Polarity graphs and Ramsey numbers for C4 versus stars

Discrete Mathematics 340 (2017) 655–660

Contents lists available at ScienceDirect

Discrete Mathematics journal homepage: www.elsevier.com/locate/disc

Polarity graphs and Ramsey numbers for C4 versus stars Xuemei Zhang a,b , Yaojun Chen a, *, T.C. Edwin Cheng c a b c

Department of Mathematics, Nanjing University, Nanjing 210093, PR China Department of Mathematics, Zhejiang Normal University, Jinhua 321004, PR China Department of Logistics and Maritime Studies, The Hong Kong Polytechnic University, Hung Kom, Kowloon, Hong Kong, PR China

article

info

Article history: Received 23 November 2015 Received in revised form 30 November 2016 Accepted 4 December 2016 Available online 3 January 2017 Keywords: Finite field Polarity graph Ramsey number Star Quadrilateral

a b s t r a c t For two given graphs G1 and G2 , the Ramsey number R(G1 , G2 ) is the smallest integer N such that for any graph of order N, either G contains a copy of G1 or its complement contains a copy of G2 . Let Cm be a cycle of√length m and K1,n a star of order n + 1. Parsons (1975) shows that R(C4 , K1,n ) ≤ n + ⌊ n − 1⌋ + 2 and if n is the square of a prime power, then the equality holds. In this paper, by discussing the properties of polarity graphs whose vertices are points in the projective planes over Galois fields, we prove that q q R(C4 , K1,q2 −t ) = q2 + q − (t − 1) if q is an odd prime power, 1 ≤ t ≤ 2⌈ 4 ⌉ and t ̸ = 2⌈ 4 ⌉ − 1, which extends a result on R(C4 , K1,q2 −t ) obtained by Parsons (1976). © 2016 Elsevier B.V. All rights reserved.

1. Introduction All graphs considered in this paper are finite simple graphs. Most of our terminology is standard and can be found in many textbooks such as [1]. Let G = (V (G), E(G)) be a graph. For S ⊆ V (G), let G[S ] and G − S denote the induced by S and V (G) − S, respectively. Let N(v ) be the set of the neighbors of a vertex v in G, N [v] = N(v ) ∪ {v} and d(v ) = |N(v )|. Let NS (v ) = N(v ) ∩ S be the set of the neighbors of a vertex v contained in S, dS (v ) = |NS (v )|. The minimum degree and maximum degree of G are denoted by δ (G) and ∆(G), respectively. Let X and Y be sets of vertices (not necessarily disjoint) of a graph G. We denote by E [X , Y ] the set of edges of G with one end in X and the other end in Y . If X = {x} or Y = {y}, then E [X , Y ] is denoted by E [x, Y ] or E [X , y], respectively. We write E [X ] for E [X , X ]. Sometimes, we refer a vertex of degree d as a d-vertex. A cycle of length m and a star of order n + 1 are denoted by Cm and K1,n , respectively. A wheel Wn is a graph obtained from a cycle Cn and a new vertex v by connecting v to all vertices of Cn . For two given graphs G1 and G2 , the Ramsey number R(G1 , G2 ) is the smallest integer N such that for any graph G of order N, either G contains a copy of G1 or G contains a copy of G2 , where G is the complement of G. The general upper bound for Ramsey number R(C4 , K1,n ) was first studied by Parsons [6]. For each positive integer n, we let

Gn = {G | G has no C4 and G has no K1,n }. Parsons [6] proved that if n > 1 and G ∈ Gn , then |G| ≤ n + bound for R(C4 , K1,n ).

√

√ n − 1 + 1. In fact, he established the following tight upper

2 Theorem √ 1 (Parsons [6]). R(C4 , K1,n ) ≤ n + ⌊ n − 1⌋ + 2 for all n ≥ 2, and if n = q + 1 and q ≥ 1, then R(C4 , K1,n ) ≤ n + ⌊ n − 1⌋ + 1.

*

Corresponding author. E-mail address: [email protected] (Y. Chen).

http://dx.doi.org/10.1016/j.disc.2016.12.005 0012-365X/© 2016 Elsevier B.V. All rights reserved.

656

X. Zhang et al. / Discrete Mathematics 340 (2017) 655–660

Table 1 Distribution of the values of R(C4 , K1,n ) for 6 ≤ n ≤ 50. n

6

7–8

9–10

11–12

13–15

16–17

18–20

21

22

23

24

R(C4 , K1,n ) References

n+3 [10]

n+4 [10]

n+4 [6]

n+5 [10]

n+5 [10]

n+5 [6]

n+5 [10]

n+6 [7]

?

n+6 [7]

n+6

n

25–26

27–33

34–36

37–42

43

44

45

46

47

48

49–50

R(C4 , K1,n ) References

n+6 [6]

?

n+7 [10]

?

n+8 [10]

?

n+8 [7]

?

n+8 [7]

n+8

n+8 [6]

∗

∗

Moreover, Parsons [6] determined Ramsey numbers R(C4 , K1,n ) in the cases when n = q2 + 1 or q2 , where q is a prime power. Theorem 2 (Parsons [6]). For every prime power q, R(C4 , K1,q2 +1 ) = q2 + q + 2 and R(C4 , K1,q2 ) = q2 + q + 1. Furthermore, by considering the structural properties of the polarity graphs Gq given in [2,5], where q is a prime power, Parsons obtained the following result. 2 Theorem 3 (Parsons [7]).⌈ R(C ⌉ 4 , K1,q2 −t ) = q + q − (t − 1) if q is an even prime power, 1 ≤ t ≤ q + 1 and t ̸= q, or q is an odd q prime power, 0 ≤ t ≤ 2 4 and t is even.

√

√

Noting that if 1 ≤ t ≤ q + 1 and n = q2 − t, then q2 + q − (t − 1) = (q2 − t) + ⌊ (q2 − t) − 1⌋ + 2 = n + ⌊ n − 1⌋ + 2, by Theorem 1, we know that the main work for proving Theorem 3 is to construct (C4 , K1,q2 −t )-Ramsey graphs. In [7], Parsons showed some subgraphs of the polarity graph Gq are (C4 , K1,q2 −t )-Ramsey graphs as needed in Theorem 3. In this paper, q q we will extend Theorem 3 for q being an odd prime power and all t with 1 ≤ t ≤ 2⌈ 4 ⌉ and t ̸ = 2⌈ 4 ⌉ − 1. Our main idea is to construct (C4 , K1,q2 −t )-Ramsey graphs based on some local structures of the polarity graph Gq . In particular, the (C4 , K1,q2 −t )-Ramsey graphs we construct for odd t are not the subgraphs of the polarity graph Gq . The main result of this paper is as follows. Theorem 4. Let q be an odd prime power. Then R(C4 , K1,q2 −t ) = q2 + q − (t − 1) if 1 ≤ t ≤ 2⌈ 4 ⌉ and t ̸ = 2⌈ 4 ⌉ − 1. q

q

Now, let us summarize all known values of R(C4 , K1,n ) for 6 ≤ n ≤ 50. For n = 9, 10, 16, 17, 25, 26, 49, 50, the values of R(C4 , K1,n ) are given by Theorem 2. With the help of computers, Tse [8] determined R(C4 , Wn ) for 3 ≤ n ≤ 12; Dybizbánski and Dzido [4] calculated R(C4 , Wn ) for 13 ≤ n ≤ 16; Wu et al. [9] evaluated R(C4 , Wn ) for 17 ≤ n ≤ 20, 34 ≤ n ≤ 36 and n = 43. Noting that R(C4 , Wn ) = R(C4 , K1,n ) for n ≥ 6, which was proved by Zhang et al. [10], we see that the values of R(C4 , K1,n ) are known for 6 ≤ n ≤ 20, 34 ≤ n ≤ 36 and n = 43. By Theorem 3, R(C4 , K1,23 ) = 29, R(C4 , K1,47 ) = 55, R(C4 , K1,21 ) = 27 and R(C4 , K1,45 ) = 53. By Theorem 4, we can obtain that R(C4 , K1,24 ) = 30 and R(C4 , K1,48 ) = 56. These known values of R(C4 , K1,n ) are listed in Table 1, where the reference ‘‘∗’’ represents this paper.

√

√

Observe all known values of R(C4 , K1,n ) for n ≥ 6, we can find that R(C4 , K1,n ) = n + ⌊ n − 1⌋ + 1 or n + ⌊ n − 1⌋ + 2. Our question is:

√

√

Question 1. Is it true that R(C4 , K1,n ) = n + ⌊ n − 1⌋ + 1 or n + ⌊ n − 1⌋ + 2? If the answer to Question 1 is affirmative, then it would give a negative answer for the following conjecture due to Burr et al. [3], for which, Erdős, one of the authors, offered $100 for a proof or disproof. Conjecture 1 (Burr et al. [3]). R(C4 , K1,n ) < n +

√

n − c holds infinitely often, where c is an arbitrary constant.

2. Polarity graphs and their properties The construction of polarity graphs is due to Brown [2], and independently to Erdős, Rényi and Sós [5]. Polarity graphs provide infinite many graphs which have no C4 , and so they sometimes can be used to establish the lower bound of the Ramsey numbers for a C4 versus a star, see for instance [6]. 2.1. Construction of polarity graphs Let q be a prime power and Fq = GF (q) the Galois field with q elements. Define an equivalence relation on Fq3 \ {(0, 0, 0)} by letting (a, b, c) ∼ (a′ , b′ , c ′ ) if there is a nonzero λ ∈ Fq such that (λa, λb, λc) = (a′ , b′ , c ′ ). The equivalence class of (a, b, c) is denoted by the ordered triple abc. Let V be the set of all equivalence classes. If there is no danger of ambiguity, we also use letters u, v and so on to denote the equivalence classes in V . The polarity graph ˆ Gq is the graph with vertex set V ,

X. Zhang et al. / Discrete Mathematics 340 (2017) 655–660

657

two vertices (equivalence classes) abc and xyz in V being adjacent if and only if ax + by + cz = 0. The simple polarity graph Gq is the graph obtained from ˆ Gq by deleting all q + 1 loops. By the definition of Gq , for any vertex v := abc ∈ V (Gq ), N(v ) = {xyz | ax + by + cz = 0} \ {v}. Thus, N(v ) = {xyz | ax + by + cz = 0} − {v} if a2 + b2 + c 2 = 0 and N(v ) = {xyz | ax + by + cz = 0} if a2 + b2 + c 2 ̸= 0. This means that v := abc is a q-vertex or (q + 1)-vertex in Gq if and only if a2 + b2 + c 2 = 0 or a2 + b2 + c 2 ̸ = 0. The graph Gq has q2 + q + 1 vertices, in which q + 1 vertices are of degree q and all other vertices are of degree q + 1. Furthermore, Gq has diameter two, but it contains no C4 and no two adjacent q-vertices. By the definition of the equivalence relation, we have

abc =

⎧ 001, ⎪ ⎪ ⎪ ⎪ (c ) ⎨ 01 ,

if a, b = 0 and c ̸ = 0; if a = 0 and b ̸ = 0;

b

( )( ) ⎪ ⎪ b c ⎪ ⎪ 1 , ⎩ a

if a ̸ = 0.

a

So we will use 001, 01( bc ), 1( ba )( ac ) and so on to denote the vertices of Gq . Let Fq∗ be the set of nonzero elements in Fq . We say that a vertex a′ b′ c ′ in Gq is of type 1 ∗ ∗ if a′ = 1 and b′ , c ′ ∈ Fq∗ . 2.2. Structural properties of polarity graphs In this section, we will discuss some structural properties of the polarity graph Gq , which will be used to obtain the main result of this paper. Lemma 1. Let q be an odd prime power and b ∈ Fq∗ with b2 ̸ = −1. Then (i) Gq has exactly two or no q-vertices of type 1 ∗ ∗ which are adjacent to 1b0; (ii) 1b0 is adjacent to some q-vertex of type 1 ∗ ∗ if and only if 1( −b1 )0 is adjacent to some q-vertex of type 1 ∗ ∗. Proof. It is clear that 1b0 is a (q + 1)-vertex since b2 ̸ = −1. Let 1yz be a q-vertex of type 1 ∗ ∗. (i) Obviously, a q-vertex 1yz is adjacent to 1b0 if and only if (y, z) is a solution to the system of equations 1 · 1 + b · y + 0 · z = 0, 1 + y2 + z 2 = 0, that is, y=

−1 b

(

) ( 1 , z2 = − 1 + 2 .

Let A = − 1 +

b

1 b2

)

(1)

. Then A ∈ Fq∗ . If A is a square in Fq∗ , then the equations system (1) has exactly two solutions, and so there

are exactly two q-vertices of type 1 ∗ ∗, which are adjacent to 1b0. If A is not a square in Fq∗ , then the equations system has no solutions, and thus there are no q-vertices of type 1 ∗ ∗ adjacent to 1b0. (ii) As seen in (i), a q-vertex 1yz is adjacent to 1b0 if and only if (y, z) is a solution to the equations system (1). Replacing b by −b1 , we find that a q-vertex 1yz is adjacent to 1( −b1 )0 if and only if (y, z) is a solution to the system of equations y = b, z 2 = − 1 + b2 .

(

)

(2) B , b2

and so both A and B are squares or both not. Therefore, both the equations systems Set B = −(1 + b2 ). Obviously, A = (1) and (2) have solutions or both not, and it follows that 1b0 is adjacent to some q-vertex of type 1 ∗ ∗ if and only if 1( −b1 )0 is adjacent to some q-vertex of type 1 ∗ ∗. ■ Lemma 2. Let u be any (q + 1)-vertex in Gq and v ∈ N(u). If d(v ) = q, then dN(u) (v ) = 0, and if d(v ) = q + 1 then dN(u) (v ) = 1. In other words, each q-vertex v in N(u) is adjacent to no other vertex in N(u), and each (q + 1)-vertex v in N(u) is adjacent to exactly one vertex in N(u), which is of course of degree q + 1. Proof. Let u := au bu cu be (q + 1)-vertex, v := av bv cv ∈ N(u). Obviously, the system of equations av x + bv y + cv z = 0, au x + bu y + cu z = 0

(3)

has one unique solution in V , denoted it by abc. By the definition of Gq , NN(u) (v ) = N(v ) ∩ N(u) = {abc } \ {v, u}.

(4)

If d(v ) = q, then a2v + b2v + cv2 = 0. Moreover, we have au av + bu bv + cu cv = 0 since v ∈ N(u). So v := av bv cv is the unique solution abc of Eqs. (3). By (4), NN(u) (v ) = ∅, and so dN(u) (v ) = 0.

658

X. Zhang et al. / Discrete Mathematics 340 (2017) 655–660

If d(v ) = q + 1, then a2v + b2v + cv2 ̸ = 0, that is, v := av bv cv is not satisfied with av x + bv y + cv z = 0. Similarly, the (q + 1)-vertex u := au bu cu is not satisfied with au x + bu y + cu z = 0. Thus, neither v nor u is the solution abc of Eqs. (3). By (4), NN(u) (v ) = {abc }, and so dN(u) (v ) = 1. ■ Lemma 2 can also be stated as follows. Lemma 2′ . Let u be any (q + 1)-vertex in Gq and v ∈ N(u). Then dN(u) (v ) = d(v ) − q. Remark 1. Let u be any (q + 1)-vertex in Gq and v ∈ N(u). By Lemma 2, no other vertex in N(u) is adjacent to v if v is a q-vertex, and there is a unique vertex adjacent to v in N(u) if v is a (q + 1)-vertex. Throughout the rest of this paper, we will use v to denote the unique vertex adjacent to v in N(u). Of course, if d(v ) = q + 1, then d(v ) = q + 1 and v = v . It is clear that E [N(u)] is a matching of Gq [N(u)], which covers all (q + 1)-vertices but no q-vertices in N(u). Lemma 3. Let u be a (q + 1)-vertex in Gq . For v ∈ N(u), let Av = N(v ) \ N [u]. Then the following hold: (i) |Av | = q − 1 for any vertex v ∈ N(u); (ii) V is the disjoint union of {u}, N(u), and all Av , v ∈ N(u). Proof. (i) For any v ∈ N(u), it is easy to see that |Av |=|N(v ) \ N [u]| = d(v ) − dN [u] (v ) = d(v ) − dN(u) (v ) + 1 . By Lemma 2′ , we can get that |Av | = q − 1. (ii) Let v, v ′ be any two distinct vertices in N(u). Since Gq has no C4 , we can see that Av ∩ Av ′ = ∅. Thus, these vertex sets Av (v ∈ N(u)) are mutually disjoint. That is to say that V contains the disjoint sets {u}, N(u), and all Av , v ∈ N(u). On the other hand,

(

∑

|{u}| + |N(u)| +

)

|Av | = 1 + (q + 1) + (q − 1)(q + 1) = q2 + q + 1 = |V |.

v∈N(u)

So, V is the disjoint union of {u}, N(u), and all Av , v ∈ N(u).

■

Lemma 4. Let u be a (q + 1)-vertex in Gq . For any v ∈ N(u), let Av = N(v ) \ N [u]. If v is a (q + 1)-vertex in N(u) and v ′ ̸ = v is a vertex nonadjacent to v in N(u), then E [Av , Av ′ ] is a perfect matching of the subgraph of Gq induced by Av ∪ Av ′ . Proof. Let v and v ′ be any two nonadjacent vertices in N(u) and d(v ) = q + 1. Obviously, E [Av , Av ′ ] is a matching of Gq since Gq has no C4 . Noting |Av | = |Av ′ | (as seen in Lemma 3(i)), to show that E [Av , Av ′ ] is a perfect matching of the subgraph of Gq induced by Av ∪ Av ′ , one need merely to check that |E [x, Av ]| ≥ 1 for any x ∈ Av ′ . Since the diameter of Gq is 2 and x, v are not adjacent, we have the distance d(x, v ) between x and v is 2. It implies that x is adjacent to some vertex in N(v ). Noting that N(v ) = Av ∪ {u, v}, where v is the unique vertex adjacent to v in N(u), and x is not adjacent to u and v , we have that x is adjacent to some vertex in Av , i.e., |E [x, Av ]| ≥ 1. ■ 3. Proof of Theorem 4 In this section, we will give the proof of our main result. Proof of Theorem 4. Let g be a generator of Fq∗ . It is easy to see that an element of Fq∗ is the square of another one if and only q −1

if it is an even power of g. Since −1 = g 2 , we can see that it is not a square in Fq∗ if q ≡ 3 (mod 4), and it is a square in Fq∗ if q ≡ 1 (mod 4). In the latter case, we denote by r , −r the square roots of −1, respectively. We first establish the lower bound for R(C4 , K1,q2 −t ). To do this, we need to discuss some local structure of the polarity graph Gq . Claim 1. Let v be any q-vertex in Gq . Then v is of type 1 ∗ ∗ if q ≡ 3 (mod 4), and v ∈ {1r0, 1(−r)0, 10r , 10(−r), 01r , 01(−r)} if q ≡ 1 (mod 4) and v is not of type 1 ∗ ∗. Proof. If q ≡ 3 (mod 4), it suffices to show that a vertex not being of type 1 ∗ ∗ must be of degree q + 1. Let W be the set of all vertices not being of type 1 ∗ ∗, then W = {001, 010, 100} ∪ {01c | c ∈ F ∗ } ∪ {10a | a ∈ F ∗ } ∪ {1b0 | b ∈ F ∗ }. Since −1 is not a square in Fq∗ , it is easy to check that x2 + y2 + z 2 ̸ = 0 for each vertex xyz ∈ W . By the definition of Gq , xyz is of degree q + 1, and so v is of type 1 ∗ ∗. If q ≡ 1 (mod 4), then since −1 has two square roots i, −i, it is easy to check that if v is not of type 1 ∗ ∗, then v ∈ {1r0, 1(−r)0, 10r , 10(−r), 01r , 01(−r)}. Claim 2. If q ≡ 3 (mod 4), then E [N(001)] is a perfect matching of Gq [N(001)]; and if q ≡ 1 (mod 4), then E [N(001) − {1r0, 1(−r)0}] is a perfect matching of Gq [N(001) − {1r0, 1(−r)0}]. Moreover, 1r0, 1(−r)0 are isolated vertices in Gq [N(001)].

X. Zhang et al. / Discrete Mathematics 340 (2017) 655–660

659

Fig. 1. Gq [N [001] ∪ {all q-vertices}] for q ≡ 3 (mod 4).

Proof. Obviously, 001 is a vertex of degree q + 1 in Gq and N(001) = {xyz | 0 · x + 0 · y + z = 0} = {010, 100} ∪ {1b0 | b ∈ Fq∗ }. By Claim 1, all vertices in N(001) are (q + 1)-vertices if q ≡ 3 (mod 4), and all vertices in N(001) − {1r0, 1(−r)0} are (q + 1)-vertices if q ≡ 1 (mod 4). For any (q + 1)-vertex v ∈ N(001), by Lemma 2, there is a unique vertex v in N(001) adjacent to v . Thus, if q ≡ 3 (mod 4), then we have 010 = 100 and 1b0 = 1( −b1 )0 for any b ∈ Fq∗ , and so E [N(001)] is a perfect matching of Gq [N(001)]; and if q ≡ 1 (mod 4), then we have 010 = 100 and 1b0 = 1( −b1 )0 for any b ∈ Fq∗ − {r , −r }, and so E [N(001) − {1r0, 1(−r)0}] is a perfect matching of Gq [N(001) − {1r0, 1(−r)0}]. Since 1r0, 1(−r)0 are q-vertices, by Lemma 2, they are isolated vertices in Gq [N(001)]. Claim 3. If q ≡ 3 (mod 4), then each q-vertex is adjacent to exactly one vertex in {1b0 | b ∈ Fq∗ }, and if q ≡ 1 (mod 4), then each q-vertex of type 1 ∗ ∗ is adjacent to exactly one vertex in {1b0 | b ∈ Fq∗ − {r , −r }}. Proof. For any v ∈ N(001), let Av = N(v ) \ N [001]. By Lemma 2, V is the disjoint union of {001}, N(001), and all Av , v ∈ N(001). Since Gq has no C4 and N(001) = {010, 100} ∪ {1b0 | b ∈ Fq∗ }, each q-vertex not in N(001) is adjacent to exactly one vertex of N(001). If q ≡ 3 (mod 4), then all vertices in N(001) are (q + 1)-vertices. By Claim 1, no q-vertex is adjacent to 010 and 100, and hence each q-vertex is adjacent to exactly one vertex in {1b0 | b ∈ Fq∗ }. If q ≡ 1 (mod 4), then all vertices except 1r0, 1(−r)0 in N(001) are (q + 1)-vertices. Since all q-vertices in Gq form an independent set, each q-vertex except 1r0, 1(−r)0 is adjacent to a (q + 1)-vertex in N(001). Clearly, no q-vertex of type 1 ∗ ∗ is adjacent to 010 and 100, and so each q-vertex of type 1 ∗ ∗ is adjacent to exactly one vertex in {1b0 | b ∈ Fq∗ − {r , −r }}. Claim 4. For any v ∈ N(001), v is adjacent to exactly two or no q-vertices. Moreover, v is adjacent to some q-vertex if and only if v is adjacent to some q-vertex. Proof. Let v ∈ N(001) = {010, 100} ∪ {1b0 | b ∈ Fq∗ }. By the definition of Gq , the vertices 010 and 100 are not adjacent to the vertices of type 1 ∗ ∗. Thus, if v ∈ {010, 100}, then by Claim 1, v is adjacent to no q-vertices if q ≡ 3 (mod 4), and exactly two q-vertices if q ≡ 1 (mod 4), that is, the vertex 010 is adjacent to r01 and (−r)01, and the vertex 100 is adjacent to 01r and 01(−r). Noting that 010 = 100, we can see that the result is true for v ∈ {010, 100}. If v ∈ {1b0 | b ∈ Fq∗ and b2 = −1}, then v ∈ {1r0, 1(−r)0} is of degree q. Since all q-vertices of Gq form an independent set, v is adjacent to no other q-vertices. If v ∈ {1b0 | b ∈ Fq∗ and b2 ̸ = −1}, then the result follows by Lemma 1. We now begin to give the lower bound for R(C4 , K1,q2 −t ) by using Gq to construct a graph on q2 + q − t vertices which is in Gq2 −t . If q ≡ 3 (mod 4), then by Claim 2, we may number the vertices of N(001) as: v1 , v2 , v3 , . . . , vq+1 such that v1 = v2 , v3 = v4 , . . . , vq = vq+1 . Since the number of all q-vertices is q + 1, by Claims 3 and 4, we may assume that each q-vertex is adjacent q+3 q+1 to some vertex vi , where 2 ≤ i ≤ q + 1. Thus, all vertices in Avi , i = 1, 2, . . . , 2 , are of degree q + 1. Now, the subgraph of Gq , induced by the set of all vertices in N [001] and all q-vertices in Gq , is shown in Fig. 1, where the q-vertices are indicated by solid points and the (q + 1)-vertices are indicated by hollow ones. If q ≡ 1 (mod 4), then by Claim 2, we may number the vertices of N(001) as: v1 , v2 , v3 , . . . , vq+1 such that v1 = 1r0, v2 = 1(−r)0, v3 = v4 , . . . , vq = vq+1 . Since the number of all q-vertices except 1r0 and 1(−r)0 is q − 1, by Claims 3 and 4, q+5 we may assume that each q-vertex except 1r0 and 1(−r)0 is adjacent to some vertex vi , where 2 ≤ i ≤ q + 1. Thus, all q+3 vertices in Avi , i = 1, 2, . . . , 2 , are of degree q + 1. Now, the subgraph of Gq , induced by the set of all vertices in N [001] and all q-vertices in Gq , is shown in Fig. 2, where the q-vertices are indicated by solid points and the (q + 1)-vertices are indicated by hollow ones. q+1 q−1 q+3 q+1 Let 1 ≤ t ≤ 2 and t ̸ = 2 if q ≡ 3 (mod 4), and 1 ≤ t ≤ 2 and t ̸ = 2 if q ≡ 1 (mod 4), and set G∗ = Gq − {001, v1 , v2 , . . . , vt }. Define

{ Ht =

G∗ , G∗ + {vt +1 vt +2 } − E [Avt +1 , Avt +2 ],

if t is ev en, if t is odd.

Since Gq has q2 + q + 1 vertices, Ht is a graph on q2 + q − t vertices. We now show that Ht ∈ Gq2 −t . By Lemma 3(ii), V (Ht ) is the disjoint union of {vi | t + 1 ≤ i ≤ q + 1}, Av1 , Av2 , . . . , Avq+1 .

660

X. Zhang et al. / Discrete Mathematics 340 (2017) 655–660

Fig. 2. Gq [N [001] ∪ {all q-vertices}] for q ≡ 1 (mod 4).

If t is even, then NHt (vi ) = Avi ∪ {vi } for t + 1 ≤ i ≤ q + 1. By Lemma 3(i), dHt (vi ) = |Avi | + 1 = q; For 1 ≤ i ≤ t and v ∈ Avi , NHt (v ) = NGq (v ) − {vi }, and so dHt (v ) = dGq (v ) − 1 = q; For t + 1 ≤ i ≤ q + 1 and v ∈ Avi , NHt (v ) = NGq (v ), and hence dHt (v ) = dGq (v ) ≥ q. Thus we have δ (Ht ) = q. Noting that Ht is a subgraph of Gq and Gq has no C4 , we see that Ht ∈ Gq2 −t . If t is odd, then because NHt (vt +1 ) = Avt +1 ∪ {vt +2 }, we have dHt (vt +1 ) = q by Lemma 3(i). For t + 2 ≤ i ≤ q + 1, since Avi ∪ {vi } ⊆ NHt (vi ), by Lemma 3(i), dHt (vi ) ≥ |Avi | + 1 = q; For 1 ≤ i ≤ t and v ∈ Avi , NHt (v ) = NGq (v ) − {vi }, and so q+1 q−1 q+3 q+1 dHt (v ) = dGq (v ) − 1 = q; Noting that 1 ≤ t ≤ 2 and t ̸ = 2 if q ≡ 3 (mod 4), and 1 ≤ t ≤ 2 and t ̸ = 2 if q ≡ 1 (mod q+1 q−1 4), we have t + 2 ≤ 2 in the former case and t + 2 ≤ 2 in the latter case, and so dGq (v ) = q + 1 for any v ∈ Avt +1 ∪ Avt +2 . Noting that E [Avt +1 , Avt +2 ] is a matching covering all vertices in Avt +1 ∪ Avt +2 by Lemma 4, we get that dHt (v ) = q for any v ∈ Avt +1 ∪ Avt +2 . For t + 3 ≤ i ≤ q + 1 and v ∈ Avi , we have NHt (v ) = NGq (v ), and hence dHt (v ) = dGq (v ) ≥ q. Thus, δ (Ht ) = q. To show that Ht ∈ Gq2 −t when t is odd, we need also to prove that Ht has no C4 . It is obvious that if Ht has a C4 , then it must contain the edge vt +1 vt +2 . Denote by C4 the set of all 4-cycles in Ht . Then we have

C4 = {C4 | C4 is a 4-cycle in Ht }

= = = =

{C4 | C4 is a 4-cycle containing vt +1 vt +2 in Ht } {C4 | C4 − vt +1 vt +2 is a path of length 3 in G∗ − E [Avt +1 , Avt +2 ]} {C4 | C4 − vt +1 vt +2 = vt +1 uwvt +2 in Ht for some u ∈ Avt +1 , w ∈ Avt +2 } ∅.

This implies that Ht ∈ Gq2 −t in this case. Therefore, we have R(C4 , K1,q2 −t ) ≥ |Ht | + 1 = q2 + q − (t − 1). On the other hand, by Theorem 1, we have

√

R(C4 , K1,q2 −t ) ≤ q2 − t + ⌊ q2 − t − 1⌋ + 2 = q2 + q − (t − 1), and hence R(C4 , K1,q2 −t ) = q2 + q − (t − 1). ■ Acknowledgments Many thanks to the anonymous referees for their careful comments on our earlier version of this paper. This research was supported by NSFC under Grant Nos. 11371193, 11671198 and 11571320. Cheng was also supported in part by The Hong Kong Polytechnic University under the Fung Yiu King-Wing Hang Bank Endowed Professorship in Business Administration. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

J.A. Bondy, U.S.R. Murty, Graph Theory, in: Graduate Texts in Mathematics, vol. 244, Springer, 2008. W.G. Brown, On graphs that do not contain a Thomsen graph, Canad. Math. Bull. 9 (1966) 281–285. S.A. Burr, P. Erdős, R.J. Faudree, C.C. Rousseau, R.H. Schelp, Some complete bipartite graph-tree Ramsey numbers, Ann. Discrete Math. 41 (1989) 79–89. J. Dybizbánski, T. Dzido, On some Ramsey numbers for quadrilaterals versus wheels, Graphs Combin. 30 (2014) 573–579. P. Erdős, A. Rényi, V.T. Sós, On a problem of graph theory, Studia Sci. Math. Hungar 1 (1966) 215–235. T.D. Parsons, Ramsey graphs and block designs I, Trans. Amer. Math. Soc. 209 (1975) 33–44. T.D. Parsons, Graphs from projective planes, Aequationes Math. 14 (1976) 167–189. K.K. Tse, On the Ramsey number of the quadrilateral versus the book and the wheel, Australas. J. Combin. 27 (2003) 163–167. Y.L. Wu, Y.Q. Sun, S.P. Radziszowski, Wheel and star-critical Ramsey numbers for quadrilateral, Discrete Appl. Math. 11 (2015) 260–271. Y.B. Zhang, H. Broersma, Y.J. Chen, A remark on star-C4 and wheel-C4 Ramsey numbers, Electron. J. Graph Theory Appl. 2 (2014) 110–114.