Testing Maximal 1-planarity of Graphs with a Rotation System in Linear Time

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1 Testing Maximal 1-planarity of Graphs with a Rotation System in Linear Time Peter Eades 1, Seok-Hee Hong 1, Naoki Katoh 2, Giuseppe Liotta 3, Pascal Schweitzer 4, and Yusuke Suzuki 5 1 University of Sydney, Australia {peter,shhong}@it.usyd.edu.au 2 Kyoto University, Japan naoki@archi.kyoto-u.ac.jp 3 Università di Perugia, Italy liotta@diei.unipg.it 4 Australian National University, Australia pascal.schweitzer@anu.edu.au 5 Niigata University, Japan y-suzuki@math.sc.niigata-u.ac.jp Abstract. A 1-planar graph is a graph that can be embedded in the plane with at most one crossing per edge. A 1-planar graph on n vertices can have at most 4n 8 edges. It is known that testing 1-planarity of a graph is NP-complete. A 1-planar embedding of a graph G is maximal, if no edge can be added without violating the 1-planarity of G. In this paper, we study combinatorial properties of maximal 1-planar embeddings. In particular, we show that in a maximal 1- planar embedding, the graph induced by the non-crossing edges is spanning and biconnected. Using the properties, we show that the problem of testing maximal 1-planarity of a graph G can be solved in linear time, if a rotation system Φ (i.e., the circular ordering of edges for each vertex) is given. We also prove that there is at most one maximal 1-planar embedding ξ of G that preserves the given rotation system Φ, and our algorithm also produces such an embedding in linear time, if it exists. In addition, we establish new bounds on the minimum number of edges of maximal 1-plane graphs, showing that for graphs on n vertices this number is between 9n and 7n Introduction A recent research topic in topological graph theory is the study of graphs that are almost planar in some sense. Examples of such almost planar graphs are 1-planar graphs, which can be embedded in a plane with at most one crossing per edge. The class of 1-planar graphs was first considered by Ringel [14] in the context of simultaneously coloring vertices and faces of planar graphs. Subsequently, various aspects of 1-planar graphs have been investigated. For example, Borodin showed that 1-planar graphs are 6-colorable [3] and there is a linear time algorithm that 7 colors Result of the Port Douglas Workshop on Geometric Graph Theory, June, 2011, Australia.

2 1-planar graphs [6]. Furthermore acyclic colorability [4] and edge colorablility [18] have been investigated. Pach and Toth proved that a 1-planar graph with n vertices has at most 4n 8 edges, which is a tight upper bound [13]. Other researches study the existence of vertices of small degree and edges with endpoints of small degree in classes of 1-planar graphs [7] [10]. The class of 1-planar graphs is not closed under edge contraction, which makes various computational tasks for them hard. Korzhik and Mohar proved that testing 1- planarity of a graph is NP-complete [12]. More recently, it was reported that, surprisingly, the problem remains NP-hard, even if the input graph is near-planar (i.e., deletion of an edge makes the resulting graph planar) [5], or a rotation system (i.e., the circular ordering of edges for each vertex) is given [1]. A 1-planar embedding of a graph G is maximal, if no edge can be added without violating the 1-planarity of G [15, 16, 11]. In this paper, we study combinatorial properties of maximal 1-planar embeddings. Based on these properties, we also study algorithmic aspects of maximal 1-planar graphs with a given rotation system. The fact that some algorithmic problems are solvable when a rotation system is given is in accordance with other results from the literature, showing that difficult NP-hard problems can be solvable in polynomial time, if we consider special classes of graphs or add some topological constraints such as a fixed embedding. For example, the well-known upward planarity testing problem is NP-hard in general [8], however a polynomial time algorithm is known if an embedding of the graph is given [2]. Similarly, testing rectilinear planarity of a graph is NP-hard in general, however it can be solvable in polynomial time if an embedding of a graph is fixed [8, 17]. More recently, a linear time algorithm for constructing a straight-line drawing of 1-planar graphs with a given embedding was presented [9]. Our contribution: We prove various combinatorial properties of maximal 1-planar graphs. We first show that in any maximal 1-planar embedding, the subgraph induced by the edges that do not intersect any other edge (also called the red graph ) is spanning and biconnected. We apply these combinatorial properties to investigate maximal 1-planarity of a graph G with a given rotation system Φ. We show that if such an embedding exists, then it is unique. The combinatorial properties that we prove enable us to develop a linear time algorithm to test maximal 1-planarity. Our algorithm also produces a maximal 1-planar embedding ξ of G consistent with Φ in linear time, if it exists 6. In addition, we study the density of maximal 1-planar graphs. A well-known upper bound of Pach and Toth shows that an n-vertex 1-planar graph can have at most 4n 8 edges. However, a maximal 1-planar graph may have far fewer edges. We establish new bounds for the minimum number of edges of maximal 1-plane graphs. More precisely, we show that the minimum number of edges of a maximal 1-plane graph on n vertices lies between 9n and 7n Independently, Auer et al. report that testing maximal 1-planarity of a graph with a rotation system can be solved in polynomial time. They also give a lower bound of 5n 4 on the 2 minimal number of edges of maximal 1-plane graphs [1]. 2

3 2 Preliminaries A drawing D(G) of a graph G is a geometric representation of a graph in the plane, such that each vertex of G is mapped to a point in the plane, and each edge of G is drawn as a curve. As customary, in this paper we only allow proper drawings: no two edges incident to a vertex cross each other; two edges cross at most once; edges may not touch each other tangentially; no three edges cross at a single point. A drawing D(G) of a graph G induces an embedding ξ(g) of a graph G. More precisely, an embedding ξ(g) of a graph G is defined by a circular ordering of the edges around each vertex in G, as well as the combinatorial configuration of the crossings in a drawing D(G). We say that the drawing D(G) realizes ξ(g). An embedding ξ(g) is called planar, if there is no crossing in any drawing D(G) realizing ξ(g). A graph G is called planar, if it admits a planar embedding. A plane graph G ξ is a graph G embedded in a plane with a planar embedding ξ. An embedding ξ(g) is called 1-planar, if each edge crosses at most one other edge in a drawing D(G) realizing ξ(g). A graph G is called 1-planar, if it admits an 1-planar embedding. A 1-plane graph G ξ is a graph G embedded in a plane with a 1-planar embedding ξ. Given a 1-planar embedding ξ(g) of a 1-planar graph G, we can obtain a planar embedding by replacing each crossing in ξ(g) with a virtual vertex. We call the resulting planar graph the planarization of G, and denote it by G P. Note that ξ(g P ) preserves ξ(g). The original vertices of G are called real vertices. The planarization of G uniquely determines the embedding ξ(g). A 1-planar embedding ξ(g) is called maximal, if no edge can be added without violating 1-planarity of G [15, 16, 11]. A maximal 1-plane graph G ξ is a graph embedded in a plane with a maximal 1-planar embedding ξ. A rotation system Φ(G) of a graph G defines a circular ordering of edges around each vertex in G. Without loss of generality, we assume that the circular ordering is clockwise, unless otherwise stated. Let G Φ denote a graph G with a given rotation system Φ. If an embedding ξ(g) of a graph G induces a rotation system Φ, we say that ξ(g) realizes Φ or that ξ(g) is consistent with Φ. Note that a rotation system Φ does not define crossings between edges. Thus, in general, there may be several 1-planar embeddings of a graph realizing a given rotation system. However, for a planar graph, a rotation system uniquely determines a planar embedding (i.e., there is at most one planar embedding realizing a given rotation system). Thus, to determine a 1-planar embedding, it is sufficient to determine a rotation system of the planarization of the 1-planar embedding. An embedding ξ(g) of a graph G in the plane defines the crossing-free edges as well as the crossing edges. We call the crossing-free edges red edges, and call the edges that are crossed blue edges. The red graph G R of a graph G is the subgraph induced by the red edges. When we define a subgraph G of a graph G, we assume that G inherits the topology of G. For example, if G is an embedded graph, then we treat the subgraph G of G as an embedded graph, preserving the embedding of G (i.e., ξ(g ) is induced by ξ(g)). Similarly, if G has a rotation system Φ(G), then the subgraph G of G inherits the rotation system of G (i.e., Φ(G ) is induced by Φ(G)). 3

4 3 Properties of Maximal 1-planar Embeddings This section presents various combinatorial properties of maximal 1-planar embeddings. In particular, we show that the red subgraph of a maximal 1-planar embedding is biconnected and spanning. In the next section, we will use this property to design a linear time maximal 1-planarity testing algorithm for a graph G with a given rotation system Φ. We first present two lemmas to prove the main theorem of this section. Lemma 1. Let ξ(g) be a 1-planar embedding of a 1-planar graph G on at least 2 vertices, and G P be the planarization of G. Each face f of G P has at least two real vertices. Proof. Without loss of generality G p is connected and has at least 3 vertices. Since G P is a simple plane graph, each face f contains at least 3 vertices. Note that virtual vertices are not adjacent in G P, since G is a 1-planar graph. Thus, f contains at least two real vertices. Lemma 2. Let G be a maximal 1-plane graph. If v 1 and v 2 are consecutive neighbors of a virtual vertex w in the planarization G P of G, then there are two internally vertex disjoint red paths (that is, paths consisting of red edges) from v 1 to v 2. Proof. See the Appendix. We now present the main theorem of this section. Theorem 1. In a maximal 1-planar embedding ξ(g) of a graph G with at least 3 vertices, the red graph G R is biconnected and spans all vertices. Proof. We apply Lemma 2 to show that the red graph G R spans all the vertices. Let ξ(g) be a maximal 1-planar embedding of a graph G. We first argue that G is connected. If this were not the case, then the embedding of each component of G would be a maximal 1-planar embedding. By Lemma 1 there would be a face f that is incident with two real vertices from different components. These could be joined by an edge through f, contradicting the maximality of ξ(g). We conclude that every maximal 1-plane graph is connected. To show that the red graph is spanning, it suffices now to show that for two vertices v 1 and v 2, which are adjacent via a blue edge, there is a red path from v 1 to v 2. The blue edge (v 1, v 2 ) is crossed in a virtual vertex w by some other edge (u 1, u 2 ). Applying Lemma 2 to the triangles formed by {v 1, u 1, w} and {v 2, u 1, w} shows that a red path from v 1 to v 2 exists. To prove biconnectivity, we first show that G is biconnected. By Lemma 2, if two real vertices are adjacent to the same virtual vertex in G P then there are two vertex disjoint paths between the two real vertices. Thus to show biconnectivity of G, it suffices to show the following: If v 1, v 2 and v 3 are real vertices that, ignoring virtual vertices, appear consecutively in this order on a face of the planarization G P of G, then there is a path from v 1 to v 3 that avoids v 2. This is clear, since v 1 and v 3 are incident with the same face, and therefore are adjacent. Finally, we show that the red graph G R is biconnected. Suppose that v is a cut vertex whose removal separates v 1 from v 2 in the red graph. Since G is biconnected, there is 4

5 a path P in G v from v 1 to v 2. It suffices to argue that, for two consecutive vertices on P, there is a red path joining them that avoids v. If they are adjacent via a red edge this is clear. Otherwise this follows from Lemma 2. 4 Testing Maximal 1-planarity of a Graph with a Fixed Rotation System In this section, we present a linear time algorithm to test maximal 1-planarity of a graph with a given rotation system. More precisely, we prove the following theorem. Theorem 2. There exists a linear time algorithm that tests whether graph G with a given rotation system Φ has a maximal 1-planar embedding realizing Φ. In case such an embedding exists, it is unique and the algorithm computes this embedding. To design the algorithm mentioned in the theorem we proceed in three phases. In each phase we reject the input if the combinatorial properties of the results do not match those of a maximal 1-planar embedding consistent with the given rotation system. (a) We first design a linear time algorithm that computes a coloring of the edges of G into red and blue. This coloring is the coloring of the edges in the unique 1-planar embedding of G in which the red graph is biconnected and spanning, if such an embedding exists. (b) We then design a linear time algorithm that, given this coloring, computes a 1- planar embedding consistent with the coloring and the rotation system Φ, if it exists. (c) Finally we design a linear time algorithm that tests whether a given 1-planar embedding is maximal. More specifically, the overview of the maximal 1-planarity testing algorithm consists of five steps. See Algorithm 1. Steps 1 and 3 can be done in linear time using existing standard algorithms for testing biconnectivity of a graph and testing planarity of a graph. In the following, we explain Steps 2, 4 and 5 in detail. Algorithm 1 Testing Maximal 1-Planarity Input: G Φ, a graph G with a rotation system Φ. Output: 1-planar embedding ξ(g Φ ) and planar embedding ξ(g P ) or no in case no such embedding exists. 1: If E(G) > 4n 8 or G is not biconnected, then return( no ). 2: Compute the red subgraph G R of G Φ. 3: If G R is not planar or not biconnected, then return( no ). 4: Test 1-planarity of G Φ and compute ξ(g Φ) and ξ(g P ). 5: Test maximality of ξ(g Φ ). 5

6 4.1 Computing the red subgraph G R We first define necessary terminology. It is convenient to regard the input graph G as a directed graph, with two directed edges (u, v) and (v, u) for each pair u, v of adjacent vertices. We say that a directed edge (v 2, v 3 ) is the rightmost continuation of a directed edge (v 1, v 2 ), if the vertex v 3 is the vertex that precedes v 1 in the circular ordering of v 2. We say that a walk v 1,..., v t is a completed rightmost walk, if the following two conditions hold (where indices are taken modulo t): 1. for every i {1,..., t}, the directed edge (v i, v i+1 ) is the rightmost continuation of the directed edge (v i 1, v i ), and 2. for all i, j {1,..., t}, if (v i, v i+1 ) = (v j, v j+1 ), then i = j. We now show that completed rightmost walks can be used to characterize the red edges of G Φ. Lemma 3. Let G be a 1-plane graph with a given rotation system Φ, whose red graph G R is spanning and biconnected. An edge e of G is red if and only if there is a completed rightmost walk on G Φ that traverses e only in one direction. Proof. We first show that a red edge is traversed only once in a completed rightmost walk. Since the red graph G R is biconnected and spanning, every face of G R is a simple cycle. Let the sequence of vertices u 1,..., u t be a clockwise facial cycle of G R. We take indices modulo t. For i {1,..., t }, we say that a directed edge (u i, v) is inside the cycle, if v = u i+1 or the vertices u i+1, v and u i 1 are encountered (not necessarily consecutively) in this order in the circular ordering of u i. Note that, since the red graph G R is spanning, each edge inside the cycle has both end vertices on the cycle. Since the red cycle is a Jordan curve, the rightmost continuation of an edge inside the cycle is an edge inside the cycle. Since an edge of the form (u i, u i 1 ) is not inside the cycle, a red edge cannot be traversed in both directions. We now show that if a blue edge is traversed in one direction in a completed rightmost walk, then it is also traversed in the opposite direction. Let w be a virtual vertex, and let u 1, u 2, u 3, u 4 be the circular ordering of the neighbors of w in G P, induced by the embedding of G. Consider the 1-planar graph G w formed from G, by inserting the virtual vertex w, and replacing the two edges crossing at w by four new edges incident with w. The embedding ξ of G gives rise to an embedding ξ w of G w. We define a rotation system Φ w on G w by taking the rotation system induced by the embedding of G w. Note that the edges incident with w are red in G w, and thus the red graph of G w is biconnected and spanning. Thus, for the two neighbors u 1, u 2 of w, there is a red face which contains the edges (w, u 1 ) and (w, u 2 ), but no other edge incident with w. We direct the face to form a directed cycle C, so that the edges inside C incident with w are exactly (u 2, w) and (w, u 1 ). Since (w, u 1 ) is the rightmost continuation of (u 2, w), any completed rightmost walk in G w that contains (w, u 1 ) must contain (u 2, w). This implies that any completed rightmost walk in G that contains (u 3, u 1 ) must contain (u 2, u 4 ). By repeating the same arguments, we conclude that any completed rightmost walk in G that contains (u 2, u 4 ) must contain (u 3, u 1 ). Thus the edge (u 1, u 3 ) is traversed in both directions. Since the vertex w and the choice of u 1 among its neighbors are arbitrary, this proves the lemma. 6

7 Lemma 3 implies that the color of an edge is invariant over all 1-planar embeddings that realize G Φ and have a red graph G R that is spanning and biconnected. We now describe an algorithm that takes G Φ, a graph G with a rotation system Φ, as input, and computes the color of the edges in such embeddings. Lemma 4. Let G Φ be a graph G with a rotation system Φ. There is a linear time algorithm that computes a coloring of the edges of G Φ into red and blue which is equal to the coloring in every 1-planar embeddings that realizes G Φ with a spanning and biconnected red graph. Proof. The algorithm follows the proof of Lemma 3 to color the edges of G Φ as either red or blue. Let e and e 1 be the two directed edges with opposite directions between the same pair of vertices of G. If e and e 1 are both traversed during a completed rightmost walk, then we color the corresponding non-oriented edge of G blue; otherwise, it is colored red. This simple traversal is described in Algorithm Edge Coloring in the Appendix. The correctness of the algorithm follows directly from Lemma 3. By maintaining a list of all marked edges, the amount of work per edge is constant, and thus we have linear time complexity. 4.2 Computing a 1-planar embedding of G Φ Given a rotation system Φ, and a coloring of the edges of G into red and blue such that the red graph is spanning and biconnected, we now test whether there exists a 1-planar embedding of G Φ consistent with the colors. If such an embedding exists, we compute a planar embedding of the planarization G P of G Φ. After testing biconnectivity and planarity of G R in Step 3 of Algorithm 1, we have a planar embedding ξ(g R ) of G R which preserves the given rotation system Φ of G. This embedding is used as input for Step 4 of Algorithm 1, testing 1-planarity of G Φ and producing a 1-planar embedding ξ(g Φ ) and planar embedding ξ(g P ). The following lemma shows that the 1-planar embedding of G Φ is unique, if it exists. Lemma 5. A graph G with a given rotation system Φ can be realized by at most one 1-planar embedding consistent with Φ for which the red graph G R is spanning and biconnected. Proof. By Lemma 3, the color of each edge is invariant over all possible 1-planar embeddings realizing G Φ for which the red subgraph G R is spanning and biconnected. Since G R is planar and its rotation system is the restriction of the rotation system Φ, there is at most one planar embedding ξ(g R ) of the red graph that preserves Φ. Given a face of the red graph G R and a blue edge e, the rotation system determines whether e lies inside the face. Note that two blue edges that lie inside a red face cross if and only if their end vertices are encountered alternatingly when traversing the red facial cycle. This implies that the rotation system determines the crossing edge pairs. For a virtual vertex w, the rotation system Φ also determines the circular ordering of its neighbors. More specifically, the circular ordering is the same as the ordering of the neighbors 7

8 of w when traversing the facial cycle containing w clockwise. Together this implies that we can construct the planarization G P of G Φ and its rotation system (i.e., planar embedding). Since G P is a simple planar graph, given its rotation system, there is only one planar embedding for G P and thus only one 1-planar embedding for G Φ consistent with the coloring of the edges. Using Lemma 5, we can test 1-planarity of G Φ in linear time. Lemma 6. Let G be a graph with a rotation system Φ, and suppose that G R is a planar, spanning and biconnected subgraph of G. There exists a linear time algorithm that tests whether there is a 1-planar embedding of G Φ that is consistent with Φ such that G R is the red subgraph. In case such an embedding ξ(g Φ ) exists, it is unique and the algorithm also produces the planar embedding ξ(g P ) of the planarization of ξ(g Φ ). Proof. Due to the uniqueness of the 1-planar embedding of G Φ and the planar embedding of G R from Lemma 5, the rotation system Φ decides the embeddings of the blue edges inside each red facial cycle f of G R. Thus, to test 1-planarity of G Φ, we need to do the following: for each red face f of G R with a set of blue edges inside the facial cycle, we test whether the blue edges have exactly one crossing. Intuitively, the algorithm proceeds as follows. We perform a walk along a red facial cycle f, and detect crossings between two blue edges. Note that two blue edges cross inside f if and only if we encounter the four end vertices of the two edges alternatingly along the cycle. More specifically, we use a stack S to detect the crossings between the blue edges inside f. We push a blue edge onto S whenever it is encountered in the clockwise traversal of f. Let (v 1, v 3 ) and (v 2, v 4 ) be the two blue edges inside f. If the top four elements of S form (v 1, v 3 ), (v 2, v 4 ), (v 3, v 1 ), (v 4, v 2 ) in this order, then the two blue edges (v 1, v 3 ) and (v 2, v 4 ) cross each other inside f. When we detect a crossing, we create a new virtual vertex w with four incident edges (w, v 1 ), (w, v 2 ), (w, v 3 ), (w, v 4 ), and define the circular ordering of the new vertex w as v 1, v 2, v 3, v 4. We then update the circular orderings of v 1, v 2, v 3 and v 4 to include w accordingly. The last step of the algorithm is to test whether the embedding ξ(g Φ ) is 1-planar. The planar embedding ξ(g P ) of G P defines an embedding ξ(g Φ ) of G Φ. When we detect a crossing between two blue edges (v 1, v 3 ) and (v 2, v 4 ), we pop the four items (v 1, v 3 ), (v 2, v 4 ), (v 3, v 1 ), (v 4, v 2 ) from S. Note that at the end of the walk, the stack is empty if and only if every blue edge is crossed exactly once inside the face f. See Algorithm 1-planar embedding in the Appendix, for details. Using standard data structures for graph representation, including a doubly connected edge list for ξ(g P ), one can implement the algorithm to run in linear time. 4.3 Testing maximality of 1-planar embedding In this section, we show that maximality of a 1-planar embedding of a graph G with a given rotation system Φ can be tested in linear time. We first present some properties of maximal 1-planar graphs. 8

9 Lemma 7. Suppose that ξ(g) is a maximal 1-planar embedding of a graph G. If two vertices v 1 and v 2 of G are incident with the same face f in the planarization G P of ξ(g), then v 1 and v 2 are adjacent in G. Proof. If the two vertices v 1 and v 2 were not adjacent, then we could join them by routing the edge (v 1, v 2 ) through f without crossing any edge. This contradicts the maximality. Lemma 8. Let ξ(g) be a maximal 1-planar embedding of a graph G. Let G P be the planarization of G, and ξ(g P ) be the planar embedding of G P induced by ξ(g). Let f be a facial cycle in ξ(g P ). (a) Each crossing in ξ(g) induces a 4-clique. (b) The face f has at most 4 real vertices. (c) The face f has at most 8 vertices (real plus virtual). Proof. We first show Property (a). Each crossing in ξ(g) is defined by a pair of edges and four vertices. Suppose that a pair u, v of these vertices is not adjacent. Since G is 1-planar, there is exactly one face in ξ(g P ) where u and v appear consecutively. One can join u and v, routing the edge through the face. This contradicts the maximality of G. Property (b) follows from Lemma 7, by observing that K 5 has no 1-planar embedding where all vertices share a face. Property (c) follows from Property (b), together with the observation that it is impossible for two virtual vertices to occur consecutively on f (since the embedding is 1-planar). Based on Lemmas 7 and 8, we now present a linear time algorithm that tests whether a 1-planar embedding is maximal. Lemma 9. Let ξ(g Φ ) be a 1-planar embedding of G Φ, and ξ(g P ) be a planar embedding of G P, the planarization of G Φ. There is a linear time algorithm to test whether G Φ is maximal 1-planar. Proof. The main task is to check whether we can add new edges while preserving 1- planarity of the 1-planar embedding ξ(g Φ ). Consider the planar embedding ξ(g P ) of the planarization G P of G Φ. Note that G Φ is a maximal 1-planar graph if and only if the following two conditions hold: 1. for each face f of G P, any two non-adjacent real vertices of f are adjacent in G Φ ; and 2. for each red edge e of G P incident with distinct faces f 1 and f 2, all pairs of real distinct vertices v 1, v 2, with v 1 f 1 and v 2 f 2, are adjacent in G Φ. To test maximality of a 1-planar embedding, we thus proceed as follows: If there is a face of the planarization G P with more than 8 vertices, then the embedding is not maximal 1-planar. Otherwise, for each vertex v, we perform the following. For every face f incident with v, we mark all real vertices incident with f. For every face f that 9

10 shares a red edge with f, we mark all real vertices incident with f. Next we unmark v and all its neighbors. If a marked vertex v remains, then v and v are non-adjacent and violate one of the properties above. Since the size of the faces in G P is bounded (see Lemma 8), the amount of work performed for vertex v is linear in the number of faces incident with v, and thus linear in the number of edges incident with v. The total amount of work is thus linear in the number of edges, which proves the lemma. 5 Bounds on the Number of Edges of Maximal 1-plane Graphs Recall that the tight upper bound of Pach and Tóth [13] on the number of edges in an n-vertex 1-planar graph is 4n 8. However, a maximal 1-plane graph may have fewer edges. Let m(n) denote the minimum number of edges over all possible maximal 1- planar embeddings of graphs on n vertices. In this section, we present a better upper bound on m(n), as well as a new lower bound. We first present two combinatorial properties of maximal 1-plane graphs that will allow us to bound the number of edges from below. Lemma 10. Let G be a maximal 1-plane graph. If v is a vertex of G of degree at most 5 which is not incident with a blue edge, then G {v} is a maximal 1-plane graph. Proof. First note that all neighbors of v are adjacent: Indeed if v 1 and v 2 are consecutive neighbors of v, then v 1 and v 2 are incident with a common face of G P and thus adjacent. Otherwise, there is a vertex v such that v 1, v and v 2 are consecutive neighbors of v in this order. In this case, removal of the red edge (v, v ) merges two faces f 1 and f 2 for which f 1 is incident with v 1 and f 2 is incident with v 2. Since G is maximal, v 1 and v 2 must be neighbors. Now suppose that G {v} is not a maximal 1-plane graph. Then there are two vertices u and u that can be joined by an edge, without violating 1-planarity, which cannot be joined in G. Since neighbors of v are adjacent, one of two vertices, say u, is not a neighbor of v. Let M be the set of points in the plane consisting of the vertex v and the edges incident with v. Since u and u cannot be joined in G, the edge that joins them in G {v} must intersect M. For each vertex of M, each face incident with M is also incident with v. Thus we can redirect the edge coming from u to end in v, without crossing any other vertex in M. Therefore, we can add the edge (u, v) in G, which shows that G is not maximal and gives a contradiction. Lemma 11. Let G be a maximal 1-plane graph. If u 1 and u 2 are adjacent vertices of degree 3 both incident with a blue edge, then G {u 1, u 2 } is a maximal 1-plane graph. Proof. See the Appendix. The two previous lemmas show restrictions in the local combinatorics of maximal 1-planar graphs, which allows us to bound the number of edges as follows. Theorem 3. Let m(n) be the minimum number of edges over all possible maximal 1- planar embeddings of graphs on n vertices. Then 9n m(n) 7n

11 Proof. For the upper bound, we construct a maximal 1-plane graph G n with at most 7n 3 2 edges. If n = 3k, for some integer k, then the construction of G n is as shown in Figure 1. To construct the graph G n for n not divisible by 3, we delete 1 or 2 vertices incident only with red edges from G n, where n is the smallest integer larger than n divisible by 3. By Lemma 10, the graph remains maximal 1-planar when deleting these vertices. Fig. 1. The upper bound construction for m(n). We now show the lower bound. Let c = 18/5. Suppose that G is a vertex-minimal maximal 1-plane graph that has less than 9n 5 c edges, that is, a minimal counterexample. Note that G must have at least 3 vertices. We first argue that G does not have a vertex of degree at most 5 incident only with red edges: Suppose that v is such a vertex. Vertex v has degree at least 2, since the red graph is biconnected by Theorem 1. By Lemma 10, G v is a maximal 1-planar graph on n 1 vertices with less than 9n 5 c 2 9(n 1) 5 c edges, and therefore G is not a minimal counterexample. Since the red graph is biconnected, every vertex is incident with at least two red edges. We conclude that every vertex in G has degree at least 3. We next argue that G does not have two adjacent vertices of degree 3. Suppose that u and u are adjacent vertices of degree 3. Then G {u, u } is maximal 1-planar by Lemma 11. The graph G {u, u } has n 2 vertices and less than 9n 9(n 2) 5 c 5 5 c edges, which gives a contradiction. Now we argue that a vertex u of degree larger than 3 has two neighbors of degree larger than 3. If u does not have a neighbor u of degree 3, then the statement is clear. Otherwise, u is adjacent to both other neighbors of u, which must be of degree larger than 3. 11

12 Let m be the number of edges of G. Suppose now that G has t vertices of degree 3. Since vertices of degree 3 are not adjacent, there are at least 3t edges incident with a vertex of degree 3. For each vertex of degree larger than 3, at least two of its neighbors have degree also larger than 3. Thus, there are at least n t edges, whose both endpoints have degree larger than 3. Thus, m 3t+n t. On the other hand, by double counting, G has at least (4(n t) + 3t)/2 edges. Since 0 t n, the bound on the number of edges minimizes when 3t + n t = (4(n t) + 3t)/2, which yields t = 2/5n and gives a bound of 9n 5 on the number of edges. References 1. C. Auer, F. J. Brandenburg, A. Gleiner, and J. Reislhuber. On 1-planar graphs with rotation systems. manuscript, 24 April, P. Bertolazzi, G. D. Battista, G. Liotta, and C. Mannino. Upward drawings of triconnected digraphs. Algorithmica, 12(6): , O. V. Borodin. Solution of the Ringel problem on vertex-face coloring of planar graphs and coloring of 1-planar graphs. Metody Diskret. Analiz., (41):12 26, 108, O. V. Borodin, A. V. Kostochka, A. Raspaud, and E. Sopena. Acyclic colouring of 1-planar graphs. Discrete Applied Mathematics, 114(1-3):29 41, S. Cabello and B. Mohar. Adding one edge to planar graphs makes crossing number and 1-planarity hard. CoRR, abs/ , Z.-Z. Chen and M. Kouno. A linear-time algorithm for 7-coloring 1-planar graphs. In MFCS 2003, volume 2747 of LNCS, pages Springer, Berlin, I. Fabrici and T. Madaras. The structure of 1-planar graphs. Discrete Mathematics, 307(7-8): , A. Garg and R. Tamassia. On the computational complexity of upward and rectilinear planarity testing. SIAM J. Comput., 31(2): , S.-H. Hong, P. Eades, G. Liotta, and S.-H. Poon. Fary s theorem for 1-planar graphs. Proceedings of COCOON 2012, To appear. 10. D. Hudák and T. Madaras. On local properties of 1-planar graphs with high minimum degree. Ars Math. Contemp., 4(2): , D. Hudak, T. Madaras, and Y. Suzuki. On properties of maximal 1-planar graphs. manuscript, 24 Febuary, V. P. Korzhik and B. Mohar. Minimal obstructions for 1-immersions and hardness of 1- planarity testing. In Graph Drawing, pages , J. Pach and G. Tóth. Graphs drawn with few crossings per edge. Combinatorica, 17(3): , G. Ringel. Ein Sechsfarbenproblem auf der Kugel. Abh. Math. Sem. Univ. Hamburg, 29: , Y. Suzuki. Optimal 1-planar graphs which triangulate other surfaces. Discrete Mathematics, 310(1):6 11, Y. Suzuki. Re-embeddings of maximum 1-planar graphs. SIAM J. Discrete Math., 24(4): , R. Tamassia. On embedding a graph in the grid with the minimum number of bends. SIAM J. Comput., 16(3): , X. Zhang and J.-L. Wu. On edge colorings of 1-planar graphs. Inform. Process. Lett., 111(3): ,

13 6 Appendix 6.1 Proof of Lemma 2 Proof. Suppose that the other end vertices of the crossing edges that originate from v 1 and from v 2 are v 3 and v 4 respectively. See Figure 2. v 4 v 3 w u 1 v 1 w v 2 u 2 Fig. 2. The inside of the triangle T (gray shaded area) used in the induction proof of Lemma 2. Straight-lines represent blue edges, whereas as the curves represent red paths. The inside of T contains two smaller triangles, and the red paths exist by induction. First note that v 1 and v 2 are adjacent, since G is a maximal 1-planar graph. The triangle T induced by the vertices v 1, v 2 and w forms a Jordan curve. We define the inside of T be the area bounded by T, which does not contain edges incident with w other than (v 1, w) and (v 2, w). We will show that the edge (v 1, v 2 ) is red or there is a red path from v 1 to v 2 inside the triangle. This suffices to prove the lemma, since by symmetry we can then conclude the existence of internally vertex disjoint red paths from v 1 to v 4, from v 2 to v 3 and from v 3 to v 4. We now show that the edge (v 1, v 2 ) is red or that there exists a red path from v 1 to v 2 inside of T by induction on the number of vertices k inside of T. If k = 0, then the edge (v 1, v 2 ) is red and there is nothing to show. Thus, suppose that k > 0. If the edge (v 1, v 2 ) is red, there is again nothing to show. So suppose otherwise. Then, there is an edge (u 1, u 2 ) that crosses the edge (v 1, v 2 ). Let w be the virtual vertex on the edge (v 1, v 2 ). Since T forms a Jordan curve, exactly one of the vertices of the crossing edge, say u 1, is inside of T. Consider the triangle T 1 induced by the vertex set {v 1, u 1, w }. The inside of T 1 is contained in the inside of T. Moreover, the vertex u 1 is not inside of T 1, and thus the inside of T 1 contains less vertices than the inside of T. By induction, there is a red path from v 1 to u 1. With a symmetric argument for the triangle T 2 induced by the vertex set {v 2, u 1, w }, we conclude that there is a red path from v 2 to u 1. These two paths together show that there is a red path from v 1 to v 2 which is inside of T. 13

14 6.2 Proof of Lemma 11 Proof. First note that all neighbors of a vertex of degree 3 are adjacent. Suppose that u 1 and u 2 are adjacent vertices of degree 3. Note that there are two adjacent vertices that are neighbors of u 1 and u 2, and that u 1 and u 2 do not have any other neighbors. There are two possible situations depending on the color of the edge (u 1, u 2 ). See Figure 3. Let w be the virtual vertex adjacent to u 1 and u 2. Note first that, due to biconnectivity of G, there cannot be a vertex inside of the triangle formed by u 1, w and any neighbor v of u 1. Otherwise, v separates the vertices inside the triangle from the other vertices. The analogous statement holds for triangles involving u 2. v 2 u 2 v 2 v 1 w w u 1 v 1 u 1 u 2 Fig. 3. Two possibilities how two adjacent vertices u 1, u 2 of degree 3 both incident with a blue edge can be situated in a maximal 1-plane graph. Due to biconnectivity, the gray area cannot contain other vertices. Let M be the set of points inside the triangles incident with w and u 1 or u 2 together with the vertices u 1, u 2 and all points on edges incident with u 1 or u 2. Suppose that there are two vertices x and x non-adjacent in G that can be connected in G {u 1, u 2 }. The edge connecting x with x must cross the set M. The boundary of M contains v 1 and v 2. We say that a vertex v is reachable from a boundary point p, if the boundary vertex v and p are both incident with a common face that lies outside of M. Without loss of generality, we may assume that the edge enters and exits M exactly once, and that u 1 or u 2 is reachable from the entry point. Note that from any point on the boundary of M, the vertices v 1 and v 2 are reachable. This implies that x and x are different from v 1 and v 2. Since u 1 or u 2 is reachable from the entry point, it is possible to connect x or x with one of the vertices u 1 or u 2, which gives a contradiction. 14

15 6.3 Algorithm Edge Coloring Algorithm 2 Edge Coloring Input: G Φ such that G is biconnected. Output: A coloring of all edges into red and blue which is the coloring in every 1-planar embeddings that realizes G Φ with a red graph that is spanning and biconnected. 1: Unmark all edges. 2: E {(u, v), (v, u) (u, v) E(G)} 3: while there is a directed edge e in E do 4: e 0 e. 5: while e 0 e or e 0 is unmarked do 6: Mark e 0. 7: e 0 the rightmost continuation of e 0. 8: end while 9: for all marked directed edges e in E do 10: if e is marked and e 1 is marked then 11: Color e blue. 12: else 13: Color e red. 14: end if 15: end for 16: Remove all marked directed edges from E. 17: Unmark all directed edges. 18: end while 15

16 6.4 Algorithm 1-planar Embedding Algorithm 3 1-planar Embedding Input: G Φ, G R : a graph G with rotation system Φ and a subgraph G R that is planar, spanning and biconnected. Output: The planar embedding ξ(g P ) of the planarization of G, which preserves the rotation system of G Φ and for which G R is the red subgraph or no if no such embedding exists. 1: G Φ G Φ. 2: E {(u, v) (u, v) E(G)}. 3: Let S be an empty stack. 4: Unmark all edges in E. 5: while there is an unmarked red edge e in E do 6: e 0 e. 7: while e 0 e or e 0 is unmarked do 8: Mark e 0. 9: Let e be the rightmost continuation of e 0. 10: if e is blue then 11: Push e to the top of S. 12: while the top four items of the stack S are of the form (v 1, v 3), (v 2, v 4), (v 3, v 1), (v 4, v 2) do 13: Add a new virtual vertex w to V (G ). 14: E(G ) ( E(G ) \ {(v 1, v 3 ), (v 2, v 4 )} ) { (v i, w) i {1,..., 4} } 15: Define the ordering of the neighbors of w in Φ as {v 1, v 2, v 3, v 4 }. 16: For i {1,..., 4}, replace v (i+2 mod 4) by w in the ordering of neighbors of v i. 17: Pop four items from S. 18: end while 19: e 0 (e ) 1. 20: else 21: e 0 e. 22: end if 23: end while 24: end while 25: if S is empty then 26: Return (G Φ, ξ(g Φ )). 27: else 28: Return( no ). 29: end if 16