On fan-wheel and tree-wheel Ramsey numbers

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Discrete Mathematics 339 (2016) 2284–2287

Contents lists available atScienceDirect

Discrete Mathematics

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

On fan–wheel and tree–wheel Ramsey numbers

Yanbo Zhang

a,b

, Hajo Broersma

b

, Yaojun Chen

a,∗

a_{Department of Mathematics, Nanjing University, Nanjing 210093, PR China}

b_{Faculty of Electrical Engineering, Mathematics and Computer Science, University of Twente, P.O. Box 217, 7500 AE Enschede,}

The Netherlands

a r t i c l e i n f o

Article history:

Received 26 January 2015

Received in revised form 5 March 2016 Accepted 18 March 2016 Keywords: Ramsey number Fan Tree Wheel

a b s t r a c t

For graphs G1and G2, the Ramsey number R(G1,G2)is the smallest integer N such that, for

any graph G of order N, G contains G1as a subgraph or the complement of G contains G2

as a subgraph. Let Tndenote a tree of order n, Wna wheel of order n+1 and Fna fan of

order 2n+1. We establish Ramsey numbers for fans and trees versus wheels of even order, thereby extending several known results. In particular, we prove that R(Fn,Wm) =6n+1

for odd m ≥3 and n ≥(5m+3)/4, and that R(Tn,Wm) = 3n−2 for odd m≥3 and

n≥m−2, and Tnbeing a tree for which the Erdős–Sós Conjecture holds.

1. Introduction

For graphs G1and G2, the Ramsey number R

(

G1

,

G2

)

is the smallest integer N such that, for any graph G of order N,

G contains G1as a subgraph or G contains G2as a subgraph, where G is the complement of G. Let

χ(

G

)

be the chromatic

number of G and s

(

G

)

the chromatic surplus of G, i.e., the minimum number of vertices in some color class under all proper vertex colorings with

χ(

G

)

colors. Let H be a connected graph of order p. Burr [6] established the following general lower bound for R

(

H

,

G

)

when p

≥

s

(

G

)

:

R

(

H

,

G

) ≥ (

p

−

1

)(χ(

G

) −

1

) +

s

(

G

).

He also defined H to be G-good in case equality holds in this inequality.

A fan Fnconsists of n triangles sharing exactly one common vertex. Thus,

|

V

(

Fn

)| =

2n

+

1. A wheel Wmis the graph obtained from Cmand an additional vertex by joining it to every vertex of Cm. In this paper, our main aim is to consider when a fan Fnor a tree Tnis Wm-good for odd m. In order to do so, we first establish the following two auxiliary theorems. Theorem 1. R

(

nK2

,

Wm

) =

max

{

2n

+ ⌈

m

/

2

⌉

,

n

+

m

}

.

Theorem 2. R

(

nK₂

,

C_m

) =

max

{

2n

+ ⌈

m

/

2

⌉ −

1

,

n

+

m

−

1

}

.

Remark. Even thoughTheorem 2is not an immediate consequence ofTheorem 1, the proofs of the two theorems use basically the same method. We therefore omit the proof ofTheorem 2. The proof ofTheorem 1is postponed to Section3.

For Ramsey numbers of fans versus wheels of even order, Surahmat et al. proved that Fnis W3-good for n

≥

3, and

obtained the following result.

∗_{Corresponding author.}

E-mail address:yaojunc@nju.edu.cn(Y. Chen).

http://dx.doi.org/10.1016/j.disc.2016.03.013

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Y. Zhang et al. / Discrete Mathematics 339 (2016) 2284–2287 2285

Theorem 3 (Surahmat et al. [16]). R

(

Fn

,

W3

) =

6n

+

1 for n

≥

3.

In our first main result we generalize the above result.

Theorem 4. R

(

Fn

,

Wm

) =

6n

+

1 for odd m

≥

3 and n

≥

(

5m

+

3

)/

4. The proof ofTheorem 4is postponed to Section3.

For Ramsey numbers of trees versus odd cycles, Burr et al. confirmed that an arbitrary tree Tnon n vertices is Cm-good for odd m

≥

3 and n

≥

756m10_.

Theorem 5 (Burr et al. [7]). R

(

Tn

,

Cm

) =

2n

−

1 for odd m

≥

3 and n

≥

756m10.

As a tribute to Erdős and Sós, a tree T of order n is called an ES-tree if every graph G

=

(

V

,

E

)

with

|

E

|

> |

V

|

(

n

−

2

)/

2 contains T as a subgraph. In 1963, Erdős and Sós conjectured that every tree is an ES-tree. Even though this conjecture is still open, it has been shown that many trees are indeed ES-trees. These results can be found in [10,15,9,17,13,11]. Two additional results were announced without being published. As mentioned in [14], Perles proved in 1990 that caterpillars are ES-trees, where a caterpillar is a tree in which a single path is incident to (or contains) every edge. Ajtai et al. [1] announced that all sufficiently large trees are ES-trees.

In the following, we first present a result on Ramsey numbers of ES-trees versus odd cycles. This turns out to be easy to prove, as is clear from the proof in Section3. Using this result, we establish a theorem on Ramsey numbers of ES-trees versus wheels of even order. By this result we generalize (to some extent) results on Ramsey numbers of wheels versus special trees, including stars, paths and star-like trees. All proofs are postponed to Section3.

Theorem 6. R

(

Tn

,

Cm

) =

2n

−

1 for every ES-tree Tn, odd m

≥

3 and n

≥

m

−

1. Theorem 7. R

(

Tn

,

Wm

) =

3n

−

≥

3 and n

≥

m

−

2.

FromTheorem 7a more general result can be obtained by induction.

Theorem 8. R

(

Tn

,

Kℓ

+

Cm

) = (ℓ +

2

)(

n

−

1

) +

≥

3,

ℓ ≥

1 and n

≥

m

−

2.

We conjecture that the above three theorems hold for all trees. If so, it provides more evidence that the Erdős–Sós Conjecture is true.

Terminology will in general follow that used in [3]. In particular, the length of a longest and shortest cycle of G is denoted by c

(

G

)

and g

(

G

)

, respectively. A graph G is weakly pancyclic if it contains cycles of every length between g

(

G

)

and c

(

G

)

. For two vertex-disjoint graphs G1and G2, G1

∪

G2denotes the disjoint union, and the join G1

+

G2is the graph obtained from

G1

∪

G2by joining every vertex of G1to every vertex of G2by an edge. We use mG to denote m vertex-disjoint copies of G.

2. Lemmas

In order to prove our main theorems, we need the following results.

Lemma 1 (Brandt [4]). Every nonbipartite graph G of order n with e

(

G

) > (

n

−

1

)

2

_/

₄

₊

_{1 is weakly pancyclic with g}

₍

_G

_{) =}

_3.

Lemma 2 (Brandt et al. [5]). Every nonbipartite graph G of order n with

δ(

G

) ≥ (

n

+

2

)/

3 is weakly pancyclic with g

(

G

) =

3 or 4.

Lemma 3 (Dirac [8]). Let G be a graph with

δ(

G

) ≥

2. Then c

(

G

) ≥ δ(

G

) +

1. Moreover, if

δ(

G

) ≥ |

V

(

G

)|/

2, then G has a Hamilton cycle.

Lemma 4 (Erdős and Gallai [10]). Let G be a graph of order n and 3

≤

c

≤

n. If e

(

G

) ≥

1

/

2

((

c

−

1

)(

n

−

1

) +

1

)

, then c

(

G

) ≥

c. Lemma 5 (Erdős and Gallai [10]). Let G be a graph of order n. If e

(

G

) >

n

(

k

−

2

)/

2, then G contains a path on k vertices. Lemma 6 (Hasmawati et al. [12]). If m is odd and 3

≤

m

≤

2n

−

1, then R

(

K1,n−1

,

Wm

) =

3n

−

2.

Lemma 7 (Baskoro et al. [2]). For odd n

≥

3, let G be a graph of order n which is obtained from K_nby removing

⌊

n

/

2

⌋

independent edges. Then G contains all trees on n vertices.

The following lemma is an established result that is straightforward to prove using a Greedy algorithm.

Lemma 8. Let G be a graph with

δ(

G

) ≥

k, and let u

∈

V

(

G

)

. Let T be a tree of order k

+

1 with

v ∈

V

(

T

)

. Then T can be embedded in G such that

v

is mapped to u.

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2286 Y. Zhang et al. / Discrete Mathematics 339 (2016) 2284–2287 3. Proofs of the main results

Proof of Theorem 1. It is clear that K2n−1

∪ ⌈

m

/

2

⌉

K1contains no nK2. Suppose its complement contains a Wm. Then, since the cardinality of a maximum independent set of Wmis (at most)

⌊

m

/

2

⌋

, at most

⌊

m

/

2

⌋

vertices of Wmare in V

(

K2n−1

)

,

hence

|

V

(

Wm

)| ≤

m, a contradiction. This shows that R

(

nK2

,

Wm

) ≥

2n

+ ⌈

m

/

2

⌉

. It is easy to check that Km

∪

(

n

−

1

)

K1

contains no Wmand that its complement contains no nK2. Thus, R

(

nK2

,

Wm

) ≥

n

+

m. Let N

=

max

{

2n

+ ⌈

m

/

2

⌉

,

n

+

m

}

. It remains to prove that R

(

nK2

,

Wm

) ≤

N.

For this, let G be a graph of order N and suppose, to the contrary, that G contains no nK2and G contains no Wm. Let M

⊆

E

(

G

)

be a maximum matching in G and let X

=

V

(

G

)−

V

(

M

)

. Then G

[

X

]

is an edgeless graph, and

|

M

| ≤

n

−

1; otherwise G contains nK2, a contradiction. Moreover, because of the choice of M, for any edge yz

∈

M, we have min

{

dX

(

y

),

dX

(

z

)} ≤

1. Suppose first that n

≥ ⌊

m

/

2

⌋

. Then N

=

2n

+ ⌈

m

/

2

⌉

. Since

|

M

| ≤

n

−

1 and 2

|

M

| + |

X

| =

2n

+ ⌈

m

/

2

⌉

, we have

|

X

| ≥ ⌈

m

/

2

⌉ +

2 and

|

M

| + |

X

| ≥

m

+

1. Thus, G

[

X

]

is a complete graph of order at least

⌈

m

/

2

⌉ +

2, and V

(

M

)

contains a subset Y of m

+

1

− |

X

|

vertices such that each vertex of Y has at least

|

X

| −

1 adjacent vertices of X in G. Thus, there is a vertex x

∈

X such that x is adjacent to every vertex of X

∪

Y

−

x in G. ByLemma 3, G

[

X

∪

Y

−

x

]

contains a Cm, which together with x forms a Wm, a contradiction.

Next suppose that n

≤ ⌊

m

/

2

⌋ −

1. Then N

=

n

+

m. Since

|

M

| ≤

n

−

1 and 2

|

M

| + |

X

| =

n

+

m, we have

|

X

| ≥ ⌈

m

/

2

⌉ +

3 and

|

M

| + |

X

| ≥

m

+

1. We may deduce a contradiction as before. This completes the proof ofTheorem 1.

Proof of Theorem 4. Because 3K2ncontains no Fnand its complement contains no Wmfor odd m, R

(

Fn

,

Wm

) ≥

6n

+

1. It remains to show that R

(

Fn

,

Wm

) ≤

6n

+

1 for m odd and n

≥

(

5m

+

3

)/

4.

For this, let G be a graph of order 6n

+

1 with n

≥

(

5m

+

3

)/

4 and with m odd. It is sufficient to show that either G contains an Fnor G contains a Wm. Suppose to the contrary that neither G contains an Fnnor G contains a Wm.

Suppose∆

(

G

) ≥

2n

+

(

m

+

1

)/

2. Let

v

be a vertex with d

(v) =

∆

(

G

)

. ByTheorem 1, either G

[

N

(v)]

contains nK2, which

together with

v

forms an Fn, or G

[

N

(v)]

contains a Wm. Hence,∆

(

G

) ≤

2n

+

(

m

−

1

)/

2, implying that

δ(

G

) ≥

4n

−

(

m

−

1

)/

2. ByTheorem 3, G contains a W3. Now we choose a vertex u from the W3and let H be the subgraph of G formed by the set of

nonadjacent vertices of u, that is, H

=

G

[

N_G

(

u

)]

. Hence,

|

H

| ≥

4n

−

(

m

−

1

)/

2, g

(

H

) =

3 and H is nonbipartite.

First assume

δ(

H

) ≥ (|

H

| +

2

)/

3. Then byLemma 2, H is weakly pancyclic. ByLemma 3, c

(

H

) ≥ δ(

H

) +

1

≥

m. Thus, H contains a Cm, which together with u forms a Wmin G, a contradiction.

Next assume

δ(

H

) < (|

H

| +

2

)/

3. Then H contains a vertex

w

such that d_H

(w) =

∆

(

H

) > (

2

|

H

| −

5

)/

3, that is, d_H

(w) ≥ (

2

|

H

| −

4

)/

3. Since

|

H

| ≥

4n

−

(

m

−

1

)/

2 and n

≥

(

5m

+

3

)/

4, we have d_H

(w) ≥

2n

+

(

m

−

1

)/

2. ByTheorem 2, either G

[

N_H

(w)]

contains nK2, which together with

w

forms an Fn, a contradiction; or G

[

NH

(w)]

contains a Cm, which together with u forms a Wmin G, also a contradiction. This completes the proof ofTheorem 4.

Proof of Theorem 6. Since 2Kn−1 contains no Tnand its complement contains no Cm for odd m, R

(

Tn

,

Cm

) ≥

2n

−

1. It remains to show that R

(

Tn

,

Cm

) ≤

2n

−

≥

3 and n

≥

m

−

1.

−

1 with odd m

≥

3 and n

≥

m

−

1.

If e

(

G

) > (

2n

−

1

)(

n

−

2

)/

2, then by the definition of ES-trees, G contains every ES-tree Tn. Thus, we may assume e

(

G

) ≤ (

2n

−

1

)(

n

−

2

)/

2.

If e

(

G

) <

1

/

2

(

n

(

2n

−

2

) +

1

)

, then it is easy to calculate that e

(

G

) +

e

(

G

) < (

2n

−

1

)(

2n

−

2

)/

2, a clear contradiction. So e

(

G

) ≥

1

/

2

(

n

(

2n

−

2

) +

1

)

. Then, byLemma 4, c

(

G

) ≥

n

+

1

≥

m. We may assume that G is a nonbipartite graph; otherwise, if G is a bipartite graph, say V

(

G

) = (

X

,

Y

)

with

|

X

| ≥ |

Y

|

, then G

[

X

]

is a complete graph of order at least n, clearly containing Tnas a subgraph. Using e

(

G

) ≥

1

/

2

(

n

(

2n

−

2

) +

1

)

, it is easy to check that e

(

G

) ≥ (

2n

−

2

)

2

/

4

+

1, and byLemma 1, G is weakly pancyclic with g

(

G

) =

3. Thus, G contains Cm. This completes the proof ofTheorem 6. Proof of Theorem 7. It is clear that 3Kn−1contains no Tnand its complement contains no Wmfor odd m. Thus, R

(

Tn

,

Wm

) ≥

3n

−

2. It remains to prove that R

(

Tn

,

Wm

) ≤

3n

−

≥

3 and n

≥

m

−

2. Since this is trivial for n

=

1

,

2, we may assume that n

≥

3.

−

2 and suppose to the contrary that G does not contain some ES-tree Tnand G contains no Wm.

If

δ(

G

) ≥

n

−

1, then byLemma 8, G contains Tn, a contradiction. Thus,

δ(

G

) ≤

n

−

2 and∆

(

G

) ≥

2n

−

1. Let

v

be a vertex with d_G

(v) =

∆

(

G

) ≥

2n

−

1. If n

≥

m

−

1, then, byTheorem 6either G

[

N_G

(v)]

contains a Tn, which is a contradiction; or G

[

N_G

(v)]

contains a Cm, which together with

v

forms a Wm, also a contradiction. Thus, for the remainder it is sufficient to consider the case when n

=

m

−

2.

If there exists a vertex u such that d_G

(

u

) ≥

2n

+

1, then, byTheorem 6either G

[

N_G

(

u

)]

contains a Tn+1, which contains

T_nas a subgraph, a contradiction; or G

[

N_G

(

u

)]

contains a C_m, which together with u forms a W_m, also a contradiction. Thus, ∆

(

G

) ≤

2n and

δ(

G

) ≥

n

−

3. ByLemma 6, G contains a K1,n−1. Let

w

be the vertex of K1,n−1with dK1,n−1

(w) =

n

−

1.

If n

=

3, then T3

=

K1,2is contained in G, a contradiction. Since n

=

m

−

2 and m is odd, we have n

=

m

−

2

≥

5.

Suppose that Tncontains a vertex x with at least two degree 1 neighbors. Then we may get a tree T′of order n

−

2 from Tnby deleting exactly two degree 1 neighbors of x. ByLemma 8, T′can be embedded in G such that x is mapped to

w

. There are at least two adjacent vertices of

w

in G that are not in the embedding of T′_{, showing that T}

ncan be embedded in G, a contradiction.

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Y. Zhang et al. / Discrete Mathematics 339 (2016) 2284–2287 2287

Hence, we may assume that every vertex of Tnhas at most one degree 1 neighbor. This implies that Tncontains a vertex y with degree 2 which has exactly one degree 1 neighbor in Tn. Let y′be the degree 1 neighbor of y, and let z be the other neighbor of y in Tn. Then we may get a tree T′′of order n

−

2 from Tnby deleting y and y′.

We claim that T′′can be embedded in G such that z is mapped to

w

and there are at least three vertices of N

(w)

which are not in the embedding of T′′_{. If d}

G

(w) ≥

n, then this follows directly fromLemma 8. Thus, we may assume that dG

(w) =

n

−

1. We first show there is a vertex z′_{such that the distance between z and z}′_{in T}′′_{is two. Supposing this is not the case, every}

vertex of T′′

−

z is adjacent to z in T′′. Since n

≥

5, then

|

T′′

| ≥

3 and z has at least two degree 1 neighbors, contrary to our assumptions.

Now let z′′_{be the common neighbor of z and z}′_{in T}′′_{. Suppose first that there is an edge}

_w

′′

_w

′_{such that}

_w

′′

_∈

_N

_(w)

_and

w

′

_∈

V

(

G

) −

N

[

w]

. Then we may first map z to

w

, z′′to

w

′′and z′to

w

′. After that, since

δ(

G

) ≥

n

−

3, T′′can be easily embedded in G by using a Greedy algorithm. Since at most n

−

3 vertices of N

[

w]

are in the embedding of T′′, at least three vertices of N

(w)

are not in the embedding of T′′_{. This proves our claim in this subcase.}

Suppose next that there is no edge between G

[

N

[

w]]

and G

−

N

[

w]

. We are going to reach a contradiction in this subcase. Since

|

N

[

w]| =

n, we have

|

V

(

G

) −

N

[

w]| =

2n

−

2. We also have∆

(

G

[

N

[

w]]) ≥

2; otherwise G

[

N

[

w]]

contains Tnby

Lemma 7, a contradiction. Let x1x2x3be a path in G

[

N

[

w]]

. Since

|

V

(

G

)−

N

[

w]| =

2n

−

2, if e

(

G

−

N

[

w]) > (

2n

−

2

)(

n

−

2

)/

2,

then by the definition of ES-trees, G

−

N

[

w]

contains every ES-tree Tn, a contradiction. Thus, e

(

G

−

N

[

w]) ≤ (

2n

−

2

)(

n

−

2

)/

2 and e

(

G

−

N

[

w]) ≥ (

2n

−

2

)(

n

−

1

)/

2. ByLemma 5, G

−

N

[

w]

contains a Pn. Set Pn

=

y1y2

. . .

yn. Since there is no edge between G

[

N

[

w]]

and G

−

N

[

w]

, then G contains a Cn+2

=

x1y1y2

. . .

yn−1x3ynx1, which together with x2forms a Wn+2

=

Wm in G, a contradiction. This completes the proof of our claim that T′′can be embedded in G such that z is mapped to

w

and there are at least three vertices of N

(w)

which are not in the embedding of T′′_.

Let z1

,

z2

,

z3 be three vertices of N

(w)

which are not in the embedding of T′′. If z1has a neighbor that is not in the

embedding of T′′, say s, then we may map y to z1, and y′to s. Then G contains Tn, a contradiction. This implies that all vertices of N

(

z1

)

are in the embedding of T′′. For the same reason, all vertices of N

(

zi

)

are in the embedding of T′′for i

=

2

,

3. Let H be the graph obtained from G by deleting the embedding of T′′_{and z}

1

,

z2

,

z3. Then

|

H

| =

2n

−

3. If e

(

H

) > (

2n

−

3

)(

n

−

2

)/

2,

then by the definition of ES-trees, H contains every ES-tree Tn, a contradiction. Thus, e

(

H

) ≤ (

2n

−

3

)(

n

−

2

)/

2 and e

(

H

) ≥ (

2n

−

3

)(

n

−

2

)/

2. ByLemma 5, H contains a Pn−1. Set Pn−1

=

s1s2

. . .

sn−1, and let snbe a vertex of H not in Pn−1.

Then G contains a Cn+2

=

z1s1s2

. . .

sn−1z2snz1, which together with z3forms a Wn+2

=

Wmin G, our final contradiction. This completes the proof ofTheorem 7.

Proof of Theorem 8. ByTheorem 7, the result holds for

ℓ =

1. Assume that k

≥

2 and that the result holds for all

ℓ

with 1

≤

ℓ <

k. It suffices to prove that it also holds for

ℓ =

k.

Since

(

k

+

2

)

Kn−1 contains no Tnand its complement contains no Kk

+

Cmfor odd m, we have R

(

Tn

,

Kk

+

Cm

) ≥

(

k

+

2

)(

n

−

1

) +

1. Let G be a graph of order

(

k

+

2

)(

n

−

1

) +

1. If

δ(

G

) ≥

n

−

1, then, byLemma 8G contains Tn and the proof is done. Let now

δ(

G

) ≤

n

−

2. Then∆

(

G

) ≥ (

k

+

1

)(

n

−

1

) +

1. Let

v

be a vertex such that d_G

(v) =

∆

(

G

)

. By the induction hypothesis, either G

[

N_G

[

v]]

contains a Tn, or G

[

NG

[

v]]

contains a Kk−1

+

Cm, which together with

v

forms a Kk

+

Cmin G. This completes the proof ofTheorem 8.

Acknowledgments

This research was supported by NSFC under grant numbers 11371193 and 11101207, and in part by the Priority Academic Program Development of Jiangsu Higher Education Institutions.

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