Strong Ramsey Numbers of All Sets of Three Graphs of Size 3

Abstract

The strong Ramsey number $\overline{R}(F,H)$ of two graphs F and H without isolated vertices is the minimum positive integer n such that every red-blue (edge) coloring of $K_n$ results in two edge-disjoint monochromatic subgraphs, one isomorphic to F and the other isomorphic to H. These numbers have been determined for all such pairs $F,H$ of graphs of size 3. For a set S of $k\ge 2$ graphs without isolated vertices, the strong Ramsey number $\overline{R}\left(S\right)$ of S is the minimum positive integer n such that every red-blue coloring of $K_n$ results in k pairwise edge-disjoint monochromatic subgraphs isomorphic to the graphs in S. The number $\overline{R}\left(S\right)$ is determined for all sets of three graphs of size 3. Other results and questions on strong Ramsey numbers are also presented.

1. Introduction

For two graphs F and H without isolated vertices, the Ramsey number$R(F,H)$ of F and H is the minimum positive integer n such that for every red-blue (edge) coloring of the complete graph $K_n$ of order n, there exists a subgraph isomorphic to F all of whose edges are colored red (a red F) or a subgraph isomorphic to H all of whose edges are colored blue (a blue H). The (diagonal) Ramsey number $R\left(F\right)$ of a graph F without isolated vertices is the minimum positive integer n such that every red-blue coloring of $K_n$ results in a subgraph of $K_n$ isomorphic to F all of whose edges are colored the same (a monochromatic F). Due to the numerous intriguing and highly nontrivial problems occurring in Ramsey theory, the study of Ramsey numbers has attracted the attention of many researchers and has been one of most popular areas of research in graph theory (see refs. [1,2,3,4,5,6], for example). Many variations of Ramsey numbers have been introduced and studied which provide different approaches to the investigation of Ramsey numbers (see [7,8,9,10,11,12], for example).

The Ramsey number $R(F,H)$ of two graphs F and H without isolated vertices is the smallest positive integer n of a complete graph $K_n$ such that every red-blue coloring of $K_n$ yields a monochromatic subgraph that is either isomorphic to F or isomorphic to H (not necessarily both). This observation gave rise to a variation of Ramsey number $R(F,H)$ in [13] that requires a strong property. The strong Ramsey number $\overline{R}(F,H)$ of two non-isomorphic graphs F and H without isolated vertices is the minimum positive integer n such that every red-blue (edge) coloring of $K_n$ results in two edge-disjoint monochromatic subgraphs, one isomorphic to F and the other isomorphic to H. Consequently, the strong Ramsey number $\overline{R}(F,H)$ requires that every red-blue coloring of $K_{\overline{R}(F,H)}$ results in two edge-disjoint monochromatic subgraphs either of different colors or the same color, one isomorphic to F and the other isomorphic to H. As mentioned in [13], since the Ramsey number $R(F,H)$ exists for every pair $F,H$ of graphs by a result of Ramsey [14], the strong Ramsey number $\overline{R}(F,H)$ exists as well. In [13], the strong Ramsey number is determined for all pairs of graphs of size 2 or 3 without isolated vertices; while in [15], the strong Ramsey number is determined for all pairs of graphs of size 4 without isolated vertices.

The concept of the strong Ramsey number of two non-isomorphic graphs can be extended to sets of two or more non-isomorphic graphs. For a set $S=\{F_1,F_2,\dots ,F_k\}$ of $k\ge 2$ pairwise non-isomorphic graphs without isolated vertices, the strong Ramsey number $\overline{R}\left(S\right)$ of S is the minimum positive integer n such that every red-blue coloring of $K_n$ results in k pairwise edge-disjoint monochromatic subgraphs $F_1,F_2,\dots ,F_k$. In particular, if $k=2$, then $\overline{R}\left(S\right)=\overline{R}(F_1,F_2)$ is the strong Ramsey number of $F_1$ and $F_2$. First, we present some preliminary information.

Proposition 1. 

For every set S of pairwise non-isomorphic graphs without isolated vertices, the strong Ramsey number $\overline{R}\left(S\right)$ of S exists and $\overline{R}\left(S\right)\ge max\{R\left(F\right):F\in S\}.$

Proof. 

Let $S=\{F_1,F_2,\dots ,F_k\}$ be a set of $k\ge 2$ pairwise non-isomorphic graphs without isolated vertices. For $1\le i\le k$, let $R\left(F_i\right)=n_i$ and let $n={\sum }_{i=1}^k{n}_i$. Let there be given a red-blue coloring of $K_n$. Because $n\ge n_k$, there is a monochromatic subgraph $F_k$ in $K_n$. Since deletion of the $n_k$ vertices of $F_k$ from $K_n$ results in $K_{n-n_k}$, where $n-n_k\ge n_{k-1}$, there is a monochromatic subgraph $F_{k-1}$ in $K_{n-n_k}$ that is edge-disjoint from $F_k$. Continuing in this manner, we obtain pairwise edge-disjoint monochromatic subgraphs $F_1,F_2,\dots ,F_k$. Thus, $\overline{R}\left(S\right)$ exists.

Next, let $\overline{R}\left(S\right)=n$. Since every red-blue coloring of $K_n$ produces pairwise edge-disjoint monochromatic subgraphs $F_1,F_2,\dots ,F_k$, it follows that $R\left(F_i\right)\le n$ for $1\le i\le k$ and so $max\left\{R\right(F):F\in S\}\le n$.   □

Observation 1. 

If $S_1$ and $S_2$ are two sets of pairwise non-isomorphic graphs without isolated vertices such that $S_1\subseteq S_2$, then $\overline{R}\left(S_1\right)\le \overline{R}\left(S_2\right)$.

Since strong Ramsey numbers have been determined in [13] for all pairs of graphs of size 2 or 3 without isolated vertices, it follows that strong Ramsey numbers have been determined for all sets of two graphs of size 3. The goal here is to determine the strong Ramsey numbers of all sets of three graphs of size 3. There are five graphs of size 3 without isolated vertices, namely the triangle $K_3$, the star $K_{1,3}$, the 4-path $P_4$, the union $P_3+K_2$ of the 3-path $P_3$ and the graph $K_2$, and the matching $3K_2$, shown in Figure 1. Here, $R\left(P_4\right)=5$, $R\left(K_{1,3}\right)=R\left(K_3\right)=R(P_3+K_2)=6$, and $R\left(3K_2\right)=8$ (see [4,16]).

Figure 1. The five graphs of size 3 without isolated vertices.

We assume that all graphs in each set under consideration are pairwise non-isomorphic and have no isolated vertices. Notation and terminology not defined here may be found in the book [17] and additional information on Ramsey numbers can be found in [2,4,5].

2. Acyclic Graphs of Size 3

The first 3-element sets of graphs we consider consist of acyclic graphs. A graph is acyclic if it has no cycles. The four acyclic graphs of size 3 without isolated vertices are $P_4,P_3+K_2$, $3K_2$, and $K_{1,3}$. It is useful to determine $\overline{R}\left(S\right)$ for the set $S=\{P_4,P_3+K_2,3K_2,K_{1,3}\}$ of all four of these graphs. First, we present a lemma.

Lemma 1. 

Every red-blue coloring of $K_7$ that results in the two edge-disjoint monochromatic subgraphs $3K_2$ and $P_4$ also results in the three pairwise edge-disjoint monochromatic subgraphs $3K_2,P_4$, and $P_3+K_2$.

Proof. 

Let there be given a red-blue coloring of $G=K_7$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_7\}$ that results in edge-disjoint monochromatic subgraphs $F_1=3K_2$ and $F_2=P_4$, where say $E\left(F_1\right)=\{v_i{v}_{i+3}:i=1,2,3\}$. Since $|V\left(F_1\right)\cup V\left(F_2\right)|\in \{6,7\}$, there are two possibilities to consider.

Case 1. $|V\left(F_1\right)\cup V\left(F_2\right)|=6$. Then $V\left(F_1\right)\cup V\left(F_2\right)=\{v_1,v_2,\dots ,v_6\}$. We may assume that $F_2=P_4$ is one of the four 4-paths $(v_1,v_2,v_4,v_5)$, $(v_1,v_2,v_4,v_3)$, $(v_1,v_2,v_3,v_4)$, $(v_1,v_2,v_3,v_5)$. We consider each of these four possibilities shown in Figure 2, where an edge in $F_1=3K_2$ is indicated by a dotted line and an edge in $F_2=P_4$ is indicated by a dashed line.

Figure 2. The four possible subgraphs $F_2=P_4$ in Case 1.

• If $F_2=(v_1,v_2,v_4,v_5)$ or $F_2=(v_1,v_2,v_4,v_3)$, then at least two of the three edges $v_7{v}_1,v_7{v}_2,v_7{v}_4$ have the same color (by the Pigeonhole Principle), say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{1,2,4\}$.
• If $F_2=(v_1,v_2,v_3,v_4)$ or $F_2=(v_1,v_2,v_3,v_5)$, then at least two of the three edges $v_7{v}_1,v_7{v}_2,v_7{v}_3$ have the same color, say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{1,2,3\}$.

In either case, let $G^{\prime }=G[\{v_1,v_2,\dots ,v_7\}-\{v_i,v_j,v_7\}]-[E\left(F_1\right)\cup E\left(F_2\right)]$. If $G^{\prime }$ contains a red edge $xy$, then there is a red $P_3+K_2=(v_i,v_7,v_j)+(x,y)$ that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $G^{\prime }$ is blue. First, suppose that $F_2=(v_1,v_2,v_4,v_5)$. In this situation, at least two of the three edges $v_7{v}_1,v_7{v}_2,v_7{v}_4$ have the same color.

⋆

If $v_7{v}_1$ and $v_7{v}_2$ are red, then $G^{\prime }=G\left[\{v_3,v_4,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. Thus, $F_2^{\prime }=(v_3,v_5,v_6,v_4)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_1,v_7,v_2)+(v_4,v_5)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_3,v_4,v_5)+(v_1,v_2)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. This is illustrated in the first graph in Figure 3, where the solid vertices represent the four vertices of $G^{\prime }$ and a solid line represents a red edge.

Figure 3. Illustrating steps in the proof of Case 1 where $F_2=(v_1,v_2,v_4,v_5)$.

⋆

If $v_7{v}_1$ and $v_7{v}_4$ are red, then $G^{\prime }=G\left[\{v_2,v_3,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. By the argument above, we may assume that $v_7{v}_2$ is blue. Thus, $F_2^{\prime }=(v_3,v_2,v_6,v_5)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_2,v_4,v_5)+(v_1,v_7)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_1,v_2,v_7)+(v_4,v_5)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. This is illustrated in the second graph in Figure 3 where the solid vertices indicate the four vertices of $G^{\prime }$.

⋆

If $v_7{v}_2$ and $v_7{v}_4$ are red, then $G^{\prime }=G\left[\{v_1,v_3,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. We may assume that $v_7{v}_1$ is blue. Then $F_2^{\prime }=(v_3,v_1,v_6,v_5)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_1,v_2,v_7)+(v_4,v_5)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_7,v_1,v_2)+(v_4,v_5)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. This is illustrated in the third graph in Figure 3 where the solid vertices indicate the four vertices of $G^{\prime }$.

If $F_2$ is one of the other three 4-paths $(v_1,v_2,v_4,v_3)$, $(v_1,v_2,v_3,v_4)$, and $(v_1,v_2,v_3,v_5)$, then we apply the proof technique employed in the situation where $F_2=(v_1,v_2,v_4,v_5)$ and use a similar argument to create pairwise edge-disjoint monochromatic subgraphs $3K_2,P_4$ and $P_3+K_2$. For example, if $F_2=P_4=(v_1,v_2,v_3,v_4)$, then at least two of the three edges $v_7{v}_1,v_7{v}_2,v_7{v}_3$ have the same color, say red.

⋆

First, suppose that $v_7{v}_1$ and $v_7{v}_2$ are red. Let $G^{\prime }=G\left[\{v_3,v_4,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$. If $G^{\prime }$ contains a red edge $xy$, then there is a red $P_3+K_2=(v_1,v_7,v_2)+(x,y)$ that is edge-disjoint from $F_1$ and $F_2$. If $G^{\prime }$ is blue, then $F_2^{\prime }=(v_3,v_5,v_6,v_4)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_1,v_7,v_2)+(v_4,v_3)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_3,v_4,v_5)+(v_1,v_2)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$.

⋆

Next, suppose that $v_7{v}_1$ and $v_7{v}_3$ are red. Let $G^{\prime }=G\left[\{v_2,v_4,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$. If $G^{\prime }$ contains a red edge $xy$, then there is a red $P_3+K_2=(v_1,v_7,v_3)+(x,y)$ that is edge-disjoint from $F_1$ and $F_2$. If $G^{\prime }$ is blue, then $F_2^{\prime }=(v_2,v_4,v_5,v_6)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_7,v_1,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_3,v_4,v_6)+(v_1,v_2)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$.

⋆

Finally, suppose that $v_7{v}_2$ and $v_7{v}_3$ are red. Let $G^{\prime }=G\left[\{v_1,v_4,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$. If $G^{\prime }$ contains a red edge $xy$, then there is a red $P_3+K_2=(v_2,v_7,v_3)+(x,y)$ that is edge-disjoint from $F_1$ and $F_2$. If $G^{\prime }$ is blue, then $F_2^{\prime }=(v_4,v_5,v_6,v_1)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_1,v_2,v_7)+(v_4,v_3)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_4,v_3,v_6)+(v_1,v_2)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$.

Case 2. $|V\left(F_1\right)\cup V\left(F_2\right)|=7$. We may assume that $F_2=P_4$ is one of the five 4-paths $(v_7,v_1,v_2,v_4)$, $(v_7,v_1,v_2,v_3)$, $(v_1,v_7,v_4,v_2)$, $(v_1,v_7,v_2,v_4)$, $(v_1,v_7,v_2,v_3)$. We consider each of these five possible 4-paths for $F_2$ shown in Figure 4.

Figure 4. The five possible subgraphs $F_2=P_4$ in Case 2.

• If $F_2=(v_7,v_1,v_2,v_4)$ or $F_2=(v_7,v_1,v_2,v_3)$, then at least two of the three edges $v_7{v}_2,v_7{v}_4,v_7{v}_5$ have the same color, say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{2,4,5\}$.
• If $F_2=(v_1,v_7,v_4,v_2)$, then at least two of the three edges $v_7{v}_2,v_7{v}_3,v_7{v}_6$ have the same color, say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{2,3,6\}$.
• If $F_2=(v_1,v_7,v_2,v_4)$, then at least two of the three edges $v_7{v}_3,v_7{v}_4,v_7{v}_6$ have the same color, say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{3,4,6\}$.
• If $F_2=(v_1,v_7,v_2,v_3)$, then at least two of the three edges $v_7{v}_3,v_7{v}_4,v_7{v}_5$ have the same color, say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{3,4,5\}$.

In either case, let $G^{\prime }=G[\{v_1,v_2,\dots ,v_7\}-\{v_i,v_j,v_7\}]-[E\left(F_1\right)\cup E\left(F_2\right)]$. If $G^{\prime }$ contains a red edge $xy$, then there is a red $P_3+K_2=(v_i,v_7,v_j)+(x,y)$ that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $G^{\prime }$ is blue. As in Case 1, an argument is given only for $F_2=(v_1,v_7,v_4,v_2)$ since the proof technique and the arguments for other choices of $F_2$ are similar. In this situation, at least two of the three edges $v_7{v}_2,v_7{v}_3,v_7{v}_6$ have the same color.

⋆

If $v_7{v}_2$ and $v_7{v}_3$ are red or $v_7{v}_2$ and $v_7{v}_6$ are red, say the former, then $G^{\prime }=G\left[\{v_1,v_4,v_5,v_6\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. Thus, $F_2^{\prime }=(v_4,v_5,v_1,v_6)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_1,v_7,v_3)+(v_2,v_4)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. If $F_2$ is blue, then there is a blue $P_3+K_2=(v_5,v_6,v_4)+(v_1,v_7)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. This is illustrated in the first graph in Figure 5 where the four solid vertices represent the vertices of $G^{\prime }$.

Figure 5. Illustrating steps in the proof of Case 2 when $F_2=(v_1,v_7,v_4,v_2)$.

⋆

If $v_7{v}_3$ and $v_7{v}_6$ are red, then $G^{\prime }=G\left[\{v_1,v_2,v_4,v_5\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. We may assume that $v_7{v}_2$ is blue. Thus, $F_2^{\prime }=(v_2,v_1,v_5,v_4)$ is a blue $P_4$ in $G^{\prime }$ that is edge-disjoint from $F_1$ and $F_2$. If $F_2$ is red, then there is a red $P_3+K_2=(v_1,v_7,v_3)+(v_2,v_4)$ that is edge-disjoint from $F_1$ and $F_2^{\prime }$. Thus, we may assume that $F_2$ is blue. If $v_5{v}_6$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_5,v_6,v_7,v_3)$ is red and $P_3+K_2=(v_7,v_1,v_2)+(v_4,v_5)$ is blue, that are edge-disjoint from $F_1$. If $v_5{v}_6$ is blue, then there is a blue $P_3+K_2=(v_1,v_7,v_2)+(v_5,v_6)$, that is edge-disjoint from $F_1$ and $F_2^{\prime }$. This is illustrated in the second graph in Figure 5 where the four solid vertices represent the vertices of $G^{\prime }$.

□

We are now prepared to determine the strong Ramsey number of the set of the four acyclic graphs of size 3 without isolated vertices.

Theorem 1. 

If $S=\{P_4,P_3+K_2,3K_2,K_{1,3}\},$ then $\overline{R}\left(S\right)=8$.

Proof. 

Since $max\{R\left(F\right):F\in S\}=R\left(3K_2\right)=8$, it follows by Proposition 1 that $\overline{R}\left(S\right)\ge 8$. It remains to show that $\overline{R}\left(S\right)\le 8$. Let there be given a red-blue coloring of $G=K_8$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_8\}$. Since $\overline{R}(P_4,3K_2)=8$ by Theorem 2, there are edge-disjoint monochromatic subgraphs $F_1=3K_2$ and $F_2=P_4$ in G, where say $E\left(F_1\right)=\{v_i{v}_{i+3}:i=1,2,3\}$. If $|V\left(F_1\right)\cup V\left(F_2\right)|\le 7$, where say $v_8\notin V\left(F_1\right)\cup V\left(F_2\right)$, then the red-blue coloring of $K_7$ with $V\left(K_7\right)=\{v_1,v_2,\dots ,v_7\}$ contains edge-disjoint monochromatic subgraphs $3K_2$ and $P_4$. By Lemma 1, there are pairwise edge-disjoint monochromatic subgraphs $3K_2,P_4$, and $P_3+K_2$ in G. Since at least three of the seven edges incident with $v_8$ in G have the same color, there is a monochromatic $K_{1,3}$ centered at $v_8$ that is edge-disjoint from $3K_2,P_4$, and $P_3+K_2$ and so $\overline{R}\left(S\right)\le 8$. Thus, we may assume that $|V\left(F_1\right)\cup V\left(F_2\right)|=8$. First, we verify the following claim.

• Claim: There are pairwise edge-disjoint monochromatic subgraphs $3K_2,P_4$, and $P_3+K_2$ in G such that some vertex v of G is incident with at most two edges in $E\left(3K_2\right)\cup E\left(P_4\right)\cup E(P_3+K_2)$.

To verify this claim, we may assume that $F_2=P_4$ is one of the six 4-paths $(v_7,v_8,v_1,v_2)$, $(v_7,v_1,v_8,v_2)$, $(v_7,v_1,v_8,v_4)$, $(v_7,v_1,v_2,v_8)$, $(v_1,v_7,v_8,v_4)$, $(v_1,v_7,v_8,v_2)$ shown in Figure 6.

Figure 6. The six possible subgraphs $F_2=P_4$ in G.

We now consider these six possible 4-paths for $F_2$. In each of these cases, at least two of the three edges $v_7{v}_2$, $v_7{v}_4$, $v_7{v}_5$ have the same color, say $v_7{v}_i$ and $v_7{v}_j$ are red where $i,j\in \{2,4,5\}$. Let $G^{\prime }=G[\{v_1,v_2,\dots ,v_8\}-\{v_i,v_j,v_7\}]-[E\left(F_1\right)\cup E\left(F_2\right)]$. If $G^{\prime }$ contains a red edge $xy$, then there is a red $P_3+K_2=(v_i,v_7,v_j)+(x,y)$ that is edge-disjoint from $F_1$ and $F_2$ and $v\in \{v_6,v_7\}$. Thus, we may assume that $G^{\prime }$ is blue. First, suppose that $F_2=P_4=(v_7,v_8,v_1,v_2)$. At least two of the three edges $v_7{v}_2,v_7{v}_4,v_7{v}_5$ have the same color, say red.

⋆

If $v_7{v}_2$ and $v_7{v}_4$ are red, then $G^{\prime }=G\left[\{v_1,v_3,v_5,v_6,v_8\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_3,v_5,v_1)$ is blue and $P_3+K_2=(v_4,v_7,v_2)+(v_1,v_8)$ is red, that are edge-disjoint from $F_1$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_3,v_1,v_5)$ and $P_3+K_2=(v_8,v_1,v_2)+(v_3,v_5)$, that are edge-disjoint from $F_1$. In either situation, let $v=v_6$. The first graph in Figure 7 illustrates this situation where the four solid vertices represent the vertices of $G^{\prime }$.

Figure 7. Illustrating the steps in the proof of Theorem 1.

⋆

If $v_7{v}_2$ and $v_7{v}_5$ are red, then $G^{\prime }=G\left[\{v_1,v_3,v_4,v_6,v_8\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_4,v_8,v_3,v_1)$ is blue and $P_3+K_2=(v_2,v_7,v_5)+(v_1,v_8)$ is red, that are edge-disjoint from $F_1$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_3,v_1,v_6)$ and $P_3+K_2=(v_7,v_8,v_4)+(v_1,v_2)$, that are edge-disjoint from $F_1$. In either situation, let $v=v_6$. The second graph in Figure 7 illustrates this situation where the four solid vertices represent the vertices of $G^{\prime }$.

⋆

If $v_7{v}_4$ and $v_7{v}_5$ are red, then $G^{\prime }=G\left[\{v_1,v_2,v_3,v_6,v_8\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_1,v_3,v_2,v_8)$ is blue and $P_3+K_2=(v_4,v_7,v_5)+(v_1,v_2)$ is red, that are edge-disjoint from $F_1$. Let $v=v_6$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_2,v_6,v_1)$ and $P_3+K_2=(v_7,v_8,v_1)+(v_2,v_3)$, that are edge-disjoint from $F_1$. Let $v=v_7$. The third graph in Figure 7 illustrates this situation where the four solid vertices represent the vertices of $G^{\prime }$.

If $F_2$ is one of the other five 4-paths $(v_7,v_1,v_8,v_2)$, $(v_7,v_1,v_8,v_4)$, $(v_7,v_1,v_2,v_8)$, $(v_1,v_7,v_8,v_4)$, $(v_1,v_7,v_8,v_2)$, then we apply the proof technique employed in the situation when $F_2=P_4=(v_7,v_8,v_1,v_2)$ and use a similar argument to create pairwise edge-disjoint monochromatic subgraphs $3K_2,P_4$ and $P_3+K_2$ such that some vertex v of G is incident with at most two edges in $E\left(3K_2\right)\cup E\left(P_4\right)\cup E(P_3+K_2)$. For example, if $F_2=P_4=(v_7,v_1,v_8,v_2)$, then at least two of the three edges $v_7{v}_2,v_7{v}_4,v_7{v}_5$ have the same color, say red.

⋆

If $v_7{v}_2$ and $v_7{v}_4$ are red, then $G^{\prime }=G\left[\{v_1,v_3,v_5,v_6,v_8\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_3,v_1,v_5)$ is blue and $P_3+K_2=(v_4,v_7,v_2)+(v_1,v_8)$ is red, that are edge-disjoint from $F_1$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_3,v_1,v_5)$ and $P_3+K_2=(v_1,v_8,v_2)+(v_3,v_5)$, that are edge-disjoint from $F_1$. In either situation, let $v=v_6$.

⋆

If $v_7{v}_2$ and $v_7{v}_5$ are red, then $G^{\prime }=G\left[\{v_1,v_3,v_4,v_6,v_8\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_4,v_8,v_3,v_1)$ is blue and $P_3+K_2=(v_2,v_7,v_5)+(v_1,v_8)$ is red, that are edge-disjoint from $F_1$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_4,v_8,v_3,v_1)$ and $P_3+K_2=(v_7,v_1,v_8)+(v_3,v_4)$, that are edge-disjoint from $F_1$. In either situation, let $v=v_6$.

⋆

If $v_7{v}_4$ and $v_7{v}_5$ are red, then $G^{\prime }=G\left[\{v_1,v_2,v_3,v_6,v_8\}\right]-[E\left(F_1\right)\cup E\left(F_2\right)]$ is blue. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_2,v_1,v_3,v_8)$ is blue and $P_3+K_2=(v_4,v_7,v_5)+(v_2,v_8)$ is red, that are edge-disjoint from $F_1$. Let $v=v_6$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_8,v_2,v_6,v_1)$ and $P_3+K_2=(v_7,v_1,v_8)+(v_2,v_3)$, that are edge-disjoint from $F_1$. Let $v=v_7$.

This completes the verification of the claim. By the claim, there is a vertex v that is incident with at least five edges in $E\left(G\right)-[E\left(3K_2\right)\cup E\left(P_4\right)\cup E(P_3+K_2)]$. Since three of these five edges have the same color, there is a monochromatic $K_{1,3}$ that is edge-disjoint from $3K_2,P_4$, and $P_3+K_2$. Hence, there are pairwise edge-disjoint monochromatic subgraphs $3K_2,P_4$, $P_3+K_2$, and $K_{1,3}$ in G. Therefore, $\overline{R}\left(S\right)\le 8$ and so $\overline{R}\left(S\right)=8$.    □

3. The 3-Element Sets of Graphs of Size 3

There are five graphs of size 3 containing no isolated vertices, namely $K_3,K_{1,3},P_4,P_3+K_2$, and $3K_2$ and so ten 3-element sets S of graphs of size 3. If $3K_2\in S$, then $max\{R\left(F\right):F\in S\}=R\left(3K_2\right)=8$, while if $3K_2\notin S$, then $max\left\{R\right(F):F\in S\}=6$. We show that $\overline{R}\left(S\right)=max\{R\left(F\right):F\in S\}$ for every 3-element set S of graphs of size 3 without isolated vertices. It is useful to know the strong Ramsey numbers of all pairs of graphs of size 3 without isolated vertices.

Theorem 2 

([13]). If $F,H\in \{K_{1,3},P_4,K_3,P_3+K_2,3K_2\},$ then

$$\overline{R}(F,H)=\left\{\begin{array}{cc}6\hfill & if F,H\in \{K_3,K_{1,3},P_4,P_3+K_2\}\hfill \\ 8\hfill & if F\in \{K_3,K_{1,3},P_4,P_3+K_2\}and H=3K_2.\hfill \end{array}\right.$$

We begin with those sets S containing $3K_2$.

3.1. The 3-Element Sets of Graphs of Size 3 Containing 3K₂

There are six sets S of graphs of size 3 containing no isolated vertices such that $3K_2\in S$, three of which consist of the acyclic graphs of size 3 and the remaining three contain the triangle $K_3$. The following is a consequence of Observation 1, Proposition 1, and Theorem 1.

Corollary 1.

If $S=\{F_1,F_2,3K_2\}$ where $F_1,F_2\in \{P_4,P_3+K_2,K_{1,3}\}$, then $\overline{R}\left(S\right)=8$.

Theorem 3. 

If $S=\{F,K_3,3K_2\}$ where $F\in \{K_{1,3},P_4,P_3+K_2\}$, then $\overline{R}\left(S\right)=8$.

Proof. 

Since $\overline{R}\left(S\right)\ge R\left(3K_2\right)=8$ by Proposition 1, it remains to show that $\overline{R}\left(S\right)\le 8$. Let there be given a red-blue coloring of $G=K_8$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_8\}$. Since $\overline{R}(K_3,3K_2)=8$ by Theorem 2, G contains edge-disjoint monochromatic subgraphs $F_1=K_3$ and $F_2=3K_2$. We may assume that $E\left(F_2\right)=\{v_i{v}_{i+3}:i=1,2,3\}$ and $F_1=K_3$ is one of the three triangles $(v_1,v_2,v_3,v_1)$, $(v_1,v_2,v_7,v_1)$, and $(v_1,v_7,v_8,v_1)$ shown in Figure 8. Since $F\in \{K_{1,3},P_4,P_3+K_2\}$, we consider these three cases.

Figure 8. The three possible subgraphs $F_1=K_3$ in G.

Case 1. $F=K_{1,3}$. Let $G_1=G[E\left(F_1\right)\cup E\left(F_2\right)]$. Since $\delta \left(G_1\right)\le 1$, there is a vertex $v\in V\left(G\right)$ that is incident with at least five edges in $E\left(G\right)-[E\left(F_1\right)\cup E\left(F_2\right)]$ and at least three of these five edges have the same color. Hence, there is a monochromatic $K_{1,3}$ that is edge-disjoint from $F_1$ and $F_2$.

Case 2. $F=P_4$. At least two of the three edges $v_8{v}_4,v_8{v}_5,v_8{v}_6$ have the same color, say $v_8{v}_4$ and $v_8{v}_5$ are red (since the arguments are similar for other choices of two red edges). If $v_4{v}_2$ or $v_4{v}_3$ is red, then there is a red $P_4$ that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $v_4{v}_2$ and $v_4{v}_3$ are blue. Regardless of the color of the edge $v_3{v}_5$, there is a monochromatic $P_4$ that is edge-disjoint from $F_1$ and $F_2$.

Case 3. $F=P_3+K_2$. First, suppose that $K_3=(v_1,v_2,v_3,v_1)$ or $K_3=(v_1,v_2,v_7,v_1)$. At least two of the three edges $v_8{v}_1,v_8{v}_2,v_8{v}_3$ have the same color, say $v_8{v}_i$ and $v_8{v}_j$ are red where $i,j\in \{1,2,3\}$. If one of the edges in $G^{\prime }=G\left[\{v_4,v_5,v_6\}\right]=K_3$ is red, then there is a red $P_3+K_2$ in G that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $G^{\prime }$ is a blue $K_3$. Let $k\in \{1,2,3\}-\{i,j\}$. If $v_8{v}_k$ is blue, then there is a blue $P_3+K_2=(v_4,v_5,v_6)+(v_k,v_8)$ that is edge-disjoint from $F_1$ and $F_2$. Hence, we may assume that $v_8{v}_k$ is red and so all of $v_8{v}_1,v_8{v}_2,v_8{v}_3$ are red. If one of $v_1{v}_5$ and $v_3{v}_5$ is red, say $v_1{v}_5$, then $P_3+K_2=(v_2,v_8,v_3)+(v_1,v_5)$ is red that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $v_1{v}_5$ and $v_3{v}_5$ are blue. Then $P_3+K_2=(v_1,v_5,v_3)+(v_4,v_6)$ is blue that is edge-disjoint from $F_1$ and $F_2$.

Next, suppose that $K_3=(v_1,v_7,v_8,v_1)$. At least two of the three edges $v_8{v}_4,v_8{v}_5,v_8{v}_6$ have the same color, say $v_8{v}_4$ and $v_8{v}_5$ are red (since the arguments are similar for other choices of two red edges). If one of the edges in $G^{\prime }=G\left[\{v_1,v_2,v_3\}\right]=K_3$ is red, then there is a red $P_3+K_2$ in G that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $G^{\prime }$ is a blue $K_3$. If one of the edges in $G^{″}=G\left[\{v_4,v_5,v_6\}\right]=K_3$ is blue, then there is a blue $P_3+K_2$ in G that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that $G^{″}$ is a red $K_3$. If $v_8{v}_3$ is red, then there is a red $P_3+K_2=(v_4,v_5,v_6)+(v_3,v_8)$ that is edge-disjoint from $F_1$ and $F_2$. Hence, we may assume that $v_3{v}_8$ is blue. For $\{i,j\}=\{4,5\}$, this is illustrated in Figure 9, where a red edge is indicated by a solid line and a blue edge is indicated by a thin line. Regardless of the color of $v_1{v}_5$, there is a monochromatic $P_3+K_2$ that is edge-disjoint from $F_1$ and $F_2$.

Figure 9. A step in the proof in Case 3 of Theorem 3.

Hence, for each $F\in \{K_{1,3},P_4,P_3+K_2\}$, there are pairwise edge-disjoint monochromatic subgraphs $F,K_3,3K_2$ in G. Therefore, $\overline{R}\left(S\right)\le 8$ and so $\overline{R}\left(S\right)=8$.   □

In summary, we have the following.

Theorem 4. 

If S is a set of three graphs of size 3 without isolated vertices containing $3K_2$, then $\overline{R}\left(S\right)=max\{R\left(F\right):F\in S\}=8$.

3.2. The 3-Element Sets of Graphs of Size 3 Not Containing 3K₂

We now turn to the four 3-element sets S of graphs of size 3 without isolated vertices such that $3K_2\notin S$. In this case, $max\left\{R\right(F):F\in S\}=6$. We show that $\overline{R}\left(S\right)=6$ for each such set S. We begin with $S=\{K_3,K_{1,3},P_4\}$ consisting of the three connected graphs of size 3.

Theorem 5. 

If $S=\{K_3,K_{1,3},P_4\},$ then $\overline{R}\left(S\right)=6$.

Proof. 

Since $max\{R\left(F\right):F\in S\}=R\left(K_3\right)=6$, it follows by Proposition 1 that $\overline{R}\left(S\right)\ge 6$. It remains to show that $\overline{R}\left(S\right)\le 6$. Let there be given a red-blue coloring of $G=K_6$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_6\}$. Since $\overline{R}(K_3,P_4)=6$ by Theorem 2, there are edge-disjoint monochromatic subgraphs $F_1=K_3$ and $F_2=P_4$ in G. If $|V\left(F_1\right)\cup V\left(F_2\right)|\le 5$, where say $v_6\notin V\left(F_1\right)\cup V\left(F_2\right)$, then at least three of the five edges $v_6{v}_i$ ($1\le i\le 5$) have the same color and so there is a monochromatic subgraph $F_3=K_{1,3}$ that is edge-disjoint from $F_1$ and $F_2$. We may therefore assume that $V\left(F_1\right)\cup V\left(F_2\right)=V\left(G\right)$. Further, we may assume that $F_1=K_3=(v_1,v_2,v_3,v_1)$ and $F_2=P_4$ is either $(v_3,v_4,v_5,v_6)$ or $(v_6,v_3,v_4,v_5)$, as shown in Figure 10. We consider these two cases.

Figure 10. The two possible subgraphs $F_2=P_4$ in G.

Case 1. $F_2=P_4=(v_3,v_4,v_5,v_6)$. Let $X=\{v_6{v}_i:i=1,2,3,4\}$. If three edges in X have the same color, then there is a monochromatic $K_{1,3}$ in G that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that exactly two edges in X are red, say $v_6{v}_1$ and $v_6{v}_i$ are red where $i\in \{2,3,4\}$. Hence, there are three possibilities. An argument is given only for $i=2$ since the proof technique is the same and the arguments for the choices $i=3$ and $i=4$ are similar.

Suppose that $v_6{v}_1$ and $v_6{v}_2$ are red. If both $v_1{v}_4$ and $v_1{v}_5$ are red, then there is a red $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_1{v}_i:i=4,5,6\}$ in G that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that at least one of $v_1{v}_4$ and $v_1{v}_5$ is blue, which results in three possibilities.

⋆

First, suppose that $v_1{v}_4$ is blue and $v_1{v}_5$ is red. Then $(v_5,v_1,v_6,v_2)$ is a red $P_4$ and $(v_1,v_4,v_6,v_3)$ is a blue $P_4$. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ where $P_4=(v_1,v_4,v_6,v_3)$ is blue and $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=1,2,5\}$ that are edge-disjoint from $F_1$. If $F_2$ is blue, then there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ where $P_4=(v_5,v_1,v_6,v_2)$ is red and $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=3,4,5\}$ that are edge-disjoint from $F_1$.

⋆

Next, suppose that $v_1{v}_4$ is red and $v_1{v}_5$ is blue. Then $(v_4,v_1,v_6,v_2)$ is a red $P_4$. If $F_2$ is red, then regardless of the color of $v_3{v}_5$, there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ that are edge-disjoint from $F_1$. For example, if $v_3{v}_5$ is red, then $P_4=(v_1,v_4,v_3,v_5)$ is red and $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=1,2,5\}$. If $v_3{v}_5$ is blue, then $P_4=(v_4,v_6,v_3,v_5)$ is blue and $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=1,2,5\}$. If $F_2$ is blue, then there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ where $P_4=(v_4,v_1,v_6,v_2)$ is red and $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=3,4,5\}$ that are edge-disjoint from $F_1$.

⋆

Finally, suppose that both $v_1{v}_4$ and $v_1{v}_5$ are blue. Then $F_2^{\prime }=(v_1,v_4,v_6,v_3)$ is a blue $P_4$. If $F_2$ is red, then $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=1,2,5\}$ is edge-disjoint from $F_2^{\prime }$ and $F_1$. If $F_2$ is blue, then $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=1,4,6\}$ is edge-disjoint from $F_2^{\prime }$ and $F_1$.

Case 2. $F_2=P_4=(v_6,v_3,v_4,v_5)$. Let $Y=\{v_6{v}_i:i=1,2,4,5\}$. If three edges in Y have the same color, then there is a monochromatic $K_{1,3}$ in G that is edge-disjoint from $F_1$ and $F_2$. Thus, we may assume that exactly two edges in Y are red, say $v_6{v}_1$ and $v_6{v}_i$ are red, where $i\in \{2,4,5\}$, which results in three possibilities. An argument is given only for $i=2$ since the proof technique is the same and the arguments for the choices $i=4$ and $i=5$ are similar.

Suppose that $v_6{v}_1$ and $v_6{v}_2$ are red. If both $v_1{v}_4$ and $v_1{v}_5$ are red, then there is a red $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_1{v}_i:i=4,5,6\}$ in G that is edge-disjoint from $F_1$ and $F_2$, for otherwise at least one of $v_1{v}_4$ and $v_1{v}_5$ is blue, which results in three possibilities.

⋆

First, suppose that $v_1{v}_4$ is blue and $v_1{v}_5$ is red. Then $(v_5,v_1,v_6,v_2)$ is a red $P_4$ and $(v_1,v_4,v_6,v_5)$ is a blue $P_4$. Regardless of the color of $F_2$, there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ that are edge-disjoint from $F_1$.

⋆

Next, suppose that $v_1{v}_4$ is red and $v_1{v}_5$ is blue. Then $(v_4,v_1,v_6,v_2)$ is a red $P_4$ and $(v_1,v_5,v_6,v_4)$ is a blue $P_4$. Regardless of the color of $F_2$, there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ that are edge-disjoint from $F_1$.

⋆

Finally, suppose that both $v_1{v}_4$ and $v_1{v}_5$ are blue. Then $(v_1,v_4,v_6,v_5)$ is a blue $P_4$. If $F_2$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $K_{1,3}$ where $P_4=(v_1,v_4,v_6,v_5)$ is blue and $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=1,2,3\}$ that are edge-disjoint from $F_1$. If $F_2$ is blue, then there are edge-disjoint blue subgraphs $P_4$ and $K_{1,3}$ where $P_4=(v_1,v_5,v_4,v_3)$ and $E\left(K_{1,3}\right)=\{v_6{v}_i:i=3,4,5\}$ that are edge-disjoint from $F_1$.

Hence, G contains pairwise edge-disjoint monochromatic subgraphs $K_3,K_{1,3},P_4$. Therefore, $\overline{R}\left(S\right)\le 6$ and so $\overline{R}\left(S\right)=6$.   □

Theorem 6. 

If $S=\{K_3,K_{1,3},P_3+K_2\},$ then $\overline{R}\left(S\right)=6$.

Proof. 

Since $max\{R\left(F\right):F\in S\}=R\left(K_3\right)=6$, it follows by Proposition 1 that $\overline{R}\left(S\right)\ge 6$. Thus, it remains to show that $\overline{R}\left(S\right)\le 6$. Let there be given a red-blue coloring of $G=K_6$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_6\}$. Since $\overline{R}(K_3,P_3+K_2)=6$ by Theorem 2, there are edge-disjoint monochromatic subgraphs $F_1=K_3$ and $F_2=P_3+K_2$ in G. We may assume that $V\left(F_1\right)=\{v_1,v_2,v_3\}$ and $F_2=P_3+K_2$ is one of the following three subgraphs $(v_3,v_4,v_5)+(v_1,v_6)$, $(v_4,v_3,v_5)+(v_1,v_6)$, $(v_1,v_6,v_2)+(v_4,v_5)$ in G shown in Figure 11. We consider these three possibilities.

Figure 11. The three possible subgraphs $F_2=P_3+K_2$ in G.

In each situation, at least two of the three edges $v_5{v}_1,v_5{v}_2,v_5{v}_6$ have the same color, say (1) $v_5{v}_1$ and $v_5{v}_6$ are red, (2) $v_5{v}_1$ and $v_5{v}_2$ are red, or (3) $v_5{v}_2$ and $v_5{v}_6$ are red. We provide an argument only for (1) since the proof technique is the same and the arguments for (2) and (3) are similar.

First, suppose that $F_2=P_3+K_2=(v_3,v_4,v_5)+(v_1,v_6)$. If $v_5{v}_1$ and $v_5{v}_6$ are red, then $v_5{v}_2$ and $v_5{v}_3$ are blue; for otherwise there is a red $K_{1,3}$ that is edge-disjoint from $F_1$ and $F_2$.

⋆

If $F_2$ is red, then there is a red $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_5{v}_i:i=1,4,6\}$. If one of $v_2{v}_6$ and $v_2{v}_4$ is red, say $v_2{v}_6$, then there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. If $v_2{v}_6$ and $v_2{v}_4$ are blue, then there is a blue $P_3+K_2=(v_6,v_2,v_4)+(v_3,v_5)$ that is edge-disjoint from $F_1$ and $F_3$.

⋆

If $F_2$ is blue, then there is a blue $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_5{v}_i:i=2,3,4\}$. If one of $v_2{v}_6$ and $v_2{v}_4$ is blue, say $v_2{v}_6$, then there is a blue $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. If $v_2{v}_6$ and $v_2{v}_4$ are red, then there is a red $P_3+K_2=(v_6,v_2,v_4)+(v_1,v_5)$ that is edge-disjoint from $F_1$ and $F_3$.

Next, suppose that $F_2=P_3+K_2=(v_4,v_3,v_5)+(v_1,v_6)$. If $v_5{v}_1$ and $v_5{v}_6$ are red, then $v_5{v}_2$ and $v_5{v}_4$ are blue; for otherwise there is a red $K_{1,3}$ that is edge-disjoint from $F_1$ and $F_2$.

⋆

If $F_2$ is red, then there is a red $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_5{v}_i:i=1,3,6\}$. If one of $v_2{v}_6$ and $v_2{v}_4$ is red, say $v_2{v}_6$, then there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. If $v_2{v}_6$ and $v_2{v}_4$ are blue, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_2{v}_i:i=4,5,6\}$ and $P_3+K_2=(v_6,v_1,v_5)+(v_3,v_4)$ is red, that are edge-disjoint from $F_1$.

⋆

If $F_2$ is blue, then there is a blue $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_5{v}_i:i=2,3,4\}$. If one of $v_2{v}_6$ and $v_2{v}_4$ is blue, say $v_2{v}_6$, then there is a blue $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. If $v_2{v}_6$ and $v_2{v}_4$ are red, then there is a red $P_3+K_2=(v_6,v_2,v_4)+(v_1,v_5)$ that is edge-disjoint from $F_1$ and $F_3$.

Finally, suppose that $F_2=P_3+K_2=(v_1,v_6,v_2)+(v_4,v_5)$. If $v_5{v}_1$ and $v_5{v}_6$ are red, then $v_5{v}_2$ and $v_5{v}_3$ are blue; for otherwise there is a red $K_{1,3}$ that is edge-disjoint from $F_1$ and $F_2$.

⋆

If $F_2$ is red, then there is a red $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_5{v}_i:i=1,4,6\}$. Then $v_4{v}_6$ is red, for otherwise there is a blue $P_3+K_2=(v_2,v_5,v_3)+(v_4,v_6)$ that is edge-disjoint from $F_1$ and $F_3$. If $v_3{v}_4$ is red, then there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. Thus, we may assume that $v_3{v}_4$ is blue. If $v_3{v}_6$ is blue, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_3{v}_i:i=4,5,6\}$ and $P_3+K_2=(v_1,v_6,v_2)+(v_4,v_5)$ is red, that are edge-disjoint from $F_1$. If $v_3{v}_6$ is red, then there are edge-disjoint red subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=3,4,5\}$ and $P_3+K_2=(v_1,v_6,v_2)+(v_4,v_5)$ that are edge-disjoint from $F_1$. This procedure is illustrated in Figure 12.

Figure 12. Illustrating the arguments when $F_2$ is red.

⋆

If $F_2$ is blue, then there is a blue $F_3=K_{1,3}$ with $E\left(F_3\right)=\{v_5{v}_i:i=2,3,4\}$. Regardless of the color of $v_3{v}_4$, there is monochromatic $P_3+K_2$ that is edge-disjoint from $F_1$ and $F_3$.

Thus, there are pairwise edge-disjoint subgraphs $K_3,K_{1,3},P_3+K_2$ in G. Therefore, $\overline{R}\left(S\right)\le 6$ and so $\overline{R}\left(S\right)=6$.   □

Theorem 7. 

If $S=\{K_3,P_4,P_3+K_2\},$ then $\overline{R}\left(S\right)=6$.

Proof. 

Since $max\{R\left(F\right):F\in S\}=R\left(K_3\right)=6$, it follows by Proposition 1 that $\overline{R}\left(S\right)\ge 6$. Thus, it remains to show that $\overline{R}\left(S\right)\le 6$. Let there be given a red-blue coloring of $G=K_6$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_6\}$. Since $\overline{R}(K_3,P_4)=6$ by Theorem 2, there are edge-disjoint monochromatic subgraphs $F_1=K_3$ and $F_2=P_4$ in G. We may assume that $V\left(F_1\right)=\{v_1,v_2,v_3\}$ and $F_2=P_4$ is one of the following four subgraphs $(v_1,v_4,v_5,v_2)$, $(v_4,v_1,v_5,v_2)$, $(v_1,v_4,v_5,v_6)$, $(v_4,v_1,v_5,v_6)$ in G shown in Figure 13. We consider these four possibilities.

Figure 13. The four possible subgraphs $F_2=P_4$ in G.

In each situation, at least two of the three edges $v_6{v}_1,v_6{v}_2,v_6{v}_3$ have the same color, namely either (1) $v_6{v}_1$ and $v_6{v}_2$ are red, (2) $v_6{v}_1$ and $v_6{v}_3$ are red, or (3) $v_6{v}_2$ and $v_6{v}_3$ are red. We provide an argument only for (1) since the proof technique is the same and the arguments for (2) and (3) are similar.

First, suppose that $F_2=P_4=(v_1,v_4,v_5,v_2)$. If $v_6{v}_1$ and $v_6{v}_2$ are red, then $v_3{v}_4$ and $v_3{v}_5$ are blue; for otherwise, say $v_3{v}_4$ is red, there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_2$.

⋆

First, suppose that $F_2$ is red. If one of $v_4{v}_2$, $v_4{v}_6,v_5{v}_1,v_5{v}_6$ is blue, say $v_4{v}_2$, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_2,v_4,v_3,v_5)$ is blue and $P_3+K_2=(v_1,v_4,v_5)+(v_2,v_6)$ is red, that are edge-disjoint from $F_1$. If $v_4{v}_2$, $v_4{v}_6,v_5{v}_1,v_5{v}_6$ are all red, then there are edge-disjoint red subgraphs $P_4=(v_4,v_2,v_5,v_6)$ and $P_3+K_2=(v_1,v_4,v_5)+(v_2,v_6)$ that are edge-disjoint from $F_1$.

⋆

Next, suppose that $F_2$ is blue. If $v_1{v}_5$ or $v_2{v}_4$ is red, say $v_1{v}_5$, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_5,v_1,v_6,v_2)$ is red and $P_3+K_2=(v_1,v_4,v_3)+(v_2,v_5)$ is blue, that are edge-disjoint from $F_1$. If $v_1{v}_5$ and $v_2{v}_4$ are blue, there are edge-disjoint blue subgraphs $P_4=(v_1,v_5,v_4,v_2)$ and $P_3+K_2=(v_1,v_4,v_3)+(v_2,v_5)$ that are edge-disjoint from $F_1$.

Next, suppose that $F_2=P_4=(v_4,v_1,v_5,v_2)$. If one of $v_3{v}_4,v_3{v}_5,v_4{v}_5$ is red, say $v_3{v}_4$, then there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. Thus, we may assume that $v_3{v}_4,v_3{v}_5,v_4{v}_5$ are blue.

⋆

First, suppose that $F_2$ is red. If $v_5{v}_6$ or $v_3{v}_6$ is blue, say the former, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_6,v_5,v_4,v_3)$ is blue and $P_3+K_2=(v_4,v_1,v_5)+(v_2,v_6)$ is red, that are edge-disjoint from $F_1$. If $v_5{v}_6$ and $v_3{v}_6$ are red, then there are edge-disjoint red subgraphs $P_4=(v_1,v_6,v_2,v_5)$ and $P_3+K_2=(v_4,v_1,v_5)+(v_3,v_6)$ that are edge-disjoint from $F_1$.

⋆

Next, suppose that $F_2$ is blue. If $v_2{v}_4$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_4,v_2,v_6,v_1)$ is red and $P_3+K_2=(v_1,v_4,v_3)+(v_2,v_5)$ is blue that are edge-disjoint from $F_1$. If $v_2{v}_4$ is blue, then there are edge-disjoint blue subgraphs $P_4=(v_2,v_4,v_5,v_3)$ and $P_3+K_2=(v_1,v_4,v_3)+(v_2,v_5)$ that are edge-disjoint from $F_1$.

We now assume that $F_2=P_4=(v_1,v_4,v_5,v_6)$. If $v_3{v}_4$ or $v_3{v}_5$ is red, say the former, then there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. Thus, we may assume that $v_3{v}_4$ and $v_3{v}_5$ are blue.

⋆

First, suppose that $F_2$ is red. If one of $v_4{v}_2$, $v_4{v}_6,v_5{v}_1,v_5{v}_2$ is blue, say $v_4{v}_2$, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_2,v_4,v_3,v_5)$ is blue and $P_3+K_2=(v_1,v_4,v_5)+(v_2,v_6)$ is red, that are edge-disjoint from $F_1$. If $v_4{v}_2$, $v_4{v}_6,v_5{v}_1,v_5{v}_2$ are all red, then there are edge-disjoint red subgraphs $P_4=(v_1,v_5,v_2,v_4)$ and $P_3+K_2=(v_1,v_4,v_5)+(v_2,v_6)$ that are edge-disjoint from $F_1$.

⋆

Next, suppose that $F_2$ is blue. If $v_2{v}_4$ is red, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_4,v_2,v_6,v_1)$ is red and $P_3+K_2=(v_1,v_4,v_3)+(v_5,v_6)$ is blue, that are edge-disjoint from $F_1$. If $v_2{v}_4$ is blue, then there are edge-disjoint blue subgraphs $P_4=(v_2,v_4,v_5,v_3)$ and $P_3+K_2=(v_1,v_4,v_3)+(v_5,v_6)$ that are edge-disjoint from $F_1$.

Finally, suppose that $F_2=P_4=(v_4,v_1,v_5,v_6)$. If one of $v_3{v}_4,v_3{v}_5,v_4{v}_5$ is red, say $v_3{v}_4$, then there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_4)$ that is edge-disjoint from $F_1$ and $F_3$. Thus, we may assume that $v_3{v}_4,v_3{v}_5,v_4{v}_5$ are blue.

⋆

First, suppose that $F_2$ is red. If $v_3{v}_6$ is blue, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_5,v_4,v_3,v_6)$ is blue and $P_3+K_2=(v_4,v_1,v_5)+(v_2,v_6)$ is red, that are edge-disjoint from $F_1$. If $v_3{v}_6$ is red, then there are edge-disjoint red subgraphs $P_4=(v_5,v_1,v_6,v_2)$ and $P_3+K_2=(v_5,v_6,v_3)+(v_1,v_4)$ that are edge-disjoint from $F_1$.

⋆

Next, suppose that $F_2$ is blue. If $v_2{v}_5$ or $v_2{v}_4$ is red, say the former, then there are edge-disjoint monochromatic subgraphs $P_4$ and $P_3+K_2$, where $P_4=(v_5,v_2,v_6,v_1)$ is red and $P_3+K_2=(v_1,v_4,v_3)+(v_5,v_6)$ is blue, that are edge-disjoint from $F_1$. If $v_2{v}_5$ and $v_2{v}_4$ are blue, then there are edge-disjoint blue subgraphs $P_4=(v_1,v_5,v_4,v_2)$ and $P_3+K_2=(v_1,v_4,v_3)+(v_5,v_6)$ that are edge-disjoint from $F_1$.

Hence, there are pairwise edge-disjoint monochromatic subgraphs $K_3,P_4,P_3+K_2$ in G. Therefore, $\overline{R}\left(S\right)\le 6$ and so $\overline{R}\left(S\right)=6$.   □

Theorem 8. 

If $S=\{P_4,P_3+K_2,K_{1,3}\},$ then $\overline{R}\left(S\right)=6$.

Proof. 

Since $max\{R\left(F\right):F\in S\}=R\left(K_{1,3}\right)=6$, it follows by Proposition 1 that $\overline{R}\left(S\right)\ge 6$. Thus, it remains to show that $\overline{R}\left(S\right)\le 6$. Let there be given a red-blue coloring of $G=K_6$ with $V\left(G\right)=\{v_1,v_2,\dots ,v_6\}$. Since $\overline{R}(P_4,K_{1,3})=6$ by Theorem 2, there are edge-disjoint monochromatic subgraphs $F_1=P_4$ and $F_2=K_{1,3}$ in G. We may assume that $F_1=P_4=(v_1,v_2,v_3,v_4)$. For the location of $F_2$ in G, we consider two cases, according to the location of the center of $F_2$.

Case 1. The center of $K_{1,3}$ is either $v_5$ or $v_6$, say the former. By symmetry, we may assume that $E\left(F_2\right)$ is one of the following six sets $\{v_5{v}_i:i=1,2,3\}$, $\{v_5{v}_i:i=1,2,4\}$, $\{v_5{v}_i:i=1,2,6\}$, $\{v_5{v}_i:i=1,3,6\}$, $\{v_5{v}_i:i=1,4,6\}$, $\{v_5{v}_i:i=2,3,6\}$ shown in Figure 14.

Figure 14. The six possible subgraphs for $F_2=K_{1,3}$ in Case 1.

We provide an argument only for $E\left(F_2\right)=\{v_5{v}_i:i=1,2,3\}$ since the proof technique is the same and the arguments for other choices of $F_2$ are similar. At least two of $v_6{v}_i$ ($i=1,2,3$) have the same color, say red. There are three possibilities.

• First, suppose that $v_6{v}_1$ and $v_6{v}_2$ are red. Then $v_4{v}_5$ is blue, for otherwise there is a red $P_3+K_2=(v_1,v_6,v_2)+(v_4,v_5)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  If $F_2$ is red, then $v_5{v}_6$ is blue, for otherwise there are edge-disjoint red subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=1,2,6\}$ and $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_5)$ that are edge-disjoint from $F_1$. Then $v_1{v}_3$ is red, for otherwise there is a blue $P_3+K_2=(v_4,v_5,v_6)+(v_1,v_3)$ that is edge-disjoint from $F_1$ and $F_2$. Then $v_1{v}_4$ and $v_2{v}_4$ are blue, for otherwise, say $v_1{v}_4$ is red, then there are edge-disjoint red subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_1{v}_i:i=3,4,5\}$ and $P_3+K_2=(v_1,v_6,v_2)+(v_3,v_5)$ that are edge-disjoint from $F_1$. Then $P_3+K_2=(v_1,v_4,v_2)+(v_5,v_6)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  We now assume that $F_2$ is blue. First, suppose that $v_5{v}_6$ is red. Then $v_1{v}_4$ and $v_2{v}_4$ are blue, for otherwise, say $v_1{v}_4$ is red, then $P_3+K_2=(v_2,v_6,v_5)+(v_1,v_4)$ is edge-disjoint from $F_1$ and $F_2$. Then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=1,2,5\}$ and $P_3+K_2=(v_1,v_5,v_3)+(v_2,v_4)$ is blue, that are edge-disjoint from $F_1$. Next, suppose that $v_5{v}_6$ is blue. Then $v_3{v}_6$ and $v_4{v}_6$ are red, for otherwise, say $v_3{v}_6$ is blue, then there are edge-disjoint blue subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=3,4,6\}$ and $P_3+K_2=(v_1,v_5,v_2)+(v_3,v_6)$ that are edge-disjoint from $F_1$. If $v_1{v}_4$ is red, there is a red $P_3+K_2=(v_2,v_6,v_3)+(v_1,v_4)$ that is edge-disjoint from $F_1$ and $F_2$. If $v_1{v}_4$ is blue, then there are blue subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=1,4,6\}$ and $P_3+K_2=(v_2,v_5,v_3)+(v_1,v_4)$ that are edge-disjoint from $F_1$.

• Next, suppose that $v_6{v}_1$ and $v_6{v}_3$ are red. By the argument above, we may assume that $v_6{v}_2$ is blue. Then $v_2{v}_4$ and $v_5{v}_4$ are blue, for otherwise, say $v_2{v}_4$ is red, there is a red $P_3+K_2=(v_1,v_6,v_3)+(v_2,v_4)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  If $F_2$ is red, then $v_1{v}_3$ and $v_1{v}_4$ are red, for otherwise, say $v_1{v}_3$ is blue, there is a blue $P_3+K_2=(v_2,v_4,v_5)+(v_1,v_3)$ that is edge-disjoint from $F_1$ and $F_2$. Then there are edge-disjoint red subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_1{v}_i:i=3,4,6\}$ and $P_3+K_2=(v_1,v_5,v_2)+(v_3,v_6)$ that are edge-disjoint from $F_1$.

  ⋆

  If $F_2$ is blue, then there are edge-disjoint blue subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=2,3,4\}$ and $P_3+K_2=(v_4,v_2,v_6)+(v_1,v_5)$ that are edge-disjoint from $F_1$.

• Finally, suppose that $v_6{v}_2$ and $v_6{v}_3$ are red. We may assume that $v_6{v}_1$ is blue. Then $v_1{v}_4$ and $v_5{v}_4$ are blue, for otherwise, say $v_1{v}_4$ is red, there is a red $P_3+K_2=(v_2,v_6,v_3)+(v_1,v_4)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  First, suppose that $F_2$ is red. If one of $v_4{v}_6,v_4{v}_2,v_1{v}_3$ is blue, say $v_4{v}_6$, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_4{v}_i:i=1,5,6\}$ and $P_3+K_2=(v_1,v_5,v_2)+(v_3,v_6)$ is red, that are edge-disjoint from $F_1$. If all $v_4{v}_6,v_4{v}_2,v_1{v}_3$ are red, then there is a red $P_3+K_2=(v_2,v_4,v_6)+(v_1,v_3)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  Next, suppose that $F_2$ is blue. If $v_5{v}_6$ is blue, then there are edge-disjoint blue subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=1,4,6\}$ and $P_3+K_2=(v_2,v_5,v_3)+(v_1,v_4)$ that are edge-disjoint from $F_1$. If $v_5{v}_6$ is red, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=2,3,5\}$ and $P_3+K_2=(v_2,v_5,v_3)+(v_1,v_4)$ is blue, that are edge-disjoint from $F_1$.

Case 2. The center of $K_{1,3}$ belongs to $\{v_1,v_2,v_3,v_4\}$. By symmetry, we may assume that $E\left(F_2\right)$ is one of the following four sets $\{v_1{v}_i:i=3,5,6\}$, $\{v_1{v}_i:i=4,5,6\}$, $\{v_1{v}_i:i=3,4,5\}$, $\{v_2{v}_i:i=4,5,6\}$, shown in Figure 15.

Figure 15. The four possible subgraphs for $F_2=K_{1,3}$ in Case 2.

We provide an argument only for $E\left(F_2\right)=\{v_1{v}_i:i=3,5,6\}$ since the proof technique is the same and the arguments for other choices of $F_2$ are similar. At least two edges in $\{v_6{v}_i:i=2,3,4\}$ have the same color, say red. There are three possibilities.

• First, suppose that $v_6{v}_2$ and $v_6{v}_3$ are red. Then $v_4{v}_1$ and $v_4{v}_5$ are blue, for otherwise, say $v_4{v}_1$ is red, there is a red $P_3+K_2=(v_2,v_6,v_3)+(v_1,v_4)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  If $F_2$ is red, then there is a red $F_2^{\prime }=K_{1,3}$ with $E\left(F_2^{\prime }\right)=\{v_6{v}_i:i=1,2,3\}$. If $v_2{v}_4$ is red, then there is a red $P_3+K_2=(v_3,v_1,v_5)+(v_2,v_4)$ that are edge-disjoint from $F_1$ and $F_2^{\prime }$. If $v_2{v}_4$ is blue, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is blue with $E\left(K_{1,3}\right)=\{v_4{v}_i:i=1,2,5\}$ and $P_3+K_2=(v_2,v_6,v_3)+(v_1,v_5)$ is red, that are edge-disjoint from $F_1$.

  ⋆

  If $F_2$ is blue, then there is blue $F_3=P_3+K_2=(v_3,v_1,v_6)+(v_4,v_5)$ that is edge-disjoint from $F_1$. If $v_4{v}_6$ or $v_5{v}_6$ is red, say the former, then there is a red $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=2,3,4\}$ that is edge-disjoint from $F_1$ and $F_3$. Thus, we may assume that $v_4{v}_6$ and $v_5{v}_6$ are blue. If $v_2{v}_4$ or $v_2{v}_5$ is blue, say the former, then there is a blue $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_4{v}_i:i=1,2,6\}$ that is edge-disjoint from $F_1$ and $F_3$. If $v_2{v}_4$ and $v_2{v}_5$ are red, then there is a red $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_2{v}_i:i=4,5,6\}$ that is edge-disjoint from $F_1$ and $F_3$.

• Next, suppose that $v_6{v}_2$ and $v_6{v}_4$ are red. By the argument above, we may assume that $v_6{v}_3$ is blue. Then $v_3{v}_5$ is blue, for otherwise there is a red $P_3+K_2=(v_2,v_6,v_4)+(v_3,v_5)$ that is edge-disjoint from $F_1$ and $F_2$. This implies that $v_1{v}_4$ and $v_2{v}_4$ are red, for otherwise, say $v_1{v}_4$ is blue, there is a blue $P_3+K_2=(v_5,v_3,v_6)+(v_1,v_4)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  If $F_2$ is red, then there are edge-disjoint red subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_1{v}_i:i=3,4,6\}$ and $P_3+K_2=(v_4,v_2,v_6)+(v_1,v_5)$ that are edge-disjoint from $F_1$.

  ⋆

  We now assume that $F_2$ is blue. If $v_2{v}_5$ is red, there is a red $P_3+K_2=(v_1,v_4,v_6)+(v_2,v_5)$ is edge-disjoint from $F_1$ and $F_2$. If $v_2{v}_5$ is blue, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_4{v}_i:i=1,2,6\}$ and $P_3+K_2=(v_1,v_3,v_6)+(v_2,v_5)$ is blue, that are edge-disjoint from $F_1$.

• Finally, suppose that $v_6{v}_3$ and $v_6{v}_4$ are red. We may assume that $v_6{v}_2$ is blue. Then $v_2{v}_5$ is blue, for otherwise there is a red $P_3+K_2=(v_3,v_6,v_4)+(v_2,v_5)$ that is edge-disjoint from $F_1$ and $F_2$. This implies that $v_1{v}_4$ is red, for otherwise there is a blue $P_3+K_2=(v_5,v_2,v_6)+(v_1,v_4)$ that is edge-disjoint from $F_1$ and $F_2$.

  ⋆

  If $F_2$ is red, then there are edge-disjoint red subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_1{v}_i:i=3,4,6\}$ and $P_3+K_2=(v_3,v_6,v_4)+(v_1,v_5)$ that are edge-disjoint from $F_1$.

  ⋆

  We now assume that $F_2$ is blue. If $v_5{v}_6$ or $v_2{v}_4$ is red, say the former, then there are edge-disjoint monochromatic subgraphs $K_{1,3}$ and $P_3+K_2$, where $K_{1,3}$ is red with $E\left(K_{1,3}\right)=\{v_6{v}_i:i=3,4,5\}$ and $P_3+K_2=(v_5,v_1,v_3)+(v_2,v_6)$ is blue, that are edge-disjoint from $F_1$. If $v_5{v}_6$ and $v_2{v}_4$ are blue, then there are edge-disjoint blue subgraphs $K_{1,3}$ with $E\left(K_{1,3}\right)=\{v_5{v}_i:i=1,2,6\}$ and $P_3+K_2=(v_4,v_2,v_6)+(v_1,v_3)$ that are edge-disjoint from $F_1$.

Hence, there are pairwise edge-disjoint monochromatic subgraphs $P_4,P_3+K_2,K_{1,3}$ in G. Therefore, $\overline{R}\left(S\right)\le 6$ and so $\overline{R}\left(S\right)=6$.   □

In summary, we have the following.

Theorem 9. 

If S is a set of three graphs of size 3 without isolated vertices not containing $3K_2$, then $\overline{R}\left(S\right)=max\{R\left(F\right):F\in S\}=6$.

4. Closing Remarks

By Theorems 2, 4, and 9, we have the following result.

Theorem 10. 

For every set S of two or three graphs of size 3 without isolated vertices,

$$\overline{R}\left(S\right)=max\{R\left(F\right):F\in S\}.$$

By Theorem 1, if S is the set of the four acyclic graphs of size 3 without isolated vertices, then $\overline{R}\left(S\right)=max\{R\left(F\right):F\in S\}$. This brings up the question as to whether this fact holds for the remaining 4-element subsets or the entire set $\{K_3,K_{1,3},P_4,P_3+K_2,3K_2\}$ of the five graphs of size 3 without isolated vertices. Furthermore, there is the more general question as to whether this fact can be extended to broader sets of graphs.