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11institutetext: WSI, Tübingen, Germany
11email: {foersth,mk}@informatik.uni-tuebingen.de
22institutetext: NTUA, Athens, Greece
22email: [email protected]

Recognizing and Embedding
Simple Optimal 22-Planar Graphsthanks: The work of C. Raftopoulou is funded by the National Technical University of Athens research program Π\mathrm{\Pi}EBE 2020. H. Förster and M. Kaufmann are supported by DFG grant KA812-18/2.

Henry Förster 11 0000-0002-1441-4189    Michael Kaufmann 11 0000-0001-9186-3538    Chrysanthi N. Raftopoulou 22 0000-0001-6457-516X
Abstract

In the area of beyond-planar graphs, i.e. graphs that can be drawn with some local restrictions on the edge crossings, the recognition problem is prominent next to the density question for the different graph classes. For 1-planar graphs, the recognition problem has been settled, namely it is NP-complete for the general case, while optimal 1-planar graphs, i.e. those with maximum density, can be recognized in linear time. For 2-planar graphs, the picture is less complete. As expected, the recognition problem has been found to be NP-complete in general. In this paper, we consider the recognition of simple optimal 2-planar graphs. We exploit a combinatorial characterization of such graphs and present a linear time algorithm for recognition and embedding.

Keywords:
22-planar graphs recognition algorithms beyond-planarity

1 Introduction

In the field of graph drawing beyond planarity, researchers study graphs that admit drawings with crossings only in restrictive local configurations [6, 14, 15, 18, 24]. Mostly studied are 1-planar graphs, which admit drawings where each edge is crossed at most once. Classic results on these graphs concern mostly the density [7, 28], recently, other problems such as generation [29], characterization [20], recognition [2, 9, 10, 11, 13], coloring [8] and page number [3, 16] have been studied. The most natural extension of 11-planar graphs, is the family of kk-planar graphs, that is graphs that admit drawings where each edge is crossed at most kk times. While there are plenty of results on 11-planarity, kk-planarity turns out to be more challenging. The first outstanding results on the density of simple kk-planar graphs came with the work of Pach and Tóth [27] for k4k\leq 4. In particular, they proved that simple 22-planar graphs have at most 5n105n-10 edges. Later on, improved upper bounds of 5.5n115.5n-11 [26] and 6n126n-12 [1] have been shown for simple 33- and 44-planar graphs, respectively. The bounds for 22- and 33-planar graphs are tight even for non-simple graph without homotopic parallel edges and self-loops [5].

While testing for planarity can be done in linear time, and several classes of beyond-planar graphs have a linear number of edges, the corresponding recognition problems are NP-hard, which impedes the practical application of related results. In particular, it was recently proven that the recognition of kk-planar graphs is NP-hard for every value of k1k\geq 1  [30]. Still, optimal 1-planar graphs, which form a prominent subclass of 11-planar graphs, can be recognized in linear time [11], while recognition of maximal 11-planar graphs and 44-maps takes cubic time [17, 12]. Previously, efficient recognition algorithms have been developed for more restricted subclasses of 11-planar graphs, see e.g. [2, 10, 13, 19].

For k2k\geq 2, there are significantly fewer results. A notable result is the linear time recognition algorithm for full outer-2-planar graphs [21], while for k3k\geq 3 the literature still lacks similar results.

Our contribution. In this paper, we extend the study on the recognition of subclasses of kk-planar graphs. Namely, we prove that simple optimal 22-planar graphs, that is simple 22-planar graphs with 5n105n-10 edges, can be efficiently recognized in 𝒪(n)\mathcal{O}(n) time. In our strategy we first remove a set of edges that must be involved in crossings in any optimal 22-planar drawing of the given graph. We unwrap specific properties of the remaining graph that allow us to find a unique drawing and insert the deleted crossing edges.

Paper organization. In Section 2 we give preliminary notations and properties of optimal 22-planar graphs. Section 3 presents CROSS-BAT, an important 22-planar substructure, and examines its structural properties. Section 4 focuses on the insertion of the deleted crossing edges. Our recognition algorithm in Section 5 combines all previous results. We conclude in Section 6 with further research directions. Due to space constraints, several proofs and details, indicated by ()(\ast), are deferred to an appendix.

2 Preliminaries

The degree of a vertex vVv\in V is denoted by d(v)d(v) and its set of neighbors by N(v)N(v). For u,vVu,v\in V we write N(u,v)N(u,v) to denote the set of common neighbors of uu and vv, that is N(u,v)=N(u)N(v)N(u,v)=N(u)\cap N(v). A drawing of a graph maps the vertices to points in the plane and the edges to simple open Jordan curves connecting the points corresponding to their endpoints. A drawing is kk-planar if each curve representing an edge is crossed at most kk times by the other edges together. A graph GG is kk-planar if it admits a kk-planar drawing.

The topology of a planar drawing is described by a rotation scheme. For kk-planarity, a kk-planar rotation scheme specifies the counter-clockwise circular order of edges around each vertex and, additionaly, a sequence of at most kk crossings along each edge. Hence, a kk-planar rotation scheme defines an equivalence class of kk-planar drawings.

For a graph GG, let HH a subgraph of GG and let H\mathcal{R}_{H} be a (kk-planar) rotation scheme of HH. We say that H\mathcal{R}_{H} is a partial (kk-planar) rotation scheme of GG. We further say that H\mathcal{R}_{H} can be extended to a (kk-planar) rotation scheme \mathcal{R} of GG if the restriction of \mathcal{R} on HH is H\mathcal{R}_{H}. Conversely, we say that \mathcal{R} extends H\mathcal{R}_{H}. Let CC be a cycle of graph GG, whose edges define a simple closed curve in a drawing Γ\Gamma of GG. We refer to the region to the left of this curve in a counter-clockwise walk along CC in Γ\Gamma as the interior of the region bounded by cycle CC. Note that given drawing Γ\Gamma, the interior of a region may be unbounded.

A 22-planar graph that reaches the upper bound of 5n105n-10 edges is called optimal 22-planar. Note that the value of 5n105n-10 edges can be reached only when n2mod3n\equiv 2\mod 3. Each induced subgraph HH of a 22-planar graph GG is also 2-planar and thus has at most 5|V(H)|105|V(H)|-10 edges. This implies that if GG is 2-planar then it is also 9-degenerate, i.e. there is a vertex ordering v1,,vnv_{1},\ldots,v_{n} such that, for each i=1,,ni=1,\ldots,n, vertex viv_{i} of the induced subgraph Gi=G[v1,,vi]G_{i}=G[v_{1},\ldots,v_{i}] has degree at most 9 in GiG_{i}. We call such a vertex ordering a 9-degenerate sequence of GG.

In this work, we consider only simple graphs which we assume from now on without explicitly stating it. The structure of simple optimal 22-planar graphs is well established. Namely, an optimal 22-planar rotation scheme \mathcal{R} of a simple optimal 22-planar graph specifies a crossing-free 33-connected111The 3-connectivity derives from the fact that the graph is simple. spanning pentangulation 𝒫\mathcal{P} and a set of crossing edges [4, 5]. The edges around each vertex are cyclically ordered so that every crossing-free edge (belonging to 𝒫\mathcal{P}) is followed by two crossing edges, which are succeeded by one crossing-free edge, and so on. Hence the degree of every vertex is a multiple of three. Moreover, the minimum degree of a simple optimal 22-planar graph is 99 since 𝒫\mathcal{P} is 33-connected, i.e., every vertex is incident to at least 33 crossing-free edges. Now, each pentangular face ff of 𝒫\mathcal{P} contains exactly five crossing edges of \mathcal{R} which form a 55-clique in GG with the boundary edges of ff; we call such a 55-clique a facial 55-clique. Note that not all 55-cliques of GG are facial in \mathcal{R}. This is a major difficulty for recognizing optimal 22-planar graphs. We denote a facial 55-clique of \mathcal{R} with outer cycle (c0,,c4)(c_{0},\ldots,c_{4}) as c0,,c4\langle c_{0},\ldots,c_{4}\rangle and say that the crossing edges (ci,cj)(c_{i},c_{j}) with (ij)mod5{2,3}(i-j)\mod 5\in\{2,3\} belong to c0,,c4\langle c_{0},\ldots,c_{4}\rangle. As 𝒫\mathcal{P} is 33-connected, it suffices to detect all facial 55-cliques to find the optimal 22-planar rotation scheme of a given input graph, if one exists.

In the following, we state some preliminary observations and properties that we will use in the remainder of the paper. In particular, Properties 2.12.4 immediately arise from the combinatorial characterization of simple optimal 2-planar graphs in [5] also summarized above. So, let GG be an optimal 22-planar graph with an optimal 22-planar rotation scheme \mathcal{R}.

Property 2.1.

A pair of crossing edges in \mathcal{R} belongs to the same facial 55-clique.

Property 2.2.

If in \mathcal{R}, edge ee is crossed by two edges e1e_{1} and e2e_{2}, then e1e_{1} and e2e_{2} share a common endpoint. Also, ee, e1e_{1} and e2e_{2} belong to the same facial 55-clique.

Property 2.3.

Any two facial 55-cliques in \mathcal{R} share at most two vertices.

Property 2.4.

A crossing edge of \mathcal{R} belongs to exactly one facial 55-clique and a crossing-free edge of \mathcal{R} belongs to exactly two facial 55-cliques.

Combining the above properties we can further prove the following:

Corollary 2.1 (\mathbf{\ast}).

For two edges (u,v)(u,v) and (u,w)(u,w) of an optimal 22-planar graph GG let S=N(u,v)N(u,w){u,v,w}S=N(u,v)\cup N(u,w)\cup\{u,v,w\}. If |S|<8|S|<8, then (u,v)(u,v) and (u,w)(u,w) belong to the same facial 55-clique in any optimal 22-planar rotation scheme of GG.

Corollary 2.2.

Let GG be an optimal 22-planar graph with an optimal 22-planar rotation scheme \mathcal{R}. Let (u,v)(u,v) be a crossing edge inside a facial 55-clique CC in \mathcal{R}. Further, let uCu^{\prime}\not\in C be a neighbor of uu and vCv^{\prime}\notin C be a neighbor of vv. Then, edges (u,u)(u,u^{\prime}) and (v,v)(v,v^{\prime}) do not cross in \mathcal{R}; see Fig. 1.

Refer to caption
Figure 1: Forbidden configuration excluded by Corollary 2.2.

Consider an optimal 22-planar rotation scheme \mathcal{R} of GG. If edge (u,v)(u,v) is crossing-free in \mathcal{R}, then by Property 2.4 uu and vv belong to two facial 55-cliques. Hence, |N(u,v)|6|N(u,v)|\geq 6 holds. Although this condition is not sufficient to conclude that an edge is crossing-free, we can use it to identify some crossing edges:

Corollary 2.3.

Let GG be an optimal 2-planar graph and (u,v)E(G)(u,v)\in E(G). Edge (u,v)(u,v) is crossing in any optimal 22-planar rotation scheme of GG if |N(u,v)|<6|N(u,v)|<6.

Our recognition algorithm relies on the identification of crossing edges based on Corollary 2.3. We say that an edge (u,v)(u,v) is clearly crossing if and only if |N(u,v)|<6|N(u,v)|<6. Otherwise (u,v)(u,v) is potentially planar. Note that in an optimal 22-planar rotation scheme, a potentially planar edge is not necessarily crossing-free. However, every crossing-free edge of any optimal 22-planar rotation scheme is potentially planar. We define as GpG_{p} the subgraph of GG formed by all potentially planar edges. Graph GpG_{p} is 33-connected and spans GG as the corresponding crossing-free pentangulation of each optimal 22-planar rotation scheme is a subgraph of GpG_{p}.

3 The CROSS-BAT configuration

In this section, we study the 3-connected spanning subgraph GpG_{p} formed by the potentially planar edges of GG. We want to compute a rotation scheme of GpG_{p} which is extendable to an optimal 22-planar rotation scheme of GG, if it exists. If in each optimal 22-planar rotation scheme of GG subgraph GpG_{p} is plane, then the rotation scheme of GpG_{p} is unique and easy to compute. Though we cannot assure this property, we prove that in any optimal 22-planar rotation scheme of GG, crossings between edges of GpG_{p} occur in restricted configurations: A CROSS-BAT instance is an induced subgraph HH of GG isomorphic to the graph GCBG_{CB} shown in Fig. 2(a), so that (i) for edge (uH,uH)(u_{H},u^{\prime}_{H}) of HH isomorphic to (u,u)(u,u^{\prime}) of GCBG_{CB}, it holds that V(H)=N(uH)N(uH)V(H)=N(u_{H})\cup N(u^{\prime}_{H}), and (ii) the isomorphism between HH and GCBG_{CB} preserves the classification of edges to clearly crossing or potentially planar.

In particular, CROSS-BAT has ten vertices, named as in Fig. 2(a). Vertices uu and uu^{\prime} have degree 99 in GG and they have 88 common neighbors (the remaining vertices of CROSS-BAT). The edges of CROSS-BAT form four 55-cliques that pairwise share two vertices, and are shown as facial 55-cliques in Fig. 2(a). We call edge (u,u)(u,u^{\prime}) the base edge of CROSS-BAT. No other edges with both endpoints in CROSS-BAT exist, except possibly for edges (v,v)(v,v^{\prime}) and (w,w)(w,w^{\prime}). Fig. 2(a) shows the classification of edges of CROSS-BAT as potentially planar and clearly crossing.

Let \mathcal{R} be an optimal 22-planar rotation scheme of GG, such that two potentially planar edges cross each other. These two edges must belong to the same facial 55-clique and, in particular, also to an instance of CROSS-BAT.

Refer to caption
(a)
Refer to caption
(b)
Refer to caption
(c)
Figure 2: (a) Graph GCBG_{CB}: red and black edges are clearly crossing and potentially planar, resp., blue edges might be both, dashed edges may be absent. (b)–(c) the two possible rotation schemes: in (b) edges (u,x)(u,x^{\prime}), (u,y)(u,y^{\prime}), (u,x)(u^{\prime},x) and (u,y)(u^{\prime},y) are crossing-free; in (c) edges (u,x)(u,x), (u,y)(u,y), (u,x)(u^{\prime},x^{\prime}) and (u,y)(u^{\prime},y^{\prime}) are crossing-free.
Lemma 3.1 (\mathbf{\ast}).

Let \mathcal{R} be an optimal 22-planar rotation scheme of GG and let C=c0,,c4C=\langle c_{0},\ldots,c_{4}\rangle be a facial 55-clique in \mathcal{R} such that (c1,c3)(c_{1},c_{3}) and (c2,c4)(c_{2},c_{4}) are potentially planar edges. Then, vertices c2c_{2} and c3c_{3} have degree 99 in GG, and the induced subgraph HH of GG with vertex set V(H)=N(c2)N(c3)V(H)=N(c_{2})\cup N(c_{3}) is an instance of CROSS-BAT where CC is the 55-clique v,x,u,u,x\langle v,x^{\prime},u^{\prime},u,x\rangle of Fig. 2(a).

Next, we show that CROSS-BAT has only two rotation schemes:

Lemma 3.2 (\mathbf{\ast}).

Any instance of CROSS-BAT in an optimal 22-planar rotation scheme \mathcal{R} has one of the two rotation schemes shown in Figs. 2(b) and 2(c).

Sketch.

Vertices are named as in Fig. 2(a). First, we prove that there are four facial 55-cliques, namely (i) CvC_{v}that contains vv, (ii) CwC_{w}that contains ww, (iii) Cx,yC_{x,y}that contains both xx and yy, and, (iv) Cx,yC_{x^{\prime},y^{\prime}}that contains both xx^{\prime} and yy^{\prime}. It is then easy to argue that x,xCvx,x^{\prime}\in C_{v}, while y,yCwy,y^{\prime}\in C_{w}. Finally, we show that ww is contained in the crossing-free cycle (u1,u2,y,w,y)(u_{1},u_{2},y^{\prime},w^{\prime},y) and vv is contained in the crossing-free cycle (u2,u1,x,v,x)(u_{2},u_{1},x,v^{\prime},x^{\prime}) where {u1,u2}={u,u}\{u_{1},u_{2}\}=\{u,u^{\prime}\}. The two choices for u1u_{1} and u2u_{2} give the two different rotation schemes of Figs. 2(b) and 2(c). ∎

As it is evident in Figs. 2(b) and 2(c), if an optimal 22-planar graph GG contains an instance of CROSS-BAT as subgraph, then it admits two different optimal 22-planar rotation schemes \mathcal{R} and \mathcal{R}^{\prime}, that only differ in the choice of the rotation scheme of CROSS-BAT. Hence for any instance of CROSS-BAT, we may arbitrarily choose one of its two rotation schemes. Next, we formalize this observation:

Lemma 3.3.

Let \mathcal{R} be an optimal 2-planar rotation scheme of GG, and let HH be an instance of CROSS-BAT in GG. If HH has the rotation scheme of Fig. 2(b) in \mathcal{R}, then there exists another optimal 2-planar rotation scheme \mathcal{R}^{\prime} of GG, in which HH has the rotation scheme of Fig. 2(c), and vice-versa.

Both rotation schemes of a CROSS-BAT instance contain two crossings between potentially planar edges:

Lemma 3.4 (\mathbf{\ast}).

Let \mathcal{R} be an optimal 22-planar rotation scheme of GG. If GG contains an instance HH of CROSS-BAT, then there exist exactly two pairs of potentially planar edges that belong to HH and cross in \mathcal{R}.

Using Lemma 3.2 we can fix a rotation scheme for each instance HH of CROSS-BAT identified in GG by reclassifying two crossing potentially planar edges of HH to clearly crossing edges. After performing this reclassification for all instances, if GG is optimal 22-planar, then the subgraph GpG_{p} induced by the potentially planar edges is planar. Furthermore, in this case, Lemma 3.3 guarantees that the unique planar embedding of GpG_{p} is part of an optimal 22-planar rotation scheme of GG.

4 Identifying Facial 55-Cliques

Assume that we have fixed the rotation scheme for every identified instance of CROSS-BAT. Let GpG_{p} be the spanning subgraph of GG formed by all potentially planar edges after the reclassification process. As discussed in Section 3, GpG_{p} is planar and 33-connected, i.e. it has a unique planar rotation scheme p\mathcal{R}_{p}. Furthermore, by Lemma 3.3, if GG is optimal 22-planar, it admits an optimal 22-planar rotation scheme that extends p\mathcal{R}_{p} which we call p\mathcal{R}_{p}-compliant. For any p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R}, GpG_{p} contains the corresponding spanning pentangulation 𝒫\mathcal{P} as a subgraph. Hence each face of p\mathcal{R}_{p} has length 33, 44 or 55 and is part of a facial 55-clique. Hence, we can arbitrarily triangulate them and assume from now on that GpG_{p} is triangulated. Triangulating a face of p\mathcal{R}_{p} corresponds to reclassifying some chords from clearly crossing to potentially planar.

Let (f1,f,f2)(f_{1},f,f_{2}) be a path in the dual GpG_{p}^{*} of GpG_{p} so that f1=(u,w1,v1)f_{1}=(u,w_{1},v_{1}), f=(u,w2,w1)f=(u,w_{2},w_{1}) and f2=(u,v2,w2)f_{2}=(u,v_{2},w_{2}), as shown in Fig. 3. If the subgraph induced by the vertices {u,v2,w2,w1,v1}\{u,v_{2},w_{2},w_{1},v_{1}\} is a 55-clique in GG, we call T=f1,f,f2T=\langle f_{1},f,f_{2}\rangle a triplet. TT contains vertices u,v2,w2,w1,v1u,v_{2},w_{2},w_{1},v_{1}, faces f1,f,f2f_{1},f,f_{2} and the three clearly crossing edges e1=(v1,w2)e_{1}=(v_{1},w_{2}), e2=(v2,w1)e_{2}=(v_{2},w_{1}) and e=(v1,v2)e=(v_{1},v_{2}), as shown in Fig. 3. We say that e1e_{1}, e2e_{2} and ee belong to triplet TT.

In any p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} (if it exists), faces and clearly crossing edges of p\mathcal{R}_{p} are partitioned into triplets, such that every face and clearly crossing edge belongs to exactly one of these triplets. Furthermore, the triplets are in 1-1 correspondence with the facial 55-cliques of \mathcal{R}. We say that ee, e1e_{1} and e2e_{2} are assigned to the triplet T=f1,f,f2T=\langle f_{1},f,f_{2}\rangle w.r.t. \mathcal{R} if TT induces a facial 55-clique in \mathcal{R}. Similarly, we say that faces f1f_{1}, ff and f2f_{2} of TT are assigned to TT. If such an assignment is not possible, then GG is not optimal 22-planar.

Refer to caption
Figure 3: Illustration of a triplet f1,f,f2\langle f_{1},f,f_{2}\rangle, along with its clearly crossing edges e1e_{1}, e2e_{2} and ee. Edges of GpG_{p} are drawn black, clearly crossing edges of GG are red.

Let 𝒯\mathcal{T} be a set of triplets such that any two triplets in 𝒯\mathcal{T} are face-disjoint and contain different clearly crossing edges. Consider the partial 22-planar rotation scheme 𝒯\mathcal{R}_{\mathcal{T}} of GG that (i) extends p\mathcal{R}_{p}, (ii) the clearly crossing edges of each T𝒯T\in\mathcal{T} are assigned to TT, and (iii) there is no other assignment of clearly crossing edges. We say that 𝒯\mathcal{T} is bad if and only if 𝒯\mathcal{R}_{\mathcal{T}} cannot be extended to a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} of GG. Our goal is to find an assignment of all clearly crossing edges of GG to a set of triplets 𝒯\mathcal{T} such that 𝒯\mathcal{T} is not bad if GG is optimal 2-planar. We actually prove a stronger result, namely that the set 𝒯\mathcal{T} is unique. We will use two observations that follow from the simplicity of GG:

Observation 4.1.

Let f=(u,v,w)f=(u,v,w) be a triangular face of p\mathcal{R}_{p} and let TT be a triplet that contains vertices uu, vv and ww. Then TT contains face ff.

Observation 4.2.

Let f=(u,v,w)f=(u,v,w) and f=(u,w,v)f^{\prime}=(u^{\prime},w,v) be two adjacent faces of p\mathcal{R}_{p} and (u,u)(u,u^{\prime}) a clearly crossing edge. Let \mathcal{R} be a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme and TT be the triplet that (u,u)(u,u^{\prime}) is assigned to w.r.t. \mathcal{R}. Then, either (i) (u,u)(u,u^{\prime})is drawn inside ff and ff^{\prime} in \mathcal{R} and TT contains both ff and ff^{\prime}, or (ii) (u,u)(u,u^{\prime})is drawn outside ff and ff^{\prime} in \mathcal{R} and TT contains neither ff nor ff^{\prime}.

If any of e1e_{1}, e2e_{2} or ee belongs only to one triplet TT, then TT is a facial 55-clique in any p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} and e1e_{1}, e2e_{2} and ee can be assigned to TT. So, we assume that each of e1e_{1}, e2e_{2} and ee belong to at least one triplet different from TT. Let T1T_{1} and T2T_{2} be two triplets that contain edges e1e_{1} and e2e_{2} respectively and are different from TT. Note that T1=T2T_{1}=T_{2} might hold. Let 𝒯={T}\mathcal{T}=\{T\}. If for each such pair of triplets T1T_{1} and T2T_{2} we can conclude that the set 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} is bad, then in any p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} of GG (if it exists) edges e1e_{1}, e2e_{2} and ee must be assigned to TT. In the following, we compare 𝒯\mathcal{T} against all possible sets 𝒯\mathcal{T}^{\prime} and prove that at least one of 𝒯\mathcal{T} and 𝒯\mathcal{T}^{\prime} is bad. This allows to decide, for each triplet TT, if TT forms a facial 55-clique in every p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG or in none. Note that when we write 𝒯={T}\mathcal{T}=\{T\} and 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\}, we assume that T=f1,f,f2T=\langle f_{1},f,f_{2}\rangle as shown in Fig. 3 and triplets T1T_{1} and T2T_{2} contain edges e1e_{1} and e2e_{2}, respectively.

We first restrict how triplets T1T_{1} and T2T_{2} relate to triplet TT. Observation 4.2 applied to T1T_{1} implies that if T1T_{1} shares either ff or f1f_{1} with TT, then, in the presence of edge e1e_{1}, T1T_{1} shares both ff and f1f_{1} with TT. A symmetric argument applies for T2T_{2}. It follows that one of the following cases holds for {i,j}={1,2}\{i,j\}=\{1,2\}:

  1.   C.1

    TiT_{i} is face-disjoint with TT, or

  2.   C.2

    TiT_{i} shares only face fjf_{j} with TT, or,

  3.   C.3

    TiT_{i} shares only faces ff and fif_{i} with TT.

In the next two lemmas, we show that every 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} is bad if Case C.1 applies either for none or for both of T1T_{1} and T2T_{2}.

Lemma 4.1 (\mathbf{\ast}).

Let 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\}. If none of T1T_{1} and T2T_{2} is face-disjoint from TT, then 𝒯\mathcal{T}^{\prime} is bad.

Lemma 4.2 (\mathbf{\ast}).

Let 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\}. If both T1T_{1} and T2T_{2} are face-disjoint from TT, then 𝒯\mathcal{T}^{\prime} is bad.

Sketch.

Assuming that 𝒯\mathcal{T}^{\prime} is not bad, we arrive at the configuration shown in Fig. 4(a). Since (w1,w2)(w_{1},w_{2}) is crossing, Corollary 2.2 is violated. ∎

By Lemmas 4.1 and 4.2, Case C.1 applies for exactly one of the two triplets T1T_{1} and T2T_{2} and T1T2T_{1}\neq T_{2}. Then, the other triplet complies with Case C.2 or with Case C.3. For each possible combination we first prove structural properties arising from the assumption that 𝒯\mathcal{T^{\prime}} is not bad, and then show that these new restrictions make 𝒯\mathcal{T} bad. In the next two lemmas, we consider the setting, where one of T1T_{1} and T2T_{2} complies with Case C.1 while the other one complies with C.2.

Refer to caption
(a)
Refer to caption
(b)
Figure 4: Illustrations for the proofs of (a) Lemma 4.2, and, (b) Lemma 4.3.
Lemma 4.3 (\mathbf{\ast}).

Let 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\}, such that T1T_{1} is face-disjoint from TT and T2T_{2} shares only face f1f_{1} with TT. For uV(T)u\in V(T), if (i) d(u)>9d(u)>9, or, (ii) for every vertex xS=N(u)V(T)x\in S=N(u)\setminus V(T) it holds |N(x)S|2|N(x)\cap S|\geq 2, then 𝒯\mathcal{T}^{\prime} is bad.

Sketch.

Assuming that 𝒯\mathcal{T}^{\prime} is not bad and d(u)=9d(u)=9, we arrive at the configuration shown in Fig. 4(b). We can conclude that not all conditions of the lemma hold. ∎

Lemma 4.4 (\mathbf{\ast}).

Let 𝒯={T}\mathcal{T}=\{T\} and let vV(T)v\in V(T) such that d(v)=9d(v)=9. If there exists a vertex xS=N(v)V(T)x\in S=N(v)\setminus V(T) such that |N(x)S|1|N(x)\cap S|\leq 1, then 𝒯\mathcal{T} is bad.

By Lemmas 4.3 and 4.4 it remains to consider the case where one of T1T_{1} and T2T_{2} complies with Case C.1 while the other complies with Case C.3. So, we assume w.l.o.g. that T1T_{1} is face-disjoint with TT, while T2T_{2} shares faces ff and f2f_{2} with TT. Recall that clearly crossing edge ee belongs to TT. If 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} is not bad ee belongs to a triplet TeTT_{e}\neq T such that 𝒯{Te}\mathcal{T}^{\prime}\cup\{T_{e}\} is not bad. Note that TeT2T_{e}\neq T_{2} as otherwise vertex v1v_{1} belongs to T2T_{2} and T2=TT_{2}=T. Hence, either TeT1T_{e}\neq T_{1} or Te=T1T_{e}=T_{1}.

The following four lemmas investigate the subcase TeT1T_{e}\neq T_{1}. Note that T1T_{1} and TeT_{e} are face-disjoint as otherwise 𝒯={T1,T2,Te}\mathcal{T}^{\prime}=\{T_{1},T_{2},T_{e}\} is bad. The first two lemmas examine the scenario where T1T_{1} contains vertex w1w_{1} or TeT_{e} contains vertex uu of TT.

Lemma 4.5 (\mathbf{\ast}).

Let 𝒯={T1,T2,Te}\mathcal{T}^{\prime}=\{T_{1},T_{2},T_{e}\}, such that T1T_{1} is face-disjoint from TT, T2T_{2} shares faces ff and f2f_{2} with TT, and TeT_{e} is face-disjoint from T1T_{1}. Then:

  • If T1T_{1} contains vertex w1w_{1} of TT and additionally (i) d(w1)>9d(w_{1})>9, or, (ii) there is a vertex xS=N(w1)V(T)x\in S=N(w_{1})\setminus V(T) such that |N(x)S|4|N(x)\cap S|\geq 4 , then 𝒯\mathcal{T}^{\prime} is bad.

  • If TeT_{e} contains vertex uu of TT and additionally (i) d(u)>9d(u)>9, or, (ii) there is a vertex xS=N(u)V(T)x\in S=N(u)\setminus V(T) such that |N(x)S|4|N(x)\cap S|\geq 4 , then 𝒯\mathcal{T}^{\prime} is bad.

Sketch.

Assuming that 𝒯\mathcal{T}^{\prime} is not bad and d(w1)=9d(w_{1})=9, in the first case, we arrive at the configuration shown in Fig. 5(a). We conclude that not all conditions hold. ∎

Refer to caption
(a)
Refer to caption
(b)
Refer to caption
(c)
Figure 5: Illustration for the proofs of Lemmas (a) 4.5, (b) 4.8, and, (c) 4.9.
Lemma 4.6 (\mathbf{\ast}).

Let 𝒯={T}\mathcal{T}=\{T\} and let vV(T)v\in V(T) with d(v)=9d(v)=9. If for every vertex xS=N(v)V(T)x\in S=N(v)\setminus V(T) it holds |N(x)S|3|N(x)\cap S|\leq 3, then 𝒯\mathcal{T} is bad.

The next two lemmas consider the scenario, where T1T_{1} does not contain vertex w1w_{1} and TeT_{e} does not contain vertex uu of triplet TT.

Lemma 4.7 (\mathbf{\ast}).

Let 𝒯={T1,T2,Te}\mathcal{T}^{\prime}=\{T_{1},T_{2},T_{e}\}, such that T1T_{1} is face-disjoint from TT, T2T_{2} shares faces ff and f2f_{2} with TT, and TeT_{e} is face-disjoint from T1T_{1}. Assume that T1T_{1} does not contain vertex w1w_{1} and TeT_{e} does not contain vertex uu of TT. If there is no triplet Tf1T_{f_{1}} that contains f1f_{1} such that (i) Tf1T_{f_{1}}shares only vertices of f1f_{1} with TT, and, (ii) Tf1T_{f_{1}}is face-disjoint from all of T1T_{1}, T2T_{2} and TeT_{e} , then 𝒯\mathcal{T}^{\prime} is bad.

Lemma 4.8 (\mathbf{\ast}).

Let triplets T1T_{1}, T2T_{2}, TeT_{e} be pairwise disjoint such that T1T_{1} is face-disjoint from TT, T2T_{2} shares faces ff and f2f_{2} with TT, TeT_{e} is face-disjoint from T1T_{1}, T1T_{1} does not contain vertex w1w_{1} and TeT_{e} does not contain vertex uu of TT. If there is a triplet Tf1T_{f_{1}} that contains f1f_{1} such that (i) Tf1T_{f_{1}}shares only vertices of f1f_{1} with TT, and, (ii) Tf1T_{f_{1}}is face-disjoint from all of T1T_{1}, T2T_{2} and TeT_{e}, then 𝒯={T}\mathcal{T}=\{T\} is bad.

Sketch.

Assuming that 𝒯\mathcal{T} is not bad, we arrive at the configuration shown in Fig. 5(b) where the position of zz is not fixed. Edges (u,z)(u,z) and (w1,z)(w_{1},z) belong to Tf1T_{f_{1}} but at least one has three crossings, e.g. edge (u,z)(u,z) in Fig. 5(b). ∎

For the second subcase, where T1=TeT_{1}=T_{e}, Property 2.3 assures that T1T_{1} and T2T_{2} share only vertices v2v_{2} and w2w_{2}, as otherwise 𝒯={T1=Te,T2}\mathcal{T}^{\prime}=\{T_{1}=T_{e},T_{2}\} would be bad. As indicated by the next lemma, no further restrictions (imposed by the assumption that 𝒯\mathcal{T}^{\prime} is not bad) are needed to prove that 𝒯={T}\mathcal{T}=\{T\} is bad.

Lemma 4.9 (\mathbf{\ast}).

Let 𝒯={T}\mathcal{T}=\{T\}. Assume that T1T_{1} is face-disjoint from TT, that T2T_{2} shares faces ff and f2f_{2} with TT, and Te=T1T_{e}=T_{1}. If T2T_{2} has exactly two common vertices with T1T_{1}, then 𝒯\mathcal{T} is bad.

Sketch.

Assuming that 𝒯\mathcal{T} is not bad, we obtain the configuration in Fig. 5(c). Here, two facial 55-cliques have three common vertices contradicting Property 2.3. ∎

In the following two lemmas, we summarize our findings from this section.

Lemma 4.10 (\mathbf{\ast}).

Let GG be optimal 22-planar and let 𝒯={T}\mathcal{T}=\{T\}. 𝒯\mathcal{T} is not bad if and only if every set 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} is bad.

Sketch.

As GG is optimal 22-planar, at least one of 𝒯\mathcal{T} or 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} for some triplets T1T_{1} and T2T_{2} is not bad. If 𝒯\mathcal{T} is bad, the lemma holds. For the reverse direction, assume there is a set 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} that is not bad, i.e. none of Lemmas 4.3, 4.5 and 4.7 applies for 𝒯{Te}\mathcal{T}^{\prime}\cup\{T_{e}\}. If 𝒯\mathcal{T} is not bad from the corresponding Lemmas 4.4, 4.6 and 4.8, the conditions of Lemma 4.9 are satisfied and 𝒯\mathcal{T} is bad. ∎

Lemma 4.11 (\mathbf{\ast}).

Let GG be optimal 22-planar and let 𝒯={T}\mathcal{T}=\{T\}. 𝒯\mathcal{T} is bad if and only if the conditions of at least one of Lemmas 4.4, 4.6, 4.8 and 4.9 are met.

Sketch.

For a contradiction we assume that 𝒯\mathcal{T} is bad. Since GG is optimal 22-planar, by Lemma 4.10, there exists a set 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} that is not bad. We go through all lemmas and check their conditions. For all the cases, we argue that if the specific lemma can not apply on 𝒯\mathcal{T}, then 𝒯\mathcal{T}^{\prime} is bad. ∎

5 The Recognition and Embedding Algorithm

Now that we have all required ingredients, we are ready to state our main theorem for the recognition of optimal 22-planar graphs.

Theorem 5.1.

Let GG be a simple graph on nn vertices. It can be decided in 𝒪(n)\mathcal{O}(n) time whether GG is optimal 22-planar. If the instance is positive, an optimal 22-planar rotation scheme of GG is reported.

The remainder of this section contains the proof of Theorem 5.1 which is split into two parts. In Section 5.1 we describe our algorithm and prove its correctness, while an efficient implementation is discussed in Section 5.2.

5.1 Recognition Algorithm

Our recognition algorithm is formalized in Algorithm 1.

Input: A 99-degenerate graph G=(V,E)G=(V,E) with |E|=5|V|10|E|=5|V|-10
Output: An optimal 22-planar rotation scheme of GG if it exists,
otherwise, false
1 Classify each edge eEe\in E as potentially planar or clearly crossing
2 Identify all instances of CROSS-BAT in GG and fix their partial rotation schemes
3 Create subgraph GpG_{p}
4 if GpG_{p} is not 33-connected or is not planar then
5       return false
6      
7Compute the unique planar rotation scheme p\mathcal{R}_{p} of GpG_{p}
8 if p\mathcal{R}_{p} contains a face of length greater than 55
9 or if a face does not induce a complete subgraph in GG then
10       return false
11      
12Augment GpG_{p} to maximal planar by triangulating the faces of p\mathcal{R}_{p}
13 Compute the set of triplets of p\mathcal{R}_{p}
14 for every triplet TT of p\mathcal{R}_{p} do
15       Label TT as facial 55-clique or as non-facial 55-clique
16      
17Let 𝒯\mathcal{T}^{*} be the set of triplets labelled as facial 55-cliques
18 if 𝒯\mathcal{T}^{*} covers all faces of p\mathcal{R}_{p} and all clearly crossing edges of GG exactly once then
19       return rotation scheme obtained by making each T𝒯T\in\mathcal{T}^{\ast} a facial 55-clique
20      
21return false
22
Algorithm 1 Recognition of Simple Optimal 22-planar Graphs

There are four main steps in the process. The first step is the classification of all edges of GG as potentially planar and clearly crossing (line 1). Second, is the identification of the CROSS-BAT instances and the creation of GpG_{p} and its planar rotation scheme p\mathcal{R}_{p} (lines 2–10). Third, we identify all triplets of p\mathcal{R}_{p} (line 11). Finally, we decide which of the triplets are not bad (lines 12–13) determining an optimal 22-planar rotation scheme of GG if one exists. Having the set of triplets that are facial 55-cliques in any p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG, we decide if GG is optimal 22-planar (lines 14–15) and compute an optimal 22-planar rotation scheme of GG if it exists (line 16). The details of the steps are given in Theorem D.1 in Appendix D.

5.2 Implementation

First, we check that the input graph has at most 5n105n-10 edges and that it is 99-degenerate. This process takes 𝒪(n)\mathcal{O}(n) time and is described in Details 1 in Appendix D, together with standard techniques and data structures that we use. In the following, we discuss in more detail the time complexity of the involved steps of Algorithm 1. The first step of the process, described in line 1 of Algorithm 1, can be performed in 𝒪(n)\mathcal{O}(n) time as stated in the following lemma:

Lemma 5.1 (\mathbf{\ast}).

All edges of GG can be classified as potentially planar or clearly crossing in 𝒪(n)\mathcal{O}(n) time.

We proceed with the second step given in lines 2–10 of Algorithm 1. First, we compute the CROSS-BAT instances of GG (line 2).

Lemma 5.2 (\mathbf{\ast}).

All CROSS-BAT instances of GG can be identified in 𝒪(n)\mathcal{O}(n) time if all edges are already classified as clearly crossing or potentially planar.

For each CROSS-BAT instance, we choose the rotation scheme of Fig. 2(b) where (u,y)(u,y) crosses (u,y)(u^{\prime},y^{\prime}) and (u,x)(u,x) crosses (u,x)(u^{\prime},x^{\prime}) by reclassifying (u,y)(u,y) and (u,x)(u,x) as clearly crossing. We proceed with lines 3–10 of Algorithm 1. We check the necessary conditions for GpG_{p} and compute the planar rotation scheme p\mathcal{R}_{p} and the dual GpG_{p}^{*} after augmenting GpG_{p} to maximal planar; see Details 2 in Appendix D.

Next, we compute the set of triplets of p\mathcal{R}_{p} (line 11) in 𝒪(n)\mathcal{O}(n) time; see Details 3 in Appendix D. During this process, for each face and each clearly crossing edge of GG, we store the set of triplets that contain it. The next lemma describes how to implement the labeling of each triplet as facial 55-clique or not; see line 13.

Lemma 5.3 (\mathbf{\ast}).

Let 𝒯={T}\mathcal{T}=\{T\} and 𝒮1\mathcal{S}_{1}, 𝒮2\mathcal{S}_{2} and 𝒮e\mathcal{S}_{e} be the set of triplets containing e1e_{1}, e2e_{2} and ee, respectively. It can be decided in 𝒪(1)\mathcal{O}(1) time if 𝒯\mathcal{T} is bad or not.

After labeling all triplets as facial 55-cliques or non-facial 55-cliques, we consider the set 𝒯\mathcal{T}^{*} of triplets labelled as facial 55-cliques. Checking whether 𝒯\mathcal{T}^{*} covers all faces and clearly crossing edges of GG exactly once, in line 15 of Algorithm 1, can be done in 𝒪(|𝒯|)=𝒪(n)\mathcal{O}(|\mathcal{T}^{*}|)=\mathcal{O}(n) time. Finally, we augment the planar rotation scheme p\mathcal{R}_{p} of GpG_{p} to an optimal 22-planar rotation scheme of GG in 𝒪(n)\mathcal{O}(n) time by inserting the clearly crossing edges of GG and identifying the crossings; refer to Details 4 in Appendix D. The overall time required for the last step is in 𝒪(n)\mathcal{O}(n).

Every step described above takes time 𝒪(n)\mathcal{O}(n), hence Algorithm 1 can be implemented to run in linear time. We point out that after fixing p\mathcal{R}_{p}, the optimal 22-planar rotation scheme of GG is uniquely defined if it exists. In particular, if there are cc instances of CROSS-BAT, there exist at most 2c2^{c} optimal 22-planar rotation schemes which can be easily enumerated.

6 Conclusions

We showed that simple optimal 22-planar graphs can be recognized and embedded in linear time. It remains open to extend our result also for simple 22-planar graphs with the maximum number of edges when n2mod3n\not\equiv 2\mod 3, as well as to non-simple optimal 22-planar graphs. Another reasonable attempt would be recognizing 55-maps, similarly to [12]. Finally, a natural question would be if a recognition algorithm for non-simple optimal 22-planar graphs could be adopted for optimal 33-planar graphs that are non-simple, or if our approach could work for simple optimal 33-planar graphs.

Acknowledgement.

We thank Michael Bekos for his valuable suggestions and endless discussions on this topic.

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Appendix 0.A Ommited Proofs from Section 2

See 2.1

Proof.

Assume for a contradiction that (u,v)(u,v) and (u,w)(u,w) belong to two different facial 55-cliques CC and CC^{\prime} of an optimal 22-planar rotation scheme \mathcal{R} of GG, respectively. Since CN(u,v)C\subseteq N(u,v) and CN(u,w)C^{\prime}\subseteq N(u,w), it follows that CCSC\cup C^{\prime}\subseteq S and thus |CC|<8|C\cup C^{\prime}|<8. Since |C|=|C|=5|C|=|C^{\prime}|=5, it follows that |CC|>2|C\cap C^{\prime}|>2; contradiction to Property 2.3. Hence, (u,v)(u,v) and (u,w)(u,w) belong to the same facial 55-clique. ∎

Appendix 0.B Ommited Proofs from Section 3

See 3.1

Proof.

Since (c1,c3)(c_{1},c_{3}) is potentially planar, vertices c1c_{1} and c3c_{3} have at least six common neighbors. Out of these common neighbors, exactly three belong to CC (namely, c0c_{0}, c2c_{2} and c4c_{4}). It follows that at least three of the common neighbors of c1c_{1} and c3c_{3}, call them uu, vv and ww, do not belong to CC; see Fig. 6(a). By Corollary 2.2, the edges connecting c1c_{1} to uu, vv and ww do not cross the edges that connect c3c_{3} to uu, vv and ww in \mathcal{R}. Assume w.l.o.g. that c1,c2,u,v,w,c4,c0c_{1},c_{2},u,v,w,c_{4},c_{0} appear in this circular order around c3c_{3} in \mathcal{R}; see Fig. 6(a).

Refer to caption
(a)
Refer to caption
(b)
Refer to caption
(c)
Refer to caption
(d)
Refer to caption
(e)
Refer to caption
(f)
Figure 6: (a) Vertices c1c_{1}, c3c_{3} have three common neighbors {u,v,w}\{u,v,w\} outside CC. (b) Illustration for Proposition 3.1.1. (c) Illustration for Proposition 3.1.2. (d) Illustration for Proposition 3.1.3. (e) Illustration for Proposition 3.1.4. (f) Summary of constraints imposed by Propositions 3.1.13.1.4.

A symmetric argument for c2c_{2} and c4c_{4} implies the existence of three common neighbors xx, yy and zz of c2c_{2} and c4c_{4} that do not belong to CC. Since CC is a facial 55-clique, vertices x,y,zx,y,z appear between c3c_{3} and c0c_{0} in the circular order around c4c_{4} in \mathcal{R}. In particular, we can assume w.l.o.g. that vertices c2,c3,x,y,z,c0,c1c_{2},c_{3},x,y,z,c_{0},c_{1} occur in this order around c4c_{4} in \mathcal{R}. Note that xx, yy and zz are not necessarily different from uu, vv and ww. In particular, we will prove in the following that |{u,v,w}{x,y,z}|=2|\{u,v,w\}\cap\{x,y,z\}|=2.

Proposition 3.1.1.

None of the edges (c4,x)(c_{4},x), (c4,y)(c_{4},y) and (c4,z)(c_{4},z) crosses (c1,w)(c_{1},w).

Proof.

Assume that edge (c4,x)(c_{4},x) crosses (c1,w)(c_{1},w) as in Fig. 6(b). This contradicts Corollary 2.2, since (c1,c4)(c_{1},c_{4}) is a crossing edge inside CC while edges (c4,x)(c_{4},x) and (c1,w)(c_{1},w) do not belong to CC. A symmetric argument applies for edges (c4,y)(c_{4},y) and (c4,z)(c_{4},z). ∎

Proposition 3.1.2.

None of vertices {x,y,z}\{x,y,z\} lies in the interior of the region delimited by the cycle (c3,w,c1,c0,c4)(c_{3},w,c_{1},c_{0},c_{4}).

Proof.

Assume for a contradiction that vertex zz lies in the interior of this region (refer to the gray-shaded region in Fig. 6(c)); an analogous argument holds for xx and yy. Then, edge (c2,z)(c_{2},z) crosses at least three edges, in particular, one edge of (c1,u)(c_{1},u) and (c3,u)(c_{3},u), one edge of (c1,v)(c_{1},v) and (c3,v)(c_{3},v), and one edge of (c1,w)(c_{1},w) and (c3,w)(c_{3},w); a contradiction to the fact that \mathcal{R} is a optimal 22-planar rotation scheme. ∎

Proposition 3.1.3.

w{x,y}w\notin\{x,y\} and edges (c4,x)(c_{4},x) and (c4,y)(c_{4},y) cross edge (c3,w)(c_{3},w).

Proof.

Assume for a contradiction, that w=yw=y; an analogous argument holds if we assume w=xw=x. Since yy, zz and c0c_{0} appear in this circular order around c4c_{4} in \mathcal{R} and since, by Proposition 3.1.1, edge (c4,z)(c_{4},z) does not cross (c1,w)(c_{1},w), vertex zz is located in the region delimited by the cycle (c4,w=y,c1,c0)(c_{4},w=y,c_{1},c_{0}); see Fig. 6(d). This contradicts Proposition 3.1.2, and hence w{x,y}w\notin\{x,y\}.

Now, since CC is a facial 55-clique, by Proposition 3.1.2, vertices xx and yy must be in the interior of the region delimited by the cycle (c3,c2,c1,w)(c_{3},c_{2},c_{1},w). Hence, each of edges (c4,x)(c_{4},x) and (c4,y)(c_{4},y) crosses either (c3,w)(c_{3},w) or (c1,w)(c_{1},w). However, by Proposition 3.1.1, the two edges must cross (c3,w)(c_{3},w) as claimed. ∎

Proposition 3.1.4.

z=wz=w.

Proof.

Assume for a contradiction that zwz\neq w. By Proposition 3.1.2, it follows that zz lies in the interior of the region delimited by the cycle (c3,c2,c1,w)(c_{3},c_{2},c_{1},w). By Proposition 3.1.1, it follows that (c4,z)(c_{4},z) crosses edge (c3,w)(c_{3},w). However, by Proposition 3.1.3, (c3,w)(c_{3},w) is also crossed by (c4,x)(c_{4},x) and (c4,y)(c_{4},y) as in Fig. 6(e); a contradiction to 22-planarity. ∎

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(a)
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(b)
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(c)
Figure 7: (a)–(b) Illustrations for Proposition 3.1.5. (c) Summary of constraints imposed by Propositions 3.1.13.1.5.

The constraints discussed so far lead to the configuration shown in Fig. 6(f). Since z=wz=w is a common neighbor of c2c_{2} and c4c_{4}, we turn our attention to edge (c2,z)=(c2,w)(c_{2},z)=(c_{2},w). We prove the following:

Proposition 3.1.5.

Edge (c2,z)(c_{2},z) crosses (c1,u)(c_{1},u) and (c1,v)(c_{1},v) and c2,c1,z,v,u\langle c_{2},c_{1},z,v,u\rangle is a facial 55-clique.

Proof.

By the circular order of edges around c1c_{1} and c3c_{3}, edge (c2,z)(c_{2},z) must cross one of edges (c1,u)(c_{1},u) and (c3,u)(c_{3},u) as well as one of edges (c1,v)(c_{1},v) and (c3,v)(c_{3},v). First we prove that it must either cross (c1,u)(c_{1},u) and (c1,v)(c_{1},v) or (c3,u)(c_{3},u) and (c3,v)(c_{3},v). Indeed if (c2,z)(c_{2},z) crosses (c1,u)(c_{1},u) and (c3,v)(c_{3},v) as in Fig. 7(a) (the case where (c2,z)(c_{2},z) crosses (c1,v)(c_{1},v) and (c3,u)(c_{3},u) is symmetric), edge (c2,z)(c_{2},z) is crossed by two independent edges, contradicting Property 2.2.

It remains to prove that (c2,z)(c_{2},z) does not cross edges (c3,u)(c_{3},u) and (c3,v)(c_{3},v). Assume the contrary; see Fig. 7(b). Then, by Property 2.2, c3,c2,u,v,z\langle c_{3},c_{2},u,v,z\rangle is a facial 55-clique with (c3,z)(c_{3},z) being one of its crossing-free boundary edges. However, by Proposition 3.1.3 edges (c4,x)(c_{4},x) and (c4,y)(c_{4},y) cross with edge (c3,z)(c_{3},z), as w=zw=z; a contradiction. We conclude that (c2,z)(c_{2},z) crosses (c1,u)(c_{1},u) and (c1,v)(c_{1},v) forming the facial 55-clique c2,c1,z,v,u\langle c_{2},c_{1},z,v,u\rangle. ∎

To summarize so far, Propositions 3.1.2 and 3.1.3 imply that vertices xx and yy lie in the interior of the region delimited by the cycle (c3,c4,c0,c1,z=w)(c_{3},c_{4},c_{0},c_{1},z=w). Since c0,c1,c2,c3,c4\langle c_{0},c_{1},c_{2},c_{3},c_{4}\rangle and c2,c1,z,v,u\langle c_{2},c_{1},z,v,u\rangle are facial 55-cliques, we can further restrict xx and yy in the interior of the region delimited by the cycle (c3,c2,u,v,z)(c_{3},c_{2},u,v,z); see Fig. 7(c). In the following proposition, we restrict xx and yy even more.

Refer to caption
(a)
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(b)
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(c)
Figure 8: (a) Illustration for Proposition 3.1.6. (b) Illustration for Proposition 3.1.8. (c) Summary of constraints imposed by Propositions 3.1.13.1.8.
Proposition 3.1.6.

Neither xx nor yy lies in the interior of the region delimited by the cycle (c3,v,z)(c_{3},v,z).

Proof.

Assume to the contrary that vertex xx lies in the interior of the region delimited by the cycle (c3,v,z)(c_{3},v,z).; an analogous argument applies for yy. Then, edge (c2,x)(c_{2},x) crosses (c3,v)(c_{3},v); see Fig. 8(a). This contradicts Corollary 2.2, since (c2,v)(c_{2},v) is a crossing edge in the facial 55-clique c2,c1,z,v,u\langle c_{2},c_{1},z,v,u\rangle. ∎

Proposition 3.1.7.

vxv\neq x and edge (c4,x)(c_{4},x) crosses (c3,v)(c_{3},v).

Proof.

First assume for a contradiction that v=xv=x. Due to circular order of the edges around c4c_{4}, vertex yy can only be located in the interior of the region delimited by the cycle (c3,v,z)(c_{3},v,z). This contradicts Proposition 3.1.6. Hence vxv\neq x. The crossing between (c4,x)(c_{4},x) and (c3,v)(c_{3},v) is imposed by Proposition 3.1.6. ∎

Proposition 3.1.8.

v=yv=y.

Proof.

Assume for a contradiction that vyv\neq y. By Propositions 3.1.6 and 3.1.7, it follows that both edges (c4,x)(c_{4},x) and (c4,y)(c_{4},y) cross both of (c3,z)(c_{3},z) and (c3,v)(c_{3},v); see Fig. 8(b). Then, by Property 2.1, the six vertices c4,c3,x,y,v,zc_{4},c_{3},x,y,v,z must form a facial 55-clique; a contradiction. ∎

The constraints imposed by Propositions 3.1.13.1.8 imply the configuration shown in Fig. 8(c). It remains to decide whether u=xu=x or if uu is located in the interior of cycle (c3,u,y)(c_{3},u,y).

Proposition 3.1.9.

xux\neq u.

Proof.

Assume for a contradiction that x=ux=u. Recall from Proposition 3.1.3 that edge (c4,x)(c_{4},x) crosses edge (c3,z)(c_{3},z); see Fig. 9(a). Consider now edge (z,x)(z,x) which is part of the facial 55-clique c2,c1,z,y,x\langle c_{2},c_{1},z,y,x\rangle. By Corollary 2.2, edge (c4,x)(c_{4},x) is not allowed to cross edge (c3,z)(c_{3},z), which however is the case; a contradiction. ∎

By Proposition 3.1.7 and by 22-planarity, we conclude that xx is located inside the cycle (c3,u,y)(c_{3},u,y); see Fig. 9(b). Since by Propositions 3.1.3 and 3.1.7, edge (c4,x)(c_{4},x) crosses with edges (c3,w)(c_{3},w) and (c3,v)(c_{3},v), respectively, it follows by Property 2.2 that c3,x,y,z,c4\langle c_{3},x,y,z,c_{4}\rangle is a facial 55-clique. In addition, (c4,z)(c_{4},z) is a crossing-free edge on the boundary of this facial 55-clique.

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(a)
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(b)
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(c)
Figure 9: (a) Illustration for Proposition 3.1.9. (b) Summary of constraints imposed by Propositions 3.1.13.1.9. (c) The crossing between (c3,u)(c_{3},u) and (c2,x)(c_{2},x) implies the facial 55-clique x,c3,c2,u,x\langle x,c_{3},c_{2},u,x^{\prime}\rangle.

Now, since edges (c2,x)(c_{2},x) and (c3,u)(c_{3},u) cross, they belong to a facial 55-clique that contains vertices c3,c2,u,xc_{3},c_{2},u,x. Let xx^{\prime} be the last vertex of this facial 55-clique. It must be xyx^{\prime}\neq y since edge (c3,y)(c_{3},y) is a crossing edge inside the facial 55-clique c3,x,y,z,c4\langle c_{3},x,y,z,c_{4}\rangle.

The final configuration is depicted in Fig. 9(c). Vertices c2c_{2} and c3c_{3} have degree nine in GG, with eight common neighbors N(c2)N(c3)={c0,c1,c4,u,v,w,x,x}N(c_{2})\cap N(c_{3})=\{c_{0},c_{1},c_{4},u,v,w,x,x^{\prime}\}. Let HH be the subgraph induced by the vertex set V(H)=N(c2)N(c3)V(H)=N(c_{2})\cup N(c_{3}). Since cycles (c0,c4,z,c1)(c_{0},c_{4},z,c_{1}) and (x,x,u,y)(x,x^{\prime},u,y) are crossing-free cycles and since the two edges (c1,c4)(c_{1},c_{4}) and (x,u)(x,u) are already part of a facial 55-clique, HH contains only the edges of the four identified facial 55-cliques and possibly one or both of edges (c0,z)(c_{0},z) and (x,y)(x^{\prime},y).

Consider a relabeling of the vertices of HH based on the following table:
Figure 9(c)a ac0c_{0}a ac1c_{1}a ac2c_{2}a ac3c_{3}a ac4c_{4}a auua av=yv=ya aw=zw=za axxa axx^{\prime}a GCBG_{CB} a vv xx^{\prime} uu^{\prime} uu xx yy^{\prime} ww^{\prime} vv^{\prime} yy ww

We obtain the CROSS-BAT substructure of Fig. 2(a) where CC is the facial 55-clique v,x,u,u,x\langle v,x^{\prime},u^{\prime},u,x\rangle as claimed. Since the boundaries of the four identified facial 55-cliques are crossing-free in \mathcal{R}, we can easily verify that the edges of HH are classified to clearly crossing and potentially planar according to the definition of CROSS-BAT; see also Fig. 2(a). ∎

See 3.2

Proof.

Consider an instance HH of CROSS-BAT in \mathcal{R}. In the following, we use the naming of vertices according to Fig. 2(a) for the vertices of HH. First we prove that the four 55-cliques of HH are always facial 55-cliques of \mathcal{R}.

Proposition 3.2.10 ().

Edges (u,v)(u,v) and (u,v)(u^{\prime},v) belong to the same facial 55-clique CvC_{v} in \mathcal{R} with V(Cv){u,u,v}{x,x,v}V(C_{v})\setminus\{u,u^{\prime},v\}\subset\{x,x^{\prime},v^{\prime}\}. Also, edges (u,w)(u,w) and (u,w)(u^{\prime},w) belong to another facial 55-clique CwC_{w} with V(Cw){u,u,w}{y,y,w}V(C_{w})\setminus\{u,u^{\prime},w\}\subset\{y,y^{\prime},w^{\prime}\}.

Proof.

By definition, vertices uu and uu^{\prime} have degree 9 in GG and V(H)=N(u)N(v)V(H)=N(u)\cup N(v). In particular N(u)=V(H){u}N(u)=V(H)\setminus\{u\} and N(u)=V(H){u}N(u^{\prime})=V(H)\setminus\{u^{\prime}\}. On the other hand, the neighborhood of vv in HH consists of uu, uu^{\prime}, xx and xx^{\prime} (which form a 55-clique with vv) and potentially vv^{\prime} (if the edge (v,v)(v,v^{\prime}) exists). We conclude that N(u,v){u,x,x,v}N(u,v)\subseteq\{u^{\prime},x,x^{\prime},v^{\prime}\}, and N(u,v){u,x,x,v}N(u^{\prime},v)\subseteq\{u,x,x^{\prime},v^{\prime}\}. Then N(u,v)N(u,v){u,u}{x,x,v}N(u,v)\cup N(u^{\prime},v)\setminus\{u,u^{\prime}\}\subseteq\{x,x^{\prime},v^{\prime}\}. By Corollary 2.1, it follows that (u,v)(u,v) and (u,v)(u^{\prime},v) belong to the same facial 55-clique CvC_{v}. The vertex set of CvC_{v} contains vertices uu, uu^{\prime} and vv, and two vertices from {x,x,v}\{x,x^{\prime},v^{\prime}\}. By symmetry, the same holds for edges (u,w)(u,w) and (u,w)(u^{\prime},w), that is, they belong to the same facial 55-clique CwC_{w}, whose vertex set contains vertices uu, uu^{\prime}, ww, and two vertices from {y,y,w}\{y,y^{\prime},w^{\prime}\}. ∎

Proposition 3.2.11 ().

Let {u1,u2}={u,u}\{u_{1},u_{2}\}=\{u,u^{\prime}\}. Edges (y,x)(y,x), (y,v)(y,v^{\prime}) and (x,w)(x,w^{\prime}) belong to a facial 55-clique Cx,y=u1,y,w,v,xC_{x,y}=\langle u_{1},y,w^{\prime},v^{\prime},x\rangle in \mathcal{R}. Also, edges (y,x)(y^{\prime},x^{\prime}) and (y,v)(y^{\prime},v^{\prime}) and (x,w)(x^{\prime},w^{\prime}) belong to another facial 55-clique Cx,y=u2,x,v,w,yC_{x^{\prime},y^{\prime}}=\langle u_{2},x^{\prime},v^{\prime},w^{\prime},y^{\prime}\rangle.

Proof.

Consider edge (y,x)(y,x). Since (y,x)(y,x) is clearly crossing it belongs to a facial 55-clique Cx,yC_{x,y} of \mathcal{R}. We have that N(y,x)={u,u,v,w}N(y,x)=\{u,u^{\prime},v^{\prime},w^{\prime}\}. By Proposition 3.2.10, the edge (u,u)(u,u^{\prime}) belongs to two facial 55-cliques, that contain vertices vv and ww respectively. Since N(y,x)N(y,x) contains neither vv nor ww, it follows that Cx,yC_{x,y} contains exactly one of uu or uu^{\prime}. Then V(Cx,y)={x,y,u1,v,w}V(C_{x,y})=\{x,y,u_{1},v^{\prime},w^{\prime}\}, where u1{u,u}u_{1}\in\{u,u^{\prime}\}. Hence all three edges (y,x)(y,x) and (y,v)(y,v^{\prime}) and (x,w)(x,w^{\prime}) belong to the same facial 55-clique, namely Cx,yC_{x,y}. By symmetry, edges (y,x)(y^{\prime},x^{\prime}) and (y,v)(y^{\prime},v^{\prime}) and (x,w)(x^{\prime},w^{\prime}) belong to another facial 55-clique Cx,yC_{x^{\prime},y^{\prime}} where V(Cx,y)={x,y,u2,v,w}V(C_{x^{\prime},y^{\prime}})=\{x^{\prime},y^{\prime},u_{2},v^{\prime},w^{\prime}\} and u2{u,u}u_{2}\in\{u,u^{\prime}\}. It remains to argue that u1u2u_{1}\neq u_{2}. Recall that vertices uu and uu^{\prime} have degree nine in GG and therefore belong to three facial 55-cliques in \mathcal{R}. Already, by Proposition 3.2.10 we have identified two facial 55-cliques. Clearly if u1=u2=uu_{1}=u_{2}=u, vertex uu would belong to two more facial 55-cliques, namely Cx,yC_{x,y} and Cx,yC_{x^{\prime},y^{\prime}}, that is a total of four facial 55-cliques. ∎

As initially mentioned, we identified four facial 55-cliques, CvC_{v}, CwC_{w}, Cx,yC_{x,y} and Cx,yC_{x^{\prime},y^{\prime}}. Recall that CvC_{v} contains two vertices from {x,x,v}\{x,x^{\prime},v^{\prime}\}. We show now, that CvC_{v} does not contain vv^{\prime}. Assume for a contradiction that CvC_{v} contains vv^{\prime} and w.l.o.g. vertex xx. Then V(Cv)V(Cx,y)={u1,x,v}V(C_{v})\cap V(C_{x,y})=\{u_{1},x,v^{\prime}\}; a contradiction to Property 2.3. Hence, CvC_{v} contains both xx and xx^{\prime}. Analogously, CwC_{w} contains both of yy and yy^{\prime}.

By Proposition 3.2.11, Cy,xC_{y,x} and Cy,xC_{y^{\prime},x^{\prime}} are two facial 55-cliques that share the crossing-free edge (v,w)(v^{\prime},w^{\prime}). Moreover, by Property 2.4, edge (u1,u2)=(u,u)(u_{1},u_{2})=(u,u^{\prime}) is crossing-free as it belongs to the two facial 55-cliques CvC_{v} and CwC_{w} by Proposition 3.2.10; refer to Figs. 2(b) and 2(c). Hence, ww is contained in the crossing-free separating cycle (u1,u2,y,w,y)(u_{1},u_{2},y^{\prime},w^{\prime},y) while vv is contained in the crossing-free separating cycle (u2,u1,x,v,x)(u_{2},u_{1},x,v^{\prime},x^{\prime}). Since (u1,w)(u_{1},w) and (u2,w)(u_{2},w) are clearly crossing, edges (w,y)(w,y) and (w,y)(w,y^{\prime}) are on the boundary of CwC_{w}, the same holds for edges (v,x)(v,x) and (v,x)(v,x^{\prime}) being on the boundary of CvC_{v}. Now, by selecting u1=uu_{1}=u or u1=uu_{1}=u^{\prime}, we yield the two different embeddings of Figs. 2(b) and 2(c). ∎

See 3.4

Proof.

By Lemma 3.3, we can assume w.l.o.g. that HH has the rotation scheme of CROSS-BAT shown in Fig. 2(b) in \mathcal{R}. Then, it is not hard to see that edges (u,x)(u^{\prime},x^{\prime}) and (u,x)(u,x) are potentially planar and cross each other. The same holds for edges (u,y)(u^{\prime},y^{\prime}) and (u,y)(u,y). It is evident that no other pair of potentially planar edges that belong to HH cross. ∎

Appendix C Ommited Proofs from Section 4

See 4.1

Proof.

Assume for a contradiction that 𝒯\mathcal{T}^{\prime} is not bad. In 𝒯\mathcal{R}_{\mathcal{T}^{\prime}}, edge e1e_{1} is assigned to T1T_{1} and edge e2e_{2} to T2T_{2}. By the assumptions of the lemma, none of T1T_{1} and T2T_{2} is face-disjoint from TT (i.e., Case C.1 applies neither for T1T_{1} nor for T2T_{2}). We consider two cases, depending on whether Case C.3 applies for one of T1T_{1} and T2T_{2} or Case C.2 applies for both T1T_{1} and T2T_{2}.

First, consider the case where w.l.o.g. T1T_{1} shares faces ff and f1f_{1} with TT (i.e., Case C.3 applies for T1T_{1}); the reasoning for T2T_{2} is symmetric. We claim that T2T_{2} is face-disjoint from TT, contradicting our assumption that none of T1T_{1} and T2T_{2} is face-disjoint from TT. Indeed, T2T_{2} cannot share faces ff and f2f_{2} with TT, as otherwise e1e_{1} and e2e_{2} cross each other in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} and thus T1T_{1} and T2T_{2} would be identical with TT. On the other hand, if T2T_{2} contains face f1f_{1} (which is also contained in T1T_{1}), then it follows that T1T_{1}, T2T_{2} and TT coincide; a contradiction. Hence, T2T_{2} is necessarily face-disjoint with TT, contradicting our assumption that Case C.1 does not apply for T2T_{2}.

Second, consider the case where Case C.2 applies for both T1T_{1} and T2T_{2}, that is, for {i,j}={1,2}\{i,j\}=\{1,2\} triplet TiT_{i} shares only face fjf_{j} with TT. Clearly T1T2T_{1}\neq T_{2}. In particular, in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} the facial 55-clique formed by T1T_{1} contains vertices u,v2,w2u,v_{2},w_{2} (the three vertices of face f2f_{2}) and vertex v1v_{1} (which is endpoint of edge e1e_{1}). Similarly, in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} the facial 55-clique formed by T2T_{2} contains vertices u,v1,w1,v2u,v_{1},w_{1},v_{2}. Hence, the two facial 55-cliques share three vertices; a contradiction to Property 2.3. This concludes the proof. ∎

See 4.2

Proof.

Assume for a contradiction that 𝒯\mathcal{T}^{\prime} is not bad, and each of T1T_{1} and T2T_{2} is face-disjoint from TT. Recall that in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} edge e1e_{1} is assigned to T1T_{1} and edge e2e_{2} to T2T_{2}, where T1=T2T_{1}=T_{2} might hold. Since 𝒯\mathcal{T}^{\prime} is not bad, 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} can be extended to an p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R}^{\prime} of GG. This implies that there exists (at least) one triplet TfT_{f} that contains face ff, such that 𝒯{Tf}\mathcal{T}^{\prime}\cup\{T_{f}\} is not bad. If TfT_{f} contains f1f_{1}, then by Observation 4.2 edge e1e_{1} is assigned to TfT_{f}, i.e., Tf=T1T_{f}=T_{1}. This contradicts our assumption that T1T_{1} is face-disjoint with TT. Hence, TfT_{f} does not contain f1f_{1}. Similarly, we argue that TfT_{f} does not contain f2f_{2}.

By the assumptions of the lemma, T1T_{1} and T2T_{2} are face-disjoint from TT. Thus, in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} both edges e1e_{1} and e2e_{2} are drawn outside the faces of TT, which implies that they cross each other. Let u,w2,x2,x1u,w_{2},x_{2},x_{1} and w1w_{1} be the vertices of TfT_{f} as shown in Fig. 10. Now edge (w1,w2)(w_{1},w_{2}) is a crossing edge of the facial 55-clique formed by TfT_{f} in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}}, which, by Corollary 2.2, implies that e1e_{1} and e2e_{2} do not cross each other; a contradiction. ∎

Refer to caption
Figure 10: Illustration for the proof of Lemma 4.2.

See 4.3

Proof.

Assume that 𝒯\mathcal{T}^{\prime} is not bad. Then 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} can be extended to a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R}^{\prime} of GG. Consider the facial 55-cliques of T1T_{1} and T2T_{2} in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}}. Since the boundary edges of facial 55-cliques are crossing-free and T2T_{2} shares only face f1f_{1} with TT, edge e2e_{2} must cross edge (v1,u)(v_{1},u); see Fig. 11. T2T_{2} contains vertices v1v_{1}, w1w_{1}, uu (since these are vertices of f1f_{1}) and vertex v2v_{2} (as an endpoint of edge e2e_{2}). Let zz be the last vertex of T2T_{2}. If zV(T)z\in V(T) then T2=TT_{2}=T by Observation 4.1, contradicting the assumption that T2T_{2} shares only face f1f_{1} with TT. Since edges ee and e2e_{2} are both incident to v2v_{2} and are crossing edges in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}}, edges (v2,z)(v_{2},z) and (v2,u)(v_{2},u) are crossing-free. Similarly, edges (v1,z)(v_{1},z) and (v1,w1)(v_{1},w_{1}) are crossing-free. Hence, the facial 55-clique of T2T_{2} is w1,v1,z,v2,u\langle w_{1},v_{1},z,v_{2},u\rangle; see Fig 11.

Refer to caption
Figure 11: Illustration for the proof of Lemma 4.3.

Let TfT_{f} and T3T_{3} be the triplets that faces ff and f2f_{2} are assigned to w.r.t. \mathcal{R}^{\prime}. It must be the case that TfT3T_{f}\neq T_{3} as otherwise, TfT_{f} would contain faces ff and f2f_{2} of TT and, as a consequence, edge e2e_{2}. Then TfT_{f} would be identical to T2T_{2} contradicting the fact that T2T_{2} shares only face f1f_{1} with TT. Hence we conclude that there exist triplets TfT_{f} and T3T_{3} such that 𝒯{Tf,T3}\mathcal{T}^{\prime}\cup\{T_{f},T_{3}\} is not bad. Observe that uu is surrounded by the three facial 55-cliques formed by T2T_{2}, TfT_{f} and T3T_{3} in \mathcal{R}^{\prime}, that pair-wise share an uncrossed edge incident to uu. This implies that d(u)=9d(u)=9. Hence, if d(u)>9d(u)>9, then 𝒯\mathcal{T}^{\prime} is bad as stated in the lemma. It remains to prove that there is a vertex xS=N(u)V(T)x\in S=N(u)\setminus V(T) with |N(x)S|<2|N(x)\cap S|<2.

The facial 55-cliques formed by triplets TfT_{f} and T3T_{3} share vertices uu and w2w_{2}, thus, by Property 2.4, edge (u,w2)(u,w_{2}) is crossing-free in \mathcal{R}^{\prime}. Similarly, T2T_{2} and T3T_{3} share vertices uu and v2v_{2}, while T2T_{2} and TfT_{f} share vertices uu and w1w_{1}. Hence, edges (u,v2)(u,v_{2}) and (u,w1)(u,w_{1}) are also crossing-free in \mathcal{R}^{\prime}.

Thus, the facial 55-clique formed by TfT_{f} is bounded by the crossing-free 55-cycle (u,w2,x2,x1,w1)(u,w_{2},x_{2},x_{1},w_{1}) while the facial 55-clique formed by TfT_{f} is bounded by the crossing-free 55-cycle (u,v2,y2,y1,w2)(u,v_{2},y_{2},y_{1},w_{2}) where x1,x2,y1,y2V(T)x_{1},x_{2},y_{1},y_{2}\not\in V(T); see Fig 11. Observe that since TfT_{f} and T3T_{3} share vertices uu and w2w_{2} we can conclude that {x1,x2}{y1,y2}=\{x_{1},x_{2}\}\cap\{y_{1},y_{2}\}=\emptyset by Property 2.3. Similarly, each of TfT_{f} and T3T_{3} shares two vertices with T2T_{2}, thus z{x1,x2,y1,y2}z\not\in\{x_{1},x_{2},y_{1},y_{2}\}.

Since triplets T1T_{1} and TfT_{f} share vertex w2w_{2}, at least one vertex among x1x_{1} and x2x_{2}, say x1x_{1}, does not belong to T1T_{1} by Property 2.3. Let S=N(u)V(T)={x1,x2,y1,y2,z}S=N(u)\setminus V(T)=\{x_{1},x_{2},y_{1},y_{2},z\}. We claim that |N(x1)S|1|N(x_{1})\cap S|\leq 1. Since T1T_{1} is a facial 55-clique in 𝒯\mathcal{R}_{\mathcal{T}^{\prime}}, there exist two crossing-free paths P1P_{1} and P1P_{1}^{\prime} that connect v1v_{1} and w2w_{2}. Assume w.l.o.g. that the edges of P1P_{1} and P1P_{1}^{\prime} incident to v1v_{1} appear in this order and between edges (v1,w1)(v_{1},w_{1}) and (v1,z)(v_{1},z) in the circular order of edges around v1v_{1}222It might be that the edge of P1P_{1} incident to v1v_{1} coincides with (v1,w1)(v_{1},w_{1}) or that the edge of P1P_{1}^{\prime} incident to v1v_{1} coincides with (v1,z)(v_{1},z).; refer to Fig. 11. The cycle (w2,P1,v1,w1,u)(w_{2},P_{1},v_{1},w_{1},u) is crossing-free in \mathcal{R}^{\prime}. Since x1x_{1} lies in the interior of this cycle, while vertices zz, y1y_{1} and y2y_{2} lie in its exterior, it follows that vertex x1x_{1} is not adjacent to vertices zz, y1y_{1} or y2y_{2}, i.e., x1x_{1} is adjacent only to vertex x2x_{2} of SS.

To summarize, assuming that 𝒯\mathcal{T}^{\prime} is not bad, vertex uu has degree nine and there exists at least one neighbor of uu that does not belong to V(T)V(T), namely x1x_{1}, such that |N(x1)S|1|N(x_{1})\cap S|\leq 1. The lemma follows. ∎

See 4.4

Proof.

Assume for a contradiction that 𝒯\mathcal{T} is not bad. Let \mathcal{R} be a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG that extends 𝒯\mathcal{R}_{\mathcal{T}}. In \mathcal{R}, the triplet TxT_{x} that contains edge (v,x)(v,x) contains three more neighbors of vv. Since |N(x)S|1|N(x)\cap S|\leq 1, TxT_{x} contains at least two vertices from V(T){v}V(T)\setminus\{v\}, contradicting Property 2.3. ∎

See 4.5

Proof.

Assume that 𝒯\mathcal{T}^{\prime} is not bad and that T1T_{1} contains vertex w1w_{1}. Then 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} can be extended to a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R}^{\prime} of GG. Since e1e_{1} belongs to T1T_{1}, it follows that T1T_{1} also contains vertices v1v_{1} and w2w_{2} of TT. Let x1x_{1} and x2x_{2} be the other two vertices of T1T_{1}. First we claim that neither x1x_{1} nor x2x_{2} is a vertex of TT. Assume the contrary, say for x1x_{1}. Then, x1x_{1} is either vertex uu or vertex v2v_{2} of TT. If x1x_{1} is vertex uu, then by Observation 4.1, triplet T1T_{1} would contain face f1f_{1} of TT contradicting the assumption that T1T_{1} is face-disjoint with TT. On the other hand, if x1x_{1} is vertex v2v_{2}, then T1T_{1} would contain edge e2e_{2}, contradicting the fact that T1T_{1} is different from T2T_{2}.

Now, consider a counter-clockwise walk around the boundary of the facial 55-clique formed by T1T_{1} in GpG_{p}. If edge (v1,w1)(v_{1},w_{1}) is not on the boundary of T1T_{1}, then face f1f_{1} would belong to T1T_{1}; a contradiction to the fact that T1T_{1} is face-disjoint with TT. Similarly, (w1,w2)(w_{1},w_{2}) is also on the boundary of T1T_{1}, since face ff does not belong to T1T_{1}. This implies that vertices v1v_{1}, w1w_{1} and w2w_{2} appear consecutively along the boundary of T1T_{1}, followed by vertices x1x_{1} and x2x_{2}.

Refer to caption
Figure 12: Illustration for the proof of Lemma 4.5.

Recall that w.r.t. 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} edges e1e_{1}, e2e_{2} and ee are assigned to T1T_{1}, T2T_{2} and TeT_{e} respectively, as shown in Fig. 12. Note that f1f_{1} is not contained in any of T1T_{1}, T2T_{2} or TeT_{e}. Let T3T_{3} be the triplet that forms the facial 55-clique of \mathcal{R}^{\prime} where f1f_{1} belongs to. Since 𝒯\mathcal{T}^{\prime} is not bad, we conclude that 𝒯{T3}\mathcal{T}^{\prime}\cup\{T_{3}\} is also not bad. Observe that w1w_{1} is surrounded by the three facial 55-cliques formed by T1T_{1}, T2T_{2} and T3T_{3} in \mathcal{R}^{\prime}, that pair-wise share an uncrossed edge incident to w1w_{1}. Thus, vertex w1w_{1} is a degree-9 vertex. Clearly, if d(w1)>9d(w_{1})>9, 𝒯\mathcal{T}^{\prime} is bad as stated in the lemma. It remains to prove that there is no vertex xS=N(w1)V(T)x\in S=N(w_{1})\setminus V(T) with |N(x)S|4|N(x)\cap S|\geq 4.

By the assumptions of the lemma, triplet T2T_{2} contains faces ff and f2f_{2} of TT, hence it contains vertices w1w_{1}, w2w_{2}, uu and v2v_{2} of TT. Let zz be the last vertex of T2T_{2}, which is different from v1v_{1}. Now, along the boundary of T3T_{3} vertices uu, w1w_{1} and v1v_{1} appear consecutively, followed by two more vertices y1,y2V(T)y_{1},y_{2}\not\in V(T) that are different from vertices x1x_{1}, x2x_{2} and zz. Hence the neighbors of w1w_{1} in circular order are {v1,y1,y2,u,z,v2,w2,x1,x2}\{v_{1},y_{1},y_{2},u,z,v_{2},w_{2},x_{1},x_{2}\} or {v1,y1,y2,u,v2,z,w2,x1,x2}\{v_{1},y_{1},y_{2},u,v_{2},z,w_{2},x_{1},x_{2}\} depending on whether vertex zz appears before of after vertex v2v_{2}. Let S={x1,x2,y1,y2,z}S=\{x_{1},x_{2},y_{1},y_{2},z\} be the set of neighbors of w1w_{1} that do not belong to TT. We claim that every vertex of SS has at most 33 neighbors in SS.

Recall that the boundary edges of the facial 55-cliques in 𝒯{T3}\mathcal{R}_{\mathcal{T}^{\prime}\cup\{T_{3}\}} are crossing-free. Since TeT_{e} is a facial 55-clique in 𝒯{T3}\mathcal{R}_{\mathcal{T}^{\prime}\cup\{T_{3}\}}, it follows that there exist two crossing-free paths PeP_{e} and PeP_{e}^{\prime} that connect vertices v1v_{1} and v2v_{2}. Assume w.l.o.g. that the edges of PeP_{e}^{\prime} and PeP_{e} incident to v1v_{1} appear in this order and between edges (v1,x2)(v_{1},x_{2}) and (v1,y1)(v_{1},y_{1}) in the circular order of edges around v1v_{1}333It might be that the edge of PeP_{e}^{\prime} incident to v1v_{1} coincides with (v1,x2)(v_{1},x_{2}) or that the edge of PeP_{e} incident to v1v_{1} coincides with (v1,y1)(v_{1},y_{1}).; refer to Fig. 12. Then PeP_{e} along with the crossing-free path v2,u,w1,v1v_{2},u,w_{1},v_{1} forms a crossing-free cycle in 𝒯{T3}\mathcal{R}_{\mathcal{T}^{\prime}\cup\{T_{3}\}}. Vertex y2y_{2} lies in the interior of this cycle (note that y1y_{1} might be a vertex of this cycle), while vertex x1x_{1} lies in its exterior. Thus, x1x_{1} and y2y_{2} can’t be adjacent vertices. Also, any other vertex of SS can be adjacent to at most one of x1x_{1} and y2y_{2}, proving the claim. To summarize, assuming that 𝒯\mathcal{T}^{\prime} is not bad, and that vertex w1w_{1} has degree nine, we conclude that for every neighbor xx of w1w_{1} in SS it holds |N(x)S|3|N(x)\cup S|\leq 3, proving the first part of the lemma. The case where 𝒯\mathcal{T}^{\prime} is not bad and TeT_{e} contains vertex uu is symmetric. ∎

See 4.6

Proof.

Assume for a contradiction that 𝒯\mathcal{T} is not bad. Thus there is a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} of GG that extends 𝒯\mathcal{R}_{\mathcal{T}}. In \mathcal{R}, the neighbors of vv belong to three facial 55-cliques, one of them being TT. Let TT^{\prime} and T′′T^{\prime\prime} be two other facial 55-cliques which include vv so that TT^{\prime} and T′′T^{\prime\prime} share vertices vv and cvc\neq v. Clearly, cSc\in S, and |N(v)N(c)|6|N(v)\cap N(c)|\geq 6. Since each of TT^{\prime} and T′′T^{\prime\prime} can contain at most one vertex of TT different from vv, this implies that cc must be adjacent to at least 44 vertices of SS; a contradiction. ∎

See 4.7

Proof.

Assume that 𝒯\mathcal{T}^{\prime} is not bad, i.e. there exists a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R}^{\prime} of GG that extends 𝒯\mathcal{R}_{\mathcal{T}^{\prime}}. By assumption, T2T_{2} shares only faces ff and f2f_{2} with TT in \mathcal{R}^{\prime}. In \mathcal{R}^{\prime}, face f1f_{1} is part of a facial 55-clique formed by a triplet Tf1T_{f_{1}} which is necessarily face-disjoint from all of T1T_{1}, T2T_{2} and TeT_{e}. Since f1f_{1} already shares vertices uu and w1w_{1} with the facial 55-clique formed by T2T_{2}, triplet Tf1T_{f_{1}} contains neither vertex v2v_{2} nor w2w_{2} of TT, i.e. Tf1T_{f_{1}} shares only the three vertices of f1f_{1} with TT. The lemma follows. ∎

See 4.8

Proof.

Assume for a contradiction that 𝒯\mathcal{T} is not bad. The boundary of triplet T1T_{1} consists of two paths in p\mathcal{R}_{p}, say P1P_{1} and P1P^{\prime}_{1}, that connect vertices v1v_{1} and w2w_{2}. Similarly, the boundary of TeT_{e} consists of two paths in p\mathcal{R}_{p}, say PeP_{e} and PeP^{\prime}_{e}, that connect vertices v1v_{1} and v2v_{2}. Note that P1P_{1} and P1P^{\prime}_{1} are edge-disjoint. The same holds for paths PeP_{e} and PeP^{\prime}_{e}. Assume w.l.o.g. that in the counter-clockwise order of edges around v1v_{1}, we encounter the edges of P1P_{1}, P1P_{1}^{\prime}, PeP_{e}^{\prime} and PeP_{e} incident to v1v_{1} in this order between edges (v1,w1)(v_{1},w_{1}) and (v1,u)(v_{1},u); see Fig. 13. Thus, P1P_{1} and PeP_{e} are edge-disjoint. On the other hand, paths P1P^{\prime}_{1} and PeP^{\prime}_{e} might either be edge-disjoint, as in Fig. 13(a), or they share an edge, as shown in Figs. 13(b) and 13(c). Note that in the second case the edge of P1P^{\prime}_{1} incident to v1v_{1} is the same as the corresponding edge of PeP^{\prime}_{e}.

Refer to caption
(a)
Refer to caption
(b)
Refer to caption
(c)
Figure 13: Illustration for the proof of Lemma 4.8.

Consider triplet Tf1T_{f_{1}} of the lemma that contains face f1f_{1} of TT. Since by assumption Tf1T2T_{f_{1}}\neq T_{2}, triplet Tf1T_{f_{1}} must contain at least one of the triangular faces that share edge (v1,w1)(v_{1},w_{1}) or (v1,u)(v_{1},u) with face f1f_{1}. Assume w.l.o.g. that Tf1T_{f_{1}} contains the face sharing edge (v1,w1)(v_{1},w_{1}) with f1f_{1}; the case where Tf1T_{f_{1}} contains the face sharing edge (v1,u)(v_{1},u) with f1f_{1} is symmetric. Let uu^{\prime} be the third vertex of this face that is different from vertices v1v_{1} and w1w_{1}; see Fig. 13. Since both uu and uu^{\prime} are vertices of triplet Tf1T_{f_{1}}, edge (u,u)(u,u^{\prime}) is an edge of GG. Furthermore uu^{\prime} lies in the interior of the cycle (w2,P1,v1,w1)(w_{2},P_{1}^{\prime},v_{1},w_{1}). In particular, the presence of path P1P^{\prime}_{1} in GpG_{p} implies that edge (u,u)(u,u^{\prime}) is a clearly crossing edge of GG.

Since 𝒯\mathcal{T} is not bad there exists a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} of GG that extends 𝒯\mathcal{R}_{\mathcal{T}}. We focus on edge (u,u)(u,u^{\prime}) in \mathcal{R}, and consider two cases, depending on whether paths P1P^{\prime}_{1} and PeP^{\prime}_{e} are edge-disjoint or not. If P1P^{\prime}_{1} and PeP^{\prime}_{e} are edge-disjoint, the edge (u,u)(u,u^{\prime}) in \mathcal{R} must cross an edge of P1P^{\prime}_{1}, an edge of PeP_{e} and an edge of PeP^{\prime}_{e}, as shown in Fig. 13(a). This implies that \mathcal{R} cannot exist, since (u,u)(u,u^{\prime}) has at least three crossings; a contradiction to the assumption that 𝒯\mathcal{T} is not bad. If P1P^{\prime}_{1} and PeP^{\prime}_{e} share edge (v1,v)(v_{1},v^{\prime}), then in \mathcal{R}, edge (u,u)(u,u^{\prime}) must cross the edge of PeP_{e} that is incident to v1v_{1}, say (v1,w)(v_{1},w^{\prime}), and the common edge (v1,v)(v_{1},v^{\prime}) of paths P1P^{\prime}_{1} and PeP^{\prime}_{e} (otherwise, we can conclude as before that 𝒯\mathcal{T} is bad). Furthermore, vertex uu^{\prime} belongs to path P1P_{1} as otherwise (u,u)(u,u^{\prime}) also crosses an edge of P1P_{1} creating three crossings along (u,u)(u,u^{\prime}). This implies that in \mathcal{R}, v1,u,v,w,u\langle v_{1},u^{\prime},v^{\prime},w^{\prime},u\rangle is a facial 55-clique, and edge (u,w)(u^{\prime},w^{\prime}) is a crossing edge of this facial 55-clique; see Figs. 13(b) and 13(c).

Now, triplet Tf1T_{f_{1}} contains vertices uu, v1v_{1}, w1w_{1} and uu^{\prime}. Let zz be the last vertex of Tf1T_{f_{1}} which by the assumptions of the lemma is not a vertex of TT, and it is adjacent to vertices uu, w1w_{1} and v1v_{1}.

Since by the assumption of the lemma, Tf1T_{f_{1}} is face-disjoint from T1T_{1} and TeT_{e}, it follows that zz is in the interior of the region bounded by cycle (w2,P1,v1,w1)(w_{2},P^{\prime}_{1},v_{1},w_{1}) or the region bounded by cycle (v1,Pe,v2,u)(v_{1},P^{\prime}_{e},v_{2},u). In the first case; see Fig. 13(b), edge (u,z)(u,z) can’t be drawn since it should cross paths P1P_{1}^{\prime}, PeP_{e}^{\prime} and PeP_{e}, having at least three crossings; a contradiction. Similarly, in the second case; see Fig. 13(c), edge (w1,z)(w_{1},z) can’t be drawn since it should cross paths P1P_{1}, P1P_{1}^{\prime} and PeP_{e}^{\prime}, having at least three crossings; a contradiction. The lemma follows. ∎

See 4.9

Refer to caption
(a)
Refer to caption
(b)
Figure 14: Illustration for the proof of Lemma 4.9.
Proof.

From the assumptions stated in the lemma, edges ee and e1e_{1} belong to both triplets TT and T1T_{1}. This implies that TT and T1T_{1} share vertices v1v_{1}, w2w_{2} and v2v_{2}. If TT and T1T_{1} share another vertex, then T2T_{2} and T1T_{1} would have at least three common vertices, contradicting the assumptions of the lemma.

Hence, TT and T1T_{1} form the configuration shown in Fig. 14(a) which is unique up to a renaming of vertices w1w_{1} and uu and a renaming of vertices w1w_{1}^{\prime} and uu^{\prime}. Triplet T2T_{2} contains vertices w1w_{1}, w2w_{2}, uu, v2v_{2} and a vertex zz which by the assumptions of the lemma does not belong to TT or T1T_{1}.

Assume that 𝒯\mathcal{T} is not bad, i.e., there exists a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} of GG that extends 𝒯\mathcal{R}_{\mathcal{T}} such that TT is a facial 55-clique in \mathcal{R}. Assume w.l.o.g. that zz lies in the exterior of the cycle Z=(v1,u,v2,u)Z=(v_{1},u,v_{2},u^{\prime}) of GpG_{p} as shown in Fig. 14(b). Since T2T_{2} is a triplet, edges (z,w1)(z,w_{1}) and (z,w2)(z,w_{2}) belong to GG. Furthermore, (z,w1)(z,w_{1}) and (z,w2)(z,w_{2}) are clearly crossing edges, as w1w_{1} and w2w_{2} lie in the interior of ZZ and zz in its exterior; see Fig. 14(b).

In \mathcal{R}, edge (z,w1)(z,w_{1}) must cross both edges (v1,u)(v_{1},u^{\prime}) and (v1,w1)(v_{1},w^{\prime}_{1}). This implies that C=z,v1,w1,w1,uC=\langle z,v_{1},w_{1},w_{1}^{\prime},u\rangle is a facial 55-clique in \mathcal{R}; see Fig. 14(b). Similarly, edge (z,w2)(z,w_{2}) must cross edges (v2,u)(v_{2},u^{\prime}) and (v2,w1)(v_{2},w_{1}^{\prime}) (the latter crossing is unavoidable in the presence of the facial 55-cliques formed by TT and CC), hence C=z,u,w1,w2,v2C^{\prime}=\langle z,u^{\prime},w_{1}^{\prime},w_{2},v_{2}\rangle is another facial 55-clique of \mathcal{R}. Then CC and CC^{\prime} form two facial 55-cliques in \mathcal{R} and share vertices zz, uu^{\prime} and w1w^{\prime}_{1}, contradicting Property 2.3. ∎

See 4.10

Proof.

Since GG is optimal 22-planar, then at least one of 𝒯\mathcal{T} or 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} for some triplets T1T_{1} and T2T_{2} is not bad. Hence if 𝒯\mathcal{T} is bad, the lemma holds. It remains to argue that if 𝒯\mathcal{T} is not bad, then every set 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} is bad. Assume that this is not the case, and there exist triplets T1T_{1} and T2T_{2} such that 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} is not bad.

By Lemmas 4.1 and 4.2, exactly one of T1T_{1} and T2T_{2} is face-disjoint with TT. Assume, w.l.o.g. that T1T_{1} is face-disjoint with TT, i.e. Case C.1 applies for T1T_{1}. Then either T2T_{2} shares only face f1f_{1} with TT or it shares only faces ff and f2f_{2} with TT, i.e either Case C.2 or Case C.3 applies for T2T_{2}, respectively. Thus, T1T2T_{1}\neq T_{2} also holds.

In the first case, by Lemma 4.3 it follows that vertex uu of TT has degree 99 and that there exists a vertex xS=N(u)V(T)x\in S=N(u)\setminus V(T) such that |N(x)S|1|N(x)\cap S|\leq 1. On the other hand, applying Lemma 4.4 for vertex uV(T)u\in V(T), we conclude that 𝒯\mathcal{T} is bad, contradicting our assumption that both 𝒯\mathcal{T} and 𝒯\mathcal{T}^{\prime} are not bad. Hence, we can assume that T2T_{2} shares only faces ff and f2f_{2} with TT, i.e. Case C.3 applies for T2T_{2}. Consider edge ee of TT. Since 𝒯\mathcal{T}^{\prime} is not bad, 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} can be extended to at least one p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG. Let \mathcal{R}^{\prime} be such a rotation scheme and let triplet TeT_{e} be the triplet that edge ee is assigned to w.r.t. \mathcal{R}^{\prime}. We observe first that TeT2T_{e}\neq T_{2}. Indeed, since T2T_{2} shares vertices uu, w1w_{1}, w2w_{2} and v2v_{2} with TT, Te=T2T_{e}=T_{2} would imply that T2=TT_{2}=T; a contradiction. We distinguish two cases, depending on whether TeT1T_{e}\neq T_{1} or Te=T1T_{e}=T_{1}.

Case 1: TeT1T_{e}\neq T_{1}. Since TeT1T_{e}\neq T_{1} and TeT2T_{e}\neq T_{2} it follows that ee belongs neither to T1T_{1} nor to T2T_{2}. Furthermore, the existence of \mathcal{R}^{\prime} implies that 𝒯′′=𝒯{Te}\mathcal{T}^{\prime\prime}=\mathcal{T}^{\prime}\cup\{T_{e}\} is not bad. Applying Lemma 4.6 for 𝒯\mathcal{T}, we conclude that for every vertex vV(T)v\in V(T) with degree nine, there exists a vertex xS=N(v)V(T)x\in S=N(v)\setminus V(T) such that |N(x)S|4|N(x)\cap S|\geq 4. In particular, this property holds for vertices w1w_{1} and uu if any of them has degree nine. Since 𝒯′′\mathcal{T}^{\prime\prime} is not bad, we conclude from Lemma 4.5 that T1T_{1} does not contain w1w_{1} while TeT_{e} does not contain uu. Hence, by Lemma 4.7 for 𝒯′′\mathcal{T}^{\prime\prime}, there exists a triplet Tf1T_{f_{1}} that contains face f1f_{1} of TT, shares only vertices of f1f_{1} with TT, and is face-disjoint from all triplets of 𝒯′′\mathcal{T}^{\prime\prime}. Then, Lemma 4.8 implies that 𝒯\mathcal{T} is bad; a contradiction.

Case 2: Te=T1T_{e}=T_{1}. Both triplets TT and T1T_{1} contain edges e1e_{1} and ee, hence they contain their endpoints, i.e. TT and T1T_{1} share at least three vertices. Since 𝒯\mathcal{T}^{\prime} is not bad, and since triplets T1T_{1} and T2T_{2} both contain vertices v2v_{2} and w2w_{2}, triplets T1T_{1} and T2T_{2} have exactly two common vertices by Property 2.3. Now the conditions of Lemma 4.9 are fulfilled and we conclude that 𝒯\mathcal{T} is bad; a contradiction. This completes the proof of the lemma. ∎

See 4.11

Proof.

Clearly, if we can apply at least one of these lemmas for 𝒯\mathcal{T} then 𝒯\mathcal{T} is bad. So, assume that 𝒯\mathcal{T} is bad, and that for each of these lemmas not all conditions hold for 𝒯\mathcal{T}. Since GG is optimal 22-planar, by Lemma 4.10, there exists a set 𝒯={T1,T2}\mathcal{T}^{\prime}=\{T_{1},T_{2}\} that is not bad.

By Lemmas 4.1 and 4.2, exactly one of T1T_{1} and T2T_{2} is face-disjoint with TT. Assume, w.l.o.g. that T1T_{1} is face-disjoint with TT. Then either T2T_{2} shares only face f1f_{1} with TT or it shares only faces ff and f2f_{2} with TT.

In the first case, since 𝒯\mathcal{T} does not fulfill all conditions of Lemma 4.4, we conclude that, in particular for vertex uV(T)u\in V(T), d(u)>9d(u)>9 holds or every vertex xS=N(u)V(T)x\in S=N(u)\setminus V(T) has |N(x)S|2|N(x)\cap S|\geq 2. Then, the conditions of Lemma 4.3 are satisfied, and 𝒯\mathcal{T}^{\prime} is bad; a contradiction.

Hence, we can assume that T2T_{2} shares only faces ff and f2f_{2} with TT. Consider edge ee of TT. Since 𝒯\mathcal{T}^{\prime} is not bad, 𝒯\mathcal{R}_{\mathcal{T}^{\prime}} can be extended to at least one p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG. Let \mathcal{R}^{\prime} be such an embedding and let triplet TeT_{e} be the triplet that ee is assigned to w.r.t. \mathcal{R}^{\prime}.

Since 𝒯\mathcal{T} does not satisfy all conditions of Lemma 4.9, either TeT1T_{e}\neq T_{1} or T2T_{2} does not have exactly two common vertices with T1T_{1}. If T2T_{2} has at least three common vertices with T1T_{1}, then by Property 2.3, set 𝒯\mathcal{T}^{\prime} is bad. On the other hand, if T2T_{2} has exactly one common vertex with T1T_{1}, then TeT1T_{e}\neq T_{1}, as otherwise, T2T_{2} and T1T_{1} would share vertices w2w_{2} and v2v_{2}. Hence, it suffices to consider the case where TeT1T_{e}\neq T_{1}. Note that TeT2T_{e}\neq T_{2} also holds. Indeed, since T2T_{2} shares vertices uu, w1w_{1}, w2w_{2} and v2v_{2} with TT, Te=T2T_{e}=T_{2} would imply that T2=TT_{2}=T also holds; a contradiction.

Then 𝒯′′={T1,T2,Te}\mathcal{T}^{\prime\prime}=\{T_{1},T_{2},T_{e}\} is not bad, which implies that TeT_{e} is face-disjoint from T1T_{1} and T2T_{2}. Since 𝒯\mathcal{T} does not satisfy all conditions of Lemma 4.6, in particular for vertex w1V(T)w_{1}\in V(T), d(w1)>9d(w_{1})>9 holds or there exists a vertex xS=N(w1)V(T)x\in S=N(w_{1})\setminus V(T) such that |N(x)S|4|N(x)\cap S|\geq 4. If T1T_{1} contains vertex w1w_{1}, then Lemma 4.5 implies that 𝒯′′\mathcal{T}^{\prime\prime} is bad; a contradiction. Hence, T1T_{1} does not contain vertex w1w_{1}. Using the same argument for vertex uV(T)u\in V(T), we conclude that TeT_{e} does not contain vertex uu. Already all conditions of Lemma 4.8 are satisfied except for the constraints related to Tf1T_{f_{1}}. Since by assumption not all conditions of Lemma 4.8 are met, we conclude that every triplet Tf1T_{f_{1}} that contains f1f_{1} (i) shares at least four vertices with TT, or, (ii) is not face-disjoint with at least one triplet of 𝒯′′\mathcal{T}^{\prime\prime}. Then all conditions of Lemma 4.5 are satisfied and 𝒯′′\mathcal{T}^{\prime\prime} is bad. The lemma follows. ∎

Appendix D Ommited Proofs and Details from Section 5

Theorem D.1.

Algorithm 1 is correct.

Proof.

We assume that the input graph GG has exactly 5|V|105|V|-10 edges and is 99-degenerate. We proceed with the four steps of our algorithm as described in Section 5.1 and formalized in Algorithm 1. In the first step, we use Corollary 2.3 to classify each edge of GG as clearly crossing or potentially planar. Recall that an edge is potentially planar if and only if its two endpoints have at least six common neighbors.

From Lemma 3.1 we get that if two potentially planar edges of GG cross each other in some optimal 22-planar rotation scheme, then they belong to an instance of CROSS-BAT. Hence, in the second step, we identify all such instances and by Lemma 3.2 we can fix the partial rotation scheme of the subgraph induced by the vertices of each CROSS-BAT instance. By doing this, we change the classification of two potentially planar edges of each instance to clearly crossing edges. Note that by Lemma 3.4 for every CROSS-BAT instance, there exist exactly two pairs of potentially planar edges that cross, hence the reclassification process takes place only when potentially planar edges cross in some optimal 22-planar rotation scheme.

As a result, the subgraph GpG_{p} defined by the set of potentially planar edges is planar. Furthermore, if GG is optimal 2-planar, then GpG_{p} must also be 3-connected and spanning GG, as it contains as subgraph the 3-connected pentangulation of every p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG. Hence if GpG_{p} is not planar or if GpG_{p} is not 3-connected we can conclude that GG is not optimal-2-planar. If GpG_{p} is planar and 3-connected, we compute its unique planar rotation scheme p\mathcal{R}_{p}. Note that by Lemma 3.3, if GG is optimal 2-planar, then there exists a p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme \mathcal{R} of GG. As mentioned at the beginning of Section 4, all faces of p\mathcal{R}_{p} must have length 33, 44 or 55. In particular non-triangular faces belong to facial 55-cliques of \mathcal{R} and hence they induce complete subgraphs in GG. If these properties are violated, we conclude that \mathcal{R} does not exist and GG is not optimal 2-planar. In order to proceed with the next step, we first need to augment GpG_{p} to maximal planar. To achieve this, we arbitrarily triangulate every non-triangular face of p\mathcal{R}_{p} (this process can not create parallel edges or self-loops in GpG_{p} as it is 3-connected). Also, since non-triangular faces induce complete subgraphs in GG, the added edges belong to GG and are classified as clearly crossing. We change their classification to potentially planar so that they belong to GpG_{p}.

The next step is to calculate the triplets of p\mathcal{R}_{p}, which can be accomplished by computing the paths of length two of the dual GpG^{*}_{p} of GpG_{p} and check whether the vertices of the three faces induce a 55-clique in GG.

Now, for each triplet TT we can use Lemma 4.11 to decide whether 𝒯={T}\mathcal{T}=\{T\} is bad or not. If 𝒯\mathcal{T} is not bad, we mark triplet TT as a facial 55-clique. By definition, if TT is a facial 55-clique of any p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme of GG, then the three clearly crossing edges of TT are assigned to TT and to no other triplet by Lemma 4.10. Hence if any clearly crossing edge of GG is not assigned to exactly one triplet, then we can conclude that GG is not optimal-2-planar. Similarly, a face of p\mathcal{R}_{p} is assigned to exactly one triplet. If this property does not hold, then GG is not optimal 2-planar.

Having reached this point, GG is an optimal 2-planar graph and it remains to compute its p\mathcal{R}_{p}-compliant optimal 22-planar rotation scheme by combining the embedding p\mathcal{R}_{p} of GpG_{p} and the triplets marked as facial 55-cliques. To do this, for each triplet T=f1,f,f2T=\langle f_{1},f,f_{2}\rangle as depicted in Fig. 3, for every vertex vv of TT we insert the clearly crossing edges of TT that are incident to vv between the corresponding edges of GpG_{p} in the cyclic order of edges around vv. Also, we define the order of crossings along each edge of TT as indicated by Fig. 3.

The correctness of Algorithm 1 follows from the above analysis. ∎

Details 1.

In our algorithm, we assume that the given graph has 5n105n-10 edges and is 99-degenerate. Checking whether a graph has these properties can be done in linear time by using standard algorithms [25].

For checking whether GpG_{p} is planar and 3-connected, and for computing the planar rotation scheme p\mathcal{R}_{p} of GpG_{p} (the cyclic order of the edges around each vertex, and the faces corresponding to the embedding), as well as the dual graph GpG^{*}_{p} we use standard algorithms that run in linear time and data structures that consume linear space [22, 23].

In all steps, it is required to check whether two vertices are adjacent or to find the edge between two given vertices. Since GG is 99-degenerate, for each vertex viv_{i} in the degeneracy sequence v1,,vnv_{1},\ldots,v_{n}, we separately store its edges in subgraphs Gi=G[v1,,vi]G_{i}=G[v_{1},\ldots,v_{i}] and Gi=G[vi,,vn]G_{i}^{\prime}=G[v_{i},\ldots,v_{n}]. For j>ij>i, edge (vi,vj)(v_{i},v_{j}) belongs to GjG_{j}. Hence, it suffices to only check the stored edges of vjv_{j} in GjG_{j}, which are at most nine. Hence, the cost of these operations is 𝒪(1)\mathcal{O}(1).

See 5.1

Proof.

First, we compute a 9-degenerate sequence L=v1,,vnL=v_{1},\ldots,v_{n} of the vertices of GG in linear time [25]. In order to classify the edges, for each edge (u,v)(u,v) we count the number of common neighbors of uu and vv as follows. For i=n,,1i=n,\ldots,1 we process vertex viv_{i}. Recall that since GG is 9-degenerate, vertex viv_{i} has at most nine neighbors in GiG_{i}. For each of the at most (92)=36{9\choose 2}=36 pairs of neighbors u,wu,w of viv_{i} in GiG_{i}, we check whether they are connected by an edge in 𝒪(1)\mathcal{O}(1) time. If so, we update the number of common neighbors for the edges (u,vi)(u,v_{i}), (vi,w)(v_{i},w) and (u,w)(u,w). After this process, we check whether the number of common neighbors computed for each edge (u,v)(u,v) is at least 66 or not. If this is the case we mark (u,v)(u,v) as potentially planar, otherwise as clearly crossing. Since for every vertex, we process at most 3636 pairs of neighbors in 𝒪(1)\mathcal{O}(1) time each, the classification of all edges of GG takes 𝒪(n)\mathcal{O}(n) time in total. ∎

See 5.2

Proof.

We iterate over all edges of GG trying to identify first the base edge of each CROSS-BAT instance. Recall that each base edge is potentially planar and both its endpoints have degree 99. So, in our iteration we explicitly focus on such edges. Let (u,u)(u,u^{\prime}) be such an edge. If (u,u)(u,u^{\prime}) is the base edge of a CROSS-BAT instance, then |N(u,u)|=8|N(u,u^{\prime})|=8. Since uu and uu^{\prime} are degree-9 vertices, we can check in 𝒪(1)\mathcal{O}(1) time if N(u,u)N(u,u^{\prime}) has eight elements. If this is not true, then (u,u)(u,u^{\prime}) is not the base edge of a CROSS-BAT instance. Otherwise, let HH be the subgraph induced by N(u,u){u,u}N(u,u^{\prime})\cup\{u,u^{\prime}\} and let N(u,u)={u1,,u8}N(u,u^{\prime})=\{u_{1},\ldots,u_{8}\}. In the following, we try to find an isomorphism between u1,,u8u_{1},\ldots,u_{8} and the vertices of CROSS-BAT as named in Fig. 2(a).

Clearly, if HH is an instance of CROSS-BAT, then {u1,,u8}\{u_{1},\ldots,u_{8}\} must have the same degree-sequence in decreasing order as CROSS-BAT. If this is not true, then HH is not an instance of CROSS-BAT.

Since vertices vv and ww of CROSS-BAT are the only vertices with 44 or 55 neighbors in CROSS-BAT (depending on whether they are adjacent to vv^{\prime} and ww^{\prime}, respectively), we can find the corresponding two vertices of HH; assume w.l.o.g. that {u1,u2}={v,w}\{u_{1},u_{2}\}=\{v,w\}. From the set of the remaining vertices S={x,x,y,y,v,w}S=\{x,x^{\prime},y,y^{\prime},v^{\prime},w^{\prime}\} of CROSS-BAT, only vv^{\prime} and ww^{\prime} have five neighbors in SS. W.l.o.g., let u3=vu_{3}=v^{\prime} and u4=wu_{4}=w^{\prime}.

Now xx and xx^{\prime} are connected to vv^{\prime} with potentially planar edges, while yy and yy^{\prime} are connected to vv^{\prime} with clearly crossing edges. This allows to distinguish xx and xx^{\prime} from yy and yy^{\prime}. So, let w.l.o.g. {u5,u6}={x,x}\{u_{5},u_{6}\}=\{x,x^{\prime}\} and {u7,u8}={y,y}\{u_{7},u_{8}\}=\{y,y^{\prime}\}. Also, xx and xx^{\prime} are connected to vv and not connected to ww. Hence, we can identify vv and ww; w.l.o.g. let u1=vu_{1}=v and u2=wu_{2}=w.

It remains to identify xx, xx^{\prime}, yy and yy^{\prime}. Since the pairs {x,y}\{x,y\} and {x,y}\{x^{\prime},y^{\prime}\} are symmetric in CROSS-BAT, we arbitrarily choose x=u5x=u_{5}. Since xx is connected to yy but not to yy^{\prime}, this allows to identify yy; w.l.o.g. let y=u7y=u_{7}. Then we can conclude that x=u6x^{\prime}=u_{6} and y=u8y^{\prime}=u_{8}. After the mapping of the vertices of HH to the ones of CROSS-BAT, it remains to check that HH is isomorphic to CROSS-BAT, i.e., that it contains exactly the edges of CROSS-BAT, and that the edges of HH have the same classification (clearly crossing or potentially planar) as the corresponding edges of the CROSS-BAT.

If the above process fails at any step, then we can conclude that HH is not an instance of CROSS-BAT. After identifying the two adjacent vertices uu and uu^{\prime} with degree 99, all steps related to determining whether (u,u)(u,u^{\prime}) is a base edge of a CROSS-BAT instance can be performed in 𝒪(1)\mathcal{O}(1) time. ∎

Details 2.

If GG is optimal 22-planar, GpG_{p} is 3-connected, planar and spans GG. These properties can be checked in linear time with algorithms that also compute the planar rotation scheme p\mathcal{R}_{p} of GpG_{p}. Note that we ensured that GpG_{p} is 33-connected, hence, p\mathcal{R}_{p} is uniquely defined (see line 6 of Algorithm 1). After computing p\mathcal{R}_{p}, we can iterate over all 𝒪(n)\mathcal{O}(n) faces of p\mathcal{R}_{p} and check for each face in 𝒪(1)\mathcal{O}(1) time, if its length is at most five and, if so, check in 𝒪(1)\mathcal{O}(1) time, if it induces a complete subgraph in GG (see lines 7–8). If any face violates these constraints, we reject the instance in line 9.

Next, we triangulate all faces of length four or five in line 10 of Algorithm 1. In fact, adding edges in GpG_{p} corresponds to reclassifying some clearly crossing edges of GG as potentially planar. This process takes time 𝒪(n)\mathcal{O}(n). Now that GpG_{p} is maximal planar, we recompute its planar rotation scheme p\mathcal{R}_{p} and its dual GpG^{*}_{p} in linear time.

Details 3.

Recall that a triplet T=f1,f,f2T=\langle f_{1},f,f_{2}\rangle consists of three faces f1f_{1},ff and f2f_{2} of p\mathcal{R}_{p} which form the path (f1,f,f2)(f_{1},f,f_{2}) of length 22 in the dual GpG_{p}^{*}. All vertices of GpG_{p}^{*} have degree 3, hence identifying all paths of length two takes linear time in the order of GpG_{p}^{*}, i.e. 𝒪(|V(Gp)|)=𝒪(n)\mathcal{O}(|V(G_{p}^{*})|)=\mathcal{O}(n) time. For each path we check in 𝒪(1)\mathcal{O}(1) time, if the vertices of the three faces induce a 55-clique in GG. If this is the case, the path is a triplet of p\mathcal{R}_{p}. During this process, for each face and each clearly crossing edge of GG, we store the set of triplets that contain it and count their cardinalities. Note that this information can be stored in linear space, since each triplet contains three faces and three clearly crossing edges of GG.

See 5.3

Proof.

By Lemma 4.11 it suffices to check whether the conditions of one of Lemmas 4.4, 4.6, 4.8 and 4.9 are met. For the first two lemmas, namely Lemmas 4.4 and 4.6, for every vertex vV(T)v\in V(T) with d(v)=9d(v)=9, and every neighbor xV(T)x\notin V(T) of vv, we compute the set S=N(v,x)V(T)S=N(v,x)\setminus V(T) and its cardinality. Since |N(v)|=9|N(v)|=9 and SN(v)S\subseteq N(v), this computation requires to check a constant number of times for the existence of some edge. Hence, it takes 𝒪(1)\mathcal{O}(1) time to check whether the conditions of Lemmas 4.4 and 4.6 hold for 𝒯\mathcal{T}.

The next two lemmas, namely Lemmas 4.8 and 4.9, assume that T1T_{1} is face-disjoint from TT and T2T_{2} contains faces ff and f2f_{2} of TT. The case where T2T_{2} is face-disjoint from TT and T1T_{1} contains faces ff and f1f_{1} of TT can be handled analogously. In a brute-force approach for Lemma 4.8, one would have to consider all triplets Tf1T_{f_{1}} that contain face f1f_{1} and all possible sets 𝒯={T1,T2,Te}\mathcal{T}^{\prime}=\{T_{1},T_{2},T_{e}\}, where edges e1e_{1}, e2e_{2} and ee belong to T1T_{1}, T2T_{2} and TeT_{e} respectively. Since Tf1T_{f_{1}} contains face f1f_{1} of TT, there exist at most five ways to extend f1f_{1} to a path of three faces in the dual GpG^{*}_{p} of GpG_{p}. Checking whether these paths of faces form triplets of p\mathcal{R}_{p} and whether they share only the vertices of f1f_{1} with TT can be performed in 𝒪(1)\mathcal{O}(1) time. Similarly, since T2T_{2} shares faces ff and f2f_{2} with TT, there are at most three choices for its third face (face f1f_{1} is excluded). For each choice we can identify in 𝒪(1)\mathcal{O}(1) time if this is a triplet of p\mathcal{R}_{p}. Clearly if we cannot identify Tf1T_{f_{1}} or T2T_{2} then the conditions of Lemma 4.8 are not met.

Now, in this brute-force approach, for triplets T1T_{1} and TeT_{e}, there might exist a linear number of triplets that contain edge e1e_{1} (or ee), thus we would have to consider a quadratic number of sets 𝒯\mathcal{T}^{\prime}. Ideally we would like to have a bounded number of triplets that contain edge e1e_{1} or ee. In fact we can prove that if this is not the case, then we can always find triplets T1T_{1} and TeT_{e} that satisfy the conditions of Lemma 4.8, i.e. it applies for 𝒯\mathcal{T}. In order to achieve this, we need the following:

Proposition 5.3.1.

Let TT be a triplet and consider a set 𝒮\mathcal{S} of triplets with at least 20 elements. Then, at least one triplet of the set is face-disjoint with TT.

Proof.
Refer to caption
Figure 15: Illustration for the proof of Proposition 5.3.1.

We want to count the triplets that contain at least one face of TT. Each triplet corresponds to a unique path of length 22 in the dual GpG_{p}^{*}; see Fig. 15. There exist five triplets that contain face f1f_{1} but not face ff, and four triplets that contain both f1f_{1} and ff. Similarly there exist nine triplets that contain face f2f_{2} (four of them also include face ff), and two triplets that contain ff but not f1f_{1} and f2f_{2}. In total we count 20 triplets, having counted TT twice: once for face f1f_{1} and once for face f2f_{2}. Hence there exist at most 19 different triplets that contain at least one face of TT. Then if the set 𝒮\mathcal{S} contains at least 20 elements, there is at least one triplet that is face-disjoint with TT. ∎

Along the same lines we can argue the following:

Proposition 5.3.2.

Let TT be a triplet, with vertices uu and vv that are consecutive along its boundary in p\mathcal{R}_{p}. Let 𝒮\mathcal{S} be a set of at least six triplets that are face-disjoint from TT and contain vertex uu. Then, at least one triplet of the set does not contain vertex vv.

Proof.

Let ff and ff^{\prime} be the two faces of p\mathcal{R}_{p} that share edge (u,v)(u,v). Assume w.l.o.g. that ff belongs to TT. Note that any triplet that is face-disjoint with TT and contains both vertices uu and vv, must contain face ff^{\prime} (and not ff). There exist at most five such triplets. Hence, if 𝒮\mathcal{S} has at least six elements, then it has at least one triplet that does not contain ff^{\prime}, i.e. does not contain vertex vv as claimed. ∎

Consider the set 𝒮1\mathcal{S}_{1} of triplets that contain edge e1e_{1} and the set 𝒮e\mathcal{S}_{e} of triplets that contain edge ee. Assume w.l.o.g. that |𝒮1||𝒮e||\mathcal{S}_{1}|\leq|\mathcal{S}_{e}|. If |𝒮e|<82|\mathcal{S}_{e}|<82 then we can check for every T1𝒮1T_{1}\in\mathcal{S}_{1} and every Te𝒮eT_{e}\in\mathcal{S}_{e} whether the conditions of Lemma 4.8 hold. Checking whether T1T_{1} and TeT_{e} are face-disjoint and whether they contain vertex w1w_{1} and uu, respectively, takes 𝒪(1)\mathcal{O}(1) time.

So assume that |𝒮e|82|\mathcal{S}_{e}|\geq 82. If we can find a triplet T1T_{1} that is face-disjoint from all of TT, T2T_{2} and Tf1T_{f_{1}}, and does not contain vertex w1w_{1}, then we claim that Lemma 4.8 applies and 𝒯\mathcal{T} is bad. Indeed, assuming a valid T1T_{1}, since |𝒮e|82|\mathcal{S}_{e}|\geq 82 then Proposition 5.3.1 assures that at least 6363 triplets of 𝒮e\mathcal{S}_{e} are face-disjoint from TT. Among these triplets, at least 4444 of them are also face-disjoint from T2T_{2}, at least 2525 of these 4444 are also face-disjoint from T1T_{1}, and at least six of them are additionally face-disjoint from Tf1T_{f_{1}}. Then by Proposition 5.3.2 we can find a triplet TeT_{e} that, in addition to the previous properties, does not contain vertex uu. Then all conditions of Lemma 4.8 are met for TeT_{e}, which implies that 𝒯\mathcal{T} is bad as claimed. Hence we only need to identify triplet T1T_{1}. If |𝒮1|<63|\mathcal{S}_{1}|<63, then we check all elements of 𝒮1\mathcal{S}_{1}, whether they are face-disjoint from all of TT, T2T_{2} and Tf1T_{f_{1}}, and do not contain vertex w1w_{1} in 𝒪(1)\mathcal{O}(1) time. If we find triplet T1T_{1}, then 𝒯\mathcal{T} is bad, otherwise the conditions of Lemma 4.8 are not met. If on the other hand, |𝒮1|63|\mathcal{S}_{1}|\geq 63, Propositions 5.3.1 and 5.3.2 assure that there exists a triplet T1T_{1} with the aforementioned properties, hence the conditions of Lemma 4.8 are satisfied.

Finally, for Lemma 4.9, first we need to find triplets T1=TeT_{1}=T_{e} and T2T_{2}, if they exist. Note that if T1T_{1} exists, then it contains the face opposite of f2f_{2} along edge (v2,w2)(v_{2},w_{2}). This face, considered as a vertex in the dual GpG^{*}_{p} of GpG_{p}, can be extended to at most five different paths of length 22 of GpG^{*}_{p}. We can check for each path in 𝒪(1)\mathcal{O}(1) time if its vertices create a 55-clique, i.e. if this is a triplet of p\mathcal{R}_{p}. On the other hand, since T2T_{2} contains faces ff and f2f_{2} of TT, there are at most three choices for its third face (face f1f_{1} is excluded). Again, we check if these options define triplets of p\mathcal{R}_{p} in 𝒪(1)\mathcal{O}(1) time. Now, for every triplet T1T_{1} and T2T_{2} we check in 𝒪(1)\mathcal{O}(1) time if they are face-disjoint and whether they have exactly two common vertices. Since there exist at most five triplets T1T_{1} and at most three triplets T2T_{2}, checking whether Lemma 4.9 applies for 𝒯\mathcal{T} can be done in 𝒪(1)\mathcal{O}(1) time. ∎

Details 4.

Here we augment the planar rotation scheme p\mathcal{R}_{p} of GpG_{p} to an optimal 22-planar rotation scheme of GG by inserting the clearly crossing edges of GG appropriately. Let triplet T=f1,f,f2𝒯T=\langle f_{1},f,f_{2}\rangle\in\mathcal{T}^{*} as depicted in Fig. 3. For every vertex vv of TT we insert the clearly crossing edges of TT that are incident to vv between the corresponding edges of p\mathcal{R}_{p} in the cyclic order of edges around vv. This can be done in 𝒪(1)\mathcal{O}(1) time for the clearly crossing edges incident to a specific vertex vv in a triplet as the triplet consists of the faces that provide access to the correct position in the cyclic neighbor list of vv. Since the triplet also defines the crossings contained in the facial 55-clique, we can define and store the order of crossings along each edge of TT in 𝒪(1)\mathcal{O}(1) time.