On the Edge-Connectivity and Restricted Edge-Connectivity of Optimal 1-Planar Graphs

Abstract

A graph is called 1-planar if it can be drawn on the plane (or on the sphere) such that each edge is crossed at most once. It is known that a 1-planar graph G with at least three vertices has at most $4|V(G)|- 8$ edges, so G is optimal if G has exactly $4|V(G)| - 8$ edges. The vertex-connectivity of an optimal 1-planar graph has been proven to be either 4 or 6. According to the Whitney inequality, the edge-connectivity of an optimal 1-planar graph is at least 4. In this paper, we provide a more precise result. We prove that the edge-connectivity of every optimal 1-planar graph is 6. Additionally, we prove that any minimum edge-cut of an optimal 1-planar graph is a set of all the edges incident with a vertex of degree 6. In addition, we prove that an optimal 1-planar graph has the restricted edge-connectivity 8, 10 or 12.

1 Introduction

A graph is planar if it can be drawn on the plane (or on the sphere) such that no edges cross each other. A graph is 1-planar if it can be drawn on the plane (or on the sphere) such that each edge is crossed at most once. A graph, together with a 1-planar drawing (resp. planar drawing) is called a 1-plane graph (resp. plane graph). Clearly, 1-planar graphs are a natural generalization of planar graphs. 1-planar graphs were first studied by Ringel (1965) [28] in the connection with the problem of the simultaneous coloring of the vertices and faces of plane graphs. Since then, many properties of 1-planar graphs have been widely investigated; see [1, 4, 7, 9, 18, 19, 23, 36], for examples, see also [24] for a survey, and see some sections in [22].

It is well known that a 1-planar graph with $n \ge 3$ vertices has at most $4n-8$ edges. A 1-planar graph with n vertices is called optimal if it has exactly $4n-8$ edges. Several articles [5, 6, 30] demonstrate that there are no n-vertex optimal 1-planar graphs with $n \le 7$ or $n = 9$ vertices, while n-vertex optimal 1-planar graphs always exist with $n = 8$ and $n \ge 10$ vertices. Optimal 1-planar graphs have also been studied extensively due to their interesting properties. Suzuki [30] proved that a 1-planar graph G is an optimal 1-planar graph if and only if G can be obtained from a 3-connected quadrangulation Q on the sphere by adding a pair of crossing edges to each face of Q. Didimo [11] proved that no optimal 1-planar graph has a 1-planar straight-line drawing. Hudák, Madaras and Suzuki [20] proved that every optimal 1-planar graph with even order is Hamiltonian and each large enough optimal 1-planar graph contains a k-vertex path whose weight (i.e., the sum of degrees of its vertices) is at most $8k-1$. Moreover, the matching extendability of optimal 1-planar graphs was characterized in [14, 34]. Suzuki [31] proved that every optimal 1-planar graph has a $K_6$-minor and characterized optimal 1-planar graphs having no $K_7$-minor. Lenhart et al. [25] proved that every optimal 1-plane graph has a red–blue edge coloring such that the blue subgraph is maximal plane while the red subgraph has vertex degree at most four. Brandenburg [8] gave a linear-time algorithm for recognizing optimal 1-planar graphs. Recently, Zhang and Huang [35] characterized the reducibility of optimal 1-planar graphs with respect to the lexicographic product.

It is worth noting that although optimal 1-planar graphs have been widely studied, there are still many open problems surrounding them that await further research. For example, Bekos et al. [3] conjectured that the book thickness of every optimal 1-planar graph is 4. In the recent article [15], Gollin et al. raised an open problem of finding upper bounds of k-cliques $(3\le k \le 5)$ among optimal 1-planar graphs with n vertices.

Connectivity is a classic concept in graph theory, which asks for the minimum number of elements (vertices or edges) that need to be removed to separate the remaining vertices into two or more connected components. Let G be a connected graph. A vertex-cut (resp. edge-cut) of a connected graph G is a set of vertices (edges) whose removal makes G disconnected. A minimum vertex-cut (resp. minimum edge-cut) of a graph is a vertex-cut (resp. edge-cut) of smallest possible size. An edge-cut is a trivial edge-cut if it consists of exactly all the edges having a common vertex. For a non-complete graph G, the vertex-connectivity $\kappa (G)$ is the size of the minimum vertex-cut of G; the vertex-connectivity of the complete graph $K_n$ is defined to be $n-1$. The edge-connectivity $\lambda (G)$ is the size of a minimum edge-cut. A graph is called k-vertex-connected (resp. k-edge-connected) if its vertex-connectivity (resp. edge-connectivity) is at least k. For simplicity, we usually shorten “k-vertex-connected graph” to “k-connected graph.”

Let G be a 1-planar graph with a 1-planar drawing D. We define the planar skeleton $\mathcal {S}(D)$ of D to be the spanning subgraph of G containing exactly the edges of G which are not crossed. Throughout the argument in [14, 30], the vertex-connectivity of an optimal 1-planar graph is either 4 or 6. If the vertex-connectivity of an optimal 1-planar graph is 4 (resp. 6), then there exists a separating 4-cycle (resp. 6-cycle) of its planar skeleton. It is not difficult to prove that the minimum degree of an optimal 1-planar graph is 6 [22, 27]. Thus, if an optimal 1-planar graph G has the vertex-connectivity 6, then by the well-known Whitney inequality (i.e., for every graph H, $\kappa (H) \le \lambda (H) \le \delta (H)$ where $\delta (H)$ is the minimum degree of H), the edge-connectivity of G is 6. But for an optimal 1-planar graph with vertex-connectivity 4, we can only roughly say that its edge-connectivity is either 4, 5 or 6. In this paper, we provide a more precise result. We prove that the edge-connectivity of every optimal 1-planar graph is 6. Furthermore, we also prove that any minimum edge-cut of an optimal 1-planar graph is formed by all edges incident with a vertex of degree 6. That is to say, every minimum edge-cut of an optimal 1-planar graph is trivial. In some literature, a graph G is called super-edge-connected if every minimum edge-cut of G is trivial. Hence, our result means that every optimal 1-planar graph is super-edge-connected.

The edge-connectivity has many interesting generalizations. In 1983, Harary [16] proposed the concept of conditional edge-connectivity. The P-edge-connectivity of a connected graph G equals the minimum cardinality of a set S of edges such that $G-S$ is disconnected and every component of $G-S$ has a given property P. As a specific example, Esfahanian and Hakimi [12] proposed in 1988 the notion of restricted edge-cut and restricted edge-connectivity. An edge-cut F of a graph G is called a restricted edge-cut if $G - F$ contains no isolated vertices. The restricted edge-connectivity$\xi (G)$ of a graph G is the minimum cardinality over all its restricted edge-cuts. A restricted edge-cut F of G is called minimum if no other restricted edge-cut with fewer edges exists, hence $|F|=\xi (G)$. Readers may refer to references [13, 17, 26, 33] for related studies on the restricted edge-connectivity of graphs. Our second work in this paper is to prove that an optimal 1-planar graph has the restricted edge-connectivity 8, 10 or 12.

The remaining sections of this paper are organized as follows: Sect. 2 contains preliminary notions and auxiliary lemmas on the properties of optimal 1-planar graphs and their minimum edge-cuts (among them, the upper bound on edge-connectivity and possible cardinality of minimal edge-cuts). In Sect. 3, we prove the main result of the paper that the edge-connectivity of every optimal 1-planar graph is 6, and each minimum 6-edge-cut is trivial (that is, consisting of all edges around a 6-valent vertex). The final Sect. 4 deals with the concept of restricted edge-connectivity of optimal 1-planar graphs, stating its possible values of 8, 10 or 12 with examples demonstrating that the values 10 and 12 are attained, while conjecturing that the remaining value 8 is never attained.

2 Preliminaries

Only simple graphs are considered in this paper unless otherwise stated. Let $G=(V,E)$ be a graph where V is its vertex set and E is its edge set.

If $e=uv$ is an edge of G, then we say u and v are adjacent, u and v are neighbors, and e and u are incident. The neighborhood$N_G(v)$ of a vertex v in G is the set of all neighbors of v. The degree$d_G(v)$ of a vertex v in G is the number of edges incident to it. A subgraph of a graph G is a graph H such that $V(H) \subseteq V(G)$ and $E(H) \subseteq E(G)$. A spanning subgraph of G is a subgraph H such that $V(H)=V(G)$. An induced subgraph of G, induced by vertex set $S \subseteq V(G)$, is the subgraph H with vertex set S such that $E(H)=\{u v \in E(G): u, v \in S\}$; we write this as $H=G[S]$. Subgraphs may also be induced by sets of edges. If S is a set of edges, the edge-induced subgraph G[S] is the subgraph of G whose edge set is S and whose vertex set consists of all ends of edges of S. We denote by $G-E'$ the edge-induced graph $G[E\setminus E']$, and for simplicity, we denote $G- \{e\}$ as $G-e$.

A walk is a sequence $v_0, e_1, v_1, \ldots , e_k, v_k$ of vertices and edges such that, for $1 \le i \le k$, the edge $e_i$ has end-vertices $v_{i-1}$ and $v_i$. A trail is a walk with no repeated edge. An Eulerian trail is a trail in a graph that visits every edge exactly once (allowing for revisiting vertices). A closed Eulerian trail is an Eulerian trail that starts and ends in the same vertex. An Eulerian graph is a graph containing a closed Eulerian trail.

Let G be a graph. Let X and Y be disjoint sets of vertices of a graph G. We denote by [X, Y] the set of edges of G with one end-vertex in X and the other end-vertex in Y, and by |[X, Y]| their number.

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 (or on the sphere) and to each edge uv in E a continuous arc connecting D(u) and D(v). If no confusion arises, we do not distinguish between a graph–theoretical object (such as a vertex or an edge) and its drawing. All drawings considered in this paper are such that no edge crosses itself, no two edges cross more than once, and no two incident edges cross.

A plane graph is a particular drawing of a planar graph in the Euclidean plane. Let G be a connected plane graph with vertex set V(G), edge set E(G) and face set F(G). Faces of G are open 2-cells. The boundary $\partial (f)$ of a face f is the boundary in the usual topological sense. It is a collection of all edges and vertices contained in the closure of f that can be organized into a closed walk in G traversing along a simple closed curve lying just inside the face f. This closed walk is unique up to the choice of initial vertex and direction, and is called the boundary walk of the face f (see [10]). The degree of a face f is the length (i.e., the number of edges) of its boundary walk. We write $f=[u_1u_2\cdots u_tu_1]$ if $u_1, u_2,\ldots , u_t$ are the vertices of $\partial (f)$ in a cyclic ordering. A face is said to be incident with the vertices and edges in its boundary, and two faces are adjacent if their boundaries have an edge in common. A face is called k-face if its degree is k.

The terms not defined here can be found in [2] and [32].

In this following, we provide some lemmas.

Lemma 1

( [30]) Let G be an optimal 1-planar graph with a 1-planar drawing D. Then, D is obtained from a 3-connected quadrangulation by inserting a pair of crossing edges into each quadrangular face.

Lemma 2

Let G be an optimal 1-planar graph with a 1-planar drawing D. If F is an edge-cut of G, then F contains at least three edges of $\mathcal {S}(D)$.

Proof

By Lemma 1, $\mathcal {S}(D)$ is a 3-connected spanning subgraph of G. Hence, the edge-cut F contains at least three edges of $\mathcal {S}(D)$. $\square$

An edge-cut is called minimal if it does not contain any other edge-cut. Observe that every minimum edge-cut is also minimal.

The following two lemmas (without proofs) are well known.

Lemma 3

Let G be a connected graph. If F is a minimal edge-cut, then $G-F$ has exactly two components.

Lemma 4

Let G be an Eulerian graph, and let F be a minimal edge-cut. Then, |F| is even.

Lemma 5

Let G be a 3-connected plane graph. Let f and g be two faces of G. Then, the boundaries $\partial (f)$ and $\partial (g)$ either do not share any vertices, or share exactly one vertex, or share exactly one edge.

Proof

Since G is 3-connected, the boundary of every face of G is a cycle. (In fact, “2-connected” is sufficient to guarantee this property.) Suppose that the boundaries $\partial (f)$ and $\partial (g)$ have at least three vertices. Then, we can choose three common vertices $u_1$, $u_2$ and $u_3$, which are sorted in a clockwise order on $\partial (f)$ (or $\partial (g)$). Now we add a Jordan curve in each of f and g connecting $u_1$ and $u_3$, denoted by $l_1$ and $l_2$, respectively. Then, $L:=l_1\cup l_2$ is a simple closed Jordan curve passing only through $u_1$ and $u_3$, and both the interior and exterior of L contain vertices of G. Therefore, $u_1$ and $u_3$ form a 2-cut of G, a contradiction. Thus, the boundaries $\partial (f)$ and $\partial (g)$ have at most two vertices in common.

If $\partial (f)$ and $\partial (g)$ have exactly two common vertices u and v, and u and v are not adjacent, then similarly, we can still construct the aforementioned Jordan closed curve, and then we can see that u and v are also a 2-vertex-cut of G, a contradiction.

Therefore, we have proved the lemma. $\square$

Lemma 6

Let G be a 3-connected quadrangulation. Let v be a vertex of G with degree k and $v_0,v_1, \cdots , v_{k-1}$ be its neighbors in clockwise order. Let $f_i$ be the 4-face of G incident with edges $vv_i$ and $vv_{i+1}$ where indices are taken modulo k. Then, the following statements hold.

• (i) If two faces $f_i$ and $f_j$ are adjacent where $0\le i <j\le k-1$, then $\partial (f_i)$ and $\partial (f_j)$ share exactly one edge $vv_{i+1}$.

• (ii) If two faces $f_i$ and $f_j$ are not adjacent where $0\le i<j\le k-1$, then $\partial (f_i)$ and $\partial (f_j)$ share exactly one vertex v.

Proof

For (i), if $f_i$ is adjacent with $f_j$, then $f_j$ and $f_i$ share the edge $vv_i$. Furthermore, by Lemma 5, $f_i$ and $f_j$ cannot share any other vertices or edges. Thus, (i) holds.

For (ii), if $f_i$ and $f_j$ are not adjacent, then clearly $\partial (f_j)$ and $\partial (f_i)$ share the vertex v. Based on the definition of $f_i$, we can assume without loss of generality that $f_i:= [vv_iw_iv_{i+1}v]$ and $f_j:= [vv_jw_jv_{j+1}v]$. Note that v and $w_j$ are not adjacent. Now we claim that $w_i\ne w_j$ otherwise, since $f_i$ and $f_j$ share two vertices v and $w_j$, by Lemma 5, v and $w_j$ are adjacent, which is a contradiction. If $\partial (f_i)$ and $\partial (f_j)$ share $v_i$ or $v_{i+1}$, then $f_i$ and $f_j$ will be adjacent, which contradicts the assumption. $\square$

Based on the analysis of (i) and (ii) of Lemma 6, the local structure around v is shown in Fig. 1 (i).

Corollary 1

Let G be an optimal 1-planar graph with a 1-planar drawing D. Let v be a vertex. Then, the structure around v is shown in Fig. 1 (ii) where v is incident to k 4-faces $[vv_iw_iv_{i+1}v]$ in the planar skeleton $\mathcal {S}(D)$ and $v_iv_{i+1}$ crosses $vw_i$ where the indices are taken modulo k. In particular, $w_0,w_1,\cdots , w_{k-1}$ are all distinct from each other.

Fig. 1

(i) Structure around a vertex v in a 3-connected quadrangulation; (ii) structure around a vertex v in an optimal 1-planar graph

Proof

By Lemma 1 and Lemma 6, we immediately obtain this corollary. $\square$

Lemma 7

Let G be an optimal 1-planar graph. Then, the following statements hold.

• (i) Every vertex has even degree;

• (ii) the minimum degree $\delta (G)=6$;

• (iii) the edge-connectivity $\lambda (G)\le 6$;

• (iv) if F is a minimal edge-cut of G, then |F| is even.

Proof

For (i), by Corollary 1, the crossing edges and non-crossing edges incident with vertex v appear alternately, and then their numbers are equal. Therefore, the degree of vertex v is even.

For (ii), let D be a 1-planar drawing of G. By Lemma 1, $\mathcal {S}(D)$ is a 3-connected quadrangulation. Combining this with Euler’s formula for planar graphs, we can easily prove that $\mathcal {S}(D)$ has a vertex, denoted by v, with minimum degree 3. Then by Corollary 1, v will be incident to exactly three crossing edges, and thus we have $d_G(v)=6$. Clearly, v is also a vertex with minimum degree of G, as desired.

For (iii), by (ii), G has a vertex with degree 6, and thus all the incident edges of a vertex with degree 6 form a 6-edge-cut. So we have $\lambda (G)\le 6.$

For (iv), it is clear that G is connected by Lemma 1. It is well known that a connected graph is Eulerian if and only if every vertex has even degree. So (i) implies that G is Eulerian. Furthermore, we can obtain (iv) from Lemma 4. $\square$

We denote the $V(G)\setminus X$ by $\overline{X}$. We call a graph with one edge removed from the complete graph $K_4$ a diamond, denoted by $K_4^-$.

Lemma 8

Let $K_4^-$ be a diamond with vertices $\{u,v,s,t\}$ where two vertices u and v are not adjacent. Let X be a set of vertices of $K_4^-$. If $u\in X$ and $v\in \overline{X}$, then $|[X,\overline{X}]|\ge 2$.

Proof

If $|\{s,t\} \cap X|=0$ or $|\{s,t\} \cap X|=2$, then $[X,\overline{X}]=\{us,ut\}$ or $[X,\overline{X}]=\{vs,vt\}$, and thus $|[X,\overline{X}]|= 2$. If $|\{s,t\} \cap X| =1$, then $[X,\overline{X}]=\{st,sv,ut\}$ or $[X,\overline{X}]=\{su,st,vt\}$, and thus $|[X,\overline{X}]|=3$, as desired. $\square$

3 Edge-Connectivity of Optimal 1-Planar Graphs

In this section, we will prove the edge-connectivity of every optimal 1-planar graph is 6. We also characterize minimum edge-cuts of an optimal 1-planar graph.

Theorem 1

Let G be an optimal 1-planar graph. Then, $\lambda (G)=6$. Furthermore, if an edge-cut is minimum, then it consists of all the edges incident with a vertex of degree 6.

Proof

Let D be a 1-planar drawing of G. Let F be a minimum edge-cut. By Lemma 7, $|F|\le 6$. Furthermore, by Lemma 2, there is an edge, namely, xy, in F that belongs to $\mathcal {S}(D)$. By Lemma 3, the vertex set of G can be partitioned into two parts, X and $\overline{X}$, such that G[X] and $G[\overline{X}]$ are connected and $F=[X,\overline{X}]$. Without loss of generality, we assume that $x \in X$ and $y \in \overline{X}$.

First we give the following claim.

Claim 1

$d_G(x)=6$ or $d_G(y)=6$.

Proof

By Lemma 7 (i), we can suppose for a contradiction that $d_G(x)\ge 8$ and $d_G(y)\ge 8$. By Corollary 1, the structure around x is as shown in Fig. 2 (it is possible that $h_2 = w_2$).

Fig. 2

The structure around x with degree at least 8

We first note the following basic property.

Property 1

The diamond $G[\{x,u,h_1,h_2\}]-xu$ does not share any edges with both $G[\{u,v,x,y\}]$ and $G[\{u',v',x,y\}]$.

Proof

Combining the fact that $d_G(u)\ge 8$ with Lemma 6 (or Corollary 1), the property is immediately obtained. $\square$

Next, in figures, we use a square and a circle to represent a vertex belonging to X and $\overline{X}$, respectively. In addition, we use a solid black node to represent a vertex is in undetermined state.

We discuss the cases where $|\{u,v,u',v'\}\cap X|$ takes certain values.

Case 1. $|\{u,v,u',v'\}\cap X|=0.$

In this case, $F\supseteq \{xu,xv,xy,xv',xu'\}$. We observe that $x\in X$ and ${ u} \in \overline{X}$ in the diamond $G[\{x,u,h_1,h_2\}]-xu$. By Lemma  8, at least two edges in the diamond belong to F. By Property 1, these edges belonging to F in the diamond are distinct from any edges in $\{xu,xv,xy,xv',xu'\}$. Hence, $|F|\ge 7$, a contradiction.

Case 2. $|\{u,v,u',v'\}\cap X|=1.$

Considering symmetry, we only need to consider the following two subcases: $u \in X$ and $\{u',v',v\}\subseteq \overline{X}$, or $v\in X$ and $\{u,u',v'\}\subseteq \overline{X}$. In the former subcase, observe that $F\supseteq \{uv,uy,xv,xy,xu',xv'\}$. Similarly, $G[\{x,u',w_1,w_2\}]-xu'$ is a diamond where $x\in X$ and $u'\in \overline{X}$. By Lemma 8, at least two edges belong to F in the diamond. By Property 1, they are distinct from any other edges in $\{uv,uy,xv,xy,xu',xv'\}$. Hence, $|F|\ge 8$. In the latter case, we similarly find the same diamond $G[\{x,u',w_1,w_2\}]-xu'$, and further get $|F| \ge 8$, and the details are left to the reader. Thus, both subcases lead to contradictions.

Case 3. $|\{u,v,u',v'\}\cap X|=2.$

Considering symmetry, we only need to consider the following three subcases.

Subcase 3.1. $\{u,u'\}\subseteq X$ and $\{v,v'\}\subseteq \overline{X}.$ In this subcase, it follows that $F\supseteq \{uv,uy,xv,xy,xv',u'y,u'v'\}$. Hence, $|F|\ge 7$, a contradiction.

Subcase 3.2. $\{u,v\}\subseteq X$ and $\{u',v'\}\subseteq \overline{X}.$ Now, we have $F\supseteq \{uy,vy,xy,xu',xv'\}$. Similarly, as $G[x,u',w_1,w_2]-xu'$ is a diamond where $x\in X$ and $u'\in \overline{X}$, at least two edges in the diamond belong to F. By Property 1, these edges are distinct from any edges in $\{uy,vy,xy,xu',xv'\}$. So $|F|\ge 7$, a contradiction.

Subcase 3.3. $\{u,v'\}\subseteq X$ and $\{v,u'\}\subseteq \overline{X}.$ In this subcase, it follows that $F \supseteq \{uv,uy,xv,xy,xu',u'v',yv'\}$, and hence $|F|\ge 7$, a contradiction.

By the symmetry of x and y, the cases $|\{u,v,u',v'\}\cap X|=3$ and $|\{u,v,u',v'\}\cap X|=4$ are analogous to Case 2 and Case 1, respectively. They will also lead to contradictions.

Since each of the above cases leads to a contradiction, the claim holds. $\square$

By Claim 1, we can assume without loss of generality that $d_G(x)=6$. By Corollary 1, the structure around x is shown in Fig. 3. We denote $f_1:=[xyvux]$, $f_2:=[xyv'u'x]$ and $f_3:=[xu'hux]$ as three 4-faces incident with x in $\mathcal {S}(D)$, respectively.

Fig. 3

Structure around the 6-degree vertex x

Since $v'y\in \mathcal {S}(D)$, besides being incident with the 4-face $f_2$, $v'y$ is also incident with another 4-face $g=[v'z_1z_2yv']$ in $\mathcal {S}(D)$.

Claim 2

The vertex $z_1\notin N_G(x)\cup \{x\}$ and $z_2 \notin (N_G(x){\setminus } \{h,v\})\cup \{x\}$.

Proof

Clearly, $z_1\ne v',x$. If $z_1= u'$, then faces g and $f_2$ share three common vertices $y,v'$ and $u'$ in $\mathcal {S}(D)$, a contradiction to Lemma 5. Since every face of $\mathcal {S}(D)$ is a 4-face, it can be easily obtained that $\mathcal {S}(D)$ is bipartite. If $z_1= h$, then $v'u'z_1v'$ forms a 3-cycle in $\mathcal {S}(D)$, a contradiction. Similarly, $z_1\ne v$. If $z_1= u$, then g and $f_1$ share two non-adjacent vertices u and y in $\mathcal {S}(D)$, a contradiction to Lemma 5. So $z_1\notin N_G(x)$. As for $z_2$, clearly, $z_2\ne x,y,v'$. If $z_2=u'$, then $yv'z_2y$ is a 3-cycle in $\mathcal {S}(D)$, a contradiction. If $z_2=u$, then g and $f_1$ share two non-adjacent vertices u and y in $\mathcal {S}(D)$, a contradiction to Lemma 5. Therefore, we have $z_2 \notin (N_G(x){\setminus } \{h,v\})\cup \{x\}$. $\square$

We observe that $G[\{uv,vy,yv',v'u',u'h,hu\}]$ and $G[\{uu',u'y,yu\}]$ is a 6-cycle and a 3-cycle, respectively. For short, we denote $G[\{uv,vy,yv',v'u',u'h,hu\}]$ by $\mathcal {C}^1$ and $G[\{uu',u'y,yu\}]$ by $\mathcal {C}^2$.

Claim 3

Either $N_G(x) \cap X=\emptyset$ or $N_G(x) {\setminus } \{y\} \subseteq X$. Furthermore, if $N_G(x) \cap X=\emptyset$, then F consists of all six edges incident with x. If $N_G(x) {\setminus } \{y\} \subseteq X$, then $d_G(y)=6$ and F consists of all edges incident with y.

Proof

First, if $N_G(x) \cap X=\emptyset$, then clearly F contains all the edges incident with x. Since $|F|\le 6$, F consists of all the six edges incident with x. As for $N_G(x) \cap X \ne \emptyset$, we will prove that $N_G(x) {\setminus } \{y\} \subseteq X$, that is to say, $|(N_G(x) {\setminus } \{y\}) \cap X|=5$. It suffices to show that each of the four cases below leads to a contradiction.

Case 1. $|(N_G(x)\setminus \{y\}) \cap X|=1$.

Since $|(N_G(x)\setminus \{y\}) \cap X|=1$, x has five incident edges belonging to F. Without loss of generality, we denote the neighbor of x belonging to X as z. Both neighbors of z on the cycle $\mathcal {C}^1$ belong to $\overline{X}$, so two edges incident with z belong to F. Clearly, the five edges incident with x belonging to F are distinct from the two edges incident with z on the cycle. So $|F|\ge 7$, a contradiction.

Case 2. $|(N_G(x)\setminus \{y\}) \cap X|=2$.

Since $|(N_G(x)\setminus \{y\}) \cap X|=2$, there are four edges incident with x belonging to F. Without loss of generality, we denote two neighbors of x belonging to X as $z_1$ and $z_2$, respectively.

If $z_1$ and $z_2$ are not adjacent in $\mathcal {C}^1$, then both two edges incident with $z_1$ on cycle $\mathcal {C}^1$ will belong to F. The same holds for $z_2$. Furthermore, all these four edges are distinct. Combining this with the fact that four edges incident with x belong to F, we have $|F|\ge 8$, a contradiction.

As for $z_1$ and $z_2$ being adjacent in $\mathcal {C}^1$, it is not difficult to find that one of $z_1$ and $z_2$ is u or $u'$. Without loss of generality, let $z_1=u$. Hence, apart from $z_1$, all other vertices of $\mathcal {C}^2$ belong to $\overline{X}$, which implies that both edges incident with $z_1$ on the cycle $\mathcal {C}^2$ belong to F. Additionally, it is easy to see that $z_1(=u)h$ or $z_1(=u)v$ belongs to F. Thus, $|F|\ge 7$, a contradiction.

Case 3. $|(N_G(x)\setminus \{y\}) \cap X|=3$.

First, we consider the case where the three vertices in $(N_G(x)\setminus \{y\}) \cap X$ appear consecutively on $\mathcal {C}^1$. Without loss of generality, we assume that the three vertices are v, u, h or $u, h, u'$. In the former case, observe that $F \supseteq \{xy,xv',xu', hu',vy,uu',uy\}$, so $|F| \ge 7$. In the latter case, observe that $F \supseteq \{xv,xy,xv',uv,uy,u'y,u'v'\}$. Both cases contradict the assumption that $|F| \le 6$.

If the three vertices in $(N_G(x)\setminus \{y\}) \cap X$ do not appear consecutively on $\mathcal {C}^1$, then we divide into two cases. The first case is that the three vertices are not adjacent to each other on $\mathcal {C}^1$, and in this case, it is easy to see that the three vertices are v, h and $v'$. Now $F\supseteq \{xu,xu',xy,hu,hu',u'v',v'y,vy,uv\}$ and thus $|F|\ge 9$, contradiction. The second case is that two of the three vertices are adjacent on $\mathcal {C}^1$ and the third vertex is not adjacent to them on $\mathcal {C}^1$. Without loss of generality, we can assume that they are u, v and $u'$. Now, all the edges $uy,vy,hu,hu'$ on $\mathcal {C}^1$ belong to F. Additionally, note that three edges incident with x, xh, $xv'$ and xy, also belong to F, and thus we have $|F|\ge 7$, a contradiction.

Case 4. $|(N(x)\setminus \{y\}) \cap X|=4$.

Now, x has only one other neighbor besides y in $\overline{X}$. We consider the following three subcases.

Subcase 4.1 $u ~(\text {or}~ u') \in \overline{X}$. By symmetry, let $u\in \overline{X}$. It is easy to see that $F\supseteq \{uv,vy,ux,xy,u'y,hu,v'y\}$, and thus $|F|\ge 7$, a contradiction.

Subcase 4.2 $h \in \overline{X}$. Clearly, $F\supseteq \{xh,xy,hu,hu',yv,yu', yv'\}$, and thus $|F|\ge 7$, a contradiction.

Subcase 4.3 $v ~(\text {or}~ v') \in \overline{X}$. By symmetry, let $v\in \overline{X}$. First, we have $F\supseteq \{uv,uy,xv,xy,u'y,v'y\}$. Recall that $v'y$ is incident to the face $g~(=[v'z_1z_2yv'])$ in $\mathcal {S}(D)$. By Claim 2, $z_1 \in V(G){\setminus } (N_G(x)\cup \{x\})$ and $z_2 \notin (N_G(x){\setminus } \{h,v\})\cup \{x\}$. So $z_2$ can be either h, v or a vertex in $V(G)\setminus (N_G(x)\cup \{x\})$ (of course, $z_1\ne z_2$.). If $z_2=h$, then $yz_2\in F$, and thus $|F|\ge 7$, a contradiction. If $z_2=v$, then $v'v\in F$, and thus $|F|\ge 7$, a contradiction. Hence, $z_2$ is a vertex in $V(G)\setminus (N_G(x)\cup \{x\})$. Now we note that $G[\{v',y,z_1,z_2\}]-v'y$ is a diamond such that $v'\in X$ and $y\in \overline{X}$. Then by Lemma 8, two edges in the diamond belong to F, and according to the previous analysis, these edges are distinct from any edge in $\{uv,uy,xv,xy,u'y,v'y\}$. Therefore, $|F| \ge 8$, a contradiction.

From the above analysis, it follows that $N_G(x) {\setminus } \{y\} \subseteq X$. Next, we can find that $F\supseteq \{yv',yu',yx,yu,yv\}$. Similar to Subcase 4.3, we only need to consider three cases. The first case is $\{z_1,z_2\} \cap (N_G(x)\cup \{x\}) =\emptyset$. The second case is $z_1\notin N_G(x)\cup \{x\}$ and $z_2=h$. The final case is $z_1\notin N_G(x)\cup \{x\}$ and $z_2=v$. Now, we consider the first case. We note that $G[\{v',y,z_1,z_2\}]-v'y$ is a diamond. By Lemma 8, we know that there are at least two edges in the diamond that belong to F, and these edges are distinct from any edges in $\{yv',yu',yx,yu,yv\}$. So $|F|\ge 7$, a contradiction. Similar, we can also obtain $|F| \ge 7$ for the second case, a contradiction. We now consider the final case where $z_1\notin N_G(x)\cup \{x\}$ and $z_2=v$ (see Fig. 4). By Lemma 1, we have $d_G(y)=6$. If $z_1\in \overline{X}$, then $\{v'z_1,vz_1\}\subseteq F$, and thus $|F|\ge 7$, a contradiction. So $z_1\in X$, and hence $z_1y\in F$. Combining this with the minimum property of F, we have $F= \{yv',yu',yx,yu,yv, yz_1\}$.

Therefore, the claim holds. $\square$

Fig. 4

The case of $N_G(x) {\setminus } \{y\} \subseteq X$

By Claim 3, F is an edge-cut with size 6, and it consists of all the incident edges of a vertex with degree 6. Therefore, we have proved the theorem. $\square$

4 Restricted Edge-Connectivity of Optimal 1-Planar Graphs

First, we note that Esfahanian and Hakimi [12] gave upper and lower bounds on the restricted edge-connectivity of a connected graph.

Lemma 9

([12]) If G is a connected graph with at least 4 vertices and it is not a star $K_{1,m}$ where $m\ge 3$, then $\lambda (G)\le \xi (G)\le \min _{uv\in E(G)} \{ d_G(u)+d_G(v)-2\}$.

Let $e=uv$ be an edge of a graph G. We say that e is an (a, b)-edge if $d_G(u)= a$ and $d_G(v)= b$, and e is an $(a,b^-)$ -edge if $d_G(u)= a$ and $d_G(v)\le b$.

Lemma 10

Let G be an optimal 1-planar graph. Then, G contains an edge of type (6, 6) or (6, 8).

Proof

We notice that Hudák and Šugerek [21] proved that every 1-planar graph of minimum degree 6 contains an edge of type $(6,8^-)$ or (7, 7). Combining this with Lemma 7 (i) and (ii), the lemma is obtained immediately. $\square$

Theorem 2

Let G be an optimal 1-planar graph. Then, $\xi (G)$ is either 8, 10 or 12.

Proof

Note that every optimal 1-planar graph has at least 8 vertices [30], and clearly it cannot be a star. By Lemma 10, G contains an edge of type (6, 6) or (6, 8). Furthermore, by Lemma 9, it follows that $\xi (G)\le 12$. Combining this with Theorem 1, we have $7\le \xi (G)\le 12$. Lemma 7 (i) implies that G is an Eulerian graph. Moreover, it is readily seen that every minimum restricted edge-cut is a minimal edge-cut. Thus, by Lemma 7 (iv), $\xi (G)$ must be even. Hence, $\xi (G)$ is either 8, 10 or 12. $\square$

Moreover, we will see some optimal 1-planar graphs have the restricted edge-connectivity 10 or 12. It is well known that the edge-connectivity of a graph can be computed by using a max-flow (or min-cut) algorithm, which is solvable in polynomial time [29]. Based on a max-flow algorithm, Esfahanian and Hakimi [12] gave a polynomial-time algorithm for computing the restricted edge-connectivity of a graph. We wrote a Maple program based on their algorithm, and the source code is available from the authors on request. Using this program on the optimal 1-planar graph (constructed in [21]) on Fig. 5 (i), we obtain that its restricted edge-connectivity is 12. The program also calculated the restricted edge-connectivity of the optimal 1-planar graph $K_{2,2,2,2}$ (see Fig. 5 (ii)) to be 10.

Fig. 5

Optimal 1-planar graphs with restricted edge-connectivity 12 and 10, respectively (in both graphs, all dashed edges represent a minimum restricted edge-cut)

However, we have not yet found an optimal 1-planar graph with the restricted edge-connectivity 8, and we even believe that such graphs do not exist. Therefore, we propose the following conjecture that is stronger than Theorem 2.

Conjecture 1

Every optimal 1-planar graph has the restricted edge-connectivity either 10 or 12.

In Theorem 1, we characterize minimum edge-cuts of an optimal 1-planar graph, and a similar problem below is also interesting.

Problem 1

Characterize minimum restricted edge-cuts of an optimal 1-planar graph.