On the sizes of bipartite 1-planar graphs ^††thanks: This work is supported by MOE-LCSM, School of Mathematics and Statistics, Hunan Normal University, Changsha, Hunan 410081, P. R. China

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 are such ones that no edge crosses itself, no two edges cross more than once, and no two edges incident with the same vertex cross. The crossing number of a graph $G$ is the smallest 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 a graph has a 1-planar drawing, then it is 1-planar. The notion of 1-planarity was introduced in 1965 by Ringel [12], and since then many properties of 1-planar graphs have been studied (e.g. see the survey paper [9]).

It is well-known that any simple planar graph with $n$ ($n\geq 3$) vertices has at most $3n-6$ edges, and a simple and bipartite graph with $n$ ($n\geq 3$) vertices has at most $2n-4$ edges. Determining the maximum number of edges in 1-planar graphs with a fixed number of vertices has aroused great interest of many authors (see, for example, [3], [5],[7], [11], [14]). It is known that [3, 7, 11] any 1-planar graph with $n$ $(\geq 3)$ vertices has at most $4n-8$ edges. For bipartite 1-planar graphs, an analogous result was due to Karpov [8].

Note that Karpov’s upper bound on the size of a bipartite 1-planar graph is in terms of its vertex number. When the sizes of partite sets in a bipartite 1-planar graph are taken into account, Czap, Przybylo and S̆krabuláková [6] obtained another upper bound for its size (i.e., Corollary 2 in [6]).

For any graph $G$, let $V(G)$ and $E(G)$ denote its vertex set and edge set.

For each pair of integers $x$ and $y$ with $x\geq 3$ and $y\geq 6x-12$, the authors in [6] constructed a bipartite 1-planar graph $G$ with partite sets of sizes $x$ and $y$ such that $|E(G)|=2|V(G)|+4x-12$ holds. Moreover, they believed this lower bound is optimal for such graphs and thus posed the following conjecture.

In this paper we obtain the following result which proves Conjecture 1.

The result in [6, Lemma 4] shows that the upper bound for $|E(G)|$ in Theorem 3 is tight. Also, if $x\leq\frac{1}{3}(y+4)$, Theorem 3 provides a better upper bound for $|E(G)|$ than Theorem 1.

The authors in [6] mentioned a question of Sopena [13]: How many edges we have to remove from the complete bipartite graph with given sizes of the partite sets to obtain a 1-planar graph? It is not hard to see that Theorem 3 implies the follow corollary which answers the problem.

The remainder of this paper is arranged as follows. In Section 2, we explain some terminology and notation used in this paper. In Section 3, under some restrictions, we present several structural properties on an extension of $D^{\times}$ for a 1-planar drawing $D$ of a bipartite 1-planar graph $G$, where $D^{\times}$ is a plane graph introduced in Section 2. Some important lemmas for proving Theorem 3 are given in Section 4, while the proof of this theorem is completed in Section 5. Finally, we give some further problems in Section 6.

All graphs considered here are simple, finite and undirected, unless otherwise stated. For terminology and notation not defined here, we refer to [2]. For any graph $G$ and $A\subseteq V(G)$, let $G[A]$ denote the subgraph of $G$ with vertex set $A$ and edge set $\{e\in E(G):e\mbox{ joins two vertices in }A\}$. $G[A]$ is called the subgraph of $G$ induced by $A$. For a proper subset $A$ of $V(G)$, let $G-A$ denote the subgraph $G[V(G)\setminus A]$.

A walk in a graph is alternately a vertex-edge sequence; the walk is closed if its original vertex and terminal vertex are the same. A path (respectively, a cycle ) of a 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 $C$ is composed of two paths $P_{1}$ and $P_{2}$, we sometimes write $C=P_{1}\cup P_{2}$.

A plane graph is a planar graph together with a drawing without crossings, and at this time we say that $G$ is embedded in the plane. A plane graph $G$ partitions the plane into a number of connected regions, each of which is called a face of $G$. If a face is 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 walks of $G$. The size of a face is the number of the edges contained in its the boundary with 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 bounded one (i.e., the interior of $C$) and the unbounded one (i.e., the exterior of $C$). We denote by $int(C)$ and $ext(C)$ the interior and exterior of $C$, respectively, and their closures by $INT(C)$ and $EXT(C)$. Clearly, $INT(C)\cap EXT(C)=C$. A cycle $C$ of a plane graph $G$ is said to be separating if both $int(C)$ and $ext(C)$ contain at least one vertex of $G$.

Let $D$ be a 1-planar drawing of a graph $G$. The associated plane graph $D^{\times}$ is the plane graph that is obtained from $D$ by turning all crossings of $D$ into new vertices of degree four; these new vertices of degree four are called the crossing vertices of $D^{\times}$.

Throughout this section, we always assume that the considered graph $G$ (possibly disconnected) is a bipartite 1-planar graph with partite sets $X$ and $Y$, where $3\leq|X|\leq|Y|$. Let $D$ be a 1-planar drawing of $G$ with the minimum number of crossings, and $D^{\times}$ be the associated graph of $D$ with the crossing vertex set $W$.

Note that subsets $X,Y$ and $W$ form a partition of $V(D^{\times})$. We color the vertices in $X,Y$ and $W$ by the black color, white color and red color respectively. As stated in [6], $D^{\times}$ can be extended to a plane graph, denoted by $D^{\times}_{W}$, by adding edges joining black vertices as described below:

for each vertex $w$ in $W$, it is adjacent to two black vertices in $D^{\times}$, say $x_{1}$ and $x_{2}$, and we draw an edge, denoted by $e_{w}$, joining $x_{1}$ and $x_{2}$ which is “most near” one side of the path $x_{1}wx_{2}$ of $D^{\times}$ such that it does not cross with any other edge, as shown in Figure 1 (b).

Observe that $D^{\times}_{W}$ is a plane graph with $D^{\times}$ as its spanning subgraph and the edge set of $D^{\times}_{W}$ is the union of $E(D^{\times})$ and $\{e_{w}:w\in W\}$. Although $D^{\times}$ is a simple graph, $D^{\times}_{W}$ might contain parallel edges (i.e., edges with the same pair of ends), as there may exist two edges in $\{e_{w}:w\in W\}$ with the same pair of ends. An example is shown in Figure 2 (c), where $D$ is a $1$-planar drawing of $K_{3,6}$.

Let $F_{D}$ (or simply $F$) and $H_{D}$ (or simply $H$) denote the subgraphs $D^{\times}_{W}[W\cup X]$ and $D^{\times}_{W}[X]$ respectively. Obviously, $H$ is a subgraph of $F$ and its edge set is $\{e_{w}:w\in W\}$, while the edge set of $F$ is the union of $E(H)$ and $\{wx_{1},wx_{2}\in E(D^{\times}):w\in W\ \&\ x_{1},x_{2}\in X\}$.

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 degree $2$ in $F$. The edges in $F$ incident with red vertices are called red edges. Thus, each edge in $F$ is either black or red, as shown in Figure 2 (d).

We have the following facts on $D^{\times}$, $F$ and $H$:

1. (1)

   $D^{\times}$, $F$ and $H$ are simultaneously embedded in the plane;

2. (2)

   $F$ and $H$ are obviously loopless, but they are possibly disconnected;

3. (3)

   $w\rightarrow e_{w}$ is a bijection from $W$ to $E(H)$, where $w$ is a red vertex, and thus the number of crossings of $D$ equals to $|E(H)|$; and

4. (4)

   $e_{w}\rightarrow x_{1}wx_{2}$ is a bijection from $E(H)$ to the set of $2$-paths in $F$ whose ends are black, where $w$ is a red vertex and $x_{1}$ and $x_{2}$ are the black vertices in $D^{\times}$ adjacent to $w$.

Moreover we have the following propositions.

Proof.  By the definition of $D^{\times}_{W}$, the proposition follows directly from the fact that the drawing of edge $e_{w}$ is most near one side of the 2-path $x_{1}wx_{2}$ in $D^{\times}$ without crossings with edges in $D^{\times}$. $\Box$

Proof.  As $H$ has no separating 2-cycles, either $int(C)$ or $ext(C)$ contains no black vertices. Assume that $int(C)$ does not contain black vertices.

Suppose that $int(C)$ contains red vertices. Then, $int(C)$ contains white vertices of $D^{\times}$. As $int(C)$ does not contain black vertices, each white vertex in $int(C)$ is of degree at most $2$ in $D^{\times}$. Thus, we can redraw the edges of $D$ in $int(C)$ such that these edges make no crossings, and then obtain a 1-planar drawing of $G$ with fewer crossings than $D$, contradicting to the choice of $D$. Hence $int(C)$ does not contain red vertices and the conclusion holds. $\Box$

Proof.  Assume to contrary that $H$ has three multiple edges $e_{1}$, $e_{2}$ and $e_{3}$ which join the same pair of black vertices $x_{1}$ and $x_{2}$. Then these three edges divide the plane into three regions, denoted by $\alpha$, $\beta$ and $\gamma$, as shown in Figure 4 (a). By Proposition 2, at least two of these three regions contain neither red vertices nor black vertices, except on its boundary. We may assume $\alpha$ and $\gamma$ are such two regions.

Let $P=x_{1}wx_{2}$ be the 2-path of $F$ that corresponds to edge $e_{3}$, where $w$ is a red vertex. Thus, this path must be within region $\beta$, as shown in Figure 4 (b).

As $P$ is within region $\beta$, black edges $e_{1}$ and $e_{2}$ are in different sets $int(e_{3}\cup P)$ and $ext(e_{3}\cup P)$, a contradiction to Proposition 1. The proof is then completed. $\Box$

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 separating 2-cycles, then each partnered edge $e$ in $H$ is parallel to a unique partnered edge $e^{\prime}$ in $H$.

Let $C$ be a cycle and $P$ be a path in $H$ such that the end vertices of $P$ are the only vertices in both $C$ and $P$. When we say that $P$ lies in $int(C)$ (resp. $ext(C)$), it means that all edges and internal vertices of $P$ lie in $int(C)$ (resp. $ext(C)$).

Proof.  Let $C_{0}$ denote the 2-cycle of $H$ consisting of edges $e$ and $e^{\prime}$. By Proposition 2, we may assume that $int(C_{0})$ contains neither black vertices nor red vertices. Thus, both $w$ and $w^{\prime}$ are in $ext(C_{0})$.

Let $C_{1}$ denote the $3$-cycle of $F$ consisting of edge $e$ and path $P$ and $C^{\prime}_{1}$ the $3$-cycle of $F$ consisting of edge $e^{\prime}$ and path $P^{\prime}$. By Proposition 1, both $int(C_{1})$ and $int(C^{\prime}_{1})$ are empty. Thus, the subgraph $F[\{x_{1},x_{3},w,w^{\prime}\}]$ is as shown in Figure 5 (a).

As these three regions $int(C_{0})$, $int(C_{1})$ and $int(C^{\prime}_{1})$ do not contain black vertices, $x_{2}$ must be in $ext(C_{2})$, where $C_{2}$ is the $4$-cycle of $F$ consisting of paths $P=x_{1}wx_{3}$ and $P^{\prime}=x_{1}w^{\prime}x_{3}$. As $F$ is a plane graph, path $x_{1}x_{2}x_{3}$ must lies in $ext(C_{2})$, as shown in Figure 5 (b).

Hence the conclusion holds. $\Box$

Proof.  Let $e_{1},e_{2}$ and $e_{3}$ be the three edges on $C$. Suppose that $e_{i}$ is not a simple edge of $H$, where $1\leq i\leq 3$. Then $e_{i}$ is parallel to another partnered edge $e^{\prime}_{i}$ of $H$. Let $P_{i}$ and $P^{\prime}_{i}$ be 2-paths in $F$ which correspond to edges $e_{i}$ and $e^{\prime}_{i}$ respectively. Since $H$ has no separating 2-cycles, by Proposition 4, $int(C)$ contains a red vertex that is on $P_{i}$ or $P^{\prime}_{i}$.

The above conclusion implies that the number of red vertices in $int(C)$ is not less than the number of partnered edges on $C$. Thus, the result holds. $\Box$

Let $G$ be a bipartite graph with partite sets $X$ and $Y$ and ${\cal O}$ 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 ${\cal O}$ and all vertices of $Y$ and all edges of $G$ in the interior of ${\cal O}$, then we say that $D$ is a $1$-disc ${\cal O}_{X}$ drawing of $G$.

Proof.  Assume that $|X|=3$. For any integer $i\geq 0$, let $Y_{i}$ be the set of vertices $y$ in $Y$ with $d_{G}(y)=i$. As $|X|=3$ and $Y$ is independent in $G$, $Y_{i}=\emptyset$ holds for all $i\geq 4$.

It can be checked easily that, for each vertex $y$ in $\in Y$, if $y\notin Y_{3}$, then $y$ is not incident with any crossed edge. Thus, $G-\bigcup_{i\leq 2}Y_{i}$ has exactly $k$ crossings, and it suffices to show that $|Y_{3}|\leq 3$ and

$$k=\left\{\begin{array}[]{ll}0,&\mbox{if }|Y_{3}|\leq 1;\\ 1,&\mbox{if }|Y_{3}|=2;\\ 3,&\mbox{if }|Y_{3}|=3.\end{array}\right.$$

The rest of the proof will be completed by showing the following claims.

Claim (a): $|Y_{3}|\leq 3$.

Suppose that $|Y_{3}|\geq 4$. Then, there exists a bipartite $1$-planar drawing $D^{\prime}$ isomorphic to $K_{3,2|Y_{3}|}$ obtained from $D[X\cup Y_{3}]$ by copying all vertices and edges in the interior of ${\cal O}_{X}$ to its exterior, implying that $K_{3,8}$ is 1-planar. It is a contradiction to the fact that $K_{3,7}$ is not 1-planar due to Czap and Hudák [4].

Thus, Claim (a) holds.

Claim (b): If $|Y_{3}|\leq 1$, then $G-\bigcup_{i\leq 2}Y_{i}$ has no crossings, i.e., $k=0$.

Claim (b) can be verified easily.

Claim (c): For any two vertices $y_{1},y_{2}\in Y_{3}$, some edge incident with $y_{1}$ crosses with some edge incident with $y_{2}$, as shown in Figure 6 (a).

If Claim (c) fails, then $G[X\cup\{y_{1},y_{2}\}]$ is a plane graph and we can get a drawing of $K_{3,3}$ from $G[X\cup\{y_{1},y_{2}\}]$ by adding a new vertex $y^{\prime}$ and three edges joining $y^{\prime}$ to all vertices in $X$ in the exterior of ${\cal O}_{X}$ without any crossing, implying that $K_{3,3}$ is planar, a contradiction.

Claim (d): $k=1$ when $|Y_{3}|=2$, and $k=3$ when $|Y_{3}|=3$.

By Claim (c), $k\geq{|Y_{3}|\choose 2}$. By the drawings in Figure 6, $k\leq 1$ when $|Y_{3}|\leq 2$, and $k\leq 3$ when $|Y_{3}|\leq 3$. Thus, Claim (d) holds.

The result follows from Claims (a), (b) and (d). $\Box$

Proof.  If $c=1$, since $G$ is simple, each face of $H$ is a cellular face and has size at least 3. Then, in this case, the conclusion can be proved easily by applying the Euler’s formula.

Now we assume that $c\geq 2$. We can obtain a simple and connected plane graph $G^{\prime}$ from $G$ by adding $c-1$ edges.

For every noncellular face $F$ of $G$, we assume that its boundary consists of $\ell$ disjoint closed walks of $G$, and then we can add $\ell-1$ new edges (not add the vertex) by appropriately drawing these new edges within $F$ so that $F$ is transformed into a cellular face of size at least 4 because $|V(G)|\geq 3$. Therefore, the resulting graph $G^{\prime}$ is a simple and connected plane, and all faces of $G^{\prime}$ are cellular.

Note that adding the $c-1$ new edges does not produce new cellular 3-faces, and thus $G^{\prime}$ has exactly $t$ faces of size 3. The conclusion for connected plane graphs implies that

$$|E(G^{\prime})|\leq 2|V(G^{\prime})|-4+t/2.$$

As $V(G^{\prime})=V(G)$ and $|E(G^{\prime})|=|E(G)|+c-1$, the above inequality implies that $|E(G)|\leq 2|V(G)|+t/2-3-c$. $\Box$

Proof.  If $G$ is connected (i.e. $c=1$), since $G$ is bipartite and simple, then each face of $G$ is a cellular face, and has the size at least 4. Because $G$ has $t$ faces of size at least 6, it follows from the Euler’s formula that $|E(G)|\leq 2|V(G)|-4-t$.

Now assume that $c\geq 2$. We can obtain a simple and connected bipartite plane graph $G^{\prime}$ from $G$ by adding $c-1$ edges.

For every noncellular face $F$ of $G$ consisting of $\ell$ distinct closed walks, similar to the proof of Lemma 2, we can add $\ell-1$ new edges within this noncellular face so that $F$ is transformed into a cellular face. We can ensure that those new added edges join the vertices in different partite sets of $G$. Hence the resulting plane graph $G^{\prime}$ is simple, bipartite and connected. Clearly, all faces of $G^{\prime}$ are cellular, and $G^{\prime}$ has at least $t$ faces whose boundaries are of length at least $6$. The conclusion for bipartite and connected plane graphs implies that

$$|E(G^{\prime})|\leq 2|V(G^{\prime})|-4-t.$$

As $V(G^{\prime})=V(G)$ and $|E(G^{\prime})|=|E(G)|+c-1$, the above inequality implies that $|E(G)|\leq 2|V(G)|-3-c-t$. $\Box$

Remark: Lemmas 2 and 3 can be strengthened when $G$ contains isolated vertices. Let $V_{\geq 1}(G)$ be the set of non-isolated vertices in $G$. Then, under the condition $|V_{\geq 1}(G)|\geq 3$, the conclusions of both Lemmas 2 and 3 still hold after $|V(G)|$ is replaced by $|V_{\geq 1}(G)|$.

The whole section contributes to the proof of Theorem 3.

Proof of Theorem 3. Suppose that Theorem 3 fails and $\chi$ is the minimum integer with $\chi\geq 2$ such that for some bipartite 1-planar graph $G$ with partite sets $X$ and $Y$, where $\chi=|X|\leq|Y|$, $|E(G)|>2|V(G)|+4|X|-12$ holds.

We will prove the following claims to show that this assumption leads to a contradiction.

Claim 1: $\chi\geq 4$.

Proof. Let $G$ be any bipartite $1$-planar graph with bipartitions $X$ and $Y$, where $2\leq|X|\leq|Y|$. If $|X|=2$, obviously,

$$|E(G)|\leq 2|Y|=2(2+|Y|)+4\times 2-12=2|V(G)|+4|X|-12.$$

Now assume that $|X|=3$. Let $Y_{i}$ be the set of vertices $y$ in $Y$ with $d_{G}(y)=i$. Then $|E(G)|\leq 2|Y|+|Y_{3}|$. Since the complete bipartite graph $K_{3,7}$ is not 1-planar (see [4]), we have $|Y_{3}|\leq 6$, implying that $|E(G)|\leq 2|Y|+6$. As $x=3$, we have

$$2|V(G)|+4|X|-12=6|X|+2|Y|-12=2|Y|+6.$$

Thus, $|E(G)|\leq 2|V(G)|+4|X|-12$.

By the assumption of $\chi$, we have $\chi\geq 4$. $\Box$

In the following, we assume that $G$ is a bipartite $1$-planar graph with bipartitions $X$ and $Y$, where $\chi=|X|\leq|Y|$, such that

$$|E(G)|>2|V(G)|+4|X|-12.$$

(1)

Let $D$ be a 1-planar drawing of $G$ with the minimum number of crossings and $W$ be the set of its crossings. Introduced in Section 3, $D^{\times}_{W}$ is a plane graph extended from $D^{\times}$, and $F$ and $H$ are the subgraphs $D^{\times}_{W}[X\cup W]$ and $D^{\times}_{W}[X]$ of $D^{\times}_{W}$ respectively. All vertices in $X$ are black vertices, all vertices in $Y$ are white vertices and all vertices in $W$ 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 two parallel edges $e_{1}$ and $e_{2}$ joining black vertices $x_{1}$ and $x_{2}$ (see Figure 3 (b)), such that both $int(C)$ and $ext(C)$ contain black vertices.

Let $G_{1}=G\bigcap INT(C)$ and $G_{2}=G\bigcap EXT(C)$. Obviously, both $G_{1}$ and $G_{2}$ are bipartite 1-planar subgraphs of $G$. Moreover, we can see that $G_{1}\cup G_{2}=G$, and $G_{1}\cap G_{2}=\{x_{1},x_{2}\}$.

For $i=1,2$, $G_{i}$ has a bipartition $X_{i}$ and $Y_{i}$, where $X_{i}=X\cap V(G_{i})$ and $Y_{i}=Y\cap V(G_{i})$. Clearly, $|X_{1}|+|X_{2}|=|X|+2=\chi+2$ and $|Y_{1}|+|Y_{2}|=|Y|$.

Since $C$ is a separating cycle of $H$, both $int(C)$ and $ext(C)$ contain black vertices, implying that $|X_{i}|\geq 3$ for both $i=1,2$. As $|X_{1}|+|X_{2}|=|X|+2$, we have $|X_{i}|<|X|=\chi$ and so $\min\{|X_{i}|,|Y_{i}|\}\leq|X_{i}|<\chi$.

For $i=1,2$, if $|Y_{i}|\leq 1$, then $|E(G_{i})|\leq|Y_{i}|\cdot|X_{i}|$ and $|E(G_{i})|\leq 2|V(G_{i})|+4|X_{i}|-12$ holds; if $|Y_{i}|\geq 2$, then $2\leq\min\{|X_{i}|,|Y_{i}|\}\leq|X_{i}|<\chi$ and the assumption on $\chi$ implies that the conclusion of Theorem 3 holds for $G_{i}$, i.e.,

$\displaystyle|E(G_{i})|\leq 2|V(G_{i})|+4\min\{|X_{i}|,|Y_{i}|\}-12\leq 2|V(G_{i})|+4|X_{i}|-12.$

(2)

Thus, by (2),

	$\displaystyle|E(G)|$	$\displaystyle=$	$\displaystyle|E(G_{1})|+|E(G_{2})|$		(3)
		$\displaystyle\leq$	$\displaystyle 2(|V(G_{1})|+|V(G_{2})|)+4(|X_{1}|+|X_{2}|)-24$
		$\displaystyle=$	$\displaystyle 2(|V(G)|+2)+4(|X|+2)-24$
		$\displaystyle=$	$\displaystyle 2|V(G)|+4|X|-12,$

which contradicts to the assumption in (1).

Hence Claim 2 holds. $\Box$

It is known from Proposition 3 that the edge multiplicity of each edge in $H$ is at most 2. Then, there exists a subset $A$ of $E(H)$ such that both $H\langle A\rangle$ and $H-A$ are simple graphs and each edge in $A$ is parallel to some edge in $E(H)-A$, where $H\langle A\rangle$ is the spanning subgraph of $H$ with edge set $A$ and $H-A$ is the graph obtained from $H$ by removing all edges in $A$. Clearly, $|A|$ is the number of pairs of edges $e$ and $e^{\prime}$ in $H$ which are parallel.

Let $H^{\prime}$ denote $H-A$. Obviously, $|A|\leq|E(H^{\prime})|$, and

$$|E(H)|=|E(H^{\prime})|+|A|.$$

(4)

Claim 3: $H^{\prime}$ contains at least one cellular 3-face.

Proof. Suppose that $H^{\prime}$ has no cellular 3-faces. As $H^{\prime}$ is a simple plane graph, by Lemma 2, $|E(H^{\prime})|\leq 2|V(H^{\prime})|-4=2|X|-4$. Because $|A|\leq|E(H^{\prime})|$, it follows from (4) that $|E(H)|\leq 2|E(H^{\prime})|\leq 4|X|-8$.

Since each edge of $H$ is in 1-1 correspondence with a crossing of a drawing $D$ of $G$, we can obtain a simple bipartite plane graph (possibly disconnected), denoted by $G^{\prime}$, by removing $|E(H)|$ edges from $G$ each of which is a crossed edge of $G$. By Lemma 3, $|E(G^{\prime})|\leq 2|V(G^{\prime})|-4=2|V(G)|-4$. Therefore,

$$|E(G)|=|E(G^{\prime})|+|E(H)|\leq|E(G^{\prime})|+4|X|-8=2|V(G)|+4|X|-12,$$

a contradiction to the assumption in (1).

Hence Claim 3 holds. $\Box$

Now we assume that $H^{\prime}$ has exactly $t$ cellular 3-faces, where $t\geq 1$. Let $\mathscr{T}(H^{\prime})$ denote the set of cellular 3-faces in $H^{\prime}$. So $t=|\mathscr{T}(H^{\prime})|$.

For each $\Delta\in\mathscr{T}(H^{\prime})$, for convenience we also use “$\Delta$” to represent the 3-cycle corresponding the boundary of $\Delta$ if there is no confusions in the context. Let $G_{\Delta}=G\bigcap INT(\Delta)$. Since $\Delta$ is a cellular 3-face of $H^{\prime}$, there are no black vertices lying in $int(\Delta)$, and thus $G_{\Delta}$ is a bipartite with with exactly three black vertices, which lie on the boundary of the face $\Delta$.

Let $\Delta\in\mathscr{T}(H^{\prime})$. Since $D$ is a 1-planar drawing of $G$ with minimal number of crossings, the induced subdrawing of $D$ of $G_{\Delta}$ is a 1-disc ${\cal O}_{X_{\Delta}}$ drawing of $G_{\Delta}$ with the minimum number of crossings, where $X_{\Delta}$ is the set of three black vertices on the boundary of $\Delta$. Otherwise, we redraw the edges of $G$ lying in the interior of $\Delta$, and obtain a 1-planar drawing of $G$ with fewer crossings than $D$, contradicting to the choice of $D$. By Lemma 1, the number of crossings of $D$ in $int(\Delta)$ is a number in the set $\{0,1,3\}$.

For any $j\in\{0,1,3\}$, let $\mathscr{T}^{(j)}(H^{\prime})$ be the set of members $\Delta$ in $\mathscr{T}(H^{\prime})$ such that $int(\Delta)$ contains exactly $j$ crossings of $D$. Assume that $\mathscr{T}^{(j)}(H^{\prime})=\{\Delta^{(j)}_{i}:1\leq i\leq t_{j}\}$, where $t_{j}=|\mathscr{T}^{(j)}(H^{\prime})|$. Thus, $t_{0}+t_{1}+t_{3}=t$.

For each $\Delta^{(j)}_{i}\in\mathscr{T}^{(j)}(H^{\prime})$, let $e^{(j)}_{i}$ be the number of the edges of the graph $G_{\Delta^{(j)}_{i}}$ and $y^{(j)}_{i}$ be the number of white vertices in $int(\Delta^{(j)}_{i})$.

Claim 4: $\sum\limits_{j\in\{0,1,3\}}\sum\limits_{i=1}^{t_{j}}e^{(j)}_{i}\leq 2\sum\limits_{j\in\{0,1,3\}}\sum\limits_{i=1}^{t_{j}}y^{(j)}_{i}+(t_{0}+2t_{1}+3t_{3})$.

Proof. By Lemma 1, $e^{(j)}_{i}\leq 2y^{(j)}_{i}+1+\big{\lceil}\sqrt{j}\big{\rceil}$ holds for any $j\in\{0,1,3\}$ and $1\leq i\leq t_{j}$. Thus, Claim 4 holds. $\Box$

Claim 5: $|E(H)|\leq 4|X|-8+t-({3t_{0}}+2t_{1})/2$.