1 Introduction
Visualization of nonplanar graphs is one of the most studied graphdrawing problems in recent years. In this context, an emerging topic is hybrid representations (see, e.g., [DBLP:conf/gd/AngeliniLBFPR15, Batagelj_Visual_2011, DBLP:conf/gd/LozzoBFP16, dlptntptsc17, hfmdhvsn07]). A hybrid representation simplifies the visual analysis of a nonplanar graph by adopting different visualization paradigms for different portions of the graph. The graph is divided into (typically dense) subgraphs called clusters which are restricted to limited regions of the plane. Edges between vertices in the same cluster are called intracluster edges, and edges between vertices in different clusters are called intercluster edges. Intercluster edges are represented according to the classical nodelink graph drawing paradigm, while the clusters and their intracluster edges are represented by adopting alternative paradigms. A hybrid representation thus reduces the number of intercluster edges and the visual complexity of much of the drawing at the cost of creating cluster regions of high visual complexity. As a result, a hybrid representation provides an easy to read overview of the graph structure and it admits a “drilldown” approach when a more detailed analysis of some of its clusters is needed.
Different representation paradigms for clusters give rise to different types of hybrid representations. For example, Angelini et al. [DBLP:conf/gd/AngeliniLBFPR15] introduce intersectionlink representations, where clusters are represented as intersection graphs of sets of rectangles, while Henry et al. [hfmdhvsn07] introduce NodeTrix representations, where dense subgraphs are represented as adjacency matrices (see Fig. 1). Batagelj et al. employ hybrid representations in the clustering model [Batagelj_Visual_2011], where and define the desired topological properties of the clusters and of the graph connecting the clusters, respectively. For instance, in a clustering of a graph each cluster is a clique and the graph obtained by contracting each cluster into a single node (called the graph of clusters) is planar. Given a graph and a hybrid representation paradigm , the hybrid planarity problem asks whether can be represented according to with no intercluster edge crossings. Variants of the problem may or may not assume that the clustering is given as part of the input.
In this paper, we present a general hybrid representation paradigm that relaxes the described hybrid paradigms. Given a graph , a (k,p) representation of is a hybrid representation in which: (i) each cluster of contains at most vertices and is identified with a closed, bounded planar region; (ii) cluster regions are pairwise disjoint, (iii) each vertex is represented by at most distinct points, called ports, on the boundary of its cluster region; (iv) each intercluster edge is represented by a Jordan arc connecting a port of to a port of . A representation is planar if edge curves do not cross and do not intersect cluster regions except at their endpoints. We say that a graph is planar if it can be clustered so that it admits a planar representation.
The definition of a representation leaves the representation of clusters and intracluster edges intentionally unspecified. It is thus a relaxation of hybrid representation paradigms where the number of ports used by the intercluster edges depends on the geometry of the cluster regions. For example, in a NodeTrix representation, the squared boundary of each matrix allows four ports for every vertex except for the vertex in the first row/column of the matrix and the vertex in the last row/column of the matrix, which both have only three ports. Hence, a NodeTrix representation can be regarded as a constrained representation (four ports for every vertex except for two, the vertices appear in the order imposed by the matrix); see Fig. 1. Similarly, a representation relaxes an intersectionlink representation with clusters represented as isothetic unit squares with their upperleft corners along a common line with slope ; see Fig. 1. We also remark that the use of different ports to represent a vertex can be regarded as an example of vertex splitting [DBLP:conf/gd/EadesM95, Eppstein2018]; however, while in the papers that use vertex splitting to remove crossings the multiple copies of each vertex can be placed anywhere in the drawing, in our model they are forced to lay within the boundary of the same cluster region.
The results of this paper are the following:

In Section 2, we give an upper bound on the edge density of a planar graph and prove that this bound is tight for .

In Section 3, we observe that the class of planar graphs coincides with the class of ICplanar graphs, from which the NPcompleteness of testing planarity follows. We then prove that testing planarity is NPcomplete. These results imply that computing the minimum such that a graph is planar is NPhard for both and . Recall that a graph is 1planar if it admits a drawing where every edge is crossed at most once, and that an ICplanar graph is a planar graph that admits a drawing where no two pairs of crossing edges share a vertex.

The NPcompleteness of the planarity testing problem naturally suggests to further investigate the combinatorial properties of planar graphs. In Section 4, we ask whether every planar graph admits a planar representation (see, e.g. Fig. 6). We prove the existence of planar graphs that are not planar and of planar graphs that are not planar. We also give a sufficient condition for planar graphs to be planar.
Sections of certain proofs are removed to the appendix. These statements are marked with [*].
2 Edge Density of Planar Graphs
In this section we give a tight bound on the number of edges of a planar graph when . First, given a planar representation , we define a skeleton of to be a planar drawing obtained by the following transformation. We first replace each port in with a vertex. Each cluster region of is now an empty convex space surrounded by up to vertices. We connect these vertices in a cycle and triangulate the interior. For our purposes any triangulation is equivalent. The resulting representation is . Figure 2 illustrates a skeleton of the planar representation of Fig. 2.
Theorem 2.1
[*] Let be a planar graph with vertices. has edges. This bound is tight for any positive integers , and such that and , where .
Proof
Let be a planar representation of and let be the number of clusters of . As each cluster contains at most vertices, has at most intracluster edges.
Let be a cluster region in with ports in total. Let be a skeleton of , and let and denote the number of vertices and the number of edges of , respectively. When is created, is replaced with vertices and edges if , or 0 edges if . Letting be the number of intercluster edges in and be the number of clusters in containing a single vertex, we have,
(1) 
In other words, the total number of edges in is equal to the number of intercluster edges in plus the number of edges added for each cluster. Note that , as is a planar drawing. As , rearranging generates and thus,
(2) 
As is equal to the sum of the number of intercluster and intracluster edges in , we have
(3) 
If all clusters contain vertices, then and Theorem 1 holds. Appendix 0.A completes the proof that in the case where some clusters contain fewer than vertices.
In order to show that the bound is tight for , we describe a planar representation with clusters and intercluster edges. is possible for any pair of positive integers and such that and for any . has clusters each with vertices and thus ports. Let and be two cluster regions. We say that and are kpconnected if they are connected by edges as shown in Fig. 3(a). (Note that, since the number of intercluster edges between two clusters is at most , we can create edges between and only if ). More precisely, , which we refer to as the small end of the connection, is connected by means of consecutive ports; the first ports have incident edges each, and the last port has an additional edge. , which we refer to as the large end of the connection, is connected by means of consecutive ports, each connected to one or two edges. Notice that, since we use ports for the large end, for the small end and two ports can be shared by the two ends, each cluster region can be the small end of one connection and the large end of another connection. Thus, we can create a cycle with clusters as shown in Fig. 3(b). In the resulting representation there are two faces of degree : One is the outer face and the other one is inside the cycle. By triangulating these two faces with edges for each face, we obtain the representation . The number of intercluster edges of is thus .
3 Recognition of Planar Graphs
This section considers the problem of testing planarity for the cases in which and .
Theorem 3.1 ()
[*] planarity testing can be performed in linear time for , and it is NPcomplete for .
Proof
The first part of Theorem 3.1 follows from the fact that the class of planar graphs coincides with the class of planar graphs for . The second part follows from the fact that the planar graphs coincide with the ICplanar graphs [Zhang2013]. Testing ICplanarity is known to be NPcomplete [DBLP:journals/tcs/BrandenburgDEKL16]. Appendix B proves both equivalencies.
Corollary 1
The problem of computing the minimum value of such that a graph is planar is NPhard.
We now focus on the planarity testing problem, hereafter referred to as Planarity. We show that Planarity is NPcomplete by a reduction from the NPcomplete problem Planar Monotone 3SAT [DBLP:journals/ijcga/BergK12]. We say that an instance of 3SAT is monotone if every clause consists solely of positive literals (a positive clause) or solely of negative literals (a negative clause). A rectilinear representation of a 3SAT instance is a planar drawing where each variable and clause is represented by a rectangle, all the variable rectangles are drawn along a horizontal line, and vertical segments connect clauses with their constituent variables. A rectilinear representation is monotone if it corresponds to a monotone instance of planar 3SAT where positive clauses are drawn above the variables and negative clauses are drawn below the variables, as shown in Fig. 4. Given a monotone rectilinear representation corresponding to a boolean formula , the problem Planar Monotone 3SAT asks if has a satisfying assignment.
We denote by the graph created by removing two adjacent edges from the complete graph . In our reduction we make use of the following transformation. Let be a vertex of . we replace with a copy of by identifying with the vertex of with degree . After performing this operation we say that is a vertex. The following lemma, whose proof is in Appendix 0.C, states a useful property of the vertices.
Lemma 1 ()
[*] Let be a vertex of a graph and let be the subgraph associated with . In any planar representation of , each vertex of is clustered with another vertex in .
Theorem 3.2 ()
[*] Planarity is NPcomplete.
Proof
Planarity is trivially in NP. We prove the NPhardness of Planarity by reduction from Planar Monotone 3SAT. Given an instance of Planar Monotone 3SAT, we construct a graph that is a Yes instance of Planarity if and only if is a Yes instance of Planar Monotone 3SAT. For convenience, figures show the construction of the graph corresponding to the Planar Monotone 3SAT instance in Fig. 4. In figures, we represent vertices and their associated subgraphs with solid dots, while ordinary vertices are represented with hollow dots.
For each variable of (with ) create in a vertex and connect such vertices in a cycle, in the order implied by (refer to Fig. 4). Split each edge (, ) of the cycle with a vertex . Split the edge (, ) with the vertices and . Finally, duplicate the edge and split the duplicated edges with the vertices and . We refer to this subgraph as the variable cycle. Given a variable , let be the number of positive clauses and be the number of negative clauses of in which appears. For , connect to with a path of ordinary vertices of length equal to . We refer to these paths as false literal boundaries.
For each clause in , create a corresponding clause gadget in . Create ordinary vertices , , and , create a vertex , and add an edge between any pair of vertices, as in Fig. 5. Observe that in any planar representation of a clause gadget, two of the four vertices and must be arranged in one cluster of size . This is due to the fact that by Lemma 1, must be clustered within its subgraph. If and were all clustered separately, the graph of clusters of would contain a minor. Also, any clustering of a clause gadget in which a literal vertex is clustered with is planar, as shown in Fig. 5.
Now, connect the clause gadgets with a tree structure corresponding to the positions of clause rectangles in . Let be a clause rectangle in with , , and corresponding to the vertical segments descending from from left to right. If is nested between vertical segments corresponding to literals and of another clause rectangle , split the edges and with vertices and connect the new vertices with an edge. If is nested under no other clause rectangle, split with a vertex and connect the new vertex to if corresponds to a positive clause and to otherwise. This procedure leads to a configuration consisting of two trees of clause gadgets connected as in Fig. 5. This concludes the construction of . Appendix 0.D proves that is planar if and only if has a satisfying assignment.
Corollary 2
The problem of computing the minimum value of such that a graph is planar is NPhard.
4 Planarity and Planarity
The NPcompleteness of Planarity suggests further investigation into the combinatorial properties of planar graphs. In this section, we study the relationship between planarity and planarity. This is partly motivated by general interest in planar graphs (see, e.g., [DBLP:journals/csr/KobourovLM17]) and partly by the following observation. Since a planar graph admits a drawing where each edge is crossed by at most one other edge, it seems reasonable to remove each crossing of the drawing by clustering two of the vertices that are involved in the crossing as shown in Fig. 6. An vertex planar graph has at most edges [DBLP:journals/combinatorica/PachT97]. By Theorem 2.1, a planar graph with vertices has at most edges, so it is not immediately clear that there are planar graphs that are not planar.
As we are going to show, however, there is an infinite family of planar graphs that are not planar for any value of . On the positive side, we demonstrate a large family of planar graphs that are planar.
Theorem 4.1
For every , there exists a planar graph with vertices and edges that is not planar, for any .
Proof
We define a recursive family of plane graphs as follows. Graph consists of a single kite , which is a plane graph isomorphic to drawn so that all the vertices are on the boundary of the outer face. Graph , for , has kites in addition to ; these kites form a cycle in the outer face of , and each kite contains a vertex of the boundary of the outer face of (note that has vertices on the boundary of the outer face). See Fig. 7 for an example. The kites of are called the external kites of . The embedding of described in the definition will be called the canonical embedding of . We also consider another possible embedding, called the reversed embedding. Let be the boundary of the outer face in the canonical embedding of ; in the reversed embedding of the cycle is the boundary of an inner face and all the rest of the graph is embedded outside . See Fig. 7 and Fig. 7 for an example. For any , let be a copy of with a canonical embedding, and let be a copy of with a reversed embedding. The graph obtained by identifying the external kites of with the external kites of is denoted as . Fig. 7 shows the graph . By construction, has vertices and edges. Hence, has vertices and edges.
We show that is not planar for any . Suppose that has a planar representation for some and let be the graph of clusters of . Since is planar, must be planar. can be obtained from by contracting each pair of vertices that is assigned to each cluster region (and removing multiple edges). Contracting a pair of vertices and , the number of vertices reduces by one and the number of edges reduces by the number of paths of length at most connecting and (for each path we remove one edge). In , there are at most such paths between any pair of vertices. Hence, if we contract pairs of vertices, the number of vertices in is , while the number of edges is . If is planar, and thus it must be , which gives , i.e. we must contract at least pairs of vertices. Since there are vertices, we can contract at most pairs. Thus, it must be , i.e., , which can be satisfied only for .
Note that our argument is independent of the planar embedding of . This implies that the result holds for planar graphs, not just for plane graphs.
Theorem 4.1 motivates further investigation of the relationship between planar and planar graphs. Note that there are infinitely many planar graphs that are not planar. For example, observe that every graph obtained by connecting with an edge a planar graph and has such a property, because is not planar (it has more than edges) but it is planar, as depicted in Fig. 8.
In what follows, we describe a nontrivial family of planar graphs that are also planar.
Let be a plane graph, and let and be a pair of crossing edges of . Any pair , with , is a representative pair of the edge crossing defined by . An independent set of distinct representatives (ISDR for short) of is a set of representative pairs such that there is exactly one representative pair per crossing and no two representative pairs in the set have a common vertex. Fig. 9 shows an ISDR for the graph of Fig. 9.
We want to show that if a plane graph has an ISDR then it is planar. The crossing edges graph of , called graph for short and denoted as , is the subgraph of induced by the crossing pairs of . is pseudoforestal if is a pseudoforest (i.e. it has at most one cycle in each connected component). For example, the planar graph of Fig. 9 is pseudoforestal, as shown in Fig. 9. The pseudoforestal planar graphs include nontrivial subfamilies of planar graphs, such as ICplanar graphs (whose graph has maximum degree one), or the planar graphs such that each vertex is shared by at most two crossing pairs (whose graph has maximum degree two).
Theorem 4.2
A pseudoforestal plane graph is planar.
Proof
We start by proving that a plane graph contains an ISDR if and only if is pseudoforestal. It is known that a graph can be oriented such that the maximum indegree is if and only if its pseudoarboricity is (i.e. the edges of can be partitioned into pseudoforests) [DBLP:conf/isaac/Kowalik06]. Thus, is pseudoforestal if and only if can be oriented so that the maximum indegree is one. We now show that this is a necessary and sufficient condition for the existence of an ISDR in . Assume that an ISDR exists. Let and be two crossing edges and let () be the representative pair of and . Direct towards and towards . Doing this for each pair of crossing edges defines an orientation for all edges of . In this orientation each vertex of has indegree at most , since no two pairs in share a vertex. Now suppose that has an orientation such that each vertex has indegree at most . For each pair of directed crossing edges in , we add the pair to . Since each vertex in has indegree at most , is a vertex of at most one pair in . Thus, the pairs selected for different crossing pairs are distinct and no two of them share a vertex.
We now describe how to use an ISDR of to construct a planar representation of where each pair in is represented as a cluster that has copies for each of its vertices. Let be a planar drawing of that respects the planar embedding of . Consider any two crossing edges and and denote by the point where they cross in . Without loss of generality, assume that is the representative pair of and (see Fig. 10 for an illustration). Subdivide the edge with a copy of placed between and along ; analogously, subdivide the edge with a copy of . Add a curve connecting to and a curve connecting to . By walking very close to the two edges and , these two curves can be drawn without crossing any existing edge and so that the closed curve formed by and together with the portion of from to and the portion of from to does not contain any vertex of . Curve defines the cluster region for the cluster containing and . Replace the edge with a curve connecting to and the edge with a curve connecting to . Again, by walking very close to the two edges and , and can be drawn without crossing existing edges and without crossing each other. The replacements of with and of with remove the crossing between and . Repeating the described procedure for every pair of crossing edges, all crossings are removed. Since for each pair of crossing edges there is a distinct representative pair and no two pair share a vertex, the result is a planar representation of .
5 Open Problems
The results in this paper suggest the following open problems: (i) Tightly bound the edge density of planar graphs for ; (ii) Study the complexity of planarity testing for larger values of and ; (iii) Further study the relationship between planar graphs and planar graphs.
Acknowledgements
The authors wish to thank Maurizio Patrignani for useful discussions.
References
Appendix
Appendix 0.A Supplement for Proof of Theorem 2.1
In this section, we complete the proof of Theorem 2.1 in the case where some clusters contain fewer than vertices. Let be a planar graph, a planar representation of , and the number of clusters of . In Section 2 we showed that if all clusters contain exactly vertices.
Denote by the clusters of and let be the size of cluster . We first add noncrossing intercluster edges so that the faces of external to the cluster regions are triangles. Let be the resulting planar representation. Notice that can have multiple edges. We then construct a sequence of planar representations so that has all clusters of size and each is obtained from by taking into account the cluster . We denote by and the number of vertices and edges of , respectively. If cluster is not modified and we set . If we remove the single vertex in and we triangulate the face that is created by this removal (see Fig. 11 and Fig.11). Also in this case multiple edges can be introduced. The number of vertices of is then , while the number of edges is . If , we augment the cluster with dummy vertices, we add ports in between two consecutive ports associated with two different vertices of (see Fig. 11 and Fig. 11). We then add edges internally to and edges externally to to triangulate the face enlarged by the insertions (again multiple edges can be created). The number of vertices of is , while the number of edges of is . We now prove the following claim that together with the fact that (because has all clusters of size ) implies that . Since and , the statement follows.
Claim 1
If then .
Clearly nothing has to be proven for . If , we have which gives . In order to prove that Claim 1 holds in this case, we show that , which can be rewritten as . Since we have , which is greater than or equal to for any integer value of . Consider now the case ; notice that this case is possible only for . We have , which in turn gives . Again, we prove that . Rearranging, we obtain ; since , we have , which holds for every .
Appendix 0.B Supplement for Proof of Theorem 3.1
In this section, we complete the proof of Theorem 3.1 by showing that the class of planar graphs coincides with the class of planar graphs for and that the class of planar graphs coincides with the class of ICplanar graphs.
If is planar, is trivially planar for all positive integers . Let be a planar graph for some , and let be a planar representation of . Replace each cluster of of size with an clique. Since the obtained drawing is planar.
Recall that an ICplanar graph admits a 1planar embedding in which no two pairs of crossing edges share a vertex. Let be an ICplanar graph, and let be an ICplanar embedding of . can be transformed into a planar representation of by replacing the vertices incident to each pair of crossing edges with a cluster.
Let be a planar graph and let be a planar representation of . Each cluster of is a subgraph of a clique and therefore each cluster region in can be replaced with a drawing that contains at most one pair of crossing edges. As contains no crossing intercluster edges, the resulting embedding is ICplanar.
Appendix 0.C Proof of Lemma 1
See 1
Proof
Suppose there exists a planar representation of that leaves unclustered or clustered with a vertex outside of . If the remaining vertices of the subgraph are grouped into at least five clusters, does not admit a planar representation because its graph of clusters includes a subgraph.
Alternatively, suppose the remaining vertices of are grouped into four clusters, in which case consists of three clusters and two vertices which may or may not be clustered with additional vertices outside of . For the purpose of our analysis, we may ignore any vertices outside of , as their presence cannot affect the possibility of a planar representation of .
Each cluster can contain at most intracluster edge, so any planar representation of has intercluster edges. However, by Equation 2, we have that in any planar representation of a graph , where is the number of clusters consisting of a single vertex and is the total number of vertices in the skeleton of . When applied to , Equation 2 implies that , a contradiction. Thus any planar representation of creates four clusters as shown in Fig. 12.
Appendix 0.D Supplement for Proof of Theorem 3.2
In this section, we complete the proof of Theorem 3.2 by proving that our constructed graph is planar if and only if the corresponding instance of Planar Monotone 3SAT is a Yes instance.
Let be a Yes instance of Planar Monotone 3SAT, and let be an assignment function satisfying . We show that the graph corresponding to is planar by constructing a planar representation of using as a template.
Replace each variable rectangle in with the corresponding vertex of and draw the variable cycle. We refer to the region defined by the variable cycle and the () vertex as the positive side (negative side). For each variable , draw its false literal boundary on the negative side if and on the positive side if . Fig. 12 illustrates a drawing of the variable cycle and false literal boundaries of according to the assignment of and to and and to .
Let be the literal vertex corresponding to clause and variable . Place at the point of intersection between the rectangle associated with and the vertical segment connecting the rectangles and .
Connect the three literal vertices of to form a face, and insert and on the interior, creating one necessary crossing. Insert the tree structure edges, which by construction can be added without creating crossings. Connect literal vertices to variable vertices, which creates a crossing on a false literal boundary precisely when the value assigned to a variable by does not match the literal. Fig. 13 illustrates such a drawing of .
Resolve each crossing at a false literal boundary by clustering the literal vertex with a vertex on the boundary. The specification that each false literal boundary has at least vertices ensures that this operation can be performed. Because satisfies , each clause gadget has at least one literal vertex that can be connected to its variable vertex without crossing a false literal boundary. Cluster this vertex with to resolve each clause gadget crossing. The result of this process is a representation of as illustrated in Fig. 13.
Let be a Yes instance of Planarity corresponding to an instance of Planar Monotone 3SAT. We show that is a Yes instance of Planar Monotone 3SAT.
Let be a planar representation of . First, note that any vertices and connected by an edge in must be drawn on the same side of the variable cycle in any planar representation of . This follows from Lemma 1, as neither nor can be clustered with any Kvertex in the variable cycle. Thus the positive (negative) clause gadgets must all be drawn on the same side of the variable cycle as they are connected by the tree structure to () and the variable vertices. We refer to the sides of the cycle with the positive and negative clause gadgets as the and sides of the cycle. As a consequence of Lemma 1, each false literal boundary is drawn either on the positive or on the negative side of the cycle as well.
Define an assignment function by setting to () if the false literal boundary for is drawn on the () side of the vertex cycle in . We claim that at least one literal vertex of each positive (negative) clause gadget is connected in to a variable vertex with set to True (False).
Without loss of generality, consider the case of a positive clause gadget with literals , , and connected to variables , , and . Assume for contradiction that every literal vertex of is connected in to a variable with , which means that the false literal boundaries of , , and are drawn on the positive side of the variable cycle. We show that any placement of the Kvertex creates an edge crossing in , contradicting our assumption.
Suppose first that is placed outside the false literal boundaries. Then each of , , and must be clustered with a boundary vertex and the clause gadget does not admit a planar representation (see Fig. 14). Now suppose that is drawn inside the false literal boundary of one constituent variable, for example. In this case, the path (, , ) intersects two false literal boundaries. Because and are vertices, only can be clustered with a false literal boundary vertex and thus this placement creates at least one necessary crossing (see Fig. 14). Likewise, suppose that is drawn inside the false literal boundary of two constituent variables, for example, and . In this case, the path (, , ) crosses three false literal boundaries and creates a necessary crossing (see Fig. 14). Finally, suppose that is drawn inside all three false literal boundaries (see Fig. 14). In this case, the path (, , ) crosses two false literal boundaries and creates a necessary crossing. Thus, regardless of the position of the a vertex in , at least one of the literal vertices of must match the assignment of its associated variable vertex. This concludes the proof of our claim, i.e., that at least one literal vertex of each clause gadget in is connected to a variable with . Thus is a satisfying assignment for , and is a Yes instance of Planar Monotone 3SAT.
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