Star-critical Ramsey numbers for cycles versus the complete graph on 5 vertices

TL;DR

This paper determines the star-critical Ramsey numbers for cycles versus the complete graph on 5 vertices, providing exact values and characterizations for these numbers in various cases.

Contribution

The paper establishes exact star-critical Ramsey numbers for cycles versus K_5 and characterizes all Ramsey critical graphs for these parameters.

Findings

r_*(C_n,K_5)=3n-1 for n>3

r_*(C_4,K_5)=13

Complete characterization of Ramsey critical graphs for these parameters

Abstract

Let $G$, $H$ and $K$ represent three graphs without loops or parallel edges and $n$ represent an integer. Given any red blue coloring of the edges of $G$, we say that $K \rightarrow (G,H)$, if there exists red copy of $G$ in $K$ or a blue copy of $H$ in $K$. Let $K_n$ represent a complete graph on $n$ vertices, $C_n$ a cycle on $n$ vertices and $S_n=K_{1,n}$ a star on $n+1$ vertices. The Ramsey number $r(G, H)$ is defined as $\min\{n \mid K_n\rightarrow (G,H)\}$. Likewise, the star-critical Ramsey number $r_*(H, G)$ is defined $\min\{k \mid K_{r(G,H)-1} \sqcup K_{1,k} \rightarrow (H, G) \}$. When $n >3$, in this paper we show that $r_*(C_n,K_5)=3n-1$ except $r_*(C_4,K_5)=13$. We also characterize all Ramsey critical $r(C_n,K_5)$ graphs.

Equations

$$r_{*}(C_{n},K_{5})=\begin{cases}$ $13&\text{if }n=4,\\ \hskip 20.0pt&\\ $ $3n-1&\text{if }n\geq 5.\\ \end{cases}$$

Full text

Keywords: Ramsey theory, Star-critical Ramsey numbers

Mathematics Subject Classification : 05C55, 05D10, 05C38

1 INTRODUCTION

In its classical form, Ramsey’s theorem ensures the existence of the Ramsey numbers $r(n,m)$ defined as $r(n,m)=r(K_{n},K_{m})$. The exact determination of these numbers progresses rapidly in difficulty from the nearly trivial $r(3,3)=6$ to stubbornly resistant $r(5,5)$ [at present known to be between 43 and 48]. One new branches of classical Ramsey numbers namely star -critical Ramsey numbers were introduced by Hook and Isaak in 2010 (see [5, 6]). Using the arrow notation, $r_{*}(H,G)$ can be also be defined as the smallest positive integer $k$ such that $K_{n}\setminus K_{1,n-k-1}\rightarrow(H,G)$ where $n=r(G,H)$ (as $K_{n-1}\sqcup K_{1,k}=K_{n}\setminus K_{1,n-k-1}$). Obviously, the rational behind finding star critical Ramsey numbers is to get a more in-dept understanding of classical case. Recently many papers have tried to find star critical Ramsey numbers for such as fans vs. complete graphs and cycles versus $K_{4}$ (see [2, 4, 5, 6, 7, 11]).

In this paper we concentrate on finding $r_{*}(C_{n},K_{5})$. Most of the work related to this paper, originated with the attempt to solve Erdös and Bondy conjecture which states that $r(C_{n},K_{m})=(n-1)(m-1)+1$ for all $n\geq m\geq 3$, except when $n=m=3$ (see [1]). However, up to now, the conjecture is unsolved and has been only proved for $n\geq m$ and $n\leq 7$ and some other special cases.

2 NOTATION

All graphs $G=(V,E)$ considered in this paper are finite graphs without loops and multiple edges. The order of the graph $G=(V,E)$ is denoted by $|V|$ and the number of edges in the graph is denoted by $|E|$. For a graph $G=(V,E)$ the degree of $v$ is defined as the cardinality of the set of vertices adjacent to $v$. A set $I$ of $V(G)$, is said to be an independent set of a graph $G$ if no pair of vertices of $I$ are connected by an edge in $G$.

Suppose that a graph $G$ contains an $n$ cycle $(u_{1},u_{2},...,u_{n},u_{1})$ and a vertex (say $y_{1}$) outside of the cycle such that $y_{1}$ is adjacent to exactly two vertices (say $u_{i}$ and $u_{j}$) of the $n$ cycle. In such a situation, we say that $y_{1}$ is adjacent to two vertices of the $\{u_{1},u_{2},...,u_{n}\}$ which are length $k$ apart where $k=\min\{(i-j)\mod n,(j-i)\mod n\}$. Throughout the paper, in any graph which is colored by red and blue, we will denote the red edges by a unbroken line and the blue edges by a broken line. For a red/blue coloring of a graph $G$, and vertices $u,v\in V(G)$ such that $(u,v)\in E(G)$, we say that $u$ is a red (resp. blue) neighbor of $v$ if $(v,u)$ is colored red (resp. blue). Any 2-coloring of $K_{r(G,H)-1}$ that does not contain a red $G$ or a blue $H$, is called a critical coloring.

3 PROPERTIES OF $(C_{4},K_{5})$ RAMSEY CRITICAL GRAPHS

In order to prove the main result of this paper, namely finding $r_{*}(C_{n},K_{5})$, we try to utilize the the critical graphs of $r(C_{n},K_{4})$ (see [7]) and the fact that $r(C_{n},K_{5})=4n-3$ for $n\geq 4$ (see [3, 10]). In addition, we use the following lemmas to arrive at the main result.

The following four lemmas is a direct consequence of [8, 9, 7], written by Jayawardene et al.

Lemma 1

(**[8]**, Lemma 2). If $G$ is a graph of order $N$ that contains no $C_{m}$ and the independent number is less than or equal to $n-1$ then the minimal degree is greater than or equal to $N-r(C_{m},K_{n-1})$.

Lemma 2

(**[9]**, Lemma 5). Any $C_{5}$-free graph of order 11 with no independent set of 4 vertices is isomorphic to one of the graphs $R_{11,1}$, $...$, $R_{11,18}$ (see Figure 1) or $R_{11,19}\cong 2K_{4}\cup K_{3}$.

P_{11,1}$$P_{11,2}$$P_{11,3}

P_{11,4}$$P_{11,5}$$P_{11,6}

P_{11,7}$$P_{11,8}$$P_{11,9}$$P_{11,10}$$P_{11,11}$$P_{11,12}

P_{11,13}$$P_{11,14}$$P_{11,15}$$P_{11,16}$$P_{11,17}$$P_{11,18}

Figure 1. $R_{11,k},\ 1\leq k\leq 18$.

Lemma 3

(**[9]**, Lemma 4). Any $C_{5}$-free graph of order 12 with no independent set of 4 vertices is isomorphic to one of the graphs $R_{12,1}$, $R_{12,2}$, $R_{12,3}$, $R_{12,4}$, $R_{12,5}$ (see the following figure) or $R_{12,6}\cong 3K_{4}$.

R_{12,1}$$R_{12,2}$$R_{12,3}$$R_{12,5}

R_{12,4}$$R_{12,5}

Figure 2. $R_{12,k},\ 1\leq k\leq 5$.

Lemma 4

(**[7]**, Lemma 7). Any $C_{6}$-free graph of order 15 with no independent set of 4 vertices is isomorphic to one of the graphs five critical graphs corresponding to $K_{15}$, denoted by $R_{15,1}$, $R_{15,2}$, $R_{15,3}$, $R_{15,4}$ or $R_{15,5}$, where $R_{15,4}\cong 3K_{5}+e$ and $R_{15,5}\cong 3K_{5}$. The other three red graphs, namely $R_{15,1}$, $R_{15,2}$, $R_{15,3}$, are illustrated in Figure 3.

R_{15,1}$$R_{15,2}$$R_{15,3}

Figure 3. The graphs $R_{15,k},\ 1\leq k\leq 3$.

The following lemmas is a direct consequence of [3], written by Bollabás et al.

Lemma 5

Suppose $G$ contains the cycle $(u_{1},u_{2},...,u_{n-1},u_{1})$ of length $n-1$ but no cycle of length $n$. Let $Y=V(G)\setminus\{u_{1},u_{2},...,u_{n-1}\}$. Then,

(a)* No vertex $x\in Y$ is adjacent to two consecutive vertices on the cycle.*

(b)* If $x\in Y$ is adjacent to $u_{i}$ and $u_{j}$ then $u_{i+1}u_{j+1}\notin E(G)$.*

(c)* If $x\in Y$ is adjacent to $u_{i}$ and $u_{j}$ then no vertex $x^{\prime}\in Y$ is adjacent to both $u_{i+1}$ and $u_{j+2}$.*

(d)* Suppose $\alpha(G)=m-1$ where $m\leq\frac{n+3}{2}$ and $\{x_{1},x_{2},...,x_{m-1}\}\subseteq Y$ is an $(m-1)$-element independent set. Then no member of this set is adjacent to $m-2$ or more vertices on the cycle.*

Lemma 6

A $C_{5}$ -free graph of order $16$ with no independent set of 5 vertices contains a isomorphic copy of $4K_{4}$.

Proof.

Let $G$ be a $C_{5}$-free graph of order 16 with no independence set of 5 vertices. First note that $\delta(G)\geq 3$ by lemma 3 as $r(C_{5},K_{4})=13$.

Remark 1: Suppose that $G$ contains a $K_{4}$, then any vertex outside of $K_{4}$ can be adjacent to at most one vertex of the $K_{4}$. Further any two adjacent vertices together can be adjacent to at most one vertex of a $K_{4}$.

Case 1: The minimum degree is 3.

Say $w$ is a vertex of minimum degree. Then $G[V(G)\setminus\overline{\Gamma(w)}]$ will satisfy the conditions of lemma 3. Thus, using all possible cases and using remark 1, any two non adjacent vertices of $G[\Gamma(w)]$ will give rise to an independent set of size five containing them. Therefore, $G[\Gamma(w)]\cong K_{3}$. This will result in the required $4K_{4}$.

Case 2: The minimum degree greater than or equal to 4.

Say $w$ is a vertex of minimum degree. Then $G[V(G)\setminus V[\overline{\Gamma(w)]}]$ will satisfy the conditions of lemma 2. Thus, using all possible cases and using remark 1 one sees that there are no possible extensions.

Case 3: The minimum degree greater than or equal to 5.

Say $w$ is a vertex of minimum degree. Then $G[\Gamma(w)]$ will contain a $C_{3}\cup K_{2}$, $P_{3}\cup K_{2}$, $K_{1,4}$, $K_{1,3}\cup K_{1}$, $2K_{2}$ or at least two isolated vertices. It is worth noting that, we will get $C_{5}$ or else an independent set of size 5 directly in all cases, other than when $G[\Gamma(w)]=C_{3}\cup K_{2}$. In this case, let $H=G[V(G)\setminus\overline{\Gamma(w)}]$. Then, $H$ has 10 vertices. In order to avoid a $C_{5}$ each these three vertices of the $C_{3}$ in $G[\Gamma(w)]$ will each have to be adjacent to two vertices of $H$ and these neighborhoods will have to be non overlapping. Select $v_{1}$, $v_{2}$ and $v_{3}$ from the three neighborhoods. Moreover, in order to avoid a $C_{5}$ the two vertices of the $K_{2}$ in $G[\Gamma(w)]$ will share a common neighbor in $H$, say $v$, as illustrated in the following figure.

v_{1}$$v$$v_{1}$$v_{2}$$v_{3}$$v$$w$$\Gamma(w)

Figure 4. The only option left in case 3

Then in order to avoid a $C_{5}$, $\{w,v_{1},v_{2},v_{3},v\}$ will have to be an independent set contrary to the assumption.

Lemma 7

A $C_{6}$ -free graph of order $20$ with no independent set of 5 vertices contains a isomorphic copy of $4K_{5}$.

Proof.

Let $G$ be a $C_{5}$-free graph of order 20 with no independence set of 5 vertices. First note that, by lemma 1, $\delta(G)\geq 4$ as $r(C_{6},K_{4})=16$.

Remark 2: Any vertex can be adjacent to one vertex of a disjoint $K_{5}$. Further any two adjacent vertices together can be adjacent to at most one vertex of a $K_{5}$.

Case 1: The minimum degree is 4.

Say $w$ is a vertex of minimum degree. Then $G[V(G)\setminus\overline{\Gamma(w)}]$ will satisfy the conditions of lemma 4. Thus, using all possible cases and using remark 2, it follows that any two non adjacent vertices of $G[\Gamma(w)]$ will give rise to an independent set of size five containing them. Therefore, $G[\Gamma(w)]\cong K_{4}$. This will result in the required $4K_{5}$.

Case 2: The minimum degree greater than or equal to 5.

Suppose that $G$ is a $C_{6}$ -free graph on $20$ with no independent set of 5 vertices. Then as $r(C_{5},K_{5})=17$ (see [3, 10]) there exists a cycle $U=(u_{l},u_{2},...,u_{5},u_{1})$ of length $5$. Let $X=\{u_{l},u_{2},...,u_{5}\}$. Define $H=G[X^{c}]$ as the induced subgraph of $G$ not containing the vertices of the cycle and $H_{1}=G[X]$. Then, $|V(H)|=15$ and $|V(H_{1})|=5$.

Suppose that there exists an independent set $Y$ in $H$ of size 4 consisting of the four vertices $y_{1}$, $y_{2}$, $y_{3}$ and $y_{4}$. In order to avoid an independent set of size 5, each vertex of $X$ must be adjacent to at least one vertex of $Y$. Clearly, no vertex of $Y$ can be adjacent to three vertices of $X$ as two of these three adjacent vertices will have to be consecutive vertices of the $C_{5}$ (which will result in a $C_{6}$ containing all the vertices of $X$). Therefore, each vertex of $Y$ is adjacent to at most two vertices of $X$ and if they are adjacent to two vertices of $X$ they must be length 2 apart. As $|X|=5$, we get that without loss of generality $y_{1}$ will have exactly two red neighbors in $X$ length 2 apart (say $u_{1}$ and $u_{3}$). Then without loss of generality $u_{2}$ is adjacent to $y_{2}$ and in order to avoid a red $C_{6}$, $y_{2}$ cannot be adjacent to any other vertex of $X$.

Next without loss of generality $\{y_{3},y_{4}\}$ will have either two vertices or one vertex or no vertices that are adjacent to two vertices of $X$. Furthermore, any vertex of $\{y_{3},y_{4}\}$ adjacent to 2 vertices of $X$ must be adjacent to either $u_{3}$ and $u_{5}$ or $u_{1}$ and $u_{4}$ or $u_{1}$ and $u_{3}$. Moreover, if $y_{3}$ is adjacent to $u_{3},u_{5}$ and $y_{4}$ is adjacent to $u_{4},u_{1}$ then $(u_{1},u_{5},y_{3},u_{3},u_{4},y_{4},u_{1})$ will be a cycle of length $6$, contrary to our assumption. Also if, $y_{3}$ is adjacent to $u_{1},u_{3}$ we will get that $y_{4}$ will be forced to be adjacent to $u_{4},u_{5}$ and then $(u_{1},u_{2},u_{3},u_{4},y_{4},u_{5},u_{1})$ will be a cycle of length $6$, contrary to our assumption. Therefore, without loss of generality, by symmetry, we are left with the two possibilities where $y_{3}$ is adjacent to $u_{5}$ and $y_{4}$ is adjacent to $u_{4}$ or else $y_{3}$ is adjacent to $u_{3},u_{5}$ and $y_{4}$ is adjacent to $u_{4}$. In the first option, in order to avoid an independent set of size 5, $u_{1}$ must be adjacent to $u_{3}$. As $y_{2}$ can not be adjacent to any other vertex of $X$ it must be adjacent to 4 vertices (as minimum degree of $G$ is greater than 5) outside of $X\cup Y$ (say $w_{1}$,$w_{2}$,$w_{3}$ and $w_{4}$), as illustrated in the following figure.

y_{1}$$y_{2}$$y_{3}$$y_{4}$$u_{2}$$u_{3}$$u_{4}$$u_{5}$$u_{1}$$w_{1}$$w_{2}$$w_{3}$$w_{4}

Figure 5. The first possibility

In order to avoid a independent set of size 5 consisting of $\{w_{1},u_{2},y_{1},y_{3},y_{4}\}$ we get $w_{1}$ must be adjacent to $u_{2}$. Similarly, we can conclude that all other vertices of $W=\{w_{1},w_{2},w_{3},w_{4}\}$ too have to be adjacent to $u_{2}$. But then $G[W]=K_{4}$ as otherwise any two non adjacent vertices of $W$ together with $\{y_{1},y_{3},y_{4}\}$ will form an independent set of 5 vertices, contrary to the assumption. This will give us a $C_{6}$ containing all vertices $W$, contrary to our assumption.

y_{1}$$y_{2}$$y_{3}$$y_{4}$$u_{2}$$u_{3}$$u_{4}$$u_{5}$$u_{1}$$w

Figure 6: The second possibility

In the second option, as $u_{5}$ can not be adjacent to $u_{2}$ (may be adjacent to $u_{3}$), it will have to be adjacent to some other vertex (since the minimum degree of $G$ is greater than 5) outside of $X\cup Y$ (say $w$), as illustrated in the above figure. But then, $w$ can not be adjacent to any vertex of $Y$ as any such occurrence will lead to some $C_{6}$. Therefore, $Y\cup\{w\}$ will be an independent set of size 5, contrary to our assumption.

Thus, we can conclude that the initial assumption is false. That is, there is no independent set of order 4 in $H$. By Lemma 4, we can conclude that $H$ is equal to one of the five graphs $R_{15,1}$, $R_{15,2}$, $R_{15,3}$, $R_{15,4}$ or $R_{15,5}$. Now consider any two vertices of $U$, say $u$ and $v$, and suppose that $\left\{u,v\right\}\not\in E(G)$. Since there is no $C_{6}$ in $G$, each of the vertices $u$ and $v$ must be adjacent to at most one vertex of each copy of $K_{5}$ in $H$. Therefore, we can select vertex $x_{1}$ in the first $K_{5}$, vertex $x_{2}$ in the second $K_{5}$ and vertex $x_{3}$ in the third $K_{5}$ such that $x_{1}$, $x_{2}$ and $x_{3}$ are independent and not adjacent to $u$ or $v$. This gives us that $\{u,v,x_{1},x_{2},x_{3}\}$ is an independent set of order 5, a contradiction. Therefore, $\left\{u,v\right\}\in E(G)$. Since $u,v$ are arbitrary vertices in $U$, we can conclude that $X$ induces a $K_{5}$ as required. $\square$

Lemma 8

A $C_{n}$ -free graph (where $n\geq 7$) of order $4(n-1)$ with no independent set of 5 vertices contains a isomorphic copy of $4K_{n-1}$.

Proof.

Suppose that $G$ is a $C_{n}$ -free graph on $4(n-1)$ with no independent set of 5 vertices. Then as $r(C_{n-1},K_{5})=4n-7$ (see [3, 10]) there exists a cycle $(u_{l},u_{2},...,u_{n-1},u_{1})$ of length $n-1$. Let $X=\{u_{l},u_{2},...,u_{n-1}\}$. Define $H=G[X^{c}]$ as the induced subgraph of $G$ not containing the vertices of the cycle and $H_{1}=G[X]$. Then, $|V(H)|=3(n-1)$ and $|V(H_{1})|=n-1$.

Suppose that there exists an independent set $Y$ in $H$ of size 4 consisting of the four vertices $y_{1}$, $y_{2}$, $y_{3}$ and $y_{4}$. That is, $\alpha(G)=4$. From lemma 5(d) (as $5\leq\frac{n+3}{2}$), it follows that every vertex $y$ is adjacent to at most two vertex of the cycle $C_{n-1}$.

Case 1: $n\geq 10$

Then as $n-1>8$, we will get a independent set of size 5, containing $Y$; a contradiction.

Case 2: $n=9$

In order to avoid an independent set of size 5, each vertex of $X$ must be adjacent to at least one vertex of $Y$. Therefore as $|X|=8$, we get that each $y_{i}$ ($1\leq i\leq 4$) will have exactly two disjoint neighbors in $X$ and without loss of generality then we have the following subcases. In the first subcase, we assume that without loss of generality $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 apart in the $C_{8}$. In the second subcase we assume that without loss of generality $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 3 apart in the $C_{8}$. In the third subcase we assume that without loss of generality all vertices of $Y$ are adjacent to two vertices of $X$ which are length 4 apart in the $C_{8}$.

Subcase 2.1: $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 apart in the $C_{8}$.

Without loss of generality assume that $y_{1}$ is adjacent in to $u_{1}$ and $u_{3}$ and $y_{2}$ is adjacent in to $u_{2}$. In order to avoid an independent set of size 5 consisting of $\{u_{1},u_{3},y_{2},y_{3},y_{4}\}$, we get that $(u_{1},u_{3})$ is an edge. Without loss of generality, then $y_{2}$ is adjacent to one vertex of $\{u_{8},u_{7},u_{6}\}$. In the first possibility, as $(u_{2},u_{8})$ must be an edge and therefore we will get a cycle $(u_{3},...,u_{8},y_{2},u_{2},u_{1},u_{3})$ of length $9$, contrary to the assumption. In the next possibility, as $(u_{2},u_{7})$ must be an edge and therefore we will get a cycle $(u_{3},u_{4},u_{5},u_{6},u_{7},y_{2},u_{2},u_{1},y_{1},u_{3})$ of length $9$, contrary to the assumption. In the last possibility without loss of generality gives rise to two scenarios as illustrated in the following two figures.

u_{1}$$u_{4}$$v_{1,3}$$u_{2}$$u_{3}$$v_{2,3}$$v_{2,1}$$u_{5}$$u_{6}$$u_{7}$$u_{8}$$y_{1}$$y_{2}$$y_{4}$$y_{3}

Figure 7. The first scenario

In the first scenario, without loss of generality, $y_{3}$ is adjacent to $u_{5}$ and $u_{7}$ and $y_{4}$ is adjacent to $u_{4}$ and $u_{8}$. Moreover, $(u_{2},u_{6})$, $(u_{4},u_{8})$ and $(u_{5},u_{7})$ must be edges. Thus, we get a cycle $(u_{2},u_{1},y_{1},u_{3},u_{4},u_{8},u_{7},u_{5},u_{6},u_{2})$ of length $9$, contrary to the assumption.

In the second scenario, without loss of generality, $y_{3}$ is adjacent to $u_{5}$ and $u_{8}$ and $y_{4}$ is adjacent to $u_{4}$ and $u_{7}$. Moreover, $(u_{2},u_{6})$, $(u_{5},u_{8})$ and $(u_{4},u_{7})$ must be edges. Thus, we get a cycle $(u_{1},y_{1},u_{3},u_{4},u_{7},u_{8},u_{5},u_{6},u_{2},u_{1})$ of length $9$, contrary to the assumption.

u_{1}$$u_{4}$$v_{1,3}$$u_{2}$$u_{3}$$v_{2,3}$$v_{2,1}$$u_{5}$$u_{6}$$u_{7}$$u_{8}$$y_{1}$$y_{2}$$y_{4}$$y_{3}

Figure 8. The second scenario

**Subcase 2.2: ** $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 3 apart in the $C_{8}$

Without loss of generality assume that $y_{1}$ is adjacent to $u_{1}$ and $u_{4}$ and $y_{2}$ is adjacent to $u_{3}$. In order to avoid an independent set of size 5 consisting of $\{u_{1},u_{4},y_{2},y_{3},y_{4}\}$, we get that $(u_{1},u_{4})$ is an edge. Without loss of generality, then $y_{2}$ must be adjacent to one vertex of $\{u_{2},u_{7},u_{8},u_{6}\}$.

The first possibility, is impossible as we get a cycle $(u_{1},u_{2},y_{2},u_{3},...,u_{8},u_{1})$ of length $9$, contrary to the assumption. In the second possibility we get a cycle $(u_{7},y_{2},u_{3},u_{2},u_{1},$ $y_{1},u_{4},u_{5},u_{6},u_{7})$ of length $9$, contrary to the assumption. In the third possibility we get a cycle $(u_{5},u_{4},u_{1},u_{2},u_{3},y_{2},u_{8},u_{7},u_{6},u_{5})$ of length $9$, contrary to the assumption. In the fourth possibility is without loss of generality $y_{3}$ is adjacent to $u_{2}$. But then the only options for $y_{3}$ to be adjacent to is to two vertices of $\{u_{5},u_{8},u_{7}\}$. However, if $y_{3}$ is adjacent to $u_{2}$ and $u_{5}$ it leads to $(u_{1},u_{4},u_{3},u_{2},y_{3},u_{5},u_{6},u_{7},u_{8},u_{1})$, a cycle of length 9 and if $y_{3}$ is adjacent in red to $u_{2}$ and $u_{8}$ it leads to $(u_{8},y_{3},u_{2},u_{1},u_{4},u_{3},y_{2},u_{6},u_{7},u_{8})$ a cycle of length 9, contrary to the assumption. Therefore, we are left with the scenario when $y_{3}$ is adjacent to $u_{2}$ and $u_{7}$. But then in order to avoid an independent set of size 5, we get that $(u_{2},u_{7})$ and $(u_{3},u_{6})$ are edges. This is illustrated in the following figure.

u_{1}$$u_{4}$$v_{1,3}$$u_{2}$$u_{3}$$v_{2,3}$$v_{2,1}$$u_{5}$$u_{6}$$u_{7}$$u_{8}$$y_{1}$$y_{2}$$y_{4}$$y_{3}

Figure 9. The only remaining scenario

However, even in this scenario too we get a cycle $(u_{1},u_{8},u_{7},u_{2},u_{3},u_{6},u_{5},u_{4},y_{1},u_{1})$ of length $9$, contrary to the assumption.

Subcase 2.3: All vertices of $Y$ are adjacent to two vertices of $X$ which are length 4 apart in the $C_{8}$

Without loss of generality assume that $y_{1}$ is adjacent in to $u_{1}$ and $u_{5}$ and $y_{2}$ is adjacent to $u_{2}$ and $u_{6}$ and $y_{3}$ is adjacent to $u_{4}$ and $u_{8}$

In order to avoid an independent set of size 5, we get that $(u_{1},u_{5})$, $(u_{2},u_{6})$ and $(u_{4},u_{8})$ are edges. But then we get cycle $(u_{3},u_{4},u_{8},u_{7},u_{6},u_{5},y_{1},u_{1},u_{2},u_{3})$ of length $9$, contrary to the assumption.

Case 3: $n=8$

In order to avoid an independent set of size 5, each vertex of $X$ must be adjacent to at least one vertex of $Y$. Therefore as $|X|=7$, we get that without loss of generality each $y_{i}$ ($1\leq i\leq 2$) will have exactly two disjoint neighbors in $X$ such that $\Gamma_{R}(x)\cap\Gamma_{R}(x)\cap X=\emptyset$ for all $x\in\{y_{1},y_{2}\}$ and $y\in\{y_{3},y_{4}\}$. This will give rise to the following subcases. In the first subcase, we assume that without loss of generality $\{y_{1},y_{2}\}$ has a vertex which is adjacent to two vertices of $X$ which are length 2 apart in the $C_{7}$. In the second subcase we assume that without loss of generality both $\{y_{1},y_{2}\}$ are adjacent to two vertices of $X$ which are length 3 apart in the $C_{7}$.

Subcase 3.1: $\{y_{1},y_{2}\}$ has a vertex which is adjacent to two vertices of $X$ which are length 2 apart in the $C_{7}$.

Without loss of generality assume that $y_{1}$ is adjacent to $u_{1}$ and $u_{3}$. Moreover, $y_{2}$ is adjacent to $u_{2}$ and $u_{4}$ or $y_{2}$ is adjacent to $u_{2}$ and $u_{5}$ or else $y_{2}$ is adjacent to $u_{4}$ and $u_{6}$. In order to avoid an independent set of size 5 consisting of $\{u_{1},u_{3},y_{2},y_{3},y_{4}\}$, we get that $(u_{1},u_{3})$ is a edge. In the first possibility we will get a cycle $(u_{1},u_{3},u_{2},y_{2},u_{4},u_{5},u_{6},u_{7},$ $u_{1})$ of length $8$, contrary to the assumption. In the second possibility, we will get a cycle $(u_{1},y_{1},u_{3},u_{2},y_{2},u_{5},u_{6},u_{7},u_{1})$ of length $8$, contrary to the assumption. Therefore, we may assume the only remaining option that $y_{2}$ is adjacent to $u_{4}$ and $u_{6}$. In order to avoid an independent set of size 5 consisting of $\{u_{1},u_{3},y_{2},y_{3},y_{4}\}$, we get that $(u_{4},u_{6})$ is a edge. Since without loss of generality, $y_{3}$ is adjacent to two vertices distinct from $\{u_{1},u_{3},u_{4},u_{6}\}$, we get three possibilities given by $y_{3}$ is adjacent to $u_{2}$ and $u_{5}$ or $y_{3}$ is adjacent to $u_{2}$ and $u_{7}$ or else $y_{3}$ is adjacent to $u_{5}$ and $u_{7}$. This will give a cycles $(u_{1},u_{3},u_{2},y_{3},u_{5},u_{4},u_{6},u_{7},u_{1})$, $(u_{4},u_{3},u_{1},u_{2},y_{3},u_{7},u_{6},u_{5},u_{4})$, $(u_{1},u_{2},u_{3},u_{4},u_{6},u_{5},y_{3},u_{7},u_{1})$ respectively in the three cases, contrary to the assumption.

Subcase 3.2: Both $\{y_{1},y_{2}\}$ are adjacent to two vertices of $X$ which are length 3 apart in the $C_{7}$.

Without loss of generality assume that $y_{1}$ is adjacent to $u_{1}$ and $u_{4}$. By symmetry, we may assume that without loss of generality $y_{2}$ is adjacent to $u_{2}$ and $u_{5}$ or $u_{2}$ and $u_{6}$. In order to avoid an independent set of size 5 consisting of $\{u_{1},u_{4},y_{2},y_{3},y_{4}\}$, we get that $(u_{1},u_{4})$ is an edge. In the first possibility, $y_{2}$ is adjacent to $u_{2}$ and $u_{5}$. However, in order to avoid an independent set of size 5, $(u_{2},u_{5})$ is an edge and thus we will get a cycle $(u_{1},u_{4},u_{3},u_{2},y_{2},u_{5},u_{6},u_{7},u_{1})$ of length $8$, contrary to the assumption. In the second possibility, $y_{2}$ is adjacent to $u_{2}$ and $u_{6}$. In order to avoid an independent set of size 5 consisting of $\{u_{2},u_{6},y_{1},y_{3},y_{4}\}$, we get that $(u_{2},u_{6})$ is a edge. This scenario is illustrated in the following figure 10.

v_{1,3}$$v_{2,3}$$v_{2,1}$$y_{1}$$y_{2}$$y_{4}$$y_{3}$$u_{4}$$u_{3}$$u_{5}$$u_{2}$$u_{4}$$u_{6}$$u_{1}$$u_{7}

Figure 10. The only remaining scenario of Subcase 3.2

Thus we will get a cycle $(u_{1},y_{1},u_{4},u_{3},u_{2},y_{2},u_{6},u_{7},u_{1})$ of length $8$, contrary to the assumption.

Case 4: $n=7$

In order to avoid an independent set of size 5, each vertex of $X$ must be adjacent to at least one vertex of $Y$. Note that in order to avoid a $C_{7}$, no vertex of $Y$ can be adjacent to two vertices of $X$ which are length 1 apart in the $C_{6}$. Thus, we assume that without loss of generality there are two main subcases generated by when $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 or 3 apart in the $C_{6}$ and no other vertex of $Y$ is adjacent to these two neighbors (subcase 4.1) and when such a situation doesn’t exist (subcase 4.2).

Subcase 4.1: $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 or 3 apart in the $C_{6}$ and no other vertex of $Y$ is adjacent inv to these two neighbors

Subcase 4.1.1: $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 apart in the $C_{6}$ and no other vertex of $Y$ is adjacent to these two neighbors

As $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 apart in the $C_{6}$, without loss of generality, assume that $y_{1}$ is adjacent to $u_{1}$ and $u_{3}$ and no other vertex of $Y$ is adjacent to $u_{1}$ and $u_{3}$. Then, either $y_{2}$ is adjacent to $u_{2}$ and $u_{4}$ or $y_{2}$ is adjacent to $u_{2}$ and $u_{5}$ or else $y_{2}$ is adjacent to $u_{4}$ and $u_{6}$ . In order to avoid an independent set of size 5 consisting of $\{u_{1},u_{3},y_{2},y_{3},y_{4}\}$, we get that $(u_{1},u_{3})$ is an edge. In the first possibility, we will get a cycle $(u_{1},u_{3},u_{2},y_{2},u_{4},u_{5},u_{6},u_{1})$ of length $7$, contrary to the assumption. In the second possibility we get a cycle $(u_{1},y_{1},u_{3},u_{2},y_{2},u_{5},u_{6},u_{1})$ of length $7$, contrary to the assumption. In the last option, we get that $y_{2}$ is adjacent to $u_{4}$ and $u_{6}$. Next note that, $y_{3}$ can not be adjacent to both $u_{2}$ and $u_{5}$ as then we will get a cycle $(u_{1},y_{1},u_{3},u_{2},y_{2},u_{5},u_{6},u_{1})$ of length $7$, contrary to the assumption. Similarly, $y_{4}$ can not be adjacent to both $u_{2}$ and $u_{5}$. Therefore, without loss of generality (excluding the posibilities we have already discused in case 4.1.1), we get that $y_{3}$ is adjacent to $u_{5}$ and $y_{4}$ is adjacent to $u_{2}$. Moreover, one can observe that in order to avoid a $C_{7}$ and the cases already discussed in case 4.1.1 both $y_{3}$ and $y_{4}$ can not be adjacent to no more vertices of $X$. In order to avoid an independent set of size 5, we get that $(u_{4},u_{6})$ is an edge. Also in the induced subgraph of $X$, in order to avoid a $C_{7}$, the only other possible edge is either $(u_{1},u_{4})$ or $(u_{3},u_{6})$, but not both. Therefore, in the graph of $G[X\cup Y]$ illustrated in the following figure 11. Without loss of generality $(u_{3},u_{6})$ is marked as a broken edge to indicate that it is a possible edge (see figure 5) and its worth noting that $\{y_{1},y_{2},u_{2},u_{5}\}$ forms an independent set of size 4.

v_{1,3}$$v_{2,3}$$v_{2,1}$$y_{1}$$y_{2}$$y_{4}$$y_{3}$$u_{4}$$u_{3}$$u_{5}$$u_{2}$$u_{4}$$u_{6}$$u_{1}

Figure 11. The structure of the graph induced by $G[X\cup Y]$

The graph $G$ is an order 24 graph with no $C_{7}$ and no independent set of size 5. Therefore, by applying lemma 1, we get that the degree of each of the vertices of $G$ is greater than or equal to 5. Therefore, $u_{4}$ has to be adjacent to some vertex outside $X\cup Y$ (say $w$). In order to avoid an independent set of size 5 consisting of $\{w,y_{1},y_{2},u_{2},u_{5}\}$, we get that $w$ must be adjacent to one of the vertices of $\{y_{1},y_{2},u_{2},u_{5}\}$. This will give a $C_{7}$ containing $w$, contrary to the assumption.

Subcase 4.1.2: $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 3 apart in the $C_{6}$ and no other vertex of $Y$ is adjacent to these two neighbors

Without loss of generality, assume that $y_{1}$ is adjacent to $u_{1}$ and $u_{4}$ and no other vertex of $Y$ is adjacent $u_{1}$ and $u_{4}$. In order to avoid an independent set of size 5, we get that $(u_{1},u_{4})$ is an edge. Hence, one of the vertices in $\{y_{2},y_{3},y_{4}\}$ (say $y_{2}$), must be adjacent to either $u_{2}$ and $u_{5}$ or $u_{2}$ and $u_{6}$ or $u_{3}$ and $u_{5}$ or else $u_{3}$ and $u_{6}$. However, if $y_{2}$ is adjacent to $u_{2}$ and $u_{5}$ (see figure 12), we will get a cycle $(u_{1},u_{4},u_{3},u_{2},y_{2},u_{5},u_{6},u_{1})$ of length $7$, contrary to the assumption. Similarly, if $y_{2}$ is adjacent to $u_{2}$ and $u_{6}$ , we will get a cycle $(u_{4},y_{1},u_{1},u_{6},y_{2},u_{2},u_{3},u_{4})$ and if $y_{2}$ is adjacent to $u_{3}$ and $u_{5}$, we will get a cycle $(u_{1},y_{1},u_{4},u_{3},y_{2},u_{5},u_{6},u_{1})$ and if $y_{2}$ is adjacent to $u_{3}$ and $u_{6}$, we will get a cycle $(u_{1},u_{4},u_{5},u_{6},y_{2},u_{3},u_{2},u_{1})$, contrary to the assumption.

v_{1,3}$$v_{2,3}$$v_{2,1}$$y_{1}$$y_{2}$$y_{4}$$y_{3}$$u_{4}$$u_{3}$$u_{5}$$u_{2}$$u_{4}$$u_{6}$$u_{1}

Figure 12. In subcase 4.2 when $y_{2}$ is adjacent to $u_{2}$ and $u_{5}$

Subcase 4.2: $Y$ has a vertex which is adjacent to two vertices of $X$ which are length 2 or 3 apart in the $C_{6}$ and some other vertex of $Y$ is adjacent in to at least one of these two neighbors

Without loss of generality, assume that $y_{1}$ is adjacent to $u_{1}$ and $w$ length 2 or 3 apart. If one other vertex in $Y$ is adjacent to only one vertex $\{u_{1},w\}$ or both vertices of $\{u_{1},w\}$ and not adjacent to any other vertex in $X$ outside $\{u_{1},w\}$ it would lead to one of the previous two subcases. Therefore, without loss of generality we may assume that, $y_{1}$ is adjacent to $u_{1}$ and $u_{3}$ and $y_{2}$ is adjacent to $u_{1}$ and $u_{4}$ or else $y_{1}$ is adjacent to $u_{1}$ and $u_{3}$ and $y_{2}$ is adjacent to $u_{1}$ and $u_{5}$. In the first possibility, in order to avoid the previous subcases, without loss of generality the only option left is for $y_{3}$ to be adjacent to $u_{2}$ and $u_{5}$ or $y_{3}$ is adjacent to $u_{2}$ and $u_{6}$. If $y_{3}$ to be adjacent to $u_{2}$ and $u_{5}$, we get a cycle $(u_{1},u_{6},u_{5},y_{3},u_{2},u_{3},y_{1},u_{1})$ of length $7$ and if $y_{3}$ to be adjacent to $u_{2}$ and $u_{6}$, we get a cycle $(u_{4},y_{2},u_{1},u_{6},y_{3},u_{2},u_{3},u_{4})$ of length $7$, contrary to the assumption. In the second possibility, in order to avoid a $C_{7}$, we get $\{u_{2},u_{4},u_{6}\}$ will induce an independent set of size 3. But then $\{y_{1},y_{2},u_{2},u_{4},u_{6}\}$ will induce an independent set of size 5 contrary to the assumption.

Now continuing with the proof of the lemma for all four cases we may assume that $H$ is a $C_{n}$ -free graph of order $3(n-1)$ with no independent set of size 4. By [7], we can deduce that $H$ is equal to $3K_{3n-3,1}$, $3K_{3n-3,2}$, …,$2K_{3n-3,6}$, as $n\geq 5$. In any case $H$ contains a copy of a $3K_{n-1}$. Now consider any two vertices of $V(C_{n-1})$ say $u$ and $v$. Since there is no $C_{n}$ in $G$, each of the vertices $u$ and $v$ must be adjacent to at most one vertex each of the three copies of $K_{4}$ in $H$. Also there can be at most two edges connecting a given $K_{n-1}$ in the copy of the $3K_{n-1}$ to the remaining $2K_{n-1}$ subgraphs. Therefore, as $n>5$, we can select three vertices $x_{1}$, $x_{2}$ and $x_{3}$ in first $K_{n-1}$, second $K_{n-1}$ and third $K_{n-1}$ respectively, such that $x_{1}$, $x_{2}$ and $x_{3}$ and are not adjacent to $u$ or $v$. As $\{u,v,x_{1},x_{2},x_{3}\}$ cannot be an independent set of size 5, this will force $(u,v)\in E(G)$. Therefore, $V(C_{n})$ will induce a $K_{n-1}$, since $u,v$ are arbitrary elements of $V(C_{n})$. Hence the lemma.

4 MAIN RESULT

Theorem 1

If $n\geq 4$, then

[TABLE]

Proof. We break up the proof into two cases.

Case $n=4$

Let $R_{13}$ (see Figure 13) represent the unique Ramsey critical $(C_{4},K_{5})$ graph (cf.[8]).

x$$y_{1}$$y_{2}$$y_{3}

Figure 13. The unique $r(C_{4},K_{5})$ - critical graph $R_{13}$

Let $R_{13}^{*}$ represent the graph of order 14, obtained from $R_{13}$ by adding a vertex $y$ and connecting it to the vertices $y_{1},y_{2},y_{3}$. Color the edges of $K_{13}\sqcup K_{1,12}\cong K_{14}-e$, using red and blue such that the red graph is isomorphic to $R_{13}^{*}$. Then the blue graph will be is isomorphic to the complement of the graph $R_{13}^{*}$ in $K_{14}-e$ where $e=(x,y)$.

x$$y$$y_{1}$$y_{2}$$y_{3}

Figure 14. A graph isomorphic to the red coloring of $K_{14}-e$ which contains no $C_{4}$ and which contains no $K_{5}$ in the complement with respect to $K_{14}-e$. Notice the missing edge between the nodes labeled $y$ and $x$ is represented by $e$.

Case $n\geq 5$

Color the graph $K_{4(n-1)+1}\setminus K_{n-2}$ using red and blue colors, such that the red graph consists of a $3K_{n-1}\cup(K_{n-1}\sqcup K_{1,1})$ as illustrated in the following figure.

$x_{1},x_{2},x_{3},$ $...$ ,x_{n-1}$$x_{n},...$$x_{2n-1},...red degree 1blue degree3(n-1)$$K_{n-1}$$K_{n-1}$$K_{n-1}redredred$K_{n-1}$red$x_{3n-2},...,x_{4(n-1)}$

Figure 15. A coloring of $K_{4(n-1)+1}-K_{n-2}$ which contains no red $C_{n}$ and no blue $K_{5}$

Therefore, $r_{*}(C_{n},K_{5})\geq 3n-1$. Next to show that, $r_{*}(C_{n},K_{5})\leq 3n-1$, consider any red/blue coloring of a graph $G=K_{4(n-1)+1}\setminus K_{n-2}$ contains no red $C_{n}$ and no blue $K_{5}$. Let $H$ be the graph obtained by deleting the vertex of degree $3n-1$ (say $v$) from $G$.

Then $H$ is a graph on $4(n-1)$ vertices such that it contains no red $C_{n}$ or a blue $K_{4}$. Therefore, by lemma 6, lemma 7 and lemma 8 we get that $H$ with a red $4K_{n-1}$. Let us denote the sets of vertices of the four connected components by $V_{1}$, $V_{2}$, $V_{3}$ and $V_{4}$ respectively. Since there is no red $C_{n}$ in the coloring, $v$ has at most one red neighbor in each of the four sets $V_{1}$, $V_{2}$, $V_{3}$ and $V_{4}$.

Case 1: $v$ is adjacent to at exactly three vertices of some $V_{i}$ ($1\leq i\leq 4$).

Without loss of generality, we may assume that $v$ is adjacent to all the vertices of $V_{1}$, $V_{2}$ and $V_{3}$. That is, $v$ is adjacent to at least 5 vertices of each $V_{i}$ ($1\leq i\leq 3$). Select $v_{1}\in V_{1}$, $v_{2}\in V_{2}$ and $v_{3}\in V_{3}$ such that $v_{1}$ has no red neighbors in $G[V_{2}\cup V_{3}\cup V_{4}\cup\{v\}]$, $v_{2}$ has no red neighbors in $G[V_{1}\cup V_{3}\cup V_{4}\cup\{v\}]$ and $v_{3}$ has no red neighbors in $G[V_{1}\cup V_{2}\cup V_{4}\cup\{v\}]$ (this is possible because each $V_{i}$ can have at most 4 vertices with red neighbors outside $V_{i}$). Because there are no red $C_{n}$ ’s in the coloring, we can find a $v_{4}\in V_{4}$ such that $\{v,v_{4}\}$ is colored blue. Then, $\{v,v_{1},v_{2},v_{3},v_{4}\}$ will induce a blue $K_{5}$, a contradiction.

Case 2: $v$ is adjacent to exactly two vertices of some $V_{i}$ (say $V_{4}$).

Without loss of generality, we may assume that $v$ is adjacent to all the vertices of $V_{1}$, $V_{2}$ and all but one vertex of $V_{3}$. That is, $v$ is adjacent to at least 5 vertices of each $V_{i}$ ($1\leq i\leq 2$) and at least 4 vertices of $V_{3}$. Since there are at least two vertices of $V_{4}$ adjacent to $v$ in blue, select $v_{4}\in V_{4}$ such that $v_{3}$ has no red neighbors in $G[V_{3}\cup\{v\}]$. Next select any vertex $v_{3}\in V_{3}$ such that $(v_{3},v)$ is colored blue. Then, $\{v,v_{3},v_{4}\}$ will induce a blue $K_{3}$. Finally select $v_{1}\in V_{1}$ and $v_{2}\in V_{2}$ such that $v_{1}$ has no red neighbors in $G[V_{2}\cup V_{3}\cup V_{4}\cup\{v\}]$ and $v_{2}$ has no red neighbors in $G[V_{1}\cup V_{3}\cup V_{4}\cup\{v\}]$ (this is possible because each $V_{i}$ can have at most 4 vertices with red neighbors outside $V_{i}$). Then, $\{v,v_{1},v_{2},v_{3},v_{4}\}$ will induce a blue $K_{5}$, a contradiction.

Case 3: Both case 1 and 2, do not hold.

Without loss of generality, we may assume that $v$ is adjacent to at least 5 vertices of each $V_{1}$ and at least 4 vertices of ($2\leq i\leq 4$). Since there are at least three vertices of $V_{4}$ adjacent to $v$ in blue, select $v_{4}\in V_{4}$ such that $v_{4}$ has no red neighbors in $G[V_{2}\cup V_{3}\cup\{v\}]$. Next select $v_{2}\in V_{2}$ and $v_{3}\in V_{3}$ such that $(v_{2},v)$, $(v_{3},v)$ $(v_{2},v_{3})$ are colored blue. Then, $\{v,v_{2},v_{3},v_{4}\}$ will induce a blue $K_{4}$. Finally select $v_{1}\in V_{1}$ such that $v_{1}$ has no red neighbors in $G[V_{2}\cup V_{3}\cup V_{4}\cup\{v\}]$ (this is possible because each $V_{i}$ can have at most 4 vertices with red neighbors outside $V_{i}$). Then, $\{v,v_{1},v_{2},v_{3},v_{4}\}$ will induce a blue $K_{5}$, a contradiction.

Hence $r_{*}(C_{n},K_{4})\leq 3n-1$. Therefore, $r_{*}(C_{n},K_{4})=3n-1$.

5 ALL RAMSEY $(C_{n},K_{5})$ CRITICAL GRAPHS WHEN $n\geq 4$

The thirty graphs $R_{4n-4,1}$, $...$ ,$R_{4n-4,30}$ are illustrated in the following figure.

R_{4n-4,3}$$R_{4n-4,4}$$R_{4n-4,5}

R_{4n-2,6}$$R_{4n-2,7}$$R_{4n-2,8}

R_{4n-4,9}$$R_{4n-4,10}$$R_{4n-4,11}

R_{4n-4,13}$$R_{4n-4,12}$$R_{4n-4,14}

R_{4n-4,15}$$R_{4n-4,16}$$R_{4n-4,17}

R_{4n-4,18}$$R_{4n-4,19}$$R_{4n-4,20}

R_{4n-4,21}$$R_{4n-4,22}$$R_{4n-4,23}

R_{4n-4,24}$$R_{4n-4,25}$$R_{4n-4,26}

R_{4n-4,27}$$R_{4n-4,28}$$R_{4n-4,29}

$R_{4n-4,30}$

Figure 16. The red graphs $R_{4n-4,1}$, $R_{4n-4,2}$, $...$ ,$R_{4n-4,30}$ when $n\geq 6$

Lemma 9

When $n>3$, all $r(C_{n},K_{5})$ critical graphs will be generated by one of the following.

(a)* One critical generated when $n=4$, with the red graph, of the red/blue coloring, corresponding to $R_{13}$ (see Figure 13).*

(b)* Thirty critical graph generated when $n=5$, with the red graph, of the red/blue coloring, corresponding to $R_{16,1}$, $R_{16,2}$, $...$ ,$R_{16,30}$ (Figure 15).*

(c)* Nineteen critical graph generated when $n=6$, with the red graph, of the red/blue coloring, corresponding to $R_{20,1}$, $R_{20,2}$, $...$ ,$R_{20,19}$ (Figure 15).*

(d)* Eighteen critical graph generated when $n=7$, with the red graph, of the red/blue coloring, corresponding to $R_{24,1}$, $R_{24,2}$, $...$ ,$R_{24,18}$ (Figure 15).*

(e)* Seventeen critical graph generated when $n\geq 8$, with the red graph, of the red/blue coloring, corresponding to $R_{4n-4,1}$, $R_{4n-4,2}$, $...$ ,$n_{4n-4,16}$ or $R_{4n-4,17}$ (Figure 15).*

Proof. (a) When $n=4$ the result follows from ([8], Lemma 11) and the fact $r(C_{4},K_{5})=14$ (cf. [8]).

(b)-(e) When $n\geq 8$, for critical $r(C_{n},K_{5})$ the red graph of the red/blue coloring corresponding to $K_{4(n-1)}$, must contain a $4K_{n-1}$ (by Lemma 8). Note that in order to avoid a $C_{n}$ there can be at most one red edge between any two of the red $K_{n-1}$ graphs. We see that these graphs are generated by the 6 connected subgraphs on 4 vertices ($R_{4n-4,k}$ for $k\in\{5,10,11,12,16,17\}$), 5 disconnected subgraphs on 5 vertices ($R_{4n-4,k}$ for $k\in\{1,2,3,8,9\}$), 5 possible spliting of vertices of trees ($R_{4n-4,k}$ for $k\in\{4,6,7,13,14\}$), and one possible splitting of vertices of a $K_{1,3}+e$ ($R_{4n-4,15}$). Thus, we see that there are only 17 distinct colorings and the corresponding red graphs are given by $R_{4n-4,k}$ for $1\leq k\leq 17$. Next, note that $R_{24,18}$ is the only possible $C_{7}$-free graph which contains a $C_{8}$ and has independence number less than 5. It is generated by splitting of 4 distinct vertices of the $C_{4}$ ($R_{24,18}$). Thus, for $n=7$ there are only 18 distinct colorings and the corresponding red graphs are given by $R_{24,k}$ for $1\leq k\leq 18$. Next, note that $R_{20,19}$ is the only possible $C_{6}$-free graph which contains a $C_{7}$ and has independence number less than 5. It is generated by splitting of 3 distinct vertices of the $C_{4}$ ($R_{20,19}$). Thus, we see that for $n=6$ there are only 19 distinct colorings and the corresponding red graphs are given by $R_{20,k}$ for $1\leq k\leq 19$. Finally, note that $R_{16,k}$ for $20\leq k\leq 30$ are the only possible $C_{5}$-free graphs which contains a $C_{6}$ and has independence number less than 5 (each one of them is generated by appropriate vertex splittings of subgraphs containing a 3 or 4 cycle). Thus, we see that for $n=5$ there are 30 distinct colorings and the corresponding red graphs are given by $R_{16,k}$ for $1\leq k\leq 30$.