Regular Graphs with Small Skewness and Crossing Numbers

Abstract

Let $G$ be a graph. The crossing number of $G$, denoted $cr(G)$, is the minimum number of pairwise intersections of its edges when $G$ is drawn in the plane. The skewness of $G$, denoted $sk(G)$, is defined to be the minimum number of edges in $G$ whose removal results in a planar graph. We show, by constructions, the existence of $5$-regular graphs on $n$ vertices with skewness $s$ and crossing number $c$ for (i) $s=1, c=3$ and $n \equiv 2 \ (\hbox {mod } 4) \ge 10$, (ii) $s=2, c=2$ and $n \ge 8$ is even, and (iii) $s=1, c=1$ and $n \ge 14$ is even.

Let $G$ be a graph. The crossing number of $G$, denoted $cr(G)$, is the minimum number of pairwise intersections of its edges when $G$ is drawn in the plane. The skewness of $G$, denoted $sk(G)$, is defined to be the minimum number of edges in $G$ whose removal results in a planar graph. The skewness of a graph $G$ can be regarded as a measure of non-planarity of $G$. Some recent works on these parameters are contained in the papers [1–4, 6]. Clearly, $cr(G) \ge sk(G)$ and that $G$ is planar if and only if $sk(G) = 0$.

In [7], Owens determined all values of $r$ and $n$ for which there exists an $r$-regular planar graph with $n$ vertices. In particular, it was shown that for every even natural number $n \not \in \{6, 8, 10, 14\}$, there exists a $5$-regular planar graph on $n$ vertices.

In this note, we consider the corresponding problem mentioned above with planar graphs replaced by nearly planar graphs. More precisely, we prove by construction that (i) for every natural number $n \equiv 2 \ (\hbox {mod } 4) \ge 10$, there exists a $5$-regular graph $G$ on $n$ vertices with $sk(G) = 1$ and $cr(G) = 3$, (ii) for every even number $n \ge 8$, there exists a $5$-regular graph $H$ on $n$ vertices with $sk(H)=2$ and $cr(H)=2$, and (iii) for every even number $n \ge 14$, there exists a $5$-regular graph $J$ on $n$ vertices with $sk(J)=1$ and $cr(J)=1$.

We first note that there is no $5$-regular graph $G$ on $n$ vertices with $sk(G) =1$, where $n$ is an even natural number no more than $12$ unless $G$ is isomorphic to the graph $G_{10}$ depicted in Fig. 1(a) whose crossing number is $3$ (see [5]).

FormalPara Theorem 1

For every natural number $n \equiv 2 \ (\hbox {mod } 4) \ge 10$, there exists a 5-regular graph $G$ on $n$ vertices with $sk(G) = 1$ and $cr(G) = 3$.

FormalPara Proof

Let $n=4m+2$, where $m\ge 2$ is a natural number.

Let $G_n$ denote the $5$-regular graph which consists of two $(m+1)$-cycles $u_0u_1u_2 \dots u_m u_0$ and $v_0 v_1 v_1 \dots v_m v_0$ together with a path on $2m$ vertices $x_0 x_1 x_2 \dots x_{2m-1}$ that are joined by the following sets of edges:

• (i) $u_ix_{2i-1}, \ u_i x_{2i}, \ u_i x_{2i+1}, \ v_ix_{2m-2i-2}, \ v_i x_{2m-2i-1}, \ v_i x_{2m-2i}$, $i = 1, \ldots , m-1$,

• (ii) $u_0x_0, \ u_0x_1, \ u_0v_0, \ u_mx_{2m-1}, \ u_mx_0, \ u_mv_m, \ v_0x_{2m-2}, \ v_0x_{2m-1}, \ v_mx_0, \ v_mx_{2m-1}.$

The graphs $G_{10}$, $G_{14}$, and $G_{18}$ are depicted in Fig. 1.

We shall show, by induction on m, that $sk(G_{n})=1$, $cr(G_{n})=3$, and that $G_n - \{e_1, e_2\}$ is planar if and only if for some $i \in \{1, 2\}$, $e_{i}$ is the edge $e = u_0v_0$.

In [5], it has been shown that the graph $G_{10}$ of Fig. 2 has the property that $sk(G_{10})=1$, $cr(G_{10})=3$, and that $G_{10} - \{e_1, e_2\}$ is planar if and only if for some $i \in \{1, 2\}$, $e_{i}$ is the edge $e = u_0v_0$. That is, the result is true for $m=2$.

Let A denote the subgraph of $G_{n}$ induced by the set of vertices $\{ u_0, u_m, v_0, v_m, x_0\}$. By contracting all the edges not incident to $A$, we obtain a non-planar graph. Hence, $G_{n}$ is non-planar, and therefore, $sk(G_{n})=1$ as we can get a planar graph by deleting the edge $u_0v_0$.

Fig. 1

The graph $G_{n}$

We now show that $cr(G_n)=3$.

Assume on the contrary that $cr(G_{n}) = t$, where $t \in \{1,2\}$.

If $t=1$, let $f_1$ and $f_2$ be the two edges which intersect each other in an optimal drawing of $G_n$. Then clearly, at least one of these two edges, say $f_1$, is not the edge $u_0v_0$. Then $G_n - f_1$ is planar.

If $t=2$, then either the two crossings are made by some four edges or they are made by some three edges where one of them intersects the other two. In either case, we can always delete two edges, none of which is the edge $u_0v_0$, so that the resulting graph is planar.

Hence, we assert that there exist two edges $c_1, c_2$ such that $G_{n}-\{c_1, c_2\}$ is planar and neither $c_1$ nor $c_2$ is the edge $u_0v_0$.

Now, we make the following observation: Let $E_1=\{v_0v_1, x_{2m-1}x_{2m-2}, x_{2m-2}x_{2m-3} \}$ and $E_2 = \{u_0u_1, x_0x_1, x_1x_2\}$. Let $X_n$ denote the graph obtained from $G_n$ by contracting all the edges of $E_1$ and replacing all multiple edges by a single edge (see Fig. 2 for the case $n=18$). In so doing, the vertices $x_{2m-1}, x_{2m-2}$, and $x_{2m-3}$ are all been identified to become a single vertex. The same happens to the vertices $v_0$ and $v_1$. Then $X_n$ with the edge $u_{m-1}x_{2m-3}$ deleted is isomorphic to the graph $G_{n-4}$ with one edge subdivided. Hence, it follows that the induction hypothesis on $G_{n-4}$ can be carried over to the graph $X_n$. That is,

Fig. 2

The graph $X_{18}$

(O1): $sk(X_n) =1$, $cr(X_n) = 3$, and that $X_n - \{e_1, e_2\}$ is planar if and only if for some $i \in \{1, 2\}$, $e_{i}$ is the edge $e = u_0v_0$.

Let $H$ be the subgraph of $G_n$ which consists of the cycle $u_0x_0v_{m-1}x_2 u_1x_1u_0$ together with the edge $v_{m-1}x_1$. Let $P_1$ denote the path $u_1 u_2 \cdots u_m v_{m}x_0$ and $P_2$ the path $u_0v_0x_{2m-2}v_1v_2 \cdots v_{m-2}x_2$.

Then $H'=H \cup P_1 \cup P_2$ is a non-planar subgraph of $G_n$ because it is a subdivision of $K_{3,3}$.

Since $G_{n}-\{c_1,c_2\}$ is planar, at least one of $c_1$ or $c_2$ is an edge of $H'$. But this implies that at least one of $c_1$ or $c_2$, say $c_1$, is such that $c_1 \notin E_1$ or $c_1 \notin E_2$.

If $c_1 \notin E_1$, then in the planar graph $G_{n} - \{c_1, c_2\}$, we can contract all the edges of $E_1$ and obtain the planar graph $X_n -\{c_1, c_2\}$. However this contradicts (O1) and the fact that neither $c_1$ nor $c_2$ is the edge $u_0v_0$.

If $c_1 \notin E_2$, then again, in the planar graph $G_n -\{c_1, c_2\}$, we do the same to the set of edges $E_2$ and obtain a similar contradiction.

Hence, $cr(G_{n})=3$.

This completes the proof. $\square$

FormalPara Theorem 2

For every even number $n \ge 8$, there exists a 5-regular graph $G$ on $n$ vertices with $sk(G)=2$ and $cr(G)=2$.

FormalPara Proof

We first show that the result is true for all $n \equiv 0 \ (\hbox {mod } 4) \ge 8$.

Let $m \ge 2$ be a natural number and let $C_{4m}$ denote the $4m$-cycle $x_0 x_1 \dots x_{4m-1} x_0$.

Let $C_{4m}^2$ denote the graph obtained from $C_{4m}$ by adding a new edge to every two vertices which have a common neighbor. To $C_{4m}^2$, we add the following set of edges:

$$\begin{aligned} x_0x_3, \ x_{2m} x_{3+2m}, \ x_{2i} x_{-2i}, \ x_{3+2i} x_{3-2i},\quad i=1,2,\ldots , m-1 \end{aligned}$$

with the operations on the subscripts reduced modulo $4m$. Call the resulting graph $H_{4m}$. The graphs $H_{8}$ and $H_{24}$ are depicted in Fig. 3, where for simplicity, the vertex $x_i$ is labeled as $i$, for each $i = 0,1, 2, \dots , 4m-1$.

Fig. 3

The graph $H_{4m}$

We shall show that $sk(H_{4m})=2=cr(H_{4m})$ for each $m \ge 2$.

Let $A_1$ (respectively $A_2$) denote the subgraph of $H_{4m}$ induced by the set of vertices $\{x_0, x_1, x_2, x_3\}$ (respectively, $\{x_{2m}, x_{2m+1}, x_{2m+2}, x_{2m+3}\}$). Then $A_i$ is a complete subgraph $K_4$ of $H_{4m}$, $i=1, 2$. By contracting all edges not incident to $A_1$, we obtain a complete subgraph on $5$ vertices. Hence, $H_{4m}$ is non-planar.

If the two edges $x_0x_3$ and $x_{2m} x_{3+2m}$ are deleted from $H_{4m}$ (see Fig. 3), we obtain a planar graph. Hence, $sk(H_{4m}) \le cr(H_{4m})\le 2$.

Assume that $sk(H_{4m})=1$. Then there exists an edge $e$ in $H_{4m}$ such that $H_{4m}-e$ is planar. Since the edge $e$ belongs to at most one of the $K_4$, either $A_1$ or $A_2$, say $A_1$, is a subgraph of $H_{4m}-e$. By contracting all edges in $H_{4m}-e$ not incident to $A_1$, we obtain the complete subgraph on $5$ vertices. This contradiction shows that $cr(H_{4m})\ge sk(H_{4m})\ge 2$.

Let $n=4m+2$, where $m \ge 2$, and let $J_n$ denote the $5$-regular graph which consists of two $(m+1)$-cycles $u_0u_1u_2 \cdots u_m u_0$ and $v_0 v_1 v_1 \cdots v_m v_0$ together with a path on $2m$ vertices $x_0 x_1 x_2 \cdots x_{2m-1}$ that are joined by the following sets of edges:

• (i) $u_ix_{2i-1}, \ u_i x_{2i}, \ u_i x_{2i+1}, \ v_ix_{2m-2i-2}, \ v_i x_{2m-2i-1}, \ v_i x_{2m-2i}$,    $i = 1,2, \ldots , m-1$,

• (ii) $u_0x_0, \ u_0x_1, \ u_0v_m, \ u_mx_{2m-1}, \ u_mx_0, \ u_mv_0, \ v_0x_{2m-2}, \ v_0x_{2m-1}, \ v_mx_0, \ v_mx_{2m-1}.$

The graphs $J_{10}$, $J_{14}$, and $J_{18}$ are depicted in Fig. 4.

Fig. 4

The graph $J_n$

It has been shown in [5] that $sk(J_{10})=2=cr(J_{10})$. We now show that $sk(J_{n})=2=cr(J_{n})$ for all $n \ge 10$.

Let $A_1$ (respectively, $A_2$) denote the subgraph of $J_{n}$ induced by the set of vertices $\{u_0, x_1, x_0, v_m, v_{m-1} \}$ (respectively, $\{v_0, x_{2m-1}, x_{2m-2}, u_{m-1}, u_m \}$). By contracting all edges not incident to $A_1$, we obtain a non-planar graph, and so $J_{n}$ is non-planar.

If the two edges $u_0v_m$ and $u_mv_0$ are deleted from $J_n$, we get a planar graph (see for example, Fig. 4 for the case $n=18$). Hence, we have $sk(J_{n}) \le cr(J_n)\le 2$.

Assume that $sk(J_n)=1$. Then there exists an edge $e$ in $J_n$ such that $J_n -e$ is planar. Since the edge $e$ belongs to at most one of the $A_i's$, it follows that either $A_1$ or $A_2$, say $A_1$, is a subgraph of $J_n-e$. By contracting all edges in $J_n-e$ not incident to $A_1$, we obtain the complete bipartite graph $K_{3,3}$. This contradiction shows that $cr(J_n) \ge sk(J_n)\ge 2$, and the proof is complete. $\square$

FormalPara Theorem 3

For every even number $n \ge 14$, there exists a 5-regular graph $G$ on $n$ vertices with $sk(G)=1$ and $cr(G)=1$.

FormalPara Proof

The graph $L_1$ (respectively, $L_2$) in Fig. 5 is a 5-regular graph on 14 (respectively, 16) vertices. Note that each $L_i$ has a complete subgraph $K_4$ induced by the vertices $a, b, c$, and $d$. By contracting all edges not incident to the $K_4$, we obtain a complete graph on 5 vertices. Hence, $L_i$ is non-planar for $i \in \{1, 2\}$. From the drawings in Fig. 5, we can then conclude that $sk(L_i)=1 = cr(L_i)$ for $i \in \{1, 2\}$.

We shall now apply the following expansion procedure to the graph $L_1$ (respectively, $L_2$) to obtain a 5-regular graph $L_1'$ (respectively, $L_2'$) on 18 (respectively, 20) vertices such that $sk(L_i') = 1$ and $cr(L_i') = 1$ for $i \in \{1, 2\}$.

Delete the edges $ab, bd, dc, ca$ from the $K_4$ and then add 4 vertices $a', b', c', d'$ which form a new $K_4$ together with 8 new edges $aa', bb', cc', dd', a'b, b'c, c'd, d'a$ joining the new $K_4$ to the vertices $a, b, c$, and $d$ (see Fig. 6).

Fig. 5

The graphs $L_1$ and $L_2$

Fig. 6

The expansion procedure

On repeating the expansion procedure to the resulting graph, after a finite number of times, we obtain a $5$-regular graph $G$ on $n$ vertices with $sk(G)=1$ and $cr(G)=1$ for every even number $n \ge 14$.

This completes the proof. $\square$