Twisted Ways to Find Plane Structures in Simple Drawings of Complete Graphs

Simple drawings are drawings of graphs in which the edges are Jordan arcs and each pair of edges share at most one point (a proper crossing or a common endpoint). A simple drawing is c-monotone if there is a point O such that each ray emanating from O crosses each edge of the drawing at most once. We introduce a special kind of c-monotone drawings that we call generalized twisted drawings. A c-monotone drawing is generalized twisted if there is a ray emanating from O that crosses all the edges of the drawing. Via this class of drawings, we show that every simple drawing of the complete graph with n vertices contains \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega (n^{\frac{1}{2}})$$\end{document}Ω(n12) pairwise disjoint edges and a plane cycle (and hence path) of length \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega (\frac{\log n }{\log \log n})$$\end{document}Ω(lognloglogn). Both results improve over best previously published lower bounds. On the way we show several structural results and properties of generalized twisted and c-monotone drawings, some of which we believe to be of independent interest. For example, we show that a drawing D is c-monotone if there exists a point O such that no edge of D is crossed more than once by any ray that emanates from O and passes through a vertex of D.


Introduction
Simple drawings are drawings of graphs in the plane such that vertices are distinct points in the plane, edges are Jordan arcs connecting their endpoints, and edges intersect at most once either in a proper crossing or in a shared endpoint.The edges and vertices of a drawing partition the plane (or, more exactly, the plane minus the drawing) into regions, which are called the cells of the drawing.If a simple drawing is plane (that is, crossing-free), then its cells are classically called faces.
In the past decades, there has been significant interest in simple drawings.Questions about plane subdrawings of simple drawings of the complete graph on n vertices, K n , have attracted particularly close attention.
Rafla [20] conjectured that every simple drawing of K n contains a plane Hamiltonian cycle.The conjecture has been shown to hold for n ≤ 9 [1], as well as for several special classes of simple drawings, like straight-line, monotone, and cylindrical drawings, but remains open in general.If Rafla's conjecture is true, then this would immediately imply that every simple drawing of the complete graph contains a plane perfect matching.However, to-date even the existence of such a matchging is still unknown.
Pach, Solymosi, and Tóth [17] showed that every simple drawing of K n contains a subdrawing of K c log , for some constant c, that is either convex or twisted 1 .They further showed that every simple drawing of K n contains a plane subdrawing isomorphic to any fixed tree with up to c log 1 6 n vertices, for some constant c.This implies that every simple drawing of K n contains a plane path of length Ω((log n) 1 6 ), which has been the best lower bound known prior to this paper.
Concerning general plane substructures, it follows from a result of Ruiz-Vargas [22] that every simple drawing of K n contains a plane subdrawing with at least 2n − 3 edges.Further, García, Pilz, and Tejel [14] showed that every maximal plane subdrawing of a simple drawing of K n is biconnected.Note that, in contrast to straight-line drawings, simple drawings of K n in general do not contain triangulations, that is, plane subdrawings where all faces (except at most one) are 3-cycles.
In this paper, we introduce a new family of simple drawings, which we call generalized twisted drawings.The name stems from the fact that one can show that any twisted drawing is weakly isomorphic to a generalized twisted drawing (but not every generalized twisted drawing is weakly isomorphic to a twisted drawing).It follows, that for any n there exists a generalized twisted drawing.Two drawings D and D ′ are weakly isomorphic if there is a bijection between the vertices and edges of D and D ′ such that a pair of edges in D crosses exactly when the corresponding pair of edges in D ′ crosses.

▶ Definition 1.
A simple drawing D is c-monotone (short for circularly monotone) if there is a point O such that any ray emanating from O intersects any edge of D at most once.A simple drawing D of K n is generalized twisted if there is a point O such that D is c-monotone with respect to O and there exists a ray r emanating from O that intersects every edge of D.
We label the vertices of c-monotone drawings v 1 , . . ., v n in counterclockwise order around O. In generalized twisted drawings, they are labeled such that the ray r emerges from O between the ray to v 1 and the one to v n .Figure 1 shows an example of a generalized twisted drawing of K 5 .
Generalized twisted drawings turn out to have quite surprising structural properties.We show some crossing properties of generalized twisted drawings in Section 2 and with that also prove that they always contain plane Hamiltonian paths (Theorem 3).This result is an essential ingredient for showing that any simple drawing of K n contains Ω( √ n) pairwise disjoint edges (Theorem 9 in Section 3), as well as a plane path of length Ω( log n log log n ) (Theorem 10 in Section 4).In Section 5, we present different characterizations of generalized twisted drawings that are of independent interest.We conclude with an outlook on further work and open problems in Section 6.

Twisted Preliminaries
In this section, we show some properties of generalized twisted drawings, which will be used in the following sections.
▶ Lemma 2. Let D be a generalized twisted drawing of K 4 , with vertices {v 1 , v 2 , v 3 , v 4 } labeled counterclockwise around O. Then the edges v 1 v 3 and v 2 v 4 do not cross.
The full proof of Lemma 2 can be found in Appendix A.
Proof Sketch.Assume, for a contradiction, that the edge v 1 v 3 crosses the edge v 2 v 4 .There are (up to strong isomorphism) two possibilities to draw the crossing edges v 1 v 3 and v 2 v 4 , depending on whether v 1 v 3 crosses the (straight-line) segment from O to v 4 or not; cf. Figure 2. In both cases, there is only one way to draw v 1 v 2 such that the drawing stays generalized twisted, yielding two regions bounded by all drawn edges (v 1 v 3 , v 2 v 4 , v 1 v 2 ).The vertices v 3 and v 4 lie in the same region.It is well-known that every simple drawing of K 4 has at most one crossing.Thus, the edge v 3 v 4 cannot leave this region.However, it is impossible Figure 2 The two possibilities to draw v1v3 and v2v4 crossing and generalized twisted.
to draw v 3 v 4 without leaving the region such that it is c-monotone and crosses the ray r (see the dotted arrows in Figure 2 for necessary emanating directions of v 3 v 4 ).◀ Using the crossing property of Lemma 2, it follows directly that generalized twisted drawings always contain plane Hamiltonian paths.
▶ Theorem 3. Every generalized twisted drawing of K n contains a plane Hamiltonian path.
Proof of Theorem 3. Let D be a generalized twisted drawing of K n .Consider the Hamilto- , v n if n is even.See for example the Hamiltonian path v 1 , v 4 , v 2 , v 5 , v 3 in Figure 1.Take any pair of edges (v i , v j ) and (v k , v l ) of the path, where we can assume without loss of generality that i < j and k < l.If the two edges share an endpoint, they are adjacent and do not cross.Otherwise, if they do not share an endpoint, either i < k < j < l or k < i < l < j by definition of the path.In any of the two cases, (v i , v j ) and (v k , v l ) cannot cross by Lemma 2. Therefore, no pair of edges cross, and the Hamiltonian path is plane.◀ Analogous to the proof of Theorem 3, one can argue that in every generalized twisted drawing of K n with n odd, the Hamiltonian cycle , v 1 is plane.We strongly conjecture that every generalized twisted drawing of K n contains a plane Hamiltonian cycle, but its structure for even n is still an open problem.Theorem 3 will be used heavily in the next two sections.Further, the following statement, which has been implicitly shown in [11] and [13], will be used in all remaining sections.For completeness, we include a proof in Appendix B.
▶ Lemma 4. Let D be a simple drawing of a complete graph containing a subdrawing D ′ , which is a plane drawing of K 2,n .Let A = {a 1 , a 2 , . . ., a n } and B = {b 1 , b 2 } be the sides of the bipartition of D ′ .Let D A be the subdrawing of D induced by the vertices of A. Then D A is weakly isomorphic to a c-monotone drawing.Moreover, if all edges in D A cross the edge b 1 b 2 , then D A is weakly isomorphic to a generalized twisted drawing.

Disjoint Edges in Simple Drawings
In this section, we show that every simple drawing of K n contains at least n 48 pairwise disjoint edges, improving the previously known best bound of Ω(n 1 2 −ε ), for any ε > 0, by Ruiz-Vargas [22].In addition to the properties of generalized twisted drawings from Section 2, we use the following theorems and observations to prove this new lower bound.
The following theorem is a direct consequence of Corollary 5 in [21].
▶ Theorem 6.Let D be a simple drawing of K n with n ≥ 3. Let H be a connected plane subdrawing of D containing at least two vertices, and let v be a vertex in D \ H. Then D contains two edges incident to v that connect v with H and do not cross any edges of H. ▶ Observation 7.For any n ≥ 3, the number of edges in a planar graph with n vertices is at most 3n − 6.
A drawing is outerplane if it is plane, and all vertices lie on the unbounded face of the drawing.A graph is outerplanar if it can be drawn outerplane.Outerplanar graphs have a smaller upper bound on their number of edges than planar graphs. .We will show how to find another plane matching, whose size is at least ⌊ n 48 ⌋.The overall idea is the following: Let H be a maximal plane subdrawing of D whose vertex set is exactly the vertices matched in M and that contains M .We will find a face f in H that contains much more unmatched vertices inside than matched vertices on its boundary.Then we will show that there exists a subset of the vertices inside that face, which induces a subdrawing of D that is weakly isomorphic to a generalized twisted drawing and contains enough vertices to guarantee the desired size of the plane matching.
We start towards finding the face f .By Theorem 5, H is biconnected.Thus, H partitions the plane into faces, where the boundary of each face is a simple cycle.Note that the vertices of H are exactly the vertices that are matched in M , and the vertices inside faces are the vertices that are unmatched in M .Let U be the set of vertices of D that are not matched by any edge of M .We denote the set of vertices of U inside a face f i by U (f i ), the number of vertices in U (f i ) by u(f i ), and the number of vertices on the boundary of the face f i by |f i |.
We next show that there exists a face Assume for a contradiction that for every face f i it holds that There are exactly n − 2m unmatched vertices.As every unmatched vertex is in the interior of a face of H (that might be the unbounded face), we can count the unmatched vertices by summing over the number of vertices in each face (including the unbounded face).Thus, The number of edges in H is Since H is plane, we can use Observation 7 to bound the number of edges of H by 3n ′ − 6, where n ′ is the number of vertices in H.As the vertices of H are exactly the matched vertices, their number is n ′ = 2m.Hence, and Putting equations ( 1) to (3) together we obtain that However, this inequality cannot be fulfilled by any n ≥ 0. Thus, there exists at least one face 12 |f i |.We call that face f .(If there are several such faces, we take an arbitrary one of them and call it f .) As a next step, we will find two vertices on the boundary of f to which many vertices inside f are connected via edges that do not cross each other or H. From f and the set U (f ), we construct a plane subdrawing H ′ as follows; cf. Figure 3 (left).We add the vertices and edges on the boundary of f .Then we iteratively add all the vertices in U (f ), where for each added vertex v we also add two edges of D incident to v such that the resulting drawing stays plane.Two such edges exist by Theorem 6.Since the matching M is maximal, any edges between two unmatched vertices must cross at least one edge of M and thus must cross the boundary of f .Hence, no edge in H ′ can connect two vertices of U (f ) (as they are unmatched).Consequently, every vertex in U (f ) is connected in H ′ to exactly two vertices that both lie on the boundary of f .We consider the edges in H ′ that connect a vertex in U (f ) as a pair of edges.Every edge in such a pair is contained in exactly one pair, since it is incident to exactly one unmatched vertex.Thus, we can see every such pair of edges as one long edge incident to two vertices on the boundary of f .If several of those long edges have the same endpoints, we call them a bundle of edges; see Figure 3

(right).
From the long edges, we can define a graph G ′ as follows.The vertices of G ′ are the vertices of D that lie on the boundary of f .Two vertices u and v are connected in G ′ if there is at least one long edge in H ′ that connects them.By the definition of long edges, G ′ is outerplanar (as can be observed in Figure 3 (right)).Note that every unmatched vertex in U (f ) defines a long edge, so the number of long edges is u(f ) ≥ √ 48n 12 |f |.From Observation 8, it follows that G ′ has at most 2|f | − 3 edges.As a consequence, there is a pair of vertices on the boundary of f such that the number of long edges in its bundle is at least This implies that there are two vertices, say v and w, to which more than √ 48n 24 vertices inside f have two plane incident edges.We call the set of vertices in U (f ) that have plane edges to both vertices v and w the set U vw .This set is marked in Figure 4 (left).We denote the subdrawing of D induced by U vw by D vw ; see Figure 4 The subdrawing D ′ induced by Uvw and the edges in Dvw.Left: The set Uvw. Right: The edges adjacent to the leftmost vertex, v1, are drawn (in red).
We show that all edges between vertices in U vw cross the edge vw.Let x and y be two vertices of D vw .Let R 1 be the region bounded by the edges xv, vy, yw, and wx that lies inside the face f ; see Figure 5.We show that xy and vw lie completely outside R 1 .The edge xy has to lie either completely inside or completely outside R 1 , because it is adjacent to all edges on the boundary of R 1 .As M is maximal and the edge xy connects two unmatched vertices, it has to cross at least one matching edge.Thus, xy has to lie completely outside R 1 .(There can be no matching edges in R 1 , as R 1 is contained inside the face f .)As H is a maximal plane subdrawing, vw cannot lie inside the face f and thus has to be outside R 1 .Since both edges vw and xy lie completely outside R 1 and the vertices along the boundary of R 1 are sorted vxwy, the two edges have to cross.Thus, all edges of D vw cross the edge vw.
Since the edges from vertices in U vw to v and w are plane, it follows from Lemma 4 that D vw is weakly isomorphic to a generalized twisted drawing.Thus, D vw contains at least ⌊

Plane Paths in Simple Drawings
In the previous section, we used generalized twisted drawings to improve the lower bound on the number of disjoint edges in simple drawings of K n .In this section, we show that generalized twisted drawings are also helpful to improve the lower bound on the length of the longest path in such drawings, where the length of a path is the number of its edges, to Ω( log n log log n ).This improves the previously known best bound of Ω((log n) ), which follows from a result of Pach, Solymosi, and Tóth [17].
▶ Theorem 10.Every simple drawing D of K n contains a plane path of length Ω( log n log log n ).To prove the new lower bound, we first show that all c-monotone drawings on n vertices contain either a generalized twisted drawing on √ n vertices or a drawing weakly isomorphic to an x-monotone drawing on √ n vertices.We know that drawings weakly isomorphic to generalized twisted drawings or x-monotone drawings contain plane Hamiltonian paths (by Theorem 3 and Observation 11 below).We conclude that c-monotone drawings contain plane paths of the desired size.We then show that every simple drawing of the complete graph contains either a c-monotone drawing or a plane d-ary tree.With easy observations about the length of the longest path in d-ary trees and by putting all results together, we obtain that every simple drawing D of K n contains a plane path of length Ω( log n log log n ).

Plane Paths in C-Monotone Drawings
A simple drawing is x-monotone if any vertical line intersects any edge of the drawing at most once (see Figure 6b).This family of drawings has been studied extensively in the literature (see for example [2,5,7,12,19]).By definition, c-monotone drawings in which there exists a ray emanating from O, which crosses all edges of the drawing, are generalized twisted.
In contrast, consider a c-monotone drawing D such that there exists a ray r emanating from O that crosses no edge of D. Then it is easy to see that D is strongly isomorphic to an x-monotone drawing.(A c-monotone drawing on the sphere can be cut along the ray r and the result drawn on the plane such that all rays are vertical lines and the ray r is to the very left of the drawing.) Figure 6a shows a c-monotone drawing D of K 5 where no edge crosses the ray r, and Figure 6b shows an x-monotone drawing of K 5 strongly isomorphic to D. We will call simple drawings that are strongly isomorphic to x-monotone drawings monotone drawings.In particular, any c-monotone drawing for which there exists a ray emanating from O that crosses no edge of the drawing is monotone.It is well-known that any x-monotone drawing of K n contains a plane Hamiltonian path.For instance, assuming that the vertices are ordered by increasing x-coordinates, the set of edges v 1 v 2 , v 2 v 3 . . ., v n−1 v n form a plane Hamiltonian path.▶ Observation 11.Every monotone drawing of K n contains a plane Hamiltonian path.
We will show that c-monotone drawings contain plane paths of size √ n, by showing that any c-monotone drawing of K n contains a subdrawing of K √ n that is either generalized twisted or monotone.To do so, we will use Dilworth's Theorem on chains and anti-chains in partially ordered sets.A chain is a subset of a partially ordered set such that any two distinct elements are comparable.An anti-chain is a subset of a partially ordered set such that any two distinct elements are incomparable.
▶ Theorem 12 (Dilworth's Theorem, [9]).Let P be a partially ordered set of at least (s−1)(t−1)+1 elements.Then P contains a chain of size s or an antichain of size t.
▶ Theorem 13.Let s, t be two integers, 1 ≤ s, t ≤ n, such that (s − 1)(t − 1) + 1 ≤ n.Let D be a c-monotone drawing of K n .Then D contains either a generalized twisted drawing of K s or a monotone drawing of K t as subdrawing.In particular, if s = t = ⌈ √ n⌉, D contains a complete subgraph K s whose induced drawing is either generalized twisted or monotone.
The full proof of Theorem 13 can be found in Appendix C Proof Sketch.Without loss of generality we may assume that the vertices of D appear counterclockwise around O in the order v 1 , v 2 , . . ., v n .Let r be a ray emanating from O, keeping v 1 and v n on different sides.We define an order, ⪯, in this set of vertices as follows: v i ⪯ v j if and only if either i = j or i < j and the edge (v i , v j ) crosses r.
We show that ⪯ is a partial order.The relation is clearly reflexive and antisymmetric.Besides, if v i ⪯ v j and v j ⪯ v k , then v i ⪯ v k , because i < j and j < k imply i < k, and if v i v j and v j v k cross r, then v i v k also crosses r (see Figure 7).Hence, the relation is transitive.
In this partial order ⪯, a chain consists of a subset v i1 , . . ., v is−1 of pairwise comparable vertices, that is, a subset of vertices such that their induced subdrawing is generalized twisted (all edges cross r).An antichain, v j1 , . . ., v jt−1 , consists of a subset of pairwise incomparable vertices, that is, a subset of vertices such that their induced subdrawing is monotone (no edge crosses r).Therefore, the first part of the theorem follows from applying Theorem 12 to the set of vertices of D and the partial order ⪯.
If edges vivj and vjv k cross r in a c-monotone drawing, then viv k must also cross r.
Combining Theorems 3 and 13 with Observation 11, we obtain the following theorem.
▶ Theorem 14.Every c-monotone drawing of K n contains a plane path of length Ω( √ n).

Plane Paths in Simple Drawings
To show that any simple drawing of K n contains a plane path of length Ω( log n log log n ), we will use d-ary trees.A d-ary tree is a rooted tree in which no vertex has more than d children.It is well-known that the height of a d-ary tree on n vertices is Ω( log n log d ).
Proof of Theorem 10.Let v be a vertex of D and let S(v) be the star centered at v, that is, the set of edges of D incident to v. S(v) can be extended to a maximal plane subdrawing H that must be biconnected by Theorem 5. See Figure 8 for a depiction of S(v) and H.  Assume first that there is a vertex w in H \ v that has degree at least (log n) 2 in H. Let U vw be the set of vertices neighboured in H to both, v and w.Note that |U vw | ≥ (log n) 2 .The subdrawing H ′ of H consisting of the vertices in U vw , the vertices v, and w, and the edges from v to vertices in U vw , and from w to vertices in U vw is a plane drawing of K 2,|Uvw| .From Lemma 4, it follows that the subdrawing of D induced by U vw is weakly isomorphic to a c-monotone drawing.Therefore, by Theorem 14, the subdrawing induced by U vw contains a plane path of length Ω( |U vw |) = Ω(log n).
Assume now that the maximum degree in H \v is less than (log n) 2 .Since H is biconnected, H \ v contains a plane tree T of order n − 1 whose maximum degree is at most (log n) 2 .Thus, considering that T is rooted, the diameter of T is at least Ω( log n log log n ).Therefore, since T is plane, it contains a plane path of length at least Ω( log n log log n ) and the theorem follows.◀

Characterizing Generalized Twisted Drawings
In previous sections, we have seen how generalized twisted drawings were used to make progress on open problems of simple drawings.In addition to this, generalized twisted drawings are also interesting in their own right and have some quite surprising structural properties.Despite the fact that research on generalized twisted drawings is rather recent and still ongoing, there are already several interesting characteristics and structural results.Some of them will be presented in this section.
One characterization involves curves crossing every edge once.From the definition of generalized twisted drawing (see Figure 1), there always exists a simple curve that crosses all edges of the drawing exactly once (for instance, a curve that starts at O and follows r until it reaches a point Z on r in the unbounded cell).In Theorem 15, we show that the converse is also true.That is, every simple drawing D of K n in which we can add a simple curve that crosses every edge of D exactly once is weakly isomorphic to a generalized twisted drawing.
Another characterization is based on what we call antipodal vi-cells.For any three vertices in a simple drawing D of K n , the three edges connecting them form a simple cycle which we call a triangle.Every such triangle partitions the plane (or sphere) into two disjoint regions which are the sides of the triangle (in the plane a bounded and an unbounded one).Two cells of D are called antipodal if for each triangle of D, they lie on different sides.Further, we call a cell with a vertex on its boundary a vertex-incident-cell or, for short, a vi-cell.
By definition, every generalized twisted drawing D contains two antipodal cells, namely, the cell containing the starting point of the ray r and the unbounded cell.This follows from the fact that the ray r crosses every edge exactly once.Hence, r crosses the boundary of any triangle exactly three times, so the cells containing the "endpoints" of r must be on different sides of the triangle.It turns out that the converse (existence of two antipodal cells implies weakly isomorphic to generalized twisted) is not true.Figure 9 (left) shows a drawing of K 6 that contains two antipodal cells, but no antipodal vi-cells.From Theorem 15 bellow it will follow that such drawings cannot be weakly isomorphic to a generalized twisted drawing.However, we observed that for all generalized twisted drawings of K n with n ≤ 6, both, the cell containing the startpoint of the ray r and the unbounded cell, are vi-cells.Figure 10 shows all (up to strong isomorphism) simple drawings of K 6 that are weakly isomorphic to generalized twisted drawings.We show that this is true in general.More than that, we show in Theorem 16 that every drawing of K n that is weakly isomorphic to a generalized twisted drawing contains a pair of antipodal vi-cells.In the other direction, we show in Theorem 15 that every simple drawing containing a pair of antipodal vi-cells is weakly isomorphic to a generalized twisted drawing.
All different generalized twisted drawings of K6 (up to weak isomorphism).The rightmost drawing is twisted.
The final characterization is based on the extension of a given drawing of the complete graph to a drawing containing a spanning, plane bipartite graph that has all vertices of the original drawing on one side of the bipartition.From the definition of generalized twisted drawings, it follows that any genereralized twisted drawing D of K n can be extended to a simple drawing D ′ of K n+2 including new vertices O and Z such that D ′ contains a plane drawing of a spanning bipartite graph.One side of the bipartition consists of all vertices in D and the other side of the bipartition consists of the new vertices O and Z.Moreover, the edge OZ crosses all edges of D. One way to add the new vertices and edges incident to them is to draw (1) the vertex O at point O, (2) the vertex Z in the unbounded cell on the ray r, (3) the edge OZ straight-line (along the ray r), (4) edges from O to the vertices of D straight-line (along the inner segment of the rays crossing through the vertices), and (5) edges from Z to the vertices of D first far away in a curve and the final part straight-line (along the outer segment of the rays crossing through the vertices).The converse, that every drawing that can be extended like this is weakly isomorphic to a generalized twisted drawing, has already been shown in Lemma 4.
We show the following characterizations.
▶ Theorem 15 (Characterizations of generalized twisted drawings).Let D be a simple drawing of K n .Then, the following properties are equivalent.To prove Theorem 15, we will first show that Property 1 implies Property 2 (Theorem 16).We next show that Property 2 implies Property 3 (Theorem 17).Then, we show that Property 3 implies Property 4 (Theorem 18).By Lemma 4, Property 4 implies Property 1 .Thus, all properties are equivalent.In a full version of this work, we will extend the theorem to show that also strong isomorphism to a generalized twisted drawing is equivalent to the properties of Theorem 15.We show this by proving that any simple drawing of K n fulfilling Property 4 is strongly isomorphic to a generalized twisted drawing.However, the reasoning for strong isomorphism is quite lengthy and would exceed the space constraints of this submission.
In the remaining parts of this section, we will show sketches of the proofs of the above mentioned theorems.The full proofs can be found in the Appendix (Theorem 16 in Appendix D, Theorem 17 in Appendix E, and Theorem 18 in Appendix F).
▶ Theorem 16.Every simple drawing of K n which is weakly isomorphic to a generalized twisted drawing of K n , with n ≥ 3, contains a pair of antipodal vi-cells.In generalized twisted drawings the cell containing O and the unbounded cell form such a pair.

Figure 11
Left: If there is a vertex v l in R, it cannot be connected to vi without crossing r before x.Right: If the edge vjv k crosses the segment Ovi and the edge v j ′ v k ′ crosses the segment Ovi+1, then there is no way of connecting vi+1 and v j ′ .
Proof sketch.We first show that every generalized twisted drawing D of K n , with n ≥ 3, contains a pair of antipodal vi-cells, where O lies in a cell of that pair.Let c be the segment OZ, where Z is a point on r in the unbounded cell.By definition of generalized twisted, c crosses every edge of D once, so O and Z are in two antipodal cells C 1 and C 2 , respectively.
To prove that C 1 is a vi-cell, we use the following properties.First, if we take the first edge v i v k that crosses c (as seen from O) at point x, then we can prove that k = i + 1 and the bounded region R defined by the edge v i v i+1 and the segments Ov i and Ov i+1 is empty (see Figure 11, left).Second, using this empty region we can prove that D cannot contain simultaneously an edge v j v k crossing Ov i and another edge v j ′ v k ′ crossing Ov i+1 (see Figure 11,right).Therefore, at least one of the segments Ov i and Ov i+1 is uncrossed, and O necessarily lies in a vi-cell (with either v i or v i+1 on the boundary).Finally, arguing on the last edge crossing c and the unbounded cell, we can show that Z also lies in a vi-cell.
To show that also every drawing which is weakly isomorphic to a generalized twisted drawing contains a pair of antipodal vi-cells, we use Gioan's Theorem [6,15].By Gioan's Theorem, any two weakly isomorphic drawings of K n can be transformed into each other with a sequence of triangle-flips and at most one reflection of the drawing.A triangle-flip is

Figure 12
Building a curve such that it crosses every edge of D once and its endpoints do not lie on any edges or vertices of D.
Decreasing the number of crossings between c and the edge w2w3.
an operation which transforms a triangular cell △ that has no vertex on its boundary by moving one of its edges across the intersection of the two other edges of △.We show that if a drawing D 1 contains two antipodal vi-cells, then after performing a triangle flip on D 1 , the resulting drawing D 2 still has two antipodal vi-cells.The main argument is that triangle-flips are only applied to cells without vertices on their boundary, and thus the antipodality of the vi-cells cannot change.◀ ▶ Theorem 17.In any simple drawing D of K n that contains a pair of antipodal vi-cells, it is possible to draw a curve c that crosses every edge of D exactly once.
Proof sketch.Let (C 1 , C 2 ) be a pair of antipodal vi-cells of D. Let v 1 be a vertex on the boundary of C 1 and v 2 a vertex on the boundary of C 2 .We construct the curve as follows: First, we draw a simple curve c from C 1 to C 2 such that (1) it emanates from v 1 in C 1 and ends in C 2 very close to v 2 , (2) does not cross any edge incident to v 1 , (3) only intersects edges of D in proper crossings, and (4) has the minimum number of crossings with edges of D among all curves that fulfill (1), ( 2) and ( 3).This curve c always exists since S(v 1 ) is a plane drawing that has only a face in which both v 1 and v 2 lie (see Figure 12, left).Then, we prove that c crosses every edge w 2 w 3 in D that is not incident to v 1 exactly once.On the one hand, since c connects two antipodal cells, the endpoints of c have to be on two different sides of the triangle T formed by v 1 , w 2 and w 3 .Thus, c has to cross w 2 w 3 an odd number of times because it does not cross S(v 1 ) and must cross the boundary of T an odd number of times.On the other hand, if c crosses w 2 w 3 at least three times, then we can prove that c can be redrawn as shown in Figure 13 which contradicts (4).Therefore, c crosses every edge w 2 w 3 at most twice and, consequently, only once.
Finally, we change the end of c from v 1 to a point in C 1 in the following way (see Figure 12, right).From some point of c sufficiently close to v 1 and inside C 1 , we reroute c by going around v 1 such that only the edges incident to v 1 are crossed, and end at a point in C 1 .◀ ▶ Theorem 18.Let D be a simple drawing of K n in which it is possible to draw a simple curve c that crosses every edge of D exactly once.Then, D can be extended by two vertices O and Z (at the position of the endpoints of the curve), and edges incident to those vertices such that the obtained drawing is a simple drawing of K n+2 , no edge incident to O crosses any edge incident to Z, and all edges in D cross the edge OZ.
Proof sketch.Let c = OZ be the curve crossing every edge of D once, oriented from O to Z. Let wu be an edge of D, oriented from w to u, crossing OZ at a point x.We say that wu is a top (respectively bottom) edge if the clockwise order of w, Z, u and O around x is w, Z, u, O (respectively w, O, u, Z).See Figure 14.With these definitions, we can prove that there is a vertex w 1 in D such that all the oriented edges emanating from w 1 are top in relation to c. Thus, by removing w 1 and all its incident edges from D, there is a vertex w 2 in the new drawing such that all its incident edges are top, and so on.As a consequence, there is a natural order w 1 , w 2 , . . ., w n of the vertices of D such that for any vertex w i , the edges w i w j with j > i are top, and the edges w i w j with j < i are bottom.
Given the natural order w 1 , w 2 , . . ., w n , our construction of the extended drawing is as follows.Let D ′ 0 be the simple drawing formed by the vertices and edges of D, O and Z as new vertices, and c as the edge connecting O and Z. From D ′ 0 , we build new drawings by adding in step i the edges w i O and w i Z.These two edges are added very close to some edges in D ′ i−1 .Figure 15 illustrates how these two edges are added in each step.
In the first step, the edge Ow 1 follows the curve OZ until the crossing point between OZ and the first top edge w 1 u emanating from w 1 , and then it follows this top edge until reaching w 1 .The edge Zw 1 is built in an analogous way, taking the last top edge emanating from w 1 .See Figure 15  and w i−1 Z, say w i−1 Z (see Figure 15 bottom-left), then Ow i follows OZ until the crossing point between OZ and the last bottom edge of S(w i ), and then it follows this bottom edge until w i .The edge Zw i is built as in the first step, using the last top edge of S(w i ).In the last step, we build Ow n and Zw n as in the first step, but using the first and the last bottom edges of S(w n ) instead of the first and last top edges.See Figure 15 bottom-right.By a detailed analysis of cases, we can prove for i = 1, . . ., n that D ′ i is a simple drawing such that no edge incident to O crosses any edge incident to Z. Therefore, D ′ n is the drawing of K n+2 satisfying the required properties.◀

Conclusion and Outlook
Generalized twisted drawings have a suprisingly rich structure and many useful properties.We showed several of those properties in Section 2 and different characterizations of generalized twisted drawings in Section 5. We have proven in Section 2 that every generalized twisted drawing on an odd number of vertices contains a plane Hamiltonian cycle, and therefore one especially interesting open question is the following.

▶ Conjecture 19. Every generalized twisted drawing of K n contains a plane Hamiltonian cycle.
Using properties of generalized twisted drawings has turned out to be helpful for investigating simple drawings in general.We first improved the lower bound on the number of disjoint edges in simple drawings of K n to Ω( √ n) (Section 3).Then generalized twisted drawings played the central role to improve the lower bound on the length of plane paths contained in every simple drawing of K n to Ω( log n log log n ) (Section 4).On the other hand, from Theorem 17 it immediately follows that no drawing that is weakly isomorphic to a generalized twisted drawing can contain three interior-disjoint triangles (since the endpoints of the curve crossing every edge once must be on opposite sides of every triangle, the maximum number of interior-disjoint triangles is two).Up to strong isomorphism, there are only two simple drawings of K 4 .The plane drawing contains three interior-disjoint triangles.Thus, (up to strong isomorphism) the only drawing of K 4 that is weakly isomorphic to a generalized twisted drawing, is the drawing with a crossing.Hence, in every generalized twisted drawing all subdrawings induced by 4 vertices contain a crossing and thus every generalized twisted drawing is crossing maximal.Up to strong isomorphism, there are two crossing maximal drawings of K 5 : the convex drawing of K 5 and the twisted drawing of K 5 .Since the convex drawing contains three interior-disjoint triangles, the only (up to strong isomorphism) drawing of K 5 that is weakly isomorphic to a generalized twisted drawing is the twisted drawing of K 5 (that is drawn generalized twisted in Figure 1).
It is part of our ongoing work to show that for n ≥ 7, a drawing is weakly isomorphic to a generalized twisted drawing if and only if all subdrawings induced by five vertices are weakly isomorphic to the twisted K 5 .Interestingly, the n ≥ 7 is necessary as there is a drawing with 6 vertices that contains only twisted drawings of K 5 but is not weakly isomorphic to a generalized twisted drawing (see the drawings in Figure 9).There are (up to strong isomorphism) three more simple drawings of K 6 that consist of only twisted drawings of K 5 and they are all weakly isomorphic to generalized twisted drawings (see Figure 10).

A
Proof of Lemma 2 For every edge, this determines in which direction it emanates from its vertices.Hence there are (up to strong isomorphism) two possibilities how the crossing edges v 1 v 3 and v 2 v 4 can be drawn in D, depending on whether v 1 v 3 crosses the ray from O through v 4 at a point x 3 before or after v 4 ; cf. Figure 16.In both cases, v 1 v 2 has to cross the ray from O through v 4 at a point x 2 .This point x 2 has to lie after v 4 in the first case and before v 4 in the second case.In both cases, as the edge v 3 v 4 has to cross r, it must emanate from v 4 in the interior of the triangular region bounded by the segment x 2 x 3 , the portion v 1 x 3 of v 1 v 3 , and the portion v 1 x 2 of v 1 v 2 .However, the vertex v 3 is in the exterior of that triangular region, and therefore v 3 v 4 would have to cross the segment x 2 x 3 , contradicting that D is c-monotone, or one of v 1 v 2 and v 1 v 3 , contradicting the simplicity of D.
Figure 16 The two possibilities to draw v1v3 and v2v4 crossing and generalized twisted.

B Proof of Lemma 4
Lemma 4 has been implicitly shown in [11] and [13].For completeness, we include a detailed proof of the lemma in this appendix.We remark that the proof presented here is in parts similar to the one in [13].Proof.We call the pair of edges in D ′ incident to a i , 1 ≤ i ≤ n, the long edge r i .Let R A be the set of long edges.We first show that any edge between vertices in A crosses any long edge at most once.Then we show how to draw D ′ such that b 1 can be taken as origin O in a c-monotone drawing weakly isomorphic to D A , where the long edges of R A , as well as the edge b 1 b 2 , emanate as rays to infinity.
We now show that every edge between two vertices of A crosses every edge of R A at most once.Let a 1 , a 2 , and a 3 be vertices in A. Let R 1 be the region bounded by the edges b 1 a 1 , a 1 b 2 , b 2 a 2 and a 2 b 1 that does not contain a 3 .Let R 2 be the region bounded by the edges b 1 a 2 , a 2 b 2 , b 2 a 3 and a 3 b 1 that does not contain a 1 .Since D ′ is plane, these regions are disjoint.
As the edge e = a 1 a 2 is incident to all edges on the boundary of R 1 , it cannot cross it.Thus, e has to lie either completely inside or completely outside R 1 (and meet the boundary only in its endvertices).If e lies inside R 1 , it can cross neither a 3 b 1 nor a 3 b 2 .If it lies outside R 1 , it has to cross the boundary of R 2 an odd number of times.(Since e must begin at a 1 outside R 2 and finish at a 2 inside R 2 , and passing through R 1 is not possible.)As e cannot cross edges incident to a 2 , this means it has to cross exactly one of the edges a 3 b 1 or a 3 b 2 .Thus, e crosses the long edge r 3 at most once, for any vertex a 3 .
We can draw D ′ such that b 1 is functioning as the origin and R A as rays emerging from it by doing the following transformations; see Figure 17.We draw the subdrawing induced by the vertices of D ′ on the sphere such that b 1 and b 2 are antipodes, and the long edges of R A , as well as the edge b 1 b 2 , are meridians.By the general Jordan-Schoenflies theorem [16,8], the drawing on the sphere is homeomorphic to the original drawing on the plane.We then apply a stereographic projection from b 2 onto the plane.This way, the long edges in R A and the edge b 1 b 2 correspond to rays emerging from vertex b 1 , where the long edges in R A are exactly the rays through the vertices of D ′ .
Finally, we can obtain a c-monotone drawing that is weakly isomorphic to D A .We consider the stereographic projection.As all edges of D A cross the long edges in R A only once, they cross in between two long edges (or rays in the projection) r 1 and r 2 if and only if their order along the rays changes (that is, the edge closer to b 1 at r 1 is further away from b 1 at r 2 ).Consequently, we can draw the edge-segments between every two rays as straight-lines and obtain a c-monotone drawing that is weakly isomorphic to D A .If all edges of D A cross the edge b 1 b 2 , they cross a ray to infinity in the weakly isomorphic c-monotone drawing, and thus the c-monotone drawing is also generalized twisted.◀

C Proof of Theorem 13
▶ Theorem 13.Let s, t be two integers, 1 ≤ s, t ≤ n, such that (s − 1)(t − 1) + 1 ≤ n.Let D be a c-monotone drawing of K n .Then D contains either a generalized twisted drawing of K s or a monotone drawing of K t as subdrawing.In particular, if s = t = ⌈ √ n⌉, D contains a complete subgraph K s whose induced drawing is either generalized twisted or monotone.
Proof.Without loss of generality we may assume that the vertices of D appear counterclockwise around O in the order v 1 , v 2 , . . ., v n .Let r be a ray emanating from O, keeping v 1 and v n on different sides.We define an order, ⪯, in this set of vertices as follows: v i ⪯ v j if and only if either i = j or i < j and the edge (v i , v j ) crosses r.
We show that ⪯ is a partial order.The relation is clearly reflexive and antisymmetric.Besides, if v i ⪯ v j and v j ⪯ v k , then i < j and j < k imply i < k, so for the transitive property, we only have to prove that if v i v j and v j v k cross r, then v i v k also crosses r.We denote by r i , r j , r k the rays emanating from O and passing trough v i , v j , v k , respectively.We have two cases depending on where v j v i crosses the ray r k at a point x k ; in the first case, x k is located before v k on r k , while in the second one it is located after v k .Then v j v k has to cross the ray r i at a point x i , which is after v i in the first case and before v i in the second case (see Figure 18).Let Q be the region bounded by the segments Ox i , Ox k and the portions v j x i , v j x k of the edges v j v k , v j v i , respectively.In both cases, the edge v k v i cannot be contained in the counterclockwise wedge from r i to r k , because v i v k should connect a vertex placed outside Q with points placed inside that region, contradicting either If edges vivj and vjv k cross r in a c-monotone drawing, then viv k must also cross r.
the simplicity or the c-monotonicity of D. Therefore, v i v k must be in the clockwise wedge from r i to r k and thus crosses the ray r.
In this partial order ⪯, a chain consists of a subset v i1 , . . ., v is−1 of pairwise comparable vertices, that is, a subset of vertices such that their induced subdrawing is generalized twisted (all edges cross r).An antichain, v j1 , . . ., v jt−1 , consists of a subset of pairwise incomparable vertices, that is, a subset of vertices such that their induced subdrawing is monotone (no edge crosses r).Therefore, the first part of the theorem follows from applying Theorem 12 to the set of vertices of D and the partial order ⪯.
Finally, observe that if n⌉ whose induced subdrawing is either generalized twisted or monotone.◀

D Generalized twisted drawings contain a pair of antipodal vi-cells
In this section, we will show that every drawing weakly isomorphic to a generalized drawing of K n contains a pair of antipodal vi-cells (Theorem 16).Before proving the theorem, we will see some useful properties of generalized twisted drawings.Recall that in a generalized twisted drawing, vertices are labeled v 1 , v 2 , . . ., v n counterclockwise around the origin O, the ray emanating from O and passing through a vertex v i is denoted by r i , and the ray r that emanates from O and crosses every edge once is between r n and r 1 , counterclockwise from r n .
▶ Lemma 20.Let D be a generalized twisted drawing of K n with n ≥ 4. Suppose the two edges v i v j and v k v l of D cross, and i < j < k < l.Then the crossing point between these two edges is in the wedge W defined by r j and r k , counterclockwise from r j to r k .
Proof.Assume for contradiction that the crossing point is not in W , so it is in the wedge defined by r l and r i , counterclockwise from r l .There are four cases, depending on whether v k and v l are to the left or the right of the directed edge v j v i ; see Figure 19.In any of the four cases, there is no way of connecting v k and v j without crossing either v i v j or v k v l , which is a contradiction.◀ ▶ Lemma 21.For every generalized twisted drawing D with n ≥ 3 vertices, the following statements hold.
i) There exists a vertex v i , with 1 ≤ i ≤ n − 1, such that the bounded region RB defined by the edge v i v i+1 and the segments Ov i and Ov i+1 is empty.
Illustrating the proof of Lemma 20.
ii) There exists a vertex v j ̸ = v i , with 1 ≤ j ≤ n − 1, such that the unbounded region RU defined by the edge v j v j+1 and the segments Ov j and Ov j+1 is empty.
Proof.We show statement i), we take the first edge v i v k (with i < k) that crosses r and we show that v i satisfies i).Let x be the crossing point between v i v k and r, and let R be the bounded region defined by the edge v i v k and the segments Ov i and Ov k .Suppose that there is a vertex v l inside R. See Figure 20, left.Then there is no way of connecting v l and v i without crossing r before x, which contradicts that v i v k is the first edge crossing r.Thus, R must be empty.Suppose now that k ̸ = i + 1, so there is a vertex v l with i < l < k.See Figure 20, right.The edge v l v k must cross r at a point after x.But then there is no way of adding the edge v i v l without crossing r before x or without crossing v l v k .Therefore, k = i + 1 and i) follows.The proof of ii) is analogous by taking the last edge crossing r.In addition, v i and v j must be different since a same edge v i v i+1 cannot be at the same time the first and the last edge crossing r. ◀ ▶ Lemma 22.Let D be a generalized twisted drawing of K n with n ≥ 3 vertices.Then the cell containing O and the unbounded cell have at least one vertex on their boundaries.
Proof.We show that the cell containing O is a vi-cell.The proof for the unbounded cell follows analogously.By Lemma 21 i), there exists a vertex v i , with 1 ≤ i ≤ n − 1, such that the bounded region RB defined by the edge v i v i+1 and the segments Ov i and Ov i+1 is empty.We now show that either the segment Ov i or the segment Ov i+1 is uncrossed.Thus, it follows immediately that O lies in a vi-cell (with either v i or v i+1 on the boundary).Suppose to the contrary that neither segments are uncrossed, so there is an edge v j v k , with j < k, crossing the segment Ov i , and another edge v j ′ v k ′ , with j ′ < k ′ , crossing the segment Ov i+1 .
Observe now the following.First, since RB is empty, an edge v i+1 v l cannot cross Ov i for any l, so neither v j nor v k can be v i+1 .Second, suppose that j, k > i + 1.Since both vertices are outside RB, if v j v k crosses Ov i , then it must also cross v i v i+1 (see Figure 21a).Hence, by Lemma 20, the crossing point between v j v k and v i v i+1 is in the wedge defined by r i+1 and r j .But then, after crossing v i v i+1 and Ov i , the edge v j v k must cross v i v i+1 a second time to reach v k , which is a contradiction.Therefore, j, k < i.
Using an analogous reasoning, we also obtain that if an edge v j ′ v k ′ crosses Ov i+1 , then j ′ , k ′ > i + 1.As a consequence, the relative position of these three edges of D is as shown in Figures 21b and 21c.After emanating in v k , the edge v k v j crosses v i v i+1 at a point in the wedge defined by r k and r i ; then it crosses Ov i , and reaches v j surrounding v i v i+1 by the exterior.The edge v j ′ v k ′ must do the same but in opposite direction, crossing first v i v i+1 at a point in the wedge defined by r i+1 and r j ′ , then crossing Ov i+1 and reaching v k ′ surrounding v i v i+1 by the exterior.Note that v j v k and v j ′ v k ′ necessarily cross, and that the crossing point must be in the wedge defined by r i and r i+1 .Hence, v j ′ must be to the left of the oriented edge v k v j and v k must be to the right of the oriented edge For the vertices v j and v k ′ , there are two possibilities, depending on whether v k ′ is to the right (see Figure 21b) or to the left (see Figure 21c) of the oriented edge v k v j .But in the first case, the edge v j ′ v i+1 would cross v j v k twice, and in the second case, the edge We extend Lemma 23 to drawings weakly isomorphic to generalized twisted drawings using Gioan's Theorem [6,15] that any two weakly isomorphic drawings of K n can be transformed into each other with a sequence of triangle-flips and at most one reflection of the drawing.A triangle-flip is the operation that transforms a triangular cell △ that has no vertex on its boundary, by moving one of its edges across the intersection of the two other edges of △ (see Figure 22).▶ Theorem 16.Every simple drawing of K n which is weakly isomorphic to a generalized twisted drawing of K n , with n ≥ 3, contains a pair of antipodal vi-cells.In generalized twisted drawings the cell containing O and the unbounded cell form such a pair.
Proof.Let D be a simple drawing of K n that is weakly isomorphic to a generalized twisted drawing D ′ .By Lemma 23, the cell in which O lies in D and the unbounded cell of D are antipodal vi-cell pairs.Without loss of generality, we can assume that O is very close to a vertex v on the boundary of C 1 and Z is very close to a vertex w on the boundary of C 2 .Using Gioan's Theorem, it is enough to show that afer every triangle flip that can be transformed on a drawing D containing antipodal vi-cells, the resulting drawing D2 still contains antipodal vi-cells.
As triangle-flips are only applied to cells without vertices on its boundary, and points O and Z are close enough to vertices on the boundary of their cell in D, they stay in vi-cells (with the vertices they are close to) after every triangle-flip.What remains to be shown is that the vi-cells stay antipodal.
Let T be a triangle of D. (Note that a triangle of the drawing is the simple cycle formed by the three edges connecting three vertices of the graph, and not the triangular cells on which we perform triangle flips.)Whenever a flip is performed on a cell △, the cell △ disappears and a new cell appears.The new cell might (but not has to) be on the other side of T , but no other cells are affected.In particular, since triangle flips are never applied on vi-cells, any pair of vi-cells that was antipodal before stays antipodal after the flip.◀

E Characterizing via antipodal vi-cells
In this section, we prove that in any simple drawing D of K n containing antipodal vi-cells it is possible to add a simple curve crossing every edge of D exactly once (Theorem 17).To this end, we will use the following lemmata that show some properties of antipodal vi-cells.
▶ Lemma 24.Let D be a drawing of K n with n ≥ 4, and let (C 1 , C 2 ) be a pair of antipodal vi-cells.Then there is no vertex that lies on the boundary of both cells.
Proof.Assume, for a contradiction, that a vertex v 1 lies on the boundary of both cells C 1 and C 2 .See Figure 23.Consider another cell C ′ different from C 1 and C 2 , with v 1 on its boundary.Since v 1 has degree n − 1 ≥ 3, this cell exists.Let u 1 and u 2 be the two vertices whose edges to v 1 are on the boundary of that cell C ′ .Then the triangle formed by vertices u 1 , u 2 , and v 1 always have the cells C 1 and C 2 on the same side, contradicting that C 1 and C 2 are antipodal.
Edges of the star S(v1) incident to v1 are drawn black; the antipodal cells are filled purple; the additional cell is indicated in cyan; the different ways to draw the edge u1u2 are drawn dashed in blue.
▶ Lemma 25.Let D be a simple drawing of K n that contains two antipodal vi-cells C 1 and C 2 .Let v 1 be a vertex on the boundary of C 1 .Let T be a triangle formed by v 1 and two other vertices u 2 and u 3 .If there is a vertex u that lies on the same side of T as C 1 , then the edge v 1 u lies completely on that side.
Proof.Assume, for a contradiction, that the edge v 1 u does not lie completely on the same side of T as C 1 , and thus crosses the boundary of T .See Figure 24.The only edge it can cross is u 2 u 3 , as the other edges are incident to v 1 .Thus, the drawing induced by v 1 , u 2 , u 3 and u contains a crossing between v 1 u and u 2 u 3 .Since any simple drawing of K 4 contains at most one crossing, the edges uu 2 and uu 3 cannot cross the boundary of T .Thus, the triangle T ′ formed by u, u 2 and u 3 has to lie on the same side of T as C 1 , but keeping C 1 and C 2 on one of its sides, which is a contradiction to the definition of antipodal.
24 An illustration Lemma 25.The vertex v2 is placed as an example in one of the possible faces and could be at another place, but C1 and C2 lie on different sides of the triangle v1u2u3 by definition.Thus, by construction of the triangle uu2u3, the cells C2 and C1 lie on the same side of uu2u3.Now, we can prove Theorem 17.

▶ Theorem 17. In any simple drawing D of K n that contains a pair of antipodal vi-cells, it is possible to draw a curve c that crosses every edge of D exactly once.
Proof.Let (C 1 , C 2 ) be a pair of antipodal vi-cells of D, v 1 a vertex on the boundary of C 1 , v 2 a vertex on the boundary of C 2 , and S(v 1 ) the star of v 1 .Note that by Lemma 24, v 1 and v 2 are different.We draw a simple curve c from v 1 to v 2 such that it emerges from v 1 in the cell C 1 and ends in the cell C 2 very close to v 2 , and the following holds: 1.The curve c does not cross any edges of S(v 1 ).2. All intersections of c with edges of D are proper crossings.3.Over all curves for which 1 and 2 hold, the curve c has the minimum number of crossings with edges of D.
Since S(v 1 ) is a plane drawing that has only one face in which both v 1 and v 2 lie, drawing c is always possible.See for example Figure 25.We will prove that c crosses every edge of D \ S(v 1 ) exactly once.To show that c crosses an arbitrary edge w 2 w 3 exactly once, we will first show that c crosses w 2 w 3 an odd number of times and then show that c crosses w 2 w 3 at most twice.
To observe that c has to cross w 2 w 3 an odd number of times, consider the triangle T formed by v 1 , w 2 and w 3 .Since c connects two antipodal cells, the endpoints of c have to be on two different sides of T .Thus, c has to cross the boundary of T an odd number of times.Since c does not cross S(v 1 ), it has to cross w 2 w 3 an odd number of times.
We show now that c crosses w 2 w 3 at most twice.Assume to the contrary that c crosses w 2 w 3 at least three times.Without less of generality, we may assume that C 1 is inside T and C 2 is outside.Given two crossing points x and y between c and w 2 w 3 that are consecutive on c when going from v 1 to v 2 , a lens is the region to the left of the cycle formed by the arc xy on c and the arc yx on w 2 w 3 .See Figure 26 for an illustration.The pairs of two consecutive crossing points between c and w 2 w 3 on c define a set of lenses on both sides of T , possibly nested (see Figure 26).Note that since c crosses w 2 w 3 at least 3 times, there is at least one lens on each side of T .Among all the lens on the same side of T as C 1 , we take one that does not contain any other lens in its interior.This lens L always exists by taking the "innermost" one in a set of nested lenses.The curve c crossing several times the edge w2w3.If there are nested lenses in T , we take a minimal one that does not contain any nested lenses.This minimal lens L is shaded green.
Let x and y be the two crossing points defining the lens, so the boundary of L consists of arc xy on c and arc yx on w 2 w 3 .We claim that there is no vertex in L. Assume that there is a vertex u inside L (see Figure 27).By Lemma 25, the edge uv 1 cannot cross the edge w 2 w 3 .Thus, it has to cross c in order to get from u inside the lens to v 1 outside the lens.This is a contradiction to c being drawn such that it does not cross S(v 1 ).Thus, there is no vertex in L, as claimed.
Since L does not contain any vertex, then every edge that crosses the arc yx on w 2 w 3 has to also cross at least once the arc xy on the curve c (as there is no vertex in the lens where it could stop and it cannot cross the edge w 2 w 3 more than once); see Figure 28 (left).Thus, c can be drawn such that it stops before x, follows an arc very close to the arc xy on w 2 w 3 until a point very close to y, and then continues as c did before; see Figure 28 (right).This way the new drawing of the curve is still simple and does not have any crossings that the original one did not, but two less crossings with w 2 w 3 , which is a contradiction to the Therefore, since c crosses all edges in D \ S(v 1 ) an odd number of times and at most twice, it follows that it crosses all edges in D \ S(v 1 ) exactly once, while (per construction) it does not cross any edges of S(v 1 ).
We can transform c to a curve that crosses all edges exactly once in the following way: Instead of the starting point being v 1 we remove an ε of the curve on this end such that it starts very close to v 1 in the cell C 1 (and consequently still crosses exactly the edges it crossed before).Then, on that start in C 1 , we extend the curve by going around the vertex v 1 so close to v 1 that the extension crosses exactly the edges of S(v 1 ), and then ending again in C 1 ; see Figure 29.This way the extension crosses all edges of S(v 1 ) exactly once and consequently, we obtained a curve crossing all edges exactly one, with its endpoints not lying on any edges or vertices of D. ◀

F Characterizing via a Curve Crossing Everything
In this section we prove Theorem 18.
▶ Theorem 18.Let D be a simple drawing of K n in which it is possible to draw a simple curve c that crosses every edge of D exactly once.Then, D can be extended by two vertices O and Z (at the position of the endpoints of the curve), and edges incident to those vertices such that the obtained drawing is a simple drawing of K n+2 , no edge incident to O crosses any edge incident to Z, and all edges in D cross the edge OZ.

Figure 29
The resulting curve after the extension.The last part crossing S(v1) is drawn bold.
We first show several properties for drawings such that there is a simple curve c = OZ crossing every edge once.Then, we show that we can extend the drawing D = D n to a drawing D n+2 of K n+2 by adding O and Z as vertices, the curve c as an edge, and edges from O and Z to each vertex w of D n in such a way that D n+2 fulfills the following properties.as the drawings are shown on the plane, a given region can be bounded or unbounded.For example, region wx 1 y 1 is bounded in the left drawing of Figure 30 and unbounded in the right drawing.But in both drawings, the arcs, vertices and crossings inside region wx 1 y 1 are the same.
In our constructions, we are going to draw new arcs that are close to (or glued to) arcs of D n .A new arc a ′ is close or glued to the arc ux if (a 2 ) Notice that z 1 can be either x 1 or y k ′ .Similarly, z n−1 can be x k or y 1 .In any case, the only way to connect w to a vertex v placed inside the triangular region z 1 z n−1 w and crossing OZ, is crossing the arc z 1 z n−1 from its bottom part.For the same reason, all the vertices placed outside that region are precisely the endpoints of the top edges of S(w).See Proof.(b 1 ) Suppose e = u i u j is a bottom edge; the other case follows analogously.Since the two endpoints u i , u j are outside the triangular region x i x j w, the edge u i u j cannot enter in that region without breaking the simplicity, and therefore u i u j cannot cross the arc x i x j .
If the edge e crosses OZ at a point x on x j Z, just after crossing OZ, e is outside the region u i xx j w, but vertex u j is inside that region, so it is impossible to reach u j without either breaking the simplicity of D n or crossing the curve OZ twice.Hence, e must cross OZ on the arc Ox i .See Figure 32  (b 2 ) Suppose that the edge e = vv ′ crosses both wx 1 and wx k .
We first analyze the case when both endpoints of e, vertices v, v ′ , are inside the triangular region x 1 x k w, and therefore, by Property (a 2 ), both are endpoints of bottom edges of S(w).Let y be the crossing point of wv with OZ.Then, e = vv ′ cannot cross the arc x 1 x k because otherwise the boundary of x 1 x k w is crossed three times, contradicting that both v, v ′ are in that region.So, vv ′ has to first cross either the arc wx 1 or wx k .In the first case, after crossing wx 1 , the edge is in the region wyx 1 and the vertex v ′ is outside that region, hence the edge cannot leave that region keeping the simplicity.See Figure 33 left.Similarly, if vv ′ first crosses wx k , it enters in the region wx k y, but the vertex v ′ is outside that region, therefore the edge cannot leave that region without breaking the simplicity.
Figure 33 Property b2: Edge e cannot cross both wx1, wx k if its endpoints are inside x1x k w or when one of them is an endpoint of a top edge.
Suppose now that v is inside the region x 1 x k w, and v ′ is an endpoint of a top edge wv ′ crossing OZ at a point x.As above, let y be the crossing point of wv with OZ, and suppose that y is placed before x on OZ.See Figure 33 centre.Then, v is in the region wyx and v ′ is outside that region, therefore the edge e = vv ′ has to cross the arc yx.On the other hand, since neither v ′ nor v are in the region xx k w and e crosses wx k , it has to cross also the arc xx k , hence e should cross OZ twice, a contradiction.A similar analysis can be done in the symmetric case, when y is placed after x.
Finally, suppose that v is outside the region x 1 x k w, and v ′ is an endpoint of a top edge wv ′ crossing OZ at point x.As both vertices v, v ′ are outside the region x 1 x k w, the edge vv ′ cannot cross three times the boundary of that region, so e cannot cross the arc x 1 x k .However, when e enters in that region, by crossing wx 1 or wx k , as it cannot cross the arc wx, it should cross x 1 x k , again a contradiction.See Figure 33 right.◀ If we consider a mirror drawing of D n on the horizontal line OZ, all the top edges become bottom and vice versa, then, (b 1 ) has a symmetric Property (b ′ 1 ): If wu i , wu j are bottom edges and the crossing point x i is placed before x j , then if e = u i u j is top, it must cross c at a point x placed on Ox i , and if it is bottom, it must cross c at a point y placed on x j Z. ▶ Lemma 28.There is one vertex w 1 such that all the edges emanating from w 1 are top.
Lemma 28 is depicted in Figure 34.Proof.We prove the existence of such a vertex by induction on the number of vertices n.For n = 2 the lemma obviously is true.Now, consider a simple drawing D n of K n and assume that the lemma holds for any simple drawing of K n−1 (with all their edges crossed by a curve).By removing a vertex w ′ of D n , we obtain a simple drawing of K n−1 , by induction, containing a vertex w such that all the edges of S(w) are top.If ww ′ is also top in D n , all the edges of S(w) are top in D n , and w is the sought vertex.Suppose now that ww ′ is bottom and let x 1 , x k be the crossing points of the first and last edges of S(w).If ww ′ crosses the arc x 1 x k , w ′ has to be inside the region x 1 x k w.Then, as all the other vertices are outside that region, by Property(b 2 ), all the edges of S(w ′ ) must cross first x 1 x k and therefore they are top.Finally, if ww ′ is bottom and does not cross x 1 x k it has to cross OZ through a point y on the arc x k Z or on the arc Ox 1 .Suppose y is on x k Z (the other case is symmetric), then w ′ is in the triangular region x 1 yw, and all the other vertices out of that region.Thus, again by Property(b 2 ), all edges incident to w ′ have to cross first the top part of OZ, and therefore w ′ is the sought vertex.◀ By removing w 1 , we obtain a subdrawing with one vertex w 2 of which all the edges are top, by removing w 2 another vertex w 3 , and so on.Thus, we obtain an order of the vertices w 1 , . . ., w n such that for each vertex w i the edges w i w j with j > i are top edges (n − i edges) and the edges w i w l , with l < i are bottom edges (i − 1 edges).

Construction of the drawing D n+2
Beginning with the simple drawing D ′ 0 formed by D n and the curve c = OZ, we build new drawings D ′ 1 , . . ., D ′ i , . . ., D ′ n , where drawing D ′ i is obtained from drawing D ′ i−1 by adding two simple curves Ow i , w i Z, following the order w 1 , w 2 , . . .w n explained above.Hence, we start adding Ow 1 , w 1 Z, where w 1 is the unique vertex such that all the edges of S(w 1 ) are top, and we finish by adding Ow n , w n Z, where w n is the unique vertex such that all the edges of S(w n ) are bottom.We build the new drawings in such a way that in D ′ i , the following invariants are satisfied: 1.The curves OZ, Zw i , w i O form a well defined triangular region (region R i = OZw i ) containing the triangular region R i−1 = OZw i−1 , and containing no vertex w j with j > i.
Notice that this invariant implies that R i contains precisely the vertices w l with l < i, and that neither Ow i nor Zw i can properly cross any edge Ow l or Zw l with l < i.

The drawing D ′
i is a simple drawing.
The last drawing obtained, D ′ n , taking O and Z as vertices, provides the sought drawing D n+2 , since Invariants 1 and 2 imply that it satisfies properties (P1) and (P2).
To prove that D ′ i satisfies Invariants 1 and 2, we suppose that D ′ i−1 satisfies these two invariants.Then, the crossing points of the edges of D ′ i−1 with the boundary of OZw i−1 must satisfy the properties given in the following lemma.On the other hand, suppose that the edge w s w t crosses w i−1 Z at a point z.Then, the vertex w i−1 is in the region xZz, like the vertices w s , w t .Therefore, if w t w i−1 (or w s w i−1 ) leaves that region crossing the arc xZ, it cannot reach w i−1 without crossing again the boundary of that region, breaking the simplicity of D ′ i−1 .Therefore, x s , x t have to be placed before x on OZ.Finally, w t w i−1 cannot cross the arc x s w s (on the edge w i−1 w s ) or the arc w s x (on the edge w s w t ), therefore it cannot cross OZ between x s and x, and the order of the crossing points must be x t , x s , x.The same arguments can be used when the edge w s w t crosses Ow i−1 (c 3 ) By Invariant 1, w s is inside the region OZw i−1 and w t is outside, so the edge w s w t has to cross the boundary of OZw i−1 an odd number of times.Suppose that it crosses the three curves of the boundary, curves OZ, Ow i−1 , w i−1 Z.It cannot cross first Ow i−1 (or w i−1 Z) then OZ because w s w t is a top edge.It cannot cross first OZ then Ow i−1 and finally w i−1 Z because then, by (c 1 ), it has to cross the top edge w i−1 w t .By the same reason it cannot cross the boundary in the order first OZ, next w i−1 Z and then Ow i−1 .Finally, if it crosses first Ow i−1 , next w i−1 Z and then OZ, then, again by (c 1 ), it has to cross the bottom edge w i−1 w s .
Besides, since w i−1 w t is a top edge, the arc w i−1 x t divides the region OZw i−1 into two disjoint regions Ox t w i−1 and x t Zw i−1 .Then, if w s is in the region Ox t w i−1 , the edge w s wi − 1 has to cross by the arc Ox t .And then, necessarily the crossing point x must be between x s and x t .The same argument can be used when w s is in the region x t Zw i−1 .(c 4 ) Since i − 1 < s < t, both vertices w s , w t are outside region OZw i−1 , and we use the same reasonings as in (c 2 ): The boundary of OZw i−1 has to be crossed twice, and since w s w t is a top edge, first one of Ow i−1 or Zw i−1 is crossed, then OZ.Now, if w s w t crosses w i−1 Z at a point z, then none of the edges w i−1 w s , w i−1 w t can cross the arc xz (placed on w s w t ), so they cannot cross the arc xZ either.Therefore the crossing points x s , x t have to be placed before x on OZ.Finally, w i−1 and w t are in different sides of the region x s xw s and the edge w i−1 w t cannot cross the arc w s x s (on edge w i−1 w s ) or the arc w s x (on the edge w s w t ), therefore it has to cross OZ between x s and x, and the order of the crossing points must be x s , x t , x.The same arguments can be used when the edge w s w t crosses Ow i−1 .◀ Observe that, as we suppose that D ′ i−1 satisfies Invariants 1 and 2, properties (c 2 ), (c 3 ), (c 4 ) imply that no edge of D n can cross simultaneously the curve Ow i−1 and the curve Figure 35 Step 1: How to draw Ow1, Zw1.
For the construction of the drawings D ′ i , we will make a first step (for D ′ 1 ), a generic step (for D ′ i with 1 < i < n), and a final step (for D ′ n ).
Step 1.Let x 1 , . . ., x n−1 denote the crossing points of S(w 1 ) with OZ.We draw Ow 1 following (slightly counterclockwise) the curve OZ until the point x 1 , then we turn counterclockwise following the arc x 1 w 1 ; see Figure 35.We draw Zw 1 analogously, following ZO until x n−1 , then following the arc x n−1 w 1 .If we denote the arcs Ox 1 , x 1 w 1 , Zx n−1 , x n−1 w 1 by a 1 , b 1 , a 2 , b 2 respectively, then Ow 1 has a first part glued to a 1 , then a second part glued to b 1 , and in the same way, Zw 1 consists of two parts glued to a 2 and b 2 respectively.Clearly, OZw 1 is a well defined region and by Property (a 2 ), there are no vertices in the triangular region w 1 x 1 x n−1 , hence region OZw 1 does not contain any vertices.That establishes Invariant 1.
Suppose now that the edge w s w t , 1 < s < t ≤ n − 1 crosses the arc a 1 , and let x s , x t be the crossings points on OZ of w 1 w s , w 1 w t , respectively.Notice that x s can be placed before or after x t , but x is placed before these two points.Then, by Property (b 1 ), since the bottom edge is w t w s , x t has to be placed before x s , and the order of the crossing points on OZ must be x, x 1 , x t , x s .Besides, the arc b 1 must start at w 1 by outside the triangular region w 1 w t w s (because it is the first top arc), and it finishes at x 1 , a point also placed outside w 1 w t w s .Therefore, the arc b 1 cannot cross w s w t , the unique edge of the triangle w 1 w t w s not incident to w 1 .A symmetric argument can be used to prove that an edge w s w t , 1 ≤ s < t < n − 1 crossing a 2 cannot cross b 2 .This establishes Invariant 2 for D ′ 1 .
Step i (2 ≤ i ≤ n − 1).We will draw Ow i in two different ways, Way 1 and Way 2, depending on whether the first top edge of S(w i ), edge e 1 , first crosses Ow i−1 or w i−1 Z.When e 1 first crosses Ow i−1 at a point z ′ 1 , we draw Ow i in Way 1, which is the following: Ow i follows the curve Ow i−1 (very close to that curve, slightly counterclockwise) until it reaches the crossing point z ′ 1 , then Ow i continues close to z ′ 1 w i until it reaches its endpoint w i ; see Figure 36 left.
When e 1 crosses first w i−1 Z, by Property (c 4 ) applied to e 1 (e 1 is some w i w t with i − 1 < i < t), the edge w i−1 w i has to cross OZ before x ′ 1 (the crossing point of e 1 with OZ), and therefore, the last bottom edge of S(w i ), edge e ′ k ′ , crosses OZ at a point y ′ k ′ placed before x ′ 1 (or e ′ k ′ coincides with w i w i−1 ).In this case, we draw Ow i in Way 2, which is the following: Ow i follows the curve OZ (very close to that curve, slightly clockwise) until it reaches the crossing point y ′ k ′ , then Ow i continues close to the arc y ′ k ′ w i on e ′ k ′ until it reaches the endpoint w i ; see Figure 36   To prove that Invariants 2 holds for D ′ i , we will see that in both Case 1 and Case 2, each edge of D n crosses at most one of the arcs a 1 , b 1 and at most one of the arcs a 2 , b 2 of D ′ i .Case 1.-By construction, the curves Ow i and w i Z do not cross each other, the triangular region OZw i contains the region OZw i−1 , and Ow i , w i Z cannot properly cross any edge Ow l , w l Z, with l < i, because these last edges are inside OZw i .Notice that the order of the edges around O and Z are counterclockwise OZ, Ow 1 , . . ., Ow i and ZO, Zw i , . . ., Zw 1 , respectively.Moreover, by Property (a 2 ), all the vertices in the triangular region x ′ 1 x ′ k w i must be reached from w i via a bottom edge (where x ′ 1 , . . ., x ′ k are the crossing points of the top edges of S(w i ) with curve OZ).On the other hand, all the vertices w s , s ≤ i − 1 are in OZw i−1 (w i−1 on the boundary), and these are precisely the endpoints of the bottom edges of S(w i ).Therefore, the subregion z ′ 1 w i−1 z ′ k w i must be empty, and OZw i contains all w s with s < i and does not contain vertices w t , t > i.
To prove the simplicity of D ′ i , it is enough to prove that for any edge w s w t of D n , the drawing formed by Ow i , w i Z, w s w t is simple.The following subcases 1, 2, 3, 4, 5 prove that simplicity for edges w s w t when s   Suppose that the edge w s w t crosses both a 2 and b 2 .By Property (c 2 ), this edge first crosses OZ at a point x, then reaches a 2 at a point z before finishing at w t .Since b 2 = w i z ′ k is outside OZw i−1 the crossing of b 2 with w s w t must be at a point in the arc xz, as shown in Figure 37 left.Observe that after crossing b 2 , which is an arc on the last top edge of S(w i ), the edge w s w t enters in OZw i−1 crossing a 2 , therefore it cannot cross again the last top edge of S(w i ), and thus w t has to be placed inside the region x ′ k Zz ′ k .A totally symmetric case occurs when the edge w s w t crosses both a 1 and b 1 , then w t must be inside the region Ox ′ 1 z ′ 1 , see Figure 37 right.Let us analyze only the first case, when e = w s w t crosses OZ, b 2 , a 2 in that order, because the other case is totally symmetric.We are going to prove that the edge w t w i−1 cannot be drawn without breaking the simplicity.By Property (c 2 ) applied to w s w t , the crossing point of w t w i−1 with OZ, must be placed before x on OZ.On the other hand, if x ′ is the crossing point of w i−1 w i with OZ, by Property (c 4 ) applied to e k (the last top edge of S(w i )), the point x ′ has to be placed before x ′ k .Then, by Property (c 3 ) applied to w t w i (t < i − 1 < i), as w t is in region x ′ Zw i−1 , the edge w t w i−1 has to cross OZ after x ′ .Therefore, the edge w t w i−1 should cross OZ after x ′ and before x, and this is not possible when x is placed before x ′ ; see Figure 38 (left).Finally, if x is placed after x ′ , then x ′ is inside the region xZz, w i outside that region, and therefore the arc x ′ w i has to cross the arc xz at some point y, see Figure 38 (right).But then, the edge w t w i−1 has to cross OZ on the arc x ′ x, entering into the region x ′ xy, bounded by arcs on the edges w s w t and w i w i−1 , that cannot be crossed by w t w i−1 without breaking the simplicity of D n .-4 Any edge w s w t , s < i − 1 < i < t crosses at most one of the arcs a Case 2.-Again, by the method that the construction is done, the curves Ow i and w i Z do not cross each other and the triangular region OZw i contains the region OZw i−1 , hence Ow i , w i Z cannot properly cross any edge Ow l , w l Z, with l < i. See Figure 36 right.Moreover, by Property (a 2 ) all the vertices in region Ow i−1 z ′ k w i y ′ k ′ are endpoints of bottom edges of S(w i ), but all these endpoints w 1 , . . ., w i−1 of the bottom edges of S(w i ) must be inside the region OZw i−1 (w i−1 on the boundary).Therefore, region Ow i−1 z ′ k w i y ′ k ′ is empty and OZw i only contains inside the i − 1 endpoints w 1 , . . ., w i−1 .So, Invariant 1, holds.
Like in Case 1, to prove the simplicity of D ′ i we analyze the same five subcases.
- See Figure 39 left.Since they cannot cross a 2 , we only have to prove that one of this edges cannot cross both a 1 and b 1 .In the definition of Way 2, we have seen that the crossing point x ′ of w i w i−1 with OZ must be placed before x ′ 1 and after y ′ k ′ .Hence, if x t is the crossing point on OZ of a top edge w i w t , t > i, then x ′ is placed between y ′ k ′ and x t .But then, by Property (c 4 ) applied to w i w t , i − 1 < i < t, the crossing point of w i−1 w t on OZ must be placed between x ′ and x t , so after y ′ k ′ .Therefore, the top edges w i−1 w t , t > i − 1 of S(w i−1 ) cannot cross a 1 .Finally, suppose that an edge w s w i−1 , s < i − 1 crosses both a 1 and b 1 .Necessarily, w s w i−1 first crosses OZ at a point x, then b 1 at a point y, finishing at w i−1 .See Figure 39 right.Then, the vertices w i , w s must be both inside the region xZw i−1 .However, the bottom edge w i w s cannot cross a 1 , (remember that y ′ k ′ is the first crossing point of those bottom edges).Therefore, by Property (c 3 ), w i w s cannot cross Ow i−1 or w i−1 Z, and it has to enter in OZw i−1 crossing through the bottom of the arc y ′ k ′ Z, but then it should first cross the arc xw i−1 (on w s w i−1 ), contradicting the simplicity of D n .
-3 Any e = w s w t , s < t < i crosses at most one of the arcs a 1 , b 1 or one of a 2 , b 2 .
A scheme of this situation is shown in Figure 40, where the two blue curves mark the boundary of regions OZw s and OZw t .On the contrary, suppose that w s w t crosses both a 1 and b 1 .Necessarily, it first crosses a 1 at a point x, next b 1 at a point y, then entering in region OZw i−1 by crossing Ow i−1 or w i−1 Z at a point z, until reaching w t .See Figure 41 left.But then, by the same reasonings as in the previous Case 2-2, the edge w i w s cannot be drawn.As above, w s and w i are in the region xZz, if z is on w i−1 Z, or in the region xZw i−1 z when z is on Ow i−1 .However, w i w s cannot cross a 1 or Ow i−1 or w i−1 Z, so it has to enter in OZw i−1 crossing through the arc y ′ k ′ Z, and therefore so it should first cross the arc xz (on w s w t ), a contradiction.Finally, suppose that w s w t crosses a 2 and b 2 .We are exactly in the same situation as in Case 1-3 (see Figure 38): necessarily the edge first crosses OZ at a point x, next crosses b 2 , then reaches a 2 at a point z before finishing at w t .And we have seen that a contradiction is reached in this situation.It does not matter if the edge w s w t crosses a 1 (like in Figure 41   -4 Any edge w s w t , s < i − 1 < i < t crosses at most one of the arcs a 1 , b 1 or a 2 , b 2 .By Property (c 3 ), an edge w s w t with s < i < t cannot cross w i−1 Z, hence, it cannot cross a 2 .So we have to prove that it cannot cross both a 1 and b 1 .Let us see that in this situation the edge w s w i cannot be drawn without breaking the simplicity of the drawing.If w s w t , s < i − 1 < t crosses both a 1 and b 1 , necessarily it starts crossing OZ through a 1 , then crossing b 1 at a point y, finishing at vertex w t , see Figure 42.Then, w s has to be inside the region w i yw t .On the other hand, the bottom edge w i w s must start from w i counterclockwise after the arc b 2 (on the last top edge of S(w i )), and before the arc b 1 (on the last bottom edge of S(w i )).Therefore, w i w s should start outside the region w i yw t , and should finish at w s , placed inside that region.However, w i w s cannot cross any of the arcs of the boundary of w i yw t , because they are on edges incident to either w i or w s .This finishes the proof: The last drawing obtained, D ′ n , satisfies Invariants 1, 2. Then, taking O and Z as vertices, it is the sought drawing D n+2 , the one satisfying the Properties (P1) and (P2).

Figure 1 A
Figure 1 A generalized twisted drawing of K5.All edges cross the (red) ray r.

▶ Observation 8 .▶ Theorem 9 .
For any n ≥ 3, the number of edges in an outerplanar graph with n vertices is at most 2n − 3. Every simple drawing of K n contains at least n 48 pairwise disjoint edges.Proof.Let D be a simple drawing of K n , and let M be a maximal plane matching of D. If m := |M | ≥ n 48 , then Theorem 9 holds.So assume that |M | < n 48

Figure 3
Figure 3 Left: The face f in H containing the plane drawing H ′ (blue lines) inside.Right: We can obtain an outerplane drawing from H ′ by interpreting bundles of edge pairs incident to the same black vertices as plane edges.

1 2 √ 1 Figure 5
Figure 5The edge xy has to cross the edge vw.
A c-monotone drawing D of K5 such that the ray r crosses no edge of D.
An x-monotone drawing of K5 strongly isomorphic to D of Figure6a.

Figure 6
Figure 6Two strongly isomorphic monotone drawings of K5. v

Figure 8 A
Figure 8 A simple drawing of K7.The red edges show the star S(v), the red and blue edges together form a maximal plane subdrawing H. Dashed edges are edges of K7 that are not in H.

Figure 9
Figure 9Two weakly isomorphic drawings of K6 that are not weakly isomorphic to any generalized twisted drawing.Antipodal cells are marked in blue.

Property 1 D
is weakly isomorphic to a generalized twisted drawing.Property 2 D contains two antipodal vi-cells.Property 3 D can be extended by a simple curve c such that c crosses every edge of D exactly once.Property 4 D can be extended by two vertices, O and Z, and edges incident to the new vertices such that D together with the new vertices and edges is a simple drawing of K n+2 , the edge OZ crosses every edge of D, and no edge incident to O crosses any edge incident to Z.

Figure 14
Figure 14 Top and bottom edges.For simplicity, the curve OZ is drawn as a horizontal line.Left: A top edge wu.Centre: A bottom edge wu.Right: The (black) top and (blue) bottom edges of S(w).

1 Figure 17 ▶ Lemma 4 .
Figure17 The homeomorphisms of D ′ .Left: DA, the edges in RA and b1b2 are drawn on the sphere, such that RA and b1b2 are meridians.Right: The steographic projecton from b2.

Figure 20
Figure 20 the proof of Lemma 21.

Figure 21
Figure 21 One of the segments Ovi and Ovi+1 is uncrossed.

v 1 v 2 c C 2 C 1 Figure 25
Figure 25Edges of the star S(v1) incident to v1 are drawn black; the antipodal cells are filled purple; the curve c is drawn in red.
Figure 26The curve c crossing several times the edge w2w3.If there are nested lenses in T , we take a minimal one that does not contain any nested lenses.This minimal lens L is shaded green.

Figure 27 Figure 28
Figure 27There is no vertex in L. If there is a vertex u inside L, then the edge from u to v1 (drawn dashed) would have to cross c, which it cannot by construction.

(Figure 30
Figure 30Top edges are drawn in black, bottom edges are drawn in blue.

( g1 )▶(a 1 ) 2 )Proof. (a 1 )
An edge crosses a ′ if and only if it crosses ux.(g2) All the crossing points on a ′ and on ux have the same order.(g3) The arcs a ′ and ux do not cross each other.(They can share the endpoints, points u, x, but do not have to.)Lemma 26.Let w be a vertex of D n .The edges of S(w) satisfy the following properties: When exploring counterclockwise around w the edges of S(w), the top edges are consecutive, wu 1 , wu 2 . . ., wu k , and cross the curve c = OZ at points x 1 , . . ., x k in that order.Then the bottom edges are consecutive, wv 1 , . . ., wv k ′ (where k ′ = n − k − 1), and cross the curve c ′ = ZO at points y 1 , . . ., y k ′ in that order.See Figure 30.(a Let z 1 and z n−1 be the first and the last crossing points of S(w) on the curve c.Then the endpoints of the bottom edges of S(w) are inside the triangular region z 1 z n−1 w, and the endpoints of the top edges are outside that region.See Figure 31.Draw a new arc a glued to OZ by its top part, and another arc a ′ glued to ZO on the bottom part.Both have endpoints O and Z, and thus a, a ′ define a cycle C.The edges of the star S(w) in counterclockwise order have to reach (that is, have their first crossing point with) the cycle C at points placed in clockwise order on C. The top edges of S(w) are the ones reaching C on the arc a, and the corresponding crossing points x 1 , . . ., x k on OZ are in increasing order (from O to Z).Then come the bottom edges reaching C on the arc a ′ , their corresponding crossing points y 1 , . . ., y k ′ in this order on ZO.

▶ 1 Figure 31 (b 1 ) 2 )
Figure31Property a2: The vertices placed in the regions with a blue circle can be reached from w only with bottom edges.

Figure 32
Figure 32 Property b1: The edge uiuj must cross OZ as in the left figure or as in the right figure.

Figure 34
Figure 34Lemma 28: There is a vertex w1 such that all the edges of S(w1) are top.
right) or not.
Assume, for a contradiction, that the edge v 2 v 4 crosses the edge v 1 v 3 .Since any simple drawing of K 4 has at most one crossing, no other edges of D can cross.Recall that in any generalized twisted drawing, all edges are drawn c-monotone and intersect the ray r.
▶ Lemma 2. Let D be a generalized twisted drawing of K 4 , with vertices {v 1 , v 2 , v 3 , v 4 } labeled counterclockwise around O. Then the edges v 1 v 3 and v 2 v 4 do not cross.Proof.

▶
Lemma 29.Suppose that D ′ i−1 satisfies Invariants 1, 2, and let x s , x t , x be the crossing points of OZ with the edges w i−1 w s , w i−1 w t , w s w t , respectively.Then (c 1 ) All the top edges of S(w i−1 ) are counterclockwise between w i−1 O and w i−1 Z, and the bottom edges are counterclockwise between w i−1 Z and w i−1 O. (c 2 ) Any edge w s w t with s < t < i − 1 crosses first OZ then one of the curves Ow i−1 or Zw i−1 .Besides, if w s w t crosses Zw i−1 , the order of the above crossing points on OZ is x t , x s , x, and if w s w t crosses Ow i−1 , this order is x, x s , x t .(c 3 ) Any edge w s w t with s < i − 1 < t crosses first OZ and it does not cross any other arc of the boundary of OZw i .Besides, if w s is in the region Ox t w i−1 , the order of the above crossing points on OZ is x s , x, x t , and if w s is in the region x t Zw i−1 , the order is x t , x, x s .(c 4 ) Any edge w s w t with i − 1 < s < t first crosses one of Ow i−1 or Zw i−1 , then OZ.Besides, if w s w t crosses Zw i−1 , the order of the crossing points is x s , x t , x, and if it crosses Ow i−1 , the order is x, x t , x s .) A top edge w i−1 w t with t > i − 1 of S(w i−1 ) has to reach OZ by inside of OZw i−1 , and since D ′ i−1 is simple, it must start entering in that region, hence it must be counterclockwise between w i−1 O and w i−1 Z.The same reasoning works for bottom edges w i−1 w s with s < i − 1. (c 2 ) Since s < t < i − 1, by Invariant 1, the vertices w s , w t are both in OZw i−1 .Hence, the edge w s w t must cross the boundary of OZw i−1 an even number of times.And since it has to cross OZ, it has to cross also only one of the boundary curves Ow i−1 , w i−1 Z. Finally, since w s w t is a top edge, it has to cross first OZ then the other boundary curve.
Symmetric constructions are used for drawing Zw i .When e k , the last top edge of S(w i ), first crossesw i−1 Z at a point z ′ k , Zw i is drawnin Way 1: Zw i consists of the arcs a 2 and b 2 glued to Zz ′ k and z ′ k w i respectively.When e k first crosses Ow i−1 , then OZ at a point x ′ k , the first bottom edge of S(w i ), edge e ′ 1 , has to cross OZ at a point y ′ 1 placed after x ′ k .Then, we build Zw i in Way 2: it consists of an arc a 2 (counterclockwise) close to Zy ′ 1 , then an arc b 2 glued to y ′ 1 w i on edge e ′ 1 .As construction Way 2 can only be used in one of the two curves Ow i , Zw i , we only need to see that Invariants 1, 2 hold in two cases: Case 1, when construction Way 1 is used for both Ow i and w i Z, and Case 2, when Way 2 is used for Ow i and Way 1 for w i Z.The case when Way 2 is used for Zw i and Way 1 for Ow i is symmetric to Case 2; see Figure 36.
t and i < s < t, respectively.No edge of S(w i ) can cross any of the arcs a1 , b 1 , a 2 , b 2 .For the arcs b 1 , b 2 , this is obvious since these arcs are glued to edges of S(w i ).Besides, no top edge of S(w i ) can cross arc a 1 because this arc follows the curve Ow i−1 until precisely the first crossing point with a top edge of S(w i ).By the same reason, top edges of S(w i ) cannot cross a 2 either.Finally, by the simplicity of D ′ i−1 the bottom edge w i w i−1 cannot cross a 1 or a 2 , and by Property (c 3 ) applied to an edge w i w s with s < i − 1 < i, this bottom edge of S(w i ) cannot cross Ow i−1 or Zw i−1 , and therefore it cannot cross a 1 or a 2 .-2Any edge of S(w i−1 ) crosses at most one of the arcs a 1 , b 1 and at most one of the arcs a 2 , b 2 .
-1 Since the edges of S(w i−1 ) cannot cross the arc a 1 nor the arc a 2 , the result follows.
1 , b 1 or a 2 , b 2 .Observe that w s is inside the triangular region OZw i−1 and w t is outside that region.Then, by Property (c 3 ), the edge w s w t must first cross OZ and it cannot enter the region OZw i−1 again.Therefore, it cannot cross a 1 or a 2 .-5 Any edge w s w t , i < s < t crosses at most one of the arcs a 1 , b 1 or a 2 , b 2 .OZw i−1 crossing only one of a 1 ,z ′ 1 w i−1 ,a 2 ,w i−1 z ′ k , then leaving through OZ, therefore it cannot cross both a 1 , b 1 or both a 2 , b 2 .
First observe that by Property (b 2 ) applied to S(w i ), the edge w s w t cannot cross both b 1 and b (because w i w s and w i w t are top edges of that star).Besides, w s and w t are outside the triangular region OZw i , therefore if edge w s w t crosses b 1 or b 2 it has to cross also z ′ 1 w i−1 or w i−1 z ′ k .Finally, according to Property (c 4 ), the edge w s w t must enter to

1
No edge of S(w i ) can cross any of the arcs a 1 , b 1 , a 2 , b 2 .As in Case 1, no edge of S(w i ) can cross the arcs b 1 , b 2 , no a top edge of that star can cross a 2 .Besides, by Property (c 3 ), a bottom edge w i w s , s < i − 1 < i, cannot cross Zw i−1 , therefore it cannot cross a 2 either.Finally, a top edge cannot cross either arc a 1 , (because the crossing points x ′ 1 , . . ., x ′ k are after y ′ k ′ ), and a bottom edge cannot cross a 1 = Oy ′ k ′ because y ′ k ′ is the first crossing point of those edges.-2 Any edge of S(w i−1 ) crosses at most one of the arcs a 1 , b 1 or one of a 2 , b 2 .