1 Introduction
A drawing of a graph maps each vertex to a distinct point in the plane, each edge to a Jordan curve connecting the points of its incident vertices but not containing the point of any other vertex, and two such Jordan curves have at most one common point. In the last few years, the focus in graph drawing has shifted from exploiting structural properties of planar graphs to addressing the question of how to produce wellstructured (understandable) drawings in the presence of edge crossings, i.e., to the topic of beyondplanar graph classes. The primary approach here has been to define and study graph classes which allow some edge crossings, but restrict the crossings in various ways. Two commonly studied such graph classes are:

planar graphs, the graphs which can be drawn so that each edge (Jordan curve) is crossed by at most other edges.

quasiplanar graphs, the graphs which can be drawn so that no pairwise nonincident edges mutually cross.
Note that the planar graphs and quasiplanar graphs are precisely the planar graphs. Additionally, the quasiplanar graphs are simply called quasiplanar.
In this paper we study these two families of classes of graphs by restricting the drawings so that the points are placed in convex position and edges mapped to line segments, i.e., we apply the above two generalizations of planar graphs to outerplanar graphs and study outer planarity and outer quasiplanarity. We consider balanced separators, treewidth, degeneracy (see paragraph “Concepts” below), coloring, edge density, and recognition for these classes.
Related work.
Ringel [27] was the first to consider planar graphs by showing that planar graphs are 7colorable. This was later improved to 6colorable by Borodin [8]. This is tight since is 1planar. Many additional results on 1planarity can be found in a recent survey paper [21]. Generally, each vertex planar graph has at most edges [26] and treewidth [14].
Outer planar graphs have been considered mostly for . Of course, the outer 0planar graphs are the classic outerplanar graphs which are wellknown to be 2degenerate and have treewidth at most 2. It was shown that essentially every graph property is testable on outerplanar graphs [5]. Outer 1planar graphs are a simple subclass of planar graphs and can be recognized in linear time [4, 18]. Full outer planar graphs, which form a subclass of outer planar graphs, can been recognized in linear time [19]. General outer planar graphs were considered by Binucci et al. [7], who (among other results) showed that, for every , there is a 2tree which is not outer planar. Wood and Telle [30] considered a slight generalization of outer planar graphs in their work and showed that these graphs have treewidth ).
The quasiplanar graphs have been heavily studied from the perspective of edge density. The goal here is to settle a conjecture of Pach et al. [25] stating that every vertex quasiplanar graph has at most edges, where is a constant depending only on . This conjecture is true for [2] and [1]. The best known upper bound is [16], where is the inverse of the Ackermann function. Edge density was also considered in the “outer” setting: Capoyleas and Pach [9] showed that any quasiplanar graph has at most edges, and that there are quasiplanar graphs meeting this bound. More recently, it was shown that the semibar visibility graphs are outer quasiplanar [17]. However, the outer quasiplanar graph classes do not seem to have received much further attention.
The relationship between planar graphs and quasiplanar graphs was considered recently. While any planar graph is clearly quasiplanar, Angelini et al. [3] showed that any planar graph is even quasiplanar.
The convex (or 1page book) crossing number of a graph [29] is the minimum number of crossings which occur in any convex drawing. This concept has been introduced several times (see [29] for more details). The convex crossing number is NPcomplete to compute [23]. However, recently Bannister and Eppstein [6] used treewidthbased techniques (via extended monadic second order logic) to show that it can be computed in linear FPT time, i.e., time where is the convex crossing number and is a computable function. Thus, for any , the outer crossing graphs can be recognized in time linear in .
Concepts.
We briefly define the key graph theoretic concepts that we will study.
A graph is degenerate when every subgraph of it has a vertex of degree at most . This concept was introduced as a way to provide easy coloring bounds [22]. Namely, a degenerate graph can be inductively colored by simply removing a vertex of degree at most . A graph class is degenerate when every graph in the class is degenerate. Furthermore, a graph class which is hereditary (i.e., closed under taking subgraphs) is degenerate when every graph in that class has a vertex of degree at most . Note that outerplanar graphs are 2degenerate, and planar graphs are 5degenerate.
A separation of a graph is pair of subsets of such that , and no edge of has one end in and the other in . The set is called a separator and the size of the separation is . A separation of a graph G on vertices is balanced if and . The separation number of a graph is the smallest number such that every subgraph of has a balanced separation of size at most . The treewidth of a graph was introduced by Robertson and Seymour [28]; it is closely related to the separation number. Namely, any graph with treewidth has separation number at most and, as Dvořák and Norin [15] recently showed, any graph with separation number has treewidth at most . Graphs with bounded treewidth are wellknown due to Courcelle’s Theorem (see Theorem 4.1) [10], i.e., having bounded treewidth means many problems can be solved efficiently.
The Exponential Time Hypothesis (ETH) [20] is a complexity theoretic assumption defined as follows. For , let : there is an time algorithm to solve SAT}. ETH states that for , , e.g., there is no quasipolynomial time^{1}^{1}1i.e., with a runtime of the form . algorithm that solves 3SAT. So, finding a problem that can be solved in quasipolynomial time and is also NPcomplete, would contradict ETH. In recent years, ETH has become a standard assumption from which many conditional lower bounds have been proven [12].
Contribution.
In Section 2, we consider outer planar graphs. We show that they are degenerate, and observe that the largest outer planar clique has size , i.e., implying each outer planar graph can be colored and this is tight. We further show that every outer planar graph has separation number at most . For each fixed , we use these balanced separators to obtain a quasipolynomial time algorithm to test outer planarity, i.e., these recognition problems are not NPhard unless ETH fails.
In Section 3, we consider outer quasiplanar graphs. Specifically, we discuss the edgemaximal graphs which have been considered previously under different names [9, 13, 24]. We also relate outer quasiplanar graphs to planar graphs.
Finally, in Section 4, we restrict outer planar and outer quasiplanar drawings to closed drawings, where the sequence of vertices on the outer boundary is a cycle. For each , we express both closed outer planarity and closed outer quasiplanarity in extended monadic secondorder logic. Thus, closed outer planarity is testable in time, for a computable function .
2 Outer Planar Graphs
In this section we show that every outer planar graph is degenerate and has separation number . This provides tight bounds on the chromatic number, and allows for testing outer planarity in quasipolynomial time.
Degeneracy.
We show that every outer planar graph has a vertex of degree at most . First we note the size of the largest outer planar clique and then we prove that each outer planar graph has a vertex matching the clique’s degree. This also tightly bounds the chromatic number in terms of , i.e., Theorem 2.1 follows from Lemma 1 (proven in Appendix 0.B.1) and Lemma 2.
Lemma 1
Every outer planar clique has at most vertices.
Lemma 2
An outer planar graph can have maximum minimum degree at most and this bound is tight.
Proof
Assume that the outer planar graph has maximum minimum degree . Since we can create a clique with vertices (see Lemma 1), . Let us show that cannot be larger than .
Consider an edge that cuts vertices of the graph to one side (not counting and ), then there are at least edges crossing the edge . We will now show by induction that if there existed an outer planar graph with minimum degree , it would be too small to accommodate such a minimum degree vertex.
Any edge that cuts vertices is crossed by at least edges. Therefore, if , there is such that cannot cut vertices because then it is crossed by edges. Take the smallest such and let us show that there also cannot be an edge that cuts more than vertices. As the induction hypothesis, assume that no edge cuts between and vertices inclusive. Thus, the minimum number of edges that cross is: , where the last term accounts for the absent edges that cut more than vertices. Now, if cuts vertices, it is crossed by
edges if .
Since for the inequality is always satisfied, there cannot be an edge that cuts more then vertices in any outer planar graph with the maximum minimum degree . But then, such a graph can have at most vertices, which is not enough to accommodate the minimum degree vertex required; a contradiction.
Theorem 2.1
Each outer planar graph is colorable. This is tight.
Quasipolynomial time recognition via balanced separators.
We show that outer planar graphs have separation number at most (Theorem 2.2). Via a result of Dvořák and Norin [15], this implies they have treewidth. However, Proposition 8.5 of [30] implies that every outer planar graph has treewidth at most , i.e., a better bound on the treewidth than applying the result of Dvořák and Norin to our separators. The treewidth bound also implies a separation number of , but our bound is better. Our separators also allow outer planarity testing in quasipolynomial time (Theorem 2.3).
Theorem 2.2
Each outer planar graph has separation number at most .
Proof
Consider an outer planar drawing. If the graph has an edge that cuts vertices to one side, we can use this edge to obtain a balanced separator of size at most , i.e., by choosing the endpoints of this edge and a vertex cover of the edges crossing it. So, suppose no such edge exists. Consider a pair of vertices such that the line divides the drawing into left and right sides having an almost equal number of vertices (with a difference at most one). If the edges which cross the line also mutually cross each other, there can be at most of them. Thus, we again have a balanced separator of size at most . So, it remains to consider the case when we have a pair of edges that cross the line , but do not cross each other. We call such a pair of edges parallel. We now pick a pair of parallel edges in a specific way. Starting from , let be the first vertex along the boundary in clockwise direction such that there is an edge that crosses the line . Symmetrically, starting from , let be the first vertex along the boundary in clockwise direction such that there is an edge that crosses the line ; see Fig. 1 (left). Note that the edges and are either identical or parallel. In the former case, we see that all other edges crossing the line must also cross the edge , and as such there are again at most edges crossing the line . In the latter case, there are two subcases that we treat below. For two vertices and , let be the set of vertices that starts with and, going clockwise, ends with . Let .
Case 1. The edge cuts vertices to the top; see Fig. 1 (center).
In this case, either or has vertices. We claim that neither the line nor the line can be crossed more than times. Namely, each edge that crosses the line also crosses the edge . Similarly, each edge that crosses the line also crosses the edge . Thus, we have a separator of size at most , regardless of whether we choose or to separate the graph. As we observed above, one of them is balanced.
Case 1. The edge cuts at most vertices to the bottom.
This is symmetric to case 1.
Case 2. The edge cuts at most vertices to the bottom, and the edge cuts at most vertices to the top; see Fig. 1 (right).
We show that we can always find a pair of parallel edges such that one cuts at most vertices to the bottom and the other cuts at most vertices to the top, and no edge between them is parallel to either of them. We call such a pair close. If there is an edge between and , we form a new pair by using and if cuts at most vertices to the bottom or by using and if cuts at most vertices to the top. By repeating this procedure, we always find a close pair. Hence, we can assume that and actually form a close pair. Let , , , and ; see Fig. 1 (right).
Suppose that or . We can now use both edges and (together with any edges crossing them) to obtain a separator of size at most . The separator is balanced since and .
So, now are all distinct. Note that since each side of the line has at most vertices. We separate the graph along the line . Namely, all the edges that cross this line must also cross or . Therefore, we obtain a separator of size at most .
To see that the separator is balanced, we consider two cases. If (or ), then (or ). Otherwise and . In this case and . In both cases the separator is balanced.
Theorem 2.3
For fixed , testing the outer planarity of an vertex graph takes time.
Proof
Our approach is to leverage the structure of the balanced separators as described in the proof of Theorem 2.2. Namely, we enumerate the sets which could correspond to such a separator, pick an appropriate outer planar drawing of these vertices and their edges, partition the components arising from this separator into regions, and recursively test the outer planarity of the regions.
To obtain quasipolynomial runtime, we need to limit the number of components on which we branch. To do so, we group them into regions defined by special edges of the separators.
By the proof of Theorem 2.2, if our input graph has an outer planar drawing, there must be a separator which has one of the two shapes depicted in Fig. 2 (a) and (b). Here we are not only interested in the up to vertices of the balanced separator, but actually the set of up to vertices one obtains by taking both endpoints of the edges used to find the separator. Note: is also a balanced separator. We use a brute force approach to find such an . Namely, we first enumerate vertex sets of size up to . We then consider two possibilities, i.e., whether this set can be drawn similar to one of the two shapes from Fig. 2. So, we now fix this set . Note that since has vertices, the subgraph induced by can have at most a function of different outer planar drawings. Thus, we further fix a particular drawing of .
We now consider the two different shapes separately. In the first case, in , we have three special vertices and and in the second case we will have two special vertices and . These vertices will be called boundary vertices and all other vertices in will be called regional vertices. Note that, since we have a fixed drawing of , the regional vertices are partitioned into regions by the specially chosen boundary vertices. Now, from the structure of the separator which is guaranteed by the proof of Theorem 2.2, no component of can be adjacent to regional vertices which live in different regions with respect to the boundary vertices.
We first discuss the case of using as depicted in Fig. 2 (a). Here, we start by picking the three special vertices and from to take the role as shown in Fig. 2 (a). The following arguments regarding this shape of separator are symmetric with respect to the pair of opposing regions.
Notice that if there is a component connected to regional vertices of different regions, we can reject this configuration. From the proof of Theorem 2.2, we further observe that no component can be adjacent to all three boundary vertices. Namely, this would contradict the closeness of the parallel edges or it would contradict the members of the separator, i.e., it would imply an edge connecting distinct regions. We now consider the four possible different types of components and in Fig. 2 (a) that can occur in a region neighboring . Components of type are connected to (possibly many) regional vertices of the same region and may be connected to boundary vertices as well. In any valid drawing, they will end up in the same region as their regional vertices. Components of type are not connected to any regional vertices and only connected to one of the three boundary vertices. Since they are not connected to regional vertices, they can not interfere with other parts of the drawing, so we can arbitrarily assign them to an adjacent region of their boundary vertex. Components that are connected to two boundary vertices appear at first to have two possible placements, e.g., as or in Fig. 2 (a). However, is not a valid placement for this type of component since it would contradict the fact that this separator arose from two close parallel edges as argued in the proof of Theorem 2.2. From the above discussion, we see that from a fixed configuration (i.e., set , drawing of , and triple of boundary vertices), if the drawing of has the shape depicted in Fig. 2 (a), we can either reject the current configuration (based on having bad components), or we see that every component of is either attached to exactly one boundary vertex or it has a welldefined placement into the regions defined by the boundary vertices. For those components which are attached to exactly one boundary vertex, we observe that it suffices to recursively produce a drawing of that component together with its boundary vertex and to place this drawing next to the boundary vertex. For the other components, we partition them into their regions and recurse on the regions. This covers all cases for this separator shape.
The other shape of our separator can be seen in Fig. 2 (b). Note that we now have two boundary vertices and and thus only have two regions. Again we see the two component types and and can handle them as above. We also have components connected to both and but no regional vertices. These components now truly have two different placement options . If we have an edge (as in Fig. 2 (b)) of the separator that is not , we now observe that there cannot be more than such components. Namely, in any drawing, for each component, there will be an edge connecting this component to either or which crosses . Thus, we now enumerate all the different placements of these components as type or and recurse accordingly.
However, the separator may be exactly the pair . Note that there are no components of type and the components of type can be handled as before. We will now argue that we can have at most a function of different components of type or in a valid drawing. Consider the components of type (the components of type can be counted similarly). In a valid drawing, each type component defines a subinterval of the left region spanning from its highest to its lowest vertex such that these vertices are adjacent to one of or . Two such intervals relate in one of three ways: They overlap, they are disjoint, or one is contained in the other. We group components with either overlapping or disjoint intervals into layers. We depict this situation in Fig. 2 (c) where, for simplicity, for every component we only draw its highest vertex and its lowest vertex and they are connected by one edge.
Let be the bottommost component of type (i.e., is the clockwisefirst vertex from in a component of type ). The first layer is defined as the component together with every component whose interval either overlaps or is disjoint from the interval of . Now consider the green edge (see Fig. 2 (c)), note we may have that this edge connects to instead. Now, for every component of this layer which is disjoint from the interval of , this edge is crossed by at least one edge connecting it to . Furthermore, for every component of this layer which overlaps the interval of , there is an edge connecting to either or which is crossed by at least one edge within that component. So in total, there can only be components in this first layer. New layers are defined by considering components whose intervals are contained in . To limit the total number of layers, let be the bottommost vertex of the first component of the deepest layer and consider the purple edge . This edge is crossed by some edge of every layer above it and as any edge can only have crossings, there can only be different levels in total. This leaves us with a total of at most components per region and again we can enumerate their placements and recurse accordingly.
The above algorithm provides the following recurrence regarding its runtime. Namely, we let denote the runtime of our algorithm, and we can see that the following expression generously upper bounds its value. Here denotes the number of different outer planar drawings of a graph with vertices.
Thus, the algorithm runs in quasipolynomial time, i.e., .
3 Outer QuasiPlanar Graphs
In this section we consider outer quasiplanar graphs. We first describe some classes of graphs which are outer 3quasiplanar. We then discuss edgemaximal outer quasiplanar drawings.
Note, all subHamiltonian planar graphs are outer 3quasiplanar. One can also see which complete and bipartite complete graphs are outer 3quasiplanar.
Proposition 1
The following graphs are outer 3quasiplanar:
(a) ;
(b) ;
(c) planar tree with three complete levels;
(d) squaregrids of any size.
Proof
and are easily observed. was experimentally verified by constructing a Boolean expression and using MiniSat to check it for satisfiability; see Appendix 0.A. follows from squaregrids being subHamiltonian.
Correspondingly, we note complete and complete bipartite graphs which are not outerquasi planar. Furthermore, not all planar graphs are outer quasiplanar, e.g., the vertexminimal planar 3tree in Fig. 3 (a) is not outer quasiplanar, this was verified checking for satisfiability the corresponding Boolean expression; see Section 0.A. A drawing of the graph in Fig. 3 (b) was constructed by removing the blue vertex and drawing the remaining graph in an outer quasiplanar way.
(a) planar drawing; (b) deleting the blue vertex makes the drawing outer quasiplanar
Proposition 2
The following graphs are not outer 3quasiplanar: (a) , , ; (b) , ; (c) planar 3tree with four complete levels.
Theorem 3.1
Planar graphs and outer 3quasiplanar graphs are incomparable under containment.
Remark 1
For outer quasiplanar graphs () containment questions become more intricate. Every planar graph is outer quasiplanar because planar graphs have page number 4 [31]. We also know a planar graph that is not outer 3quasiplanar. It is open whether every planar graph is outer 4quasiplanar.
Maximal outer quasiplanar graphs.
A drawing of an outer quasiplanar graph is called maximal if adding any edge to it destroys the outer quasiplanarity. We call an outer quasiplanar graph maximal if it has a maximal outer quasiplanar drawing. Recall that Capoyleas and Pach [9] showed the following upper bound on the edge density of outer quasiplanar graphs on vertices: .
We prove (see Appendix 0.B.2) that each maximal outer quasiplanar graph meets this bound. Our proof builds on the ideas of Capoyleas and Pach [9] and directly shows the result via an inductive argument. However, while preparing the cameraready version of this paper, we learned of two other proofs of this result in the literature [13, 24]. We thank David Wood for pointing us to these results. Both papers prove a slightly stronger theorem (concerning edge flips) as their main result. Namely, for a drawing , an edge flip produces a new drawing by replacing an edge with a new edge . They [13, 24] show that, for every two maximal outer quasiplanar drawings and , there is a sequence of edge flips producing drawings such that each is a maximal quasiplanar drawing. Together with the tight example of Capoyleas and Pach [9], this implies the next theorem, and makes our proof fairly redundant.
4 Closed Convex Drawings in MSO
Here we express graph properties in extended monadic secondorder logic (MSO). This subset of secondorder logic is built from the following primitives.

variables for vertices, edges, sets of vertices, and sets of edges;

binary relations for: equality (), membership in a set (), subset of a set (), and edge–vertex incidence ();

standard propositional logic operators: , , , .

standard quantifiers () which can be applied to all types of variables.
For a graph and an MSO formula , we use to indicate that can be satisfied by in the obvious way. Properties expressed in this logic allow us to use the powerful algorithmic result of Courcelle stated next.
Theorem 4.1 ([10, 11])
For any integer and any MSO formula of length , an algorithm can be constructed which takes a graph with treewidth at most and decides in time whether where the function from this time bound is a computable function of and .
Outer planar graphs are known to have treewidth (see Proposition 8.5 of [30]). So, expressing outer planarity by an MSO formula whose size is a function of would mean that outer planarity could be tested in linear time. However, this task might be out of the scope of MSO. The challenge in expressing outer planarity in MSO is that MSO does not allow quantification over sets of pairs of vertices which involve nonedges. Namely, it is unclear how to express a set of pairs that forms the circular order of vertices on the boundary of our convex drawing. However, if this circular order forms a Hamiltonian cycle in our graph, then we can indeed express this in MSO. With the edge set of a Hamiltonian cycle of our graph in hand, we can then ask that this cycle was chosen in such a way that the other edges satisfy either planarity or quasiplanarity. With this motivation in mind, we define the classes closed outer planar and closed outer quasiplanar, where closed means that there is an appropriate convex drawing where the circular order forms a Hamiltonian cycle. Our main result here is stated next.
Theorem 4.2
Closed outer planarity and closed outer quasiplanarity can be expressed in MSO. Thus, closed outer planarity can be tested in linear time.
The formulas for our graph properties are built using formulas for Hamiltonicity (Hamiltonian), partitioning of vertices into disjoint subsets (VertexPartition) and connected induced subgraphs on sets of vertices using only a subset of the edges (Connected). They can be found in Appendix 0.C.
For a closed outer planar or closed outer quasiplanar graph , we want to express that two edges and cross. To this end, we assume that there is a Hamiltonian cycle of that defines the outer face. We partition the vertices of into three subsets , , and , as follows: is the set containing the endpoints of , whereas and are connected subgraphs on the remaining vertices that use only edges of . In this way, we partition the vertices of into two sets, one left and the other one right of . For such a partition, must cross whenever has one endpoint in and one in .
Cro  
Now we can describe the crossing patterns for closed outer planarity and closed outer quasiplanarity as follows:
Here we insist that is Hamiltonian and that, for every edge and any set of distinct other edges, at least one among them does not cross .
Again, we insist that is Hamiltonian and further that, for any set of distinct edges, there is at least one pair among them that does not cross.
We conclude this section by mentioning an intermediate concept between closed outer planarity and outer planarity, i.e., full outer kplanarity [19]. The full outer planar graphs are defined as having a convex drawing which is planar and additionally there is no crossing on the outer boundary of the drawing. Hong and Nagamochi [19] gave a lineartime recognition algorithm for full outer 2planar graphs. Clearly, the closed planar graphs are a subclass of the full planar graphs. So, one open question is whether one can generalize our MSO expressions of closed outer planarity and closed outer quasiplanarity to the full versions. If yes, this would provide lineartime recognition of full outer planar graphs for every , including the full outer planar case.
Acknowledgement.
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Appendix 0.A SAT Experiments
In this section, we describe a logical formula for testing whether a given graph is outer quasiplanar. We present the formula in firstorder logic. After transformation to Boolean logic, we solve the formula using MiniSat http://minisat.se/.
A quasi outerplanar embedding corresponds to a
circular order of the vertices.
If we cut a circular order at some vertex to turn the circular into a
linear order, the edge crossing pattern remains the same.
Therefore, we look for a linear order.
For any pair of vertices and
any pair of edges , we introduce 0–1
variables and , respectively.
The intended meaning of is that vertex
comes before .
The intended meaning of is that edge crosses
edge .
Now we list the clauses of our SAT formula.
(1)  
(2)  
(3)  
(4) 
Appendix 0.B Omitted Proofs
0.b.1 Proof of Lemma 1
Proof
Let the largest outer planar clique have vertices. An edge is crossed by the number of vertices on one side times the number of vertices on the other side of . Therefore, the edge that is crossed the most has almost equal number of vertices on both sides, so if is even and if
is odd. Therefore,
is at most .0.b.2 Proof of Theorem 3.2
For an outer quasiplanar drawing of graph we call an edge a long edge, if and are separated along the outer face of by at least vertices on both sides. In the following, long edges will always be drawn vertically with on top, dividing the graph into regions left and right of . Further, we call all edges intersecting the long edge crossing edges. All vertices incident to crossing edges will be called crossing vertices and for illustration we will label them in a toptobottom fashion like depicted in Fig. 4 (a). On the left side of , vertices will be labeled and on the right side, respectively . In the following, we will count the number of crossing edges and give an inductive argument for the maximality of the considered graph.
To do so, we are following the idea given by Capoyleas and Pach: we will construct hierarchical levels and define a replacementoperation that will use these levels to split the original graph into two subgraphs on less vertices. The hierarchical levels will be built greedily as in Algorithm 0.B.2.
Algorithm 1: BuildLevels(Outer quasiplanar Graph , long edge )
Lemma 3
For a given outer quasiplanar graph , Algorithm 0.B.2 generates hierarchical levels – subsets of the crossing edges of that form maximal crossingfree connected subgraphs of .
Proof
We need to argue that the algorithm creates levels that cover all crossing edges with respect to and that every level is connected. We order the levels by order of construction so we say level is after level or , if is constructed after . To do so, we first state two important properties:
Property (P1) follows from the construction of level : If there where no edge of level crossing edge , then would also belong to level .
Property (P2) holds by induction: For the first level there is no previous level. Edges of level two are crossed by edges of the first level due to (P1). Any edge of level must be crossed by some edge of level . Inductively we know that is crossed by an edge of every previous level. Together, they form a chain of pairwise crossings from above, and we get the following index patterns:
These patterns indicate that in fact all the considered edges are pairwise crossing. For a given edge of level , we call any set following the description in property (P2) a certificate for to be in . In fact, any edge of level crossing can be extended to some certificate for .
Let be the last level and consider the last edge taken from that level. Suppose the algorithm created too many levels, so . By property (P2), the certificate of belonging to together with and forms a set of pairwise crossing edges. This contradicts that the graph from which is taken is outer quasiplanar and so we never create more than levels.
Note, the maximality of each level comes from the way it is constructed.
To argue about the connectivity of the levels we carefully consider the way they will be generated in a maximal graph and make sure that greedily picking edges never disconnects the edge set of any level. Consider Fig. 5 (a): Assume we already generated levels to and in level we pick the edges and but not and thus would end up disconnecting . The edge is going upward with respect to the other edges of level and thus cannot cross them from above due to (P1). Thus, if is present in our graph, it would belong to level and would be connected. So further assume that is not present. As the graph is supposed to be maximal, there must be a prevention set of pairwise crossing edges that prevent its existance. Due to the edge , the prevention set contains at most crossing edges. By existence of the edges and we know that the prevention set also cannot consist of edges only locally on one side. In this context, is considered to be both locally left and locally right. This implies two possible prevention sets: crossing edges together with locally left or right edges – see Fig. 5 (b). For illustration we color the edges of the prevention set differently: black edges start between and , appearing between and but not belonging to the same level as ; red edges cross from above ending strictly below ; orange edges cross from below; blue edges are locally on one side crossing all edges of other colors. By verifying the following claim, we finish the proof.
Claim
A prevention set for can be transformed into a prevention set for (or ) – contradicting ’s existence.
For the transformation, we keep all the orange and blue edges of , as they pairwise cross and also cross . For the remaining edges, we need to make sure that, if we replace them, the new edges must also cross the blue and orange ones. Other edges in the set belong to previous levels because of the following:

Red edges all cross from above (P1).

The black edges divide into two cases: crossing or completely between and . Edges crossing do so from above (P1). Edges between and can not be crossed by edges of level , so they must have been picked before.
In fact we now see that the prevention set for consists of a certificate for – they all pairwise cross (P2) – together with blue and orange edges. We now consider the edges of the certificate for in the order of the level they belong to starting at . To see that for any such level, there is an edge crossing whose left vertex is covered by all blue edges consider the following: Suppose the edge of level does not cross . In order for an edge of level to cross , its right vertex must be below the right vertex of . Thus its left vertex can either be the left vertex of or some other vertex below it but still above ; in both cases, the new left vertex will be covered by all blue edges. After doing one replacement, in the worst case every remaining edge of the certificate for does not cross because it is below them. This we can fix be using a certificate for to be in level ; notice that every edge of this new certificate must be below its corresponding counterpart in the old certificate and thus still all starting vertices are covered by blue edges.
Thus, a prevention set for using locally left edges is not possible since it would imply a prevention set for as well. A similar argument can be made using locally right edges by again taking a prevention set for , transforming the certificate for into one for and observing that the endpoints of of the edges of the certificate are again still covered by the locally right edges.
The construction of Pach and Capoyleas [9, proof of Claim 4] yields the following:
Remark 2
In an outer quasiplanar graph with vertices labeled according to their cyclic order along the outer face (), every vertex can be adjacent to with . These frame edges are present in any maximal outer planar graph; see Fig. 5 (c).
Using our hierarchical levels, we now describe a replacement operation, which is used to split our graph into two smaller parts. Let be a maximal outer quasiplanar graph with a long edge and hierarchical levels created by Algorithm 0.B.2. Let and be the vertices of incident to edges of level left and right of , respectively. We make two graphs from . For (), delete all vertices left (right) of and for every level add a new levelvertex connected to all vertices in (), e.g., see Fig. 6 (a) and (b). Finally, add any absent frame edges to () to make it maximal.
Lemma 4
After applying the replacement operation to a maximal outer quasiplanar graph , the following relations among , and hold:

and

,
where is the set of edges of crossing and are the sets of crossing edges added to by the replacement operation.
Proof
(i) In and , we only modify the right or left side of , leaving the other side unmodified. The modification adds levelvertices to both graphs. Notice that is present in both and , so subtracting the levelvertices and one copy of and , yields
(ii) To count the edges added to both and and compare them to the number of edges removed by splitting , we consider the structure of our hierarchical levels. As before, let and be the sets of vertices incident to edges of level left (right) of . Every level is a connected and noncrossing set of edges, so there are exactly edges in level
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