On the Relationship between Planar
and Quasi Planar Graphs
Abstract
A graph is planar if it can be drawn in the plane such that no edge is crossed more than times. A graph is quasi planar if it can be drawn in the plane with no pairwise crossing edges. The families of planar and quasi planar graphs have been widely studied in the literature, and several bounds have been proven on their edge density. Nonetheless, only trivial results are known about the relationship between these two graph families. In this paper we prove that, for , every planar graph is quasi planar.
1 Introduction
Drawings of graphs are used in a variety of application domains, including software engineering, circuit design, computer networks, database design, social sciences, and biology (see e.g. [15, 16, 25, 28, 38, 40]). The aim of a graph visualization is to clearly convey the structure of the data and their relationships, in order to support users in their analysis tasks. In this respect, and independent of the specific domain, there is a general consensus that graph layouts with many edge crossings are hard to read, as also witnessed by several user studies on the subject (see e.g. [24, 35, 36, 43]). This motivation has generated lots of research on finding bounds on the number of edge crossings in different graph families (see e.g. [34, 37, 42]) and on the problem of automatically computing graph layouts with as few crossings as possible (see e.g. [4, 12]). We recall that, although it is lineartime solvable to decide whether a graph admits a planar drawing (i.e. a drawing without edge crossings) [10, 23], minimizing the number of edge crossings is a wellknown NPhard problem [21].
An emerging research area, informally recognized as beyond planarity (see e.g. [22, 26, 29]), concentrates on different models of graph planarity relaxations, which allow edge crossings but forbid specific configurations that would affect the readability of the drawing too much. Forbidden crossing configurations can be, for example, a single edge that is crossed too many times [33], a group of mutually crossing edges [20, 39], two edges that cross at a sharp angle [17], a group of adjacent edges crossed by another edge [14], or an edge that crosses two independent edges [5, 9, 27]. Different models give rise to different families of “beyond planar” graphs. Two of the most popular families introduced in this context are the planar graphs and the quasi planar graphs, which are usually defined in terms of topological graphs, i.e., graphs with a geometric representation in the plane with vertices as points and edges as Jordan arcs connecting their endpoints. A topological graph is planar if no edge is crossed more than times, while it is quasi planar if it can be drawn in the plane with no pairwise crossing edges. Figure 1 shows a crossing configuration that is forbidden in a planar topological graph. Figure 1 depicts a planar topological graph that is not planar (e.g., the thick edge is crossed three times). Figure 1 shows a crossing configuration that is forbidden in a quasi planar topological graph. Figure 1 depicts a quasi planar topological graph that is not quasi planar. A graph is planar (quasi planar) if it is isomorphic to some planar (quasi planar) topological graph. Clearly, by definition, planar graphs are also planar and quasi planar graphs are also quasi planar. This naturally defines a hierarchy of planarity and a hierarchy of quasi planarity. Also, the class of quasi planar graphs coincides with that of planar graphs. Note that, in the literature, quasi planar graphs are also called quasi planar.
The planarity and quasi planarity hierarchies have been widely explored in graph theory, graph drawing, and computational geometry, mostly in terms of edge density. Pach and Tóth [33] proved that a planar simple topological graph with vertices has at most edges. We recall that a topological graph is simple if any two edges cross in at most one point and no two adjacent edges cross. For , Pach and Tóth [33] also established a finer bound of on the edge density, and prove its tightness for . For , the best known upper bound on the edge density is , which is tight up to an additive constant [6, 30].
Concerning quasi planar graphs, a yearold conjecture by Pach, Shahrokhi, and Szegedy [32] asserts that, for every fixed , the maximum number of edges in a quasi planar graph with vertices is . However, linear upper bounds have been proven only for . Agarwal et al. [3] were the first to prove that quasi planar simple topological graphs have a linear number of edges. This was generalized by Pach et al. [31], who proved that all quasi planar graphs on vertices have at most edges. This bound was further improved to by Ackerman and Tardos [2]. For quasi planar simple topological graphs they also proved a bound of , which is tight up to an additive constant. Ackerman [1] also proved that quasi planar graphs have at most a linear number of edges. For , several authors have shown superlinear upper bounds on the edge density of quasi planar graphs (see, e.g., [13, 19, 20, 32, 41]). The most recent results are due to Suk and Walczak [39], who proved that any quasi planar simple topological graph on vertices has at most edges, where is a number that depends only on . For quasi planar topological graphs where two edges can cross in at most points, they give an upper bound of , where is the inverse of the Ackermann function, and depends only on and .
Despite the many papers mentioned above, the relationships between the hierarchies of planar and quasi planar graphs have not been studied yet and only trivial results are known. For example, due to the tight bounds on the edge density of planar and quasi planar simple graphs, it is immediate to conclude that there are infinitely many quasi planar graphs that are not planar. Also, it can be easily observed that, for , every planar graph is quasi planar. Indeed, if a planar graph were not quasi planar, any topological graph isomorphic to would contain pairwise crossing edges; but this would imply that any of these edges is crossed at least times, thus contradicting the hypothesis that is planar.
Contribution. In this paper we focus on simple topological graphs and prove the first nontrivial inclusion relationship between the planarity and the quasi planarity hierarchies. We show that every planar graph is quasi planar, for every . In other words, we show that every planar simple topological graph can be redrawn so to become a quasi planar simple topological graph (). For example, the simple topological graph of Figure 1 is planar but not quasi planar. The simple topological graph of Figure 1, on the other hand, is a quasi planar graph obtained from the one of Figure 1 by rerouting an edge (but it is no longer planar).
The proof of our result is based on the following novel methods: A general purpose technique to “untangle” groups of mutually crossing edges. More precisely, we show how to reroute the edges of a planar topological graph in such a way that all vertices of a set of pairwise crossing edges lie in the same connected region of the plane. A global edge rerouting technique, based on a matching argument, used to remove all forbidden configurations of pairwise crossing edges from a planar simple topological graph, provided that these edges are “untangled”.
The remainder of the paper is structured as follows. In Section 2 we give some basic terminology and observations that will be used throughout the paper. Section 3 describes our general proof strategy. Section 4 and Section 5 provide details about methods and , respectively. Conclusions and open problems are in Section 6.
2 Preliminaries
We only consider graphs with neither parallel edges nor selfloops. Also, we will assume our graphs to be connected, as our results immediately carry over to disconnected graphs. A topological graph is a graph drawn in the plane with vertices represented by points and edges represented by Jordan arcs connecting the corresponding endpoints. In notation and terminology, we do not distinguish between the vertices and edges of a graph, and the points and arcs representing them, respectively. Two edges cross if they share one interior point and alternate around this point. Graph is almost simple if any two edges cross at most once. Graph is simple if it is almost simple and no two adjacent edges cross each other. Graph divides the plane into topologically connected regions, called faces. The unbounded region is the outer face. Note that the boundary of a face can contain vertices of the graph and crossing points between edges.
If and are two isomorphic graphs, we write . A graph is planar (quasi planar) if there exists a planar (quasi planar) topological graph .
Given a subgraph of a graph , the arrangement of , denoted by , is the arrangement of the curves corresponding to the edges of . We denote the vertices and edges of by and , respectively. A node of is either a vertex or a crossing point of . A segment of is a part of an edge of that connects two nodes, i.e., a maximal uncrossed part of an edge of .
A fan is a set of edges that share a common endpoint. A set of vertexdisjoint mutually crossing edges in a topological graph is called a crossing. A crossing is untangled if in the arrangement of all nodes corresponding to vertices in are incident to a common face. Otherwise, it is tangled. For example, the crossing in Figure 2 is untangled, whereas the crossing in Figure 2 is tangled. We observe the following.
Observation 2.1
Let be a planar simple topological graph and let be a crossing in . An edge in cannot be crossed by any other edge in . In particular, for any two crossings in , holds.
Proof
Each edge in a crossing crosses each of the remaining edges in . Since graph is planar, edge is not crossed by any other edge in .∎
3 Edge Rerouting Operations and Proof Strategy
We introduce an edge rerouting operation that will be a basic tool for our proof strategy. Let be a planar simple topological graph and consider an untangled crossing in . Without loss of generality, the vertices in lie in the outer face of .
Let and let . Denote by the arrangement obtained from by removing all nodes corresponding to vertices in , together with their incident segments, and by removing edge . The operation of rerouting around consists of redrawing sufficiently close to the boundary of the outer face of , choosing the routing that passes close to , in such a way that crosses the fan incident to , but not any other edge in . See Figure 3 for an illustration. More precisely, let be a topological disk that encloses all crossing points of and such that each edge in crosses the boundary of exactly twice. Then, the rerouted edge keeps unchanged the parts of that go from to the boundary of and from to the boundary of . We call the unchanged parts of a rerouted edge its tips and the remaining part, which routes around , its hook.
Lemma 1
Let be a planar simple topological graph and let be an untangled crossing in . Let be the topological graph obtained from by rerouting an edge around a vertex . Let be the edge of incident to . has the following properties: Edges and do not cross; The edges that are crossed by in but not in form a fan at ; is almost simple.
Proof
Conditions and immediately follow from the definition of the rerouting operation and from the fact that edge can be drawn arbitrarily close to the boundary of the outer face of . Since is simple, in order to prove that is almost simple, we only need to show that edge does not cross any other edge more than once. The only part of that is drawn in differently than in is the one between the intersection points of and the boundary of the topological disk that () encloses all crossing points of and such that () each edge in crosses the boundary of exactly twice. By () and by the definition of the rerouting operation, the two crossing points between an edge and alternate with the two crossing points between any edge and along . Hence, by redrawing edge sufficiently close to any of the two parts of between the two intersection points of edge and , we encounter each edge in exactly once. Thus, edge crosses all the edges in exactly once. This concludes the proof.∎
Lemma 1 does not guarantee that graph is simple. Indeed, if the edge or the edge existed in , then the rerouted edge would cross such an edge. We will show in Section 5 how to fix this problem by redrawing and .
We are now ready to describe our general strategy for transforming a planar simple topological graph into a simple topological graph that is quasi planar. The idea is to pick from each crossing in an edge and a vertex not adjacent to , and to apply the rerouting operation simultaneously for all pairs , i.e., rerouting around . This operation, which we call global rerouting, is welldefined since the crossings are pairwise edgedisjoint by Observation 2.1.
There are however several constraints that have to be satisfied in order for such a global rerouting to have the desired effect. First of all, as mentioned above, the rerouting operation can only be applied to untangled crossings. Thus, as a first step, we will show that, in a planar simple topological graph, all tangled crossings can be removed, leaving the resulting graph simple and planar. More precisely, given a tangled crossing , it is possible to redraw the whole graph such that either at least two edges of do not cross or becomes an untangled crossing, and, further, any two edges cross only if they crossed before the redrawing. The technical details for this operation are described in Section 4. Notice that, even assuming that all crossings are untangled, there are further problems that can occur when performing all the rerouting operations independently of each other. Specifically, the resulting topological graph may be nonsimple and/or the rerouting may create new crossings. We explain how to overcome these issues in Section 5.
4 Removing Tangled Crossings
The proof of the next lemma describes a technique to “untangle” all crossings in a planar simple topological graph. This technique is of general interest, as it gives more insights on the structure of planar simple topological graphs.
Lemma 2
Let be a planar simple topological graph. There exists a planar simple topological graph without tangled crossings.
Proof
We first show how to untangle a crossing in a planar simple topological graph by neither creating new crossings nor introducing new crossings.
Let be a tangled crossing and let be its arrangement. For each face of , denote by the subset of vertices of incident to . Since in all vertices have degree , the sets form a partition of .
For each inner face of , denote by the subgraph of consisting of the vertices of , and of the vertices and edges of that lie in the interior of . Refer to Figure 4 for an illustration. Since is planar and is a crossing, there exists no crossing between a segment in and a segment not in . Therefore, the boundary of corresponds to the boundary of a topological disk such that is planarly embedded inside : only the vertices of lie on the boundary of . For the external face , graph consists of the vertices of , and of the vertices and edges of that lie outside . In this case, the topological disk is obtained after a suitable inversion of , if needed. We can rearrange and deform each such that: the part of the boundary of that contains all the vertices of lies on a circle ; for each face of , disks and do not intersect; the interior of circle is empty. Then, the edges of are redrawn as straightline segments inside . This construction implies that is untangled (and some of its edges may not cross anymore). Also, each subgraph remains topologically equivalent to its initial drawing. Thus, two edges cross after the transformation only if they crossed before, which ensures that the resulting graph is simple and no new crossing is created.
We iteratively apply the above transformation to each subgraph of the new topological graph until all crossings are untangled. This concludes the proof.∎
An illustration of the untangling procedure described in the proof of Lemma 2 is given in Figure 4. Figure 4 shows an example of a planar simple topological graph with a tangled crossing ; the edges of are thicker. Faces , , and are the three faces of whose union contains the vertices of . Subgraphs , , and are schematically depicted. Figure 4 shows after the transformation that untangles .
5 Removing Untangled Crossings
Let be a planar simple topological graph with . By Lemma 2, we may assume that has no tangled crossings. In Section 5.1, we show how to transform into a (possibly not almost simple) quasi planar topological graph . Then, in Section 5.2, we describe how to make simple without introducing crossings.
5.1 Obtaining quasi planarity
We first show the existence of a global rerouting such that no two edges of are rerouted around the same vertex (Lemma 5). Note that this is a necessary condition for almost simplicity; see Figure 5. Then, we show that any global rerouting of with this property yields a topological graph with no crossings (Lemma 9).
The existence of this global rerouting is proved by defining a bipartite graph composed of the vertices of and of its crossings, and by showing that a matching covering all the crossings always exists. A bipartite graph with vertex sets and is denoted by . A matching from into is a set such that each vertex in is incident to exactly one edge in and each vertex in is incident to at most one edge in . For a subset , we denote by the set of all vertices in that are adjacent to a vertex in . We recall that, by Hall’s theorem, graph has a matching from into if and only if, for each set , it is . Let be a planar simple topological graph and let be the set of crossings of . We define a bipartite graph as follows. For each crossing , set contains a vertex and set contains the endpoints of (that is, ). Also, set contains an edge between a vertex and a vertex if and only if . We have the following.
Lemma 3
Graph is a simple bipartite planar graph. Also, each vertex in has degree .
Proof
The graph is simple and bipartite, by construction. Also, for each crossing , vertex is incident to the vertices in belonging to .
We prove that is also planar by showing that a planar embedding of can be obtained from as follows. First, we remove from all the vertices and edges that are not in any crossing. Then, for each crossing of , we remove the portion of in the interior of a topological disk that encloses all crossing points of and such that each edge in crosses the boundary of exactly twice (as defined in Section 3) and add vertex inside . Finally, for each vertex in , let be the edge in incident to and let be the intersection point between and in . We complete the drawing of edge by adding a curve between and in the interior of without introducing any crossings. The resulting topological graph is planar.∎
Lemma 4
Graph has a matching from into .
Proof
Lemma 5
Let be a planar simple topological graph. It is possible to execute a global rerouting on such that no two edges are rerouted around the same vertex.
Proof
Let , with , be the set of crossings of . By Lemma 4, it is possible to assign a vertex to each crossing in such a way that no two distinct crossings are assigned the same vertex. The statement follows by considering a global rerouting such that, for each crossing , any edge in not incident to is rerouted around .∎
Denote by the topological graph obtained from by executing a global rerouting as in Lemma 5. We prove that has no crossings. To this aim, we first give the conditions under which new crossings may arise in (Lemmas 6–8).
Lemma 6
Let and be two edges that cross in but not in . Then, one of and has been rerouted around an endpoint of the other.
Proof
Since and do not cross in , we may assume that one of them, say , has been rerouted. Suppose first that the hook of crosses . We claim that has been rerouted around an endpoint of . In fact, if has not been rerouted, then the claim is trivially true; see Figure 5. On the other hand, if has been rerouted, then the crossing with must be on a tip of , and not on its hook, since no two edges have been rerouted around the same vertex in the global rerouting; see Figure 5. Thus, the claim follows. Suppose now that a tip of crosses . Then, this crossing must be with the hook of , and the same argument applies to prove that has been rerouted around an endpoint of .∎
Lemma 7
Every nonrerouted edge is crossed by at most three rerouted edges in . Further, if is crossed by exactly three rerouted edges, then two of them have been rerouted around distinct endpoints of .
Proof
Since at most one edge has been rerouted around each vertex, by construction, it suffices to prove that there exists at most one rerouted edge crossing that has not been rerouted around an endpoint of .
For this, note that any edge with this property crosses also in , by Lemma 6, and thus it belongs to the same crossing as . Since, by construction, only one edge per crossing is rerouted, the statement follows.∎
Lemma 8
If contains a crossing , then contains at most one edge that has not been rerouted.
Proof
Assume to the contrary that a crossing in contains at least two nonrerouted edges and .
We first claim that there exists an edge of that does not cross in . If has less than crossings in , then the claim trivially follows. If has crossings in but it is not part of a crossing, then none of the edges crossing in can be part of a crossing in , as otherwise they would have crossings in the crossing and an additional crossing with . Thus, the claim follows also in this case. Finally, if is part of a crossing in , the claim follows from the fact that the edges of do not form a crossing in , due to a rerouting of one of its edges.
Thus, edge has been rerouted around an endpoint of , by Lemma 6, which means that is part of a crossing in containing neither nor . Hence, and do not cross in . We prove that they do not cross in either, a contradiction to the assumption that is a crossing. Namely, by Lemma 1, all new crossings of with nonrerouted edges are on its hook; however, since and cross in (which is simple), they do not share any endpoint, and the statement follows.∎
Altogether the above lemmas can be used to prove the following.
Lemma 9
Graph does not contain any crossing.
Proof
Assume for a contradiction that contains a crossing . By Lemma 8, contains at most one nonrerouted edge.
Suppose that contains one of such edges . By Lemma 7, there are at most three rerouted edges crossing in . If they are less than , then the claim follows, as . If they are three, say , then by Lemma 7, two of them, say and , have been rerouted around (distinct) endpoints of . Thus, and do not cross in , by Observation 2.1, as they belong to different crossings. Hence, they can cross in only if one of them has been rerouted around an endpoint of the other, by Lemma 6. This is not possible since neither nor share an endpoint with , as is simple.
Suppose that contains only rerouted edges. Let be any edge of and let be the vertex used for rerouting . Since at most one edge in can be incident to and since , there are two edges in that have been rerouted around distinct endpoints of . As in the previous case, we can prove that and do not cross.∎
5.2 Obtaining simplicity
Lemmas 2, 5, and 9 imply that, for , any planar simple topological graph can be redrawn such that the resulting topological graph contains no crossings and no two edges are rerouted around the same vertex. However, the graph may be not simple, and even not almost simple. We first show how to remove from pairs of edges crossing more than once, without introducing crossings, thus resulting in a quasi planar almostsimple graph (Lemma 11). Then we show how to remove crossings between edges incident to a common vertex, still without introducing crossings (Lemma 12). We will use the following auxiliary lemma.
Lemma 10
Graph is almostsimple if and only if there is no pair of edges such that each of them is rerouted around an endpoint of the other.
Proof
Clearly the condition is necessary for almost simplicity; see Figure 6.
For the sufficiency, suppose that there exist two edges and crossing twice in . By Lemma 6 and by the simplicity of , at least one of them, say , has been rerouted. By Lemma 1, this rerouting did not introduce two crossings between and , if has not been rerouted, since the rerouting of a single edge leaves the graph almost simple. Thus, we may assume that also has been rerouted. This implies that and belong to different crossings of ; so, they do not cross in , by Observation 2.1. Hence, by Lemma 6, at least one of them has been rerouted around an endpoint of the other, say around . This introduces a single crossing between and , namely between the hook of and a tip of . Thus, the other crossing must be between the hook of and a tip of , again by Lemma 6 and by the fact that in no two edges are rerouted around the same vertex, and so there is no crossing between the hooks of two edges. The statement follows.∎
Lemma 11
There is a quasi planar almost simple topological graph .
Proof
We may assume that is not almost simple, as otherwise the statement would follow with . By Lemma 10, there exist pairs of edges in which each of the two edges has been rerouted around an endpoint of the other; see Figure 6. For each pair , we remove the two crossings by redrawing one of the two edges, say , by following between the two crossings. More precisely, we redraw the tip of crossed by the hook of by following the tip of crossed by the hook of , without crossing it; see Figure 6. In the following we prove that the graph obtained by applying this operation for all the pairs does not contain new crossings and is almost simple.
Observe first that each edge tip is involved in at most one pair, since no two edges are rerouted around the same vertex. Thus, no tip of an edge is transformed twice in and no two transformed edges cross each other. Hence, if a crossing exists in , then it contains exactly one transformed edge. We prove that this is not the possible.
Let be an edge that has been redrawn due to a double crossing with an edge , and let and be the crossings of containing and , respectively. The edges crossing in are (see Figure 6): (i) a set of edges in crossing the tip of that has been used to redraw ; (ii) a set of edges in crossing the tip of not crossed by ; (iii) a set of edges incident to the vertex around which has been rerouted (and thus cross the hook of ). Note that contains all the edges that cross in and not in . These edges do not cross those in , since they are nonrerouted edges belonging to distinct crossings of . Also, they do not cross edges in , since does not contain any edge incident to , other than . Finally, the edges in are at most , since contains edges and at least two of them do not cross , namely and the edge incident to the endpoint of around which has been rerouted. Thus, the edges in are not involved in any crossing with . To see that the same holds for the edges in and in , note that any crossing in involving these edges and would also appear in , which is however quasi planar.
To prove that is almost simple, we show that the edges in are crossed only once by . Recall that none of these edges crosses in . Also, since each tip is involved in at most one transformation, all the edges in cross the tip of (and hence the tip of that has been redrawn) only once. On the other hand, it could be that also the other tip of has been transformed by following the tip of an edge and that this transformation introduced a new crossing between and . But then crosses both and in , and hence by Lemma 6 also in . This is however not possible, since both and have been rerouted, and hence they belong to different crossings in .∎
Lemma 12
There is a quasi planar simple topological graph .
Proof
We may assume that is not simple, as otherwise the statement would follow with . Let and be two crossing edges that share an endpoint . Since is simple, at least one of them has been redrawn, say .
We distinguish two cases, based on whether has rerouted but not transformed afterwards, or has also been transformed, due to a double crossing.
In case , edge crosses with its hook, see Figure 7. We redraw by following till reaching , as in Figure 7. This guarantees that and no longer cross and that does not cross any edge twice, since crosses only edges that cross the tip of incident to . Also, no crossing is introduced. Indeed, any new crossing should contain , but then also would be part of this crossing, which is impossible since and do not cross and does not contain crossings.
In case , let be the edge that crosses twice in ; note that has not been transformed, see Figure 7. Recall that, by Lemma 10, and have been rerouted one around an endpoint of the other. Suppose that the endpoint of around which has been rerouted is , the case in which it is can be treated analogously. This implies that crosses a tip of , and therefore and are part of a crossing in , namely the one that caused the rerouting of . We redraw the part of from to its crossing point with by following , without crossing it, and leave the rest of unchanged, as in Figure 7. This guarantees that and do no cross any longer, and that any new crossing of is with an edge that also crosses . As in case , this implies that does not cross any edge twice and that no new crossing has been generated.∎
The next theorem summarizes the main result of the paper.
Theorem 5.1
Let be a planar simple topological graph. Then, there exists a quasi planar simple topological graph such that .
Proof
First recall that, by Lemma 2, we can assume that does not contain any tangled crossing. We apply Lemma 5 to compute a global rerouting for in which no two edges are rerouted around the same vertex. By Lemma 9, the resulting topological graph is quasi planar. Also, by Lemma 11, if is not almost simple, then it is possible to redraw some of its edges in such a way that the resulting topological graph is almost simple and remains quasi planar. Finally, by Lemma 12, if is not simple, then it can be made so, again by redrawing some of its edges, while maintaining quasiplanarity. This concludes the proof that there exists a quasi planar simple topological graph .∎
6 Conclusions and Open Problems
We proved that, for any , the family of planar graphs is included in the family of quasi planar graphs. This result represents the first nontrivial relationship between the planar and the quasi planar graph hierarchies, and contributes to the literature that studies the connection between different families of beyond planar graphs (see, e.g. [8, 9, 11, 18]). Several interesting problems remain open. Among them:

The main open question is whether planar graphs are quasi planar. The reason why our technique does not apply to the case of is mainly due to the possible existence of three rerouted edges that are pairwise crossing after a global rerouting (as in Figure 5). A conceivable approach to overcome this issue is by matching more than one vertex to each crossing, in order to execute a global rerouting that does not create forbidden configurations. In fact, within the lines of Lemma 4, we proved that up to vertices can be reserved for each crossing. Nonetheless, it is not clear how to control which vertices of a crossing are assigned to it in the matching, which makes it difficult to exploit these extra vertices. Note that it was recently proved that optimal 2planar graphs are ()quasi planar [7]. We recall that an vertex planar graph is optimal if it has edges.

For , one can also ask whether the family of planar graphs is included in the family of quasi planar graphs. For the answer is trivially negative, as quasi planar graphs coincide with the planar graphs. On the other hand, optimal 3planar graphs are known to be ()quasi planar [7]. We recall that an vertex planar graph is optimal if it has edges. For sufficiently large values of , one can even investigate whether every planar simple topological (sparse) graph is quasi planar, for some function .

One can study noninclusion relationships between the planar and the quasi planar graph hierarchies, other than those that are easily derivable from the known edge density results. For example, for any given , can we establish an integer function such that some planar graph is not quasi planar?
Acknowledgements.
The research in this paper started at the Dagstuhl Seminar 16452 “BeyondPlanar Graphs: Algorithmics and Combinatorics”. We thank all participants, and in particular Pavel Valtr and Raimund Seidel, for useful discussions on the topic.
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