the first vertex of with an edge incident to from above, crossing the spine once immediately before and once immediately after and arriving to from above (by property (d), is visible from above). We then connect the last vertex of to . Since is hook visible by property (d), we connect it to with an edge that starting from reaches the access interval of , crosses the spine between the last vertex of and and reaches from above. As shown in Fig. 3 the two edges and can be added without creating any crossing. Property (a) holds by construction. About properties (b) and (c), we added two arcs in the bottom page. The first one connects a point immediately before and a point immediately after it, so the segment of the spine between and its following vertex is accessible from below; also, the addition of this arc does not change the accessibility of the spine crossing of . The second arc added in the bottom page connects a point in the access interval of with a point immediately after ; by property (d) of there is no vertex or spine crossings between and . Thus the segments connecting to its preceding and to its following vertices are visible from below and property (b) holds; furthermore the addition of this arc does not change the accessibility to existing spine crossings. Since the new created spine crossings are visible from below, property (c) also holds. It is immediate to see that also property (d) holds; see for example Fig. 3.
Case 2: and are not a minimally balanced pair. In this case there exists a prefix (i.e. a subpath) of and a corresponding prefix of that are balanced. is colored path and is a colored sequence compatible with and their length is less than . By induction, admits a topological book embedding consistent with and statisfying properties (a)–(d). On the other hand, is also a colored path and is a colored sequence consistent with . Thus, also admits a topological book embedding consistent with and statisfying properties (a)–(d). Since the last vertex of is hook visible in and the first vertex of is visible from above in , the two vertices can be connected with an edge that crosses the spine twice (see Fig. 3), thus creating a topological book embedding of consistent with . Property (a) holds by construction. The only arc added in the bottom page connects a point in the access interval of and a point immediately after the first vertex of . By property (d) of there is no vertex or spine crossing between and and between and , thus properties (b) and (c) hold for . Property (d) holds because it holds for and . ∎
Lemma 9
A colored path admits a topological book embedding with at most two spine crossings per edge consistent with any compatible colored sequence.
Proof (sketch)
Let be a color distinct from the colors of the endvertices of . Let be a maximal subpath of colored . Let and be the vertices along before and after , respectively. We replace the subpath with an edge . We do the same for every maximal subpath colored . Let be the resulting colored path and be the colored sequence obtained from by removing all elements of color .
By Lemma 8, admits a topological book embedding consistent with that satisfies properties (a), (b), (c) and (d). We add to a set of points colored to represent the removed vertices that will be added back. These points must be placed so that the sequence of colors along the spine coincides with . By property (b) of all these points can be placed so that they are accessible from below. We now have to replace some edges of with paths of vertices colored . Let be an edge that has to be replaced by a path . For each vertex to be added () we add an image point to the drawing. The image points are added as follows. By property (a), the edge crosess the spine at least once. Let be the point where crosses the spine for the first time when going from to . By property (c) is visible from below. This means there is a segment of with as an endpoint that is visibile from below. We place image points inside this segment, while is the th image point (it is the leftmost if is to the left of , while it is the rightmost if is to the right of ). The first arc of the edge is replaced by an arc connecting to . Each image point is connected to the () by means of an arc in the top page. Finally, the last image point is already connected to by means of the remaining part of the original edge . Notice that the edge does not cross the spine, and the same is true for any edge , while the edge crosses the spine at most once (the original edge had at most two spine crossing one of which was at ). We have replaced the edge with a path with edges, as needed. However, the points representing the intermediate vertices of this path are not the points of the set . The idea then is to “connect” the image points to the points of . To this aim, we add matching edges in the bottom page between the image points and the points of . Since both the points of and the image points are visible from below, these matching edges do not cross any other existing edge. Moreover, by using a simple brackets matching algorithm, we can add the matching edges so that they do not cross each other. Finally the matching edges can be used to create the actual path that represent . ∎
Theorem 4.1
Every colored path admits a colored pointset embedding with curve complexity at most on any compatible colored point set.
Theorem 4.1 can be extended to a subclass of colored caterpillars.
Theorem 4.2
Every colored caterpillar with monochromatic leaves admits a colored pointset embedding with curve complexity at most on any compatible colored point set.
The above results motivate the study of colored graphs, in particular a natural question is whether colored paths admit pointset embedding on any set of points with constant curve complexity.
Theorem 4.3
Let be a colored path with vertices and let be a colored point set compatible with . If the first vertices along only have two colors and the remaining only have the other two colors, then has a colored pointset embedding on with curve complexity at most .
5 Open Problems
Motivated by the results of this paper we suggest the following open problems: (i) Investigate whether the lower bound of Theorem 3.2 is tight. We recall that an upper bound of holds for all colored planar graphs [DBLP:journals/gc/PachW01]. (ii) Study whether constant curve complexity can always by guaranteed for colored paths. (iii) Characterize the colored caterpillars that admit a colored pointset embedding with constant curve complexity on any given set of points.
References
Appendix
Appendix 0.A Omitted proofs from Section 2
Lemma 1
Let be a colored graph, let be a colored onesided convex point set compatible with . If has a topological colored pointset embedding on such that each edge crosses at most times, then admits a colored pointset embedding on with at most bends per edge.
Proof
Replace each crossing between an edge and with a division vertex on and connect any two (real or division) vertices that are consecutive along with an edge if they are not yet connected. The resulting graph is a colored Hamiltonian augmentation consistent with (since is onesided convex the circular order of the colors along coincides with ) and with at most division vertices per edge. By Theorem 2.1 admits a colored pointset embedding on with at most bends per edge.∎
Appendix 0.B Omitted proofs from Section 3
Lemma 4
Let be a sky, be an alternating point set compatible with , and be a colored topological pointset embedding of on with a root triplet. If has a leaf triplet that is crossed times () and each edge crosses at most times, then there exists a subgraph of and a subset of such that: (i) ; (ii) there exists a colored topological pointset embedding of on such that each edge crosses at most times; (iii) is an uncrossed leaf triplet of .
Proof
The idea is to make uncrossed by removing all edges that cross it. In order to keep the set of points alternating, for each removed edge we will remove the whole triplet that contains the point representing the leaf of . Suppose first that none of the triplets to be removed coincide with (i.e., no edge that crosses has an endpoint in ). In this case, we remove all edges that cross and all the triplets containing their leaves. Notice that, since has a root triplet, no root is removed. Since we always remove triplets, the final drawing has the same number of points for each color. Thus, such a drawing is a colored topological pointset embedding of a sky on an alternating point set with an uncrossed triplet . The number of triplets removed is at most , and therefore . Finally, since we have only removed edges, the number of times that an edge crosses remains .
Suppose now that at least one edge that crosses has an endpoint in . Denote by the set of these edges. We first remove all edges that cross but are not in . Then we redraw the edges of so that they do not cross . To this aim, consider a closed curve around that only intersects edges that are incident to or cross . We cut the three edges incident to the points of (these three edges include those in , possibly coinciding with them) at the point where they cross for the first time. We then connect the three points on the curve to the points of . One can observe that this can always be done by adding to each edge at most one crossing of (see Figure 4 for an illustration). Also in this case the resulting drawing is a colored topological pointset embedding of a sky on an alternating point set with an uncrossed triplet . Again, the number of triplets removed is at most , and therefore . ∎
Lemma 5
Let be a sky, be an alternating point set compatible with , and be a colored topological pointset embedding of on . If each edge of crosses at most times, then there exists a subgraph of and a subset of such that: (i) ; (ii) there exists a colored topological pointset embedding of on such that each edge crosses at most times; (iii) has a root triplet.
Proof
If has a root triplet, then coincides with and the thesis holds. Suppose then that does not have a root triplet. Consider the triplets that contain the roots. They are at most three of them. Denote them as , and . Denote by (for , indices taken modulo ) be the set of triplets that are encountered between and when moving clockwise along . The number of triplets in , , and is and thus at least one of these three sets has triplets. We remove all the triplets of the other two sets. Furthermore we remove the points of the triplets containing the roots that do not represent the roots. This removes at most six extra vertices. The three roots are now consecutive. If they form a triplet, i.e., they are colored , , in the clockwise order, then the drawing obtained after the removal is the desired . In , there are at least triplets plus the root triplet, hence .
If the three roots do not form a triplet, we locally modify the drawing to transform them into a triplet. To this aim, we consider a closed curve around the three roots that is crossed only by the edges incident to the three roots and by the edges that cross between them. It is possible to reroute the edges crossing so as to guarantee that the clockwise order of the root colors is , , . This can be done by adding to each edge at most two crossing of (see Figure 5 for an illustration). After the rerouting, we are in the same situation as in the previous case and the resulting drawing is the desired with . ∎
Lemma 6
Let be a positive integer, be a sky for , and