On the sizes of bipartite 1-planar graphs
aa r X i v : . [ m a t h . C O ] J u l On the sizes of bipartite 1-planar graphs ∗ Yuanqiu Huang † Department of Mathematics, Normal University of Hunan, Changsha 410081, [email protected]
Zhangdong Ouyang
Department of Mathematics, Hunan First Normal University , Changsha 410205, [email protected]
Fengming Dong
National Institute of Education, Nayang Technological University, Singaporedonggraph @163.com
Abstract
A graph is called 1-planar if it admits a drawing in the plane such thateach edge is crossed at most once. Let G be a bipartite 1-planar graph with n ( ≥
4) vertices and m edges. Karpov showed that m ≤ n − n ≥ m ≤ n − n ≥
7. Czap, Przybylo and ˘Skrabul´akov´aproved that if the partite sets of G are of sizes x and y , then m ≤ n + 6 x − ≤ x ≤ y , and conjectured that m ≤ n + 4 x −
12 holds for x ≥ y ≥ x −
12. In this paper, we settle their conjecture and our result iseven under a weaker condition 2 ≤ x ≤ y . Keywords : bipartite graph, drawing, face, 1-planar graph. ∗ This work is supported by MOE-LCSM, School of Mathematics and Statistics, Hunan NormalUniversity, Changsha, Hunan 410081, P. R. China † Corresponding author Introduction A drawing of a graph G = ( V, E ) is a mapping D that assigns to each vertex in V a distinct point in the plane and to each edge uv in E a continuous arc connecting D ( u ) and D ( v ). We often make no distinction between a graph-theoretical object(such as a vertex, or an edge) and its drawing. All drawings considered here aresuch ones that no edge crosses itself, no two edges cross more than once, and no twoedges incident with the same vertex cross. The crossing number of a graph G is thesmallest number of crossings in any drawing of G .A drawing of a graph is 1- planar if each of its edges is crossed at most once. If agraph has a 1-planar drawing, then it is 1- planar . The notion of 1-planarity wasintroduced in 1965 by Ringel [11], and since then many properties of 1-planar graphshave been studied (e.g. see the survey paper [8]).It is well-known that any simple planar graph with n ( n ≥
3) vertices has at most3 n − n ( n ≥
3) vertices has atmost 2 n − n ( ≥
3) vertices has at most 4 n − Theorem 1 ([7])
Let G be a bipartite -planar graph with n vertices. Then G hasat most n − edges for even n = 6 , and at most n − edges for odd n and for n = 6 . For all n ≥ , these bounds are tight. Note that Karpov’s upper bound on the size of a bipartite 1-planar graph is in termsof its vertex number. When the sizes of partite sets in a bipartite 1-planar graph aretaken into account, Czap, Przybylo and ˘Skrabul´akov´a [5] obtained another upperbound for its size (i.e., Corollary 2 in [5]).For any graph G , let V ( G ) and E ( G ) denote its vertex set and edge set. Theorem 2 ([5]) If G is a bipartite -planar graph with partite sets of sizes x and y , where ≤ x ≤ y , then | E ( G ) | ≤ | V ( G ) | + 6 x − . x and y with x ≥ y ≥ x −
12, the authors in [5]constructed a bipartite 1-planar graph G with partite sets of sizes x and y such that | E ( G ) | = 2 | V ( G ) | + 4 x −
12 holds. Moreover, they believed this lower bound isoptimal for such graphs and thus posed the following conjecture.
Conjecture 1 ([5])
For any integers x and y with x ≥ and y ≥ x − , if G is a bipartite 1-planar graph with partite sets of sizes x and y , then | E ( G ) | ≤ | V ( G ) | + 4 x − . In this paper we obtain the following result which proves Conjecture 1.
Theorem 3
Let G be a bipartite -planar graph with partite sets of sizes x and y ,where ≤ x ≤ y . Then | E ( G ) | ≤ | V ( G ) | + 4 x − , and the upper bound is bestpossible. The result in [5, Lemma 4] shows that the upper bound for | E ( G ) | in Theorem 3 istight. Also, if x ≤ ( y + 4), Theorem 3 provides a better upper bound for | E ( G ) | than Theorem 1.The authors in [5] mentioned a question of Sopena [12]: How many edges we haveto remove from the complete bipartite graph with given sizes of the partite sets toobtain a 1-planar graph?
It is not hard to see that Theorem 3 implies the followcorollary which answers the problem.
Corollary 1
Let K x,y be the complete bipartite graph with partite sets of sizes x and y , where ≤ x ≤ y . Then at least ( x − y − edges must be removed from K x,y such that the resulting graph becomes possibly a 1-planar graph, and the lowerbound on the number of removed edges is best possible. The remainder of this paper is arranged as follows. In Section 2, we explain someterminology and notation used in this paper. In Section 3, under some restrictions,we present several structural properties on an extension of D × for a 1-planar drawing D of a bipartite 1-planar graph G , where D × is a plane graph introduced in Section 2.Some important lemmas for proving Theorem 3 are given in Section 4, while theproof of this theorem is completed in Section 5. Finally, we give some furtherproblems in Section 6. 3 Terminology and notation
All graphs considered here are simple, finite and undirected, unless otherwise stated.For terminology and notation not defined here, we refer to [1]. For any graph G and A ⊆ V ( G ), let G [ A ] denote the subgraph of G with vertex set A and edge set { e ∈ E ( G ) : e joins two vertices in A } . G [ A ] is called the subgraph of G induced by A . For a proper subset A of V ( G ), let G − A denote the subgraph G [ V ( G ) \ A ].A walk in a graph is alternately a vertex-edge sequence; the walk is closed if itsoriginal vertex and terminal vertex are the same. A path (respectively, a cycle ) ofa graph is a walk (respectively, a closed walk) in which all vertices are distinct; the length of a path or cycle is the number of edges contained in it. A path (respectively,a cycle) of length k is said to be a k -path (respectively, k -cycle ). If a cycle C iscomposed of two paths P and P , we sometimes write C = P ∪ P .A plane graph is a planar graph together with a drawing without crossings, and atthis time we say that G is embedded in the plane. A plane graph G partitions theplane into a number of connected regions, each of which is called a face of G . If a faceis homeomorphic to an open disc, then it is called cellular ; otherwise, noncellular .Actually, a noncellular face is homeomorphic to an open disc with a few removed“holes”. For a cellular face f , the boundary of f can be regarded as a closed walk of G , while for a noncellular face f , its boundary consists of many disjoint closed walksof G . The size of a face is the number of the edges contained in its the boundarywith each repeated edge counts twice. A face of size k is also said to be a k - face .It is known that a plane graph G has no noncellular faces if and only if G is connected.For a connected plane graph G , the well-known Euler’s formula states that | V ( G ) | −| E ( G ) | + | F ( G ) | = 2, where F ( G ) denotes the face set of G .A cycle C of a plane graph G partitions the plane into two open regions, the boundedone (i.e., the interior of C ) and the unbounded one (i.e., the exterior of C ). Wedenote by int ( C ) and ext ( C ) the interior and exterior of C , respectively, and theirclosures by IN T ( C ) and EXT ( C ). Clearly, IN T ( C ) ∩ EXT ( C ) = C . A cycle C ofa plane graph G is said to be separating if both int ( C ) and ext ( C ) contain at leastone vertex of G .Let D be a 1-planar drawing of a graph G . The associated plane graph D × is theplane graph that is obtained from D by turning all crossings of D into new vertices4f degree four; these new vertices of degree four are called the crossing vertices of D × . D × Throughout this section, we always assume that the considered graph G (possiblydisconnected) is a bipartite 1-planar graph with partite sets X and Y , where 3 ≤| X | ≤ | Y | . Let D be a 1-planar drawing of G with the minimum number of crossings,and D × be the associated graph of D with the crossing vertex set W .Note that subsets X, Y and W form a partition of V ( D × ). We color the vertices in X, Y and W by the black color, white color and red color respectively. As stated in[5], D × can be extended to a plane graph, denoted by D × W , by adding edges joiningblack vertices as described below:for each vertex w in W , it is adjacent to two black vertices in D × , say x and x , and we draw an edge, denoted by e w , joining x and x whichis “most near” one side of the path x wx of D × such that it does notcross with any other edge, as shown in Figure 1 (b). wx x wx x (a) (b)Figure 1: The extension of D × .Observe that D × W is a plane graph with D × as its spanning subgraph and the edgeset of D × W is the union of E ( D × ) and { e w : w ∈ W } . Although D × is a simplegraph, D × W might contain parallel edges (i.e., edges with the same pair of ends), asthere may exist two edges in { e w : w ∈ W } with the same pair of ends. An exampleis shown in Figure 2 (c), where D is a 1-planar drawing of K , .Let F D (or simply F ) and H D (or simply H ) denote the subgraphs D × W [ W ∪ X ] and D × W [ X ] respectively. Obviously, H is a subgraph of F and its edge set is { e w : w ∈ } , while the edge set of F is the union of E ( H ) and { wx , wx ∈ E ( D × ) : w ∈ W & x , x ∈ X } .(a) D (b) D × (c) D × W (d) F Figure 2:
D, D × , D × W and F , where D is a 1-planar drawing of K , All vertices in H are black and the edges in H are also called black edges . Clearly, W is an independent set in F and each vertex in W (i.e., a red vertex) is of degree2 in F . The edges in F incident with red vertices are called red edges . Thus, eachedge in F is either black or red, as shown in Figure 2 (d).We have the following facts on D × , F and H :(1) D × , F and H are simultaneously embedded in the plane;(2) F and H are obviously loopless, but they are possibly disconnected;(3) w → e w is a bijection from W to E ( H ), where w is a red vertex, and thus thenumber of crossings of D equals to | E ( H ) | ; and(4) e w → x wx is a bijection from E ( H ) to the set of 2-paths in F whose endsare black, where w is a red vertex and x and x are the black vertices in D × adjacent to w .Moreover we have the following propositions. Proposition 1
Let e w be an edge of H with ends x and x and C be the -cycle of F consisting of e w and its corresponding 2-path P = x wx , where w is a red vertex(see Figure 3 (a)). Then int ( C ) contains none of black vertices, red vertices andblack edges in F ; in this sense we also say that int ( C ) is “empty”. roof . By the definition of D × W , the proposition follows directly from the fact thatthe drawing of edge e w is most near one side of the 2-path x wx in D × withoutcrossings with edges in D × . ✷ x x w x x e e e w (a) (b)Figure 3: Some 3-cycles and 2-cycles in F . Proposition 2
Assume that H has no separating 2-cycles. If C is a 2-cycle in H that consists of two multiple edges e and e joining two black vertices x and x (see Figure 3 (b)), then either int ( C ) or ext ( C ) contains neither black vertices norred vertices.Proof . As H has no separating 2-cycles, either int ( C ) or ext ( C ) contains no blackvertices. Assume that int ( C ) does not contain black vertices.Suppose that int ( C ) contains red vertices. Then, int ( C ) contains white vertices of D × . As int ( C ) does not contain black vertices, each white vertex in int ( C ) is ofdegree at most 2 in D × . Thus, we can redraw the edges of D in int ( C ) such thatthese edges make no crossings, and then obtain a 1-planar drawing of G with fewercrossings than D , contradicting to the choice of D . Hence int ( C ) does not containred vertices and the conclusion holds. ✷ Proposition 3
Assume that H contains no separating 2-cycles. Then the edgemultiplicity of H is at most 2.Proof . Assume to contrary that H has three multiple edges e , e and e which jointhe same pair of black vertices x and x . Then these three edges divide the planeinto three regions, denoted by α , β and γ , as shown in Figure 4 (a). By Proposition2, at least two of these three regions contain neither red vertices nor black vertices,except on its boundary. We may assume α and γ are such two regions.7et P = x wx be the 2-path of F that corresponds to edge e , where w is a redvertex. Thus, this path must be within region β , as shown in Figure 4 (b).As P is within region β , black edges e and e are in different sets int ( e ∪ P ) and ext ( e ∪ P ), a contradiction to Proposition 1. The proof is then completed. ✷ x x e e e αβγ x x e e e αγ w β (a) (b)Figure 4: Possible three multiple edges.An edge of H is called a simple edge if it is not parallel to another edge in H and a partnered edge otherwise. It follows from Proposition 3 that, if H has no separating2-cycles, then each partnered edge e in H is parallel to a unique partnered edge e ′ in H .Let C be a cycle and P be a path in H such that the end vertices of P are the onlyvertices in both C and P . When we say that P lies in int ( C ) (resp. ext ( C )), itmeans that all edges and internal vertices of P lie in int ( C ) (resp. ext ( C )). Proposition 4
Assume that H has no separating -cycles. Let C be a -cycle of H consisting of black vertices x , x and x , and e be the edge on C joining x and x . Assume that e ′ is a partnered edge in H which is parallel to e . If P = x wx and P ′ = x w ′ x are the -paths in F corresponding to e and e ′ respectively, thenone of P and P ′ lies in int ( C ) and the other in ext ( C ) .Proof . Let C denote the 2-cycle of H consisting of edges e and e ′ . By Proposition2, we may assume that int ( C ) contains neither black vertices nor red vertices. Thus,both w and w ′ are in ext ( C ).Let C denote the 3-cycle of F consisting of edge e and path P and C ′ the 3-cycleof F consisting of edge e ′ and path P ′ . By Proposition 1, both int ( C ) and int ( C ′ )are empty. Thus, the subgraph F [ { x , x , w, w ′ } ] is as shown in Figure 5 (a).8 x w w ′ x x w w ′ x e e ′ e e ′ P P ′ P P ′ (a) F [ { x , x , w, w ′ } ] (b) F [ { x , x , x , w, w ′ } ]Figure 5: x lies in ext ( C ), where C is the cycle x wx w ′ x As these three regions int ( C ), int ( C ) and int ( C ′ ) do not contain black vertices, x must be in ext ( C ), where C is the 4-cycle of F consisting of paths P = x wx and P ′ = x w ′ x . As F is a plane graph, path x x x must lies in ext ( C ), as shownin Figure 5 (b).Hence the conclusion holds. ✷ Proposition 5
Suppose that H has no separating 2-cycles. For any -cycle C in H , if int ( C ) contains exactly r red vertices, where ≤ r ≤ , then C contains atleast − r simple edges of F .Proof . Let e , e and e be the three edges on C . Suppose that e i is not a simpleedge of H , where 1 ≤ i ≤
3. Then e i is parallel to another partnered edge e ′ i of H . Let P i and P ′ i be 2-paths in F which correspond to edges e i and e ′ i respectively.Since H has no separating 2-cycles, by Proposition 4, int ( C ) contains a red vertexthat is on P i or P ′ i .The above conclusion implies that the number of red vertices in int ( C ) is not lessthan the number of partnered edges on C . Thus, the result holds. ✷ Let G be a bipartite graph with partite sets X and Y and O be a disk on the plane.If D is a 1-planar drawing of G that draws all vertices of X on the boundary of O Y and all edges of G in the interior of O , then we say that D isa 1 -disc O X drawing of G . Lemma 1
Let G be a bipartite graph with partite sets X and Y , and let D be a1-disc O X drawing of G with the minimum number of crossings k . If | X | = 3 , then k ∈ { , , } and | E ( G ) | ≤ | Y | + 1 + (cid:6) √ k (cid:7) , i.e., | E ( G ) | ≤ | Y | + 1 , if k = 0;2 | Y | + 2 , if k = 1;2 | Y | + 3 , if k = 3 . x x x x x x y y y y y (a) | Y | = 2 (b) | Y | = 3Figure 6: The 1-disc O X drawing of G for Y = Y and | Y | ∈ { , } Proof . Assume that | X | = 3. For any integer i ≥
0, let Y i be the set of vertices y in Y with d G ( y ) = i . As | X | = 3 and Y is independent in G , Y i = ∅ holds for all i ≥ y in ∈ Y , if y / ∈ Y , then y is notincident with any crossed edge. Thus, G − S i ≤ Y i has exactly k crossings, and itsuffices to show that | Y | ≤ k = , if | Y | ≤ , if | Y | = 2;3 , if | Y | = 3 . The rest of the proof will be completed by showing the following claims.
Claim (a) : | Y | ≤ | Y | ≥
4. Then, there exists a bipartite 1-planar drawing D ′ isomorphicto K , | Y | obtained from D [ X ∪ Y ] by copying all vertices and edges in the interior10f O X to its exterior, implying that K , is 1-planar. It is a contradiction to the factthat K , is not 1-planar due to Czap and Hud´ak [3].Thus, Claim (a) holds. Claim (b) : If | Y | ≤
1, then G − S i ≤ Y i has no crossings, i.e., k = 0.Claim (b) can be verified easily. Claim (c) : For any two vertices y , y ∈ Y , some edge incident with y crosseswith some edge incident with y , as shown in Figure 6 (a).If Claim (c) fails, then G [ X ∪ { y , y } ] is a plane graph and we can get a drawingof K , from G [ X ∪ { y , y } ] by adding a new vertex y ′ and three edges joining y ′ toall vertices in X in the exterior of O X without any crossing, implying that K , isplanar, a contradiction. Claim (d) : k = 1 when | Y | = 2, and k = 3 when | Y | = 3.By Claim (c), k ≥ (cid:0) | Y | (cid:1) . By the drawings in Figure 6, k ≤ | Y | ≤
2, and k ≤ | Y | ≤
3. Thus, Claim (d) holds.The result follows from Claims (a), (b) and (d). ✷ Lemma 2
Let G be a plane simple graph with | V ( G ) | ≥ . If G has exactly c components and t ( ≥ cellular -faces, then | E ( G ) | ≤ | V ( G ) | − − c + t .Proof . If c = 1, since G is simple, each face of H is a cellular face and has sizeat least 3. Then, in this case, the conclusion can be proved easily by applying theEuler’s formula.Now we assume that c ≥
2. We can obtain a simple and connected plane graph G ′ from G by adding c − F of G , we assume that its boundary consists of ℓ disjointclosed walks of G , and then we can add ℓ − F so that F is transformed into acellular face of size at least 4 because | V ( G ) | ≥
3. Therefore, the resulting graph G ′ is a simple and connected plane, and all faces of G ′ are cellular.Note that adding the c − G ′ has exactly t faces of size 3. The conclusion for connected plane graphs11mplies that | E ( G ′ ) | ≤ | V ( G ′ ) | − t/ . As V ( G ′ ) = V ( G ) and | E ( G ′ ) | = | E ( G ) | + c −
1, the above inequality implies that | E ( G ) | ≤ | V ( G ) | + t/ − − c . ✷ Lemma 3
Let G be a simple and bipartite plane graph with | V ( G ) | ≥ . If G hasexactly c components and t cellular faces whose boundaries are of length at least ,then | E ( G ) | ≤ | V ( G ) | − − c − t .Proof . If G is connected (i.e. c = 1), since G is bipartite and simple, then eachface of G is a cellular face, and has the size at least 4. Because G has t faces of sizeat least 6, it follows from the Euler’s formula that | E ( G ) | ≤ | V ( G ) | − − t .Now assume that c ≥
2. We can obtain a simple and connected bipartite planegraph G ′ from G by adding c − F of G consisting of ℓ distinct closed walks, similar to theproof of Lemma 2, we can add ℓ − F is transformed into a cellular face. We can ensure that those new added edgesjoin the vertices in different partite sets of G . Hence the resulting plane graph G ′ issimple, bipartite and connected. Clearly, all faces of G ′ are cellular, and G ′ has atleast t faces whose boundaries are of length at least 6. The conclusion for bipartiteand connected plane graphs implies that | E ( G ′ ) | ≤ | V ( G ′ ) | − − t. As V ( G ′ ) = V ( G ) and | E ( G ′ ) | = | E ( G ) | + c −
1, the above inequality implies that | E ( G ) | ≤ | V ( G ) | − − c − t . ✷ Remark : Lemmas 2 and 3 can be strengthened when G contains isolated vertices.Let V ≥ ( G ) be the set of non-isolated vertices in G . Then, under the condition | V ≥ ( G ) | ≥
3, the conclusions of both Lemmas 2 and 3 still hold after | V ( G ) | isreplaced by | V ≥ ( G ) | . The whole section contributes to the proof of Theorem 3.12 roof of Theorem 3. Suppose that Theorem 3 fails and χ is the minimum integerwith χ ≥ G with partite sets X and Y , where χ = | X | ≤ | Y | , | E ( G ) | > | V ( G ) | + 4 | X | −
12 holds.We will prove the following claims to show that this assumption leads to a contra-diction.
Claim 1 : χ ≥ Proof . Let G be any bipartite 1-planar graph with bipartitions X and Y , where2 ≤ | X | ≤ | Y | . If | X | = 2, obviously, | E ( G ) | ≤ | Y | = 2(2 + | Y | ) + 4 × −
12 = 2 | V ( G ) | + 4 | X | − . Now assume that | X | = 3. Let Y i be the set of vertices y in Y with d G ( y ) = i . Then | E ( G ) | ≤ | Y | + | Y | . Since the complete bipartite graph K , is not 1-planar (see[3]), we have | Y | ≤
6, implying that | E ( G ) | ≤ | Y | + 6. As x = 3, we have2 | V ( G ) | + 4 | X | −
12 = 6 | X | + 2 | Y | −
12 = 2 | Y | + 6 . Thus, | E ( G ) | ≤ | V ( G ) | + 4 | X | − χ , we have χ ≥ ✷ In the following, we assume that G is a bipartite 1-planar graph with bipartitions X and Y , where χ = | X | ≤ | Y | , such that | E ( G ) | > | V ( G ) | + 4 | X | − . (1)Let D be a 1-planar drawing of G with the minimum number of crossings and W bethe set of its crossings. Introduced in Section 3, D × W is a plane graph extended from D × , and F and H are the subgraphs D × W [ X ∪ W ] and D × W [ X ] of D × W respectively. Allvertices in X are black vertices, all vertices in Y are white vertices and all verticesin W are red vertices.We are now going to prove the following claim. Claim 2 : H has no separating 2-cycles. Proof . Assume to the contrary that H has a separating 2-cycle C consisting of twoparallel edges e and e joining black vertices x and x (see Figure 3 (b)), such thatboth int ( C ) and ext ( C ) contain black vertices.13et G = G T IN T ( C ) and G = G T EXT ( C ). Obviously, both G and G arebipartite 1-planar subgraphs of G . Moreover, we can see that G ∪ G = G , and G ∩ G = { x , x } .For i = 1 , G i has a bipartition X i and Y i , where X i = X ∩ V ( G i ) and Y i = Y ∩ V ( G i ). Clearly, | X | + | X | = | X | + 2 = χ + 2 and | Y | + | Y | = | Y | .Since C is a separating cycle of H , both int ( C ) and ext ( C ) contain black vertices,implying that | X i | ≥ i = 1 ,
2. As | X | + | X | = | X | + 2, we have | X i | < | X | = χ and so min {| X i | , | Y i |} ≤ | X i | < χ .For i = 1 ,
2, if | Y i | ≤
1, then | E ( G i ) | ≤ | Y i | · | X i | and | E ( G i ) | ≤ | V ( G i ) | + 4 | X i | − | Y i | ≥
2, then 2 ≤ min {| X i | , | Y i |} ≤ | X i | < χ and the assumption on χ implies that the conclusion of Theorem 3 holds for G i , i.e., | E ( G i ) | ≤ | V ( G i ) | + 4 min {| X i | , | Y i |} − ≤ | V ( G i ) | + 4 | X i | − . (2)Thus, by (2), | E ( G ) | = | E ( G ) | + | E ( G ) |≤ | V ( G ) | + | V ( G ) | ) + 4( | X | + | X | ) −
24= 2( | V ( G ) | + 2) + 4( | X | + 2) −
24= 2 | V ( G ) | + 4 | X | − , (3)which contradicts to the assumption in (1).Hence Claim 2 holds. ✷ It is known from Proposition 3 that the edge multiplicity of each edge in H is atmost 2. Then, there exists a subset A of E ( H ) such that both H h A i and H − A aresimple graphs and each edge in A is parallel to some edge in E ( H ) − A , where H h A i is the spanning subgraph of H with edge set A and H − A is the graph obtainedfrom H by removing all edges in A . Clearly, | A | is the number of pairs of edges e and e ′ in H which are parallel.Let H ′ denote H − A . Obviously, | A | ≤ | E ( H ′ ) | , and | E ( H ) | = | E ( H ′ ) | + | A | . (4) Claim 3 : H ′ contains at least one cellular 3-face.14 roof . Suppose that H ′ has no cellular 3-faces. As H ′ is a simple plane graph, byLemma 2, | E ( H ′ ) | ≤ | V ( H ′ ) | − | X | −
4. Because | A | ≤ | E ( H ′ ) | , it followsfrom (4) that | E ( H ) | ≤ | E ( H ′ ) | ≤ | X | − H is in 1-1 correspondence with a crossing of a drawing D of G , we can obtain a simple bipartite plane graph (possibly disconnected), denotedby G ′ , by removing | E ( H ) | edges from G each of which is a crossed edge of G . ByLemma 3, | E ( G ′ ) | ≤ | V ( G ′ ) | − | V ( G ) | −
4. Therefore, | E ( G ) | = | E ( G ′ ) | + | E ( H ) | ≤ | E ( G ′ ) | + 4 | X | − | V ( G ) | + 4 | X | − , a contradiction to the assumption in (1).Hence Claim 3 holds. ✷ Now we assume that H ′ has exactly t cellular 3-faces, where t ≥
1. Let T ( H ′ )denote the set of cellular 3-faces in H ′ . So t = | T ( H ′ ) | .For each ∆ ∈ T ( H ′ ), for convenience we also use “∆” to represent the 3-cyclecorresponding the boundary of ∆ if there is no confusions in the context. Let G ∆ = G T IN T (∆). Since ∆ is a cellular 3-face of H ′ , there are no black verticeslying in int (∆), and thus G ∆ is a bipartite with with exactly three black vertices,which lie on the boundary of the face ∆.Let ∆ ∈ T ( H ′ ). Since D is a 1-planar drawing of G with minimal number ofcrossings, the induced subdrawing of D of G ∆ is a 1-disc O X ∆ drawing of G ∆ withthe minimum number of crossings, where X ∆ is the set of three black vertices onthe boundary of ∆. Otherwise, we redraw the edges of G lying in the interior of ∆,and obtain a 1-planar drawing of G with fewer crossings than D , contradicting tothe choice of D . By Lemma 1, the number of crossings of D in int (∆) is a numberin the set { , , } .For any j ∈ { , , } , let T ( j ) ( H ′ ) be the set of members ∆ in T ( H ′ ) such that int (∆) contains exactly j crossings of D . Assume that T ( j ) ( H ′ ) = { ∆ ( j ) i : 1 ≤ i ≤ t j } , where t j = | T ( j ) ( H ′ ) | . Thus, t + t + t = t .For each ∆ ( j ) i ∈ T ( j ) ( H ′ ), let e ( j ) i be the number of the edges of the graph G ∆ ( j ) i and y ( j ) i be the number of white vertices in int (∆ ( j ) i ). Claim 4 : P j ∈{ , , } t j P i =1 e ( j ) i ≤ P j ∈{ , , } t j P i =1 y ( j ) i + ( t + 2 t + 3 t ).15 roof . By Lemma 1, e ( j ) i ≤ y ( j ) i +1+ (cid:6) √ j (cid:7) holds for any j ∈ { , , } and 1 ≤ i ≤ t j .Thus, Claim 4 holds. ✷ Claim 5 : | E ( H ) | ≤ | X | − t − (3 t + 2 t ) / Proof . Since H ′ is a simple plane graph with exactly t ≥ | V ( H ′ ) | = | X | = χ ≥
4, by Lemma 2, | E ( H ′ ) | ≤ | V ( H ′ ) | − t | X | − t . (5)On the other hand, by Proposition 5, for any j ∈ { , , } and 1 ≤ i ≤ t j , at least3 − j edges on the boundary of ∆ ( j ) i are simple edges, implying that at least 3 − j edgeson the boundary of ∆ ( j ) i are in H ′ . Because each simple edge of H ′ belongs to theboundaries of at most two different faces of H ′ , it follows that | E ( H ′ ) | ≥ (3 t +2 t ) / | A | ≤ | E ( H ′ ) | − (3 t + 2 t ) / , and therefore, by (4) and (5), | E ( H ) | = | E ( H ′ ) | + | A | ≤ | X | − t − (3 t + 2 t ) / . (6)Thus, Claim 5 holds. ✷ Let D ′ denote the drawing obtained from D by deleting all white vertices and edgesof D that lie in the interiors of all cellular 3-faces ∆ ( j ) i of H ′ , where j ∈ { , , } and1 ≤ i ≤ t j , and let G ′ denote the graph represented by D ′ .We see that the graph G ′ is a bipartite 1-planar graph with a bipartition X and Y ′ = Y ∩ V ( G ′ ), where | Y ′ | = | Y | − P j ∈{ , , } t j P i =1 y ( j ) i . Thus, | V ( G ′ ) | = | V ( G ) | − X j ∈{ , , } t j X i =1 y ( j ) i . (7)and | E ( G ′ ) | = | E ( G ) | − X j ∈{ , , } t j X i =1 e ( j ) i . (8)As the number of crossings of D equals to | E ( H ) | and D ′ has no crossings lying inthe interior of any cellular 3-face of H ′ , D ′ has exactly | E ( H ) | − ( t + 3 t ) crossings.16or each crossing of D ′ , we remove exactly one crossed edge from G ′ and obtain abipartite plane graph G ∗ . Thus, | E ( G ∗ ) | = | E ( G ′ ) | − ( | E ( H ) | − ( t + 3 t )). Then,(8) implies that | E ( G ∗ ) | = (cid:16) | E ( G ) | − X j ∈{ , , } t j X i =1 e ( j ) i (cid:17) − | E ( H ) | + ( t + 3 t ) . (9)Clearly, by (7), | V ( G ) | = | V ( G ′ ) | + X j ∈{ , , } t j X i =1 y ( j ) i = | V ( G ∗ ) | + X j ∈{ , , } t j X i =1 y ( j ) i . (10)Now, we shall obtain an upper bound of | E ( G ∗ ) | in terms of | V ( G ∗ ) | by constructinga bipartite plane graph with at least t cellular 6-faces. Claim 6 : | E ( G ∗ ) | ≤ | V ( G ∗ ) | − − t . Proof . Note that the simple and bipartite plane graph G ∗ is obtained from G byremoving all white vertices and edges of G lying in the interiors of all cellular 3-facesof H ′ and, for each crossing of D not lying in any cellular 3-face of H ′ , removingexactly one edge of G involved in this crossing.Now let G ∗∗ denote the graph obtained from G ∗ by adding all black edges in H ′ which belong to the boundary of cellular 3-faces of H ′ and then subdividing eachof these added edges. Let m be the number of edges in H ′ that belong to theboundaries of cellular 3-faces of H ′ . Then | V ( G ∗∗ ) | = | V ( G ∗ ) | + m and | E ( G ∗∗ ) | = | E ( G ∗ ) | + 2 m. (11)Because the edges of H (and thus H ′ ) are not crossed with the edges of G (and thus G ∗ ), we observe that G ∗∗ is also a simple and bipartite plane graph and has at least t cellular 6-faces. Applying Lemma 3 to G ∗∗ yields that | E ( G ∗∗ ) | ≤ | V ( G ∗∗ ) | − − t. Then, (11) implies that | E ( G ∗ ) | ≤ | V ( G ∗ ) | − − t . This proves the claim. ✷ Claim 7 : | E ( G ) | ≤ | V ( G ) | + 4 | X | − − t / Proof . By (9), we have | E ( G ) | = | E ( G ∗ ) | + | E ( H ) | + X j ∈{ , , } t j X i =1 e ( j ) i − ( t + 3 t ) . | E ( G ) | ≤ (cid:16) | V ( G ∗ ) | − − t (cid:17) + (cid:16) | X | − t − (3 t + 2 t ) / (cid:17) + (cid:16) X j ∈{ , , } t j X i =1 y ( j ) i + ( t + 2 t + 3 t ) (cid:17) − ( t + 3 t )= 2 | V ( G ) | + 4 | X | − − t / ≤ | V ( G ) | + 4 | X | − . ✷ Clearly, Claim 7 contradicts the assumption in (1). Hence Theorem 3 holds. ✷ For any x ≥ y ≥ x −
12, Czap, Przybylo and ˘Skrabul´akov´a [5, Lemma4] constructed a bipartite 1-planar graph G with partite sets X and Y such that | E ( G ) | = 2 | V ( G ) | + 4 | X | −
12. Notice that the 1-planar drawing D of this graph G given in [5] has the following property:(*) each vertex in X is incident with crossed edges in D .The proof of Theorem 3 also yields that, if | E ( G ) | = 2 | V ( G ) | + 4 | X | −
12 holds for abipartite 1-planar graph G with partite sets X and Y , where 4 ≤ | X | ≤ | Y | , and D is a 1-planar drawing of G with the minimum number of crossings, then the graph H ′ introduced in Section 3 does not have isolated vertices, i.e., property (*) aboveholds.Based on the above observations, we propose the following problem. Problem 1
For any bipartite -planar graph G with partite sets X and Y , where ≤ | X | ≤ | Y | , if | E ( G ) | = 2 | V ( G ) | + 4 | X | − , does property (*) hold for every -planar drawing D of G with the minimum number of crossings? From Claims 3 and 5 in the proof of Theorem 3, we can see that if H ′ does not haveseparating 2-cycles and | X > | ≥
3, where X > is the set of non-isolated vertices in H ′ (i.e., the set of vertices in X which are incident with crossed edges of D ), then | E ( G ) | ≤ | V ( G ) | + 4 | X > | −
12 holds. 18 roblem 2
Let G be a bipartite -planar graph with partite sets X and Y , where ≤ | X | ≤ | Y | . If D is a -planar drawing of G with the minimum number ofcrossings and | X > | ≥ , where X > is the set of vertices in X which are incidentwith crossed edges of D , does | E ( G ) | ≤ | V ( G ) | + 4 | X > | − hold? Theorem 3 shows that | E ( G ) | ≤ | V ( G ) | + 4 x −
12 holds for any bipartite 1-planargraph G with bipartite sets of sizes x and y , where 2 ≤ x ≤ y . For any x ≥ y ≥ x −
12, Czap, Przybylo and ˘Skrabul´akov´a [5] constructed a bipartite 1-planargraph G with bipartite sets of sizes x and y and | E ( G ) | = 2 | V ( G ) | + 4 x −
12. Noticethat these graphs constructed in [5] have minimum degree 3. By Theorem 1, anybipartite 1-planar graph of n vertices has at most 3 n − Problem 3
Let ≤ t ≤ and G be any bipartite -planar graph with partitesets X and Y , where t ≤ | X | ≤ | Y | . If G is t -connected (or δ ( G ) = t ), does | E ( G ) | ≤ | V ( G ) | + f ( t ) | X | + c holds for some f ( t ) < ? Let t ≥
2. A drawing of a graph is t - planar if each of its edges is crossed at most t times. If a graph has a t -planar drawing, then it is t - planar . Does Theorem 3 havean analogous result for bipartite 2-planar graphs? Problem 4
Let G be a bipartite -planar graph with partite sets X and Y , where ≤ | X | ≤ | Y | . Determine constants a, b and c such that | E ( G ) | ≤ a | V ( G ) | + b | X | + c . Lemma 1 gives an upper bound for the size of a bipartite graph G with partitesets X and Y , where | X | = 3, which has a 1-disc O X drawing. Can this result begeneralized for such a bipartite graph without the condition that | X | = 3? Problem 5
Let G be a bipartite graph with partite sets X and Y which has a 1-disc O X drawing. Is it true that | E ( G ) | ≤ | Y | + 5 | X | / − ? References [1] J. A. Bondy and U. S. R. Murty,
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