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.
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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
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.
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 \(n\ge 5\),
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 \(n\ge 6\),
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.
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.
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
for each \(T_n\in \{S_n[4],S_n(1,3),T_A(n),T_B(n)\}\) and \(T_n'\in \{T_C(n),S_n(3,1)\}\).
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.
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
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.
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.
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 \(v_2\).
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. \(\square \)
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
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. \(\square \)
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
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
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
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
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
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
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
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
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 \(n\ge 8\),
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 \(n\ge 8\),
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
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.\(\square \)
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Chng, Z.Y., Britz, T., Tan, T.S. et al. The Ramsey Numbers for Trees of Large Maximum Degree Versus the Wheel Graph \(W_8\). Bull. Malays. Math. Sci. Soc. 47, 134 (2024). https://doi.org/10.1007/s40840-024-01733-0
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DOI: https://doi.org/10.1007/s40840-024-01733-0