Abstract

It has been conjectured by Erdős, Faudree, Rousseau, and Schelp that 𝑟(𝐶𝑚,𝐾𝑛)=(𝑚1)(𝑛1)+1 for all (𝑚,𝑛)(3,3) satisfying that 𝑚𝑛3 (except 𝑟(𝐶3,𝐾3)=6). In this paper, we prove this for the case 𝑚=9 and 𝑛=8.

1. Introduction

All graphs considered in this paper are undirected and simple. 𝐶𝑚,𝑃𝑚,𝐾𝑚 and 𝑆𝑚 stand for cycle, path, complete, and star graphs on 𝑚 vertices, respectively. The graph 𝐾1+𝑃𝑛 is obtained by adding an additional vertex to the path 𝑃𝑛 and connecting this new vertex to each vertex of 𝑃𝑛. The number of edges in a graph 𝐺 is denoted by (𝐺). Further, the minimum degree of a graph 𝐺 is denoted by 𝛿(𝐺). An independent set of vertices of a graph 𝐺 is a subset of the vertex set𝑉(𝐺) in which no two vertices are adjacent. The independence number of a graph 𝐺,𝛼(𝐺), is the size of the largest independent set. The neighborhood of the vertex 𝑢 is the set of all vertices of 𝐺 that are adjacent to 𝑢, denoted by 𝑁(𝑢). 𝑁[𝑢] denote to 𝑁(𝑢){𝑢}. For vertex-disjoint subgraphs 𝐻1 and 𝐻2 of 𝐺 we let 𝐸(𝐻1,𝐻2)={𝑥𝑦𝐸(𝐺)𝑥𝑉(𝐻1),𝑦𝑉(𝐻2)}. Let 𝐻 be a subgraph of the graph 𝐺 and 𝑈𝑉(𝐺), 𝑁𝐻(𝑈) is defined as (𝑢𝑈𝑁(𝑢))𝑉(𝐻). Suppose that 𝑉1𝑉(𝐺) and 𝑉1 is nonempty, the subgraph of 𝐺 whose vertex set is 𝑉1 and whose edge set is the set of those edges of 𝐺 that have both ends in 𝑉1 is called the subgraph of 𝐺 induced by 𝑉1, denoted by 𝑉1𝐺.

The cycle-complete graph Ramsey number 𝑟(𝐶𝑚,𝐾𝑛) is the smallest integer 𝑁 such that for every graph 𝐺 of order 𝑁, 𝐺 contains 𝐶𝑚 or 𝛼(𝐺)𝑛. The graph (𝑛1)𝐾𝑚1 shows that 𝑟(𝐶𝑚,𝐾𝑛)(𝑚1)(𝑛1)+1. In one of the earliest contributions to graphical Ramsey theory, Bondy and Erdős [1] proved the following result: for all 𝑚𝑛22,𝑟(𝐶𝑚,𝐾𝑛)=(𝑚1)(𝑛1)+1. The above restriction was improved by Nikiforov [2] when he proved the equality for 𝑚4𝑛+2. Erdős et al. [3] gave the following conjecture.

Conjecture 1. 𝑟(𝐶𝑚,𝐾𝑛)=(𝑚1)(𝑛1)+1, for all 𝑚𝑛3 except 𝑟(𝐶3,𝐾3)=6.

The conjecture was confirmed by Faudree and Schepl [4] and Rosta [5] for 𝑛=3 in early work on Ramsey theory. Yang et al. [6] and Bollobás et al. [7] proved the conjecture for 𝑛=4 and 𝑛=5, respectively. The conjecture was proved by Schiermeyer [8] for 𝑛=6. Jaradat and Baniabedalruhman [9, 10] proved the conjecture for 𝑛=7 and 𝑚=7,8. Later on, Chena et. al. [11] proved the conjecture for 𝑛=7. Recently, Jaradat and Al-Zaleq [12] and Y. Zhang and K. Zhang [13], independently, proved the conjecture in the case 𝑛=𝑚=8. In a related work, Radziszowski and Tse [14] showed that 𝑟(𝐶4,𝐾7)=22 and 𝑟(𝐶4,𝐾8)=26. In [15] Jayawardene and Rousseau proved that 𝑟(𝐶5,𝐾6)=21. Also, Schiermeyer [16] proved that 𝑟(𝐶5,𝐾7)=25. For more results regarding the Ramsey numbers, see the dynamic survey [17] by Radziszowski.

Until now, the conjecture is still open. Researchers are interested in determining all the values of the Ramsey number 𝑟(𝐶𝑚,𝐾8). In this paper our main purpose is to determine the values of 𝑟(𝐶9,𝐾8) which confirm the conjecture in the case 𝑚=9 and 𝑛=8. The follwoing known theorem will be used in the sequel.

Theorem 1.1. Let 𝐺 be a graph of order 𝑛 without a path of length 𝑘(𝑘1). Then (𝐺)𝑘12𝑛.(1.1) Further, equality holds if and only if its components are complete graphs of order 𝑘.

2. Main Result

In this paper we confirm the Erdős, Faudree, Rousseau, and Schelp conjecture in the case 𝐶9 and 𝐾8. In fact, we prove that 𝑟(𝐶9,𝐾8)=57. It is known, by taking 𝐺=(𝑛1)𝐾𝑚1, that 𝑟(𝐶𝑚,𝐾𝑛)(𝑚1)(𝑛1)+1. In this section we prove that this bound is exact in the case 𝑚=9 and 𝑛=8. Our proof depends on a sequence of 8 lemmas.

Lemma 2.1. Let 𝐺 be a graph of order 57 that contains neither 𝐶9 nor an 8-elemant independent set. Then 𝛿(𝐺)8.

Proof. Suppose that 𝐺 contains a vertex of degree less than 8, say 𝑢. Then |𝑉(𝐺𝑁[𝑢])|49. Since 𝑟(𝐶9,𝐾7)=49, as a result 𝐺𝑁[𝑢] has independent set consists of 7 vertices. This set with the vertex 𝑢 is an 8-elemant independent set of vertices of 𝐺. That is a contradiction.

Throughout all Lemmas 2.2 to 2.8, we let 𝐺 be a graph with minimum degree 𝛿(𝐺)8 that contains neither 𝐶9 nor an 8-elemant independent set.

Lemma 2.2. If 𝐺 contains 𝐾8, then |𝑉(𝐺)|72.

Proof. Let 𝑈={𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7,𝑢8} be the vertex set of 𝐾8, Let 𝑅=𝐺𝑈 and 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) for each 1𝑖8. Since 𝛿(𝐺)8,𝑈𝑖 for all 1𝑖8. Since there is a path of order 8 joining any two vertices of 𝑈, as a result 𝑈𝑖𝑈𝑗= for all 1𝑖<𝑗8 (otherwise, if 𝑤𝑈𝑖𝑈𝑗 for some 1𝑖<𝑗8, then the concatenation of the 𝑢𝑖𝑢𝑗-path of order 8 with 𝑢𝑖𝑤𝑢𝑗, is a cycle of order 9, a contradiction). Similarly, since there is a path of order 7 joining any two vertices of 𝑈, as a result for all 1𝑖<𝑗8 and for all 𝑥𝑈𝑖 and 𝑦𝑈𝑗 we have that 𝑥𝑦𝐸(𝐺) (otherwise, if there are 1𝑖<𝑗8 such that 𝑥𝑈𝑖, 𝑦𝑈𝑗 and 𝑥𝑦𝐸(𝐺), then the concatenation of the 𝑢𝑖𝑢𝑗-path of order 7 with 𝑢𝑖𝑥𝑦𝑢𝑗, is a cycle of order 9, a contradiction). Also, since there is a path of order 6 joining any two vertices of 𝑈, as a result, 𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)=, 1𝑖<𝑗8 (otherwise, if there are 1𝑖<𝑗8 such that 𝑤𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗), then the concatenation of the 𝑢𝑖𝑢𝑗-path of order 6 with 𝑢𝑖𝑥𝑤𝑦𝑢𝑗, is a cycle of order 9 where 𝑥𝑈𝑖, 𝑦𝑈𝑗 and 𝑥𝑤,𝑤𝑦𝐸(𝐺), a contradiction). Therefore |𝑈𝑖𝑁𝑅(𝑈𝑖){𝑢𝑖}|𝛿(𝐺)+1. Thus, |𝑉(𝐺)|8(𝛿(𝐺)+1)(8)(9)=72.

Lemma 2.3. If 𝐺 contains 𝐾8𝑆6, then 𝐺 contains 𝐾8.

Proof. Let 𝑈={𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7,𝑢8} be the vertex set of 𝐾8𝑆6 where the induced subgraph of {𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7} is isomorphic to 𝐾7. Without loss of generality we may assume that 𝑢1𝑢8,𝑢2𝑢8𝐸(𝐺). Let 𝑅=𝐺𝑈 and 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) for each 1𝑖8. Then, as in Lemma 2.2, we have the following:(1)𝑈𝑖𝑈𝑗= for all 1𝑖<𝑗8 except possibly for 𝑖=1 and 𝑗=2.(2)𝐸(𝑈𝑖,𝑈𝑗)= for all 1𝑖<𝑗8.(3)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= for all 1𝑖<𝑗8.(4)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for all 1𝑖<𝑗8.
Since 𝛼(𝐺)7, as a result at least five of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 3𝑖8 are complete. Since 𝛿(𝐺)8, it implies that these complete graphs contain 𝐾8. Hence, 𝐺 contains 𝐾8.

Lemma 2.4. If 𝐺 contains 𝐾7, then 𝐺 contains 𝐾8𝑆6 or 𝐾8.

Proof. Let 𝑈={𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7} be the vertex set of 𝐾7. Let 𝑅=𝐺𝑈 and 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) for each 1𝑖7. Since 𝛿(𝐺)8, 𝑈𝑖 for all 1𝑖7. Now we consider the following two cases.Case 1. 𝑈𝑖𝑈𝑗. for some 1𝑖<𝑗7, say 𝑤𝑈𝑖𝑈𝑗. Then it is clear that 𝐺 contains 𝐾8𝑆6. In fact, the induced subgraph 𝑈{𝑤}𝐺 contains 𝐾8𝑆6.Case 2. 𝑈𝑖𝑈𝑗=. for all 1𝑖<𝑗7. Note that between any two vertices of 𝑈 there are paths of order 5,6 and 7. Thus, as in Lemma 2.2, for all 1𝑖<𝑗7, we have the following.(1)𝐸(𝑈𝑖,𝑈𝑗)=. (2)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)=. (3)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))=.
Since 𝛼(𝐺)7, we have that the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 1𝑖7 are complete. Since 𝛿(𝐺)8, it implies that these complete graphs contain 𝐾8. Hence, 𝐺 contains 𝐾8.

Lemma 2.5. If 𝐺 contains 𝐾1+𝑃7, then 𝐺 contains 𝐾7.

Proof. Let 𝑈={𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7,𝑢8} be the vertex set of 𝐾1+𝑃7 where 𝐾1=𝑢1 and 𝑃7=𝑢2𝑢3𝑢4𝑢5𝑢6𝑢7𝑢8. Let 𝑅=𝐺𝑈 and 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) for each 1𝑖8. Now we have the following two cases.Case 1. 𝑈4𝑈6=. Since 𝛿(𝐺)8, 𝑈𝑖 for all 1𝑖8. Now we have the following.(1)𝑈𝑖𝑈𝑗=for all 2𝑖<𝑗8 except possibly for (𝑖,𝑗){(3,5),(3,6),(3,7),(4,7),(5,7)} since otherwise a cycle of order 9 is produced, a contradiction.(2)𝐸(𝑈𝑖,𝑈𝑗)= for all 2𝑖<𝑗8.(3)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= for all 2𝑖<𝑗8.(4)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for all 2𝑖<𝑗8.
((2), (3), and (4) follows easily from being that 𝐾1+𝑃7 contains paths of order 7, 6, and 5 between any two vertices 𝑢𝑖 and 𝑢𝑗, 2𝑖<𝑗8). Since 𝛼(𝐺)7, as a result at least three of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 𝑖=2,4,5,6,8 are complete graphs. Now we have the following two assertions.
(𝑖)  |𝑁𝑅(𝑈𝑖)|7 and so |𝑈𝑖𝑁𝑅(𝑈𝑖)|8 for each 𝑖=2,8. The following is the proof of assertion (i) for 𝑖=8.
Since 𝛿(𝐺)8, |𝑈8|1. Let 𝑦𝑈8 and 𝑦 is adjacent to 𝑥{𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7}. Then we have the following.(i)If 𝑥=𝑢1, then 𝑢8𝑦𝑢1𝑢2𝑢3𝑢4𝑢5𝑢6𝑢7𝑢8 is a 𝐶9, this is a contradiction.(ii)If 𝑥=𝑢2, then 𝑢8𝑦𝑢2𝑢3𝑢4𝑢5𝑢6𝑢7𝑢1𝑢8 is a 𝐶9, this is a contradiction.(iii)If 𝑥=𝑢3, then 𝑢8𝑦𝑢3𝑢2𝑢1𝑢4𝑢5𝑢6𝑢7𝑢8 is a 𝐶9, this is a contradiction.(iv)If 𝑥=𝑢4, then 𝑢8𝑦𝑢4𝑢3𝑢2𝑢1𝑢5𝑢6𝑢7𝑢8 is a 𝐶9, this is a contradiction.(v)If 𝑥=𝑢5, then 𝑢8𝑦𝑢5𝑢4𝑢3𝑢2𝑢1𝑢6𝑢7𝑢8 is a 𝐶9, this is a contradiction.(vi)If 𝑥=𝑢1, then 𝑢8𝑦𝑢6𝑢5u4𝑢3𝑢2𝑢1𝑢7𝑢8 is a 𝐶9, this is a contradiction.(vii)If 𝑥=𝑢7, then 𝑢8𝑦𝑢7𝑢6𝑢5𝑢4𝑢3𝑢2𝑢1𝑢8 is a 𝐶9, this is a contradiction.
Since 𝛿(𝐺)8, |𝑁𝑅(𝑦)|7, and so |{𝑦}𝑁𝑅(𝑦)|8. Hence, |𝑈8𝑁𝑅(𝑈8)|8. By a similar argument as above and using the symmetry of 𝑃7+𝐾1, one can easily show that |𝑈2𝑁𝑅(𝑈2)|8.
(𝑖𝑖) If there is 𝑖{4,5,6} such that |𝑁𝑅(𝑈𝑖)|<6, then |𝑁𝑅(𝑈𝑗)|6 and so |𝑈𝑗𝑁𝑅(𝑈𝑗)|7 for any 𝑗{4,5,6} with 𝑖𝑗. The following is the proof of assertion (ii).
Assume that |𝑁𝑅(𝑈4)|<6. By (1) 𝑈4𝑈𝑖= for all 2𝑖8 except possibly 𝑖=4,7. Thus, for 𝑦𝑈4, 𝑦 is adjacent to 𝑢4 and to at most 𝑢1 and 𝑢7. Now we show that |𝑁𝑅(𝑈5)|6. Assume |𝑁𝑅(𝑈5)|<6. By (1) 𝑈5𝑈𝑖= for all 2𝑖8 except possibly 𝑖=3,5,7. Thus, for any 𝑤𝑈5, 𝑤 is adjacent to 𝑢5 and to at most 𝑢1, 𝑢3 and 𝑢7. Now, we have the following.(A)If 𝑤 adjacent to both 𝑢1 and 𝑢3, then 𝑢2𝑢3𝑤𝑢5𝑢6𝑢7𝑦𝑢4𝑢1𝑢2 is a 𝐶9.(B)If 𝑤 adjacent to both 𝑢1 and 𝑢7, then 𝑢2𝑢3𝑢4𝑦𝑢5𝑤𝑢7𝑢6𝑢1𝑢2 is a 𝐶9.(C)If 𝑤 adjacent to both 𝑢3 and 𝑢7, then 𝑢2𝑢3𝑤𝑢7𝑢6𝑢5𝑢4𝑦𝑢1𝑢2 is a 𝐶9.
Thus, 𝑤 is adjacent to at most one of 𝑢1,𝑢3, and 𝑢7, and so |𝑁𝑅(𝑈5)|6. We now show that |𝑁𝑅(𝑈6)|6. As above assume |𝑁𝑅(𝑈6)|<6. By (1), 𝑈6𝑈𝑖= for all 2𝑖8 except possibly 𝑖=3,6. Thus, for 𝑤𝑈6, 𝑤 is adjacent to 𝑢1,𝑢3, and 𝑢6. Hence, 𝑢8𝑢7𝑦𝑢4𝑢3𝑤𝑢6𝑢5𝑢1𝑢8 is a 𝐶9, which implies that 𝑤 is adjacent to at most one of 𝑢1 and 𝑢3 and so |𝑁𝑅(𝑈6)|6.
Now, by using the same argument as above and taking into account that 𝑃7+𝐾1 is symmetry, we can easily see that if |𝑁𝑅(𝑈6)|<6, then both of |𝑁𝑅(𝑈4)| and |𝑁𝑅(𝑈5)| are greater than or equal 6. So we need to consider the case when |𝑁𝑅(𝑈5)|<6. As above, 𝑈5𝑈𝑖= for all 2𝑖8 except possibly 𝑖=5,3 and 7. Thus, for any 𝑤𝑈5, 𝑤 is adjacent to 𝑢5 and to at most 𝑢1,𝑢3 and 𝑢7. Now, assume that |𝑁𝑅(𝑈4)|<6. By using (A), (B) and (C) as above and using the same arguments to get the same contradiction. Similarly, by symmetry we get that |𝑁𝑅(𝑈6)|6.
Therefore, from (i) and (ii), at least four of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 𝑖=2,4,5,6,8 contain 7 vertices and so at least two of them contain 𝐾7. Thus, 𝐺 contains 𝐾7.
Case 2. 𝑈4𝑈6, say 𝑢9𝑈4𝑈6. For simplicity, in the rest of this case we consider 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) where 𝑅=𝐺𝑈{𝑢9} and let 𝐽={2,3,5,7,8,9}. Then 𝑢2𝑢9,𝑢3𝑢9,𝑢5𝑢9,𝑢7𝑢9,𝑢8𝑢9𝐸(𝐺) (otherwise, 𝐺 contains 𝐶9) and 𝛿(𝐺)8. Hence 𝑈𝑖,forall𝑖𝐽. Now we have the following assertions (see the Appendix).(1)𝑈𝑖𝑈𝑗= for all 𝑖,𝑗𝐽 and 𝑖𝑗.(2)𝐸(𝑈𝑖,𝑈𝑗)= for all 𝑖,𝑗𝐽 and 𝑖𝑗.(3)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= for all 𝑖,𝑗𝐽 and 𝑖𝑗.(4)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for all 𝑖,𝑗𝐽 and 𝑖𝑗.
Since 𝛼(𝐺)7, as a result at least five of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 𝑖=2,3,5,7,8,9 are complete graphs. Since 𝛿(𝐺)8 and 𝐺 contains no 𝐶9,   |𝑁𝑅(𝑈𝑖)|6 and so |𝑈𝑖𝑁𝑅(𝑈𝑖)|7 for each 𝑖=2,5,8,9. Therefore at least three of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 𝑖=2,3,5,7,8,9 contain 𝐾7. Thus, 𝐺 contains 𝐾7.

Lemma 2.6. If 𝐺 contains 𝐾1+𝑃6, then 𝐺 contains 𝐾1+𝑃7 or 𝐾7.

Proof. Let 𝑈={𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7} be the vertex set of 𝐾1+𝑃6 where 𝐾1=𝑢1 and 𝑃6=𝑢2𝑢3𝑢4𝑢5𝑢6𝑢7. Let 𝑅=𝐺𝑈 and 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) for each 1𝑖7. Since 𝛿(𝐺)8,|𝑈𝑖|2 for all 1𝑖7. Now we have the following cases.Case 1. 𝑈𝑖𝑈j= for all 2𝑖<𝑗7. Then we have the following.(1)𝐸(𝑈𝑖,𝑈𝑗)= for all 2𝑖<𝑗7 except possibly for (𝑖,𝑗){(3,5),(3,6),(4,6)}.(2)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= for all 2𝑖<𝑗7.(3)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for all 2𝑖<𝑗7.
Since 𝛼(𝐺)7, as a result at least one of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 𝑖=2,4,5,7 is complete. Since 𝛿(𝐺)8, it implies that this complete graph contains 𝐾7.
Case 2. 𝑈𝑖𝑈𝑗 for some 2𝑖<𝑗7, say 𝑢8𝑈𝑟𝑈𝑠. In the rest of this case we have the following subcases:Subcase 2.1. (𝑟,𝑠){(6,7),(5,7),(4,7),(7,3),(2,7),(5,6),(4,6),(6,3),(4,5)}. For simplicity, in the rest of this subcase we consider 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) where 𝑅=𝐺𝑈{𝑢8} and let 𝐽={𝑚2𝑚8and𝑚{𝑟,𝑠,(𝑟+𝑠)/𝑠+1}}. Since 𝛿(𝐺)8, then 𝑈𝑖, for all 2𝑖8. Now we have the following assertions.(1)𝑈𝑖𝑈𝑗= for all 𝑖,𝑗𝐽 with 𝑖𝑗.(2)𝐸(𝑈𝑖,𝑈𝑗)= for all 𝑖,𝑗𝐽 with 𝑖𝑗.(3)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= for all 𝑖,𝑗𝐽 with 𝑖𝑗.(4)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for all 𝑖,𝑗𝐽 with 𝑖𝑗.
Since 𝛼(𝐺)7, as a result at least one of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺,𝑖𝐽 is complete.Since 𝛿(𝐺)8 and |𝑈𝑖|2 for each 𝑖𝐽 (because otherwise 𝐺 contains 𝐾1+𝑃7), it implies that this complete graph contains 𝐾7.
Subcase 2.2. (𝑟,𝑠){(6,7),(5,7),(4,7),(7,3),(2,7),(5,6),(4,6),(6,3),(4,5)}. Then, by the symmetry, we have a subcase similar to Subcase 2.1.

Lemma 2.7. If 𝐺 contains 𝐾6, then 𝐺 contains 𝐾1+𝑃6 or 𝐾7.

Proof. Let 𝑈={𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6} be the vertex set of 𝐾6. Let 𝑅=𝐺𝑈 and 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) for each 1𝑖6. Since 𝛿(𝐺)8,|𝑈𝑖|3 for all 1𝑖6. Now we split our work into the following two cases.Case 1. There are 1𝑖<𝑗6 such that 𝑈𝑖𝑈𝑗, then 𝐺 contains 𝐾1+𝑃6.Case 2. 𝑈𝑖𝑈𝑗= for all 1𝑖<𝑗6. Then we consider the following subcases.Subcase 2.1. 𝐸(𝑈𝑖,𝑈𝑗)= for all 1𝑖<𝑗6. Since between any two vertices of 𝑈 there are paths of order 5 and 6, as a result 𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= and 𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for each 1𝑖<𝑗6. Therefore, since 𝛼(𝐺)7, at least five of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 1𝑖6 are complete graphs. Since 𝛿(𝐺)8, these complete graphs contain 𝐾7. Thus, 𝐺 contains 𝐾7.Subcase 2.2. 𝐸(𝑈𝑖,𝑈𝑗) for some 1𝑖<𝑗6, say 𝑖=1 and 𝑗=2 and 𝑢1𝑢7𝑢8𝑢2 is a path. For simplicity, in the rest of this subcase we consider 𝑈𝑖=𝑁(𝑢𝑖)𝑉(𝑅) where 𝑅=𝐺𝑈{𝑢7,𝑢8}. Since 𝛿(𝐺)8, then 𝑈𝑖, for all 3𝑖8. Now we have the following.(1)𝑈𝑖𝑈𝑗= for all 3𝑖<𝑗8.(2)𝐸(𝑈𝑖,𝑈𝑗)= for all 3𝑖<𝑗8.(3)𝑁𝑅(𝑈𝑖)𝑁𝑅(𝑈𝑗)= for all 3𝑖<𝑗8.(4)𝐸(𝑁𝑅(𝑈𝑖),𝑁𝑅(𝑈𝑗))= for all 3𝑖<𝑗8.
Therefore, since 𝛼(𝐺)7, at least five of the induced subgraphs 𝑈𝑖𝑁𝑅(𝑈𝑖)𝐺, 3𝑖8 are complete graphs. Since 𝛿(𝐺)8, it implies that these complete graphs contain 𝐾7. Thus, 𝐺 contains 𝐾7.

Lemma 2.8. If 𝐺 be a graph of order 57, then 𝐺 contains 𝐾1+𝑃6 or 𝐾6.

Proof. Suppose that 𝐺 contains neither 𝐾1+𝑃6 nor 𝐾6. Then we have the following claims.Claim 1. |𝑁(𝑢)|28 for any 𝑢𝑉(𝐺).Proof. Suppose that 𝑢 is a vertex with |𝑁𝐺(𝑢)𝐺|29. Let 𝑁𝐺(𝑢)𝐺=𝑟𝑖=1𝐺𝑖 where 𝐺𝑖 is a component for each 𝑖. 𝑁𝐺(𝑢)𝐺 has minimum number of independent vertices if it has a maximum number of edges. Thus, by Theorem 1.1  𝐺𝑖 must be a complete graph for each 𝑖. But 𝑁𝐺(𝑢)𝐺 contains no 𝑃6. Thus, 𝐺𝑖 must be a complete graph of order at most 5. Also 𝑁𝐺(𝑢)𝐺 contains no 𝐾5, thus 𝐺𝑖 must be a complete graph of order at most 4. Hence, the minimum number of independent vertices of 𝑁𝐺(𝑢) occurs only if 𝑁𝐺(𝑢) contains either a 7 tetrahedrons and an isolated vertex or 6 tetrahedron, a triangle and a 𝐾2 or 6 tetrahedrons and 2 triangles. In any of these cases 𝛼(𝐺)8. This is a contradiction. The proof of the claim is complete.Claim 2. 𝛼(𝐺)=7. Proof. Since |𝑉(𝐺)|57 and 𝐺 contains no 𝐶9 and since 𝑟(𝐶9,𝐾7)=49,𝛼(𝐺)7. But 𝐺 has no 8-element independent set, so 𝛼(𝐺)7. Thus, 𝛼(𝐺)=7. The proof of the claim is complete.
Now, for any 7 independent vertices 𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6, and 𝑢7, set 𝑁𝑖[𝑢𝑖+1]=𝑁[𝑢𝑖+1](𝑖𝑗=1𝑁[𝑢𝑗]),1𝑖6. Analogously, we set 𝑁𝑖(𝑢𝑖+1),   1𝑖6. Let 𝐴=6𝑖=1𝑁𝑖[𝑢𝑖+1],   𝐵=6𝑖=1𝑁𝑖(𝑢𝑖+1), and 𝛽=𝛼(𝐵𝐺).
Claim 3. |𝑁(𝑢1)𝐵|50. Proof. Suppose that |𝑁(𝑢1)𝐵|49. Then |𝑁[𝑢1]𝐴|56. And so |𝐺(𝑁[𝑢1]𝐴)|5756=1. But 𝑟(𝐶9,𝐾1)=1, so 𝐺(𝑁[𝑢1]𝐴) contains a vertex, say 𝑢8, which is not adjacent to any of 𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6, and 𝑢7. Thus, {𝑢1,𝑢2,𝑢3,𝑢4,𝑢5,𝑢6,𝑢7,𝑢8} is an 8-element independent set. Therefore, 𝛼(𝐺)8. That is a contradiction. The proof of the claim is complete.

Now, by Lemma 2.1, 𝛿(𝐺)8 and so by Claim 1, we have that 8|𝑁(𝑢1)|28. Thus, if |𝑁(𝑢1)|=𝑟, then |𝐵|50𝑟. By a similar argument as in Claim 1, we have that 𝛼(𝑁(𝑢1)𝐺)𝑟/4 and 𝛽(50𝑟)/4. Note that for any 8𝑟21,(50𝑟)/4 is greater than or equal to 8. And so 𝛼(𝐺)8. Now we have the following cases.

Case 1. 22|𝑁(𝑢1)|25, then by a similar argument as in Claim 1, we have that 𝛼(𝑁(𝑢1)𝐺)6 and 𝛽7. Then, 𝐵𝐺 has an independent set which consists of 7 vertices. This set with the vertex 𝑢1 is an 8-element independent set of vertices of 𝐺. That is a contradiction.

Case 2. |𝑁(𝑢1)|=26, then |𝐵|24. By a similar argument as in Claim 1, we have that 𝛼(𝑁(𝑢1)𝐺)7 and 𝛽6. Now we have the following two subcases.Subcase 2.1. 𝛽7. Then we have a subcase similar to Case 1.Subcase 2.2. 𝛽=6. The best case of such subgraph is the graph that shown in Figure 1. Now we have the following two subcases.
Subcase 2.2.1. There is a vertex of 6𝑖=1𝑁𝑖(𝑢𝑖+1), say 𝑎1, that is not adjacent to at least one vertex of each 𝐾(𝑗)(1𝑗7), say 𝑥𝑗. Then {𝑥1,𝑥2,𝑥3,𝑥4,𝑥5,𝑥6,𝑥7,𝑎1} is an 8-elemant independent set of vertices of 𝐺. And so 𝛼(𝐺)8. That is a contradiction.
Subcase 2.2.2. For each vertex of 6𝑖=1𝑁𝑖(𝑢𝑖+1) there is 1𝑗7 such that this vertex is adjacent to all vertices of 𝐾(𝑗). Then 𝐺 contains 𝐾1+𝑃6 or 𝐶9. That is a contradiction.

Case 3. 27|𝑁(𝑢1)|28, Then by using the same argument as in Case 2, we have the same contradiction.

Theorem 2.9. 𝑟(𝐶9,𝐾8)=57. Proof. Suppose that there exists a graph 𝐺 of order 57 that contains neither 𝐶9 nor an 8-elements independent set. Then by Lemma 2.1, 𝛿(𝐺)8 and by Lemma 2.8, 𝐺 contains 𝐾1+𝑃6 or 𝐾6. Thus, by Lemmas 2.7, 2.6, 2.5, 2.4, 2.3, and 2.2, we have that |𝑉(𝐺)|72. That is a contradiction. The proof is complete.

Appendix

To show that the assertions (1)–(4) of Case 2 of Lemma 2.5 are true, it suffices to show that for any two vertices of {𝑢2,𝑢3,𝑢5,𝑢7,𝑢8,𝑢9} there are paths of order 8,7,6 and 5. The following are paths of order 8 between vertices of {𝑢3,𝑢5,𝑢7,𝑢8,𝑢9}.1- 𝑢2-𝑢3 path: 𝑢2𝑢1𝑢8𝑢7𝑢6𝑢5𝑢4𝑢3, by symmetry we find 𝑢7-𝑢8 path.2-𝑢2-𝑢5 path: 𝑢2𝑢3𝑢4𝑢1𝑢8𝑢7𝑢6𝑢5, by symmetry we find 𝑢5-𝑢8 path.3-𝑢2-𝑢7 path: 𝑢2𝑢3𝑢4𝑢5𝑢6𝑢1𝑢8𝑢7, by symmetry we find 𝑢3-𝑢8 path.4-𝑢2-𝑢8 path: 𝑢2𝑢3𝑢4𝑢5𝑢6𝑢7𝑢1𝑢8.5-𝑢2-𝑢9 path: 𝑢2𝑢3𝑢4𝑢5𝑢1𝑢7𝑢6𝑢9, by symmetry we find 𝑢8-𝑢9 path.6-𝑢3-𝑢5 path: 𝑢3𝑢4𝑢9𝑢6𝑢7𝑢8𝑢1𝑢5, by symmetry we find 𝑢5-𝑢7 path.7-𝑢3-𝑢7 path: 𝑢3𝑢4𝑢9𝑢6𝑢5𝑢1𝑢8𝑢7.8-𝑢3-𝑢9 path: 𝑢3𝑢4𝑢5𝑢1𝑢8𝑢7𝑢6𝑢9, by symmetry we find 𝑢7-𝑢9 path.3-𝑢5-𝑢9 path: 𝑢5𝑢4𝑢3𝑢2𝑢1𝑢7𝑢6𝑢9.The following are paths of order 7 between vertices of {𝑢3,𝑢5,𝑢7,𝑢8,𝑢9}.1- 𝑢2-𝑢3 path: 𝑢2𝑢1𝑢7𝑢6𝑢5𝑢4𝑢3, by symmetry we find 𝑢7-𝑢8 path.2-𝑢2-𝑢5 path: 𝑢2𝑢3𝑢4𝑢1𝑢7𝑢6𝑢5, by symmetry we find 𝑢5-𝑢8 path.3-𝑢2-𝑢7 path: 𝑢2𝑢3𝑢4𝑢5𝑢6𝑢1𝑢7, by symmetry we find 𝑢3-𝑢8 path.4-𝑢2-𝑢8 path: 𝑢2𝑢3𝑢4𝑢5𝑢6𝑢1𝑢8.5-𝑢2-𝑢9 path: 𝑢2𝑢3𝑢4𝑢5𝑢1𝑢6𝑢9, by symmetry we find 𝑢8-𝑢9 path.2-𝑢3-𝑢5 path: 𝑢3𝑢2𝑢1𝑢8𝑢7𝑢6𝑢5, by symmetry we find 𝑢5-𝑢7 path.2-𝑢3-𝑢7 path: 𝑢3𝑢2𝑢1𝑢4𝑢5𝑢6𝑢7.2-𝑢3-𝑢9 path: 𝑢3𝑢4𝑢5𝑢1𝑢7𝑢6𝑢9, by symmetry we find 𝑢7-𝑢9 path.3-𝑢5-𝑢9 path: 𝑢5𝑢4𝑢3𝑢2𝑢1𝑢6𝑢9.The following are paths of order 6 between vertices of {𝑢3,𝑢5,𝑢7,𝑢8,𝑢9}.1- 𝑢2-𝑢3 path: 𝑢2𝑢1𝑢6𝑢5𝑢4𝑢3, by symmetry we find 𝑢7-𝑢8 path.2-𝑢2-𝑢5 path: 𝑢2𝑢3𝑢4𝑢1𝑢6𝑢5, by symmetry we find 𝑢5-𝑢8 path.3-𝑢2-𝑢7 path: 𝑢2𝑢3𝑢4𝑢5𝑢1𝑢7, by symmetry we find 𝑢3-𝑢8 path.4-𝑢2-𝑢8 path: 𝑢2𝑢3𝑢4𝑢5𝑢1𝑢8.5-𝑢2-𝑢9 path: 𝑢2𝑢3𝑢4𝑢1𝑢6𝑢9, by symmetry we find 𝑢8-𝑢9 path.2-𝑢3-𝑢5 path: 𝑢3𝑢2𝑢1𝑢7𝑢6𝑢5, by symmetry we find 𝑢5-𝑢7 path.2-𝑢3-𝑢7 path: 𝑢3𝑢2𝑢1𝑢5𝑢6𝑢7.2-𝑢3-𝑢9 path: 𝑢3𝑢2𝑢1𝑢7𝑢6𝑢9, by symmetry we find 𝑢7-𝑢9 path.3-𝑢5-𝑢9 path: 𝑢5𝑢1𝑢2𝑢3𝑢4𝑢9.The following are paths of order 5 between vertices of {𝑢3,𝑢5,𝑢7,𝑢8,𝑢9}.1-𝑢2-𝑢3 path: 𝑢2𝑢1𝑢5𝑢4𝑢3, by symmetry we find 𝑢7-𝑢8 path.2-𝑢2-𝑢5 path: 𝑢2𝑢3𝑢4𝑢1𝑢5, by symmetry we find 𝑢5-𝑢8 path.3-𝑢2-𝑢7 path: 𝑢2𝑢3𝑢4𝑢1𝑢7, by symmetry we find 𝑢3-𝑢8 path.4-𝑢2-𝑢8 path: 𝑢2𝑢3𝑢4𝑢1𝑢8.1-𝑢2-𝑢9 path: 𝑢2𝑢1𝑢5𝑢4𝑢9, by symmetry we find 𝑢8-𝑢9 path.2-𝑢3-𝑢5 path: 𝑢3𝑢2𝑢1𝑢6𝑢5, by symmetry we find 𝑢5-𝑢7 path.2-𝑢3-𝑢7 path: 𝑢3𝑢2𝑢1𝑢6𝑢7.2-𝑢3-𝑢9 path: 𝑢3𝑢4𝑢5𝑢67𝑢9, by symmetry we find 𝑢7-𝑢9 path.3-𝑢5-𝑢9 path: 𝑢5𝑢1𝑢3𝑢4𝑢9.