Abstract
The 1-crossing index of a graph is the smallest integer such that the th iterated line graph of has crossing number greater than 1. In this paper, we show that the 1-crossing index of a graph is either infinite or it is at most 5. Moreover, we give a full characterization of all graphs with respect to their 1-crossing index.
1. Introduction
We only consider simple, connected, and undirected graph in this paper. For any vertex in , we denote the degree of by . A graph is -regular if for all . Denote by the maximum degree of vertices of , i.e., . A cut-vertex (resp. cut-edge) of a connected graph is a vertex (resp. edge) whose deletion results in a disconnected graph. A block of a graph is a maximal connected subgraph of that has no cut-vertex. For notations not defined here, the readers are referred to [1].
A graph is planar if it can be drawn in the plane in such a way that no two of its edges intersect. The crossing number of a graph , denoted by , is the smallest number of intersections of pairs of edges in any drawing of in the plane. Obviously, is planar if and only if . The crossing number of a graph was introduced by Turán [2], and crossing numbers of certain families of graphs were studied, see [3–8] and the references therein.
The line graph of , denoted by , is defined as the graph whose vertices are the edges of , i.e., , and two vertices in are adjacent if and only if the corresponding edges in are incident to a common vertex.
Given a graph , we denote the th iterated line graph of by , where . In particular, is the original graph and is the line graph of . Since iterated line graphs are well-suited for designing of interconnection networks, the investigation of iterated line graphs has recorded a large progress in recent years, see [9–12].
In paper [13], Sedláček obtained the following result which characterized graphs whose line graph is planar.
Lemma 1 (see [13]). The line graph of a graph is planar if and only if is planar, , and every vertex of degree 4 in is a cut-vertex.
Based on the above result, Ghebleh and Khatirinejad [14] began to study the planarity of iterated line graphs. They defined the line index of a graph , denoted by , to be the smallest such that is nonplanar. If is planar for all , then .
Corollary 1 in [15] showed that . It is noted that a graph has line index 0 if and only if it is nonplanar. For a planar graph , we have
Let be the complete bipartite graph with partite sets and . Let , and be positive integers; we define to be the graph obtained from by subdividing the edge to a path of length (). We define to be the graph obtained by adding two vertices and to and joining them to , see Figure 1.
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Ghebleh and Khatirinejad [14] showed that the line index of a graph is either infinite or it is at most 4. Moreover, with the graphs depicted in Figure 2, a full characterization of all graphs with respect to their line index was given.
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Lemma 2 (see [14]). Let be a connected graph. Then,(1) if and only if is nonplanar(2) if and only if is either a path, a cycle, or (3) if and only if is planar and either or has a vertex of degree 4 which is not a cut-vertex(4) if and only if is planar and contains one of the graphs in Figure 2 as a subgraph(5) if and only if is one of the graphs or for some (6), otherwise
In 1971, Chartrand et al. [16] obtained a necessary and sufficient condition for graphs whose line graph is outerplanar.
Lemma 3 (see [16]). The line graph of a graph is outerplanar if and only if , and if for a vertex of , then is a cut-vertex.
Using this result, Lin et al. [17] investigated the outerplanarity of line graphs and iterated line graphs; they defined the outerplanar index of a graph to be the smallest such that is nonouterplanar. They also showed that the outerplanar index of a graph is either infinite or it is at most 3; furthermore, they completely characterized all graphs with respect to their outerplanar index.
Motivated by these results, we began to consider the problem of nonplanarity of iterated line graphs. Another motivation for this work is a result of Kulli et al. [18] and a result of Jendrol’ and Klešč [15], which characterized graphs whose line graphs have crossing number one.
Lemma 4 (see [15, 18]). The line graph of a planar graph has crossing number one if and only if or holds:(1) and there is a unique noncut-vertex of degree 4(2), every vertex of degree 4 is a cut-vertex, there is a unique vertex of degree 5, and it has at most 3 incident edges in any block
Lemma 5 (see [15]). Let be a nonplanar graph. Then, if and only if the following conditions hold:(1)(2), and every vertex of degree 4 is a cut-vertex of (3)There exists a drawing of in the plane with exactly one crossing in which each crossed edge is incident with a vertex of degree 2
Here, we give similar results for the 1-crossing index of graphs.
Definition 1. The 1-crossing index of a graph , denoted by , is the smallest integer such that . If for all , we define .
It is inferred from Definition 1 that if and only if . For a graph whose crossing number is no more than one, we have
In this paper, we show that the 1-crossing index of a graph is either infinite or it is at most 5. Moreover, another purpose of this paper is to characterize all graphs with respect to their 1-crossing index.
This paper is organized as follows. In Section 2, we show that either or for any graph ; moreover, Theorems 1–3 characterize graphs with , , or , respectively. Our main efforts are dedicated to characterize graphs whose 1-crossing index is 2, 5, or 4 in Sections 3–5, respectively. Based on these results, graphs with 1-crossing index 3 can also be given since that is the only remaining case.
2. The Bound of and Graphs with = ∞ or
The following fact is clear; however, since it is used several times throughout this paper, we state it formally.
Lemma 6. If is a subgraph of , then .
Proof. Let . Then, and since , which implies that .
Lemma 7. If is a graph with , then .
Proof. If , then has as a subgraph. It is easily seen that is a 4-regular planar graph without cut-vertex, by Lemmas 1 and 4, . By Lemma 6, .
Let and denote, respectively, the path and cycle on vertices and let be the empty graph, namely, a graph with no vertices and no edges.
For , we define to be the graph obtained by identifying a vertex of a cycle with a vertex of .
Lemma 8. For , we have .
Proof. It is seen that is a planar graph with and that has a unique noncut-vertex of degree 4, thus by Lemma 4. Furthermore, since , then we have by Lemma 5. By Definition 1, .
Theorem 1. For any graph , we have . Moreover, if and only if is a path, a cycle, or .
Proof. We know for , , , and . Therefore, if is a path, a cycle, or .
Next, assume that is distinct from a path, a cycle, and . First of all, we know that . Case 1.. Then, it has according to Lemma 7 Case 2.. Then, either has or as a subgraph since is not . It can be verified that and , thus due to Lemma 5. Therefore, if has as a subgraph, then ; if has as a subgraph, then by Lemma 8.
In [19], the authors present a structure characterization of graphs with crossing number one, the result lead to an equivalent description of graphs with crossing number at least 2.
Lemma 9 (see [19]). Let be a nonplanar graph and let be a Kuratowski subgraph of . Then, if and only if, for every crossing pair of , are separated by cycles in or at least one of and is not planar.
As a corollary of Lemma 9, we can characterize graphs for which .
Theorem 2. For any graph , if and only if the following conditions hold:(1) is nonplanar(2)Let be a Kuratowski subgraph of . For every crossing pair of , are separated by cycles in or at least one of and is not planar
We end this section by characterizing graphs for which in terms of Lemmas 4 and 5.
Theorem 3. For any graph , if and only if one of the following conditions holds:(1) is a planar graph with , and there are at least two noncut-vertices of degree 4(2) is a planar graph with , and there are at least two vertices of degree 5(3) is a planar graph with ; there is a unique vertex of degree 5 and there is at least one noncut-vertex of degree 4(4) is a planar graph with , every vertex of degree 4 is a cut-vertex, and there is a unique vertex of degree 5; however, it has at least 4 incident edges in one block(5) is a planar graph with (6) and (7), , and there exists a noncut-vertex of degree 4(8), , and every vertex of degree 4 is a cut-vertex; however, there does not exist a good drawing of satisfying condition (3) in Lemma 5
3. Graphs with = 2
We use to denote the complement of a graph . Let and be two vertex-disjoint graphs; denotes the join of graphs and . In the proof of Theorem 2 in [20], we found the following result to be useful.
Proposition 1 (see [20]). Let has a vertex of degree 4 that is not a cut-vertex. Then, contains a subgraph either homeomorphic to or to .
To characterize graphs with , the graphs depicted in Figure 2 are needed. Moreover, the graph defined below is also meaningful in the following discussions.
Definition 2. Define to be the graph satisfying the following four conditions:(1)(2), and every vertex of degree 4 is a cut-vertex(3)For any noncut-edge , it has (4)For any drawing of with exactly one crossing, at least one of the crossed edge is not incident with vertices of degree 2
Remark 1. By conditions (1) and (2) of Definition 2, contains a subgraph homeomorphic to . The third condition of Definition 2 implies that there exists a drawing of with exactly one crossing in which each crossed edge is incident with a vertex of degree 2. Thus, .
Lemma 10. Let be a graph with and . Then, is either or has a subgraph isomorphic to one of the graphs ,,,, and .
Proof. Let be a graph with and . First of all, we have . Since , by Lemma 5, either has a vertex of degree at least 5; or and there is a noncut-vertex of degree 4 in ; or , and every vertex of degree 4 is a cut-vertex in ; however, there does not exist a good drawing of in the plane satisfying condition (3) of Lemma 5. Case 1. has a vertex, say , of degree at least 5. Recall that , then contains either , , , , or as a subgraph. Case 2. and there is a noncut-vertex, say , of degree 4 in . Case 2.1. is a noncut-edge of . Hence, has a cycle containing . Since has degree 4 in , contains either , , or as a subgraph. Case 2.2. is a cut-edge of . Since is a noncut-vertex of , then must be a pendant edge in . By the condition that and that ; thus, must be . Case 3., and every vertex of degree 4 in is a cut-vertex; however, there does not exist a good drawing of satisfying condition (3) of Lemma 5. Claim 1.. Proof of Claim 1. Suppose to contrary that . Then, since . Thus, has noncut-vertices of degree 4, a contradiction. Claim 2. Each vertex of degree 4 in is a cut-vertex. Proof of Claim 2. Suppose to contrary that there is a noncut-vertex of degree 4 in . W.l.o.g., let be four edges incident with in . By Proposition 1, contains a subgraph either homeomorphic to or to , thus is not a bridge of , which enforces that is a noncut-vertex of degree 4 in , a contradiction. Claim 3. For any noncut-edge , it has . Proof of Claim 3. Suppose to contrary that there is a noncut-edge with . Since , then . Thus, is a noncut-vertex of degree 4 in , a contradiction. Since , it follows from Claims 1 and 2 and Lemma 1 that . Moreover, by the assumption that , satisfies all of the conditions in Definition 1. Therefore, is .
Lemma 11. Let be a graph with and . Then, satisfies one of the following conditions:(1) has an edge that joins two vertices of degree 4(2) has at least two edges, each of whose degree sum of its two end vertices is 7(3)There is only one edge in such that the degree sum of its two end vertices is 7, and has a subgraph isomorphic to , , or (4) has at least two noncut-edges whose degree sum of its two end vertices is 6
Proof. Let be a graph with and . Firstly, Lemma 1 tells that is a planar graph with . Secondly, because . By Lemma 4, the following three cases are discussed by considering the maximum degree of . Case 1.. We have since . W.l.o.g., let be a vertex of degree 6 in . Thus, each end vertex of has degree 4 in and is the graph depicted in (1). Case 2.. Case 2.1. There are at least two vertices of degree 5 in . W.l.o.g., let and be two vertices of degree 5 in . Therefore, , for and is the graph depicted in (2). Case 2.2. There is a unique vertex of degree 5 in , and has a noncut-vertex, say , of degree 4. Then, has a subgraph isomorphic to , , or . Thus, is the graph depicted in (3). Case 2.3. Every vertex of degree 4 in is a cut-vertex; there is a unique vertex, say , of degree 5 in ; however, it has at least 4 incident edges in one block. W.l.o.g., assume that . Since and , we have that . Because has at least 4 incident edges in a block of , we conclude that is not a bridge of . Hence, has a cycle containing , which implies that contains a subgraph isomorphic to , , or . Thus, is the graph depicted in (3). Case 3.. By Lemmas 1 and 4, there are at least two noncut-vertices of degree 4 in since . W.l.o.g., let and be two noncut-vertices of degree 4 in . Therefore, for , and is a noncut-edge in . Thus, is the graph depicted in (4).
Theorem 4. For any graph , if and only if one of the conditions holds:(1), and has a subgraph either isomorphic to or to (2) is (3) is planar, and has an edge that joins two vertices of degree 4(4) is planar, and has at least two edges, each of whose degree sum of its two end vertices is 7(5) is planar, and there is only one edge in such that the degree sum of its two end vertices is 7; moreover, has a subgraph isomorphic to , , or (6) is planar, and has at least two noncut-edges whose degree sum of its two end vertices is 6
Proof. Let be a graph with . By the definition, we have and . According to Lemmas 10 and 11, is one of the graphs depicted in (1)–(6).
For the converse, we shall show that if is one of the graphs depicted in (1)–(6). Case 1. is the graph depicted in (1). Let be or one of the graphs . If , then either has a vertex of degree at least 5 or has a noncut-vertex of degree 4; if , then has a vertex of degree at least 5; finally, if , then has a vertex of degree 4 which is a noncut-vertex. Lemma 5 yields that , thus . Since , it follows that . Case 2. is the graph depicted in (2). By Remark 1 and by the forth condition of Definition 2, we have and . Therefore, . Case 3. is the graph depicted in (3). W.l.o.g., let be an edge of that joins two vertices of degree 4, then . By Lemmas 1 and 4, . Since is planar, it follows that . Case 4. is the graph depicted in (4). W.l.o.g., let and be two edges whose degree sum of its two end vertices is 7 in , then both and are vertices of degree 5 in . By Lemma 1, is a nonplanar graph. Furthermore, we have by Lemma 4. Therefore, . Case 5. is the graph depicted in (5). W.l.o.g., let be the edge whose degree sum of its two end vertices is 7 in , then . Moreover, has a subgraph isomorphic to one of , , or . Note that there is a noncut-edge in , see Figure 2. Now, we consider the degree of in . If , then is a noncut-vertex of degree 4 in and thus by Lemma 4. If , that means , then it can be verified that, in , has at least 4 incident edges in one block, thus by Lemma 4. Therefore, . Case 6. is the graph depicted in (6). W.l.o.g., let and be two noncut-edges whose degree sum of its two end vertices is 6 in . Then, both and are noncut-vertices of degree 4 in . By Lemma 1, is a nonplanar graph. Furthermore, by Lemma 4, we have no matter or . Therefore, .
4. Graphs with = 5
Remind that the graph defined in Section 1 is obtained by subdividing each edge of to a path of length (), respectively.
Lemma 12. When , we have . When and , we have .
Proof. Firstly, we prove that when . It is not difficult to see that is a planar graph with ; moreover, there are at least two noncut-vertices of degree 4 in ; therefore, according to Lemma 4. By Definition 1, when .
Now, we prove that when and . One can see that is a planar graph with ; moreover, has a unique noncut-vertex of degree 4; therefore, by Lemma 4. Furthermore, , thus by Lemma 5. By the definition, .
Theorem 5. For any graph , if and only if .
Proof. Assuming that , it has been checked in Theorem 1 that and that , thus .
For the converse, let be a graph with . It follows from Lemma 2 and Theorem 1 that . Assuming that is not a tree, by Lemmas 6 and 8, we have since contains as a subgraph. This contradiction enforces that is a tree. Moreover, Lemmas 6 and 12 indicate that cannot have nor as a subgraph. All of the analyses assert that .
5. Graphs with = 4
To characterize graphs with , the following graphs are introduced. Let be the graph obtained by identifying a vertex of with a pendent vertex of . Define to be the simple graph obtained from by adding a path of arbitrary length to connect two vertices of . Define to be the graph obtained from by adding two pendant edges, each to a vertex of . These families of graphs are shown in Figure 3.
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Let be a tree with diameter 3 such that two nonleaf vertices and both have degree 3. Denote to be the graph obtained from firstly by adding two new vertices and to and joining to and then by subdividing the edge to a path of length . These families of graphs are depicted in Figure 4.
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Lemma 13. For , we have .
Proof. When , it can be seen that is a planar graph with at least two noncut-vertices of degree 4, then by Lemma 4. By the definition, .
When , observe that, if the path added to is of length greater than one, then is a planar graph with at least two noncut-vertices of degree 4; if the path added to is of length one, then and . According to Lemmas 5 and 4, we have . Thus, .
Using the analogous arguments, we have for .
Define to be the graph obtained from by subdividing the edges and to a path of length and , respectively. Define to be the graph obtained by adding two new vertices and to and joining them to . These two families of graphs are depicted in Figure 5.
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Lemma 14. When and , we have . When and , we have .
Proof. Firstly, we show that when and . It is seen that, when , is a planar graph with two noncut-vertices of degree 4; when , we have and . This observation together with Lemmas 5 and 4 yields . By the definition, .
Now, we prove that when and . Note that is a planar graph with and that has a unique noncut-vertex of degree 4, thus by Lemma 4. Furthermore, since , it follows from Lemma 5 that . Consequently, .
Lemma 15. .
Proof. Observe that is a planar graph with maximum degree 4; moreover, has at least two noncut-vertices of degree 4. Thus, by Lemma 4, . By the definition, .
Theorem 6. For any graph , if and only if is one of the graphs , , , , or .
Proof. By Lemmas 8, 12, and 14, if is one of the graphs , , , , or , we obtain that .
For the converse, let be a graph with . By Lemma 7 and Theorem 1, .
Claim 4. Each edge of is contained in at most one cycle; moreover, there is at most one cycle in .
Proof. of Claim 4. If not, contains a subgraph either isomorphic to or to . By Lemmas 6 and 13, , a contradiction.
By Claim 4, we only need to consider the following two cases: Case 1. is unicyclic. Let be the unique cycle in . Since , then at least one vertex of has degree 3 in . If there are at least two such vertices, then contains as a subgraph; hence, by Lemmas 6 and 13. This contradiction implies that, in , there is exactly one vertex of degree 3 in . As we mentioned above, cannot contain as a subgraph. By Lemma 8, is the graph . Case 2. is a tree. By Lemmas 6, 13, and 15, we claim the following.
Claim 5. cannot contain a subgraph isomorphic to , , or to .
Claim 6. In , there are at most three vertices of degree 3.
Proof. of Claim 6. Suppose to contrary that there are at least four vertices of degree 3 in . By Claim 5, these 3-degree vertices cannot be adjacent to each other; otherwise, contains as a subgraph. Since cannot contain as a subgraph; therefore, the tree with at least four vertices of degree 3 will contain as a subgraph, a contradiction with Claim 5.
Assuming that there are three vertices of degree 3 in , by Claim 5 and by Lemma 14, is the graph .
Assume that there are two vertices of degree 3 in , and then is by Lemma 14.
Finally, assume that there is only one vertex of degree 3 in ; then, by Claim 5 and Lemma 12, is either or . The proof is completed.
Data Availability
No data were used to support this study.
Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper.
Acknowledgments
This work was supported by Hunan Provincial Natural Science Foundation (no. 2018JJ2454) and Hunan Education Department Foundation (no. 18A382).