International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 926191 | 10 pages | https://doi.org/10.5402/2011/926191

The Cycle-Complete Graph Ramsey Number π‘Ÿ ( 𝐢 9 , 𝐾 8 )

Academic Editor: R. L. Soto
Received11 May 2011
Accepted14 Jun 2011
Published21 Jul 2011

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.

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Copyright © 2011 M. S. A. Bataineh et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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