The Ramsey Numbers for Trees of Large Maximum Degree Versus the Wheel Graph $W_8$

Abstract

The Ramsey numbers $R(T_n,W_8)$ are determined for each tree graph $T_n$ of order $n\ge 7$ and maximum degree $\Delta (T_n)$ equal to either $n-4$ or $n-5$. These numbers indicate strong support for the conjecture, due to Chen, Zhang and Zhang and to Hafidh and Baskoro, that $R(T_n,W_m) = 2n-1$ for each tree graph $T_n$ of order $n\ge m-1$ with $\Delta (T_n)\le n-m+2$ when $m\ge 4$ is even.

1 Introduction

Let G and H be two simple graphs. The Ramsey number R(G, H) is the smallest integer n such that, for any graph of order n, either it contains G or its complement contains H as a subgraph. Chvátal and Harary [7] proved that $R(G,H)\ge (c(G)-1)(\chi (H)-1)+1$ where c(G) is the largest order of any connected component of G and where $\chi (H)$ is the chromatic number of H. For any tree graph $G=T_n$ of order n and the wheel graph $H = W_m$ of order $m+1$ obtained by connecting a vertex to each vertex of the cycle graph $C_m$, the Chvátal-Harary bound implies that $R(T_n,W_m)\ge 2n-1$ when m is even and $R(T_n,W_m)\ge 3n-2$ when m is odd.

Chen et al. [12] and Zhang [23] showed that $R(P_n,W_m)$ achieves these Chvátal-Harary bounds for the path graph $T_n=P_n$ of order n when m is odd and $3\le m\le n+1$ and when m is even and $4\le m\le n+1$; see also [1, 21]. Baskoro et al. [3] and Surahmat and Baskoro [22] further proved that $R(T_n,W_m)$ achieves the Chvátal-Harary bounds for $m = 4,5$ and all tree graphs $T_n$ of order $n\ge 3$, except when $m=4$ and $T_n$ is the star graph $S_n$, in which case $R(S_n,W_4) = 2n+1$. This led Baskoro et al. [3] to conjecture that $R(T_n,W_m) = 3\,m-2$ for all tree graphs $T_n$ of order n when $m\ge 5$ is odd. The conjecture is true for all sufficiently large n, according to a result of Burr et al. [5]. In contrast, the analogous equality $R(T_n,W_m) = 2n-1$ for even $m\ge 4$ is false since the star graph $T_n = S_n$ does not achieve this bound, as the following combined result of Zhang [24] and Zhang et al. [25, 26] shows; see also [8, 15, 16, 18, 20].

Theorem 1.1

[24,25,26] For $n\ge 5$,

$$\begin{aligned} R(S_n,W_8)={\left\{ \begin{array}{ll} 2n+1 &{} \text {if }n\text { is odd};\\ 2n+2 &{} \text {if }n\text { is even}. \end{array}\right. } \end{aligned}$$

Baskoro et al. [3] therefore conjectured that $R(T_n,W_m) = 2n-1$ for all non-star tree graphs $T_n$ of order n when $n\ge 4$ is even. This conjecture was disproved by Chen, Zhang and Zhang [9] who showed that $R(T_n,W_6)=2n$ for certain non-star tree graphs $T_n$. Zhang [23] further proved the following theorem which shows that the conjecture is false when n is small, even for the path graph $P_n$; see also [2, 12, 19, 21].

Theorem 1.2

[23] If m is even and $n+2\le m\le 2n$, then $R(P_n,W_m) = m+n-2$.

However, Chen, Zhang and Zhang [9] conjectured that $R(T_n,W_m) = 2n-1$ for all tree graphs $T_n$ of order $n\ge m-1$ when m is even and the maximum degree $\Delta (T_n)$ “is not too large"; see also [10, 11, 13]. Hafidh and Baskoro [14] refined this conjecture by specifying the bound $\Delta (T_n)\le n-m+2$. When n is large compared to m, $\Delta (T_n)$ is not required to be small; indeed, the refined conjecture implies that, for each fixed even integer m, all but a vanishing proportion of the tree graphs $\{T_n \,:\, n\ge m-1\}$ satisfy $R(T_n,W_m)=2n-1$.

For $m=8$, the bound is $\Delta (T_n)\le n-6$. There is exactly one tree graph $T_n$ of order n with maximum degree $\Delta (T_n) = n-1$, namely the star graph $S_n$; see Theorem 1.1. There is exactly one tree graph $T_n$ of order $n\ge m-1$ with maximum degree $\Delta (T_n) = n-2$: the graph $S_n(1,1)$ obtained by subdividing an edge of $S_{n-1}$. More generally, let $S_n(\ell ,m)$ be the tree graph of order n obtained by subdividing m times each of $\ell$ chosen edges of $S_{n-\ell m}$; see Fig. 1.

Fig. 1

Examples of $S_n(\ell ,m)$, $S_n(\ell )$ and $S_n[\ell ]$

By Theorem 1.2, $R(P_4,W_8) = 10$. Hafidh and Baskoro [14] determined the Ramsey number $R\big (S_n(1,1),W_8\big )$ as follows.

Theorem 1.3

[14] For $n\ge 5$,

$$\begin{aligned} R\big (S_n(1,1),W_8\big ) = {\left\{ \begin{array}{ll} 2n+1 &{} \text {if }n\text { is odd},\\ 2n &{} \text {if }n\text { is even}. \end{array}\right. } \end{aligned}$$

There are exactly 3 tree graphs $T_n$ of order n with maximum degree $n-3$, namely $S_n(1,2)$, $S_n(3)$ and $S_n(2,1)$, where $S_n(\ell )$ is the tree graph of order n obtained by adding an edge joining the centers of two star graphs $S_\ell$ and $S_{n-\ell }$; see Fig. 1. By Theorem 1.2, $R(P_5,W_8)=11$. Hafidh and Baskoro [14] determined the Ramsey numbers for the three other graphs as follows.

Theorem 1.4

[14] For $n\ge 6$,

$$\begin{aligned} R(S_n(1,2),W_8)&= {\left\{ \begin{array}{ll} 2n+1 &{} \hbox { if}\ n\equiv 3 \pmod {4}\,;\\ 2n &{} \text{ otherwise } \end{array}\right. }\\ R(S_n(3),W_8)&={\left\{ \begin{array}{ll} 2n-1 &{} \hbox {if} n \hbox {is odd and} n\ge 9\,;\\ 2n &{} \text {otherwise} \end{array}\right. }\\ R(S_n(2,1),W_8)&={\left\{ \begin{array}{ll} 2n-1 &{} \text {if }n\text { is odd}\,;\\ 2n &{} \text {otherwise}\,. \end{array}\right. } \end{aligned}$$

The purpose of the present paper is to determine the Ramsey numbers $R(T_n,W_8)$ for all tree graphs $T_n$ of order $n\ge 6$ with maximal degree $\Delta (T_n) \ge n-5$; see Theorems 2.1, 2.2 and 3.1 in Sects. 2 and 3. These Ramsey numbers show that the proportion of tree graphs $T_n$ that satisfy the equality $R(T_n,W_8) = 2n-1$ quickly grows as the maximal degree $\Delta (T_n)$ decreases. When $\Delta (T_n) \ge n-2$, no tree graph $T_n$ satisfies the equality. In contrast, when $\Delta (T_n) = n-3$, roughly one third of all tree graphs $T_n$ satisfy the equality; see Theorem 1.4. When $\Delta (T_n) = n-4$, more than 85% of all tree graphs $T_n$ satisfy the equality; see Theorems 2.1 and 2.2. And when $\Delta (T_n) = n-5$, roughly 94.7% of all tree graphs $T_n$ satisfy the equality; see Theorem 3.1. These results thereby lend strong support for the conjecture described above by Chen, Zhang and Zhang [9] and Hafidh and Baskoro [14].

The contents of the present paper are as follows. Sections 2 and 3 present the main results, namely Theorems 2.1, 2.2 and 3.1 mentioned above. Section 4 provides useful auxiliary results that are used in the proofs of the main results. These proofs are presented in Sects. 5, 6 and 7, respectively.

2 The Ramsey numbers $R(T_n,W_8)$ for $\Delta (T_n)=n-4$

This section presents the Ramsey numbers $R(T_n,W_8)$ for all tree graphs $T_n$ of order $n\ge 6$ with $\Delta (T_n) = n-4$. For $n=6$, there is just one such graph, namely the path graph $T_6 = P_6$. Theorem 1.2 provides the Ramsey number $R(P_6,W_8) = 12$. For $n=7$, there are five tree graphs with $\Delta (T_n) = n-4$, namely the graphs A, B, C, D and E shown in Fig. 2.

Fig. 2

Tree graphs of order 7 with $\Delta (T_n)=n-4$

The Ramsey numbers $R(T_n,W_8)$ for these tree graphs are determined as follows.

Theorem 2.1

$R(T,W_8) = 13$ for each $T\in \{A,B,C\}$, $R(D,W_8)=14$ and $R(E,W_8)=15$.

For $n\ge 8$, there are 7 tree graphs $T_n$ of order n with $\Delta (T_n) = n-4$, namely the graphs $S_n(4)$, $S_n[4]$, $S_n(1,3)$, $S_n(3,1)$, $T_A(n)$, $T_B(n)$ and $T_C(n)$ shown in Figs. 1 and 3, where $S_n[\ell ]$ is the tree graph of order n obtained by adding an edge joining the center of $S_{n-\ell }$ to a degree-one vertex of $S_\ell$; see Fig. 1.

Fig. 3

Three tree graphs with $\Delta (T_n)=n-4$

The Ramsey numbers $R(T_n,W_8)$ for these seven tree graphs are determined as follows.

Theorem 2.2

If $n\ge 8$, then

$$\begin{aligned} R(S_n(4),W_8)&={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\ge 9\,;\\ 16 &{} \hbox { if}\ n=8 \end{array}\right. }\\ R(T_n,W_8)&={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4}\,;\\ 2n &{} \text {otherwise}\\ \end{array}\right. }\\ R(T_n',W_8)&=2n-1\,, \end{aligned}$$

for each $T_n\in \{S_n[4],S_n(1,3),T_A(n),T_B(n)\}$ and $T_n'\in \{T_C(n),S_n(3,1)\}$.

Proofs of Theorems 2.1 and 2.2 are given in Sects. 5 and 6.

3 The Ramsey numbers $R(T_n,W_8)$ for $\Delta (T_n)=n-5$

This section presents the Ramsey numbers $R(T_n,W_8)$ for all tree graphs $T_n$ of order $n\ge 7$ with $\Delta (T_n) = n-5$. For $n=7$, there is just one such graph, namely the path graph $T_7 = P_7$. Theorem 1.2 provides the Ramsey number $R(P_7,W_8) = 13$. For $n=8$, there are 16 tree graphs $T_n$ of order n with $\Delta (T_n) = n-5$, namely $S_n(1,4)$, $S_n(2,2)$ and the tree graphs shown in Fig. 4. For $n=9$, there are 18 tree graphs $T_n$ of order n with $\Delta (T_n) = n-5$, namely $S_n(1,4)$, $S_n[5]$, $S_n(2,2)$, $S_n(4,1)$ and the tree graphs shown in Fig. 4. For $n\ge 10$, there are 19 tree graphs $T_n$ of order n with $\Delta (T_n) = n-5$, namely $S_n(1,4)$, $S_n(5)$, $S_n[5]$, $S_n(2,2)$, $S_n(4,1)$ and the tree graphs shown in Fig. 4.

Fig. 4

Tree graphs $T_n$ with $\Delta (T_n)=n-5$

The Ramsey numbers $R(T_n,W_8)$ for these tree graphs are determined as follows.

Theorem 3.1

If $n\ge 8$, then $R(T_n,W_8) = 2n-1$ for all

$$\begin{aligned} T_n\in \{S_n(1,4), S_n(2,2), T_D(n), \ldots , T_S(n)\} \end{aligned}$$

except when $T_n\in \{T_E(8), T_F(8), S_n(1,4), S_n(2,2), T_D(n), T_N(n)\}$ and $n\equiv 0\pmod {4}$, in which case $R(T_n,W_8) = 2n$.

Furthermore, if $n\ge 9$, then $R(T_n,W_8)=2n-1$ for each $T_n\in \{S_n[5], S_n(4,1)\}$, and if $n\ge 10$, then $R(S_n(5),W_8)=2n-1$.

A proof of this theorem is given in Sect. 7.

4 Auxiliary results

To prove the main theorems, the following auxiliary results will be used. For any simple graph $G = (V,E)$, let $\delta (G)$ be the minimum degree of any vertex in G, and let ${\overline{G}} = \big (V, \left( {\begin{array}{c}V\\ 2\end{array}}\right) \backslash E\big )$ be the complement of G.

Lemma 4.1

[4] Let G be a graph of order n. If $\delta (G)\ge \frac{n}{2}$, then either G contains $C_\ell$ for all $3\le \ell \le n$, or n is even and $G=K_{\frac{n}{2},\frac{n}{2}}$.

Lemma 4.2

[6] Let G be a graph with $\delta (G)\ge n-1$. Then G contains all tree graphs of order n.

Observation 4.3

If $G=H_1\cup H_2$ is the disjoint union of graphs $H_1$ and $H_2$, where $\overline{H_1}$ contains $S_5$ and $H_2$ is a graph of order at least 4, then ${\overline{G}}$ contains $W_8$.

Lemma 4.4

Let $H_1$ be a graph whose complement $\overline{H_1}$ contains $S_4$, and let $H_2$ be a graph of order $m\ge 5$. If $G=H_1\cup H_2$, then either ${\overline{G}}$ contains $W_8$, or $H_2$ is $K_m$ or $K_m-e$, where e is an edge in $K_m$.

Proof

If $\overline{H_2}$ has at most one edge, then $H_2$ is the complete graph $K_m$ or the graph $K_m-e$ obtained from removing an edge e from $K_m$. Suppose now that $\overline{H_2}$ has at least two edges. Consider a star $S_4$ in $\overline{H_1}$ and let $v_0$ be its center and $v_1,v_2,v_3$ its leaves. Note that each $v_i$ is adjacent to each $a\in V(H_2)$ in ${\overline{G}}$. Choose 5 vertices $a,b,c,d,e \in V(H_2)$ such that either ab and cd are independent edges, or abc is a path, in $\overline{H_2}$. In both cases, ${\overline{G}}$ contains $W_8$ with hub $v_0$. In the former case, $v_1abv_2cdv_3ev_1$ forms the $C_8$ rim; in the latter, $v_1abcv_2dv_3ev_1$ forms the $C_8$ rim. $\square$

The neighbourhood $N_G(v)$ of a vertex v in G is the set of vertices that are adjacent to v in G and $d_G(v)=|N_G(v)|$ is the degree of the vertex v. For $X, Y\subseteq V$, G[X] is the subgraph induced by X in G and $E_G(X,Y)$ is the set of edges in G with one endpoint in X and the other in Y. The following lemma provides sufficient conditions for a graph or its complement to contain $C_8$.

Lemma 4.5

Suppose that $U=\{u_1,\ldots ,u_4\}$ and $V=\{v_1,\ldots ,v_4\}$ are two disjoint subsets of vertices of a graph G for which $|N_{G[V\cup \{u\}]}(u)|\le 1$ for each $u\in U$ and $|N_{G[U\cup \{v\}]}(v)|\le 2$ for each $v\in V$. Then ${\overline{G}}[U\cup V]$ contains $C_8$.

Proof

Suppose that $N_{G[U\cup \{v\}]}(v)\le 1$ for each $v\in V$. Then ${\overline{G}}[U\cup V]$ contains a subgraph obtained by removing a matching from $K_{4,4}$ and therefore contains $C_8$. Suppose now that $N_{G[U\cup \{v_1\}]}(v_1) = \{u_1,u_2\}$, and assume without loss of generality that $v_3\notin N_{G[V\cup \{u_3\}]}(u_3)$ and $v_4\notin N_{G[V\cup \{u_4\}]}(u_4)$. Neither $u_1$ nor $u_2$ is adjacent to $v_2$, $v_3$ or $v_4$, so $v_1u_3v_3u_1v_2u_2v_4u_4v_1$ forms $C_8$ in ${\overline{G}}[U\cup V]$. $\square$

Lemma 4.6

[17] Let G(u, v, k) be a simple bipartite graph with bipartition U and V, where $|U|=u\ge 2$ and $|V|=v\ge k$, and where each vertex of U has degree of at least k. If $u\le k$ and $v\le 2k-2$, then G(u, v, k) contains a cycle of length 2u.

Corollary 4.7

Suppose that U and V are two disjoint subsets of vertices of a graph G for which $|N_{G[V\cup \{u\}]}(u)|\le 2$ for each $u\in U$. If $|U|\ge 4$ and $|V|\ge 6$, then ${\overline{G}}[U\cup V]$ contains $C_8$.

Proof

Since $|U|\ge 4$ and $|V|\ge 6$, we can choose any 4 vertices from U to form $U'$ and any 6 vertices from V to form $V'$. We have that $N_{G[V'\cup \{u\}]}(u)\le 2$ for each $u\in U'$. Then each vertex of $U'$ is adjacent to at least 4 vertices of $V'$ in ${\overline{G}}$ and ${\overline{G}}[U'\cup V']$ must contain a graph with the properties of G(4, 6, 4) in Lemma 4.6. Hence by that lemma, ${\overline{G}}[U\cup V]$ must contain $C_8$. $\square$

We will also use the following corollary whose proof is almost identical to that of Corollary 4.7.

Corollary 4.8

Suppose that U and V are two disjoint subsets of vertices of a graph G for which $|N_{G[V\cup \{u\}]}(u)|\le 3$ for each $u\in U$. If $|U|\ge 4$ and $|V|\ge 8$, then ${\overline{G}}[U\cup V]$ contains $C_8$.

5 Proof of Theorem 2.1

The proof of Theorem 2.1 is here proved as three theorems, the first of which is as follows.

Theorem 5.1

$R(T,W_8) = 13$ for each $T\in \{A,B,C\}$.

Proof

Note that $G=2K_6$ does not contain A, B or C and that ${\overline{G}}$ does not contain $W_8$. Therefore, $R(T,W_8)\ge 13$ for $T = A,B,C$.

Let G be a graph of order 13 whose complement ${\overline{G}}$ does not contain $W_8$. By Theorem 1.4, G has a subgraph $T=S_7(2,1)$. Label V(T) as in Fig. 5. Set $U=V(G)-V(T)$; then $|U|=6$.

First, suppose that $A\nsubseteq G$. Then $v_1$ is not adjacent to $v_2$ or $v_6$. Similarly, $v_2$ and $v_5$ are not adjacent.

Fig. 5

$S_7(2,1)$ and U in G

Case 1a: There is a vertex in U, say u, that is adjacent to $v_1$.

Since A is not contained in G, $v_1$ is not adjacent to $v_3$, $v_4$ or any vertex of U other than u. Let $W=\{v_2,v_3,v_4,v_6,u_1,\ldots ,u_4\}$ for any 4 vertices $u_1,\ldots ,u_4$ in U other than u. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 and, together with $v_1$ as hub, forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[W])\le 3$ and $\Delta (G[W])\ge 4$. Note that $|N_{G[\{u_1,\ldots ,u_4,v_i\}]}(v_i)|\le 1$ for $i=2,3,4,6$ since G does not contain A. It is now straightforward to check that $v_2$, $v_3$, $v_4$ and $v_6$ cannot be the vertex with degree at least 4. Without loss of generality, assume that $u_1$ has degree at least 4 in G[W]. Then $u_1$ is adjacent to at least one of $v_2,v_3,v_4,v_6$, so G contains A, a contradiction.

Case 1b: $v_1$ is not adjacent to any vertices in U.

By arguments similar to those in Case 1a, $v_2$ is not adjacent to any vertex in U. Let $W = \{v_2,v_6\}\cup U$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which, with $v_1$ as hub, forms $W_8$ in ${\overline{G}}[W]$, a contradiction. Thus, $\delta ({\overline{G}}[W])\le 3$ and $\Delta (G[W])\ge 4$. Since $v_2$ is not adjacent to any vertex in U, there are only three subcases to be considered.

Subcase 1b.1: $d_{G[W]}(v_6)\ge 4$.

Label $U = \{u_1,\ldots ,u_6\}$ so that $v_6$ is adjacent to $u_1$, $u_2$ and $u_3$ in G[W]. Since G does not contain A, vertices $u_1,u_2,u_3,v_2$ are not adjacent to $v_3$ or $v_4$ in G. Note that by arguments as in Case 1a, $u_1$, $u_2$ and $u_3$ are isolated vertices in G[U]. Then $v_1u_4u_2v_3v_2u_5u_3u_6v_1$ and $u_1$ form $W_8$ in ${\overline{G}}$, a contradiction.

Subcase 1b.2: $d_{G[W]}(v_6)\le 3$ and $v_6$ is adjacent to a vertex $u\in U$ with $d_{G[W]}(u)\ge 4$.

The graph G contains A, with u as the vertex of degree 3 in A, a contradiction.

Subcase 1b.3: $d_{G[W]}(v_6)\le 3$ and $v_6$ is not adjacent to any vertex $u\in U$ with $d_{G[W]}(u)\ge 4$.

Label $V(U) = \{u_1,\ldots ,u_6\}$ so that $u_6$ is adjacent to $u_2$, $u_3$, $u_4$ and $u_5$ in G. Since $A\nsubseteq G$, none of $v_1,\ldots ,v_7$ is adjacent in G to any of $u_2,\ldots ,u_5$. If $v_1$ is not adjacent in G to any two of the vertices $v_3,v_4,v_7$, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Therefore, $N_{G[v_3,v_4,v_7]}(v_1)\ge 2$ and, similarly, $N_{G[v_3,v_4,v_7]}(v_2)\ge 2$. Hence, one of $v_3,v_4,v_7$ is adjacent in G to both $v_1$ and $v_2$. If $v_3$ or $v_4$ is adjacent to both $v_1$ and $v_2$, then G contains A, with $v_7$ as vertex of degree 3, a contradiction. Finally, if both $v_1$ and $v_2$ are adjacent in G to $v_7$ and each of them is adjacent to a different vertex in $v_3$ and $v_4$, then G also contains A, where either $v_1$ or $v_2$ is the vertex of degree 3, a contradiction.

Therefore, $R(A,W_8)\le 13$, so $R(A,W_8) = 13$.

Now, suppose that $B\nsubseteq G$. Then $v_1,v_2,v_5,v_6$ are not adjacent to $v_3$ or $v_4$ in G, and $v_1$ and $v_2$ are not adjacent to U in G. Label the vertices $U=\{u_1,\ldots ,u_6\}$ and let $W=\{v_3,v_4\}\cup U$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which, with $v_1$ as hub, forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[W])\le 3$ and $\Delta (G[W])\ge 4$. If $v_3$ or $v_4$ is adjacent to the vertex of degree at least 4 in G[W], then B is contained in G, with $v_7$ as the vertex of degree 3. Hence, only two cases need to be considered.

Case 2a: $v_3$ or $v_4$ is the vertex of degree at least 4 in G[W].

Without loss of generality, assume that $v_3$ is the vertex of degree at least 4 in G[W]. As previously shown, $v_3$ is not adjacent to $v_4$. Therefore, it may be assumed that $v_3$ is adjacent to $u_1$, $u_2$, $u_3$ and $u_4$ in G. Since $B\nsubseteq G$, $u_1,\ldots ,u_4$ are independent in G and are not adjacent to $\{v_1,v_2,v_4,v_5,v_6\}$. Also, $v_1$ is not adjacent to $v_6$ and $v_2$ is not adjacent to $v_5$. Then $v_1v_6u_2v_2v_5u_3v_4u_4v_1$ and $u_1$ form $W_8$ in ${\overline{G}}$, a contradiction.

Case 2b: One of the vertices in U, say $u_1$, is the vertex of degree at least 4 in G[W].

As above, $u_1$ is not adjacent to $v_3$ or $v_4$ in G. It may then be assumed that $u_1$ is adjacent to $u_2$, $u_3$, $u_4$ and $u_5$. Since $B\nsubseteq G$, $v_1,\ldots ,v_7$ are not adjacent to $\{u_2,\ldots ,u_5\}$. Note that $v_3$ is not adjacent to $\{v_1,v_2,v_5,v_6\}$. By Observation 4.3, ${\overline{G}}$ contains $W_8$, a contradiction.

Therefore, $R(B,W_8)\le 13$.

Lastly, suppose that $C\nsubseteq G$. Then $v_5$ and $v_6$ are not adjacent in G to each other or to $v_3$, $v_4$ or U. Furthermore, $v_5$ is not adjacent to $v_2$ and $v_6$ is not adjacent to $v_1$. Label the vertices $U=\{u_1,\ldots ,u_6\}$ and let $W=\{v_3,v_4,v_6,u_1,\ldots ,u_5\}$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which, with $v_5$ as hub, forms $W_8$, a contradiction. Then $\delta ({\overline{G}}[W])\le 3$ and $\Delta (G[W])\ge 4$. Note that $v_6$ is not adjacent to any other vertex in G[W], $v_6$ is not the vertex of degree at least 4 in G[W]. If $v_3$ or $v_4$ is the vertex of degree 4, then G contains C, with $v_3$ or $v_4$ and $v_7$ as the vertices of degree 3. Thus, one of the vertices in U, say $u_1$, is the vertex of degree at least 4 in G[W]. Now, consider the following three cases.

Case 3a: Both $v_3$ and $v_4$ are adjacent to $u_1$ in G[W].

Suppose that $u_1$ is also adjacent to $u_2$ and $u_3$ in G[W]. Since $C\nsubseteq G$, $v_3$ is not adjacent in G to $v_4$ and neither $v_3$ nor $v_4$ is adjacent to $\{v_1,v_2,v_5,v_6,u_2,\ldots ,u_6\}$. Note that $|N_{G[\{v_1,v_2,u_i\}]}(u_i)|\le 1$ for $i=2,3$ since $C\nsubseteq G$. If $v_1$ is adjacent to $u_2$ and $u_3$ in ${\overline{G}}$, then $v_1u_2v_5u_4v_3u_5v_6u_3v_1$ and $v_4$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $v_1$ is adjacent in G to at least one of $u_2$ and $u_3$. Similarly, $v_2$ is adjacent to at least one of $u_2$ and $u_3$. Since $|N_{G[\{v_1,v_2,u_i\}]}(u_i)|\le 1$ for $i=2,3$, $v_1$ is adjacent to $u_2$ and $v_2$ is adjacent to $u_3$, or vice versa. Then neither $u_2$ nor $u_3$ is adjacent in G to $u_4, u_5, u_6$, since $C\nsubseteq G$. Therefore, $v_1v_3v_2v_5u_2u_4u_3v_6v_1$ and $v_4$ form $W_8$ in ${\overline{G}}$, a contradiction.

Case 3b: One of $v_3$ and $v_4$, say $v_3$, is adjacent to $u_1$ in G[W].

Suppose that $u_1$ is adjacent to $u_2$, $u_3$ and $u_4$ in G[W]. Then $v_1, v_2, v_4, v_5, v_6, u_2, u_3, u_4\notin N_G(v_3)$ and $|N_{G[\{v_4,u_2,u_3,u_4\}]}(v_4)|\le 1$. Without loss of generality, assume that $v_4$ is not adjacent to $u_2$ or $u_3$ in G. Now, suppose that $v_4$ is adjacent to $u_4$ in G. Since $C\nsubseteq G$, $u_4$ is not adjacent to $v_1$ or $v_2$ in G. Then $v_1u_4v_2v_5u_2v_4u_3v_6v_1$ and $v_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Otherwise, suppose that $v_4$ is not adjacent to $u_4$ in G. Then, $|N_{G[\{u_i,v_1,v_2\}]}(u_i)|\le 1$ for $i=2,3,4$ and at least two of $u_2$, $u_3$ and $u_4$ are not adjacent to $v_1$ or $v_2$ in G. Without loss of generality, assume that $u_2$ and $u_3$ are not adjacent to $v_1$ in G. In this case, $v_1u_2v_4u_4v_5u_5v_6u_3v_1$ and $v_3$ form $W_8$ in ${\overline{G}}$, again a contradiction.

Case 3c: $v_3$ and $v_4$ are both not adjacent in G[W] to $u_1$.

Assume that $u_1$ is adjacent to each of $u_2,\ldots ,u_5$ in G[W]. Since $C\nsubseteq G$, $|N_{G[\{v_1,\ldots ,v_7,u_i\}]}(u_i)|\le 1$ for $i=2,\ldots ,5$, and $|N_{G[\{u_2,\ldots ,u_5,v_j\}]}(v_j)|\le 1$ for $j=3,4$. Since $|N_{G[\{v_1,v_2,u_i\}]}(u_i)|\le 1$ for $i=2,\ldots ,5$, one of $v_1$ and $v_2$, say $v_1$, satisfies $|N_{G[\{u_2,\ldots ,u_5,v_1\}]}(v_1)|\le 2$. By Lemma 4.5, ${\overline{G}}[v_1,v_3,v_4,v_5,u_2,\ldots ,u_5]$ contains $C_8$ which, with hub $v_6$, forms $W_8$ in ${\overline{G}}$.

Therefore, $R(C,W_8)\le 13$. This completes the proof of the theorem. $\square$

Theorem 5.2

$R(D,W_8)=14$.

Proof

Let $G=K_6\cup H$ where H is the graph shown in Fig. 6.

Fig. 6

The graph H

Since G does not contain D and ${\overline{G}}$ does not contain $W_8$, $R(D,W_8)\ge 14$.

Now, let G be any graph of order 14. Suppose neither G contains D as a subgraph, nor ${\overline{G}}$ contains $W_8$ as a subgraph. By Theorem 5.1, $B\subseteq G$. Label the vertices of B as shown in Fig. 7 and set $U=\{u_1,\ldots ,u_7\}=V(G)-V(B)$. Since $D\nsubseteq G$, $v_7$ is non-adjacent to $v_6$ and U, and $v_4$ is non-adjacent to $v_1$ and $v_2$.

Fig. 7

$B\subseteq G$

Let $W=\{v_6\}\cup U$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which, with $v_7$ as hub, forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[W])\le 3$ and $\Delta (G[W])\ge 4$. Three cases will now be considered.

Case 1: $v_6$ is the vertex of degree at least 4 in G[W].

Assume that $v_6$ is adjacent to $u_1$, $u_2$, $u_3$ and $u_4$ in G[W]. Then $v_5$ is adjacent to $v_1$ and $v_2$ in ${\overline{G}}$ and $v_3$ is adjacent in ${\overline{G}}$ to $v_6$, $u_1$, $u_2$, $u_3$ and $u_4$.

Subcase 1.1: $E_G(\{u_1,\ldots ,u_4\},\{u_5,u_6,u_7\})\ne \emptyset$.

Without loss of generality, assume that $u_1$ is adjacent to $u_5$ in G. Since $D\nsubseteq G$, $\{u_2,u_3,u_4\}$ is independent in G and is adjacent to $v_1$, $v_2$, $u_6$ and $u_7$ in ${\overline{G}}$; $v_6$ is adjacent in ${\overline{G}}$ to $v_1$ and $v_2$; $v_4$ and $v_5$ are adjacent in ${\overline{G}}$ to $u_1$ and $u_5$; and $v_3$ is adjacent in ${\overline{G}}$ to $u_5$. If $v_4$ is adjacent to $u_2$ in G, then $v_5$ is adjacent in ${\overline{G}}$ to $u_3$ and $u_4$, so $v_1v_5v_2u_2u_6v_7u_7u_3v_1$ and $u_4$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $v_4$ is adjacent to $u_2$ in ${\overline{G}}$, and $v_1v_4v_2u_4u_6v_7u_7u_3v_1$ and $u_2$ form $W_8$ in ${\overline{G}}$, again a contradiction.

Subcase 1.2: $\{u_1,\ldots ,u_4\}$ is not adjacent to $\{u_5,u_6,u_7\}$ in G[W].

Suppose that $v_5$ is adjacent in G to $v_7$; then $v_7$ is not adjacent to $v_1$ or $v_2$. If $|N_{G[\{u_1,\ldots ,u_4,v_2\}]}(v_2)|\le 2$, then ${\overline{G}}[u_1,\ldots ,u_7,v_2]$ contains $C_8$ by Lemma 4.5 which with $v_7$ forms $W_8$ in ${\overline{G}}$, a contradiction. Thus, $|N_{G[\{u_1,\ldots ,u_4,v_2\}]}(v_2)|\ge 3$, so $v_1$ is not adjacent to $u_1,\ldots ,u_4$ in G. By Lemma 4.5, ${\overline{G}}[u_1,\ldots ,u_7,v_1,v_7]$ contains $W_8$, a contradiction.

Hence, $v_5$ is not adjacent to $v_7$ in G. If $|N_{G[\{u_1,\ldots ,u_4,v_5\}]}(v_5)|\le 2$, then ${\overline{G}}[u_1,\ldots ,u_7,v_5]$ contains $C_8$ by Lemma 4.5 which with $v_7$ forms $W_8$ in ${\overline{G}}$, a contradiction. Thus $|N_{G[\{u_1,\ldots ,u_4,v_5\}]}(v_5)|\ge 3$, so $v_4$ is not adjacent to $\{u_1,\ldots ,u_4\}$ in G, or else G will contain D with $v_4$ be the vertex of degree 3. By Lemma 4.5, ${\overline{G}}[u_1,\ldots ,u_7,v_1]$ contains $C_8$. If $v_4$ is not adjacent to $v_7$ in G, then ${\overline{G}}$ contains $W_8$, a contradiction. Thus, $v_4$ is adjacent to $v_7$, and since $D\nsubseteq G$, $v_1$ is not adjacent to $v_7$. If $|N_{G[\{u_1,\ldots ,u_4,v_1\}]}(v_1)|\le 2$, then ${\overline{G}}[u_1,\ldots ,u_7,v_1]$ contains $C_8$ by Lemma 4.5 which with $v_7$ forms $W_8$, a contradiction. Thus, $|N_{G[\{u_1,\ldots ,u_4,v_1\}]}(v_1)|\ge 3$, so $|N_{G[\{u_1,\ldots ,u_4,v_1\}]}(v_1)\cap N_{G[\{u_1,\ldots ,u_4,v_5\}]}(v_5)|\ge 2$, and G contains D with $v_5$ as the vertex of degree 3, a contradiction.

Case 2: $u_1$ is the vertex of degree at least 4 in G[W] and $v_6$ is adjacent to $u_1$.

Without loss of generality, suppose that $u_1$ is adjacent to $u_2$, $u_3$ and $u_4$ in G[W]. If $v_5$ is adjacent to $u_1$, then Case 1 applies with $v_6$ replaced by $u_1$. Suppose then that $v_5$ is not adjacent to $u_1$. Since $D\nsubseteq G$, $v_1$ and $v_2$ are not adjacent in G to $v_4$, $v_5$ or $v_6$; $v_3$ is not adjacent to $v_6, u_1, \ldots , u_4$; and $v_4$ is not adjacent to $u_1,\ldots ,u_4$.

Subcase 2.1: $E_G(\{u_2,u_3,u_4\},\{u_5,u_6,u_7\})\ne \emptyset$.

Without loss of generality, assume that $u_2$ is adjacent to $u_5$ in G. Then $u_3$ and $u_4$ are not adjacent to each other or to $v_1, v_2, u_6, u_7$. Also, $u_1$ is not adjacent to $v_1$ or $v_2$, and neither $u_2$ nor $u_5$ is adjacent to $v_3, v_4, v_5, v_6$.

Suppose that $v_7$ is adjacent to $v_4$ in G. If $u_1$ is adjacent to $v_1$, $u_5$, $u_6$ or $u_7$, then Case 1 can be applied through a slight adjustment of the vertex labelings. Suppose that $u_1$ is not adjacent to any of these vertices. Since $D\nsubseteq G$, $v_7$ is not adjacent to $v_1$. If $v_6$ is not adjacent to $u_6$, then $v_1u_1u_5v_6u_6u_3u_7u_4v_1$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction. Similarly, ${\overline{G}}$ contains $W_8$ if $v_6$ is not adjacent to $u_7$, a contradiction. Therefore, $v_6$ is adjacent to both $u_6$ and $u_7$ in G. Since $D\nsubseteq G$, $u_6$ is not adjacent to $u_7$, and neither $u_6$ nor $u_7$ is adjacent to $u_2$. Then $v_1u_1u_5v_6u_2u_6u_7u_3v_1$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction.

Suppose now that $v_7$ is not adjacent to $v_4$ in G. If $v_7$ is adjacent to $v_5$, then $v_7$ is not adjacent to $v_1$ or $v_2$, and $v_4$ is not adjacent to $v_6$, $u_6$ or $u_7$. Then $v_1u_1v_2u_3u_6v_4u_7u_4v_1$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $v_7$ is not adjacent to $v_5$ in G. If $v_6$ is not adjacent to $u_3$, then $u_3v_6u_2v_5u_5v_4u_4u_6u_3$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction. Similarly, ${\overline{G}}$ contains $W_8$ if $v_6$ is not adjacent to $u_4$, a contradiction. Then $v_6$ is adjacent to both $u_3$ and $u_4$ in G, so $v_6$ is not adjacent to $u_6$ and $u_7$, or else Case 1 applies. Hence, $v_4u_2v_5u_5v_6u_6u_3u_4v_4$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction.

Subcase 2.2: $\{u_2,u_3,u_4\}$ is not adjacent to $\{u_5,u_6,u_7\}$ in G[W].

If $|N_{G[\{u_2,u_3,u_4,v_6\}]}(v_6)|\ge 3$ or $|N_{G[\{u_5,u_6,u_7,v_6\}]}(v_6)|\ge 3$, then Case 1 applies, so $|N_{G[\{u_2,u_3,u_4,v_6\}]}(v_6)|$ $\le 2$ and $|N_{G[\{u_5,u_6,u_7,v_6\}]}(v_6)|\le 2$. Without loss of generality, assume that $v_6$ is not adjacent in G to $u_2$ or $u_5$.

Suppose that $v_4$ is not adjacent to $v_7$ in G. If $u_5$ is adjacent to $u_6$ or $u_7$, say $u_6$, then $v_4$ is not adjacent to $u_5$ or $u_6$, so $v_4u_2v_6u_5u_3u_7u_4u_6v_4$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction. If $u_5$ is not adjacent to $u_6$ or $u_7$, then $v_4u_2v_6u_5u_6u_3u_7u_4v_4$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction. Suppose that $v_4$ is adjacent to $v_7$ in G. By similar arguments to those in Subcase 2.1, $u_1$ is not adjacent to $v_1$, $u_5$, $u_6$ or $u_7$, and $v_7$ is not adjacent to $v_1$. Then $v_1v_6u_5u_2u_6u_3u_7u_1v_1$ and $v_7$ form $W_8$ in ${\overline{G}}$, a contradiction.

Case 3: $u_1$ is the vertex of degree at least 4 in G[W] and $v_6$ is not adjacent to $u_1$.

Assume that $u_1$ is adjacent to $u_2$, $u_3$, $u_4$ and $u_5$ in G[W]. Since $D\nsubseteq G$, $v_3$ and $v_4$ are not adjacent to $u_1$, $u_2$, $u_3$, $u_4$ or $u_5$ in G. If either $v_1$ or $v_5$ are adjacent to $u_1$ in G, then Case 1 applies, so suppose that $v_1$ and $v_5$ are not adjacent to $u_1$. In addition, $v_1$ and $v_5$ are not adjacent to $u_2$, $u_3$, $u_4$ or $u_5$ in G, or else Case 2 applies.

Subcase 3.1: $N_{G[u_2,\ldots ,u_5]}(v_6)\ne \emptyset$.

Assume that $v_6$ is adjacent to $u_2$ in G. Note that $v_4$ is not adjacent to $v_6$, $v_7$, $u_6$ or $u_7$ in G, and $v_3$ is not adjacent to $v_5$ in G, or else Case 2 applies by slight adjustment of vertex labels. Since $D\nsubseteq G$, $v_1$ and $v_2$ are not adjacent in G to $v_5$, $v_6$ or $u_2$, and $v_3$ is not adjacent to $v_6$ in G.

If $u_2$ and $u_6$ are not adjacent in G, then $v_1u_1v_6v_2u_2u_6v_7u_3v_1$ and $v_4$ form $W_8$ in ${\overline{G}}$, a contradiction. A similar contradiction arises if $u_2$ and $u_7$ are not adjacent. Therefore, $u_2$ is adjacent to both $u_6$ and $u_7$ in G, and $u_3$, $u_4$ and $u_5$ are not adjacent to $u_6$ or $u_7$ in G since $D\nsubseteq G$. Then $v_1u_1v_6v_2u_2v_7u_6u_3v_1$ and $v_4$ form $W_8$ in ${\overline{G}}$, a contradiction.

Subcase 3.2: $N_{G[u_2,\ldots ,u_5]}(v_6)=\emptyset$.

Suppose that $v_1$ is adjacent to $v_7$ in G. Then $v_2$ is not adjacent to $v_5$, $v_6$ or U since $D\nsubseteq G$. If $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)|\le 2$, then Lemma 4.5 implies that ${\overline{G}}[u_2,\ldots ,u_5,v_4,v_5,v_6,u_6]$ contains $C_8$ in ${\overline{G}}$ which with $v_2$ forms $W_8$, a contradiction. Thus, $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)|\ge 3$. Similarly, $|N_{G[\{u_2,\ldots ,u_5,u_7\}]}(u_7)|\ge 3$. By the Inclusion-exclusion Principle, $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)\cap N_{G[\{u_2,\ldots ,u_5,u_7\}]}(u_7)|\ge 2$. Without loss of generality, $u_6$ is adjacent to $u_2$, $u_3$ and $u_4$ in G, and $u_7$ is adjacent to $u_3$ and $u_4$, and $G[u_1,\ldots ,u_7]$ contains D with $u_3$ or $u_4$ being the vertex of degree 3, a contradiction.

Now suppose that $v_1$ is not adjacent to $v_7$ in G. If $v_7$ is adjacent to $v_4$ in G, then $v_2$ is not adjacent to any of $u_1, \ldots , u_5$ in G, or else either Case 1 or 2 applies. Also, $|N_{G[\{v_2,v_5,v_7\}]}(v_7)|\le 1$ since $D\subseteq G$. Assume that $v_7$ is not adjacent to $v_2$ in G. If $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)|\le 2$, then Lemma 4.5 implies that ${\overline{G}}[u_2,\ldots ,u_5,v_1,v_2,v_6,u_6]$ contains $C_8$ which with $v_7$ forms $W_8$, a contradiction. Thus, $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)|\ge 3$. Similarly, $|N_{G[\{u_2,\ldots ,u_5,u_7\}]}(u_7)|\ge 3$, so $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)\cap N_{G[\{u_2,\ldots ,u_5,u_7\}]}(u_7)|\ge 2$. By arguments similar to those in the previous paragraph, G will contain a subgraph D, a contradiction.

Thus, $R(D,W_8)\le 14$ which completes the proof of the theorem. $\square$

Theorem 5.3

$R(E,W_8)=15$.

Proof

The graph $G = K_6\cup K_{4,4}$ does not contain E and ${\overline{G}}$ does not contain $W_8$. Thus, $R(E,W_8)\ge 15$. For the upper bound, let G be any graph of order 15. Suppose that G does not contain E and that ${\overline{G}}$ does not contain $W_8$. By Theorem 1.4, G contains a $T=S_7(3)$ subgraph. Label the vertices of this subgraph as in Fig. 8 and set $U=V(G)-V(T)$. Note that $|U|= 8$.

Case 1: Some vertex u in U is adjacent to $v_6$.

Since $E\nsubseteq G$, $v_6$ is not adjacent to $v_1$, $v_2$, $v_3$, $v_7$ or any vertex of U other than u. Let $W=\{v_1,v_2,v_3,v_7,u_1,\ldots ,u_4\}$, for any vertices $u_1,\ldots ,u_4$ in U other than u. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which with $v_6$ forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[W])\le 3$ and $\Delta (G[W])\ge 4$. Since $E\nsubseteq G$, $N_{G[\{u_1,\ldots ,u_4,v_1,v_7\}]}(v_7)\le 1$ and $N_{G[\{u_1,\ldots ,u_4,v_7,v_i\}]}(v_i)\le 1$ for $i=1,2,3$, so none of $v_1,v_2,v_3,v_7$ has degree at least 4. Without loss of generality, assume that $u_1$ has degree at least 4. If $u_1$ is adjacent to $v_7$, then G contains E with $u_1$ and $v_5$ as the vertices of degree 3, a contradiction. Similarly, if $u_1$ is adjacent to $v_1$, $v_2$ or $v_3$, then G contains E with $u_1$ and $v_4$ as the vertices of degree 3, a contradiction. Therefore, $u_1$ is not adjacent to $v_1$, $v_2$, $v_3$ or $v_7$. However, then $u_1$ has degree at most 3 in G[W], a contradiction.

Case 2: $v_6$ is not adjacent to any vertices in U.

If $v_7$ is adjacent to some vertex in U, then Case 1 applies with $v_7$ replacing $v_6$, so suppose that $v_7$ is not adjacent to any vertex in U. Now, if $\delta ({\overline{G}}[U])\ge 4$, then ${\overline{G}}[U]$ contains $C_8$ by Lemma 4.1 which with $v_6$ or $v_7$ forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[U])\le 3$ and $\Delta (G[U])\ge 4$. Let $V(U)=\{u_1,\ldots ,u_8\}$. Without loss of generality, assume that $u_1$ is adjacent to $u_2$, $u_3$, $u_4$ and $u_5$. Since $E\nsubseteq G$, $v_4$ is not adjacent in G to any of $u_1,\ldots ,u_5$; $v_5$ is not adjacent to any of $v_1, v_2, v_3, u_1, \ldots , u_5$; and $u_1$ is not adjacent to $v_1$, $v_2$ or $v_3$. Furthermore, $|N_{G[\{u_2,\ldots ,u_5,v_i\}]}(v_i)|\le 1$ for $i=1,2,3$ and $|N_{G[\{v_1,v_2,v_3,u_j\}]}(u_j)|\le 1$ for $j=2,\ldots ,5$.

Suppose that $N_{G[\{v_5,u_6,u_7,u_8\}]}(v_5)=\emptyset$. If $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)|\le 1$, then ${\overline{G}}[u_2,\ldots ,u_5,v_1,v_2,v_3,u_6]$ contains $C_8$ by Lemma 4.5 which with $v_5$ forms $W_8$, a contradiction. Therefore, $|N_{G[\{u_2,\ldots ,u_6\}]}(u_6)|\ge 2$. Similarly, $|N_{G[\{u_2,\ldots ,u_5,u_7\}]}(u_7)|\ge 2$ and $|N_{G[\{u_2,\ldots ,u_5,u_8\}]}(u_8)|\ge 2$. By the Inclusion-Exclusion Principle, $u_2$, $u_3$, $u_4$ or $u_5$ is adjacent in G to at least two of $u_6,u_7,u_8$. Without loss of generality, assume that $u_2$ is adjacent to $u_6$ and $u_7$. Then $u_2$ is not adjacent to $u_3$, $u_4$ or $u_5$, Therefore, Lemma 4.5 implies that ${\overline{G}}[u_1,u_3,u_4,u_5,v_1,v_2,v_3,u_2]$ contains $C_8$ which with $v_5$ forms $W_8$, a contradiction.

On the other hand, if $N_{G[u_6,u_7,u_8]}(v_5)\ne \emptyset$, then without loss of generality assume that $u_6$ is adjacent to $v_5$ in G. Since $E\nsubseteq G$, $v_4$ is not adjacent to $v_6$, $v_7$ or $u_6$ in G. Also, $\{v_1,v_2,v_3\}$ and $\{v_6,v_7,u_6\}$ are independent in G, and $v_1, v_2, v_3, v_6, v_7, u_6\notin N_G(u_i)$ for $i=1,\ldots ,5,7,8$, or else Case 1 applies with vertex label adjustments. Now, if $u_1$ is not adjacent to both $u_7$ and $u_8$ in G, then $v_1v_2v_3u_7v_6v_7u_6u_8v_1$ and $u_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $N_{G[\{u_1,u_7,u_8\}]}(u_1)\ne \emptyset$. Without loss of generality, assume that $u_1$ is adjacent to $u_7$ in G. Note that for $E\nsubseteq G$, $|N_{G[\{v_4,v_5,u_8\}]}(u_8)|\le 1$. Assume that $u_8$ is not adjacent to $v_4$ in G. If $|N_{G[\{u_2,\ldots ,u_5,u_8\}]}(u_8)|\le 3$, then assume without loss of generality that $u_8$ is not adjacent to $u_2$ or $u_3$ in G. Then $v_6u_4v_7u_5u_6u_2u_8u_3v_6$ and $v_4$ form $W_8$ in ${\overline{G}}$, a contradiction. Similar arguments work if $u_8$ is not adjacent to $v_5$ in G, by replacing $v_4$ with $v_5$ and $v_6,v_7,u_6$ with $v_1,v_2,v_3$, respectively. Hence, $|N_{G[\{u_2,\ldots ,u_5,u_7,u_8\}]}(u_8)|\ge 4$. However, G then contains E with $u_1$ and $u_8$ of degree 3, a contradiction.

Thus, $R(E,W_8)\le 15$. This completes the proof of the theorem. $\square$

Fig. 8

$S_7(3)$ and U in G

6 Proof of Theorem 2.2

Consider the tree graphs $T_n$ of order $n\ge 8$ with $\Delta (T_n) = n-4$, namely $S_n(4)$, $S_n[4]$, $S_n(1,3)$, $S_n(3,1)$, $T_A(n)$, $T_B(n)$ and $T_C(n)$; see Figs. 1 and 3.

Lemma 6.1

Let $n\ge 8$. Then $R(T_n,W_8)\ge 2n-1$ for each $T_n \in \{S_n(4), S_n(3,1), T_C(n)\}$. Also for each $T_n \in \{S_n[4], S_n(1,3), T_A(n), T_B(n)\}$, $R(T_n,W_8)\ge 2n-1$ if $n\not \equiv 0 \pmod {4}$ and $R(T_n,W_8)\ge 2n$ otherwise.

Proof

The graph $G=2K_{n-1}$ clearly does not contain any tree graph of order n, and ${\overline{G}}$ does not contain $W_8$. Finally, if $n\equiv 0 \pmod {4}$, then the graph $G=K_{n-1}\cup K_{4,\ldots ,4}$ of order $2n-1$ does not contain $S_n[4]$, $S_n(1,3)$, $T_A(n)$ or $T_B(n)$; nor does the complement ${\overline{G}}$ contain $W_8$. $\square$

Theorem 6.2

If $n\ge 8$, then

$$\begin{aligned} R(S_n(4),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\ge 9;\\ 16 &{} \hbox { if}\ n = 8. \end{array}\right. } \end{aligned}$$

Proof

By Lemma 6.1, $R(S_n(4),W_8)\ge 2n-1$ for $n\ge 8$. For $n = 8$, observe that the graph $G=K_7\cup H_8$, where $H_8$ is the graph of order 8 as shown in Fig. 9 does not contain $S_8(4)$ and its complement ${\overline{G}}$ does not contain $W_8$. Therefore, for $n=8$, we have a better bound of $R(S_8(4),W_8)\ge 16$.

For the upper bound, let G be any graph of order $2n-1$ if $n\ge 9$, and of order 16 if $n=8$. Assume that G does not contain $S_n(4)$ and that ${\overline{G}}$ does not contain $W_8$.

If $n\ge 9$ is odd or $n=8$, then G has a subgraph $T=S_n(3)$ by Theorem 1.4. Let $V(T)=\{v_0, \ldots , v_{n-3}, w_1, w_2\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-3},v_1w_1,v_1w_2\}$. Also, let $V=\{v_2,\ldots ,v_{n-3}\}$ and $U=V(G)-V(T)$; then $|V|=n-4\ge 5$ and $|U|=n-1\ge 8$ if n is odd, while $|U|=8$ if $n=8$. Since $S_n(4)\nsubseteq G$, $v_1$ is not adjacent to any vertex of $U\cup V$ in G. Furthermore, for each $2\le i\le n-3$, $v_i$ is adjacent to at most two vertices of U in G. By Corollary 4.7, ${\overline{G}}[U\cup V]$ contains $C_8$, and together with $v_1$, gives us $W_8$ in ${\overline{G}}$, a contradiction.

For the remaining case when $n\ge 10$ is even, $S_{n-1}\subseteq G$ by Theorem 1.1. Let $v_0$ be the center of $S_{n-1}$ and set $L=N_{S_{n-1}}(v_0)=\{v_1,\ldots ,v_{n-2}\}$ and $U=V(G)-V(S_{n-1})$. Then $|U|=n$. Since G does not contain $S_n(4)$, each vertex of L is adjacent to at most two vertices of U. We consider two cases.

Case 1: $E(L,U)=\emptyset$.

If $\Delta ({\overline{G}}[U])\ge 4$, then some vertex u in U is adjacent to at least four vertices in ${\overline{G}}[U]$. These four vertices and any four vertices from L form $C_8$ in ${\overline{G}}$ which with u forms $W_8$, a contradiction. Therefore, $\Delta ({\overline{G}}[U])\le 3$ and $\delta (G[U])\ge n-4$. Suppose $\delta (G[U])=n-4+l$ for some $l\ge 0$, and let $u_0$ be a vertex in U with minimum degree in G[U]. Label the remaining vertices in U as $u_1,\ldots ,u_{n-1}$ such that $U_A=\{u_1,\ldots ,u_{n-4}\}\subseteq N_G(u_0)$, and let $U_B=\{u_{n-3},u_{n-2},u_{n-1}\}$. Since $S_n(4)\nsubseteq G$, each vertex in $U_A$ is adjacent to at most two vertices in $U_B$, and so $|E_G(U_A,U_B)|\le 2(n-4)$. On the other hand, noting that $u_0$ is adjacent to exactly l vertices in $U_B$ and letting $e_B\le 3$ be the number of edges in $G[U_B]$, we see that $|E_G(U_A,U_B)|\ge 3\delta (G[U]) - l - 2e_B = 3(n-4+l) - l - 2e_B$. Therefore, $2(n-4)\ge |E_G(U_A,U_B)|\ge 3n - 12 + 2l - 2e_B$, implying that $n+2l\le 4+2e_B\le 10$, which is only possible when $n=10$, $l=0$, $e_B=3$, and $|E_G(U_A,U_B)| = 2(n-4) = 12$. For such scenario where $n=10$, noting that $u_0$ was an arbitrary vertex with minimum degree in G[U], it is straightforward to deduce that the only possible edge set of G[U] (up to isomorphism) with $S_{10}(4)\nsubseteq G[U]$ is $\{u_0u_1,\ldots ,u_0u_6\}\cup \{u_1u_7,\ldots ,u_4u_7\}\cup \{u_1u_8,u_2u_8,u_5u_8,u_6u_8\}\cup \{u_3u_9,\ldots ,u_6u_9\}\cup \{u_1u_2,u_3u_4,u_5u_6\} \cup \{u_1u_3,u_1u_5,u_3u_5\} \cup \{u_2u_4,u_2u_6, u_4u_6\}\cup \{u_7u_8,u_7u_9,u_8u_9\}$. Observe now that ${\overline{G}}[U]$ contains $C_8$ which forms $W_8$ in ${\overline{G}}$ with any vertex in L as hub, a contradiction.

Case 2: $E(L,U)\ne \emptyset$.

Without loss of generality, assume that $v_1$ is adjacent to $u_1$ in G. Since $S_n(4)\nsubseteq G$, $v_1$ is adjacent to at most one vertex of $U\cup L{\setminus }\{u_1\}$ in G. Therefore, we can find a 4-vertex set $V'\subseteq V{\setminus } \{v_1\}$ and an 8-vertex set $U'\subseteq U{\setminus } \{u_1\}$ such that $v_1$ is not adjacent in G to any vertex of $U'\cup V'$. Note that each vertex of $V'$ is adjacent to at most two vertices of $U'$ in G, so $|E(V',U')|\le 8$. This implies that there are four vertices in $U'$ that are each adjacent in G to at most one vertex of $V'$, and so ${\overline{G}}$ contains $C_8$ by Lemma 4.5 which with $v_1$ forms $W_8$, a contradiction.

Thus, $R(S_n(4),W_8)\le 2n-1$ when $n\ge 9$ and $R(S_n(4),W_8)\le 16$ when $n=8$. This completes the proof of the theorem. $\square$

Fig. 9

The graphs $H_8$

Lemma 6.3

Let H be a graph of order $n\ge 8$ with minimum degree $\delta (H)\ge n-4$. Then either H contains $S_n[4]$ and $T_A(n)$, or $n\equiv 0 \pmod {4}$ and ${{\overline{H}}}$ is the disjoint union of $\frac{n}{4}$ copies of $K_4$, i.e., ${{\overline{H}}}=\frac{n}{4} K_4$.

Proof

Let $V(H)=\{u_0,\ldots ,u_{n-1}\}$. First, consider the case where H has a vertex of degree at least $n-3$, say $u_0$, and that $\{u_1,\ldots ,u_{n-3}\}\subseteq N_H(u_0)$.

Suppose $u_{n-2}$ is adjacent to $u_{n-1}$ in H. Since $\delta (H)\ge n-4$, $u_{n-2}$ is adjacent to at least $n-6\ge 2$ vertices of $\{u_1,\ldots , u_{n-3}\}$, say $u_1$ and $u_2$, and so H contains $S_n[4]$. Furthermore by the minimum degree condition, $u_1$ is adjacent to at least $n-7\ge 1$ vertices of $\{u_1,\ldots , u_{n-3}\}$, and so H contains $T_A(n)$.

Suppose now that $u_{n-2}$ is not adjacent to $u_{n-1}$ in H. Then by the minimum degree condition, there is a vertex in $\{u_1,\ldots , u_{n-3}\}$, say $u_1$, that is adjacent to both $u_{n-2}$ and $u_{n-1}$. The vertices $u_1$ and $u_{n-2}$ must also each be adjacent to a vertex of $\{u_2,\ldots , u_{n-3}\}$, and so H contains both $S_n[4]$ and $T_A(n)$.

For the remaining case, suppose that H is $(n-4)$-regular and that $N_H(u_0)=\{u_1,\ldots ,u_{n-4}\}$. Let $U=\{u_{n-3},u_{n-2},u_{n-1}\}$ and suppose that H[U] has an edge, say $u_{n-3}u_{n-2}$. Since $u_{n-3}$ must be adjacent in H to some vertex of $N_H(u_0)$, it follows that H contains $S_n[4]$ if $u_{n-3}$ or $u_{n-2}$ is adjacent to $u_{n-1}$. Suppose then that neither $u_{n-3}$ nor $u_{n-2}$ is adjacent to $u_{n-1}$. Then $u_{n-1}$ is adjacent to every vertex of $N_H(u_0)$. Note that $d_{H[N_H(u_0)\cup \{u_{n-3}\}]}(u_{n-3})=n-5$ and let u be the vertex of $N_H(u_0)$ that is not adjacent in H to $u_{n-3}$. Since $d_H(u)=n-4$, u is adjacent in H to some vertex in $N_H(u_{n-3})$, so H contains $S_n[4]$. Also, note that $u_{n-3}$ is adjacent in H to at least $n-6$ vertices of $N_H(u_0)$. If $u_{n-1}$ is adjacent to some vertex of $N_{H[N_H(u_0)\cup \{u_{n-3}\}]}(u_{n-3})$, then H contains $T_A(n)$. Note that this will always happen for $n\ge 9$. For $n=8$, there is a case where $|N_{H[N_H(u_0)\cup \{u_{n-3}\}]}(u_{n-3})|=|N_{H[N_H(u_0)\cup \{u_{n-1}\}]}(u_{n-1})|=2$ and $N_{H[N_H(u_0)\cup \{u_{n-3}\}]}(u_{n-3})\cap N_{H[N_H(u_0)\cup \{u_{n-1}\}]}(u_{n-1})=\emptyset$, so $u_{n-1}$ is adjacent to $u_{n-3}$ and $u_{n-2}$, giving $T_A(n)$ in H.

Now, suppose that H[U] contains no edge. Then $U_1=U\cup \{u_0\}$ is an independent set in H. Furthermore, $N_H(u)=\{u_1,\ldots , u_{n-4}\}$ for every $u\in U$, as every vertex has degree $n-4$. Therefore, ${{\overline{H}}}[U_1]$ is a $K_4$ component in ${{\overline{H}}}$. Repeating the above proof for each vertex u of H shows that either u is contained in a $K_4$ component of ${{\overline{H}}}$, or H contains both $S_n[4]$ or $T_A(n)$. In other words, either H contains both $S_n[4]$ and $T_A(n)$, or ${{\overline{H}}}$ is the disjoint union of $\frac{n}{4}$ copies of $K_4$, and so $n\equiv 0 \pmod {4}$. $\square$

Theorem 6.4

If $n\ge 8$, then

$$\begin{aligned} R(S_n[4],W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 6.1 provides the lower bounds, so it remains to prove the upper bounds. Now let G be a graph that does not contain $S_n[4]$ and assume that ${\overline{G}}$ does not contain $W_8$.

First, suppose that G has order 2n if $n\equiv 0 \pmod {4}$ and G has order $2n-1$ if n is odd. By Theorem 1.4, G has a subgraph $T=S_n(3)$. Let $V(T)=\{v_0,\ldots ,v_{n-3},w_1,w_2\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-3}\}\cup \{v_1w_1,v_1w_2\}$. Set $U=V(G)-V(T)$ and $V=\{v_2,\ldots ,v_{n-3}\}$. Then $|U|=n-j$, for $j=0$ if $n\equiv 0 \pmod {4}$ and $j=1$ if n is odd, and $|V|=n-4$. Since G does not contain $S_n[4]$, $v_1$ is not adjacent to any vertex of V in G, and each vertex of V is adjacent to at most $n-6$ vertices of $U\cup V$ in G. Noting also that $w_1$ and $w_2$ each is adjacent to at most one vertex of $\{w_1,w_2\}\cup U$ in G, we consider two cases.

Case 1: At least one of $w_1$ and $w_2$ is not an isolated vertex in $G[\{w_1,w_2\}\cup U]$.

Without loss of generality, assume that $w_1$ is adjacent to some vertex $u\in \{w_2\}\cup U$ in G. Let $Z = \big (V\cup U\cup \{w_2\}\big ){\setminus } \{u\}$ and note that $|Z|=2n-4-j$. Since $S_n[4]\nsubseteq G$, $w_1$ is not adjacent to any vertex of Z in G. If $\delta ({\overline{G}}[Z])\ge \lceil \frac{2n-4-j}{2}\rceil$, then ${\overline{G}}[Z]$ contains $C_8$ by Lemma 4.1 which with $w_1$, forms $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $\delta ({\overline{G}}[Z])\le \lceil \frac{2n-4-j}{2}\rceil -1$ and $\Delta (G[Z])\ge \lfloor \frac{2n-4-j}{2}\rfloor = n-2-j$. Since each v of V is adjacent to at most $n-6$ vertices of $U\cup V$ in G, and $w_2$ is adjacent to at most one vertex of U in G, a vertex with maximum degree in G[Z] must be a vertex of $U{\setminus }\{u\}$. So let $u_2$ be a vertex of U with $d_{G[Z]}(u_2)\ge n-2$. As $S_n[4]\nsubseteq G$, observe that $N_{G[Z]}(u_2)\subseteq U$; each vertex of V is adjacent to at most one vertex of $N_{G[Z]}(u_2)$ in G; and each vertex of $N_{G[Z]}(u_2)$ is adjacent to at most one vertex of V in G. Then by Lemma 4.5, any four vertices from V and any four vertices from $N_{G[Z]}(u_2)$ form $C_8$ in ${\overline{G}}$ which with $w_1$ forms $W_8$ in ${\overline{G}}$, a contradiction.

Case 2: $w_1$ and $w_2$ are isolated vertices in $G[\{w_1,w_2\}\cup U]$.

If $\delta ({\overline{G}}[U])\ge \frac{n-j}{2}$, then ${\overline{G}}[U]$ contains $C_8$ by Lemma 4.1 which with $w_1$ forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[U])\le \frac{n-j}{2}-1$, and $\Delta (G[U])\ge \frac{n-j}{2}$. Let $u_1$ be a vertex of U with $d_{G[U]}\ge \frac{n-j}{2}$. Since $S_n[4]\nsubseteq G$, $v_0$ is not adjacent to any vertex of $N_{G[U]}(u_1)$ in G. Now, if $v_1$ is adjacent to some vertex u of $N_{G[U]}(u_1)$ in G, then apply Case 1 with $w_1$ and u interchanged. So assume that $v_1$ is not adjacent to any vertex of $N_{G[U]}(u_1)$ in G.

If $E(V,N_{G[U]}(u_1))=\emptyset$ in G, then any four vertices of V and any four vertices of $N_{G[U]}(u_1)$ form $C_8$ in ${\overline{G}}$, and with $v_1$, form $W_8$ in ${\overline{G}}$, a contradiction. So without loss of generality, assume that $v_2$ is adjacent to some vertex $u_2$ of $N_{G[U]}(u_1)$ in G. Since $S_n[4]\nsubseteq G$, $u_2$ is not adjacent to any vertex of $U{\setminus }\{u_1\}$. Then $v_0,v_1,w_1,w_2$ and any four vertices from $U{\setminus }\{u_1,u_2\}$, at least three of which are from $N_{G[U]}(u_1){\setminus } \{u_2\}$, form $C_8$ in ${\overline{G}}$ and, with $u_2$, form $W_8$ in ${\overline{G}}$, a contradiction.

In either case, $R(S_n[4],W_8)\le 2n$ for $n\equiv 0 \pmod {4}$ and $R(S_n[4],W_8)\le 2n-1$ for odd n.

Next, suppose that $n\equiv 2 \pmod {4}$ and G has order $2n-1$. If G contains a subgraph $S_n(3)$, then the previous arguments show that $R(S_n[4],W_8)\le 2n-1$. Hence, we only need to consider the case where G does not contain $S_n(3)$. Now, by Theorem 6.2, G has a subgraph $T=S_n(4)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,v_1w_3\}$. Let $U=V(G)-V(T)$; then $|U|=n-1$. Since G does not contain $S_n(3)$ and $S_n[4]$, $v_0$ is not adjacent in G to $w_1$, $w_2$, $w_3$ or U. Now, set $U'=N_{G[U\cup \{w_1\}]}(w_1)\cup N_{G[U\cup \{w_2\}]}(w_2)\cup N_{G[U\cup \{w_3\}]}(w_3)$. Then $|U'|\le 3$ and $w_1$, $w_2$ and $w_3$ are not adjacent in G to any vertex of $U{\setminus } U'$. By Lemma 4.4, $G[U{\setminus } U']$ is either $K_{n-1-|U'|}$ or $K_{n-1-|U'|}-e$. If $d_{{\overline{G}}[U{\setminus } U']}(u')\ge 2$ for some vertex $u'$ in $U'$, then at least two vertices of $U{\setminus } U'$ are not adjacent to $u'$ in G. Let X be a set containing these two vertices and any other two vertices in $U{\setminus } U'$, and set $Y=\{w_1,w_2,w_3,u'\}$. Note that ${\overline{G}}[X\cup Y]$ contains $C_8$ by Lemma 4.5 which with $v_0$ forms $W_8$, a contradiction. Therefore, every vertex of $U'$ is adjacent in G to at least $n-2-|U'|$ vertices of $U{\setminus } U'$. Hence, $\delta (G[U])\ge n-5$, and since $S_n[4]\nsubseteq G$, $E_G(T,U)=\emptyset$. Now, if ${\overline{G}}[V(T)]$ contains $S_5$, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Therefore, $\delta (G[V(T)])\ge n-4$. By Lemma 6.3, G contains $S_n[4]$, a contradiction. Thus, $R(S_n[4],W_8)\le 2n-1$ for $n\equiv 2 \pmod {4}$. $\square$

Theorem 6.5

If $n\ge 8$, then

$$\begin{aligned} R(S_n(1,3),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 6.1 provides the lower bounds, so it remains to prove the upper bounds. Let G be any graph of order 2n if $n\equiv 0\pmod {4}$ and of order $2n-1$ if $n\not \equiv 0\pmod {4}$. Assume that G does not contain $S_n(1,3)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.4, G has a subgraph $T = S_n[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},w_1v_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$. Since $S_n(1,3)\nsubseteq G$, $w_2$ and $w_3$ are not adjacent to each other, or to any vertex in $U\cup V$. Since $C_8\nsubseteq {\overline{G}}[U\cup V]$ as $W_8\nsubseteq {\overline{G}}$, Lemma 4.1 implies that $G[U\cup V]$ has a vertex u of degree at least $n-3$ in $G[U\cup V]$. Since $S_n(1,3)\nsubseteq G$, $u\in U$ and u is not adjacent to any vertices in V. Furthermore, $E(V,N_{G[U]}(u))=\emptyset$. Finally, note that $w_3$, any 3 vertices in V and any 4 vertices in $N_{G[U]}(u)$ form $C_8$ in ${\overline{G}}$ which, with $w_2$ as hub, form $W_8$, a contradiction. $\square$

Theorem 6.6

If $n\ge 8$, then

$$\begin{aligned} R(T_A(n),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 6.1 provides the lower bounds, so it remains to prove the upper bounds. Let G be any graph of order 2n if $n\equiv 0\pmod 4$ and of order $2n-1$ if $n\not \equiv 0\pmod 4$. Assume that G does not contain $T_A(n)$ and that ${\overline{G}}$ does not contain $W_8$.

Suppose first that G has a subgraph $T=S_n(3)$. Let $V(T)=\{v_0,\ldots ,v_{n-3},w_1,w_2\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-3},v_1w_1,v_1w_2\}$. Set $V=\{v_2,\ldots ,v_{n-3}\}$ and $U=V(G)-V(T)$. Since G does not contain $T_A(n)$, $w_1$ and $w_2$ are not adjacent to any vertex of $U\cup V$ in G. Let $V'$ be the set of any $n-5$ vertices in V, and $U'$ be the set of any $n-1$ vertices in U. If $\delta ({\overline{G}}[U'\cup V'])\ge n-3$, then ${\overline{G}}[U'\cup V']$ contains $C_8$ by Lemma 4.1 which, with $w_1$ as hub, form $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U'\cup V'])\le n-4$ and $\Delta (G[U'\cup V'])\ge n-3$. Since $T_A(n)\nsubseteq G$, $d_{G[U'\cup V']}(v)\le n-6$ for each $v\in V'$. Hence, some vertex $u\in U'$ satisfies $d_{G[U'\cup V']}(u)\ge n-3$, which also implies that u is adjacent to at least two vertices of U.

Since $T_A(n)\nsubseteq G$, each vertex of V is adjacent to at most one vertex of $N_{G[U]}(u)$. If $|N_{G[U]}(u)|\ge n-4$, then each vertex of $N_{G[U]}(u)$ is adjacent to at most one vertex of V, and so ${\overline{G}}[V\cup N_{G[U]}(u)]$ contains $C_8$ by Lemma 4.1 which with $w_1$ forms $W_8$, a contradiction. Thus, at least three vertices of $V'$ (and so of V), say $v_2, v_3, v_4$, are adjacent to u in G. Let a and b be any two vertices in $N_{G[U]}(u)$. As $T_A(n)\nsubseteq G$, each of $v_2,v_3,v_4$ is not adjacent to any vertex of $V(G){\setminus }\{u,v_0\}$. Then $w_1v_5w_2v_3av_1bv_4w_1$ and $v_2$ form $W_8$ in ${\overline{G}}$, a contradiction.

By Theorem 1.4, $R(S_n(3),W_8)\le 2n$ for $n\equiv 0 \pmod {4}$. So now assume that G has order $2n-1$ with $n\not \equiv 0 \pmod {4}$ and that G does not contain $S_n(3)$. By Theorem 6.2, G has a subgraph $T=S_n(4)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,v_1w_3\}$. Then $U=V(G)-V(T)$ and $|U|=n-1$. Since $T_A(n)\nsubseteq G$, $w_1,w_2,w_3$ are not adjacent to each other in G or to any vertex of U. Since $S_3(n)\nsubseteq G$, $v_0$ is not adjacent to any vertex of $U\cup \{w_1,w_2,w_3\}$. By Lemma 4.4, G[U] is $K_{n-1}$ or $K_{n-1}-e$. Since $T_A(n)\nsubseteq G$, each vertex of T is not adjacent to any vertex of U in G, and so $\delta (G[V(T)])\ge n-4$ by Observation 4.3, which in turn implies that G[V(T)] contains $T_A(n)$ by Lemma 6.3, a contradiction.

This completes the proof of the theorem. $\square$

Theorem 6.7

If $n\ge 8$, then

$$\begin{aligned} R(T_B(n),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 6.1 provides the lower bounds, so it remains to prove the upper bounds. Let G be a graph with no $T_B(n)$ subgraph whose complement ${\overline{G}}$ does not contain $W_8$.

Suppose that $n\equiv 0 \pmod {4}$ and that G has order 2n. By Theorem 6.4, G has a subgraph $T=S_n[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n$. Since $T_B(n)\nsubseteq G$, $E_G(U,V)=\emptyset$ and neither $w_2$ nor $w_3$ is adjacent in G to V. Suppose that $n\ge 12$. If $w_2$ is non-adjacent to some 4 vertices from U, then these 4 vertices and any 4 vertices from V form $C_8$ in ${\overline{G}}$ that with $w_2$ forms $W_8$, a contradiction. Otherwise, $w_2$ must be adjacent to at least $n-3$ vertices of U in G. Since $T_B(n)\nsubseteq G$, $w_3$ must not be adjacent to these $n-3$ vertices; then any 4 vertices from these $n-3$ vertices and 4 vertices from V form $C_8$ in ${\overline{G}}$ and with $w_3$ forms $W_8$, again a contradiction. For $n=8$, $|V|=3$ and $|U|=8$. If $w_2$ is not adjacent to any vertex of U in G, then by Lemma 4.4, G[U] is $K_8$ or $K_8 - e$ which contains $T_B(8)$, a contradiction. Otherwise, suppose that $w_2$ is adjacent to $u\in U$. Since $T_B(8)\nsubseteq G$, $w_1$ must not be adjacent to $(U\cup V){\setminus } \{u\}$ in G. Now, if $w_3$ is not adjacent to $v_0$ in G, then by Observation 4.3, ${\overline{G}}$ contains $W_8$, a contradiction. Otherwise, u is not adjacent to $V\cup \{w_3\}$, and again by Observation 4.3, ${\overline{G}}$ contains $W_8$, another contradiction. Thus, $R(T_B(n),W_8)\le 2n$ for $n\equiv 0 \pmod {4}$.

Next, suppose that $n\not \equiv 0 \pmod {4}$ and that G has order $2n-1$. By Theorem 6.4, G has a subgraph $T=S_n[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_B(n)\nsubseteq G$, $E_G(U,V)=\emptyset$ and neither $w_2$ nor $w_3$ is adjacent in G to V. For $n\ge 9$, if $w_2$ is non-adjacent to some 4 vertices from U, then these 4 vertices and any 4 vertices from V form $C_8$ in ${\overline{G}}$ and with $w_2$ form $W_8$, a contradiction. Otherwise, $w_2$ is adjacent to at least $n-4$ vertices of U in G. Since $T_B(n)\nsubseteq G$, $w_3$ is not adjacent to these $n-4$ vertices, so any 4 vertices from these $n-4$ vertices and 4 vertices from V form $C_8$ in ${\overline{G}}$ which with $w_3$ form $W_8$, again a contradiction. Therefore, $R(T_B(n),W_8)\le 2n-1$ for $n\not \equiv 0 \pmod {4}$.

This completes the proof. $\square$

Theorem 6.8

For $n\ge 8$, $R(T_C(n),W_8)=2n-1$.

Proof

Lemma 6.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$ and assume that G does not contain $T_C(n)$ and that ${\overline{G}}$ does not contain $W_8$.

Suppose first that there is a subset $X\subseteq V(G)$ of size n with $\delta (G[X])\ge n-4$. If $\delta (G[X]) = n-4$, then let $x\in X$ be such that $d_{G[X]}(x) = n-4$, and set $Y = X{\setminus } (\{x\}\cup N_{G[X]}(x))$ where $|Y|=3$. Noting that $3(n-6)>n-4$ for $n\ge 8$, there must be two vertices of Y that are adjacent to a common vertex of $N_{G[X]}(x)$ in G, say to $x'\in N_{G[X]}(x)$. Then the remaining vertex of Y is not adjacent to any vertex of $N_{G[X]}(x){\setminus }\{x'\}$, as $T_C(n)\nsubseteq G$, contradicting $\delta (G[X])\ge n-4$. So $\delta (G[X]) \ge n-3$. Pick any vertex $x\in X$ and any subset $X'\subseteq N_{G[X]}(x)$ of size $n-3$. Set $Y = X{\setminus } (\{x\}\cup X')$ where $|Y|=2$. As $2(n-5)>n-3$ for $n\ge 8$, the two vertices of Y must be adjacent to a common vertex of $X'$ in G, say $x'$. Then $G[X'{\setminus }\{x'\}]$ is an empty graph as $T_C(n)\nsubseteq G$, contradicting $\delta (G[X])\ge n-3$.

Now assume that $\delta (G[X])\le n-5$ whenever $X\subseteq V(G)$ is of size n. By Theorem 1.4, G has a subgraph $T=S_{n-1}(3)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2\}$. Set $V=\{v_2,v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n$. Since $T_C(n)\nsubseteq G$, $E_G(U,V)=\emptyset$.

For the case $n=8$ such that $v_1$ is not adjacent to any vertex of U in G, or the case $n\ge 9$, there are four vertices of V(T) that are not adjacent to any vertex of U in G. Since $\delta (G[U])\le n-5$, ${\overline{G}}[U]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

For the final case $n=8$ with $v_1$ adjacent to some vertex u of U in G, observe that since $T_C(8)\nsubseteq G$, the vertex u is not adjacent to any vertex of $\{v_2, v_3, v_4\}\cup U$. By Lemma 4.4, $G[U{\setminus }\{u\}]$ is $K_7$ or $K_7 - e$, which implies that no vertex of $V(T)\cup \{u\}$ is adjacent to any vertex of $U{\setminus }\{u\}$ in G, as $T_C(8)\nsubseteq G$. Since $\delta (G[V(T)\cup \{u\}])\le n-5$, ${\overline{G}}[V(T)\cup \{u\}]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

This completes the proof of the theorem. $\square$

Theorem 6.9

For $n\ge 8$, $R(S_n(3,1),W_8)=2n-1$.

Proof

Lemma 6.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $S_n(3,1)$ and that ${\overline{G}}$ does not contain $W_8$.

Suppose first that there is a subset $X\subseteq V(G)$ of size n with $\delta (G[X])\ge n-4$. Let $x_0$ be any vertex of X, and pick a subset $X'\subseteq N_{G[X]}(x_0)$ of size $n-4$. Set $Y = X{\setminus } (\{x_0\}\cup X')$, and so $|Y|=3$. Since $\delta (G[X])\ge n-4$, each vertex of Y is adjacent to at least $n-7$ vertices of $X'$ in G. For $n\ge 10$, it is straightforward to see that there is a matching from Y to $X'$ in G; hence, G contains $S_n(3,1)$, a contradiction. For $n=9$, if $d_{G[X]}(x_0) = n-4 = 5$, then we can similarly deduce the contradiction that G contains $S_9(3,1)$, since in this case, each vertex of Y is adjacent to at least $n-6=3$ vertices of $X'$ in G. As $x_0$ was arbitrary, we may assume for the case when $n=9$ that $\delta (G[X])\ge n-3 = 6$, which again leads to the contradiction that G contains $S_9(3,1)$.

Now for $n=8$, suppose $d_{G[X]}(x_0) = 4$. Let $X' = \{x_1,x_2,x_3,x_4\}$ and $Y = \{x_5,x_6,x_7\}$. Since $\delta (G[X])\ge n-4$ and $S_8(3,1)\nsubseteq G$, G[Y] is $K_3$; all three vertices of Y are adjacent to exactly two common vertices of $X'$ in G, say to $x_1$ and $x_2$; and neither $x_3$ nor $x_4$ are adjacent to any vertex of Y in G. By the minimum degree condition, $x_3$ and $x_4$ are then adjacent in G, and each is also adjacent to both $x_1$ and $x_2$. This implies that G contains $S_8(3,1)$, with $x_1$ being the vertex with degree four, a contradiction. As $x_0$ was arbitrary, assume for the case when $n=8$ that $\delta (G[X])\ge 5$, which again leads to the contradiction that G contains $S_8(3,1)$.

Now assume that $\delta (G[X])\le n-5$ whenever $X\subseteq V(G)$ is of size n. Recall that G has order $2n-1$, and so by Theorem 1.4, G has a subgraph $T=S_{n-1}(2,1)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_2w_2\}$. Set $V=\{v_3,v_4,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-6$ and $|U|=n$. Since $S_n(3,1)\nsubseteq G$, $E_G(U,V)=\emptyset$. Now as $\delta (G[U])\le n-5$, ${\overline{G}}[U]$ contains $S_5$, and so for $n\ge 10$, ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

For $n=9$, Theorem 1.4 shows that G has a subgraph $T=S_9(2,1)$, so without loss of generality assume that $v_0$ is adjacent to some vertex u in U. Since $S_9(3,1)\nsubseteq G$, $G[V\cup \{u\}]$ is an empty graph and u is not adjacent to any vertex of U in G. By Lemma 4.4, $G[U{\setminus }\{u\}]$ is $K_8$ or $K_8 - e$, which implies that no vertex of $V(T)\cup \{u\}$ is adjacent to any vertex of $U{\setminus }\{u\}$ in G, as $S_9(3,1)\nsubseteq G$. Since $\delta (G[V(T)\cup \{u\}])\le n-5$, ${\overline{G}}[V(T)\cup \{u\}]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Finally for $n=8$, recall that G has order 15, and so G has a subgraph $T'=S_7$ by Theorem 1.1. Let $V(T')=\{v_0',\ldots ,v_6'\}$ and $E(T')=\{v_0'v_1',\ldots ,v_0'v_6'\}$. Set $V'=\{v_1',\ldots ,v_6'\}$ and $U'=V(G)-V(T')$, then $|U'| = 8$. Suppose that $v_2'$ and $v_3'$ are adjacent to a common vertex u of $U'$ in G, while $v_1'$ is adjacent to another vertex $u'\ne u$ of $U'$ in G. Then as $S_8(3,1)\nsubseteq G$, no vertex of $\{v_4',v_5',v_6'\}\cup (U'{\setminus }\{u,u'\})$ is adjacent to any vertex of $V'{\setminus }\{v_1'\}$ in G. Now $G[V'{\setminus }\{v_1'\}]$ contains $S_5$ and $|U'{\setminus }\{u,u'\}|=6$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Similar arguments lead to the same contradiction when the roles of $v_1',v_2'$, and $v_3'$ are replaced by any three vertices of $V'$. So assume that it is not the case that two vertices of $V'$ are adjacent to a common vertex of $U'$ in G while a third vertex of $V'$ is adjacent to another vertex of $U'$ in G.

For $1\le i\le 6$, let $d_i = |E_G(\{v_i'\}, U)|$ be the number of vertices of $U'$ that are adjacent to $v_i'$. Without loss of generality, assume that $d_1\ge d_2\ge \cdots \ge d_6$. Recalling that $\delta (G[U'])\le 3$ and so $S_5\subseteq {\overline{G}}[U']$, Observation 4.3 implies that $d_3\ge 1$. If $d_1\ge 3$ and $d_2\ge 2$, then it is trivial that G contains $S_8(3,1)$, a contradiction. By our assumption on the adjacencies of vertices in $V'$ to vertices of $U'$ in G, it is clear that when $(d_1,d_2,d_3)$ is of the form $(\ge 3, 1, 1)$, (2, 2, 2), or (2, 2, 1), there is a matching from $\{v_1',v_2',v_3'\}$ to $U'$ in G, as $v_2'$ and $v_3'$ are adjacent to different vertices of $U'$ in G. This implies that G contains $S_8(3,1)$, a contradiction. If $(d_1,d_2,d_3) = (2,1,1)$, then, similarly, $v_2'$ and $v_3'$ are adjacent to different vertices of $U'$ in G, say to u and $u'$, respectively, which in turn implies that $v_1'$ is adjacent to two vertices in $U'{\setminus }\{u,u'\}$. So G contains $S_8(3,1)$, again a contradiction.

For the final case when $d_1=d_2=d_3=1$, our assumption implies that $v_1'$, $v_2'$ and $v_3'$ must be adjacent to a common vertex u of $U'$ in G to avoid a matching from $\{v_1',v_2',v_3'\}$ to $U'$ in G. Furthermore, no vertex of $\{v_4',v_5',v_6'\}$ is adjacent to any vertex of $U'{\setminus }\{u\}$ in G. Now if $S_5\subseteq {\overline{G}}[V']$, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. So $\delta (G[V'])\ge 2$, and in particular, $v_4'$ is adjacent to some vertex of $V'$ in G. Without loss of generality, $v_4$ is adjacent to either $v_1$ or $v_5$ in G. Since $S_8(3,1)\nsubseteq G$, ${\overline{G}}[\{v_5',v_2',v_3',v_6'\}]$ contains $S_4$ if $v_4'$ is adjacent to $v_1'$ in G, while ${\overline{G}}[\{v_6',v_1',v_2',v_3'\}]$ contains $S_4$ if $v_4'$ is adjacent to $v_5'$ in G. By Lemma 4.4, $G[U'{\setminus }\{u\}]$ is $K_7$ or $K_7 - e$, which implies that no vertex of $V(T')\cup \{u\}$ is adjacent to any vertex of $U'{\setminus }\{u\}$ in G, as $S_8(3,1)\nsubseteq G$. Since $\delta (G[V(T')\cup \{u\}])\le 3$, ${\overline{G}}[V(T)\cup \{u\}]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Thus, $R(S_n(3,1),W_8)\le 2n-1$ for $n\ge 8$ which completes the proof. $\square$

7 Proof of Theorem 3.1

Lemma 7.1

Let $n\ge 8$. If the tree graph $T_n$ exists, then $R(T_n,W_8)\ge 2n-1$ for each

$$\begin{aligned} T_n \in \{S_n(1,4),S_n(5), S_n[5], S_n(2,2), S_n(4,1), T_D(n), \ldots , T_S(n)\}. \end{aligned}$$

Also, $R(T_n,W_8)\ge 2n$ if $n\equiv 0 \pmod {4}$ and $T_n \in \{S_n(1,4), S_n(2,2), T_D(n), T_N(n)\}$ or if $T_n \in \{T_E(8), T_F(8)\}$.

Proof

The graph $G=2K_{n-1}$ clearly does not contain any tree graph of order n, and ${\overline{G}}$ does not contain $W_8$. Furthermore, if $n\equiv 0 \pmod {4}$, then the graph $G=K_{n-1}\cup K_{4,\ldots ,4}$ of order $2n-1$ does not contain $S_n(1,4)$, $T_D(n)$ or $S_n(2,2)$; nor does the complement ${\overline{G}}$ contain $W_8$. Finally, the graph $G = K_7\cup K_{4,4}$ does not contain $T_E(8)$ or $T_F(8)$ and ${\overline{G}}$ does not contain $W_8$. $\square$

Theorem 7.2

If $n\ge 8$, then

$$\begin{aligned} R(S_n(1,4),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be a graph with no $S_n(1,4)$ subgraph whose complement ${\overline{G}}$ does not contain $W_8$. Suppose that G has order 2n if $n\equiv 0 \pmod {4}$ and that G has order $2n-1$ if $n\not \equiv 0 \pmod {4}$. By Theorem 6.5, G has a subgraph $T=S_n(1,3)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_2w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=j$ where $j=n$ if $n\equiv 0 \pmod {4}$ and $j=n-1$ if $n\not \equiv 0 \pmod {4}$. Since $S_n(1,4)\nsubseteq G$, $w_3$ is not adjacent in G to any vertex of $U\cup V$ and $d_{G[U\cup V]}(v_i)\le n-7$ for each $v_i\in V$. If $\delta ({\overline{G}}[U\cup V])\ge \lceil \frac{n-5+j}{2}\rceil \ge \frac{n-5+j}{2}$, then ${\overline{G}}[U\cup V]$ contains $C_8$ by Lemma 4.1 and thus $W_8$ with $w_3$ as hub, a contradiction. Therefore, $\delta ({\overline{G}}[U\cup V])\le \lceil \frac{n-5+j}{2}\rceil -1$ and $\Delta (G[U\cup V]) \ge n-5+j-\lceil \frac{n-5+j}{2}\rceil = \lfloor \frac{n-5+j}{2}\rfloor \ge n-3$. Since $d_{G[U\cup V]}(v_i)\le n-7$ for each $v_i\in V$, $d_{G[U\cup V]}(u)\ge n-3$ for some vertex $u\in U$. Since $S_n(1,4)\nsubseteq G$, no vertex of V is adjacent to $\{u\}\cup N_{G[U\cup V]}(u)$ in G.

For $n\ge 9$, any 4 vertices from V and any 4 vertices from $\{u\}\cup N_{G[U\cup V]}(u)$ form $C_8$ in ${\overline{G}}$ and, with $w_3$ as hub, form $W_8$, a contradiction. Suppose that $n=8$; then $V=\{v_2,v_3,v_4\}$. Let $\{u_1,\ldots ,u_4\}$ be 4 vertices in $N_{G[U\cup V]}(u)$. Since $S_8(1,4)\nsubseteq G$, $w_1$ is not adjacent to $N_{G[U\cup V]}(u)$. If $w_1$ is not adjacent to $w_3$, then $w_1u_1v_2u_2v_3u_3v_4u_4w_1$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $w_1$ is adjacent to $w_3$ in G. Then $w_2$ is not adjacent to any vertex of $U\cup V$ in G. Since $d_{G[V]}(v_i)\le 1$ for $i=2,3,4$, one of the vertices of V, say $v_2$, is not adjacent to the other two vertices of V. Then $u_1w_2u_2w_3u_3v_3u_4v_4u_1$ and $v_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $R(S_n(1,4),W_8)\le 2n$ for $n\equiv 0 \pmod {4}$ and $R(S_n(1,4),W_8)\le 2n-1$ for $n\not \equiv 0 \pmod {4}$.

This completes the proof. $\square$

Theorem 7.3

If $n\ge 10$, then $R(S_n(5),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $S_n(5)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.2, G has a subgraph $T=S_n(4)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,v_1w_3\}$. Set $V=\{v_2,v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $S_n(5)\nsubseteq G$, $v_1$ is not adjacent to any vertex of $U\cup V$ in G. Furthermore, for each $v_i$ in V, $v_i$ is adjacent to at most three vertices of U in G.

For $n\ge 10$, $|V|\ge 5>4$ and $|U|\ge 9>8$. By Corollary 4.8, ${\overline{G}}[U\cup V]$ contains $C_8$ which together with $v_1$ gives $W_8$ in ${\overline{G}}$, a contradiction. Thus, $R(S_n(5),W_8)\le 2n-1$ which completes the proof. $\square$

Theorem 7.4

If $n\ge 9$, then $R(S_n[5],W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $S_n[5]$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 7.3, G has a subgraph $T=S_{n}(5)$. Let $V(T)=\{v_0,\ldots ,v_{n-5},w_1,\ldots ,w_4\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-5},v_1w_1,\ldots ,v_1w_4\}$. Set $V=\{v_2,\ldots ,v_{n-5}\}$ and $U=V(G)-V(T)$; then $|V|=n-6$ and $|U|=n-1$. Since $S_n[5]\nsubseteq G$, $v_0$ is not adjacent to $w_1,\ldots ,w_4$ in G and $w_1,\ldots ,w_4$ are each adjacent to at most two vertices of U in G. Now, suppose that $v_0$ is non-adjacent to at least six vertices of U in G. By Corollary 4.7, six of these vertices together with $w_1,\ldots ,w_4$ contain $C_8$ in ${\overline{G}}$ which with $v_0$ gives $W_8$ in ${\overline{G}}$, a contradiction. Then, suppose that $v_0$ is adjacent to at least $n-6$ vertices of U in G. Choose a set $U'$ of $n-6$ of these vertices. Since $S_n[5]\nsubseteq G$, $v_1$ is not adjacent to any vertex of $V\cup U'$ in G. If $\delta ({\overline{G}}[V\cup U'])\ge n-6$, then by Lemma 4.1, ${\overline{G}}[V\cup U']$ contains $C_8$ which with $v_1$ gives $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $\delta ({\overline{G}}[V\cup U'])\le n-7$ and $\Delta (G[V\cup U'])\ge n-6$. However, this gives $S_n[5]$ in G with u and $v_1$ as the center of $S_{n-5}$ and $S_5$, respectively, where u is a vertex in $V\cup U'$ with $d_{G[V\cup U']}(u)\ge n-6$, a contradiction. Thus, $R(S_n[5],W_8)\le 2n-1$ which completes the proof. $\square$

Theorem 7.5

If $n\ge 8$, then

$$\begin{aligned} R(S_n(2,2),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Assume that G is a graph with no $S_n(2,2)$ subgraph whose complement ${\overline{G}}$ does not contain $W_8$. Suppose that $n\equiv 0 \pmod {4}$ and that G has order 2n. By Theorem 6.7, G has a subgraph $T=T_B(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,v_2w_3\}$. Set $V=\{v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-6$ and $|U|=n$. Since $S_n(2,2)\nsubseteq G$, $w_3$ is not adjacent in G to $U\cup V$ and $v_2$ is not adjacent to V. If $\delta ({\overline{G}}[U\cup V])\ge \frac{2n-6}{2}=n-3$, then ${\overline{G}}[U\cup V]$ contains $C_8$ by Lemma 4.1 which with $w_2$ forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U\cup V])\le n-4$, and $\Delta (G[U\cup V])\ge n-3$. Now, there are two cases to be considered.

Case 1a: One of the vertices of V, say $v_3$, is a vertex of degree at least $n-3$ in $G[U\cup V]$.

Note that in this case, there are at least 4 vertices from U, say $u_1,\ldots ,u_4$, that are adjacent to $v_3$ in G. Since $S_n(2,2)\nsubseteq G$, these 4 vertices are independent and are not adjacent to any other vertices of U. Since $n\ge 8$, U contains at least 4 other vertices, say $u_5,\ldots ,u_8$, so $u_1u_5u_2u_6u_3u_7u_4u_8u_1$ and $w_3$ forms $W_8$ in ${\overline{G}}$, a contradiction.

Case 1b: Some vertex $u\in U$ has degree at least $n-3$ in $G[U\cup V]$.

Since $S_n(2,2)\nsubseteq G$, u is not adjacent to any vertex of V in G. Therefore, u must be adjacent to at least $n-3$ vertices of U in G. Without loss of generality, suppose that $u_1,\ldots ,u_{n-3}\in N_{G[U]}(u)$. Note that V is not adjacent to $N_{G[U]}(u)$, or else there will be $S_n(2,2)$ in G, a contradiction. If $n\ge 12$, then any 4 vertices from $N_{G[U]}(u)$ and any 4 vertices from V form $C_8$ in ${\overline{G}}$ which, with $w_3$ as hub, forms $W_8$, a contradiction. Suppose that $n=8$ and let the remaining two vertices be $u_6$ and $u_7$. If $|N_{G[\{u_1,\ldots ,u_5,u_i\}}(u_i)|\le 1$ for $i=6,7$, then let $X=\{u_1,\ldots ,u_4\}$ and $Y=\{v_3,v_4,u_6,u_7\}$. By Lemma 4.5, ${\overline{G}}[X\cup Y]$ contains $C_8$ and, with $w_3$ as hub, forms $W_8$ in ${\overline{G}}$, a contradiction. Therefore, one of $u_6$ and $u_7$, say $u_6$, is adjacent to at least two of $u_1,\ldots ,u_5$, say $u_1$ and $u_2$. Since $S_8(2,2)\nsubseteq G$, $u_7$ is adjacent in ${\overline{G}}$ to at least two of $u_3,u_4,u_5$, say $u_3$ and $u_4$, and $v_0,\ldots ,v_4,w_1$ are not adjacent in G to $u,u_1,\ldots ,u_6$. Now, if $w_3$ is not adjacent to some vertex $a\in \{v_0,v_1,w_1\}$, then $u_1v_3u_2v_4u_3u_7u_4au_1$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Hence, $w_3$ is adjacent to $v_0$, $v_1$ and $w_1$ in G. Similarly, $v_2$ is not adjacent to $u_7$ and $v_2$ is adjacent to $v_1$ and $w_1$. Since $S_8(2,2)\nsubseteq G$, $w_2$ is not adjacent to $U\cup V$, and $w_1$ is not adjacent to V. Then $u_1v_2u_2w_1u_3w_2u_4w_3u_1$ and $v_3$ forms $W_8$ in ${\overline{G}}$, a contradiction.

In either case, $R(S_n(2,2),W_8)\le 2n$.

Suppose that $n\not \equiv 0 \pmod {4}$ and that G has order $2n-1$. By Theorem 6.7, G has a subgraph $T=T_B(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,v_2w_3\}$. Set $V=\{v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-6$ and $|U|=n-1$. Since $S_n(2,2)\nsubseteq G$, $w_3$ is not adjacent in G to $U\cup V$. If $\delta ({\overline{G}}[U\cup V])\ge \lceil \frac{2n-5}{2}\rceil$, then ${\overline{G}}[U\cup V]$ contains $C_8$ by Lemma 4.1 which with $w_3$ forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U\cup V])\le \lceil \frac{2n-5}{2}\rceil -1=n-3$, and $\Delta (G[U\cup V])\ge n-3$. Again, there are two cases to be considered.

Case 2a: A vertex of V, say $v_3$, has degree at least $n-3$ in $G[U\cup V]$.

There must be at least 4 vertices from U, say $u_1,\ldots ,u_4$ that are adjacent to $v_3$ in G. Since $S_n(2,2)\nsubseteq G$, $u_1,\ldots ,u_4$ are independent and are not adjacent to any other vertex of U. Since $n\ge 9$, there are at least 4 other vertices of U, say $u_5,\ldots ,u_8$, and $u_1u_5u_2u_6u_3u_7u_4u_8u_1$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction.

Case 2b: A vertex $u\in U$ has degree at least $n-3$ in $G[U\cup V]$.

Since $S_n(2,2)\nsubseteq G$, no vertex of V is adjacent to u or to $N_{G[U]}(u)$. Then u is adjacent to at least $n-3$ vertices of U in G; suppose without loss of generality that $u_1,\ldots ,u_{n-3}\subseteq N_{G[U]}(u)$. If $n\ge 10$, then any 4 vertices from $N_{G[U]}(u)$, any 4 vertices from V and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Suppose that $n=9$ and let $u_7$ be the vertex in $U{\setminus }\{u,u_1,\ldots ,u_{n-3}\}$. If $u_7$ is adjacent in ${\overline{G}}$ to at least two of $u_1,\ldots ,u_6$, say $u_1$ and $u_2$, then $u_1u_7u_2v_3u_3v_4u_4v_5u_1$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $u_7$ is adjacent in G to at least 5 of the vertices $u_1,\ldots ,u_6$, say $u_1,\ldots ,u_5$. Since $S_9(2,2)\nsubseteq G$, U is not adjacent in G to $\{v_0,v_1,v_2,w_1\}\cup V$ and $w_2$ is not adjacent to u or $u_7$. If $w_3$ is not adjacent to some vertex $a\in \{v_0,v_1,w_1,w_2\}$, then $uv_3u_1v_4u_2v_5u_7au$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Hence, $w_3$ is adjacent to $v_0$, $v_1$, $w_1$ and $w_2$ in G. Similarly, $v_2$ is adjacent to $v_1$, $w_1$ and $w_2$. Since $S_9(2,2)\nsubseteq G$, $w_2$ is non-adjacent to at least one of $v_3,v_4,v_5$, say $v_3$ without loss of generality. If $v_1$ is also not adjacent to $v_3$, then $uw_2u_7v_1u_1v_2u_2w_3u$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $v_1$ is adjacent to $v_3$, then $v_3$ is not adjacent to both $v_4$ and $v_5$, or else G contains $S_9(2,2)$. Without loss of generality, assume that $v_3$ is not adjacent to $v_4$ in G. Then $uw_2u_7v_4u_1v_2u_2w_3u$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction.

In either case, $R(S_n(2,2),W_8)\le 2n-1$ for $n\not \equiv 0 \pmod {4}$, which completes the proof. $\square$

Theorem 7.6

If $n\ge 9$, then $R(S_n(4,1),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $S_n(4,1)$ and that ${\overline{G}}$ does not contain $W_8$.

Suppose first that there is a subset $X\subseteq V(G)$ of size n with $\delta (G[X])\ge n-4$. Let $x_0$ be any vertex of X, and pick a subset $X'\subseteq N_{G[X]}(x_0)$ of size $n-5$. Set $Y = X{\setminus } (\{x_0\}\cup X')$, and so $|Y|=4$. Since $\delta (G[X])\ge n-4$, each vertex of Y is adjacent to at least $n-8$ vertices of $X'$ in G and each vertex of $X'$ is adjacent to at least one vertex of Y in G. Hence, for $n\ge 11$, it is straightforward to see that there is a matching from Y to $X'$ in G; hence, G contains $S_n(4,1)$, a contradiction.

For $n=10$ and $\delta (G[X])\ge n-4=6$, let $X=\{x_0,\ldots ,x_9\}$ and $\{x_1,\ldots ,x_6\}\subseteq N_{G[X]}(x_0)$. Since $\delta (G[X])\ge 6$, vertices $x_7$, $x_8$ and $x_9$ must each be adjacent to at least 3 vertices of $x_1,\ldots ,x_6$. It is straightforward to see that there is a matching from $\{x_7,x_8,x_9\}$ to $\{x_1,\ldots ,x_6\}$ in G; without loss of generality, assume that $x_i$ is adjacent to $x_{i+6}$ in G for $i=1,2,3$. Now, if there is any edge in $G[\{x_4,x_5,x_6\}]$, then $S_{10}(4,1)\subseteq G$, a contradiction. Otherwise, $G[\{x_4,x_5,x_6\}]$ must be independent and each of $x_4,x_5,x_6$ must be adjacent to at least two vertices of $x_7,x_8,x_9$ in G. Without loss of generality, assume that $x_4$ is adjacent to $x_7$ and $x_8$ in G. Since $S_{10}(4,1)\nsubseteq G$, $x_5$ cannot be adjacent to $x_1$ and $x_2$ in G, but this is impossible since $\delta (G[X])\ge 6$.

Now for $n=9$, suppose that $d_{G[X]}(x_0) = n-4 = 5$. Let $N_{G[X]}(x_0) = \{x_1,\ldots ,x_5\}$ and $Y = \{x_6,x_7,x_8\}$. Then, three vertices of Y are each adjacent to at least $n-6=3$ vertices of $N_{G[X]}(x_0)$ in G. Without loss of generality, assume that $x_1$ is adjacent to $x_6$, $x_2$ is adjacent to $x_7$ and $x_3$ is adjacent to $x_8$, respectively. Now, if $x_4$ is adjacent to $x_5$, then G contains $S_9(4,1)$, a contradiction. Otherwise, $x_4$ and $x_5$ must each be adjacent to at least one of $x_6$, $x_7$ and $x_8$. Assume that $x_4$ is adjacent to $x_6$. Then $x_5$ is not adjacent to $x_1$ and $x_4$ in G, or else G contains $S_9(4,1)$. If $x_5$ is adjacent to $x_6$, then $x_1,x_4,x_5$ must be independent in G, and they are each adjacent to $x_7$ or $x_8$ in G; assume that $x_1$ is adjacent to $x_7$. Then, $x_4$ and $x_5$ are not adjacent to $x_2$ in G, and since $\delta (G[X])\ge 5$, they are adjacent to $x_7$ and $x_8$ in G, and G contains $S_9(4,1)$, a contradiction. If $x_5$ is not adjacent to $x_6$, then since $d_{G[X]}(v_0) \ge 5$, $x_5$ is adjacent to $x_2$, $x_3$, $x_7$ and $x_8$ in G. Then, $x_4$ is not adjacent to $x_2$ and $x_3$ in G, and $x_4$ is adjacent to $x_1$, $x_6$, $x_7$ and $x_8$ in G, and this gives us $S_9(4,1)$ in G, a contradiction. As $x_0$ was arbitrary, assume for the case when $n=9$ that $\delta (G[X])\ge n-3 = 6$, which again leads to the contradiction that G contains $S_9(4,1)$.

Now assume that $\delta (G[X])\le n-5$ whenever $X\subseteq V(G)$ is of size n. Recall that G has order $2n-1$, and so by Theorem 6.9, G has a subgraph $S_n(3,1)$ and thus a subgraph $T=S_{n-1}(3,1)$. Let $V(T)=\{v_0,\ldots ,v_{n-5},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-5},v_1w_1,v_2w_2,v_3w_3\}$. Set $V=\{v_4,\ldots ,v_{n-5}\}$ and $U=V(G)-V(T)=\{u_1,\ldots ,u_n\}$; then $|V|=n-8$ and $|U|=n$. Since $S_n(4,1)\nsubseteq G$, V is not adjacent to any vertex of U in G. Now as $\delta (G[U])\le n-5$, ${\overline{G}}[U]$ contains $S_5$, and so for $n\ge 12$, ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Suppose that $n=11$. If $v_0$ is not adjacent to any vertex of U in G, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Assume that $v_0$ is adjacent to some vertex $u\in U$. Since $S_{11}(4,1)\nsubseteq G$, $G[V\cup \{u\}]$ is an empty graph and u is not adjacent to any vertex of U in G. By Lemma 4.4, $G[U{\setminus }\{u\}]$ is $K_{10}$ or $K_{10} - e$, so no vertex of $V(T)\cup \{u\}$ is adjacent to any vertex of $U{\setminus }\{u\}$ in G, as $S_{11}(4,1)\nsubseteq G$. Since $\delta (G[V(T)\cup \{u\}])\le n-5$, ${\overline{G}}[V(T)\cup \{u\}]$ contains $S_5$, so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Now, suppose that $n=10$. Then G has order 19, and by Theorem 2.2, G has a subgraph $T'=S_{10}(3,1)$. Let $V(T')=\{v_0',\ldots ,v_6',w_1',w_2',w_3'\}$ and $E(T')=\{v_0'v_1',\ldots ,v_0'v_6',v_1'w_1',v_2'w_2',v_3'w_3'\}$. Set $V'=\{v_4',v_5',v_6'\}$ and $U'=V(G)-V(T')=\{u_1',\ldots ,u_9'\}$. Since $S_{10}(4,1)\nsubseteq G$, $V'$ must be independent in G and is not adjacent to any vertex of $U'$ in G. If $v_0'$ is adjacent to some vertices in $U'$ in G, say $u_1'$. Since $S_{10}(4,1)\nsubseteq G$, $u_1'$ is not adjacent to any vertex of $V'$ or $U'{\setminus } \{u_1'\}$ in G. Then, by Lemma 4.4, $G[U'{\setminus }\{u_1'\}]$ is $K_8$ or $K_8 - e$, so no vertex of $V(T')$ is adjacent to any vertex of $U'{\setminus }\{u_1'\}$ in G, as $S_{10}(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le 5$, ${\overline{G}}[V(T')]$ contains $S_5$, so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Now, suppose that $v_0'$ is not adjacent to any vertex of $U'$ in G. Note that $|U'\cup \{w_1'\}|=n$; therefore, $\delta (G[U'\cup \{w_1'\}])\le 5$, and so ${\overline{G}}[U'\cup \{w_1'\}]$ contains $S_5$. If $w_1'$ is not adjacent to any vertex from $V'\cup \{v_0'\}$, then by Observation 4.3, ${\overline{G}}$ contains $W_8$, a contradiction. Otherwise, there are two cases to be considered.

Case 1a: $w_1'$ is adjacent to some vertices of $V'$ in G.

Without loss of generality, assume that $w_1'$ is adjacent to $v_4'$ in G. In this case, $v_1'$ is not adjacent to $U'\cup \{v_5',v_6'\}$. Then by Lemma 4.4, $G[U']$ is $K_9$ or $K_9 - e$, so no vertex of $V(T')$ is adjacent to any vertex of $U'$ in G, as $S_{10}(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le 5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Case 1b: $w_1'$ is non-adjacent to each vertex of $V'$ in G.

In this case, $w_1'$ is adjacent to $v_0'$ in G. Note that $w_1'$ is not adjacent to $U'$, since this would revert to the case where $v_0'$ is adjacent to some vertex of $U'$. Then again by Lemma 4.4, $G[U']$ is $K_9$ or $K_9 - e$, so no vertex of $V(T')$ is adjacent to any vertex of $U'$ in G, as $S_{10}(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le 5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Finally, suppose that $n=9$. Then G has order 17, and so G has a subgraph $T'=S_9(2,1)$ by Theorem 1.4. Let $V(T')=\{v_0',\ldots ,v_6',w_1',w_2'\}$ and $E(T')=\{v_0'v_1',\ldots ,v_0'v_6',v_1'w_1',v_2'w_2'\}$. Set $V'=\{v_3',\ldots ,v_6'\}$ and $U'=V(G)-V(T')=\{u_1',\ldots ,u_8'\}$.

Now, suppose that $E_G(V',U')\ne \emptyset$. Without loss of generality, assume that $v_3'$ is adjacent to $u_1'$ in G. Since $S_9(4,1)\nsubseteq G$, $v_4',v_5',v_6'$ are independent and not adjacent to any vertex of $U'{\setminus } \{u_1'\}$ in G.

Suppose that $v_0'$ is adjacent to some vertex of $U'{\setminus } \{u_1'\}$, say $u_2'$. Then $u_2'$ is non-adjacent to $\{v_4',v_5',v_6'\}\cup U'{\setminus } \{u_1',u_2'\}$ in G. Since $\delta (G[\{w_1',w_2'\}\cup U'{\setminus } \{u_2'\}])\le n-5$, ${\overline{G}}[\{w_1',w_2'\}\cup U'{\setminus } \{u_2'\}]$ contains $S_5$. If $v_4'$, $v_5'$, $v_6'$ and $u_2'$ are not adjacent to $w_1'$, $w_2'$ or $u_1'$ in G, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Assume that $v_4'$ is adjacent to $w_1'$ in G. In this case, $v_1'$ is not adjacent to $\{v_5',v_6'\}\cup U'{\setminus } \{u_1'\}$ in G, and $v_1'u_3'v_4'u_4'v_6'u_7'u_2'u_8'v_1'$ and $v_5'$ form $W_8$ in ${\overline{G}}$, a contradiction. Similar contradictions occur if we assume that $v_5'$, $v_6'$ or $u_2'$ are adjacent to $w_1'$, $w_2'$ or $u_1'$ in G.

Thus, $v_0'$ is not adjacent to any vertex of $U'{\setminus } \{u_1'\}$ in G. Since $\delta (G[\{w_1',w_2'\}\cup U'{\setminus } \{u_1'\}])\le n-5$, ${\overline{G}}[\{w_1',w_2'\}\cup U'{\setminus } \{u_1'\}]$ contains $S_5$. If $v_0'$, $v_4'$, $v_5'$ and $v_6'$ are not adjacent to $w_1'$ or $w_2'$ in G, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. There are two cases to be considered.

Case 2a: $v_0'$ is adjacent to $w_1'$ or $w_2'$ in G.

Without loss of generality, assume that $v_0'$ is adjacent to $w_1'$ in G. Note that $v_1'$ and $w_1'$ are not adjacent to $U'{\setminus } \{u_1'\}$, since this would revert to the case where $v_0'$ is adjacent to some vertex of $U'{\setminus } \{u_1'\}$. Again, since $\delta (G[\{w_2'\}\cup U'])\le n-5$, ${\overline{G}}[\{w_2'\}\cup U'\}]$ contains $S_5$. If $v_1'$, $v_4'$, $v_5'$ and $v_6'$ are not adjacent to $w_2'$ and $u_1'$ in G, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Suppose that $v_1'$ is adjacent to $w_2'$ or $u_1'$, say $w_2'$, in G. If $w_1'$ is not adjacent to $v_4'$, $v_5'$ or $v_6'$, then by Lemma 4.4, $G[U'{\setminus }\{u_1'\}]$ is $K_7$ or $K_7 - e$, so no vertex of $V(T')\cup \{u_1'\}$ is adjacent to any vertex of $U'{\setminus }\{u_1'\}$ in G, as $S_9(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Otherwise, $w_1'$ is adjacent to at least one of $v_4',v_5',v_6'$ in G, say $v_4'$. Then, $v_2'$ is not adjacent to $\{v_5',v_6'\}\cup U'{\setminus } \{u_1'\}$, since G does not contain $S_9(4,1)$. Similarly, by Lemma 4.4, $G[U'{\setminus }\{u_1'\}]$ is $K_7$ or $K_7 - e$, so no vertex of $V(T')\cup \{u_1'\}$ is adjacent to any vertex of $U'{\setminus }\{u_1'\}$ in G, as $S_9(4,1)\nsubseteq G$. Again, since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Now suppose that $v_1'$ is non-adjacent to both $w_2'$ and $u_1'$ in G. Then, one of $v_4',v_5',v_6'$ is adjacent to $w_2'$ or $u_1'$ in G. Without loss of generality, assume that $v_4'$ is adjacent to $w_2'$ in G. In this case, $v_2'$ is not adjacent to $\{v_5',v_6'\}\cup U'{\setminus } \{u_1'\}$. Then, again, by Lemma 4.4, $G[U'{\setminus }\{u_1'\}]$ is $K_7$ or $K_7 - e$, so no vertex of $V(T')\cup \{u_1'\}$ is adjacent to any vertex of $U'{\setminus }\{u_1'\}$ in G, as $S_9(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Case 2b: $v_0'$ is non-adjacent to both $w_1'$ and $w_2'$ in G.

In this case, one of $v_4',v_5',v_6'$ is adjacent to $w_1'$ or $w_2'$ in G, say $v_4'$ to $w_1'$ in G. Since $S_9(4,1)\nsubseteq G$, $v_1'$ is not adjacent to $\{v_5',v_6'\}\cup U'{\setminus } \{u_1'\}$ in G. By Lemma 4.4, $G[U'{\setminus }\{u_1'\}]$ is $K_7$ or $K_7 - e$, so no vertex of $V(T')\cup \{u_1'\}$ is adjacent to any vertex of $U{\setminus }\{u_1'\}$ in G, as $S_9(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Now suppose that $E_G(V',U')= \emptyset$. If $\delta (G[V'])=0$, then by Lemma 4.4, $G[U']$ is $K_8$ or $K_8 - e$, and no vertex of $V(T')$ is adjacent to any vertex of $U'$ in G, as $S_9(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Hence, $\delta (G[V'])\ge 1$, and since $S_9(4,1)\nsubseteq G$, one of the vertices in $V'$ is adjacent to other three in G. Without loss of generality, assume that $v_3'$ is adjacent to $v_4'$, $v_5'$ and $v_6'$ in G. Since G does not contain $S_9(4,1)$, $v_4',v_5',v_6'$ are independent in G. Furthermore, $v_0'$ is not adjacent to $U'$ in G or else this reverts to the case where $v_3'$ is adjacent to $u_1'$ and $v_0'$ is adjacent to any vertex of $U'{\setminus } \{u_1'\}$. Since $\delta (G[\{w_1'\}\cup U'])\le n-5$, ${\overline{G}}[\{w_1'\}\cup U']$ contains $S_5$. If $v_0'$, $v_4'$, $v_5'$ and $v_6'$ are non-adjacent to $w_1'$ in G, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction. Again, there are two cases to be considered.

Case 3a: $v_0'$ is adjacent to $w_1'$ in G.

Note that $v_1'$ and $w_1'$ are not adjacent to $U'$, or else this reverts to the case where $v_3'$ is adjacent to $u_1'$ and $v_0'$ is adjacent to any vertex of $U'{\setminus } \{u_1'\}$. Now, since $\delta (G[\{w_2'\}\cup U'])\le n-5$, ${\overline{G}}[\{w_2'\}\cup U'\}]$ contains $S_5$. If $v_0'$, $v_4'$, $v_5'$ and $v_6'$ are non-adjacent to $w_2'$ in G, then ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Suppose that $v_0'$ is adjacent to $w_2'$ in G. Again, $v_2'$ and $w_2'$ are non-adjacent to $U'$, or else else this reverts to the case where $v_3'$ is adjacent to $u_1'$ and $v_0'$ is adjacent to any vertex of $U'{\setminus } \{u_1'\}$. Now, $E_G(V(T'),U')= \emptyset$, and since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Therefore, $w_2'$ is adjacent to at least one of $v_4'$, $v_5'$ and $v_6'$ in G, say $v_4'$. Then, $v_2'$ is not adjacent to $v_5'$, $v_6'$ or $U'$, as $S_9(4,1)\nsubseteq G$, a contradiction. By Lemma 4.4, $G[U']$ is $K_8$ or $K_8 - e$, so no vertex of $V(T')$ is adjacent to any vertex of $U'$ in G, as $S_9(4,1)\nsubseteq G$. Again, since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Case 3b: $v_0'$ is not adjacent to $w_1'$ in G.

In this case, one of $v_4',v_5',v_6'$ is adjacent to $w_1'$ in G, say $v_4'$. Since $S_9(4,1)\nsubseteq G$, $v_1'$ is not adjacent to $v_5'$, $v_6'$ or $U'$ in G. By Lemma 4.4, $G[U']$ is $K_8$ or $K_8 - e$, so no vertex of $V(T')\cup \{u_1'\}$ is adjacent to any vertex of $U'$ in G, as $S_9(4,1)\nsubseteq G$. Since $\delta (G[V(T')])\le n-5$, ${\overline{G}}[V(T')]$ contains $S_5$, and so ${\overline{G}}$ contains $W_8$ by Observation 4.3, a contradiction.

Thus, $R(S_n(4,1),W_8)\le 2n-1$ for $n\ge 9$ which completes the proof. $\square$

Theorem 7.7

If $n\ge 8$, then

$$\begin{aligned} R(T_D(n),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be a graph with no $T_D(n)$ subgraph whose complement ${\overline{G}}$ does not contain $W_8$. Suppose that $n\equiv 0 \pmod {4}$ and that G has order 2n. By Theorem 6.2, G has a subgraph $T=S_n[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n$. Since $T_D(n)\nsubseteq G$, neither $w_2$ nor $w_3$ is adjacent in G to $U\cup V$.

Suppose that $n=8$. Since G does not contain $T_D(n)$, V must be independent and non-adjacent to U in G. Then for any vertices $u_1,\ldots ,u_4$ in U, $v_3u_1v_4u_2w_2u_3w_3u_4v_3$ and $v_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Suppose that $n\ge 12$. Then $|U\cup V|=2n-5$. If $\delta ({\overline{G}}[U\cup V])\ge \lceil \frac{2n-5}{2}\rceil$, then ${\overline{G}}[U\cup V]$ contains $C_8$ by Lemma 4.1 which with $w_2$ forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[U\cup V])\le \lceil \frac{2n-5}{2}\rceil -1=n-3$, and $\Delta (G[U\cup V])\ge n-3$. Now, there are two cases to consider.

Case 1: One of the vertices of V, say $v_2$, is a vertex of degree at least $n-3$ in $G[U\cup V]$.

Since $T_D(n)\nsubseteq G$, $v_1$ is not adjacent in G to $w_2$, $w_3$ or $U\cup V{\setminus } \{v_2\}$. Let $U'=\{w_2,w_3\}\cup U\cup V{\setminus } \{v_2\}$; then $|U'|=2n-4$. Now, if $\delta ({\overline{G}}[U'])\ge \frac{2n-4}{2}=n-2$, then ${\overline{G}}[U']$ contains $C_8$ by Lemma 4.1 which with $v_1$ forms $W_8$, a contradiction. Hence, $\delta ({\overline{G}}[U'])\le n-3$, and $\Delta (G[U'])\ge n-2$. Note that neither $w_2$ nor $w_3$ have degree $\Delta (G[U'])$. Therefore, $d_{G[U']}(u')\ge n-2$ for some vertex $u'\in U\cup V{\setminus } \{v_2\}$. By the Inclusion–Exclusion Principle, some vertex $a\in U\cup V{\setminus } \{v_2\}$ is adjacent in G to both $u'$ and $v_2$. Then G has a subgraph $T_D(n)$ in which $u'$ is the vertex of degree $n-5$ and $v_2$ is the vertex of degree 3, a contradiction.

Case 2: Some vertex $u\in U$ has degree at least $n-3$ in $G[U\cup V]$.

Suppose that there is at least one vertex in V that is adjacent to u in G, say $v_2$. Then G has a subgraph $T_D(n)$ in which u is the vertex of degree $n-5$ and $v_0$ is the vertex of degree 3, a contradiction. Similarly, no other vertex of V is adjacent to u. Now, since $T_D(n)\nsubseteq G$, $d_{G[N_{G[U]}(u)\cup \{v\}]}(v)\le 1$ and $d_{G[V\cup \{x\}]}(x)\le 1$, for any $v\in V$ and $x\in N_{G[U]}(u)$. Then, by Lemma 4.5, ${\overline{G}}[V\cup N_{G[U]}(u)]$ must contain $C_8$, which with $w_2$ as hub, forms $W_8$ in ${\overline{G}}$, a contradiction.

Now, suppose that $n\not \equiv 0 \pmod {4}$ and that G has order $2n-1$. By Theorem 6.4, G has a subgraph $T=S_n[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_D(n)\nsubseteq G$, neither $w_2$ nor $w_3$ is adjacent to $U\cup V$ in G. If $\delta ({\overline{G}}[U\cup V])\ge \frac{2n-6}{2}=n-3$, then ${\overline{G}}[U\cup V]$ contains $C_8$, by Lemma 4.1, which with $w_2$ forms $W_8$ in ${\overline{G}}$, a contradiction. Thus, $\delta ({\overline{G}}[U\cup V])\le n-4$, and $\Delta (G[U\cup V])\ge n-3$. The arguments of the preceding cases then lead to contradictions.

Thus, $R(T_D(n),W_8)\le 2n$, which completes the proof. $\square$

Lemma 7.8

Each graph H of order $n\ge 8$ with minimal degree at least $n-4$ contains $T_E(n)$ unless $n=8$ and $H=K_{4,4}$.

Proof

Let $V(H)=\{u_0,\ldots ,u_{n-1}\}$. First, suppose that $\Delta (H)\ge n-3$ and assume without loss of generality that $u_1,\ldots ,u_{n-3}\in N_H(u_0)$. Suppose that $u_{n-2}$ and $u_{n-1}$ are adjacent in H. Since $\delta (H)\ge n-4$, $N_H(u_0)\cap N_H(u_{n-2})\ne \emptyset$, so assume without loss of generality that $u_1$ is adjacent to $u_{n-2}$ in H. Furthermore, $u_1$ must be adjacent to at least $n-7$ vertices from $\{u_2,\ldots ,u_{n-3}\}$ in H. Without loss of generality, assume that $u_1$ is adjacent to $u_2,\ldots ,u_{n-6}$ in H. Now, if any vertex of $\{u_2,\ldots ,u_{n-6}\}$ is adjacent to $u_{n-5}$, $u_{n-4}$ or $u_{n-3}$ in H, then we have $T_E(n)$ in H. Suppose that is not the case; then each vertex of $\{u_2,\ldots ,u_{n-6}\}$ must be adjacent to each other and to $u_0$, $u_1$, $u_{n-2}$ and $u_{n-1}$ in H. Since $d_H(u_{n-3})\ge n-4$, $u_{n-3}$ is adjacent to at least one of $u_1$, $u_{n-2}$ and $u_{n-1}$ in H, so H contains $T_E(n)$, a contradiction.

Suppose that $u_{n-2}$ is not adjacent to $u_{n-1}$ in H. Since $\delta (H)\ge n-4$, $u_{n-2}$ and $u_{n-1}$ are each adjacent to at least $n-5$ vertices in $N_H(u_0)$, so at least one vertex of $N_H(u_0)$, say $u_1$, is adjacent in H to both $u_{n-2}$ and $u_{n-1}$. If $H[\{u_2,\ldots ,u_{n-3}\}]$ contains subgraph $2K_2$, then H contains subgraph $T_E(n)$. Note that this will always happens for $n\ge 11$, since $\delta (H)\ge n-4$.

Suppose that $n=10$. Since $\delta (H)\ge 6$, $u_2$ must be adjacent in H to at least two vertices of $u_3,\ldots ,u_7$, without loss of generality say $u_3$ and $u_4$. If $H[\{u_4,\ldots ,u_7\}]$ contains any edge, then H contains $T_E(10)$. Otherwise, $\{u_4,\ldots ,u_7\}$ must be independent in H and each of these vertices must be adjacent to $u_0$, $u_1$, $u_2$, $u_3$, $u_8$ and $u_9$; this also gives a subgraph $T_E(10)$ in H.

Similarly, for $n=9$, $u_2$ must be adjacent to at least one of $u_3,\ldots ,u_6$, say $u_3$, in H. If $H[\{u_4,u_5,u_6\}]$ contains any edge, then H contains $T_E(9)$. Otherwise, $\{u_4,u_5,u_6\}$ is independent in H and since $\delta (H)\ge 5$, $u_4$ is adjacent to at least one of $u_2$ and $u_3$, and $u_5$ is adjacent to at least one of $u_7$ and $u_8$. Again, this gives a subgraph $T_E(9)$ in H.

For $n=8$, if $u_2,\ldots ,u_5$ are independent in H, then they are each adjacent to $u_0$, $u_1$, $u_6$ and $u_7$ in H, which gives $T_E(8)$ in H. Otherwise, we can assume that $u_2$ is adjacent to $u_3$ in H. If $u_4$ is adjacent to $u_5$ in H, we will have $T_E(8)$ in H; otherwise, assume that $u_4$ is not adjacent to $u_5$. Now, suppose that $u_4$ is adjacent to $u_2$ or $u_3$ in H. If $u_5$ is adjacent to $u_6$ or $u_7$ in H, then H contains $T_E(8)$. Otherwise, $u_5$ must be adjacent to $u_0$, $u_1$, $u_2$ and $u_3$ since $\delta (H)\ge 4$. However, this also gives $T_E(8)$ in H. On the other hand, suppose that $u_4$ is adjacent to neither $u_2$ nor $u_3$ in H. Similarly, $u_5$ is not adjacent to $u_2$ or to $u_3$ in H. Since $\delta (H)\ge 4$, both $u_4$ and $u_5$ must be adjacent to $u_0$, $u_1$, $u_6$ and $u_7$ in H, and this also gives $T_E(8)$ in H.

Suppose that H is $(n-4)$-regular and that $N_H(u_0)=\{u_1,\ldots ,u_{n-4}\}$. By the Handshaking Lemma, this only happens when n is even.

Suppose that $n\ge 10$. Note that $u_{n-3}$, $u_{n-2}$ and $u_{n-1}$ are each adjacent to at least $n-6$ vertices of $N_H(u_0)$ in H. By the Inclusion–Exclusion Principle, at least one of $u_1,\ldots ,u_{n-4}$ is adjacent to two of $u_{n-3},u_{n-2},u_{n-1}$ in H, say $u_1$ to $u_{n-3}$ and $u_{n-2}$, and there must be another vertex, say $u_2$, that is adjacent to $u_{n-1}$ in H. Now, if there is any edge in $H[\{u_3,\ldots ,u_{n-4}\}]$, then $T_E(n)\subseteq H$, and this always happens for $n\ge 12$. For $n=10$, since $d_H(u_1)=6$, $u_1$ is non-adjacent in H to at least one of $u_3,\ldots ,u_6$, say $u_3$. Since $d_H(u_3)=6$, $u_3$ is adjacent to one of $u_4,u_5,u_6$, giving $T_E(10)$ in H.

Now suppose that $n=8$. If $u_5$, $u_6$ and $u_7$ are independent in H, then $H=K_{4,4}$. Otherwise, we can assume that $u_5$ is adjacent to $u_6$ in H. If $u_5$ is also adjacent to $u_7$ in H, then $u_5$ is adjacent in H to two vertices of $N_H(u_0)$, say $u_1$ and $u_2$. Suppose that $u_6$ is adjacent to $u_1$ or $u_2$, say $u_1$, in H. Since $d_H(u_6)=4$, $u_6$ is also adjacent to at least one of $u_2,u_3,u_4,u_7$, so $T_E(8)\subseteq H$. Otherwise, suppose that neither $u_6$ nor $u_7$ is adjacent to $u_1$ or $u_2$ in H. Since H is a 4-regular graph, $u_6$ and $u_7$ are both adjacent to $u_3$ and $u_4$ in H, and $u_1$ is adjacent to at least one of $u_3$ and $u_4$ in H. This gives $T_E(8)$ in H. On the other hand, suppose that $u_5$ is not adjacent to $u_7$ in H. Then, similarly, $u_6$ is not adjacent to $u_7$ in H, so $u_7$ is adjacent to $u_1$, $u_2$, $u_3$ and $u_4$ in H, and H contains $T_E(8)$. $\square$

Theorem 7.9

For $n\ge 8$,

$$\begin{aligned} R(T_E(n),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\ge 9;\\ 16 &{} \hbox { if}\ n=8. \end{array}\right. } \end{aligned}$$

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$ if $n\ge 9$ and of order 16 if $n = 8$. Assume that G does not contain $T_E(n)$ and that ${\overline{G}}$ does not contain $W_8$.

By Theorem 6.9, G has a subgraph $T=S_n(3,1)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_2w_2,v_3w_3\}$. Set $V=\{v_4,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$. Then $|V|=n-7$ and $|U|\ge n-1$. Since $T_E(n)\nsubseteq G$, each of $v_1,v_2,v_3$ is not adjacent to any vertex of $V\cup U$ in G, and each vertex of V is adjacent to at most one vertex of U in G. Let W be a set of $n-2$ vertices of U that are not adjacent to $v_4$ in G. By Lemma 4.4, G[W] is $K_{n-2}$ or $K_{n-2}-e$. Since $T_E(n)\nsubseteq G$, no vertex of T is adjacent to any vertex of W, and so $\delta (G[V(T)])\ge n-4$ by Observation 4.3.

Lemma 7.8 implies that G[V(T)] contains $T_E(n)$ if $n\ge 9$, a contradiction, and so $n=8$ and $G[V(T)] = K_{4,4}$. Note that $|U| = 8$, and as $T_E(8)\nsubseteq G$, no vertex of U is adjacent to any vertex of G[V(T)]. By Lemma 4.4, G[U] is $K_8$ or $K_8-e$, and thus contains $T_E(8)$, a contradiction.

Therefore, $R(T_E(n),W_8)\le 2n-1$ when $n\ge 9$ and $R(T_E(n),W_8)\le 16$ when $n=8$. $\square$

Lemma 7.10

Each graph H of order $n\ge 8$ with minimal degree at least $n-4$ contains $T_F(n)$ unless $n=8$ and $H=K_{4,4}$.

Proof

Let $V(H) = \{u_0, u_1\ldots , u_{n-1}\}$ so that $d(u_0) = \delta (H)$ and $V = \{u_1,\ldots , u_{n-4}\}\subseteq N(u_0)$. Set $U = \{u_{n-3}, u_{n-2}, u_{n-1}\}$. By the minimum degree condition, every vertex of U is adjacent to at least $n-6$ vertices of V. It is straightforward to see that some pair of vertices in U have a common neighbour in V. Moreover, for $n\ge 9$, every pair of vertices in U has a common neighbour in V.

Assume without loss of generality that $u_1$ is adjacent to both $u_{n-3}$ and $u_{n-2}$, and that $u_2$ is adjacent to $u_{n-1}$. If $u_2$ is adjacent to a vertex of $V{\setminus }\{u_1\}$, which is the case when $n\ge 10$, then H contains $T_F(n)$. Assume now that $n\le 9$ and that $u_2$ is not adjacent to any vertex of $V{\setminus }\{u_1\}$.

For the case when $n=9$, $u_{n-1}$ is adjacent to at least $n-6=3$ vertices of V, and so it is adjacent to another vertex, say to $u_3$. As above, assume that $u_3$ is not adjacent to any vertex of $V{\setminus }\{u_1\}$. By the minimum degree condition, each of $u_2$ and $u_3$ is adjacent to every vertex of $\{u_1\} \cup U$, giving $T_F(9)$ in H.

For the final case when $n=8$, the minimum degree condition implies that $u_2$ is adjacent to at least two vertices of $\{u_1,u_5,u_6\}$. If $u_2$ is adjacent to $u_1$, then H contains $T_F(8)$. Remaining is the case when $u_2$ is not adjacent to $u_1$ but is adjacent to both $u_5$ and $u_6$. Exchanging the roles of $u_1$ and $u_2$, we may assume that $u_1$ is adjacent to $u_7$ but not adjacent to any vertex of V. From the minimum degree condition on $u_3$ and $u_4$, it is easy to see that either H contains $T_F(8)$ or $H = K_{4,4}$. $\square$

Theorem 7.11

For $n\ge 8$,

$$\begin{aligned} R(T_F(n),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\ge 9;\\ 16 &{} \hbox { if}\ n=8. \end{array}\right. } \end{aligned}$$

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be a graph with no $T_F(n)$ subgraph whose complement ${\overline{G}}$ does not contain $W_8$. Suppose that $n=8$ and that G has order 16. By Theorem 6.8, G has a subgraph $T=T_C(8)$. Let $V(T)=\{v_0,\ldots ,v_4,w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_4,v_1w_1,v_2w_2,v_2w_3\}$. Set $U=V(G)-V(T)=\{u_1,\ldots ,u_8\}$; then $|U|=8$. Since $T_F(8)\nsubseteq G$, $v_1$ is not adjacent in G to $v_2,v_3,v_4$ or any vertex of U, and $d_{G[U]}(v)\le 1$ for $v=v_3,v_4,w_2,w_3$.

Suppose that $v_1$ is adjacent to $w_2$ or $w_3$, without loss of generality say $w_2$. Since $T_F(8)\nsubseteq G$, $v_2$ is not adjacent to $\{v_3,v_4\}\cup U$. If neither $v_3$ nor $v_4$ are adjacent to U, then by Lemma 4.4, G[U] is $K_8$ or $K_8-e$, so G[U] contains $T_F(8)$, a contradiction. Suppose that only one of the vertices $v_3$ and $v_4$ is adjacent to U in G, say $v_3$. By Lemma 4.4, $G[U{\setminus } \{u_1\}]$ is $K_7$ or $K_7-e$, and $G[V(T)\cup \{u_1\}]$ is not adjacent to $G[U{\setminus } \{u_1\}]$. By Observation 4.3, $\delta (G[V(T)\cup \{u_1\}])\ge 5$, and by Lemma 7.10, $G[V(T)\cup \{u_1\}]$ contains $T_F(9)$ and hence $T_F(8)$, a contradiction. Suppose that both $v_3$ and $v_4$ are adjacent to U in G and assume that $v_3$ is adjacent to $u_1$ and that $v_4$ is adjacent to $u_2$. By Lemma 4.4, $G[U{\setminus } \{u_1,u_2\}]$ is $K_6$ or $K_6-e$. At most one vertex from $G[V(T)\cup \{u_1,u_2\}]$ is adjacent to $G[U{\setminus } \{u_1,u_2\}]$ or else G contains $T_F(8)$. Therefore, 9 vertices from $G[V(T)\cup \{u_1,u_2\}]$ form a vertex set W that is not adjacent to $U{\setminus } \{u_1,u_2\}$. By Observation 4.3, $\delta (G[W])\ge 5$, and by Lemma 7.10, G[W] contains $T_F(9)$ and hence $T_F(8)$, a contradiction.

Suppose then that $v_1$ is not adjacent to $w_2$ or $w_3$. Since $d_{G[U]}(v)\le 1$ for $v=v_3,v_4,w_2,w_3$, there are 4 vertices from U that are not adjacent to $\{v_3,v_4,w_2,w_3\}$. These 8 vertices form $C_8$ in ${\overline{G}}$ and thus, with $v_1$ as hub, $W_8$, a contradiction.

Thus, $R(T_F(8),W_8)\le 16$.

Now, suppose that $n\ge 9$ and that G has order $2n-1$. By Theorem 6.8, G has a subgraph $T=T_C(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},v_4,w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_2w_2,v_2w_3\}$. Set $V=\{v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)=\{u_1,\ldots ,u_{n-1}\}$; then $|V|=n-6$ and $|U|=n-1$. Since $T_F(n)\nsubseteq G$, $v_1$ is not adjacent in G to any vertex of $U\cup V$, and $d_{G[U]}(v)\le 1$ for $v\in V$. Since $n\ge 10$, there are 4 vertices from U, 4 vertices from V and $v_1$ that form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $R(T_F(n),W_8)\le 2n-1$ for $n\ge 10$.

Suppose that $n=9$ and let m be the number of vertices of U that are adjacent in G to at least one vertex of V. Since $d_{G[U]}(v)\le 1$ for $v\in V$, $0\le m\le 3$. If $m=0$, then G[U] is $K_8$ or $K_8-e$ by Lemma 4.4, so G[V(T)] is not adjacent to G[U]. By Observation 4.3, $\delta (G[V(T)])\ge 5$, and G[V(T)] contains $T_F(9)$ by Lemma 7.10, a contradiction. Suppose that $m=1$. Assume without loss of generality that $u_1$ is adjacent to some vertex of V, and that $E_G(V,U{\setminus } \{u_1\}) = \emptyset$. By Lemma 4.4, $G[U{\setminus } \{u_1\}]$ is $K_7$ or $K_7-e$, and at most one vertex from $G[V(T)\cup \{u_1\}]$ is adjacent to $G[U{\setminus } \{u_1\}]$ or else G contains $T_F(9)$. There are then 9 vertices from $G[V(T)\cup \{u_1\}]$ that form a vertex set $W_1$ that is not adjacent to $U{\setminus } \{u_1\}$. By Observation 4.3, $\delta (G[W_1])\ge 5$, and $G[W_1]$ contains $T_F(9)$ by Lemma 7.10, a contradiction. Suppose that $m=2$. Assume that $u_1$ and $u_2$ are adjacent to some vertices of V and that $E_G(V,U{\setminus } \{u_1,u_2\})=\emptyset$. By Lemma 4.4, $G[U{\setminus } \{u_1,u_2\}]$ is $K_6$ or $K_6-e$. If at least three vertices in $U{\setminus } \{u_1,u_2\}$ are adjacent to $V(T)\cup \{u_1\}$, then $T_F(9)\subseteq G$. If at most two vertices in $U{\setminus } \{u_1,u_2\}$ are adjacent to $V(T)\cup \{u_1\}$, then there are 4 vertices in $U{\setminus } \{u_1,u_2\}$ that are not adjacent to V(T). Then Observation 4.3 gives $\delta (G[V(T)])\ge 5$, and G[V(T)] contains $T_F(9)$ by Lemma 7.10, a contradiction. Suppose that $m=3$. Assume that $u_1,u_2,u_3$ are each adjacent to some vertex of V and that $E_G(V,U{\setminus } \{u_1,u_2,u_3\})=\emptyset$. Without loss of generality, assume that $u_i$ is adjacent to $v_{i+2}$ for $i=1,2,3$. By Lemma 4.4, $G[U{\setminus } \{u_1,u_2,u_3\}]$ is $K_5$ or $K_5-e$. Since $T_F(9)\nsubseteq G$, $\{v_1,v_3,v_4,v_5\}$ is independent and $V(T){\setminus } \{w_1\}$ is not adjacent to $U{\setminus } \{u_1,u_2,u_3\}$. Then by Observation 4.3, $\delta (G[V(T){\setminus } \{w_1\}])\ge 4$, and $v_1$, $v_3$, $v_4$ and $v_5$ are each adjacent to $v_2$, $w_2$ and $w_3$ in G. This gives $T_F(9)$ in G. Therefore, $T_F(9)\le 17=2n-1$. $\square$

Theorem 7.12

If $n\ge 8$, then $R(T_G(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $T_G(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.9, G has a subgraph $T=S_n(3,1)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_2w_2,v_3w_3\}$. Set $V=\{v_4,v_5,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-7$ and $|U|=n-1$. Since $T_G(n)\nsubseteq G$, $w_1,w_2,w_3$ are not adjacent to $U\cup V$ in G, and $v_1,v_2,v_3$ are not adjacent to V.

Suppose that $n\ge 9$; then $|U|\ge 8$. If $\delta ({\overline{G}}[U])\ge \frac{n-1}{2}$, then ${\overline{G}}[U]$ contains $C_8$ by Lemma 4.1 which, with $w_2$ as hub, forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U])<\frac{n-1}{2}$, and $\Delta (G[U\cup V])\ge \frac{n-1}{2}\ge 4$. Therefore, some vertex $u\in U$ satisfies $|N_{G[U]}(u)|\ge 4$. Since $T_G(n)\nsubseteq G$, $N_{G[U]}(u)$ is not adjacent in G to $N_{G[V(T)]}(v_0)$. Hence, 4 vertices from $N_{G[U]}(u)$, $v_1,v_2,v_3,w_1$ and any vertex from V form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $R(T_G(n),W_8)\le 2n-1$ for $n\ge 9$.

Suppose that $n=8$ and let $U=\{u_1,\ldots ,u_7\}$ and $W=\{v_4\}\cup U$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}$ contains $C_8$ by Lemma 4.1 and thus $W_8$, with $w_1$ as hub, a contradiction. Therefore, $\delta ({\overline{G}}[W])\le 3$, and $\Delta (G[W])\ge 4$. Now, suppose that $d_{G[W]}(v_4)\ge 4$. Then without loss of generality, assume that $u_1,\ldots ,u_4\in N_G(v_4)$. Then $u_1,\ldots ,u_4,w_1,w_2,w_3$ are independent and are not adjacent to $u_5$, $u_6$ or $u_7$, giving $W_8$, a contradiction. On the other hand, suppose that some vertex in U, say $u_1$, satisfies $d_{G[W]}(u_1)\ge 4$. Then $v_4$ is not adjacent to $u_1$; therefore, assume that $u_2,\ldots ,u_5\in N_G(u_1)$. Then $v_1,\ldots ,v_4$ are not adjacent to $\{u_1,\ldots ,u_5\}$, so $v_1u_1v_2u_2v_3u_3w_1u_4v_1$ and $v_4$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $R(T_L(8),W_8)\le 15$. $\square$

Lemma 7.13

Each graph H of order $n\ge 8$ with minimal degree at least $n-4$ contains $T_H(n)$, $T_K(n)$ and $T_L(n)$.

Proof

Let $V(H)=\{u_0,\ldots ,u_{n-1}\}$ where $u_1,\ldots ,u_{n-4}\in N_H(u_0)$. Suppose that $u_{n-3}$, $u_{n-2}$ or $u_{n-1}$, say $u_{n-3}$, is adjacent in H to the two others.

Since $\delta (H)\ge n-4$, $u_{n-3}$ is adjacent to at least one of $u_1,\ldots ,u_{n-4}$, say $u_1$. If $u_1$ is adjacent to another vertex in $\{u_2,\ldots ,u_{n-4}\}$, then H contains $T_K(n)$. Note that this always happens for $n\ge 9$. Suppose that $n=8$ and that $u_1$ is not adjacent to any of $u_2,u_3,u_4$. Then $u_1$ is adjacent to $u_6$ and $u_7$. Since $\delta (H)\ge n-4$, $u_2$ is adjacent to at least one of $u_5, u_6, u_7$, giving $T_K(n)$ in H.

Similarly, since $\delta (H)\ge n-4$, $u_{n-2}$ is adjacent to at least $n-7$ vertices of $\{u_1,\ldots ,u_{n-4}\}$. Suppose that $u_{n-2}$ is adjacent to $u_1$. If $n\ge 10$, then at least two of $u_2,\ldots ,u_{n-4}$ are adjacent, so H contains $T_H(n)$. If $n\ge 9$, then $u_1$ is adjacent to at least one of $u_2,\ldots ,u_{n-4}$, so H contains $T_L(n)$. Now suppose that $n=9$. If any of $u_2,\ldots ,u_5$ are adjacent to each other, then H contains $T_H(9)$. Otherwise, $u_2,\ldots ,u_5$ are each adjacent to $u_6$, $u_7$ and $u_8$, and so H contains $T_H(9)$. Finally, suppose that $n=8$. If any two of $u_2,u_3,u_4$ are adjacent, then H contains $T_H(8)$; otherwise, they are each adjacent to $u_6$ or $u_7$. Now, if $u_1$ is adjacent to any of $u_2,u_3,u_4$, then H contains $T_H(8)$. Otherwise, $u_1,\ldots ,u_4$ are each adjacent to $u_5$, $u_6$ and $u_7$, and H also contains $T_H(8)$. Furthermore, if $u_1$ is adjacent to $u_2$, $u_3$ or $u_4$, then H contains $T_L(8)$. If $u_1$ is not adjacent to $u_2$, $u_3$ or $u_4$, then $u_6,u_7,u_8$ are adjacent to $u_2,u_3,u_4$, and then H contains $T_L(8)$. Now if $u_{n-2}$ is adjacent to some $u_2,\ldots ,u_{n-4}$, say $u_2$, then similar arguments apply by interchanging $u_1$ and $u_2$.

Suppose now that neither $u_{n-3}$, $u_{n-2}$ nor $u_{n-1}$ is adjacent to both of the others. Then one of these, say $u_{n-3}$, is adjacent to neither of the others. Since $\delta (H)\ge n-4$, $u_{n-3}$ is adjacent to at least $n-5$ of the vertices $u_1,\ldots ,u_{n-4}$. Without loss of generality, assume that $u_1,\ldots ,u_{n-5}\in N_H(u_{n-3})$. Then $u_{n-2}$ is adjacent to at least $n-7$ of the vertices $u_1,\ldots ,u_{n-5}$ including, without loss of generality, the vertex $u_1$. Also, $u_{n-1}$ is adjacent to at least one of $u_2,\ldots ,u_{n-4}$, so H contains $T_H(n)$. If $u_{n-2}$ is adjacent to $u_{n-1}$, then H also contains $T_L(n)$. If $u_{n-2}$ is not adjacent to $u_{n-1}$, then $u_{n-2}$ is adjacent to at least $n-6$ vertices of $u_1,\ldots ,u_{n-5}$, so H contains $T_L(n)$. Now, suppose that $n\ge 9$. Then $u_{n-2}$ and $u_{n-1}$ are each adjacent to at least 3 of $u_1,\ldots ,u_5$, and one of those vertices must be adjacent to both $u_{n-2}$ and $u_{n-1}$; thus, H contains $T_K(n)$. Finally, suppose that $n = 8$. If $u_6$ and $u_7$ are each adjacent to at least two of the vertices $u_1,u_2,u_3$, then one of those vertices must be adjacent to both $u_6$ and $u_7$; thus, H contains $T_K(8)$. Otherwise, $u_6$ or $u_7$, say $u_6$, is non-adjacent to at least two of $u_1,u_2,u_3$, say $u_1$ and $u_2$. Then $u_6$ is adjacent to $u_0$, $u_3$, $u_4$ and $u_7$, and so H contains $T_K(8)$. $\square$

Theorem 7.14

If $n\ge 8$, then $R(T_H(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$ and assume that G does not contain $T_H(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 7.12, G has a subgraph $T=T_G(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-5},w_1,\ldots ,w_4\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-5},v_1w_1,v_2w_2,v_3w_3,w_3w_4\}$. Set $U=\{u_1,\ldots ,u_{n-1}\}=V(G)-V(T)$; then $|U|=n-1$. Since $T_G(n)\nsubseteq G$, $E_G(\{w_1,w_2\},\{w_3,w_4\})=\emptyset$ and $w_4$ is not adjacent to U. Now, let $W=\{w_1\}\cup U$; then $|W|=n$. If $\delta ({\overline{G}}[W])\ge \frac{n}{2}$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which, with $w_4$ as hub, forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[W])<\frac{n}{2}$, and $\Delta (G[W])\ge \lfloor \frac{n}{2} \rfloor \ge 4$.

First, suppose that $w_1$ is a vertex with degree at least $\frac{n}{2}$ in G[W]. Assume without loss of generality that $u_1,\ldots ,u_4\in N_{G[W]}(w_1)$. Since $T_H(n)\nsubseteq G$, $u_1,\ldots ,u_4$ are independent and are not adjacent to $\{w_2,u_5,\ldots ,u_{n-1}\}$ in G. Then $w_2,u_1,\ldots ,u_4,w_4$ and any 3 vertices from $\{u_5,\ldots ,u_{n-1}\}$ form $W_8$ in ${\overline{G}}$, a contradiction. Hence, $d_{G[W]}(u')\ge \frac{n}{2}$ for some vertex $u'\in U$, say $u'=u_1$. Note that $w_1$ is not adjacent to $u_1$, or else G contains $T_H(n)$. Without loss of generality, suppose that $u_2,\ldots ,u_5\in N_{G[W]}(u_1)$. Since $T_H(n)\nsubseteq G$, $u_2,\ldots ,u_5$ are not adjacent to $V(T){\setminus } \{v_0\}$ in G. Now, if $v_0$ is not adjacent to $\{u_2,\ldots ,u_5\}$ in G, then by Observation 4.3, $\delta (G[V(T)])\ge n-4$, or else ${\overline{G}}$ contains $W_8$. By Lemma 7.13, G[V(T)] contains $T_H(n)$, a contradiction. On the other hand, suppose that $v_0$ is adjacent to at least one of $u_2,\ldots ,u_5$, say $u_2$. Then $u_3,u_4,u_5$ are independent in G and are not adjacent to $u_6$ and $u_7$ in G. Furthermore, $w_4$ is not adjacent to $v_1$ or $v_2$. Then $v_1u_3v_2u_4u_6w_1u_7u_5v_1$ and $w_4$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $R(T_H(n),W_8)\le 2n-1$. $\square$

Theorem 7.15

If $n\ge 8$, then $R(T_J(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$ and assume that G does not contain $T_J(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.8, G has a subgraph $T=T_C(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,v_2w_3\}$. Set $V=\{v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|U|=n-1$. Let $U = \{u_1,\ldots ,u_{n-1}\}$. Since $T_J(n)\nsubseteq G$, neither $w_1$ nor $w_2$ is adjacent in G to any vertex from $U\cup V$.

Let $W=\{v_3\}\cup U$; then $|W|=n$. If $\delta ({\overline{G}}[W])\ge \lceil \frac{n}{2} \rceil \ge \frac{n}{2}$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which with $w_1$ forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[W])<\lceil \frac{n}{2} \rceil$, and $\Delta (G[W])\ge \lfloor \frac{n}{2} \rfloor \ge 4$.

Suppose that $d_{G[W]}(v_3)\ge \lfloor \frac{n}{2} \rfloor \ge 4$. Without loss of generality, assume that $u_1,\ldots ,u_4\in N_G(v_3)$. Since $T_J(n)\nsubseteq G$, $u_1,\ldots ,u_4$ is independent in G and is not adjacent to any remaining vertices from U in G. Then $u_2w_1u_3u_5u_4u_6w_2u_7u_2$ and $u_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Hence, there is a vertex in U, say $u_1$, such that $d_{G[W]}(u_1)\ge \lfloor \frac{n}{2} \rfloor \ge 4$.

Now, suppose that $v_3$ is adjacent to $u_1$ in G[W]. Then $u_1$ is adjacent to at least 3 other vertices of U in G, say $u_2$, $u_3$ and $u_4$. Since $T_J(n)\nsubseteq G$, $v_3$ is not adjacent to $v_1,v_2,v_4,\ldots ,v_{n-4},w_1,w_2,w_3,u_2,u_3,u_4$ and neither $v_1$ nor $v_2$ is adjacent to $u_2$, $u_3$ or $u_4$ in G. Then $v_2u_2v_1u_3w_1v_4w_2u_4v_2$ and $v_3$ form $W_8$ in ${\overline{G}}$, a contradiction.

Thus, $v_3$ is not adjacent to $u_1$ in G. Note that $u_1$ is not adjacent to any other vertices of V in G or else previous arguments apply. Similarly, $v_0$ is not adjacent to $N_{G[W]}(u_1)$ in G. Since $T_J(n)\nsubseteq G$, neither $v_1$ nor $v_2$ is adjacent to $u_1$ or $N_{G[W]}(u_1)$ in G, and so $d_{N_{G[W]}(u_1)}(v)\le 1$ for all $v\in V$.

Suppose that $n\ge 10$; then $|V|\ge 4$ and $|N_{G[W]}(u_1)|\ge 5$. If $d_{G[V]}(u)\le 2$ for each $u\in N_{G[W]}(u_1)$, then ${\overline{G}}[V\cup N_{G[W]}(u_1)]$ contains $C_8$ by Lemma 4.5 which, with $w_1$ as hub, forms $W_8$ in ${\overline{G}}$, a contradiction. Thus, $d_V(u')\ge 3$ for some vertex $u'\in N_{G[W]}(u_1)$. Then any 4 vertices from V, of which at least 3 are in $N_{G[V]}(u')$, and any 4 vertices from $N_{G[W]}(u_1){\setminus } \{u'\}$ satisfy the condition in Lemma 4.5, so ${\overline{G}}[V\cup N_{G[W]}(u_1)]$ contains $C_8$ which with $w_1$ forms $W_8$, a contradiction.

Suppose that $n=9$; then $V=\{v_3,v_4,v_5\}$. Assume that $u_2,\ldots ,u_5\in N_{G[W]}(u_1)$. Suppose that $w_1$ is not adjacent to $w_2$ in G. Let $X=\{v_3,v_4,v_5,w_2\}$ and $Y=\{u_2,\ldots ,u_5\}$ and note that $d_{G[Y]}(x)\le 1$ for each $x\in X$. If $d_{G[X]}(y)\le 2$ for each $y\in Y$, then ${\overline{G}}[X\cup Y]$ contains $C_8$ by Lemma 4.5 which, with $w_1$ as hub, forms $W_8$, a contradiction. Thus, $d_{G[X]}(u')\ge 3$ for some $u'\in Y$, say $u'=u_2$, so X is not adjacent to $Y{\setminus } \{u_2\}$. Hence, $v_3u_1v_4u_3v_5u_4w_2u_5v_3$ and $w_1$ form $W_8$ in ${\overline{G}}$, a contradiction.

Thus, $w_1$ is adjacent to $w_2$ in G. Then $v_1$ is not adjacent to $\{v_3,v_4,v_5\}\cup U$. Suppose that $v_1$ is not adjacent to $v_2$. Then set $X=\{v_2,\ldots ,v_5\}$ and $Y=\{u_2,\ldots ,u_5\}$. If $d_{G[X]}(y)\le 2$ for each $y\in Y$, then ${\overline{G}}[X\cup Y]$ contains $C_8$ by Lemma 4.5 which, with $v_1$ as hub, forms $W_8$, a contradiction. Thus, $d_{G[X]}(u')\ge 3$ for some $u'\in Y$, say $u'=u_2$, so X is not adjacent to $Y{\setminus } \{u_2\}$, and $v_2u_1v_3u_3v_4u_4v_5u_5v_2$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $v_1$ is adjacent to $v_2$ in G. Then V is independent and is not adjacent to U in G. Since $W_8\nsubseteq {\overline{G}}$, G[U] is $K_{n-1}$ or $K_{n-1}-e$ by Lemma 4.4. Since $T_J(9)\nsubseteq G$, T is not adjacent to U and, by Observation 4.3, $\delta (G[V(T)])\ge 5$. However, this is impossible since V is independent and is not adjacent to $v_1$, $w_1$ or $w_2$.

Finally, suppose that $n=8$; then $V=\{v_3,v_4\}$. Assume that $u_2,\ldots ,u_5\in N_{G[W]}(u_1)$. If $v_3$ is adjacent to any vertex of $\{u_2,\ldots ,u_5\}$, say $u_2$, then $v_3$ is not adjacent to $\{v_1,v_2,v_4,w_3\}\cup U{\setminus } \{u_2\}$, so $v_1u_1v_2u_3w_1u_4w_2u_5v_1$ and $v_3$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $v_3$ is not adjacent to $\{u_2,\ldots ,u_5\}$. Similarly, $v_4$ is not adjacent to $\{u_2,\ldots ,u_5\}$. Now, if $w_3$ is adjacent to any of the vertices $u_2,\ldots ,u_5$, say $u_2$, then $v_2$ is not adjacent to $\{w_1, w_2, v_3, v_4\}$, so $v_3u_1v_4u_2w_1u_3w_2u_4v_3$ and $v_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $w_3$ is not adjacent to $\{u_2,\ldots ,u_5\}$. By Observation 4.3, $\delta (G[V(T)])\ge 4$. Suppose that $v_2$ is adjacent to $w_1$. Since $T_J(8)\nsubseteq G$, neither $v_3$ nor $v_4$ is adjacent to $w_3$. Since $\delta (G[V(T)])\ge 4$, $v_3$ and $v_4$ are adjacent to $v_1$ and $v_2$, and $\{w_1,w_2,w_3\}$ is not independent. However, then $T_J(8)\subseteq G[V(T)]$, a contradiction. Thus, $v_2$ is not adjacent to $w_1$ and, similarly, $v_2$ is not adjacent to $w_2$. Since $\delta (G[V(T)])\ge 4$, $w_1$ and $w_2$ are adjacent to each other and to $w_3$. Since $T_J(8)\nsubseteq G$, neither $v_3$ nor $v_4$ is adjacent to $v_1$ or $v_2$; however, this contradicts $\delta (G[V(T)])\ge 4$.

In each case, $R(T_J(8),W_8)\le 2n-1$, which completes the proof of the theorem. $\square$

Theorem 7.16

If $n\ge 8$, then $R(T_K(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be a graph of order $2n-1$ and assume that G does not contain $T_K(n)$ and that ${\overline{G}}$ does not contain $W_8$.

Suppose that $n\not \equiv 0 \pmod {4}$. By Theorem 6.5, G has a subgraph $T=S_n(1,3)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_2w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_K(n)\nsubseteq G$, $w_2$ is not adjacent in G to any vertex of $U\cup V$. Now, if $\delta (G[U])\ge \frac{n-1}{2}$, then ${\overline{G}}[U]$ contains $C_8$ by Lemma 4.1 which, with $v_1$ as hub, forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U])<\frac{n-1}{2}$, and $\Delta (G[U])\ge \lfloor \frac{n-1}{2} \rfloor$. Let $U=\{u_1,\ldots ,u_{n-1}\}$ and assume without loss of generality that $d_{G[U]}(u_1)\ge \lfloor \frac{n-1}{2} \rfloor \ge 4$. Since $T_K(n)\nsubseteq G$, $E_G(V,N_{G[U]}(u_1))=\emptyset$, so any 4 vertices from V, any 4 vertices from $N_{G[U]}(u_1)$ and $w_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $R(T_K(n),W_8)\le 2n-1$ for $n\not \equiv 0 \pmod {4}$.

Let $n=8$. By Theorem 7.14, G has a subgraph $T=T_H(8)$. Let $V(T)=\{v_0,v_1,v_2,v_3,w_1,\ldots ,w_4\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_3,v_1w_1,w_1w_2,w_2w_3,v_2w_4\}$. Set $U=V(G)-V(T)=\{u_1,\ldots ,u_7\}$; then $|U|=7$. Since $T_K(8)\nsubseteq G$, $w_2$ is not adjacent to $\{w_4\}\cup U$. Let $W=\{w_4\}\cup U$. Then $|W|=8$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which, with $w_2$ as hub, forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[W])<3$, and $\Delta (G[W])\ge 4$.

Now, suppose that $d_{G[W]}(w_4)\ge 4$ and assume without loss of generality that $w_4$ is adjacent to $u_1$, $u_2$, $u_3$ and $u_4$. Then $v_1$ is not adjacent to $\{v_3,w_2,w_3\}\cup U$ and neither $v_2$ nor $v_3$ is adjacent to $\{u_1,\ldots ,u_4\}$, since $T_K(8)\nsubseteq G$. Now, suppose that $E_G(\{u_1,\ldots ,u_4\},\{u_5,u_6,u_7\})\ne \emptyset$ and assume that $u_1$ is adjacent to $u_5$. Then $u_1$ is not adjacent to $\{w_1,w_2,w_3,u_2,\ldots ,u_7\}$ in G, and $v_1u_2v_2u_3v_3u_4w_2u_6v_1$ and $u_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $E_G(\{u_1,\ldots ,u_4\},\{u_5,u_6,u_7\})=\emptyset$, so $u_1u_5u_2u_6u_3u_7u_4v_3u_1$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction.

Now suppose that $d_{G[W]}(u')\ge 4$ for some vertex $u'\in U$, say $u'=u_1$. Since, $T_K(8)\nsubseteq G$, $w_4$ is not adjacent to $u_1$. Then without loss of generality, suppose that $u_2,\ldots ,u_5\in N_G(u_1)$. Since $T_K(8)\nsubseteq G$, $E_G(\{v_1,v_2,v_3\},\{u_2,\ldots ,u_5\})=\emptyset$. If $u_2$ is adjacent to $w_1$, then $u_2$ is not adjacent to $\{u_3,\ldots ,u_7\}$ and $v_1$ is not adjacent to $u_6$. Then $w_2u_3v_2u_4v_3u_5v_1u_6w_2$ and $u_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $u_2$ is not adjacent to $w_1$. Similarly, $u_3$, $u_4$ and $u_5$ are not adjacent to $w_1$. If $u_2$ is adjacent to $v_0$, then $v_2$ is not adjacent to $\{v_1,v_3,w_1,w_2,w_3,u_2,\ldots ,u_7\}$, and $v_1u_2v_3u_3w_1u_4w_2u_5v_1$ and $v_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $u_2$ is not adjacent to $v_0$. Similarly, $u_3$, $u_4$ and $u_5$ are not adjacent to $v_0$. By similar arguments, $u_3$, $u_4$ and $u_5$ are not adjacent to $w_3$ or $w_4$.

Hence, $u_2,\ldots ,u_5$ are not adjacent to V(T) in G, so $\delta (G[V(T)])\ge 4$ by Observation 4.3. By Lemma 7.13, G[V(T)] contains $T_K(8)$, a contradiction. Thus, $R(T_K(8),W_8)\le 15$.

Now suppose that $n\equiv 0 \pmod {4}$ and that $n\ge 12$. If G has an $S_n(1,3)$ subgraph, then the arguments above lead to contradictions. Thus, G does not contain $S_n(1,3)$ as a subgraph. Now, by Theorem 7.14, G has a subgraph $T=T_H(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-5},w_1,\ldots ,w_4\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-5},v_1w_1,w_1w_2,w_2w_3,v_2w_4\}$. Set $V=\{v_3,\ldots ,v_{n-5}\}$ and let $U=V(G)-V(T)=\{u_1,\ldots ,u_{n-1}\}$. Then $|V|=n-7$ and $|U|=n-1$. Since $T_K(n)\nsubseteq G$, $w_2$ is not adjacent in G to $\{w_4\}\cup U$. Since $S_n(1,3)\nsubseteq G$, $v_0$ is not adjacent to $\{w_4\}\cup U$.

If $\delta ({\overline{G}}[U])\ge \frac{n-1}{2}$, then ${\overline{G}}[U]$ contains $C_8$ by Lemma 4.1 which, with $w_2$, forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[U])<\frac{n-1}{2}$, and $\Delta (G[U])\ge \lfloor \frac{n-1}{2} \rfloor \ge 5$. Without loss of generality, assume that $u_2,\ldots ,u_6\in N_G(u_1)$. Since $T_K(n)\nsubseteq G$, $v_1$, $v_2$ and V are not adjacent to $\{u_2,\ldots ,u_6\}$, and $w_1$ and $w_2$ are not adjacent to $u_1$.

Now, if $u_2$ is adjacent to $w_1$, then $u_2$ is not adjacent to $\{w_3, w_4\}\cup U{\setminus } \{u_1\}$, since $T_K(n)\nsubseteq G$, so $v_0u_3v_1u_4v_2u_5v_3u_6v_0$ and $u_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $u_2$ is not adjacent to $w_1$. Similarly, $u_3,\ldots ,u_6$ are not adjacent to $w_1$. If $u_2$ is adjacent to $w_3$ in G, then $v_0$ is not adjacent to $w_1,w_2,w_3$, and $d_{G[U{\setminus } \{u_1,u_2\}]}(u_i)\le n-6$ for $i=3,\ldots ,6$, since $S_n(1,3)\nsubseteq G$. Since $T_K(n)\nsubseteq G$, $w_3$ is not adjacent to $w_1$ or $w_4$. Since $d_{G[U{\setminus } \{u_1,u_2\}]}(u_3)\le n-6$ and $d_{G[U{\setminus } \{u_1,u_2\}]}(u_4)\le n-6$, $u_3$ and $u_4$ are adjacent in ${\overline{G}}$ to at least 2 vertices in $\{u_7,\ldots ,u_{n-1}\}$. Without loss of generality, assume that $u_3$ is adjacent in ${\overline{G}}$ to $u_7$ and that $u_4$ is adjacent to $u_8$. Then $u_3u_7w_2u_8u_4w_1w_3w_4u_3$ and $v_0$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $u_2$ is not adjacent to $w_3$. Similarly, $u_3,\ldots ,u_6$ are not adjacent to $w_4$.

Hence, $u_2,\ldots ,u_6$ are not adjacent to V(T). By Observation 4.3, $\delta (G[V(T)])\ge 4$, so G[V(T)] contains $T_K(n)$ by Lemma 7.13, a contradiction. Thus, $R(T_K(n),W_8)\le 2n-1$ for $n\equiv 0 \pmod {4}$. This completes the proof. $\square$

Theorem 7.17

If $n\ge 8$, then $R(T_L(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be a graph with no $T_L(n)$ subgraph whose complement ${\overline{G}}$ does not contain $W_8$. Suppose that $n\not \equiv 0 \pmod {4}$ and that G has order $2n-1$. By Theorem 6.5, G has a subgraph $T=S_n(1,3)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_2w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_L(n)\nsubseteq G$, $v_1$ is not adjacent to $U\cup V$, and $d_{G[U]}(v_i)\le n-7$ for each $v_i\in V$. Now, if $\delta (G[U])\ge \frac{n-1}{2}$, then ${\overline{G}}[U]$ contains $C_8$ by Lemma 4.1 which, with $v_1$, forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[U])<\frac{n-1}{2}$, and $\Delta (G[U])\ge \lfloor \frac{n-1}{2} \rfloor$.

Let $U=\{u_1,\ldots ,u_{n-1}\}$ and without loss of generality assume that $d_{G[U]}(u_1)\ge \lfloor \frac{n-1}{2} \rfloor \ge 4$ and that $u_2,\ldots ,u_5\in N_{G[U]}(u_1)$. Now if $E_G(V,N_{G[U]}(u_1))=\emptyset$, then 4 vertices from V, 4 vertices from $N_{G[U]}(u_1)$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $E_G(V,N_{G[U]}(u_1))\ne \emptyset$. Assume without loss of generality that $v_2$ is adjacent to $u_2$. Since $T_L(n)\nsubseteq G$, $v_2$ is not adjacent to $U{\setminus } \{u_1,u_2\}$. Since $d_{G[U]}(v_i)\le n-7$ for each $v_i\in V$, $v_5$ is non-adjacent to at least one of $u_6,\ldots ,u_{n-1}$, say $u_6$. Now if $E_G(\{v_3,v_4,v_5\},\{u_3,u_4,u_5\})=\emptyset$, then $v_2u_3v_3u_4v_4u_5v_5u_6v_2$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus assume, say, that $v_3$ is adjacent to $u_3$ in G; then $v_3$ is not adjacent to $U{\setminus } \{u_1,u_3\}$. Again, if $E_G(\{v_4,v_5\},\{u_4,u_5\})=\emptyset$, then $v_2u_7v_3u_4v_4u_5v_5u_6v_2$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus assume, say, that $v_4$ is adjacent to $u_4$, then $v_4$ is not adjacent to $U{\setminus } \{u_1,u_4\}$. If $v_5$ is not adjacent to $u_5$, then $v_2u_7v_3u_2v_4u_5v_5u_6v_2$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $v_5$ is adjacent to $u_5$, so $v_5$ is not adjacent to $U{\setminus } \{u_1,u_5\}$, and $v_2u_7v_3u_2v_4u_3v_5u_6v_2$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction.

Hence, $R(T_L(n),W_8)\le 2n-1$ for $n\not \equiv 0 \pmod {4}$.

Now, suppose that $n\equiv 0 \pmod {4}$ and that G has order $2n-1$. Suppose first that $n=8$. By Theorem 7.14, G has a subgraph $T=T_H(8)$. Let $V(T)=\{v_0,\ldots ,v_3,w_1,\ldots ,w_4\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_3,v_1w_1,w_1w_2,w_2w_3,v_2w_4\}$. Set $U=V(G)-V(T)=\{u_1,\ldots ,u_7\}$; then $|U|=7$. Since $T_L(8)\nsubseteq G$, neither $v_1$ nor $v_2$ are adjacent to U, and $d_{G[U]}(v_3)\le 1$. Furthermore, $v_1$ is not adjacent to $w_4$, and $v_2$ is not adjacent to $w_1$ or $w_3$. Let $W=\{w_4\}\cup U$; then $|W|=8$. If $\delta ({\overline{G}}[W])\ge 4$, then ${\overline{G}}[W]$ contains $C_8$ by Lemma 4.1 which with $v_1$ forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[W])<3$ and $\Delta (G[W])\ge 4$.

Now, suppose that $d_{G[W]}(w_4)\ge 4$ and assume without loss of generality that $u_1,\ldots ,u_4\in N_G(w_4)$. Then $v_2$ is not adjacent to $v_1, v_3, w_1, w_2$ and $d_{G[U]}(u_i)\le 1$ for $1\le i\le 4$, since $T_L(8)\nsubseteq G$. Since $d_{G[U]}(v_3)\le 1$, assume without loss of generality that $v_3$ is not adjacent to $u_3$ or $u_4$. Now, suppose that $E_G(\{u_1,\ldots ,u_4\},\{u_5,u_6,u_7\})\ne \emptyset$ and assume, say, that $u_1$ is adjacent to $u_5$. Then $u_1$ is not adjacent to $\{v_3,w_1,w_2,w_3,u_2,\ldots ,u_7\}$. Since $T_L(8)\nsubseteq G$, at least one of $w_1$ and $w_2$ is adjacent in ${\overline{G}}$ to $u_2$, $u_3$ and $u_4$, say $w_1$, so $v_1u_2w_1u_3v_3u_4v_2u_6v_1$ and $u_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $E_G(\{u_1,\ldots ,u_4\},\{u_5,u_6,u_7\})=\emptyset$. Then $u_1u_5u_2u_6u_3u_7u_4v_2u_1$ and $v_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $d_{G[W]}(u')\ge 4$ for some vertex of $u'\in U$, say $u'=u_1$.

Suppose that $w_4$ is adjacent to $u_1$. Then without loss of generality, assume that $u_1$ is adjacent to $u_2$, $u_3$ and $u_4$. Since $T_L(8)\nsubseteq G$, neither $v_0$ nor $w_4$ is adjacent to $w_1$ or $w_2$, and $w_4$ is not adjacent to $\{v_1, v_3\}\cup U{\setminus } \{u_1\}$. If $E_G(\{u_2,u_3,u_4\},\{u_5,u_6,u_7\})\ne \emptyset$, then, say, $u_2$ is adjacent to $u_5$ and is thus not adjacent to $\{v_0,v_3,w_1,w_2,w_3,u_3,u_4,u_6,u_7\}$, so $w_1v_0w_2w_4u_3v_1u_4v_2w_1$ and $u_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus $E_G(\{u_1,\ldots ,u_4\},\{u_5,u_6,u_7\}=\emptyset$. Let $X=\{v_1,u_2,u_3,u_4\}$ and $Y=\{v_3,u_5,u_6,u_7\}$. Since $d_{G[U]}(v_3)\le 1$, ${\overline{G}}[X\cup Y]$ contains $C_8$ by Lemma 4.5 which, with $w_4$, forms $W_8$, a contradiction.

Thus, $u_1$ is not adjacent to $w_4$ so assume without loss of generality that $u_2,\ldots ,u_5\in N_G(u_1)$. Since G does not contain $T_L(8)$, $d_{G[V(T)]}(u_i)\le 1$ for $2\le i\le 5$. If $u_2$ is adjacent to $w_4$, then $u_2$ is not adjacent to $V(G){\setminus } \{u_1,w_4\}$ in G. Since $d_{G[U]}(v_3)\le 1$, that $v_3$ is not adjacent to, say, $u_3$ or $u_4$. Since $d_{G[V(T)]}(u_i)\le 1$ for $2\le i\le 5$, $u_4$ and $u_5$ are each adjacent in ${\overline{G}}$ to at least 2 of $w_1,w_2,w_3$, so some $w_i\in \{w_1,w_2,w_3\}$ is adjacent in ${\overline{G}}$ to both $u_4$ and $u_5$. Therefore, $u_3v_3u_4w_iu_5v_2u_6v_1u_3$ and $u_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $u_2$ is not adjacent to $w_4$. Similarly, $u_3, u_4, u_5$ are not adjacent to $w_4$. Similar arguments show that $u_2,\ldots ,u_5$ are not adjacent to $w_1$ or $w_2$.

Now, if $u_2$ is adjacent to any other vertex of V(T), then $u_2$ is not adjacent to $\{u_3,u_4,u_5\}$, so $u_3w_1u_4w_4u_5v_2u_6v_1u_3$ and $u_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Hence, $u_2$ is not adjacent to V(T) and, similarly, $u_3,u_4,u_5$ are not adjacent to V(T). Therefore, by Observation 4.3, $\delta (G[V(T)])\ge 4$. By Lemma 7.13, G[V(T)] contains $T_L(8)$, a contradiction. Thus, $R(T_L(8),W_8)\le 15$.

Now suppose that $n\ge 12$. If G contains $S_n(1,3)$, then the previous arguments above lead to contradictions. Thus, G does not contain $S_n(1,3)$. By Theorem 6.8, G has a subgraph $T=T_C(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_2w_2,v_2w_3\}$. Set $U=V(G)-V(T)=\{u_1,\ldots ,u_{n-1}\}$; then $|U|=n-1$.

Suppose that $w_2$ is not adjacent to U. If $\delta ({\overline{G}}[U])\ge \frac{n-1}{2}$, then G contains $C_8$ by Lemma 4.1 and, with $w_2$ as hub, forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U])<\frac{n-1}{2}$ and so $\Delta (G[U])\ge \lfloor \frac{n-1}{2} \rfloor \ge 5$. Without loss of generality, assume that $u_2,\ldots ,u_6\in N_G(u_1)$. Since $S_n(1,3)\nsubseteq G$, $u_2,\ldots ,u_6$ are not adjacent to $V(T){\setminus } \{v_0\}$. If $u_2$ is adjacent to $v_0$, then since $S_n(1,3)\nsubseteq G$, $u_3,\ldots ,u_6$ are not adjacent to $\{u_7,\ldots ,u_{n-1}\}$, so $u_3u_7u_4u_8u_5u_9u_6u_{10}u_3$ and $w_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Thus, $u_2$ is not adjacent to $v_0$ and, similarly, $u_3,\ldots ,u_6$ are also not adjacent to $v_0$. Hence, $u_2,\ldots ,u_6$ are not adjacent to V(T). Therefore, by Observation 4.3, $\delta (G[V(T)])\ge n-4$, so G[V(T)] contains $T_L(n)$ by Lemma 7.13, a contradiction.

Thus some vertex of U, say $u_{n-1}$, is adjacent to $w_2$. Set $U'=U{\setminus } \{u_{n-1}\}$; then $|U'|=n-2$. Since $T_L(n)\nsubseteq G$, $u_{n-1}$ is not adjacent to $U'$ in G. Now, if $\delta ({\overline{G}}[U'])\ge \frac{n-2}{2}$, then ${\overline{G}}[U']$ contains $C_8$ by Lemma 4.1 which, with $u_{n-1}$, forms $W_8$, a contradiction. Thus, $\delta ({\overline{G}}[U'])\le \frac{n-2}{2}-1$, and $\Delta (G[U'])\ge \frac{n-2}{2}\ge 5$. Without loss of generality, assume that $u_2,\ldots ,u_6\in N_G(u_1)$ and repeat the above arguments to prove that $u_2,\ldots ,u_6$ are not adjacent to V(T). Therefore, $\delta (G[V(T)])\ge n-4$ by Observation 4.3, so G[V(T)] contains $T_L(n)$ by Lemma 7.13, a contradiction.

Therefore, $R(T_L(n),W_8)\le 2n-1$ for $n\equiv 0 \pmod {4}$, which completes the proof. $\square$

Theorem 7.18

If $n\ge 9$, then $R(T_M(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $T_M(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.4, G has a subgraph $T=S_n(4)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,v_1w_3\}$. Set $V=\{v_2,v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)=\{u_1,\ldots ,u_{n-1}\}$; then $|V|=n-5$ and $|U|=n-1$. Since $T_M(n)\nsubseteq G$, $w_1$, $w_2$ and $w_3$ are not adjacent to any vertex of $U\cup V$ in G.

Now, suppose that some vertex in V is adjacent to at least 4 vertices of U in G, say $v_2$ to $u_1,\ldots ,u_4$. Then $u_1,\ldots ,u_4$ are not adjacent to other vertices in U. Then $u_1u_5u_2u_6u_3u_7u_4u_8u_1$ and $w_1$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, each vertex in V is adjacent to at most three vertices of U in G. Choose any 8 vertices of U. By Corollary 4.8, ${\overline{G}}[U\cup V]$ contains $C_8$ which together with $w_1$ gives $W_8$ in ${\overline{G}}$, a contradiction.

Thus, $R(T_M(n),W_8)\le 2n-1$ for $n\ge 9$. This completes the proof. $\square$

Theorem 7.19

If $n\ge 9$, then

$$\begin{aligned} R(T_N(n),W_8)={\left\{ \begin{array}{ll} 2n-1 &{} \hbox { if}\ n\not \equiv 0 \pmod {4};\\ 2n &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order 2n if $n\equiv 0 \pmod {4}$ and of order $2n-1$ if $n\not \equiv 0 \pmod {4}$. Assume that G does not contain $T_N(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.6, G has a subgraph $T=T_A(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,w_1w_3\}$. Set $V=\{v_2,v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)=\{u_1,\ldots ,u_j\}$, where $j=n-1$ if $n\not \equiv 0 \pmod {4}$ and $j=n$ otherwise. Since $T_N(n)\nsubseteq G$, $w_2$ is not adjacent to $U\cup V$ in G. If each $v_i\in V$ is adjacent to at most three vertices of U in G, then by Corollary 4.8, ${\overline{G}}[U\cup V]$ contains $C_8$ which with $w_2$ gives $W_8$ in ${\overline{G}}$, a contradiction. Therefore, some vertex in V, say $v_2$, is adjacent to at least four vertices of U in G, say $u_1,\ldots ,u_4$. If none of these is adjacent to other vertices of U in G, then $u_1u_5u_2u_6u_3u_7u_4u_8u_1$ and $w_2$ form $W_8$ in ${\overline{G}}$, a contradiction.

Therefore, assume that $u_1$ is adjacent to $u_5$ in G. Since $T_N(n)\nsubseteq G$, $u_2,u_3,u_4$ are not adjacent to $\{u_6,\ldots ,u_j\}$ in G. For $n=9$ and $n=10$, $\{v_3,\ldots ,v_{n-4}\}$ is not adjacent to $\{u_5,\ldots ,u_{n-1}\}$ or else G will contain $T_N(n)$ with $v_2$ and $v_0$ being the vertices of degree $n-5$ and 3, respectively. However, $v_3u_5v_4u_6u_2u_7u_3u_8v_3$ and $w_2$ form $W_8$ in ${\overline{G}}$, a contradiction. For $n\ge 11$, if $v_2$ is not adjacent to $\{u_6,\ldots ,u_j\}$ in G, then $v_2u_6u_2u_7u_3u_8u_4u_9v_2$ and $w_2$ form $W_8$ in ${\overline{G}}$, a contradiction. Therefore, assume that $v_2$ is adjacent to $u_6$ in G. Then $u_6$ is not adjacent to $\{u_7,\ldots ,u_j\}$ in G, and $u_2u_7u_3u_8u_4u_9u_6u_{10}u_2$ and $w_2$ form $W_8$ in ${\overline{G}}$, again a contradiction.

Thus, $R(T_N(n),W_8)\le 2n$ for $n\equiv 0 \pmod {4}$ and $R(T_N(n),W_8)\le 2n-1$ for $n\not \equiv 0 \pmod {4}$. $\square$

Theorem 7.20

If $n\ge 9$, then $R(T_P(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $T_P(n)$ and that ${\overline{G}}$ does not contain $W_8$. Suppose $n\not \equiv 0 \pmod {4}$. By Theorem 6.6, G has a subgraph $T=T_A(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,w_1w_3\}$. Set $V=\{v_2,v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_P(n)\nsubseteq G$, $w_1$ is not adjacent to any vertex of $U\cup V$ in G. If each $v_i$ in V is adjacent to at most three vertices of U in G, then by Corollary 4.8, ${\overline{G}}[U\cup V]$ contains $C_8$ which with $w_1$ gives $W_8$ in ${\overline{G}}$, a contradiction. Therefore, some vertex in V, say $v_2$, is adjacent to at least four vertices of U in G, say $u_1,\ldots ,u_4$. For $n=9$ and $n=10$, G contains $T_P(9)$ and $T_P(10)$ with edge set $\{u_1v_2,u_2v_2,u_3v_2,v_2v_0,v_0v_1,v_0v_3,v_1w_1,v_1w_2\}$ and $\{u_1v_2,u_2v_2,u_3v_2,u_4v_2,v_2v_0,v_0v_1,v_0v_3,v_1w_1,v_1w_2\}$, respectively. For $n\ge 11$, each of $u_1,\ldots ,u_4$ is adjacent to at most two remaining vertices in U. Then by Corollary 4.7, ${\overline{G}}[U]$ contains $C_8$ which with $w_1$ gives $W_8$ in ${\overline{G}}$, a contradiction.

Suppose that $n\equiv 0 \pmod {4}$. By Theorem 7.18, G contains a subgraph $T=T_M(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-5},w_1,\ldots ,w_4\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-5},v_1w_1,v_1w_2,v_1w_3,w_1w_4\}$. Let $V=\{v_2,v_3,\ldots ,v_{n-5}\}$ and $U=V(G)-V(T)$; then $|V|=n-6$ and $|U|=n-1$. Since $T_P(n)\nsubseteq G$, $w_1$ is not adjacent to $\{v_0,w_2,w_3\}\cup U$ in G, and so $d_{G[U]}(w_2)\le 1$, $d_{G[U]}(w_3)\le 1$ and $d_{G[U]}(v) \le n-7$ for any vertex $v\in V$. Now, if G contains a subgraph $T_A(n)$, then arguments similar to those used for the case $n\not \equiv 0 \pmod {4}$ above can be used. Therefore, G contains no $T_A(n)$. Then $v_0$ is not adjacent to $\{w_2,w_3\}\cup U$ in G.

Suppose that some vertex $v\in V$ is not adjacent to $w_1$ in G. Let X be any four vertices in U that are not adjacent to v in G and set $Y=\{v,v_0,w_2,w_3\}$. By Lemma 4.5, ${\overline{G}}[X\cup Y]$ contains $C_8$ which with $w_1$ gives $W_8$ in ${\overline{G}}$, a contradiction. Therefore, each vertex of V is adjacent to $w_1$ in G. Since $T_P(n)\nsubseteq G$, $w_4$ is adjacent to at most $n-7$ vertices of U in G. Since $T_A(n)\nsubseteq G$, $w_2$ and $w_3$ are not adjacent in G. Now, if $w_4$ is adjacent to both $w_2$ and $w_3$ in G, then $w_4$ is not adjacent to $v_0$ in G since $T_P(n)\nsubseteq G$. Let X be any four vertices of U that are not adjacent to $w_4$ in G and let $V=\{w_1,\ldots ,w_4\}$. By Lemma 4.5, ${\overline{G}}[X\cup Y]$ contains $C_8$ which with $w_1$ gives $W_8$ in ${\overline{G}}$, a contradiction. Therefore, $w_4$ is non-adjacent to either $w_2$ or $w_3$ in G, say $w_2$. Since $d_{G[U]}(w_2)\le 1$ and $d_{G[U]}(w_4)\le n-7$, there is a set X of four vertices in U that are not adjacent to both $w_2$ and $w_4$ in G. Let $Y=\{v_0,w_1,w_3,w_4\}$. By Lemma 4.5, ${\overline{G}}[X\cup Y]$ contains $C_8$ which with $w_1$ gives $W_8$ in ${\overline{G}}$, again a contradiction.

In either case, $R(T_P(n),W_8)\le 2n-1$ for $n\ge 9$ and this completes the proof. $\square$

Theorem 7.21

If $n\ge 9$, then $R(T_Q(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $T_Q(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.4, G has a subgraph $T=S_n(4)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_1w_2,v_1w_3\}$. Set $V=\{v_2,v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_Q(n)\nsubseteq G$, G[V] are independent vertices and not adjacent to U.

Suppose that $n\ge 10$. Then $|V|\ge 5$ and $|U|\ge 9$, so by Observation 4.3, ${\overline{G}}$ contains $W_8$, a contradiction. If $n=9$, then $|V|=4$ and $|U|=8$. By Lemma 4.4, G[U] is $K_8$ or $K_8-e$. Since $T_Q(9) \nsubseteq G$, T is not adjacent to U, and $\delta (G[V(T)]\ge 5$. As $v_2,\ldots ,v_5$ are independent in G, they are each adjacent to all other vertices in G[V(T)]. Therefore, G[V(T)] contains $T_Q(9)$ with $v_2$ and $v_0$ as the vertices of degree 4, a contradiction.

Thus, $R(T_Q(n),W_8)\le 2n-1$ for $n\ge 9$, which completes the proof. $\square$

Theorem 7.22

If $n\ge 9$, then $R(T_R(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $T_R(n)$ and that ${\overline{G}}$ does not contain $W_8$. By Theorem 6.8, G has a subgraph $T=T_C(n)$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,v_2w_2,v_2w_3\}$. Set $V=\{v_3,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)=\{u_1,\ldots ,u_{n-1}\}$; then $|V|=n-6$ and $|U|=n-1$. Since $T_R(n)\nsubseteq G$, $w_1$ is not adjacent in G to any vertex of $U\cup V$. If $\delta ({\overline{G}}[U\cup V])\ge \lceil \frac{2n-7}{2}\rceil$, then ${\overline{G}}[U\cup V]$ contains $C_8$ by Lemma 4.1 which with $w_3$ forms $W_8$, a contradiction. Therefore, $\delta ({\overline{G}}[U\cup V])\le \lceil \frac{2n-7}{2}\rceil -1$, and $\Delta (G[U\cup V])\ge \lfloor \frac{2n-7}{2}\rfloor =n-4$. Now, there are two cases to be considered.

Case 1: One of the vertices of V, say $v_3$, is a vertex of degree at least $n-4$ in $G[U\cup V]$.

Note that in this case, there are at least 3 vertices from U, say $u_1,u_2,u_3$, that are adjacent to $v_3$ in G. Suppose that $v_3$ is also adjacent to a in G, where a is a vertex in $U\cup V$. Since $T_R(n)\nsubseteq G$, these 4 vertices are independent and are not adjacent to any other vertices of U. Since $n\ge 9$, U contains at least 4 other vertices, say $u_5,\ldots ,u_8$, so $u_1u_5u_2u_6u_3u_7au_8u_1$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction.

Case 2: Some vertex $u\in U$ has degree at least $n-4$ in $G[U\cup V]$.

Since $T_R(n)\nsubseteq G$, u is not adjacent to any vertex of V in G. Therefore, u must be adjacent to at least $n-4$ vertices of U in G. Without loss of generality, suppose that $u_1,\ldots ,u_{n-4}\in N_{G[U]}(u)$. Note that V is not adjacent to $N_{G[U]}(u)$, or else it will form $T_R(n)$ in G, a contradiction. If $n\ge 10$, then any 4 vertices from $N_{G[U]}(u)$ and any 4 vertices from V form $C_8$ in ${\overline{G}}$ which with $w_3$ forms $W_8$, a contradiction. Suppose that $n=9$ and let the remaining two vertices be $u_6$ and $u_7$. If either $u_6$ or $u_7$ is non-adjacent to any two vertices of $\{u_1,\ldots ,u_5\}$ in G, say $u_6$ is not adjacent to $u_1$ and $u_2$ in G, then $u_1u_6u_2v_3u_3v_4u_4v_5u_1$ and $w_3$ form $W_8$ in ${\overline{G}}$, a contradiction. So, both $u_6$ and $u_7$ are adjacent to at least 4 vertices of $\{u_1,\ldots ,u_5\}$ in G. Since $T_R(9)\nsubseteq G$, T cannot be adjacent to U, and $\delta (G[V(T)]\ge 5$. As both $v_2$ and $w_3$ are not adjacent to $v_3$, $v_4$ or $v_5$ in G, they are adjacent to all other vertices in G[V(T)]. Similarly, since $v_3$ is not adjacent to $v_2$ or $w_3$ in G, $v_3$ is adjacent to $w_1$ or $w_2$ in G. Without loss of generality, assume that $v_3$ is adjacent to $w_1$. Then G[V(T)] contains $T_R(9)$ with edge set $\{v_2w_2,v_2v_1,v_2v_0,v_0v_4,v_0v_5,v_2w_3,v_2w_1,w_1v_3\}$, a contradiction.

In either case, $R(T_R(n),W_8)\le 2n-1$. $\square$

Theorem 7.23

If $n\ge 9$, then $R(T_S(n),W_8)=2n-1$.

Proof

Lemma 7.1 provides the lower bound, so it remains to prove the upper bound. Let G be any graph of order $2n-1$. Assume that G does not contain $T_S(n)$ and that ${\overline{G}}$ does not contain $W_8$. Suppose $n\not \equiv 0 \pmod {4}$. By Theorem 6.4, G has a subgraph $T=S_n[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-4},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-4},v_1w_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-4}\}$ and $U=V(G)-V(T)$; then $|V|=n-5$ and $|U|=n-1$. Since $T_S(n)\nsubseteq G$, G[V] are independent vertices and are not adjacent to U. If $n\ge 10$, then $|V|\ge 5$ and $|U|\ge 9$, so by Observation 4.3, ${\overline{G}}$ contains $W_8$, a contradiction. Suppose that $n=9$. Then $|V|=4$ and $|U|=8$. By Lemma 4.4, G[U] is $K_8$ or $K_8-e$. Since $T_S(9) \nsubseteq G$, T is not adjacent to U, and $\delta (G[V(T)]\ge 5$. As $v_2,\ldots ,v_5$ are independent in G, they are adjacent to all other vertices in G[V(T)], and so G[V(T)] contains $T_S(9)$ with edge set $\{v_0v_1,v_0v_2,v_1v_4,v_1v_5,v_2w_1,v_2w_2,v_2w_3,v_3w_1\}$.

Now suppose that $n\equiv 0 \pmod {4}$. By Theorem 6.4, G has a subgraph $T=S_{n-1}[4]$. Let $V(T)=\{v_0,\ldots ,v_{n-5},w_1,w_2,w_3\}$ and $E(T)=\{v_0v_1,\ldots ,v_0v_{n-5},v_1w_1,w_1w_2,w_1w_3\}$. Set $V=\{v_2,\ldots ,v_{n-5}\}$ and $U=V(G)-V(T)$; then $|V|=n-6$ and $|U|=n$. Since $T_S(n)\nsubseteq G$, G[V] is not adjacent to U. Since $|V|=n-6>4$, by Observation 4.3, $\Delta ({\overline{G}}[U])\le 3$ and $\delta (G[U])\ge n-4$ since $W_8\nsubseteq {\overline{G}}$. By Lemma 6.3, either G[U] is $K_{4,\ldots ,4}$ and contains $T_S(n)$ or G[U] contains $S_n[4]$ and the arguments from the $n\not \equiv 0\pmod {4}$ case above lead to a contradiction.

Thus, $R(T_S(n),W_8)\le 2n-1$ for $n\ge 9$, which completes the proof.$\square$