Extending Ramsey Numbers for Connected Graphs of Size 3

Abstract

It is well known that the famous Ramsey number $R(K_3,K_3)=6$. That is, the minimum positive integer n for which every red-blue coloring of the edges of the complete graph $K_n$ results in a monochromatic triangle $K_3$ is 6. It is also known that every red-blue coloring of $K_6$ results in at least two monochromatic triangles, which need not be vertex-disjoint or edge-disjoint. This fact led to an extension of Ramsey numbers. For a graph F and a positive integer t, the vertex-disjoint Ramsey number $V{R}_t\left(F\right)$ is the minimum positive integer n such that every red-blue coloring of the edges of the complete graph $K_n$ of order n results in t pairwise vertex-disjoint monochromatic copies of subgraphs isomorphic to F, while the edge-disjoint Ramsey number $E{R}_t\left(F\right)$ is the corresponding number for edge-disjoint subgraphs. Since $V{R}_1\left(F\right)$ and $E{R}_1\left(F\right)$ are the well-known Ramsey numbers of F, these new Ramsey concepts generalize the Ramsey numbers and provide a new perspective for this classical topic in graph theory. These numbers have been investigated for the two connected graphs $K_3$ and the path $P_3$ of order 3. Here, we study these numbers for the remaining connected graphs, namely, the path $P_4$ and the star $K_{1,3}$ of size 3. We show that $V{R}_t\left(P_4\right)=4t+1$ for every positive integer t and $V{R}_t\left(K_{1,3}\right)=4t$ for every integer $t\ge 2$. For $t\le 4$, the numbers $E{R}_t\left(K_{1,3}\right)$ and $E{R}_t\left(P_4\right)$ are determined. These numbers provide information towards the goal of determining how the numbers $V{R}_t\left(F\right)$ and $E{R}_t\left(F\right)$ increase as t increases for each graph $F\in \{K_{1,3},P_4\}$.

1. Introduction

Ramsey theory is one of the most popular areas in graph theory (see [1,2,3], for example). While there are many highly nontrivial and beautiful results in various Ramsey numbers (see [4,5,6,7,8,9,10], for example), probably the best known and often-mentioned Ramsey number in graph theory is the one denoted by $R(3,3)$ or $R(K_3,K_3)$. This is the smallest positive integer n such that for every red-blue coloring (of the edges) of the complete graph $K_n$ of order n, there is a subgraph of $K_n$ isomorphic to $K_3$, all of whose edges are colored the same (a monochromatic $K_3$). It is well known that $R(K_3,K_3)=6$. That is, every red-blue coloring of $K_6$ results in a monochromatic $K_3$, but such is not the case for every red-blue coloring of $K_5$. Indeed, the red-blue coloring of $K_5$, whose red and blue subgraphs are the cycles $C_5$ of order 5, results in no monochromatic $K_3$. In fact, every red-blue coloring of $K_6$ results in two monochromatic copies of $K_3$, although the two subgraphs may have an edge in common (see Figure 1, where a bold edge represents a red edge and a thin edge represents a blue edge).

In a red-blue coloring of a graph G, every edge of G is colored red or blue. For two graphs F and H, the well-known Ramsey number $R(F,H)$ is the minimum positive integer n such that for every red-blue coloring of the complete graph $K_n$ of order n, there is either a subgraph of $K_n$ isomorphic to F, all of whose edges are colored red (a red F), or a subgraph of $K_n$ isomorphic to H, all of whose edges are colored blue (a blue H). Therefore, for a single graph F, the Ramsey number $R(F,F)$, also denoted by $R\left(F\right)$, is the minimum positive integer n such that for every red-blue coloring of $K_n$, there is a subgraph of $K_n$ isomorphic to F, all of whose edges are colored the same (a monochromatic F). Although these numbers exist for every graph F (see [11]), it is challenging in general to determine the exact value $R\left(F\right)$ for many graphs F. For example, $R\left(K_n\right)$ is only known when $n\le 4$.

While the Ramsey number of a graph F is the smallest order n of a complete graph $K_n$ such that every red-blue coloring of $K_n$ produces a monochromatic F, the size Ramsey number is the minimum size of any graph G such that every red-blue coloring of G produces a monochromatic F (see [12,13], for example). Recently, there has been an interest shown in edge colorings of complete graphs that guarantee multiple pairwise edge-disjoint or vertex-disjoint monochromatic copies of a given graph. In particular, Ramsey numbers of graphs were extended to two multiple Ramsey numbers in [14]. We refer to the book [15] for notation and terminology in graph theory that are not defined here.

Let t be a positive integer and let F be a graph without isolated vertices. The vertex-disjoint Ramsey number $V{R}_t\left(F\right)$ is the minimum positive integer n such that for every red-blue coloring of $K_n$, there are at least t pairwise vertex-disjoint monochromatic copies of F. Hence, $V{R}_1\left(F\right)=R\left(F\right)$ for every graph F. Since the Ramsey number $R\left(tF\right)$ exists for the union $tF$ of t pairwise vertex-disjoint copies of a graph F by a result due to Ramsey [11], we have the following.

Observation 1

([14]). For every graph F without isolated vertices and every positive integer t, the number $V{R}_t\left(F\right)$ exists and $t\left|V\left(F\right)\right|\le V{R}_t\left(F\right)\le R\left(tF\right)$. Furthermore, $V{R}_t\left(F\right)\le V{R}_{t+1}\left(F\right)$.

Let t be a positive integer and let F be a graph without isolated vertices. The edge-disjoint Ramsey number $E{R}_t\left(F\right)$ of F is the minimum positive integer n such that for every red-blue coloring of $K_n$, there are at least t pairwise edge-disjoint monochromatic copies of F. Hence, $E{R}_1\left(F\right)=R\left(F\right)$ for every graph F. There is an observation for edge-disjoint Ramsey numbers similar to Observation 1.

Observation 2

([14]). For every graph F without isolated vertices and every positive integer t, the number $E{R}_t\left(F\right)$ exists and $E{R}_t\left(F\right)\le V{R}_t\left(F\right)$. Furthermore, $E{R}_t\left(F\right)\le E{R}_{t+1}\left(F\right)$.

In [14], the numbers $V{R}_t\left(F\right)$ and $E{R}_t\left(F\right)$ were investigated for the two connected graphs of order 3, namely, the complete graph $K_3$ and the path $P_3$. The following two results were obtained on vertex-disjoint Ramsey numbers of these two graphs.

Theorem 1

([14]). For an integer $t\ge 2$, $V{R}_t\left(K_3\right)=3t+2$.

Theorem 2

([14]). For every positive integer t, $V{R}_t\left(P_3\right)=3t$.

Regarding edge-disjoint Ramsey numbers of $K_3$, it was also shown in [14] that $E{R}_2\left(K_3\right)=7$, $E{R}_3\left(K_3\right)=9$, and $E{R}_4\left(K_3\right)=10$. The edge-disjoint Ramsey numbers of $P_3$ were obtained in [14] as well.

Theorem 3

([14]). For every positive integer t, $E{R}_t\left(P_3\right)=\lceil 2\sqrt{t}+1\rceil$.

The goal of this paper is to study $V{R}_t\left(F\right)$ and $E{R}_t\left(F\right)$ for the connected graphs of size 3 different from $K_3$, namely, the star $K_{1,3}$ and the path $P_4$ of order 4 as shown in Figure 2. The star $K_{1,3}$ is also referred to as a claw. We begin with vertex-disjoint Ramsey numbers.

2. Vertex-Disjoint Ramsey Numbers

We begin by establishing an upper bound for $V{R}_t\left(F\right)$ for every graph F of order at least 4 with no isolated vertices.

Proposition 1.

Let F be a graph of order $n\ge 4$ without isolated vertices. If there is a positive integer $t_0$ such that $V{R}_{t_0}\left(F\right)\le q$ for some positive integer q, then $V{R}_t\left(F\right)\le q+(t-t_0)n$ for every integer $t\ge t_0$.

Proof. 

We proceed by induction on $t\ge t_0$. Since $V{R}_{t_0}\left(F\right)\le q$, the result is true for $t=t_0$. Assume that $V{R}_k\left(F\right)\le q+(k-t_0)n$ for an integer $k\ge t_0$. We show that $V{R}_{k+1}\left(F\right)\le q+(k+1-t_0)n$. Let there be given a red-blue coloring of $G=K_{q+(k+1-t_0)n}$. Since

$$q+(k+1-t_0)n=q+(k-t_0)n+n\ge q,$$

there is a monochromatic copy of F in G. Let $H=G-V\left(F\right)=K_{q+(k-t_0)n}$. Since $V{R}_k\left(F\right)\le q+(k-t_0)n$, it follows that H contains k pairwise vertex-disjoint monochromatic copies of F. Hence, there are $k+1$ pairwise vertex-disjoint monochromatic copies of F in G. Therefore, $V{R}_{k+1}\left(F\right)\le q+(k+1-t_0)n$. □

With the aid of Proposition 1, we can determine $V{R}_t\left(P_4\right)$ for every positive integer t.

Proposition 2.

For each positive integer t, $V{R}_t\left(P_4\right)=4t+1$.

Proof. 

Since $R\left(P_4\right)=V{R}_1\left(P_4\right)=5$ and $P_4$ has order 4, it follows by Proposition 1 that $V{R}_t\left(P_4\right)\le 5+4(t-1)=4t+1$. On the other hand, the red-blue coloring of $K_{4t}$ with red subgraph $K_{1,4k-1}$ and blue subgraph $K_{4k-1}$ has no red copy of $P_4$ and $t-1$ pairwise vertex-disjoint blue copies of $P_4$. Hence, $V{R}_t\left(P_4\right)\ge 4t+1$, and so $V{R}_t\left(P_4\right)=4t+1$. □

We now turn our attention to the claw $K_{1,3}$. Since $V{R}_1\left(K_{1,3}\right)=R\left(K_{1,3}\right)=6$, it remains to determine $V{R}_t\left(K_{1,3}\right)$ for $t\ge 2$.

Theorem 4.

For each integer $t\ge 2$, $V{R}_t\left(K_{1,3}\right)=4t$.

Proof. 

Since the order of $K_{1,3}$ is 4, it follows that $V{R}_t\left(K_{1,3}\right)\ge 4t$ by Observation 1. Consequently, it remains to show that for every integer $t\ge 2$ and every red-blue coloring of $G=K_{4t}$, there are t pairwise vertex-disjoint monochromatic copies of $K_{1,3}$. Let there be given a red-blue coloring of G. By Theorem 1, $V{R}_t\left(K_3\right)=3t+2$ for each integer $t\ge 2$. Since $3t+2\le 4t$, there are t pairwise vertex-disjoint monochromatic copies $T_1,T_2,\dots ,T_t$ of $K_3$. Let $V\left(G\right)-{\bigcup }_{i=1}^{t}V\left(T_i\right)=\{v_1,v_2,\dots ,v_t\}.$ For $1\le i\le t$, consider $T_i$ and $v_i$. If $v_i$ is joined to a vertex of $T_i$ having the same color as $T_i$, then a monochromatic $K_{1,3}$ centered at that vertex of $T_i$ is produced. If this is not the case, then $v_i$ is joined to the three vertices of $T_i$ by edges whose colors are different from that of $T_i$, which produces a monochromatic $K_{1,3}$ centered at $v_i$. In either case, a monochromatic $K_{1,3}$, denoted by $G_i$, is produced where $V\left(G_i\right)=\{V\left(T_i\right)\cup \left\{v_i\right\}$. Thus, G contains t pairwise vertex-disjoint monochromatic subgraphs $G_1,G_2,\dots ,G_t$, each of which is a copy of $K_{1,3}$. Therefore, $V{R}_t\left(K_{1,3}\right)\le 4t$, and so $V{R}_t\left(K_{1,3}\right)=4t$. □

3. Edge-Disjoint Ramsey Numbers

We now investigate $E{R}_t\left(F\right)$ for the two connected graphs $K_{1,3}$ and $P_4$ of size 3 for positive integers t, beginning by determining $E{R}_t\left(F\right)$ for small values of t, namely, $1\le t\le 4$.

A vertex and an incident edge are said to cover each other. A vertex cover in a graph G is a set of vertices that covers all edges of G. The minimum number of vertices in a vertex cover of G is the vertex covering number $\beta \left(G\right)$ of G. A vertex cover of cardinality $\beta \left(G\right)$ is a minimum vertex cover in G. We mentioned that $E{R}_t\left(F\right)\le E{R}_{t+1}\left(F\right)$ for every graph F and every positive integer t in Observation 2. In fact, $E{R}_{t+1}\left(F\right)$ can never exceed $E{R}_t\left(F\right)$ by more than $\beta \left(F\right)$.

Proposition 3.

For every graph F without isolated vertices and each positive integer t,

$$E{R}_{t+1}\left(F\right)\le E{R}_t\left(F\right)+\beta \left(F\right).$$

Proof. 

Let $E{R}_t\left(F\right)=k$ and let $\beta \left(F\right)=\beta$. Let c be a red-blue coloring of $G=K_{k+\beta }$. Then, there is a monochromatic copy $F_0$ of F in G. Let X be a minimum vertex cover of $F_0$, and let $H=G\left[V\right(G)-X]$. Then, $H=K_k$ contains no edge of $F_0$. Since $E{R}_t\left(F\right)=k$, there are t pairwise edge-disjoint monochromatic copies $F_1$, $F_2$, …, $F_t$ of F in H that are edge-disjoint from $F_0$. Therefore, $F_0$, $F_1$, $F_2$, …, $F_t$ are $t+1$ pairwise edge-disjoint monochromatic copies of F in G, and so $E{R}_{t+1}\left(F\right)\le k+\beta =E{R}_t\left(F\right)+\beta \left(F\right)$. □

Since $\beta \left(K_{1,3}\right)=1$ and $\beta \left(P_4\right)=2$, the following is an immediate consequence of Proposition 3.

Corollary 1.

For each positive integer t,

$E{R}_{t+1}\left(K_{1,3}\right)\le E{R}_t\left(K_{1,3}\right)+1$ and $E{R}_{t+1}\left(P_4\right)\le E{R}_t\left(P_4\right)+2$.

In [16], the Ramsey numbers $R\left(K_{1,n}\right)$ of all stars $K_{1,n}$ were determined.

Theorem 5

([16]). For each integer $n\ge 3$, $R\left(K_{1,n}\right)=\left\{\begin{array}{cc}2n\hfill & ifnisodd\hfill \\ 2n-1\hfill & ifniseven.\hfill \end{array}\right.$

By Theorem 5, $E{R}_1\left(K_{1,3}\right)=R\left(K_{1,3}\right)=6$. We now determine $E{R}_t\left(K_{1,3}\right)$ for $t=2,3,4$.

Proposition 4.

$E{R}_2\left(K_{1,3}\right)=6$ and $E{R}_3\left(K_{1,3}\right)=E{R}_4\left(K_{1,3}\right)=7$.

Proof. 

Since $R\left(K_{1,3}\right)=6$, it follows that $E{R}_2\left(K_{1,3}\right)\ge 6$. It remains to show that $E{R}_2\left(K_{1,3}\right)\le 6$. Let c be a red-blue coloring of $G=K_6$. We show that G contains two edge-disjoint monochromatic copies of $K_{1,3}$. Since $R\left(K_{1,3}\right)=6$, there is a monochromatic copy $S_1$ of $K_{1,3}$ in G. Let $v\in V\left(G\right)-V\left(S_1\right)$. Since v is incident with five edges not in $E\left(S_1\right)$, at least three of these five edges have the same color, and so there is a monochromatic copy $S_2$ of $K_{1,3}$ centered at v that is edge-disjoint from $S_1$. Therefore, $E{R}_2\left(K_{1,3}\right)\le 6$, and so $E{R}_2\left(K_{1,3}\right)=6$.

Next, we show that $E{R}_3\left(K_{1,3}\right)=7$. Since $E{R}_3\left(K_{1,3}\right)\le E{R}_2\left(K_{1,3}\right)+1=7$ by Corollary 1, it remains to show that $E{R}_3\left(K_{1,3}\right)\ge 7$. Define a red-blue coloring of $K_6$ such that the red subgraph is $C_6$ and the blue graph is the 3-regular graph $K_3\square K_2$ (the Cartesian product of $K_3$ and $K_2$) of order 6. Since there is no red $K_{1,3}$ and the blue subgraph produces only two edge-disjoint copies of $K_{1,3}$, this coloring of $K_6$ does not produce three pairwise edge-disjoint monochromatic copies of $K_{1,3}$, and so $E{R}_3\left(K_{1,3}\right)\ge 7$. Therefore, $E{R}_3\left(K_{1,3}\right)=7$.

Finally, we show that $E{R}_4\left(K_{1,3}\right)=7$. Since $E{R}_4\left(K_{1,3}\right)\ge E{R}_3\left(K_{1,3}\right)=7$, it suffices to show that $E{R}_4\left(K_{1,3}\right)\le 7$. Let c be a red-blue coloring of $G=K_7$, and let $v\in V\left(G\right)$. Then, $H=G-v=K_6$. Let $V\left(H\right)=\{v_1,v_2,\dots ,v_6\}$. Since $E{R}_2\left(K_{1,3}\right)=6$, there are two edge-disjoint monochromatic copies $S_1$ and $S_2$ of $K_{1,3}$ in H. We may assume that $S_1$ is centered at $v_1$ and $S_2$ is centered at $v_2$. Then, $v_1\ne v_2$. There are six possibilities for the subgraph $F=H[E\left(S_1\right)\cup E\left(S_2\right)]$ of H induced by $E\left(S_1\right)\cup E\left(S_2\right)$. We may assume that $V\left(S_1\right)=\{v_1,v_3,v_4,v_5\}$ if $v_2\notin V\left(S_1\right)$ and $V\left(S_1\right)=\{v_1,v_2,v_3,v_4\}$ otherwise, up to the remaining vertices. These six possibilities for F are shown in Figure 3.

Suppose that v is incident with r red edges and b blue edges, where $r+b=6$. We may assume that $r\ge b$. If $r=6$ or $r=b=3$, then there are two edge-disjoint monochromatic copies $S_3$ and $S_4$ of $K_{1,3}$ centered at v that are edge-disjoint from $S_1$ and $S_2$. Thus, we may assume that $r\in \{4,5\}$. Hence, there is a red copy $S_3$ of $K_{1,3}$ centered at v that is edge-disjoint from $S_1$ and $S_2$ such that $v_6\notin V\left(S_3\right)$.

★

First, suppose that the order of F is 5, where then $V\left(F\right)=\{v_1,v_2,\dots ,v_5\}$. Since at least three of the five edges $v_6{v}_1,v_6{v}_2,v_6{v}_3,v_6{v}_4,v_6{v}_5$ have the same color, there is a monochromatic copy $S_4$ of $K_{1,3}$ centered at $v_6$ that is edge-disjoint from $S_1,S_2$, and $S_3$.

★

Next, suppose that $V\left(F\right)=\{v_1,v_2,\dots ,v_6\}$. Since $v_6$ is adjacent only to $v_2$ in G, it follows that none of the five edges $v_{6}v,v_6{v}_1,v_6{v}_3,v_6{v}_4,v_6{v}_5$ belong to $E\left(S_1\right)\cup E\left(S_2\right)\cup E\left(S_3\right)$. There are at least three edges incident with $v_6$ that have the same color, producing a monochromatic copy $S_4$ of $K_{1,3}$ centered at $v_6$ edge-disjoint from $S_1,S_2$, and $S_3$.

Therefore, $E{R}_4\left(K_{1,3}\right)\le 7$, and so $E{R}_4\left(K_{1,3}\right)=7$. □

With the aid of Corollary 1 and Proposition 4, we have bounds for $E{R}_t\left(K_{1,3}\right)$ for all integers $t\ge 4$.

Theorem 6.

For each integer $t\ge 4$,

$$⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉\le E{R}_t\left(K_{1,3}\right)\le t+3.$$

Proof. 

First, we verify the lower bound. Let $E{R}_t\left(K_{1,3}\right)=k$. Then, every red-blue coloring of $K_k$ produces at least t pairwise edge-disjoint copies of $K_{1,3}$. Consider the red-blue coloring of $G=K_k$ with red subgraph $G_r=C_k$ and blue subgraph $G_b=K_k-E\left(C_k\right)$. Since there is no red $K_{1,3}$, it follows that $G_b$ contains at least t pairwise edge-disjoint copies of $K_{1,3}$. This implies that $|E\left(G_b\right)|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)-k\ge 3t$ or $k^2-3k-6t\ge 0$. Consequently, $E{R}_t\left(K_{1,3}\right)=k\ge ⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉$.

To verify the upper bound for $E{R}_t\left(K_{1,3}\right)$, we proceed by induction on t. Since $E{R}_4\left(K_{1,3}\right)=7$ by Proposition 4, the result is true for $t=4$. Assume that $E{R}_t\left(K_{1,3}\right)\le t+3$ for some integer $t\ge 4$. Since $\beta \left(K_{1,3}\right)=1$, it follows by Proposition 3 and the induction hypothesis that $E{R}_{t+1}\left(K_{1,3}\right)\le E{R}_t\left(K_{1,3}\right)+1\le (t+1)+3$. □

For $t=3,4,5$, we have $⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉=t+3$. We saw that $E{R}_t\left(K_{1,3}\right)=t+3$ for $t=3,4$ by Proposition 4. By Theorem 6, $E{R}_5\left(K_{1,3}\right)=8$. For $t\ge 6$, then $⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉<t+3$. In fact, if $t=6,7$, then $⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉=t+2$; if $t=8$, then $⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉=t+1$; and if $t=9,10$, then $⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉=t+1$; while if $t\ge 11$, then $⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉\le t-1$.

We now turn to the remaining connected graph of size 3, namely, $P_4$. In [17], the Ramsey numbers $R\left(P_n\right)$ of all paths $P_n$ are determined.

Theorem 7

([17]). For each integer $n\ge 3$, $R\left(P_n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{3(n-1)}{2}}\hfill & ifnisodd\hfill \\ {\displaystyle \frac{3n}{2}}-1\hfill & ifniseven.\hfill \end{array}\right.$

By Theorem 7, $E{R}_1\left(P_4\right)=R\left(P_4\right)=5$. We now determine $E{R}_t\left(P_4\right)$ for $t=2,3,4$.

Proposition 5.

$E{R}_2\left(P_4\right)=5$.

Proof. 

Since $R\left(P_4\right)=R(P_4,P_4)=5$, it follows that $E{R}_2\left(P_4\right)\ge 5$. It therefore remains to show that $E{R}_2\left(P_4\right)\le 5$. Let c be a red-blue coloring of $G=K_5$. We show that there are two edge-disjoint monochromatic copies of $P_4$. Since $R\left(P_4\right)=5$, there is a monochromatic subgraph $H=P_4$ of G. Let $H=(v_1,v_2,v_3,v_4)$, and let v be the vertex of G that does not belong to H. Thus, v is incident with at least two edges of the same color, say, red. Without loss of generality, we consider the following four cases.

Case 1. $v{v}_1$ and $v{v}_2$ are red edges. If $v_1{v}_3$ or $v_1{v}_4$ is red, then $(v_2,v,v_1,v_3)$ or $(v_2,v,v_1,v_4)$ is a monochromatic $P_4$ that is edge-disjoint from H. If $v_2{v}_4$ is red, then $(v_1,v,v_2,v_4)$ is a monochromatic $P_4$ that is edge-disjoint from H. Thus, we may assume that all three edges $v_1{v}_3$, $v_1{v}_4$, $v_2{v}_4$ are blue, and so $(v_2,v_4,v_1,v_3)$ is a monochromatic $P_4$ that is edge-disjoint from H.

Case 2. $v{v}_2$ and $v{v}_3$ are red edges. By Case 1, we may assume that $v{v}_1$ and $v{v}_4$ are blue. If $v_1{v}_3$ is red, then $(v_1,v_3,v,v_2)$ is a monochromatic $P_4$ that is edge-disjoint from H. Thus, we may assume that $v_1{v}_3$ is blue, and so $(v_4,v,v_1,v_3)$ is a monochromatic $P_4$ that is edge-disjoint from H.

Case 3. $v{v}_1$ and $v{v}_3$ are red edges. By Cases 1 and 2, we may assume that $v{v}_2$ and $v{v}_4$ are blue. Regardless of the color of $v_1{v}_4$, there is a monochromatic $P_4$ that is edge-disjoint from H. Similarly, if $v{v}_2$ and $v{v}_4$ are red, there is a monochromatic $P_4$ that is edge-disjoint from H.

Case 4. $v{v}_1$ and $v{v}_4$ are red edges. By Cases 1 and 2, we may assume that $v{v}_2$ and $v{v}_3$ are blue. Regardless of the color of $v_1{v}_3$, there is a monochromatic $P_4$ that is edge-disjoint from H.

Therefore, $E{R}_2\left(P_4\right)\le 5$, and so $E{R}_2\left(P_4\right)=5$. □

Proposition 6.

$E{R}_3\left(P_4\right)=6$.

Proof. 

The red-blue coloring of $K_5$ with a red $C_5$ and a blue $C_5$ does not produce three pairwise edge-disjoint monochromatic copies of $P_4$. Thus, $E{R}_3\left(P_4\right)\ge 6$. It remains to show that $E{R}_3\left(P_4\right)\le 6$. Let c be a red-blue coloring of $G=K_6$. We show that there are three pairwise edge-disjoint monochromatic copies of $P_4$. Since $E{R}_2\left(P_4\right)=5$, there are two edge-disjoint monochromatic copies $Q_1$ and $Q_2$ of $P_4$ in the subgraph $H=K_5$ in G, where $V\left(H\right)=\{v_1,v_2,v_3,v_4,v_5\}$. Let $Q_1=(v_1,v_2,v_3,v_4)$.

First, suppose that $V\left(Q_2\right)=V\left(Q_1\right)=\{v_1,v_2,v_3,v_4\}$. Then, $Q_2=(v_2,v_4,v_1,v_3)$. Suppose that $Q_1$ is red. We consider the two possibilities where $Q_2$ is either red or blue. Let y be the vertex of G that is not in H.

• First, suppose that $Q_2$ is also red. Thus, $H\left[\{v_1,v_2,v_3,v_4\}\right]$ is a red $K_4$. If either $v_5$ or y, say, $v_5$, is joined to $\{v_1,v_2,v_3,v_4\}$ by three or more red edges, say, $v_5{v}_1,v_5{v}_2,v_5{v}_3$ are red, then $(v_5,v_1,v_2,v_4)$, $(v_5,v_3,v_1,v_4)$, and $(v_5,v_2,v_3,v_4)$ are three pairwise edge-disjoint red copies of $P_4$ in G. Next, suppose that each $v_5$ and y is joined to $\{v_1,v_2,v_3,v_4\}$ by two red edges and two blue edges. We may assume that $v_5{v}_1$ and $v_5{v}_2$ are red and $v_5{v}_3$ and $v_5{v}_4$ are blue. If $y{v}_1$ or $y{v}_2$ is red, say, the former, then $(y,v_1,v_5,v_2)$ is a red $P_4$ that is edge-disjoint from $Q_1$ and $Q_2$. Thus, we may assume that $y{v}_1$ and $y{v}_2$ are both blue, and so $y{v}_3$ and $y{v}_4$ are red. Regardless of the color of $v_{5}y$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Finally, we may assume that either $v_5$ or y is joined to $\{v_1,v_2,v_3,v_4\}$ by three or more blue edges, say, $v_5{v}_1,v_5{v}_2,v_5{v}_3$ are blue. Since y is joined to $\{v_1,v_2,v_3,v_4\}$ by at most two red edges, at least one of $y{v}_1,y{v}_2,y{v}_3$ is blue. Thus, G contains a blue $P_4$ edge-disjoint from $Q_1$ and $Q_2$.
• Next, suppose that $Q_2$ is a blue $P_4$. First, suppose that either $v_5$ or y, say, $v_5$, is joined to $\{v_1,v_2,v_3,v_4\}$ by four edges of the same color, say, red. Then, $(v_1,v_5,v_3,v_4)$, $(v_1,v_2,v_5,v_4)$, and $Q_2$ are three pairwise edge-disjoint monochromatic copies of $P_4$ in G. Next, suppose that either $v_5$ or y, say, $v_5$, is joined to $\{v_1,v_2,v_3,v_4\}$ by three edges of the same color. We may assume that either (i) $v_5{v}_1,v_5{v}_2,v_5{v}_3$ are red and $v_5{v}_4$ is blue or (ii) $v_5{v}_1,v_5{v}_2,v_5{v}_4$ are red and $v_5{v}_3$ is blue. For (i), if $y{v}_i$ is red for some $i\in \{1,2,3\}$, say, $y{v}_1$ is red, then $(y,v_1,v_5,v_2)$ is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Thus, $y{v}_i$ is blue for each $i=1,2,3$ and $y{v}_4$ is red. Regardless of the color of $v_{5}y$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. The argument for (ii) is similar. Finally, suppose that each of $v_5$ and y is joined to $\{v_1,v_2,v_3,v_4\}$ by two red edges and two blue edges. We may assume that (i) $v_5{v}_1$ and $v_5{v}_2$ are red, (ii) $v_5{v}_1$ and $v_5{v}_3$ are red, (iii) $v_5{v}_1$ and $v_5{v}_4$ are red, or (iv) $v_5{v}_2$ and $v_5{v}_3$ are red. For (i), if $y{v}_1$ or $y{v}_2$ is red, say, the former, then $(y,v_1,v_5,v_2)$ is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Thus, we may assume that $y{v}_1$ and $y{v}_2$ are both blue, and so $y{v}_3$ and $y{v}_4$ are red. Regardless of the color of $v_{5}y$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. The arguments for (ii)–(iv) are similar.

Thus, we may now assume that $V\left(Q_2\right)\ne V\left(Q_1\right)$, and so $V\left(Q_1\right)\cup V\left(Q_2\right)=V\left(H\right)$. Here, we show that there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Henceforth, no assumption is made on the colors of $Q_1=(v_1,v_2,v_3,v_4)$ or $Q_2$. Since $Q_2$ contains $v_5$ either as an end-vertex of $Q_2$ (and has degree 1 in $Q_2$) or as an interior vertex of $Q_2$ (and has degree 2 in $Q_2$), it follows that there are either three or two edges incident with $v_5$ in H that do not belong to $Q_2$. Furthermore, there is either one or two edges joining vertices in $V\left(Q_1\right)=\{v_1,v_2,v_3,v_4\}$ that do not belong to $Q_1$ or $Q_2$.

First, suppose that there are two edges of H incident with $v_5$ not belonging to $Q_2$ that are colored the same, say, red. That is, suppose that H contains two red edges $v_5{v}_i$ and $v_5{v}_j$ ($1\le i<j\le 4$) not belonging to $Q_2$. By symmetry, we may assume that $(i,j)\in \left\{\right(1,2),(1,3),(1,4),(2,3\left)\right\}$. If either of the two edges $y{v}_i$ and $y{v}_j$ is red, then there is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Hence, we may assume that $y{v}_i$ and $y{v}_j$ are both blue. If $(i,j)\in \left\{\right(1,2),(1,4),(2,3\left)\right\}$, then either $v_1{v}_3$ or $v_2{v}_4$ is not on $Q_2$. Regardless of the color of such an edge, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Hence, we may assume that $(i,j)=(1,3)$. In this case, either (1) $Q_2=(v_2,v_5,v_4,v_1)$, (2) $Q_2=(v_5,v_2,v_4,v_1)$, or (3) $Q_2=(v_5,v_4,v_1,v_3)$. Since $v_5{v}_1$ and $v_5{v}_3$ are red and $y{v}_1$ and $y{v}_3$ are blue, we may assume without loss of generality that $v_{5}y$ is red. Then, $y{v}_4$ is blue, for otherwise there is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Should (1) or (2) occur, then $v_1{v}_3$ is not an edge of $Q_2$ and regardless of its color, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Hence, we may assume that $Q_2=(v_5,v_4,v_1,v_3)$. In this case, $v_2{v}_4$ is not an edge of $Q_2$. We may assume that $v_2{v}_4$ is red, for otherwise there is a blue $P_4=(v_3,y,v_4,v_2)$ edge-disjoint from $Q_1$ and $Q_2$. This implies that $y{v}_2$ is blue, for otherwise there is a red $P_4=(v_5,y,v_2,v_4)$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $v_5{v}_2$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$.

Consequently, we may now assume that no two edges of H incident with $v_5$ and not belonging to $Q_2$ are colored the same. Thus, there are exactly two edges of H incident with $v_5$ and not belonging to $Q_2$, where one of which is colored red and the other is colored blue. This implies that $v_5$ is an interior vertex of $Q_2$. Let $v_5{v}_i,v_5{v}_j\in E\left(Q_2\right)$, where $1\le i<j\le 4$. By symmetry, we may assume that $(i,j)\in \left\{\right(1,2),(1,4),(2,3),(2,4\left)\right\}$. Thus, there are four possibilities for $Q_2$, namely, $Q_2=(v_1,v_5,v_2,v_4)$, $Q_2=(v_1,v_5,v_4,v_2)$, $Q_2=(v_2,v_5,v_3,v_1)$, and $Q_4=(v_2,v_5,v_4,v_1)$. We consider these four possibilities.

Case 1. $Q_2=(v_1,v_5,v_2,v_4)$. In this case, the edges $v_5{v}_3$ and $v_5{v}_4$ do not belong to $Q_2$. We may assume that $v_5{v}_3$ is red and $v_5{v}_4$ is blue (since the argument is similar when $v_5{v}_3$ is blue and $v_5{v}_4$ is red). First, suppose that $v_{5}y$ is red. Thus, $y{v}_1$ and $y{v}_2$ are both blue, for otherwise there is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $v_1{v}_3$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Next, suppose that $v_{5}y$ is blue. Thus, $y{v}_1$ and $y{v}_2$ are both red, for otherwise there is a blue $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $v_1{v}_4$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$.

Case 2. $Q_2=(v_1,v_5,v_4,v_2)$. In this case, the edges $v_5{v}_2$ and $v_5{v}_3$ do not belong to $Q_2$. We may assume that $v_5{v}_2$ is red and $v_5{v}_3$ is blue (since the argument is similar when $v_5{v}_2$ is blue and $v_5{v}_3$ is red). First, suppose that $v_{5}y$ is red. Thus, $y{v}_1$ and $y{v}_3$ are both blue, for otherwise there is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Then, $(v_1,y,v_3,v_5)$ is a blue $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Next, suppose that $v_{5}y$ is blue. Thus, $y{v}_1$ is red, for otherwise there is a blue $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $y{v}_2$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$.

Case 3. $Q_2=(v_2,v_5,v_3,v_1)$. In this case, the edges $v_5{v}_1$ and $v_5{v}_4$ do not belong to $Q_2$. We may assume that $v_5{v}_1$ is red and $v_5{v}_4$ is blue (since the argument is similar when $v_5{v}_1$ is blue and $v_5{v}_4$ is red). First, suppose that $v_{5}y$ is red. Thus, $y{v}_4$ is blue, for otherwise there is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $y{v}_2$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Next, suppose that $v_{5}y$ is blue. Thus, $y{v}_1$ is red, for otherwise there is a blue $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $y{v}_2$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$.

Case 4. $Q_4=(v_2,v_5,v_4,v_1)$. In this case, the edges $v_5{v}_1$ and $v_5{v}_3$ do not belong to $Q_2$. We may assume that $v_5{v}_1$ is red and $v_5{v}_3$ is blue (since the argument is similar when $v_5{v}_1$ is blue and $v_5{v}_3$ is red). First, suppose that $v_{5}y$ is red. Thus, $y{v}_3$ is blue, for otherwise there is a red $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $y{v}_2$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Next, suppose that $v_{5}y$ is blue. Thus, $y{v}_4$ is red, for otherwise there is a blue $P_4$ edge-disjoint from $Q_1$ and $Q_2$. Regardless of the color of $y{v}_1$, there is a monochromatic $P_4$ edge-disjoint from $Q_1$ and $Q_2$. □

Proposition 7.

$E{R}_4\left(P_4\right)=7$.

Proof. 

Since the red-blue coloring of $K_6$ for which the red subgraph is $2K_3$ and the blue subgraph $K_{3,3}$ has no red $P_4$ and only three pairwise edge-disjoint blue copies of $P_4$, it follows that $E{R}_4\left(P_4\right)\ge 7$. It remains to show that $E{R}_4\left(P_4\right)\le 7$.

Let c be a red-blue coloring of $G=K_7$. We show that there are four pairwise edge-disjoint monochromatic copies of $P_4$. Let $V\left(G\right)=\{v_1,v_2,\dots ,v_7\}$. Let $S=\{v_1,v_2,\dots ,v_6\}$, and let $H=G\left[S\right]$ be the subgraph induced by S. Thus, $H=K_6$. Since $E{R}_3\left(P_4\right)=6$ by Proposition 6, there are three pairwise edge-disjoint monochromatic copies $Q_1,Q_2,Q_3$ of $P_4$ in H. Since nine edges of H belong to $Q_1,Q_2,Q_3$, there are six edges of H that belong to none of $Q_1,Q_2,Q_3$. Let $H^{\prime }$ be the spanning subgraph of H whose edge set consists of these six edges of H that do not belong to any of $Q_1,Q_2,Q_3$. Thus, $H^{\prime }$ is a graph of order 6 and size 6 and so $H^{\prime }$ contains cycles. We consider two cases, according to whether $H^{\prime }$ contains odd cycles or $H^{\prime }$ contains no odd cycles.

Case 1. $H^{\prime }$ contains odd cycles. First, suppose that $H^{\prime }$ contains a 5-cycle C, say, $C=(v_1,v_2,\dots ,v_5,v_1)$. Thus, either $v_6$ has degree 1 or is isolated in $H^{\prime }$. Necessarily, there are two adjacent edges of C that are colored the same, say, $v_1{v}_2$ and $v_1{v}_5$ are colored red. Thus, $v_2{v}_3$, $v_4{v}_5$, $v_7{v}_2$, $v_7{v}_5$ are all blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. However, then, $(v_3,v_2,v_7,v_5)$ is a blue $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. We now assume that $H^{\prime }$ does not contain a 5-cycle. Therefore, $H^{\prime }$ contains a triangle T, say, $T=(v_1,v_2,v_3,v_1)$. Two adjacent edges of T are colored the same, say, $v_1{v}_2$ and $v_1{v}_3$ are colored red. We may assume that $v_7{v}_2$ and $v_7{v}_3$ are all blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$.

★

If $v_2$ or $v_3$ is incident with an edge in $H^{\prime }$ not belonging to T, then regardless of the color of such an edge, there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Hence, we may assume that $v_2$ and $v_3$ have degree 2 in $H^{\prime }$.

★

Suppose that $v_1$ is incident with an edge in $H^{\prime }$ not belonging to T. We may assume that $v_1{v}_4$ is such an edge. Assume first that $v_1{v}_4$ is red. Then, $v_2{v}_3$ is blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Regardless of the color of $v_7{v}_4$, there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Assume then that $v_1{v}_4$ is blue. Then, $v_7{v}_4$ is red or otherwise there is a blue $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Regardless of the color of $v_1{v}_7$, there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$.

Hence, we may assume that $v_1,v_2,v_3$ all have degree 2 in $H^{\prime }$. Since $H^{\prime }$ has order 6 and size 6, it follows that $H^{\prime }=2K_3$. Let $T_1=(v_1,v_2,v_3,v_1)$ and $T_2=(v_4,v_5,v_6,v_4)$ be the two triangles in $H^{\prime }$. At least two edges of $T_1$ are colored the same and at least two edges of $T_2$ are colored the same. We may assume that $v_1{v}_2$ and $v_1{v}_3$ are colored the same and $v_4{v}_5$ and $v_5{v}_6$ are colored the same. We may further assume, without loss of generality, that $v_1{v}_2$ and $v_1{v}_3$ are red. Thus, $v_4{v}_5$ and $v_5{v}_6$ are either both red or both blue.

• Assume first that $v_4{v}_5$ and $v_4{v}_6$ are red. Then, $v_7{v}_2$, $v_7{v}_3$, $v_7{v}_5$, $v_7{v}_6$ must be blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Therefore, $v_2{v}_3$ and $v_5{v}_6$ are red or otherwise there is a blue $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Consequently, we may assume that $v_7{v}_1$ and $v_7{v}_4$ are both blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Since $H^{\prime }=2K_3$, it follows that the complete bipartite subgraph $H^{″}=K_{3,3}$ of H with partite sets $\{v_1,v_2,v_3\}$ and $\{v_4,v_5,v_6\}$ is decomposed into $Q_1,Q_2,Q_3$. Let $Q_1=(v_i,v_j,v_k,v_{\ell })$, where $i,k\in \{1,2,3\}$ and $j,\ell \in \{4,5,6\}$. First, suppose that $Q_1$ is red. Then, there is a red $P_4$ denoted by $Q_1^{\prime }$ obtained from $v_i{v}_j$ and two edges of $T_1$ and a red $P_4$ denoted by $Q_1^{″}$ obtained from $v_j{v}_k$ and two edges of $T_2$. The monochromatic subgraphs $Q_1^{\prime },Q_1^{″},Q_2,Q_3$ are pairwise edge-disjoint copies of $P_4$. Next, suppose that $Q_1$ is blue. Then, $Q_0^{\prime }=(v_j,v_i,v_7,v_a)$, where $a\in \{1,2,3\}-\left\{i\right\}$, and $Q_0^{″}=(v_k,v_j,v_7,v_b)$, where $b\in \{4,5,6\}-\left\{j\right\}$, are edge-disjoint blue copies of $P_4$. Then, $Q_0^{\prime },Q_0^{″},Q_2,Q_3$ are pairwise edge-disjoint monochromatic copies of $P_4$.
• Assume next that $v_4{v}_5$ and $v_4{v}_6$ are blue. Then, $v_7{v}_2,v_7{v}_3$ are blue and $v_7{v}_5,v_7{v}_6$ are red or otherwise there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. If $v_7{v}_1$ is red or $v_7{v}_4$ is blue, then there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Thus, we may assume that $v_7{v}_1$ is blue and $v_7{v}_4$ is red. Then, we may further assume that $v_2{v}_3$ is red and $v_5{v}_6$ is blue or otherwise there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Here, too, the complete bipartite subgraph $H^{″}=K_{3,3}$ of H with partite sets $\{v_1,v_2,v_3\}$ and $\{v_4,v_5,v_6\}$ is decomposed into $Q_1,Q_2,Q_3$. Once again, let $Q_1=(v_i,v_j,v_k,v_{\ell })$, where $i,k\in \{1,2,3\}$ and $j,\ell \in \{4,5,6\}$. We may assume without loss of generality that $Q_1$ is red. Then, there is a red $P_4$, denoted by $Q_1^{\prime }$, obtained from $v_i{v}_j$ and two edges of $T_1$, and a red $P_4$ edge-disjoint from $Q_1^{\prime }$, denoted by $Q_1^{″}$, obtained from $v_j{v}_k$ and two of the edges $v_7{v}_4,v_7{v}_5,v_7{v}_6$. Then, $Q_1^{\prime },Q_1^{″},Q_2,Q_3$ are pairwise edge-disjoint monochromatic copies of $P_4$.

Case 2. $H^{\prime }$ contains no odd cycles. Thus, $H^{\prime }$ contains even cycles. First, suppose that $H^{\prime }=C_6$, say, $H^{\prime }=(v_1,v_2,\cdots ,v_6,v_1)$. Assume first that two adjacent edges of $H^{\prime }$ are colored the same, say, $v_1{v}_2$ and $v_1{v}_6$ are red. We may assume $v_2{v}_3$, $v_5{v}_6$, $v_7{v}_2$, $v_7{v}_6$ are all blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Then, $(v_3,v_2,v_7,v_6)$ is a blue $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Assume then that no two adjacent edges of $H^{\prime }$ are colored the same. Then, we may assume that $v_1{v}_2,v_3{v}_4,v_5{v}_6$ are red and $v_2{v}_3,v_4{v}_5,v_6{v}_1$ are blue. By symmetry, we may further assume that $v_7{v}_1$ is red. Then, $v_7{v}_3$ and $v_7{v}_5$ are blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Then, $(v_2,v_3,v_7,v_5)$ is a blue $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Next, suppose that $H^{\prime }\ne C_6$. Thus, $H^{\prime }=K_{2,3}+K_1$ (the union of $K_{2,3}$ and $K_1$) or $H^{\prime }$ is a unicyclic graph of order 6 whose only cycle is a 4-cycle. In either case, $H^{\prime }$ has a subgraph consisting of a 4-cycle with a pendant edge. Hence, we may assume that $(v_1,v_2,v_3,v_4,v_1)$ is a 4-cycle in $H^{\prime }$ and $v_1{v}_5$ is an edge in $H^{\prime }$. First, suppose that the two edges $v_1{v}_2$ and $v_1{v}_4$ incident with $v_1$ have the same color, say, red. Thus, $v_2{v}_3$ and $v_3{v}_4$ are blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Regardless of the color of $v_7{v}_2$, there is a monochromatic $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Thus, we may assume without loss of generality that $v_1{v}_2$ and $v_1{v}_5$ are red and $v_1{v}_4$ is blue. Thus, $v_2{v}_3,v_7{v}_2,v_7{v}_5$ are blue or otherwise there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. Then, $(v_3,v_2,v_7,v_5)$ is a blue $P_4$ edge-disjoint from $Q_1,Q_2,Q_3$. □

The following is a consequence of Propositions 5, 6, and 7.

Corollary 2.

If $t=2,3,4$, then $E{R}_t\left(P_4\right)=t+3$.

By Corollary 1, it follows that $E{R}_{t+1}\left(P_4\right)\le E{R}_t\left(P_4\right)+2$ for each positive integer t. We saw that $E{R}_t\left(P_4\right)=t+3$ for $t=2,3,4$ and $E{R}_{t+1}\left(P_4\right)=E{R}_t\left(P_4\right)+1$ for $t=2,3$. In fact, $E{R}_{t+1}\left(P_4\right)\le E{R}_t\left(P_4\right)+1$ for each integer $t\ge 2$ under the condition that $E{R}_t\left(P_4\right)\ge t+3$. In order to establish this fact, we first present two lemmas.

Lemma 1.

Let G be a graph of order n and size m. If G is a union of p triangles and q stars, then $m=n-q$.

Proof. 

Let $S_k$ denote a star of order $k\ge 1$. Then, $G=p{K}_3+S_{n_1}+S_{n_2}+\cdots +S_{n_q}$ for some positive integers $n_1,n_2,\dots ,n_q$ if $q\ge 1$. Therefore, $n=3p+{\sum }_{i=1}^q{n}_i$ and $m=3p+{\sum }_{i=1}^q(n_i-1)=3p+\left({\sum }_{i=1}^q{n}_i\right)-q=n-q$. □

Lemma 2.

If k and t are integers with $k\ge t+3\ge 8$, then ${\displaystyle \frac{1}{2}}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)-3t\right]\ge k+2$.

Proof. 

Since $t\ge 5$, it follows that $t\ge 3+{\displaystyle \frac{8}{t}}$, and so $t^2\ge 3t+8$. Since $k\ge t+3$, we have

$$k^2-3k=k(k-3)\ge (t+3)t=t^2+3t\ge (3t+8)+3t=6t+8.$$

Consequently, $k^2+k-4k-8\ge 6t$, and so ${\displaystyle \frac{(k+1)k}{2}}-3t\ge 2k+4$. Thus, ${\displaystyle \frac{1}{2}}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)-3t\right]\ge k+2$. □

Proposition 8.

Let $t\ge 2$ be an integer.

$$If E{R}_t\left(P_4\right)\ge t+3,thenE{R}_{t+1}\left(P_4\right)\le E{R}_t\left(P_4\right)+1.$$

Proof. 

Since the statement is true for $t=2,3$, we may assume that $t\ge 4$. First, suppose that $t=4$. We show that $E{R}_5\left(P_4\right)\le E{R}_4\left(P_4\right)+1=8$. Let c be a red-blue coloring of $G=K_8$. Then, there are four pairwise edge-disjoint monochromatic copies $Q_1,Q_2,Q_3,Q_4$ of $P_4$ in G. Let H be the spanning subgraph of G whose edge set consists of all edges of G that do not belong to any of $Q_1,Q_2,Q_3,Q_4$. Thus, H is a graph of order 8 and size $\left({\displaystyle \genfrac{}{}{0pt}{}{8}{2}}\right)-12=16$. Let $H_r$ be the red subgraph of H and $H_b$ the blue subgraph of H. We may assume that $|E\left(H_r\right)|=m_r\ge m_b=\left|E\left(H_b\right)\right|$. Thus, $m_r\ge 8$, and so $H_r$ contains cycles. Since the order of $H_r$ is 8, it follows by Lemma 1 that $H_r$ is not a union of triangles and stars. Therefore, $H_r$ contains either a cycle of order at least 4 or $(K_2+K_1)\vee K_1$ (or a graph obtained from a triangle by adding a pendant edge). In either case, there is a red $P_4$ edge-disjoint from $Q_1,Q_2,Q_3,Q_4$. Hence, $E{R}_5\left(P_4\right)\le 8$.

Next, suppose that $t\ge 5$. Let $E{R}_t\left(P_4\right)=k\ge t+3\ge 8$. Let c be a red-blue coloring of $G=K_{k+1}$. We show that there are $t+1$ pairwise edge-disjoint monochromatic copies of $P_4$ in G. Since $E{R}_t\left(P_4\right)=k$, there are t pairwise edge-disjoint monochromatic copies $Q_1,Q_2,\dots ,Q_t$ of $P_4$ in G. Let H be the spanning subgraph of G whose edge set consists of all edges of G that do not belong to any of $Q_1,Q_2,\dots ,Q_t$. Thus, H is a graph of order $k+1$ and size $\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)-3t$. Let $H_r$ be the red subgraph of H and $H_b$ the blue subgraph of H. We may assume that $|E\left(H_r\right)|=m_r\ge m_b=\left|E\left(H_b\right)\right|$. Thus, $m_r\ge {\displaystyle \frac{1}{2}}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)-3t\right]$. Since $k\ge t+3\ge 8$, it follows by Lemma 2 that ${\displaystyle \frac{1}{2}}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)-3t\right]\ge k+2$. Thus, $H_r$ has order $k+1$ and size at least $k+2$. Hence, $H_r$ contains cycles. By Lemma 1, it follows that $H_r$ is not a union of triangles and stars and so $H_r$ contains either a cycle of order at least 4 or $(K_2+K_1)\vee K_1$. Therefore, there is a red $P_4$ edge-disjoint from $Q_1,Q_2,\cdots ,Q_t$. Hence, $E{R}_{t+1}\left(P_4\right)\le k+1=E{R}_t\left(P_4\right)+1$. □

With the aid of Propositions 5 and 8, we obtain the following.

Theorem 8.

For each integer $t\ge 2$, $E{R}_t\left(P_4\right)\le t+3$.

(1)

If $E{R}_t\left(P_4\right)\cancel{\equiv }0\left(\mathrm{mod}3\right)$, then $E{R}_t\left(P_4\right)\ge ⌈{\displaystyle \frac{3+\sqrt{1+24t}}{2}}⌉$.

(2)

If $E{R}_t\left(P_4\right)\equiv 0\left(\mathrm{mod}3\right)$, then $E{R}_t\left(P_4\right)\ge ⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉$.

Proof. 

To verify the upper bound, we proceed by induction on t. Since $E{R}_2\left(P_4\right)=5$ by Proposition 5, the result is true for $t=2$. Assume that $E{R}_t\left(P_4\right)\le t+3$ for some integer $t\ge 2$. By Proposition 8 and the induction hypothesis, $E{R}_{t+1}\left(P_4\right)\le E{R}_t\left(P_4\right)+1\le (t+1)+3$.

Next, we verify the lower bound for $E{R}_t\left(P_4\right)$. Let $E{R}_t\left(P_4\right)=k$. Then, every red-blue coloring of $K_k$ produces at least t pairwise edge-disjoint monochromatic copies of $P_4$.

★

First, suppose that $k\cancel{\equiv }0\left(\mathrm{mod}3\right)$. Consider the red-blue coloring of $G=K_k$ with red subgraph $G_r=K_{1,k-1}$ and blue subgraph $G_b=K_{k-1}+K_1$. Since there is no red $P_4$, it follows that $G_b$ contains at least t pairwise edge-disjoint monochromatic copies of $P_4$. This implies that $|E\left(G_b\right)|=\left({\displaystyle \genfrac{}{}{0pt}{}{k-1}{2}}\right)\ge 3t$ or $k^2-3k+(2-6t)\ge 0$. Consequently, $E{R}_t\left(P_4\right)=k\ge ⌈{\displaystyle \frac{3+\sqrt{1+24t}}{2}}⌉$.

★

Next, suppose that $k\equiv 0\left(\mathrm{mod}3\right)$. Consider the red-blue coloring of $G=K_k$ with red subgraph $G_r={\displaystyle \frac{k}{3}}{K}_3$ and blue subgraph $G_b=K_k-E\left({\displaystyle \frac{k}{3}}{K}_3\right)$. Since there is no red $P_4$, it follows that $G_b$ contains at least t pairwise edge-disjoint copies of $P_4$. This implies that $|E\left(G_b\right)|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)-k\ge 3t$ or $k^2-3k-6t\ge 0$. Consequently, $E{R}_t\left(P_4\right)=k\ge ⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉$.

□

If $t=3,4$, then $⌈{\displaystyle \frac{3+\sqrt{1+24t}}{2}}⌉=⌈{\displaystyle \frac{3+\sqrt{9+24t}}{2}}⌉=t+3$. As we saw, $E{R}_t\left(P_4\right)=t+3$ for $t=2,3,4$. Since $E{R}_4\left(P_4\right)\le E{R}_5\left(P_4\right)\le E{R}_4\left(P_4\right)+1$, it follows that $E{R}_5\left(P_4\right)\in \{7,8\}$. Observe that if $t=5$, then ${\displaystyle \frac{3+\sqrt{1+24t}}{2}}=t+2=7$.

4. Conclusions

Vertex-disjoint and edge-disjoint Ramsey numbers were studied for the two connected graphs $K_3$ and $P_3$ of order 3 in an earlier paper [14]. More precisely, it was shown in [14] that $V{R}_t\left(K_3\right)=3t+2$ for every integer $t\ge 2$ and $V{R}_t\left(P_3\right)=3t$ for every positive integer t. Moreover, $E{R}_t\left(P_3\right)=\lceil 2\sqrt{t}+1\rceil$ for every positive integer t. In this paper, we have studied vertex-disjoint and edge-disjoint Ramsey numbers for the two connected graphs $P_4$ and $K_{1,3}$ of size 3. We have shown that $V{R}_t\left(P_4\right)=4t+1$ for every positive integer t and $V{R}_t\left(K_{1,3}\right)=4t$ for every integer $t\ge 2$. Furthermore, $E{R}_2\left(K_{1,3}\right)=6$, $E{R}_3\left(K_{1,3}\right)=E{R}_4\left(K_{1,3}\right)=7$, $E{R}_2\left(P_4\right)=5$, $E{R}_3\left(P_4\right)=6$, and $E{R}_4\left(P_4\right)=7$. While the vertex Ramsey number $V{R}_t\left(F\right)$ increases in terms of the order of F for each $F\in \{K_3,P_3,K_{1,3},P_4\}$ as t increases, the number $E{R}_t\left(F\right)$ does not increase in this manner. It would be interesting to have information about this.