Abstract

It has been conjectured by ErdΕ‘s, Faudree, Rousseau, and Schelp that π‘Ÿ(πΆπ‘š,𝐾𝑛)=(π‘šβˆ’1)(π‘›βˆ’1)+1 for all (π‘š,𝑛)β‰ (3,3) satisfying that π‘šβ‰₯𝑛β‰₯3 (except π‘Ÿ(𝐢3,𝐾3)=6). In this paper, we prove this for the case π‘š=9 and 𝑛=8.

1. Introduction

All graphs considered in this paper are undirected and simple. πΆπ‘š,π‘ƒπ‘š,πΎπ‘š and π‘†π‘š stand for cycle, path, complete, and star graphs on π‘š 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 β„°(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 𝑒 is the set of all vertices of 𝐺 that are adjacent to 𝑒, denoted by 𝑁(𝑒). 𝑁[𝑒] denote to 𝑁(𝑒)βˆͺ{𝑒}. For vertex-disjoint subgraphs 𝐻1 and 𝐻2 of 𝐺 we let 𝐸(𝐻1,𝐻2)={π‘₯π‘¦βˆˆπΈ(𝐺)∢π‘₯βˆˆπ‘‰(𝐻1),π‘¦βˆˆπ‘‰(𝐻2)}. Let 𝐻 be a subgraph of the graph 𝐺 and π‘ˆβŠ†π‘‰(𝐺), 𝑁𝐻(π‘ˆ) is defined as (β‹ƒπ‘’βˆˆπ‘ˆπ‘(𝑒))βˆ©π‘‰(𝐻). Suppose that 𝑉1βŠ†π‘‰(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 βŸ¨π‘‰1⟩𝐺.

The cycle-complete graph Ramsey number π‘Ÿ(πΆπ‘š,𝐾𝑛) is the smallest integer 𝑁 such that for every graph 𝐺 of order 𝑁, 𝐺 contains πΆπ‘š or 𝛼(𝐺)β‰₯𝑛. The graph (π‘›βˆ’1)πΎπ‘šβˆ’1 shows that π‘Ÿ(πΆπ‘š,𝐾𝑛)β‰₯(π‘šβˆ’1)(π‘›βˆ’1)+1. In one of the earliest contributions to graphical Ramsey theory, Bondy and ErdΕ‘s [1] proved the following result: for all π‘šβ‰₯𝑛2βˆ’2,π‘Ÿ(πΆπ‘š,𝐾𝑛)=(π‘šβˆ’1)(π‘›βˆ’1)+1. The above restriction was improved by Nikiforov [2] when he proved the equality for π‘šβ‰₯4𝑛+2. ErdΕ‘s et al. [3] gave the following conjecture.

Conjecture 1. π‘Ÿ(πΆπ‘š,𝐾𝑛)=(π‘šβˆ’1)(π‘›βˆ’1)+1, for all π‘šβ‰₯𝑛β‰₯3 except π‘Ÿ(𝐢3,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 BollobΓ‘s 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 π‘š=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=π‘š=8. In a related work, Radziszowski and Tse [14] showed that π‘Ÿ(𝐢4,𝐾7)=22 and π‘Ÿ(𝐢4,𝐾8)=26. In [15] Jayawardene and Rousseau proved that π‘Ÿ(𝐢5,𝐾6)=21. Also, Schiermeyer [16] proved that π‘Ÿ(𝐢5,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 π‘Ÿ(πΆπ‘š,𝐾8). In this paper our main purpose is to determine the values of π‘Ÿ(𝐢9,𝐾8) which confirm the conjecture in the case π‘š=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 π‘˜(π‘˜β‰₯1). Then β„°(𝐺)β‰€π‘˜βˆ’12𝑛.(1.1) Further, equality holds if and only if its components are complete graphs of order π‘˜.

2. Main Result

In this paper we confirm the ErdΕ‘s, Faudree, Rousseau, and Schelp conjecture in the case 𝐢9 and 𝐾8. In fact, we prove that π‘Ÿ(𝐢9,𝐾8)=57. It is known, by taking 𝐺=(π‘›βˆ’1)πΎπ‘šβˆ’1, that π‘Ÿ(πΆπ‘š,𝐾𝑛)β‰₯(π‘šβˆ’1)(π‘›βˆ’1)+1. In this section we prove that this bound is exact in the case π‘š=9 and 𝑛=8. Our proof depends on a sequence of 8 lemmas.

Lemma 2.1. Let 𝐺 be a graph of order β‰₯57 that contains neither 𝐢9 nor an 8-elemant independent set. Then 𝛿(𝐺)β‰₯8.

Proof. Suppose that 𝐺 contains a vertex of degree less than 8, say 𝑒. Then |𝑉(πΊβˆ’π‘[𝑒])|β‰₯49. Since π‘Ÿ(𝐢9,𝐾7)=49, as a result πΊβˆ’π‘[𝑒] has independent set consists of 7 vertices. This set with the vertex 𝑒 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)β‰₯8 that contains neither 𝐢9 nor an 8-elemant independent set.

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

Proof. Let π‘ˆ={𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7,𝑒8} be the vertex set of 𝐾8, Let 𝑅=πΊβˆ’π‘ˆ and π‘ˆπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(𝑅) for each 1≀𝑖≀8. Since 𝛿(𝐺)β‰₯8,π‘ˆπ‘–β‰ βˆ… for all 1≀𝑖≀8. Since there is a path of order 8 joining any two vertices of π‘ˆ, as a result π‘ˆπ‘–βˆ©π‘ˆπ‘—=βˆ… for all 1≀𝑖<𝑗≀8 (otherwise, if π‘€βˆˆπ‘ˆπ‘–βˆ©π‘ˆπ‘— for some 1≀𝑖<𝑗≀8, then the concatenation of the 𝑒𝑖𝑒𝑗-path of order 8 with 𝑒𝑖𝑀𝑒𝑗, is a cycle of order 9, a contradiction). Similarly, since there is a path of order 7 joining any two vertices of π‘ˆ, as a result for all 1≀𝑖<𝑗≀8 and for all π‘₯βˆˆπ‘ˆπ‘– and π‘¦βˆˆπ‘ˆπ‘— we have that π‘₯π‘¦βˆ‰πΈ(𝐺) (otherwise, if there are 1≀𝑖<𝑗≀8 such that π‘₯βˆˆπ‘ˆπ‘–, π‘¦βˆˆπ‘ˆπ‘— and π‘₯π‘¦βˆˆπΈ(𝐺), then the concatenation of the 𝑒𝑖𝑒𝑗-path of order 7 with 𝑒𝑖π‘₯𝑦𝑒𝑗, is a cycle of order 9, a contradiction). Also, since there is a path of order 6 joining any two vertices of π‘ˆ, as a result, 𝑁𝑅(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—)=βˆ…, 1≀𝑖<𝑗≀8 (otherwise, if there are 1≀𝑖<𝑗≀8 such that π‘€βˆˆπ‘π‘…(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—), then the concatenation of the 𝑒𝑖𝑒𝑗-path of order 6 with 𝑒𝑖π‘₯𝑀𝑦𝑒𝑗, is a cycle of order 9 where π‘₯βˆˆπ‘ˆπ‘–, π‘¦βˆˆπ‘ˆπ‘— and π‘₯𝑀,π‘€π‘¦βˆˆπΈ(𝐺), a contradiction). Therefore |π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)βˆͺ{𝑒𝑖}|β‰₯𝛿(𝐺)+1. Thus, |𝑉(𝐺)|β‰₯8(𝛿(𝐺)+1)β‰₯(8)(9)=72.

Lemma 2.3. If 𝐺 contains 𝐾8βˆ’π‘†6, then 𝐺 contains 𝐾8.

Proof. Let π‘ˆ={𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7,𝑒8} be the vertex set of 𝐾8βˆ’π‘†6 where the induced subgraph of {𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7} is isomorphic to 𝐾7. Without loss of generality we may assume that 𝑒1𝑒8,𝑒2𝑒8∈𝐸(𝐺). Let 𝑅=πΊβˆ’π‘ˆ and π‘ˆπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(𝑅) for each 1≀𝑖≀8. Then, as in Lemma 2.2, we have the following:(1)π‘ˆπ‘–βˆ©π‘ˆπ‘—=βˆ… for all 1≀𝑖<𝑗≀8 except possibly for 𝑖=1 and 𝑗=2.(2)𝐸(π‘ˆπ‘–,π‘ˆπ‘—)=βˆ… for all 1≀𝑖<𝑗≀8.(3)𝑁𝑅(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—)=βˆ… for all 1≀𝑖<𝑗≀8.(4)𝐸(𝑁𝑅(π‘ˆπ‘–),𝑁𝑅(π‘ˆπ‘—))=βˆ… for all 1≀𝑖<𝑗≀8.
Since 𝛼(𝐺)≀7, as a result at least five of the induced subgraphs βŸ¨π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)⟩𝐺, 3≀𝑖≀8 are complete. Since 𝛿(𝐺)β‰₯8, it implies that these complete graphs contain 𝐾8. Hence, 𝐺 contains 𝐾8.

Lemma 2.4. If 𝐺 contains 𝐾7, then 𝐺 contains 𝐾8βˆ’π‘†6 or 𝐾8.

Proof. Let π‘ˆ={𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7} be the vertex set of 𝐾7. Let 𝑅=πΊβˆ’π‘ˆ and π‘ˆπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(𝑅) for each 1≀𝑖≀7. Since 𝛿(𝐺)β‰₯8, π‘ˆπ‘–β‰ βˆ… for all 1≀𝑖≀7. Now we consider the following two cases.Case 1. π‘ˆπ‘–βˆ©π‘ˆπ‘—β‰ βˆ…. for some 1≀𝑖<𝑗≀7, say π‘€βˆˆπ‘ˆπ‘–βˆ©π‘ˆπ‘—. Then it is clear that 𝐺 contains 𝐾8βˆ’π‘†6. In fact, the induced subgraph βŸ¨π‘ˆβˆͺ{𝑀}⟩𝐺 contains 𝐾8βˆ’π‘†6.Case 2. π‘ˆπ‘–βˆ©π‘ˆπ‘—=βˆ…. for all 1≀𝑖<𝑗≀7. Note that between any two vertices of π‘ˆ there are paths of order 5,6 and 7. Thus, as in Lemma 2.2, for all 1≀𝑖<𝑗≀7, we have the following.(1)𝐸(π‘ˆπ‘–,π‘ˆπ‘—)=βˆ…. (2)𝑁𝑅(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—)=βˆ…. (3)𝐸(𝑁𝑅(π‘ˆπ‘–),𝑁𝑅(π‘ˆπ‘—))=βˆ….
Since 𝛼(𝐺)≀7, we have that the induced subgraphs βŸ¨π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)⟩𝐺, 1≀𝑖≀7 are complete. Since 𝛿(𝐺)β‰₯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 π‘ˆ={𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7,𝑒8} be the vertex set of 𝐾1+𝑃7 where 𝐾1=𝑒1 and 𝑃7=𝑒2𝑒3𝑒4𝑒5𝑒6𝑒7𝑒8. Let 𝑅=πΊβˆ’π‘ˆ and π‘ˆπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(𝑅) for each 1≀𝑖≀8. Now we have the following two cases.Case 1. π‘ˆ4βˆ©π‘ˆ6=βˆ…. Since 𝛿(𝐺)β‰₯8,β€‰π‘ˆπ‘–β‰ βˆ… for all 1≀𝑖≀8. Now we have the following.(1)π‘ˆπ‘–βˆ©π‘ˆπ‘—=βˆ…for all 2≀𝑖<𝑗≀8 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(π‘ˆπ‘–,π‘ˆπ‘—)=βˆ… for all 2≀𝑖<𝑗≀8.(3)𝑁𝑅(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—)=βˆ… for all 2≀𝑖<𝑗≀8.(4)𝐸(𝑁𝑅(π‘ˆπ‘–),𝑁𝑅(π‘ˆπ‘—))=βˆ… for all 2≀𝑖<𝑗≀8.
((2), (3), and (4) follows easily from being that 𝐾1+𝑃7 contains paths of order 7, 6, and 5 between any two vertices 𝑒𝑖 and 𝑒𝑗, 2≀𝑖<𝑗≀8). Since 𝛼(𝐺)≀7, as a result at least three of the induced subgraphs βŸ¨π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)⟩𝐺, 𝑖=2,4,5,6,8 are complete graphs. Now we have the following two assertions.
(𝑖)  |𝑁𝑅(π‘ˆπ‘–)|β‰₯7 and so |π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)|β‰₯8 for each 𝑖=2,8. The following is the proof of assertion (i) for 𝑖=8.
Since 𝛿(𝐺)β‰₯8, |π‘ˆ8|β‰₯1. Let π‘¦βˆˆπ‘ˆ8 and 𝑦 is adjacent to π‘₯∈{𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7}. Then we have the following.(i)If π‘₯=𝑒1, then 𝑒8𝑦𝑒1𝑒2𝑒3𝑒4𝑒5𝑒6𝑒7𝑒8 is a 𝐢9, this is a contradiction.(ii)If π‘₯=𝑒2, then 𝑒8𝑦𝑒2𝑒3𝑒4𝑒5𝑒6𝑒7𝑒1𝑒8 is a 𝐢9, this is a contradiction.(iii)If π‘₯=𝑒3, then 𝑒8𝑦𝑒3𝑒2𝑒1𝑒4𝑒5𝑒6𝑒7𝑒8 is a 𝐢9, this is a contradiction.(iv)If π‘₯=𝑒4, then 𝑒8𝑦𝑒4𝑒3𝑒2𝑒1𝑒5𝑒6𝑒7𝑒8 is a 𝐢9, this is a contradiction.(v)If π‘₯=𝑒5, then 𝑒8𝑦𝑒5𝑒4𝑒3𝑒2𝑒1𝑒6𝑒7𝑒8 is a 𝐢9, this is a contradiction.(vi)If π‘₯=𝑒1, then 𝑒8𝑦𝑒6𝑒5u4𝑒3𝑒2𝑒1𝑒7𝑒8 is a 𝐢9, this is a contradiction.(vii)If π‘₯=𝑒7, then 𝑒8𝑦𝑒7𝑒6𝑒5𝑒4𝑒3𝑒2𝑒1𝑒8 is a 𝐢9, this is a contradiction.
Since 𝛿(𝐺)β‰₯8, |𝑁𝑅(𝑦)|β‰₯7, and so |{𝑦}βˆͺ𝑁𝑅(𝑦)|β‰₯8. Hence, |π‘ˆ8βˆͺ𝑁𝑅(π‘ˆ8)|β‰₯8. By a similar argument as above and using the symmetry of 𝑃7+𝐾1, one can easily show that |π‘ˆ2βˆͺ𝑁𝑅(π‘ˆ2)|β‰₯8.
(𝑖𝑖) If there is π‘–βˆˆ{4,5,6} such that |𝑁𝑅(π‘ˆπ‘–)|<6, then |𝑁𝑅(π‘ˆπ‘—)|β‰₯6 and so |π‘ˆπ‘—βˆͺ𝑁𝑅(π‘ˆπ‘—)|β‰₯7 for any π‘—βˆˆ{4,5,6} with 𝑖≠𝑗. The following is the proof of assertion (ii).
Assume that |𝑁𝑅(π‘ˆ4)|<6. By (1) π‘ˆ4βˆ©π‘ˆπ‘–=βˆ… for all 2≀𝑖≀8 except possibly 𝑖=4,7. Thus, for π‘¦βˆˆπ‘ˆ4, 𝑦 is adjacent to 𝑒4 and to at most 𝑒1 and 𝑒7. Now we show that |𝑁𝑅(π‘ˆ5)|β‰₯6. Assume |𝑁𝑅(π‘ˆ5)|<6. By (1) π‘ˆ5βˆ©π‘ˆπ‘–=βˆ… for all 2≀𝑖≀8 except possibly 𝑖=3,5,7. Thus, for any π‘€βˆˆπ‘ˆ5, 𝑀 is adjacent to 𝑒5 and to at most 𝑒1, 𝑒3 and 𝑒7. Now, we have the following.(A)If 𝑀 adjacent to both 𝑒1 and 𝑒3, then 𝑒2𝑒3𝑀𝑒5𝑒6𝑒7𝑦𝑒4𝑒1𝑒2 is a 𝐢9.(B)If 𝑀 adjacent to both 𝑒1 and 𝑒7, then 𝑒2𝑒3𝑒4𝑦𝑒5𝑀𝑒7𝑒6𝑒1𝑒2 is a 𝐢9.(C)If 𝑀 adjacent to both 𝑒3 and 𝑒7, then 𝑒2𝑒3𝑀𝑒7𝑒6𝑒5𝑒4𝑦𝑒1𝑒2 is a 𝐢9.
Thus, 𝑀 is adjacent to at most one of 𝑒1,𝑒3, and 𝑒7, and so |𝑁𝑅(π‘ˆ5)|β‰₯6. We now show that |𝑁𝑅(π‘ˆ6)|β‰₯6. As above assume |𝑁𝑅(π‘ˆ6)|<6. By (1), π‘ˆ6βˆ©π‘ˆπ‘–=βˆ… for all 2≀𝑖≀8 except possibly 𝑖=3,6. Thus, for π‘€βˆˆπ‘ˆ6, 𝑀 is adjacent to 𝑒1,𝑒3, and 𝑒6. Hence, 𝑒8𝑒7𝑦𝑒4𝑒3𝑀𝑒6𝑒5𝑒1𝑒8 is a 𝐢9, which implies that 𝑀 is adjacent to at most one of 𝑒1 and 𝑒3 and so |𝑁𝑅(π‘ˆ6)|β‰₯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(π‘ˆ6)|<6, then both of |𝑁𝑅(π‘ˆ4)| and |𝑁𝑅(π‘ˆ5)| are greater than or equal 6. So we need to consider the case when |𝑁𝑅(π‘ˆ5)|<6. As above, π‘ˆ5βˆ©π‘ˆπ‘–=βˆ… for all 2≀𝑖≀8 except possibly 𝑖=5,3 and 7. Thus, for any π‘€βˆˆπ‘ˆ5, 𝑀 is adjacent to 𝑒5 and to at most 𝑒1,𝑒3 and 𝑒7. Now, assume that |𝑁𝑅(π‘ˆ4)|<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(π‘ˆ6)|β‰₯6.
Therefore, from (i) and (ii), at least four of the induced subgraphs βŸ¨π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)⟩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. π‘ˆ4βˆ©π‘ˆ6β‰ βˆ…, say 𝑒9βˆˆπ‘ˆ4βˆ©π‘ˆ6. For simplicity, in the rest of this case we consider π‘ˆξ…žπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(π‘…ξ…ž) where π‘…ξ…ž=πΊβˆ’π‘ˆβˆͺ{𝑒9} and let 𝐽={2,3,5,7,8,9}. Then 𝑒2𝑒9,𝑒3𝑒9,𝑒5𝑒9,𝑒7𝑒9,𝑒8𝑒9βˆ‰πΈ(𝐺) (otherwise, 𝐺 contains 𝐢9) and 𝛿(𝐺)β‰₯8. Hence π‘ˆξ…žπ‘–β‰ βˆ…,forallπ‘–βˆˆπ½. Now we have the following assertions (see the Appendix).(1)π‘ˆξ…žπ‘–βˆ©π‘ˆξ…žπ‘—=βˆ… for all 𝑖,π‘—βˆˆπ½ and 𝑖≠𝑗.(2)𝐸(π‘ˆξ…žπ‘–,π‘ˆξ…žπ‘—)=βˆ… for all 𝑖,π‘—βˆˆπ½ and 𝑖≠𝑗.(3)𝑁𝑅(π‘ˆξ…žπ‘–)βˆ©π‘π‘…(π‘ˆξ…žπ‘—)=βˆ… for all 𝑖,π‘—βˆˆπ½ and 𝑖≠𝑗.(4)𝐸(𝑁𝑅(π‘ˆξ…žπ‘–),𝑁𝑅(π‘ˆξ…žπ‘—))=βˆ… for all 𝑖,π‘—βˆˆπ½ and 𝑖≠𝑗.
Since 𝛼(𝐺)≀7, as a result at least five of the induced subgraphs βŸ¨π‘ˆξ…žπ‘–βˆͺ𝑁𝑅(π‘ˆξ…žπ‘–)⟩𝐺, 𝑖=2,3,5,7,8,9 are complete graphs. Since 𝛿(𝐺)β‰₯8 and 𝐺 contains no 𝐢9,   |𝑁𝑅(π‘ˆξ…žπ‘–)|β‰₯6 and so |π‘ˆξ…žπ‘–βˆͺ𝑁𝑅(π‘ˆξ…žπ‘–)|β‰₯7 for each 𝑖=2,5,8,9. Therefore at least three of the induced subgraphs βŸ¨π‘ˆξ…žπ‘–βˆͺ𝑁𝑅(π‘ˆξ…žπ‘–)⟩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 π‘ˆ={𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7} be the vertex set of 𝐾1+𝑃6 where 𝐾1=𝑒1 and 𝑃6=𝑒2𝑒3𝑒4𝑒5𝑒6𝑒7. Let 𝑅=πΊβˆ’π‘ˆ and π‘ˆπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(𝑅) for each 1≀𝑖≀7. Since 𝛿(𝐺)β‰₯8,|π‘ˆπ‘–|β‰₯2 for all 1≀𝑖≀7. Now we have the following cases.Case 1. π‘ˆπ‘–βˆ©π‘ˆj=βˆ… for all 2≀𝑖<𝑗≀7. Then we have the following.(1)𝐸(π‘ˆπ‘–,π‘ˆπ‘—)=βˆ… for all 2≀𝑖<𝑗≀7 except possibly for (𝑖,𝑗)∈{(3,5),(3,6),(4,6)}.(2)𝑁𝑅(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—)=βˆ… for all 2≀𝑖<𝑗≀7.(3)𝐸(𝑁𝑅(π‘ˆπ‘–),𝑁𝑅(π‘ˆπ‘—))=βˆ… for all 2≀𝑖<𝑗≀7.
Since 𝛼(𝐺)≀7, as a result at least one of the induced subgraphs βŸ¨π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)⟩𝐺, 𝑖=2,4,5,7 is complete. Since 𝛿(𝐺)β‰₯8, it implies that this complete graph contains 𝐾7.
Case 2. π‘ˆπ‘–βˆ©π‘ˆπ‘—β‰ βˆ… for some 2≀𝑖<𝑗≀7, say 𝑒8βˆˆπ‘ˆπ‘Ÿβˆ©π‘ˆπ‘ . In the rest of this case we have the following subcases:Subcase 2.1. (π‘Ÿ,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 π‘ˆξ…žπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(π‘…ξ…ž) where π‘…ξ…ž=πΊβˆ’π‘ˆβˆͺ{𝑒8} and let 𝐽={π‘šβˆΆ2β‰€π‘šβ‰€8andπ‘šβˆ‰{π‘Ÿ,𝑠,⌈(π‘Ÿ+𝑠)/π‘ βŒ‰+1}}. Since 𝛿(𝐺)β‰₯8, then π‘ˆξ…žπ‘–β‰ βˆ…, for all 2≀𝑖≀8. Now we have the following assertions.(1)π‘ˆξ…žπ‘–βˆ©π‘ˆξ…žπ‘—=βˆ… for all 𝑖,π‘—βˆˆπ½ with 𝑖≠𝑗.(2)𝐸(π‘ˆξ…žπ‘–,π‘ˆξ…žπ‘—)=βˆ… for all 𝑖,π‘—βˆˆπ½ with 𝑖≠𝑗.(3)𝑁𝑅′(π‘ˆξ…žπ‘–)βˆ©π‘π‘…β€²(π‘ˆξ…žπ‘—)=βˆ… for all 𝑖,π‘—βˆˆπ½ with 𝑖≠𝑗.(4)𝐸(𝑁𝑅′(π‘ˆξ…žπ‘–),𝑁𝑅′(π‘ˆξ…žπ‘—))=βˆ… for all 𝑖,π‘—βˆˆπ½ with 𝑖≠𝑗.
Since 𝛼(𝐺)≀7, as a result at least one of the induced subgraphs βŸ¨π‘ˆξ…žπ‘–βˆͺ𝑁𝑅′(π‘ˆξ…žπ‘–)⟩𝐺,π‘–βˆˆπ½ is complete.Since 𝛿(𝐺)β‰₯8 and |π‘ˆξ…žπ‘–|β‰₯2 for each π‘–βˆˆπ½ (because otherwise 𝐺 contains 𝐾1+𝑃7), it implies that this complete graph contains 𝐾7.
Subcase 2.2. (π‘Ÿ,𝑠)βˆ‰{(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 π‘ˆ={𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6} be the vertex set of 𝐾6. Let 𝑅=πΊβˆ’π‘ˆ and π‘ˆπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(𝑅) for each 1≀𝑖≀6. Since 𝛿(𝐺)β‰₯8,|π‘ˆπ‘–|β‰₯3 for all 1≀𝑖≀6. Now we split our work into the following two cases.Case 1. There are 1≀𝑖<𝑗≀6 such that π‘ˆπ‘–βˆ©π‘ˆπ‘—β‰ βˆ…, then 𝐺 contains 𝐾1+𝑃6.Case 2. π‘ˆπ‘–βˆ©π‘ˆπ‘—=βˆ… for all 1≀𝑖<𝑗≀6. Then we consider the following subcases.Subcase 2.1. 𝐸(π‘ˆπ‘–,π‘ˆπ‘—)=βˆ… for all 1≀𝑖<𝑗≀6. Since between any two vertices of π‘ˆ there are paths of order 5 and 6, as a result 𝑁𝑅(π‘ˆπ‘–)βˆ©π‘π‘…(π‘ˆπ‘—)=βˆ… and 𝐸(𝑁𝑅(π‘ˆπ‘–),𝑁𝑅(π‘ˆπ‘—))=βˆ… for each 1≀𝑖<𝑗≀6. Therefore, since 𝛼(𝐺)≀7, at least five of the induced subgraphs βŸ¨π‘ˆπ‘–βˆͺ𝑁𝑅(π‘ˆπ‘–)⟩𝐺, 1≀𝑖≀6 are complete graphs. Since 𝛿(𝐺)β‰₯8, these complete graphs contain 𝐾7. Thus, 𝐺 contains 𝐾7.Subcase 2.2. 𝐸(π‘ˆπ‘–,π‘ˆπ‘—)β‰ βˆ… for some 1≀𝑖<𝑗≀6, say 𝑖=1 and 𝑗=2 and 𝑒1𝑒7𝑒8𝑒2 is a path. For simplicity, in the rest of this subcase we consider π‘ˆξ…žπ‘–=𝑁(𝑒𝑖)βˆ©π‘‰(π‘…ξ…ž) where π‘…ξ…ž=πΊβˆ’π‘ˆβˆͺ{𝑒7,𝑒8}. Since 𝛿(𝐺)β‰₯8, then π‘ˆξ…žπ‘–β‰ βˆ…, for all 3≀𝑖≀8. Now we have the following.(1)π‘ˆξ…žπ‘–βˆ©π‘ˆξ…žπ‘—=βˆ… for all 3≀𝑖<𝑗≀8.(2)𝐸(π‘ˆξ…žπ‘–,π‘ˆξ…žπ‘—)=βˆ… for all 3≀𝑖<𝑗≀8.(3)𝑁𝑅(π‘ˆξ…žπ‘–)βˆ©π‘π‘…(π‘ˆξ…žπ‘—)=βˆ… for all 3≀𝑖<𝑗≀8.(4)𝐸(𝑁𝑅(π‘ˆξ…žπ‘–),𝑁𝑅(π‘ˆξ…žπ‘—))=βˆ… for all 3≀𝑖<𝑗≀8.
Therefore, since 𝛼(𝐺)≀7, at least five of the induced subgraphs βŸ¨π‘ˆξ…žπ‘–βˆͺ𝑁𝑅(π‘ˆξ…žπ‘–)⟩𝐺, 3≀𝑖≀8 are complete graphs. Since 𝛿(𝐺)β‰₯8, it implies that these complete graphs contain 𝐾7. Thus, 𝐺 contains 𝐾7.

Lemma 2.8. If 𝐺 be a graph of order β‰₯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(𝑒)|≀28 for any π‘’βˆˆπ‘‰(𝐺).Proof. Suppose that 𝑒 is a vertex with |βŸ¨π‘πΊ(𝑒)⟩𝐺|β‰₯29. Let βŸ¨π‘πΊ(𝑒)⟩𝐺=β‹ƒπ‘Ÿπ‘–=1𝐺𝑖 where 𝐺𝑖 is a component for each 𝑖. βŸ¨π‘πΊ(𝑒)⟩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 βŸ¨π‘πΊ(𝑒)⟩𝐺 contains no 𝑃6. Thus, 𝐺𝑖 must be a complete graph of order at most 5. Also βŸ¨π‘πΊ(𝑒)⟩𝐺 contains no 𝐾5, thus 𝐺𝑖 must be a complete graph of order at most 4. Hence, the minimum number of independent vertices of βŸ¨π‘πΊ(𝑒)⟩ occurs only if βŸ¨π‘πΊ(𝑒)⟩ 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)β‰₯8. This is a contradiction. The proof of the claim is complete.Claim 2. 𝛼(𝐺)=7. Proof. Since |𝑉(𝐺)|β‰₯57 and 𝐺 contains no 𝐢9 and since π‘Ÿ(𝐢9,𝐾7)=49,𝛼(𝐺)β‰₯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 𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6, and 𝑒7, set 𝑁𝑖[𝑒𝑖+1]=𝑁[𝑒𝑖+1⋃]βˆ’(𝑖𝑗=1𝑁[𝑒𝑗]),1≀𝑖≀6. Analogously, we set 𝑁𝑖(𝑒𝑖+1),   1≀𝑖≀6. Let ⋃𝐴=6𝑖=1𝑁𝑖[𝑒𝑖+1],   ⋃𝐡=6𝑖=1𝑁𝑖(𝑒𝑖+1), and 𝛽=𝛼(⟨𝐡⟩𝐺).
Claim 3. |𝑁(𝑒1)βˆͺ𝐡|β‰₯50. Proof. Suppose that |𝑁(𝑒1)βˆͺ𝐡|≀49. Then |𝑁[𝑒1]βˆͺ𝐴|≀56. And so |πΊβˆ’(𝑁[𝑒1]βˆͺ𝐴)|β‰₯57βˆ’56=1. But π‘Ÿ(𝐢9,𝐾1)=1, so πΊβˆ’(𝑁[𝑒1]βˆͺ𝐴) contains a vertex, say 𝑒8, which is not adjacent to any of 𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6, and 𝑒7. Thus, {𝑒1,𝑒2,𝑒3,𝑒4,𝑒5,𝑒6,𝑒7,𝑒8} is an 8-element independent set. Therefore, 𝛼(𝐺)β‰₯8. That is a contradiction. The proof of the claim is complete.

Now, by Lemma 2.1, 𝛿(𝐺)β‰₯8 and so by Claim 1, we have that 8≀|𝑁(𝑒1)|≀28. Thus, if |𝑁(𝑒1)|=π‘Ÿ, then |𝐡|β‰₯50βˆ’π‘Ÿ. By a similar argument as in Claim 1, we have that 𝛼(βŸ¨π‘(𝑒1)⟩𝐺)β‰₯βŒˆπ‘Ÿ/4βŒ‰ and 𝛽β‰₯⌈(50βˆ’π‘Ÿ)/4βŒ‰. Note that for any 8β‰€π‘Ÿβ‰€21,⌈(50βˆ’π‘Ÿ)/4βŒ‰ is greater than or equal to 8. And so 𝛼(𝐺)β‰₯8. Now we have the following cases.

Case 1. 22≀|𝑁(𝑒1)|≀25, then by a similar argument as in Claim 1, we have that 𝛼(βŸ¨π‘(𝑒1)⟩𝐺)β‰₯6 and 𝛽β‰₯7. Then, ⟨𝐡⟩𝐺 has an independent set which consists of 7 vertices. This set with the vertex 𝑒1 is an 8-element independent set of vertices of 𝐺. That is a contradiction.

Case 2. |𝑁(𝑒1)|=26, then |𝐡|β‰₯24. By a similar argument as in Claim 1, we have that 𝛼(βŸ¨π‘(𝑒1)⟩𝐺)β‰₯7 and 𝛽β‰₯6. Now we have the following two subcases.Subcase 2.1. 𝛽β‰₯7. 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(𝑒𝑖+1), say π‘Ž1, that is not adjacent to at least one vertex of each 𝐾(𝑗)(1≀𝑗≀7), say π‘₯𝑗. Then {π‘₯1,π‘₯2,π‘₯3,π‘₯4,π‘₯5,π‘₯6,π‘₯7,π‘Ž1} is an 8-elemant independent set of vertices of 𝐺. And so 𝛼(𝐺)β‰₯8. That is a contradiction.
Subcase 2.2.2. For each vertex of ⋃6𝑖=1𝑁𝑖(𝑒𝑖+1) there is 1≀𝑗≀7 such that this vertex is adjacent to all vertices of 𝐾(𝑗). Then 𝐺 contains 𝐾1+𝑃6 or 𝐢9. That is a contradiction.

Case 3. 27≀|𝑁(𝑒1)|≀28, Then by using the same argument as in Case 2, we have the same contradiction.

Theorem 2.9. π‘Ÿ(𝐢9,𝐾8)=57. Proof. Suppose that there exists a graph 𝐺 of order 57 that contains neither 𝐢9 nor an 8-elements independent set. Then by Lemma 2.1, 𝛿(𝐺)β‰₯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)|β‰₯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 {𝑒2,𝑒3,𝑒5,𝑒7,𝑒8,𝑒9} there are paths of order 8,7,6 and 5. The following are paths of order 8 between vertices of {𝑒3,𝑒5,𝑒7,𝑒8,𝑒9}.1- 𝑒2-𝑒3 path: 𝑒2𝑒1𝑒8𝑒7𝑒6𝑒5𝑒4𝑒3, by symmetry we find 𝑒7-𝑒8 path.2-𝑒2-𝑒5 path: 𝑒2𝑒3𝑒4𝑒1𝑒8𝑒7𝑒6𝑒5, by symmetry we find 𝑒5-𝑒8 path.3-𝑒2-𝑒7 path: 𝑒2𝑒3𝑒4𝑒5𝑒6𝑒1𝑒8𝑒7, by symmetry we find 𝑒3-𝑒8 path.4-𝑒2-𝑒8 path: 𝑒2𝑒3𝑒4𝑒5𝑒6𝑒7𝑒1𝑒8.5-𝑒2-𝑒9 path: 𝑒2𝑒3𝑒4𝑒5𝑒1𝑒7𝑒6𝑒9, by symmetry we find 𝑒8-𝑒9 path.6-𝑒3-𝑒5 path: 𝑒3𝑒4𝑒9𝑒6𝑒7𝑒8𝑒1𝑒5, by symmetry we find 𝑒5-𝑒7 path.7-𝑒3-𝑒7 path: 𝑒3𝑒4𝑒9𝑒6𝑒5𝑒1𝑒8𝑒7.8-𝑒3-𝑒9 path: 𝑒3𝑒4𝑒5𝑒1𝑒8𝑒7𝑒6𝑒9, by symmetry we find 𝑒7-𝑒9 path.3-𝑒5-𝑒9 path: 𝑒5𝑒4𝑒3𝑒2𝑒1𝑒7𝑒6𝑒9.The following are paths of order 7 between vertices of {𝑒3,𝑒5,𝑒7,𝑒8,𝑒9}.1- 𝑒2-𝑒3 path: 𝑒2𝑒1𝑒7𝑒6𝑒5𝑒4𝑒3, by symmetry we find 𝑒7-𝑒8 path.2-𝑒2-𝑒5 path: 𝑒2𝑒3𝑒4𝑒1𝑒7𝑒6𝑒5, by symmetry we find 𝑒5-𝑒8 path.3-𝑒2-𝑒7 path: 𝑒2𝑒3𝑒4𝑒5𝑒6𝑒1𝑒7, by symmetry we find 𝑒3-𝑒8 path.4-𝑒2-𝑒8 path: 𝑒2𝑒3𝑒4𝑒5𝑒6𝑒1𝑒8.5-𝑒2-𝑒9 path: 𝑒2𝑒3𝑒4𝑒5𝑒1𝑒6𝑒9, by symmetry we find 𝑒8-𝑒9 path.2-𝑒3-𝑒5 path: 𝑒3𝑒2𝑒1𝑒8𝑒7𝑒6𝑒5, by symmetry we find 𝑒5-𝑒7 path.2-𝑒3-𝑒7 path: 𝑒3𝑒2𝑒1𝑒4𝑒5𝑒6𝑒7.2-𝑒3-𝑒9 path: 𝑒3𝑒4𝑒5𝑒1𝑒7𝑒6𝑒9, by symmetry we find 𝑒7-𝑒9 path.3-𝑒5-𝑒9 path: 𝑒5𝑒4𝑒3𝑒2𝑒1𝑒6𝑒9.The following are paths of order 6 between vertices of {𝑒3,𝑒5,𝑒7,𝑒8,𝑒9}.1- 𝑒2-𝑒3 path: 𝑒2𝑒1𝑒6𝑒5𝑒4𝑒3, by symmetry we find 𝑒7-𝑒8 path.2-𝑒2-𝑒5 path: 𝑒2𝑒3𝑒4𝑒1𝑒6𝑒5, by symmetry we find 𝑒5-𝑒8 path.3-𝑒2-𝑒7 path: 𝑒2𝑒3𝑒4𝑒5𝑒1𝑒7, by symmetry we find 𝑒3-𝑒8 path.4-𝑒2-𝑒8 path: 𝑒2𝑒3𝑒4𝑒5𝑒1𝑒8.5-𝑒2-𝑒9 path: 𝑒2𝑒3𝑒4𝑒1𝑒6𝑒9, by symmetry we find 𝑒8-𝑒9 path.2-𝑒3-𝑒5 path: 𝑒3𝑒2𝑒1𝑒7𝑒6𝑒5, by symmetry we find 𝑒5-𝑒7 path.2-𝑒3-𝑒7 path: 𝑒3𝑒2𝑒1𝑒5𝑒6𝑒7.2-𝑒3-𝑒9 path: 𝑒3𝑒2𝑒1𝑒7𝑒6𝑒9, by symmetry we find 𝑒7-𝑒9 path.3-𝑒5-𝑒9 path: 𝑒5𝑒1𝑒2𝑒3𝑒4𝑒9.The following are paths of order 5 between vertices of {𝑒3,𝑒5,𝑒7,𝑒8,𝑒9}.1-𝑒2-𝑒3 path: 𝑒2𝑒1𝑒5𝑒4𝑒3, by symmetry we find 𝑒7-𝑒8 path.2-𝑒2-𝑒5 path: 𝑒2𝑒3𝑒4𝑒1𝑒5, by symmetry we find 𝑒5-𝑒8 path.3-𝑒2-𝑒7 path: 𝑒2𝑒3𝑒4𝑒1𝑒7, by symmetry we find 𝑒3-𝑒8 path.4-𝑒2-𝑒8 path: 𝑒2𝑒3𝑒4𝑒1𝑒8.1-𝑒2-𝑒9 path: 𝑒2𝑒1𝑒5𝑒4𝑒9, by symmetry we find 𝑒8-𝑒9 path.2-𝑒3-𝑒5 path: 𝑒3𝑒2𝑒1𝑒6𝑒5, by symmetry we find 𝑒5-𝑒7 path.2-𝑒3-𝑒7 path: 𝑒3𝑒2𝑒1𝑒6𝑒7.2-𝑒3-𝑒9 path: 𝑒3𝑒4𝑒5𝑒67𝑒9, by symmetry we find 𝑒7-𝑒9 path.3-𝑒5-𝑒9 path: 𝑒5𝑒1𝑒3𝑒4𝑒9.