The Crossing Numbers of Join Products of Paths and Cycles with Four Graphs of Order Five

Abstract

The main aim of the paper is to establish the crossing numbers of the join products of the paths and the cycles on n vertices with a connected graph on five vertices isomorphic to the graph $K_{1,1,3}\backslash e$ obtained by removing one edge e incident with some vertex of order two from the complete tripartite graph $K_{1,1,3}$. The proofs are done with the help of well-known exact values for the crossing numbers of the join products of subgraphs of the considered graph with paths and cycles. Finally, by adding some edges to the graph under consideration, we obtain the crossing numbers of the join products of other graphs with the paths and the cycles on n vertices.

1. Introduction

The crossing number $\mathrm{cr}\left(G\right)$ of a simple graph G with the vertex set $V\left(G\right)$ and the edge set $E\left(G\right)$ is the minimum possible number of edge crossings in a drawing of G in the plane. (For the definition of a drawing, see also Klešč [1].) One can easily verify that a drawing with the minimum number of crossings (an optimal drawing) is always a good drawing, meaning that no two edges cross more than once, no edge crosses itself, and also no two edges incident with the same vertex cross. Let D ($D\left(G\right)$) be a good drawing of the graph G. We denote by ${\mathrm{cr}}_D\left(G\right)$ the number of crossings among edges of G in the drawing D.

Let $G_i$ and $G_j$ be two edge-disjoint subgraphs of G. We denote, by ${\mathrm{cr}}_D(G_i,G_j)$, the number of crossings between the edges of $G_i$ and edges of $G_j$, and, by ${\mathrm{cr}}_D\left(G_i\right)$ and ${\mathrm{cr}}_D\left(G_j\right)$, the number of crossings among edges of $G_i$ and of $G_j$ in D, respectively. For any three mutually edge-disjoint subgraphs $G_i$, $G_j$, and $G_k$ of G by Klešč [1], the following equations hold:

$${\mathrm{cr}}_D(G_i\cup G_j)={\mathrm{cr}}_D\left(G_i\right)+{\mathrm{cr}}_D\left(G_j\right)+{\mathrm{cr}}_D(G_i,G_j),$$

$${\mathrm{cr}}_D(G_i\cup G_j,G_k)={\mathrm{cr}}_D(G_i,G_k)+{\mathrm{cr}}_D(G_j,G_k).$$

The problem of reducing the number of crossings is interesting in many areas. One of the most popular areas is the implementation of the VLSI layout, which has revolutionized circuit design and had a strong impact on parallel computing. Crossing numbers were also studied to improve the readability of hierarchical structures and automated graphs. The visualized graph should be easy to read and understand. For the sake of clarity of the graphical drawings, the reduction of crossings is likely the most important. Therefore, the investigation on the crossing number of simple graphs is a classical, but very difficult problem. Garey and Johnson [2] proved that determining $\mathrm{cr}\left(G\right)$ is an NP-complete problem. Throughout the proofs of paper, we will also use the Kleitman’s result [3] on the crossing numbers of the complete bipartite graphs $K_{m,n}$ in the form

$$\mathrm{cr}\left(K_{m,n}\right)=⌊\frac{m}{2}⌋⌊\frac{m-1}{2}⌋⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋,\mathrm{with}min\{m,n\}\le 6.$$

The join product of two graphs $G_i$ and $G_j$, denoted $G_i+G_j$, is obtained from vertex-disjoint copies of $G_i$ and $G_j$ by adding all edges between $V\left(G_i\right)$ and $V\left(G_j\right)$. For $\left|V\right(G_i\left)\right|=m$ and $\left|V\right(G_j\left)\right|=n$, the edge set of $G_i+G_j$ is the union of the disjoint edge sets of the graphs $G_i$, $G_j$, and the complete bipartite graph $K_{m,n}$. Let $P_n$ and $C_n$ be the path and the cycle on n vertices, respectively, and let $D_n$ denote the discrete graph (sometimes called empty graph) on n vertices.

Again, using Kleitman’s result [3], the crossing numbers for the join product of two paths, the join product of two cycles, and also for the join product of a path and a cycle were determined by Klešč [4]. Notice that a lot of the exact values for crossing numbers of $G+D_n$, $G+P_n$, and of $G+C_n$ for arbitrary graph G at most on four vertices were estimated in [5,6]. The crossing numbers of the join product of many graphs G on five and six vertices with $P_n$ and $C_n$ were also investigated in [1,7,8,9,10,11,12,13,14,15].

The crossings numbers of the join products of the paths and the cycles with all graphs of order at most four have been well-known for a long time, and therefore it is understandable that our immediate goal is to establish the exact values for the crossing numbers of $G+P_n$ and of $G+C_n$ also for all graphs G of order five. Especially the results of $G_6+P_n$, $G_9+P_n$, $G_{11}+P_n$, $G_{14}+P_n$ and of $G_6+C_n$, $G_9+C_n$, $G_{11}+C_n$, $G_{14}+C_n$ can be used to determine the crossing number of the join product of the most complicated graph $K_5$ with the path and the cycle on n vertices. For this purpose, we present a new technique regarding the use of knowledge from the subgraphs whose values of crossing numbers are already known. Due to several possible isomorphisms, the results on the smaller graphs are important to confirm the validity of many conjectures, e.g., Corollary 7.

Let $G_{11}$ be the connected graph on five vertices isomorphic to the graph $K_{1,1,3}\backslash e$ obtained by removing one edge e incident with some vertex of order two from the complete tripartite graph $K_{1,1,3}$, and let $V\left(G_{11}\right)=\{v_1,v_2,\cdots ,v_5\}$. The crossing number of $G_{11}+D_n$ was determined for any $n\ge 1$ by Staš [12] using the properties of cyclic permutations, where $D_n$ denotes the discrete graph on n vertices. The main aim of the paper is to establish the crossing numbers of the join products $G_{11}+P_n$ and $G_{11}+C_n$, where $P_n$ and $C_n$ are the path and the cycle on n vertices, respectively.

The proofs are done with the help of a lot of well-known exact values for the crossing numbers of the join products of five subgraphs of $G_{11}$ with paths and cycles. These subgraphs are indexed in the order originally designated by Klešč [16] (except in the case of the graph $G_0$, because it is disconnected), and their planar drawings are presented in Figure 1.

Figure 1. Planar drawings of five graphs $G_0$, $G_2$, $G_3$, $G_5$, and $G_7$, which are subgraphs of the graph $G_{11}$.

The results in Theorems 2 and 3, and in Corollaries 5 and 6 have already been claimed by Li [17] and by Yue et al. [18], respectively. Since these papers do not appear to be available in English, we were unable to verify these results. Clancy et al. [19] also placed an asterisk on a number of the results in their survey to essentially indicate that the mentioned results appeared in journals, which do not have a sufficiently rigorous peer-review process. In certain parts of the presented proofs, it is also possible to simplify the procedure with the help of software generating all cyclic permutations of five elements and its description can be found in Berežný and Buša [20].

2. Cyclic Permutations and Possible Drawings of ${\mathit{G}}_{\mathbf{11}}$

We consider the join product of the graph $G_{11}$ with the discrete graph $D_n$, which yields that the graph $G_{11}+D_n$ consists of just one copy of $G_{11}$ and of n vertices $t_1,t_2,\cdots ,t_n$. Here, each vertex $t_i$, $i=1,2,\cdots ,n$, is adjacent to every vertex of the graph $G_{11}$. Let $T^i$, $1\le i\le n$, denote the subgraph that is uniquely induced by the five edges incident with the fixed vertex $t_i$. This means that the graph $T^1\cup \cdots \cup T^n$ is isomorphic to the complete bipartite graph $K_{5,n}$ and

$$G_{11}+D_n=G_{11}\cup K_{5,n}=G_{11}\cup \left(\bigcup _{i=1}^n{T}^i\right).$$

(1)

Throughout the paper, we also use the same definitions and notation for the good drawings D of the graphs $G_{11}+P_n$ and $G_{11}+C_n$ as in [13,14]. The graph $G_{11}+P_n$ contains $G_{11}+D_n$ as a subgraph, and therefore let $P_n^{*}$ denote the path induced on n vertices of $G_{11}+P_n$ not belonging to the subgraph $G_{11}$. The path $P_n^{*}$ consists of the vertices $t_1,t_2,\cdots ,t_n$ and of the edges $\{t_i,t_{i+1}\}$ for $i=1,2,\cdots ,n-1$, and thus

$$G_{11}+P_n=G_{11}\cup K_{5,n}\cup P_n^{*}=G_{11}\cup \left(\bigcup _{i=1}^n{T}^i\right)\cup P_n^{*}.$$

(2)

Similarly, the graph $G_{11}+C_n$ contains both $G_{11}+D_n$ and $G_{11}+P_n$ as subgraphs. Let $C_n^{*}$ denote the subgraph of $G_{11}+C_n$ induced on the vertices $t_1,t_2,\cdots ,t_n$. Therefore,

$$G_{11}+C_n=G_{11}\cup K_{5,n}\cup C_n^{*}=G_{11}\cup \left(\bigcup _{i=1}^n{T}^i\right)\cup C_n^{*}.$$

(3)

Let D be a good drawing of the graph $G_{11}+D_n$. The rotation${\mathrm{rot}}_D\left(t_i\right)$ of a vertex $t_i$ in the drawing D is the cyclic permutation that records the (cyclic) counter-clockwise order in which the edges leave $t_i$, see [21]. We use the notation $\left(12345\right)$ if the counter-clockwise order the edges is incident with the vertex $t_i$ is $t_i{v}_1$, $t_i{v}_2$, $t_i{v}_3$, $t_i{v}_4$, and $t_i{v}_5$. We emphasize that a rotation is a cyclic permutation; that is, (12345), (23451), (34512), (45123), and (51234) denote the same rotation. Thus, $5!/5=24$ different ${\mathrm{rot}}_D\left(t_i\right)$ can appear in a drawing of the graph $G_{11}+D_n$.

By ${\overline{\mathrm{rot}}}_D\left(t_i\right)$, we understand the inverse permutation of ${\mathrm{rot}}_D\left(t_i\right)$. In the given drawing D, all subgraphs $T^i$, $i=1,\cdots ,n$ of the graph $G_{11}+D_n$ are divided into three mutually disjoint subsets depending on how many times the edges of the subgraph $T^i$ cross the edges of $G_{11}$ in the considered drawing D. For $i=1,\cdots ,n$, $T^i\in R_D$ if ${\mathrm{cr}}_D(G_{11},T^i)=0$, and $T^i\in S_D$ if ${\mathrm{cr}}_D(G_{11},T^i)=1$. Every other subgraph $T^i$ crosses the edges of $G_{11}$ at least twice in D. Clearly, this idea of dividing all subgraphs $T^i$ into three mentioned subsets will be also retained in all drawings of the graphs $G_{11}+P_n$ and $G_{11}+C_n$.

Due to arguments in the proofs of Theorems 2 and 3, at least one of the sets $R_D$ and $S_D$ must be nonempty in any optimal drawing D of $G_{11}+P_n$ and of $G_{11}+C_n$. For $T^i\in R_D\cup S_D$, let $F^i$ denote the subgraph $G_{11}\cup T^i$, $i\in \{1,2,\cdots ,n\}$, of $G_{11}+D_n$, and let $D\left(F^i\right)$ be its subdrawing induced by D.

According to the Lemmas 1 and 3, we suppose only two possible non isomorphic planar good subdrawings of $G_{11}$ as shown in Figure 2, and where the vertex notation of the graph $G_{11}$ will be explained later.

Figure 2. Two possible non isomorphic planar drawings of the graph $G_{11}$. (a) The planar drawing of $G_{11}$ with five vertices in one region; (b) the planar drawing of $G_{11}$ with at most four vertices in one region.

3. The Crossing Number of ${\mathit{G}}_{\mathbf{11}}+{\mathit{P}}_{\mathit{n}}$

Lemma 1.

For $n\ge 2$, if D is any good drawing of the join product $G_{11}+P_n$ with ${\mathrm{cr}}_D\left(G_{11}\right)\ge 1$, then there are at least $n(n-1)+1$ crossings in D.

Proof. 

Let us consider any good drawing D of $G_{11}+P_n$ with ${\mathrm{cr}}_D\left(G_{11}\right)\ge 1$. In the rest of the paper, suppose that let $v_1$, $v_3$, and $v_2$ be the vertex notation of two vertices of degree two and of one vertex of degree three in the considered good subdrawing of the graph $G_{11}$, respectively. Since no two edges incident with the same vertex cross, there is at least one crossing on the edge $v_1{v}_2$ or $v_2{v}_3$ in the subdrawing of $G_{11}$ induced by D. By removing both these edges from the graph $G_{11}$, we obtain a subgraph isomorphic to the graph $G_2$. The exact value for the crossing number of the graph $G_2+P_n$ is given in [12], i.e., $\mathrm{cr}(G_2+P_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+2⌊\frac{n}{2}⌋=n(n-1)$, which yields that there are at least $n(n-1)+1$ crossings in D. □

As the same argument with the removing of the edges $v_1{v}_2$ and $v_2{v}_3$ from the graph $G_{11}$ can be also applied for two possible planar subdrawings of $G_{11}$ in D, the proof of Corollary 1 can be omitted.

Corollary 1.

Let D be any good drawing of the join product $G_{11}+P_n$, $n\ge 2$, with ${\mathrm{cr}}_D\left(G_{11}\right)=0$ and also with the vertex notation of $G_{11}$ given in Figure 2a or Figure 2b. If any of the edges $v_1{v}_2$ or $v_2{v}_3$ is crossed in D, then there are at least $n(n-1)+1$ crossings in the drawing D.

In the proof of Theorem 2, several parts are based on the previous Lemma 1, Corollary 1, and on the following theorem presented in [12].

Theorem 1

(See [12] Corollary 1). $\mathrm{cr}(G_{11}+D_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+2⌊\frac{n}{2}⌋$ for $n\ge 1$.

The exact values of the crossing numbers for many small graphs can be calculated using an algorithm located on a website http://crossings.uos.de/ (accessed on 10 October 2020). This system also generates verifiable formal proofs, like those described by Chimani and Wiedera [22]. However, the capacity of this system is unfortunately limited.

Lemma 2.

$\mathrm{cr}(G_{11}+P_2)=3$ and $\mathrm{cr}(G_{11}+P_3)=7$.

Theorem 2.

$\mathrm{cr}(G_{11}+P_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+2⌊\frac{n}{2}⌋+1=n(n-1)+1$ for $n\ge 2$.

Proof. 

In Figure 3, the edges of $K_{5,n}$ cross each other $4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋$ times, each subgraph $T^i$, $i=1,\cdots ,⌈\frac{n}{2}⌉$ on the right side does not cross the edges of $G_{11}$, and each subgraph $T^i$, $i=⌈\frac{n}{2}⌉+1,\cdots ,n$ on the left side crosses the edges of $G_{11}$ exactly twice. The path $P_n^{*}$ crosses $G_{11}$ once, and thus $4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+2⌊\frac{n}{2}⌋+1$ crossings appear among the edges of the graph $G_{11}+P_n$ in this drawing. Thus, $\mathrm{cr}(G_{11}+P_n)\le n(n-1)+1$. Lemma 2 confirms this result for $n=2$ and $n=3$. We prove the reverse inequality by induction on n. Now, let us suppose that for some $n\ge 4$, there is a drawing D for which

$${\mathrm{cr}}_D(G_{11}+P_n)<n(n-1)+1,$$

(4)

and that

$$\mathrm{cr}(G_{11}+P_m)=m(m-1)+1\mathrm{for}\mathrm{any}2\le m<n.$$

(5)

Figure 3. The good drawing of $G_{11}+P_n$ with $n(n-1)+1$ crossings.

Since the graph $G_{11}+D_n$ is a subgraph of $G_{11}+P_n$, by Theorem 1, the edges of $G_{11}+P_n$ are crossed exactly $n(n-1)$ times, and therefore, no edge of the path $P_n^{*}$ is crossed in D. In addition, all vertices $t_i$ of the path $P_n^{*}$ have to be placed in the same region of the considered good subdrawing of $G_{11}$. By Lemma 1, we can only suppose planar subdrawings of the graph $G_{11}$ induced by D, that is, ${\mathrm{cr}}_D\left(G_{11}\right)=0$. If $r=|R_D|$ and $s=|S_D|$, the assumption (5) together with $\mathrm{cr}\left(K_{5,n}\right)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋$ enforces that there are at least $⌈\frac{n}{2}⌉$ subgraphs $T^i$ whose edges cross the edges of $G_{11}$ at most once in D. More precisely:

$${\mathrm{cr}}_D\left(G_{11}\right)+{\mathrm{cr}}_D(G_{11},K_{5,n})\le 2⌊\frac{n}{2}⌋,$$

i.e.,

$$0+0r+1s+2(n-r-s)\le 2⌊\frac{n}{2}⌋.$$

(6)

This implies that $2r+s\ge 2⌈\frac{n}{2}⌉$. Now, we will show that, in all subcases, a contradiction with the assumption (4) can be obtained:

Case 1: We suppose the drawing with the vertex notation of $G_{11}$ in such a way as shown in Figure 2a. Since the set $R_D\cup S_D$ is nonempty and no edge of the path $P_n^{*}$ is crossed in the drawing D, all vertices $t_i$ of $P_n^{*}$ are placed in the region of subdrawing $D\left(G_{11}\right)$ with five vertices $v_1$, $v_2$, $v_3$, $v_4$, and $v_5$ of $G_{11}$ on its boundary. By Klešč and Staš [11], it was proved that $\mathrm{cr}(G_0+P_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+⌊\frac{n}{2}⌋$, and thus there are at most $⌊\frac{n}{2}⌋$ crossings on three edges $v_1{v}_5$, $v_3{v}_5$, and $v_4{v}_5$ in D. This, also with Corollary 1, enforces that $r\ge n-⌊\frac{n}{2}⌋=⌈\frac{n}{2}⌉$, because each subgraph $T^k\notin R_D$ crosses some of these three edges at least once.

As the set $R_D$ is nonempty, our aim is to list all possible rotations ${\mathrm{rot}}_D\left(t_i\right)$ existing in D if no edge of $T^i$ cross any edge of $G_{11}$. Since there is only one subdrawing of $F^i\backslash v_5$ represented by the subrotation $\left(1234\right)$, we have only two ways to obtain the subdrawing of the subgraph $F^i$ depending on which region the edge $t_i{v}_5$ is placed in. We denote these two possibilities by ${\mathcal{A}}_1$ and ${\mathcal{A}}_3$, and they are represented by the cyclic permutations $\left(12345\right)$ and $\left(12354\right)$, respectively (in order to comply with the same notation as in [12]). One can easily determine, in five possible regions of $D(G_{11}\cup T^i)$, that ${\mathrm{cr}}_D(G_{11}\cup T^i,T^k)\ge 3$ holds for any subgraph $T^k$, $k\ne i$. Thus, by fixing the subgraph $G_{11}\cup T^i$, we have

$${\mathrm{cr}}_D(G_{11}+P_n)={\mathrm{cr}}_D\left(K_{5,n-1}\right)+{\mathrm{cr}}_D(K_{5,n-1},G_{11}\cup T^i)+{\mathrm{cr}}_D(G_{11}\cup T^i)$$

$$\ge 4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+3(n-1)+0,$$

where $4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+3(n-1)\ge n(n-1)+1$ is true only for n even. For n odd, without a loss of generality on based their symmetry, let us also consider that the number of all subgraphs with the configuration ${\mathcal{A}}_1$ is at least as much as the number of all subgraphs with the configuration ${\mathcal{A}}_3$, and let $T^i\in R_D$ be such a subgraph with the configuration ${\mathcal{A}}_1$ of $F^i$. As $r\ge ⌈\frac{n}{2}⌉\ge 3$ for n at least 5, there is at least one subgraph $T^j\in R_D$, $j\ne i$ with ${\mathrm{rot}}_D\left(t_j\right)={\mathrm{rot}}_D\left(t_i\right)$, which yields that ${\mathrm{cr}}_D(G_{11}\cup T^i,T^j)\ge 0+4=4$. This allows us to add at least one crossing in the following inequalities

$${\mathrm{cr}}_D(G_{11}+P_n)\ge 4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+3(n-1)+1\ge n(n-1)+1.$$

Both subcases confirm a contradiction with the assumption in D.

Case 2: We consider the planar drawing of $G_{11}$ in D given in Figure 2b. As no face is incident to all vertices in $D\left(G_{11}\right)$, there is no possibility to obtain a subdrawing of $G_{11}\cup T^i$ for a $T^i\in R_D$. As $r=0$, each subgraph $T^i$ crosses the edges of $G_{11}$ exactly once using the inequality (6). If all vertices $t_i$ of the path $P_n^{*}$ are placed in the region of $D\left(G_{11}\right)$ with four vertices $v_1$, $v_2$, $v_3$, and $v_5$ of $G_{11}$ on its boundary, then the edge $v_1{v}_5$ must be crossed by each subgraph $T^i\in S_D$ by Corollary 1.

This contradicts the fact that there are, at most, $⌊\frac{n}{2}⌋$ crossings on the edge $v_1{v}_5$ using the already well-known result from the previous case by [11]. Finally, if all vertices $t_i$ are placed in the region of $D\left(G_{11}\right)$ with four vertices $v_1$, $v_2$, $v_4$, and $v_5$ of $G_{11}$ on its boundary, then only one of the edges $v_1{v}_5$ and $v_2{v}_5$ can be crossed by any subgraph $T^i\in S_D$ again by Corollary 1. The authors in [10,11] proved that $\mathrm{cr}(G_5+P_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+⌊\frac{n}{2}⌋$ and $\mathrm{cr}(G_7+P_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+⌊\frac{n}{2}⌋+1$, respectively, and thus there are at most $⌊\frac{n}{2}⌋+⌊\frac{n}{2}⌋-1$ crossings on the pair of edges $v_1{v}_5$ and $v_2{v}_5$. These facts imply a contradiction with $s=n$.

We have shown, in all cases, that there are at least $n(n-1)+1$ crossings in each good drawing D of the graph $G_{11}+P_n$. The proof of Theorem 2 is done. □

4. The Crossing Number of ${\mathit{G}}_{\mathbf{11}}+{\mathit{C}}_{\mathit{n}}$

Let $S_m$ denote the star on $m+1$ vertices. Using the results of Klešč et al. [9], the crossing numbers of the graphs $S_m+C_n$ for $m=3,4,5$ and $n\ge 3$ were established. Hence, the exact value for the crossing number of the graph $G_2+C_n$ is given by $n(n-1)+2$. Given the use of arguments similar to those in the proof of Lemma 1, the proofs of Lemma 3 and Corollary 2 can be omitted.

Lemma 3.

For $n\ge 3$, if D is any good drawing of the join product $G_{11}+C_n$ with ${\mathrm{cr}}_D\left(G_{11}\right)\ge 1$, then there are at least $n(n-1)+3$ crossings in D.

Corollary 2.

Let D be any good drawing of the join product $G_{11}+C_n$, $n\ge 3$, with ${\mathrm{cr}}_D\left(G_{11}\right)=0$ and also with the vertex notation of $G_{11}$ given in Figure 2a or Figure 2b. If any of the edges $v_1{v}_2$ or $v_2{v}_3$ is crossed in D, then there are at least $n(n-1)+3$ crossings in the drawing D.

Two vertices $t_i$ and $t_j$ of the graph $G_{11}+D_n$ are said to be antipodal in a drawing of $G_{11}+D_n$ if the considered subgraphs $T^i$ and $T^j$ do not cross. A drawing with no antipodal vertices is antipode-free. Clearly, this antipode-free property is also retained in all drawings of the graph $G_{11}+C_n$.

Lemma 4.

For $n>2$, let D be a good and antipode-free drawing of the join product $G_{11}+D_n$ with ${\mathrm{cr}}_D\left(G_{11}\right)=0$ and also with the vertex notation of the graph $G_{11}$ given in Figure 2a. Let $T^i\in R_D$ be a subgraph such that $F^i$ has the configuration ${\mathcal{A}}_1$, i.e., ${\mathrm{rot}}_D\left(t_i\right)=\left(12345\right)$, and let there be no crossing on the edges $v_1{v}_2$ and $v_2{v}_3$ in D. If there is a subgraph $T^k\in S_D$ with ${\mathrm{cr}}_D(T^i,T^k)=2$, then

(a) 

${\mathrm{cr}}_D(G_{11}\cup T^i\cup T^k,T^l)\ge 6$ holds for any subgraph $T^l\in R_D$, $l\ne i$; and

(b) 

${\mathrm{cr}}_D(G_{11}\cup T^i\cup T^k,T^l)\ge 6$ holds for any subgraph $T^l\notin R_D$, $l\ne k$ such that the edge $v_1{v}_5$ of $G_{11}$ is not crossed by the edges of $T^l$.

Proof. 

Let us consider the configuration ${\mathcal{A}}_1$ of the subgraph $F^i$. If there is a subgraph $T^k\in S_D$ such that ${\mathrm{cr}}_D(T^i,T^k)=2$, then the considered vertex $t_k$ has to be placed in the quadrangular region of $D(G_{11}\cup T^i)$ with exactly three vertices $v_3$, $v_4$, and $v_5$ of $G_{11}$ on its boundary. This enforces that the edges $v_3{v}_5$ or $v_4{v}_5$ of the graph $G_{11}$ must be crossed by the edges $t_k{v}_2$ or $t_k{v}_1$, respectively. For more, see also the two mentioned subdrawings of the graph $G_{11}\cup T^i\cup T^k$ in Figure 4.

Figure 4. Two possible subdrawings of $G_{11}\cup T^i\cup T^k$ for $T^k\in S_D$ with ${\mathrm{cr}}_D(T^i,T^k)=2$, where $T^i\in R_D$ with the configuration ${\mathcal{A}}_1$ of $F^i$. (a) The subdrawing in which the edge $v_3{v}_5$ of $G_{11}$ is crossed by the edge $t_k{v}_2$; (b) the subdrawing in which the edge $v_4{v}_5$ of $G_{11}$ is crossed by the edge $t_k{v}_1$.

(a)

Let $T^k\in S_D$ be a subgraph with ${\mathrm{cr}}_D(T^i,T^k)=2$. If we suppose the drawing of subgraph $T^k$ as shown in Figure 4a, then ${\mathrm{rot}}_D\left(t_k\right)=\left(13254\right)$. For $T^l\in R_D$ with $l\ne i$, the possible configurations ${\mathcal{A}}_1$ and ${\mathcal{A}}_3$ are uniquely represented by the cyclic permutations $\left(12345\right)$ and $\left(12354\right)$, respectively. Using the distances between two cyclic permutations, we are able to determine the minimum numbers of crossings of $T^l$ with the subgraphs $T^i$ and $T^k$ in the first two columns of Table 1. The smallest value in the last column of Table 1 gives the required minimum number of crossings. Of course, the same idea for the case of ${\mathrm{rot}}_D\left(t_k\right)=\left(14235\right)$ forces the same result.

Table 1. All possibilities of the subgraph $F^l$ for $T^l\in R_D$, $l\ne i$ with ${\mathrm{cr}}_D(T^i,T^k)=2$ and $T^k\in S_D$.

$\mathbf{conf}\left({\mathit{F}}^{\mathit{l}}\right)$	${\mathbf{cr}}_{\mathit{D}}({\mathit{T}}^{\mathit{i}},{\mathit{T}}^{\mathit{l}})$	${\mathbf{cr}}_{\mathit{D}}({\mathit{T}}^{\mathit{k}},{\mathit{T}}^{\mathit{l}})$	${\mathbf{cr}}_{\mathit{D}}({\mathit{T}}^{\mathit{i}}\cup {\mathit{T}}^{\mathit{k}},{\mathit{T}}^{\mathit{l}})$	${\mathbf{cr}}_{\mathit{D}}({\mathit{G}}_{11}\cup {\mathit{T}}^{\mathit{i}}\cup {\mathit{T}}^{\mathit{k}},{\mathit{T}}^{\mathit{l}})$
${\mathcal{A}}_1$	4	2	6	6
${\mathcal{A}}_3$	3	3	6	6

(b)

For $l\ne k$, let $T^l\notin R_D$ be a subgraph with respect to the restriction that the edges of $T^l$ does not cross the edges $v_1{v}_2$, $v_2{v}_3$, and $v_1{v}_5$ of the graph $G_{11}$. Since the considered drawing D is antipode-free and $T^l$ can cross only some of edges $v_2{v}_5$, $v_3{v}_5$, and $v_4{v}_5$ of $G_{11}$, one can easily determine, in all possible regions of the subdrawing $D(G_{11}\cup T^i\cup T^k)$, that ${\mathrm{cr}}_D(G_{11}\cup T^i\cup T^k,T^l)\ge 6$ is fulfilling for such a subgraph $T^l$.

□

Again, using the algorithm on the website http://crossings.uos.de/ accessed on 10 October 2020, we can also determine the crossing numbers of two small graphs in Lemma 5.

Lemma 5.

$\mathrm{cr}(G_{11}+C_3)=9$ and $\mathrm{cr}(G_{11}+C_4)=15$.

Theorem 3.

$\mathrm{cr}(G_{11}+C_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+2⌊\frac{n}{2}⌋+3=n(n-1)+3$ for $n\ge 3$.

Proof. 

Figure 5 shows the drawing of the graph $G_{11}+C_n$ with exactly $n(n-1)+3$ crossings. Thus, $\mathrm{cr}(G_{11}+C_n)\le n(n-1)+3$. By Lemma 5, the result holds for $n=3$ and $n=4$. We prove the reverse inequality by induction on n. Now, let us suppose that, for some $n\ge 5$, there is a drawing D for which

$${\mathrm{cr}}_D(G_{11}+C_n)<n(n-1)+3,$$

(7)

and that

$$\mathrm{cr}(G_{11}+C_m)=m(m-1)+3\mathrm{for}\mathrm{any}\mathrm{integer}3\le m<n.$$

(8)

Figure 5. The good drawing of $G_{11}+C_n$ with $n(n-1)+3$ crossings.

Since the graph $G_{11}+D_n$ is also a subgraph of $G_{11}+C_n$, also by Theorem 1, the edges of $G_{11}+C_n$ are crossed at least $n(n-1)$ times. Therefore, at most two edges of the cycle $C_n^{*}$ can be crossed in D, and this also implies that the vertices $t_i$ of $C_n^{*}$ have to be placed at most in two different regions of $D\left(G_{11}\right)$. Moreover, by Theorem 2, there is at most one crossing on each edge of $C_n^{*}$. By Lemma 3, we can only suppose two possible planar subdrawings of the graph $G_{11}$ induced by D. All our assumptions on D with $\mathrm{cr}\left(K_{5,n}\right)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋$ enforce that

$${\mathrm{cr}}_D\left(G_{11}\right)+{\mathrm{cr}}_D(G_{11},K_{5,n})\le 2⌊\frac{n}{2}⌋+2,$$

i.e.,

$$0+0r+1s+2(n-r-s)\le 2⌊\frac{n}{2}⌋+2,$$

(9)

if we use the notation $r=|R_D|$ and $s=|S_D|$ again. This forces that $2r+s\ge 2⌈\frac{n}{2}⌉-2$, and if $r=0$, then $s\ge 2⌈\frac{n}{2}⌉-2$. Again, we will suppose all possibilities of obtaining some subgraph $T^i\in R_D\cup S_D$ in order to obtain a contradiction with the assumption (7) in all considered subcases in D:

Case 1: We consider the planar drawing of $G_{11}$ in D with the vertex notation in such a way as shown in Figure 2a. We claim that the drawing D must be antipode-free. For a contradiction, suppose that ${\mathrm{cr}}_D(T^k,T^l)=0$ for two different subgraphs $T^k$ and $T^l$. If at least one of $T^k$ and $T^l$, say $T^k$, does not cross $G_{11}$, it is not difficult to check in Figure 2a that the subgraph $T^l$ must cross the edges of $G_{11}\cup T^k$ at least twice, that is, ${\mathrm{cr}}_D(G_{11},T^k\cup T^l)={\mathrm{cr}}_D(G_{11},T^l)\ge 2$. Moreover, the Kleitman’s result [3] for $\mathrm{cr}\left(K_{5,3}\right)=4$ implies that each $T^m$, $m\ne k,l$ crosses the edges of the subgraph $T^k\cup T^l$ at least four times. Consequently, for the number of crossings in D holds:

$${\mathrm{cr}}_D(G_{11}+C_n)={\mathrm{cr}}_D\left(G_{11}+C_{n-2}\right)+{\mathrm{cr}}_D(K_{5,n-2},T^k\cup T^l)+{\mathrm{cr}}_D(G_{11},T^k\cup T^l)$$

$$+{\mathrm{cr}}_D(T^k\cup T^l)\ge (n-2)(n-3)+3+4(n-2)+2+0=n(n-1)+3.$$

This contradiction with (7) confirms that D is antipode-free. The authors in [11] also proved that $\mathrm{cr}(G_0+C_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+⌊\frac{n}{2}⌋+1$, and therefore there are at most $⌊\frac{n}{2}⌋+1$ crossings on three edges $v_1{v}_5$, $v_3{v}_5$, and $v_4{v}_5$ in D. This, also with Corollary 2, implies that $r\ge n-\left(⌊\frac{n}{2}⌋+1\right)=⌈\frac{n}{2}⌉-1$ provided by each subgraph $T^k\notin R_D$ crosses some of these three edges at least once. Thus, for any $T^i\in R_D$, the vertex $t_i$ have to be placed in the region of $D\left(G_{11}\right)$ with all five vertices of the graph $G_{11}$ on its boundary.

Let us turn to the possibility of an existence of vertex $t_j$ of the cycle $C_n^{*}$ in some region of $D\left(G_{11}\right)$ with three vertices of $G_{11}$ on its boundary, that is, two different edges of $C_n^{*}$ cross one of the edges $v_1{v}_5$ or $v_3{v}_5$ in D again by Corollary 2. Since there are two additional crossings on one of these two edges of the graph $G_{11}$, the mentioned result [11] enforces $r\ge ⌈\frac{n}{2}⌉+1\ge 4$ for n at least 5. Let $D^{{}^{\prime }}$ be the subdrawing of $G_{11}+D_n$ induced by D without the edges of $C_n^{*}$.

Clearly, the subdrawing $D^{{}^{\prime }}$ is some optimal drawing of the graph $G_{11}+D_n$ with exactly $n(n-1)$ crossings. Therefore, we can apply the similar idea as in the proof of Theorem 2, because ${\mathrm{cr}}_{D^{{}^{\prime }}}(G_{11}\cup T^i,T^k)\ge 3$ holds for any two different subgraphs $T^i$, $T^k$ with $T^i\in R_D$. Again, without a loss of generality, let us also consider that the number of all subgraphs with the configuration ${\mathcal{A}}_1$ is at least as much as the number of all subgraphs with the configuration ${\mathcal{A}}_3$, and let $T^i\in R_D$ be such a subgraph with this configuration ${\mathcal{A}}_1$ of $F^i$. Then, by fixing the subgraph $G_{11}\cup T^i$, we have

$${\mathrm{cr}}_{D^{{}^{\prime }}}(G_{11}+D_n)\ge 4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+3(n-1)+1\ge n(n-1)+1.$$

This contradiction with the optimality of the subdrawing $D^{{}^{\prime }}$ of $G_{11}+D_n$ confirms that all vertices $t_i$ of the cycle $C_n^{*}$ are placed in the region of $D\left(G_{11}\right)$ with five vertices $v_1$, $v_2$, $v_3$, $v_4$, and $v_5$ of $G_{11}$ on its boundary. By [11], we already know that there are, at most, $⌊\frac{n}{2}⌋+1$ crossings on the three edges $v_1{v}_5$, $v_3{v}_5$, and $v_4{v}_5$ in D. In the rest of the paper, based on their symmetry, let the edge $v_1{v}_5$ be crossed, at most, as many times as the edge $v_3{v}_5$, that is, there are at most $⌊\frac{\lfloor \frac{n}{2}\rfloor +1}{2}⌋$ crossings on the edge $v_1{v}_5$ in D. We denote, by ${\mathcal{M}}_D$, the nonempty subset of $\mathcal{M}=\{{\mathcal{A}}_1,{\mathcal{A}}_3\}$ consisting of all configurations existing in D. Now, two possible subcases may occur:

(a)

${\mathcal{A}}_1\in {\mathcal{M}}_D$. For $T^i\in R_D$ with the configuration ${\mathcal{A}}_1$ of $F^i$, there is the possibility of obtaining a subdrawing of $G_{11}\cup T^i\cup T^k$ in which ${\mathrm{cr}}_D(T^i,T^k)=2$ holds for some $T^k\in S_D$. For this case by Lemma 4, the edges of the graph $G_{11}\cup T^i\cup T^k$ are crossed by each subgraph $T^l$, $l\ne i,k$ at least six times except in cases where the edge $v_1{v}_5$ of $G_{11}$ is crossed by the edges of $T^l$. Thus, by fixing the subgraph $G_{11}\cup T^i\cup T^k$, we have

$${\mathrm{cr}}_D(G_{11}+C_n)\ge 4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋+6(n-2)-⌊\frac{\lfloor \frac{n}{2}\rfloor +1}{2}⌋+3\ge n(n-1)+3.$$

This also contradicts the assumption of D, and therefore, in the next part, suppose that ${\mathrm{cr}}_D(G_{11}\cup T^i,T^k)\ge 4$ holds for each $T^k\in S_D$. Notice that if $r\ge 3$ and there are two different subgraphs $T^i,T^j\in R_D$ such that $F^i$ and $F^j$ have configurations ${\mathcal{A}}_1$ and ${\mathcal{A}}_3$, respectively, then ${\mathrm{cr}}_D(T^i\cup T^j,T^k)\ge 3+4=7$ is fulfilling for any $T^k\in R_D$, $k\ne i,j$ and ${\mathrm{cr}}_D(G_{11}\cup T^i\cup T^j,T^k)\ge 5$ holds for any $T^k\notin R_D$. Therefore, in such a contemplated case, by fixing the graph $G_{11}\cup T^i\cup T^j$, we receive the following contradiction with the assumption in D

$${\mathrm{cr}}_D(G_{11}+C_n)\ge 4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋+7(r-2)+5(n-r)+3$$

$$\ge 4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋+5n+2r-11\ge 4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋+5n+6-11\ge n(n-1)+3.$$

Further, if there is a subgraph $T^k\notin R_D\cup S_D$ such that ${\mathrm{cr}}_D(G_{11}\cup T^i,T^k)=3$, then the edges $v_3{v}_5$ and $v_4{v}_5$ of the graph $G_{11}$ are crossed by the edges $t_k{v}_2$ and $t_k{v}_1$, respectively, which yields by the result in [11] that $r\ge ⌈\frac{n}{2}⌉\ge 3$ for n at least 5. Finally, if either ${\mathcal{M}}_D=\{{\mathcal{A}}_1,{\mathcal{A}}_3\}$ and $r=2$ or ${\mathcal{M}}_D=\left\{{\mathcal{A}}_1\right\}$, by fixing the subgraph $G_{11}\cup T^i$, we have

$${\mathrm{cr}}_D(G_{11}+C_n)\ge 4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+3(n-1)+1+1+1\ge n(n-1)+3.$$

(b)

${\mathcal{M}}_D=\left\{{\mathcal{A}}_3\right\}$. Let $T^i$ be any subgraph from the nonempty set $R_D$. Then, ${\mathrm{cr}}_D(T^i,T^k)\ge 4$ holds for each subgraph $T^k\in R_D$, $k\ne i$ provided by ${\mathrm{rot}}_D\left(t_i\right)={\mathrm{rot}}_D\left(t_k\right)$. Moreover, we can easily verify in five possible regions of $D(G_{11}\cup T^i)$ that ${\mathrm{cr}}_D(G_{11}\cup T^i,T^k)\ge 4$ is fulfilling for any $T^k\notin R_D$, if the edge $v_1{v}_5$ of $G_{11}$ is not crossed by the edges of $T^k$. Thus, by fixing the subgraph $G_{11}\cup T^i$, we have

$${\mathrm{cr}}_D(G_{11}+C_n)\ge 4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+4(n-1)-⌊\frac{\lfloor \frac{n}{2}\rfloor +1}{2}⌋+0\ge n(n-1)+3.$$

All these subcases confirm a contradiction with the assumption in D.

Case 2: We assume the planar subdrawing of $G_{11}$ with the vertex notation given in Figure 2b. The set $R_D$ is empty, and therefore there are at least $2⌈\frac{n}{2}⌉-2$ subgraphs $T^i\in S_D$ using the inequality (9). The authors in [10,11] also proved that $\mathrm{cr}(G_5+C_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+⌊\frac{n}{2}⌋+2$ and $\mathrm{cr}(G_7+C_n)=4⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+⌊\frac{n}{2}⌋+2$, and therefore there are at most $⌊\frac{n}{2}⌋$ crossings on each of the edges $v_1{v}_5$ and $v_2{v}_5$, respectively. Since each subgraph $T^i$ crosses some edge of the cycle $v_1{v}_2{v}_5{v}_1$ at least once in $D(G_{11}\cup T^i)$, and none of the edges $v_1{v}_2$ and $v_2{v}_3$ can be crossed in D due to Corollary 2, each of the edges $v_1{v}_5$ and $v_2{v}_5$ is crossed exactly $⌊\frac{n}{2}⌋$ times. This also enforces that n must be even and all vertices $t_i$ of the cycle $C_n^{*}$ are placed in the region of $D\left(G_{11}\right)$ with four vertices $v_1$, $v_2$, $v_4$, and $v_5$ of $G_{11}$ on its boundary.

Now, let us turn to list all possible rotations ${\mathrm{rot}}_D\left(t_i\right)$ that can appear in the drawing D if the edges of the graph $G_{11}$ are crossed by the edges of $T^i$ just once. For $T^i\in S_D$, based on the previous discussion, there is only one possible subdrawing of $F^i\backslash \{v_3,v_5\}$ represented by the subrotation $\left(142\right)$. This offers four ways of obtaining the subdrawing of $F^i$ depending on which of two edges of the graph $G_{11}$ can be crossed by the edge $t_i{v}_3$ and in which region of $D(F_i\backslash v_5)$ the edge $t_i{v}_5$ is placed.

We denote these four possibilities by ${\mathcal{B}}_1$, ${\mathcal{B}}_2$, ${\mathcal{B}}_3$, and ${\mathcal{B}}_4$ with the corresponding cyclic permutations $\left(13542\right)$, $\left(14532\right)$, $\left(13452\right)$, and $\left(15432\right)$, respectively. For any $T^i\in S_D$ with the configuration of either ${\mathcal{B}}_3$ or ${\mathcal{B}}_4$ of $F^i$, the reader can easily verify in five possible regions of $D(G_{11}\cup T^i)$ that ${\mathrm{cr}}_D(G_{11}\cup T^i,T^k)\ge 3$ holds for each subgraph $T^k$ with $k\ne i$. Moreover, ${\mathrm{cr}}_D(G_{11}\cup T^i,T^k)\ge 4$ is fulfilling for each subgraph $T^k$, $k\ne i$ if the edges $t_i{v}_3$ and $t_k{v}_3$ cross the same edge of $G_{11}$. Thus, by fixing the subgraph $G_{11}\cup T^i$ having the configuration either ${\mathcal{B}}_3$ or ${\mathcal{B}}_4$, we obtain

$${\mathrm{cr}}_D(G_{11}+C_n)\ge 4⌊\frac{n-1}{2}⌋⌊\frac{n-2}{2}⌋+4\left(⌊\frac{n}{2}⌋-1\right)+3⌊\frac{n}{2}⌋+1\ge n(n-1)+3.$$

This also contradicts the assumption (7) of D. Finally, suppose that there is no subgraph $T^i\in S_D$ with the configuration ${\mathcal{B}}_3$ and ${\mathcal{B}}_4$ of $F^i$. As $s\ge n-2\ge 4$ for n even of at least 6, there are two different subgraphs $T^i,T^j\in S_D$ such that $F^i$ and $F^j$ have the configurations ${\mathcal{B}}_1$ and ${\mathcal{B}}_2$, respectively. The minimum number of interchanges of adjacent elements of $\left(13542\right)$ required to produce the cyclic permutation $\overline{\left(14532\right)}=\left(12354\right)$ is one. Thus, the subgraph $T^j$ must cross the edges of $T^i$ at least $1+2m$ times for some nonnegative integer m and ${\mathrm{cr}}_D(T^i\cup T^j,T^k)\ge ⌊\frac{5}{2}⌋⌊\frac{5-1}{2}⌋-1=3$ is also fulfilling for each subgraph $T^k$, $k\ne i,j$, for more see Woodall’s results [23]. These properties of the cyclic permutations imply that ${\mathrm{cr}}_D(G_{11}\cup T^i\cup T^j,T^k)\ge 1+1+4=6$ holds for any $T^k\in S_D$, $k\ne i,j$, and ${\mathrm{cr}}_D(G_{11}\cup T^i\cup T^j,T^k)\ge 2+3=5$ is also true for any $T^k\notin S_D$. Hence, by fixing the subgraph $G_{11}\cup T^i\cup T^j$, we have

$${\mathrm{cr}}_D(G_{11}+C_n)\ge 4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋+6(s-2)+5(n-s)+2+1=4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋$$

$$+5n+s-9\ge 4⌊\frac{n-2}{2}⌋⌊\frac{n-3}{2}⌋+5n+(n-2)-9\ge n(n-1)+3.$$

We have shown, in all cases, that there are at least $n(n-1)+3$ crossings in each good drawing D of the graph $G_{11}+C_n$. This completes the proof of Theorem 3. □

5. Some Consequences of the Main Result

In Figure 6, let $G_{14}$ be the connected graph of order five obtained from $G_{11}$ by adding the edge $v_3{v}_4$ in the subdrawing in Figure 2a. Since we can add this edge $v_3{v}_4$ to the graph $G_{11}$ without additional crossings in Figure 3 and Figure 5, the drawings of the graphs $G_{14}+P_n$ and $G_{14}+C_n$ with exactly $n(n-1)+1$ and $n(n-1)+3$ crossings are obtained, respectively. Further, the graph $G_{11}$ is some subgraph of $G_{14}$, and therefore, $\mathrm{cr}(G_{14}+P_n)\ge \mathrm{cr}(G_{11}+P_n)$ and $\mathrm{cr}(G_{14}+C_n)\ge \mathrm{cr}(G_{11}+C_n)$. Therefore, the following results are obvious.

Figure 6. Three graphs $G_6$, $G_9$, and $G_{14}$.

Corollary 3.

$\mathrm{cr}(G_{14}+P_n)=n(n-1)+1$ for $n\ge 2$.

Corollary 4.

$\mathrm{cr}(G_{14}+C_n)=n(n-1)+3$ for $n\ge 3$.

Similarly, in Figure 6, let $G_9$ be the graph obtained from $G_{11}$ by adding the edge $v_3{v}_4$ and by removing the edge $v_2{v}_3$ from the subdrawing in Figure 2a, which yields the good drawing of $G_9+C_n$ with exactly $n(n-1)+2$ crossings from the optimal drawing of $G_{11}+C_n$ in Figure 5. As $G_2$ is a subgraph of the graph $G_9$, we have $\mathrm{cr}(G_9+C_n)\ge \mathrm{cr}(G_2+C_n)=n(n-1)+2$ due to the result by Klešč et al. [9]. Let $G_6$ be the graph obtained from $G_{11}$ by removing the edge $v_2{v}_3$ from the subdrawing in Figure 2a, that is, $\mathrm{cr}(G_9+C_n)\ge \mathrm{cr}(G_6+C_n)\ge \mathrm{cr}(G_2+C_n)$.

Corollary 5.

$\mathrm{cr}(G_6+C_n)=n(n-1)+2$ for $n\ge 3$.

Corollary 6.

$\mathrm{cr}(G_9+C_n)=n(n-1)+2$ for $n\ge 3$.

Notice that Staš [12] also established the results of $\mathrm{cr}(G_6+P_n)=\mathrm{cr}(G_9+P_n)=n(n-1)$ as some consequences of $\mathrm{cr}(G_2+D_n)=n(n-1)$. Finally, Staš and Valiska [14] conjectured that the crossing numbers of $W_m+P_n$ are given by $(Z\left(m\right)-1)⌊\frac{n}{2}⌋+Z(m+1)Z\left(n\right)+n+1$, for all $m\ge 3$ and $n\ge 2$, and where $W_m$ denotes the wheel on $m+1$ vertices and the Zarankiewicz’s number $Z\left(n\right)=⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋$ is defined for all positive integers n. Recently, this conjecture was proven for $W_3+P_n$ and $W_4+P_n$ by Klešč and Schrötter [6] and by Staš and Valiska [14], respectively.

On the other hand, the graphs $W_m+P_2$ and $W_m+P_3$ are isomorphic to the join product of the cycle $C_m$ with the cycle $C_3$ and with the graph $K_4\backslash e$ obtained by removing one edge from $K_4$, respectively. The exact values for the crossing numbers of the graphs $C_m+C_n$ and $K_4\backslash e+C_m$ are given by Klešč [4,5], respectively, and so the graphs $W_m+P_2$ and $W_m+P_3$ confirm the validity of this conjecture. Since the graph $W_m+P_4$ is isomorphic to the graph $G_{14}+C_m$, we establish the validity of this conjecture also for the graph $W_m+P_4$.

Corollary 7.

$\mathrm{cr}(W_m+P_4)=m(m-1)+3$ for $m\ge 3$.

6. Conclusions

We suppose that similar forms of discussions can be used to estimate the unknown values of the crossing numbers of the remaining graphs on five vertices with a much larger number of edges in the join products with the paths, and also with the cycles. We expect the same for other symmetric graphs of order six. Berežný and Staš [24] determined the crossing number of $W_5+D_n$. Using this result, it would also be useful to confirm the conjecture mentioned in Section 5 for the graph $W_5+P_n$ in the form $\mathrm{cr}(W_5+P_n)=6⌊\frac{n}{2}⌋⌊\frac{n-1}{2}⌋+n+3⌊\frac{n}{2}⌋+1$ for $n\ge 2$.