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arXiv:1111.3480v2 [math.CO] 5 Dec 2011

Oriented diameter and rainbow

connection number of a graph∗

Xiaolong Huang, Hengzhe Li, Xueliang Li, Yuefang Sun

Center for Combinatorics and LPMC-TJKLC

Nankai University, Tianjin 300071, China

huangxiaolong@mail.nankai.edu.cn; lhz2010@mail.nankai.edu.cn;

lxl@nankai.edu.cn; bruceseun@gmail.com

Abstract

The oriented diameter of a bridgeless graph G is min{diam(H) |H is an

orientation of G}. A path in an edge-colored graph G, where adjacent

edges may have the same color, is called rainbow if no two edges of the

path are colored the same. The rainbow connection number rc(G) of G is

the smallest integer k for which there exists a k-edge-coloring of G such

that every two distinct vertices of G are connected by a rainbow path.

In this paper, we obtain upper bounds for the oriented diameter and the

rainbow connection number of a graph in terms of rad(G) and η(G), where

rad(G) is the radius of G and η(G) is the smallest integer number such

that every edge of G is contained in a cycle of length at most η(G). We

also obtain constant bounds of the oriented diameter and the rainbow

connection number for a (bipartite) graph G in terms of the minimum

degree of G.

Keywords: Diameter, Radius, Oriented diameter, Rainbow connection

number, Cycle length, Bipartite graph

AMS subject classification 2010: 05C15, 05C40

∗Supported by NSFC No.11071130, and “the Fundamental Research Funds for the Central

Universities”.

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1Introduction

All graphs in this paper are undirected, finite and simple. We refer to book [2]

for notation and terminology not described here. A path u = u1,u2,...,uk= v

is called a Pu,v path. Denote by uiPuj the subpath ui,ui+1,...,uj for i ≤ j.

The length ℓ(P) of a path P is the number of edges in P. The distance be-

tween two vertices x and y in G, denoted by dG(x,y), is the length of a short-

est path between them. The eccentricity of a vertex x in G is eccG(x) =

maxy∈V (G)d(x,y). The radius and diameter of G are rad(G) = minx∈V (G)ecc(x)

and diam(G) = maxx∈V (G)ecc(x), respectively.

graph G if ecc(u) = rad(G). The oriented diameter of a bridgeless graph G

is min{diam(H) | H is an orientation of G}, and the oriented radius of a

bridgeless graph G is min{rad(H) |H is an orientation of G}. For any graph

G with edge-connectivity λ(G) = 0,1, G has oriented radius (resp. diameter) ∞.

A vertex u is a center of a

In 1939, Robbins solved the One-Way Street Problem and proved that a graph

G admits a strongly connected orientation if and only if G is bridgeless, that is,

G does not have any cut-edge. Naturally, one hopes that the oriented diameter of

a bridgeless graph is as small as possible. Bondy and Murty suggested to study

the quantitative variations on Robbins’ theorem. In particular, they conjectured

that there exists a function f such that every bridgeless graph with diameter d

admits an orientation of diameter at most f(d).

In 1978, Chv´ atal and Thomassen [5] obtained some general bounds.

Theorem 1 (Chv´ atal and Thomassen 1978 [5]). For every bridgeless graph

G, there exists an orientation H of G such that

rad(H) ≤ rad(G)2+ rad(G),

diam(H) ≤ 2rad(G)2+ 2rad(G).

Moreover, the above bounds are optimal.

There exists a minor error when they constructed the graph Gdwhich arrives

at the upper bound when d is odd. Kwok, Liu and West gave a slight correction

in [11].

They also showed that determining whether an arbitrary graph can be oriented

so that its diameter is at most 2 is NP-complete. Bounds for the oriented diameter

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of graphs have also been studied in terms of other parameters, for example, radius,

dominating number [5, 6, 11, 18], etc. Some classes of graphs have also been

studied in [6, 7, 8, 9, 14].

Let η(G) be the smallest integer such that every edge of G belongs to a cycle

of length at most η(G). In this paper, we show the following result.

Theorem 2. For every bridgeless graph G, there exists an orientation H of G

such that

rad(H) ≤

rad(G)

?

i=1

min{2i,η(G) − 1} ≤ rad(G)(η(G) − 1),

diam(H) ≤ 2

rad(G)

?

i=1

min{2i,η(G) − 1} ≤ 2rad(G)(η(G) − 1).

Note that?rad(G)

2rad(G). So our result implies Chv´ atal and Thomassen’s Theorem 1.

i=1

min{2i,η(G) − 1} ≤ rad(G)2+ rad(G) and diam(H) ≤

A path in an edge-colored graph G, where adjacent edges may have the same

color, is called rainbow if no two edges of the path are colored the same. An edge-

coloring of a graph G is a rainbow edge-coloring if every two distinct vertices of

graph G are connected by a rainbow path. The rainbow connection number rc(G)

of G is the minimum integer k for which there exists a rainbow k-edge-coloring

of G. It is easy to see that diam(G) ≤ rc(G) for any connected graph G. The

rainbow connection number was introduced by Chartrand et al. in [4]. It is of

great use in transferring information of high security in multicomputer networks.

We refer the readers to [3] for details.

Chakraborty et al. [3] investigated the hardness and algorithms for the rain-

bow connection number, and showed that given a graph G, deciding if rc(G) = 2

is NP-complete. Bounds for the rainbow connection number of a graph have also

been studies in terms of other graph parameters, for example, radius, dominating

number, minimum degree, connectivity, etc. [1, 4, 10]. Cayley graphs and line

graphs were studied in [12] and [13], respectively.

A subgraph H of a graph G is called isometric if the distance between any

two distinct vertices in H is the same as their distance in G. The size of a largest

isometric cycle in G is denoted by ζ(G). Clearly, every isometric cycle is an

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induced cycle and thus ζ(G) is not larger than the chordality, where chordality

is the length of a largest induced cycle in G. In [1], Basavaraju, Chandran,

Rajendraprasad and Ramaswamy got the the following sharp upper bound for

the rainbow connection number of a bridgeless graph G in terms of rad(G) and

ζ(G).

Theorem 3 (Basavaraju et al. [1]). For every bridgeless graph G,

rc(G) ≤

rad(G)

?

i=1

min{2i + 1,ζ(G)} ≤ rad(G)ζ(G).

In this paper, we show the following result.

Theorem 4. For every bridgeless graph G,

rc(G) ≤

rad(G)

?

i=1

min{2i + 1,η(G)} ≤ rad(G)η(G).

From Lemma 2 of Section 2, we will see that η(G) ≤ ζ(G). Thus our result

implies Theorem 3.

This paper is organized as follows: in Section 2, we introduce some new

definitions and show several lemmas. In Section 3, we prove Theorem 2 and study

upper for the oriented radius (resp. diameter) of plane graphs, edge-transitive

graphs and general (bipartite) graphs. In Section 4, we prove Theorem 4 and

study upper for the rainbow connection number of plane graphs, edge-transitive

graphs and general (bipartite) graphs.

2Preliminaries

In this section, we introduce some definitions and show several lemmas.

Definition 1. For any x ∈ V (G) and k ≥ 0, the k-step open neighborhood

is {y|d(x,y) = k} and denoted by Nk(x), the k-step closed neighborhood is

{y |d(x,y) ≤ k} and denoted by Nk[x]. If k = 1, we simply write N(x) and N[x]

for N1(x) and N1[x], respectively.

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Definition 2. Let G be a graph and H be a subset of V (G) (or a subgraph of

G). The edges between H and G \ H are called legs of H. An H-ear is a path

P = (u0,u1,...,uk) in G such that V (H)∩V (P) = {u0,uk}. The vertices u0, uk

are called the foot of P in H and u0u1, uk−1uk are called the legs of P. The

length of an H-ear is the length of the corresponding path. If u0= uk, then P

is called a closed H-ear. For any leg e of H, denote by ℓ(e) the smallest number

such that there exists an H-ear of length ℓ(e) containing e, and such an H-ear is

called an optimal (H,e)-ear.

Note that for any optimal (H,e)-ear P and every pair (x,y) ?= (u0,uk) of

distinct vertices of P, x and y are adjacent on P if and only if x and y are

adjacent in G.

Definition 3. For any two paths P and Q, the joint of P and Q are the common

vertex and edge of P and Q. Paths P and Q have k continuous common segments

if the common vertex and edge are k disjoint paths.

Definition 4. Let P and Q be two paths in G. Call P and Q independent if

they has no common internal vertex.

Lemma 1. Let n ≥ 1 be an integer, and let G be a graph, H be a subgraph of

G and ei= uivibe a leg of H and Pi= Puiwibe an optimal (G,ei)-ear, where

1 ≤ i ≤ n and ui,wiare the foot of Pi. Then for any leg ej= ujvj?= ei, 1 ≤ i ≤ n,

either there exists an optimal (H,ej)-ear Pj= Pujwjsuch that either Piand Pj

are independent for any Pi, 1 ≤ i ≤ n, or Piand Pj have only one continuous

common segment containing wjfor some Pi.

Proof. Let Pjbe an optimal (H,ej)-ear. If Piand Pjare independent for any i,

then we are done. Suppose that Piand Pjhave m continuous common segments

for some i, where m ≥ 1. When m ≥ 2, we first construct an optimal (H,ej)-

ear P∗

Pi1,Pi2,...,Pimbe the m continuous common segments of Piand Pj and they

appear in Piin that order. See Figure 1 for details. Furthermore, suppose that

xikand yikare the two ends of the path Pikand they appear in Pisuccessively.

We say that the following claim holds.

jsuch that Pi and P∗

jhas only one continuous common segment. Let

Claim 1: ℓ(ykPixk+1) = ℓ(ykPjxk+1) for any 1 ≤ k ≤ m − 1.

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Pi1

Pi2

Pik

H

ui

uj

wi

wj

ei

ej

Pi

Pj

xi1

yi1

xi2

yi2

xik

yik

Figure 1. Two H-ears Piand Pj.

If not, that is, there exists an integer k such that ℓ(ykPixk+1) ?= ℓ(ykPjxk+1).

Without loss of generality, we assume ℓ(ykPixk+1) < ℓ(ykPjxk+1). Then we shall

get a more shorter path H-ear containing ejby replacing ykPjxk+1with ykPixk+1,

a contradiction. Thus ℓ(ykPixk+1) = ℓ(ykPjxk+1) for any k.

Let P∗

jbe the path obtained from Pj by replacing ykPjxk+1with ykPixk+1,

and let Pj = P∗

contain wj. Suppose x and y are the two ends of the common segment such that

x and y appeared on P starting from uito wisuccessively. Similar to Claim 1,

ℓ(yPiwi) = ℓ(yPjwj). Let P∗

with yPiwi. Clearly, P∗

j. If the continuous common segment of Pi and Pj does not

jbe the path obtained from Pjby replacing yPjwj

jis our desired optimal (H,ujvj)-ear.

Lemma 2. For every bridgeless graph G, η(G) ≤ ζ(G).

Proof. Suppose that there exists an edge e such that the length ℓ(C) of the

smallest cycle C containing e is larger than ζ(G). Then, C is not an isometric

cycle since the length of a largest isometric cycle is ζ(G). Thus there exist two

vertices u and v on C such that dG(u,v) < dC(u,v). Let P be a shortest path

between u and v in G. Then a closed trial C′containing e is obtained from the

segment of C containing e between u and v by adding P. Clearly, the length ℓ(C′)

is less than ℓ(C). We can get a cycle C′′containing e from C′. Thus there exists

a cycle C′′containing e with length less than ℓ(C), a contradiction. Therefore

η(G) ≤ ζ(G).

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Lemma 3. Let G be a bridgeless graph and u be a center of G. For any i ≤

rad(G) − 1 and every leg e of Ni(u), there exists an optimal (Ni[u],e)-ear with

length at most min{2(rad(G) − i) + 1,η(G)}.

Proof. Let P be an optimal (Ni[u],e)-ear. Since e belongs a cycle with length

at most η(G), ℓ(P) ≤ η(G). On the other hand, if ℓ(P) ≥ 2(rad(G) − i) + 1,

then the middle vertex of P has length at least rad(G) − i + 1 from Ni[u], a

contradiction.

3Oriented diameter

At first, we have the following observation.

Observation 1. Let G be a graph and H be a bridgeless spanning subgraph of

G. Then the oriented radius (resp. diameter) of G is not larger than the oriented

radius (resp. diameter) of H.

Proof of Theorem 2: We only need to show that G has an orientation H such

that rad(H) ≤?rad(G)

i=1

min{2i,η(G)−1} ≤ rad(G)(η(G)−1). Let u be a center

of G and let H0be the trivial graph with vertex set {u}. We assert that there

exists a subgraph Giof G such that Ni[u] ⊆ V (Gi) and Gihas an orientation Hi

satisfying that rad(Hi) ≤ eccHi(u) ≤ Σi

j=1min{2(rad(G) − j),η(G) − 1}.

Basic step: When i = 1, we omit it since the proof of this step is similar to

that of the following induction step.

Induction step: Assume that the above assertion holds for i−1. Next we show

that the above assertion also holds for i. For any v ∈ Ni(u), either v ∈ V (Hi−1)

or v ∈ N(Hi) since Ni−1[u] ⊆ V (Hi−1). If Ni(u) ⊆ V (Hi−1), then let Hi= Hi−1

and we are done. Thus, we suppose Ni(u) ?⊆ V (Hi−1) in the following.

Let X = Ni(u) \ V (Hi−1). Pick x1∈ X, let y1be a neighbor of x1in Hi−1

and let P1 = Py1z1be an optimal (Hi−1,x1y1)-ear. We orient P such that P1

is a directed path. Pick x2 ∈ X satisfying that all incident edges of x2 are

not oriented. Let y2 be a neighbor of x2 in Hi−1. If there exists an optimal

(Hi−1,x2y2)-ear P2such that P1and P2are independent, then we can orient P2

such that P2is a directed path. Otherwise, by Lemma 1 there exists an optimal

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(Hi−1,x2y2)-ear P2= Py2z2such that P1and P2has only one continuous common

segment containing z2. Clearly, we can orient the edges in E(P2)\E(P1) such that

P2is a directed path. We can pick the vertices of X and oriented optimal H-ears

similar to the above method until that for any x ∈ X, at least two incident edges

of x are oriented. Let Hibe the graph obtained from Hi−1by adding vertices in

V (G) \ V (Hi−1), which has at least two new oriented incident edges, and adding

new oriented edges. Clearly, Ni[u] ⊆ V (Hi) = V (Gi).

Now we show that rad(Hi) ≤ Σi

show that for every vertex x of Hi, dHi(Hi−1,x) ≤ min{2(rad(G) − i),η(G) − 1}

and dHi(x,Hi−1) ≤ min{2(rad(G) − i),η(G) − 1}. If x ∈ V (Hi−1), then the

assertion holds by inductive hypothesis.

rected optimal (Hi,e)-ear containing x, where e is some leg of Hi−1 (such a

leg and such an ear exists by the definition of Hi.

min{2(rad(G) − i) + 1,η(G)}. Thus, dHi(x,Hi−1) ≤ min{2(rad(G) − i),η(G) −

1} and dHi(Hi−1,x) ≤ min{2(rad(G) − i),η(G) − 1}. Therefore, rad(Hi) ≤

Σi

j=1min{2(rad(G)−i),η(G)−1}. It suffices to

If x ?∈ V (Hi−1).Let P be a di-

By Lemma 3, ℓ(P) ≤

j=1min{2(rad(G) − j),η(G) − 1}.

?

Remark 1. The above theorem is optimal since it implies Chv´ atal and Thomassen’s

optimal Theorem 1. Readers can see [5, 11] for optimal examples.

The following example shows that our result is better than that of Theorem 1.

Example 1. Let H3be a triangle with one of its vertices designated as root. In

order to construct Hr, take two copies of Hr−1. Let Hrbe the graph obtained

from the triangle u0,u1,u2by identifying the root of first (resp. second) copy of

Hr−1with u1(resp. u2), and u0be the root of Hr. Let Grbe the graph obtained

by taking two copies of Hrand identifying their roots. See Figure 2 for details. It

is easy to check that Grhas radius r and every edge belongs to a cycle of length

η(G) = 3. By Theorem 1, Grhas an orientation Hrsuch that rad(Hr) ≤ r2+ r

and diam(Hr) ≤ 2r2+ 2r. But, by Theorem 2, Grhas an orientation Hrsuch

that rad(G) ≤ 2r and diam(G) ≤ 4r. On the other hand, it is easy to check that

all the strong orientations of Grhas radius 2r and diameter 4r.

We have the following result for plane graphs.

Theorem 5. Let G be a plane graph. If the length of the boundary of every face

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u

Figure 2. The graph G3which has oriented

radius 6 and oriented diameter 12.

is at most k, then G has an oriented H such that rad(H) ≤ rad(G)(k − 1) and

diam(H) ≤ 2rad(G)(k − 1).

Since every edge of a maximal plane (resp. outerplane) graph belongs to a

cycle with length 3, the following corollary holds.

Corollary 1. Let G be a maximal plane (resp. outerplane) graph. Then there ex-

ists an orientation H of G such that rad(H) ≤ 2rad(G) and rad(H) ≤ 4rad(G).

A graph G is edge-transitive if for any e1,e2∈ E(G), there exists an auto-

morphism g such that g(e1) = e2. We have the following result for edge-transitive

graphs.

Theorem 6. Let G be a bridgeless edge-transitive graph. Then G has an orienta-

tion H such that rad(H) ≤ rad(G)(g(G)−1) and diam(H) ≤ 2rad(G)(g(G)−1),

where g(G) is the girth of G, that is, the length of a smallest induced cycle.

For general bipartite graphs, the following theorem holds.

Theorem 7. Let G = (V1∪V2,E) be a bipartite graph with |V1| = n and |V2| = m.

If d(x) ≥ k > ⌈m/2⌉ for any x ∈ V1, d(y) ≥ r > ⌈n/2⌉ for any y ∈ V2, then there

exists an orientation H of G such that rad(H) ≤ 9.

Proof. It suffices to show that rad(G) ≤ 3 and η(G) ≤ 4 by Theorem 2.

First, we show that rad(G) ≤ 3. Fix a vertex x in G, and let y be any

vertex different from x in G. If x and y belong to the same part, without loss

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of generality, say x,y ∈ V1. Let X and Y be neighborhoods of x and y in V2,

respectively. If X ∩ Y = ∅, then |V2| ≥ |X| + |Y | ≥ 2k > m, a contradiction.

Thus X ∩ Y ?= ∅, that is, there exists a path between x and y of length two. If

x and y belong to different parts, without loss of generality, say x ∈ V1,y ∈ V2.

Suppose x and y are nonadjacent, otherwise there is nothing to do. Let X and Y

be neighborhoods of x and y in G, and let X′be the set of neighbors except for

x of X in G. If X′∩Y = ∅, then |V1| ≥ 1+|Y |+|X′| ≥ 1+r+(r−1) = 2r > n,

a contradiction (Note that |X′| ≥ r − 1). Thus X′∩ Y ?= ∅, that is, there exists

a path between x and y of length three in G.

Next we show that η(G) ≤ 4. Let xy be any edge in G. Let X be the set of

neighbors of x except for y in G, let Y be the set of neighbors of y except for x

in G, let X′be the set of neighbors except for x of X in G. If X′∩ Y = ∅, then

|V1| ≥ 1 + |Y | + |X′| ≥ 1 + (r − 1) + (r − 1) = 2r − 1 > n, a contradiction (Note

that |X′| ≥ r − 1). Thus X′∩ Y ?= ∅, that is, there exists a cycle containing xy

of length four in G.

Remark 2. The degree condition is optimal. Let m,n be two even numbers with

n,m ≥ 2. Since Kn/2,m/2∪ Kn/2,m/2is disconnected, the oriented radius (resp.

diameter) of Kn/2,m/2∪ Kn/2,m/2is ∞.

For equal bipartition k-regular graph, the following corollary holds.

Corollary 2. Let G = (V1∪V2,E) be a k-regular bipartite graph with |V1| = |V2| =

n. If k > n/2, then there exists an orientation H of G such that rad(H) ≤ 9.

The following theorem holds for general graphs.

Theorem 8. Let G be a graph.

(i) If there exists an integer k ≥ 2 such that |Nk(u)| > n/2−1 for every vertex

u in G, then G has an orientation H such that rad(H) ≤ 4k2and diam(H) ≤ 8k2.

(ii) If δ(G) > n/2, then G has an orientation H such that rad(H) ≤ 4 and

diam(H) ≤ 8.

Proof. Since methods of proofs of (i) and (ii) are similar, we only prove (i). For

(i), it suffices to show that rad(G) ≤ 2k and η(G) ≤ 2k + 1 by Theorem 2.

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We first show rad(G) ≤ 2k. Fix u in G, for every v ∈ V (G), if v ∈ Nk[u],

then d(u,v) ≤ k. Suppose v ?∈ Nk[u], we have Nk(u) ∩ Nk(v) ?= ∅. If not, that

is, Nk(u) ∩ Nk(v) = ∅, then |Nk(u)| + |Nk(v)| + 2 > n (a contradiction). Thus

d(u,v) ≤ 2k.

Next we show η(G) ≤ 2k + 1. Let e = uv be any edge in G. If Nk(u) ∩

Nk(v) = ∅, then |V (G)| ≥ |Nk(u)| + |Nk(v)| + 2 > n, a contradiction. Thus

Nk(u) ∩ Nk(v) ?= ∅. Pick w ∈ Nk(u) ∩ Nk(v), and let P (resp. Q) be a path

between u and w (resp. between v and w). Then e belongs a close trial uPwQvu

of length 2k + 1. Therefore, e belongs a cycle of length at most 2k + 1.

Remark 3. The above condition is almost optimal since Kn/2∪ Kn/2is discon-

nected for even n.

Corollary 3. Let G be a graph with minimum degree δ(G) and girth g(G). If

there exists an integer k such that k ≤ g(G)/2 and δ(G)(δ(G)−1)k−1> n/2−1,

then G has an orientation H such that rad(H) ≤ 4k2.

Proof. Let k be an integer such that k ≤ g(G)/2 and δ(G)(δ(G)−1)k−1> n/2−1.

For any vertex u of G, let 1 ≤ i < k be any integer and x,y ∈ Ni(u). If x and

y have a common neighbor z in Ni+1(u), then G has a cycle of length at most

2i < 2k ≤ g(G)/2, a contradiction. Thus x and y has no common neighbor in

Ni+1(u). Therefore, |Nk(u)| ≥ δ(G)(δ(G) − 1)k−1> n/2 − 1. By Theorem 2, G

has an orientation H such that rad(H) ≤ 4k2.

4Upper bound for rainbow connection number

At first, we have the following observation.

Observation 2. Let G be a graph and H be a spanning subgraph of G. Then

rc(H) ≤ rc(G).

Proof of Theorem 4: Let u be a center of G and let H0 be the trivial graph

with vertex set {u}. We assert that there exists a subgraph Hi of G such that

Ni[u] ⊆ V (Hi) and rc(Hi) ≤ Σi

j=1min{2(rad(G) − j) + 1,η(G)}.

Basic step: When i = 1, we omit it since the proof of this step is similar to

that of the following induction step.

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Induction step: Assume that the above assertion holds for i − 1 and c is a

rc(Hi−1)-rainbow coloring of Hi−1. Next we show that the above assertion holds

for i. For any v ∈ Ni(u), either v ∈ V (Hi−1) or v ∈ N(Hi) since Ni−1[u] ⊆

V (Hi−1). If Ni(u) ⊆ V (Hi−1), then let Hi= Hi−1and we are done. Thus, we

suppose Ni(u) ?⊆ V (Hi−1) in the following.

Let C1= {α1,α2,···} and C2= {β1,β2,···} be two pools of colors, none of

which are used to color Hi−1. An edge-coloring of an H-ear P = (u0,u1,··· ,uk) is

a symmetrical coloring if its edges are colored by α1,α2,··· ,α⌈k/2⌉,β⌊k/2⌋,··· ,β2,β1

in that order or β1,β2,··· ,β⌊k/2⌋,α⌈k/2⌉··· ,α2,α1in that order.

Let X = Ni(u)\V (Hi−1) and m = min{2(rad(G)−i)+1,η(G)}. Pick x1∈ X,

Let y1be a neighbor of x1in Hi−1and P1be an optimal (Hi−1,x1y1)-ear. We can

color P symmetrically with colors α1,α2,··· ,α⌈ℓ(P)/2⌉,β⌊ℓ(P)/2⌋,...,β2,β1. Pick

x2 ∈ X satisfying that all the incident edges of x2are not colored. Let y2be

a neighbor of x2 in Hi−1. If there exists an optimal (Hi−1,x2y2)-ear P2 such

that P1and P2are independent, then we can color P2symmetrically with colors

α1,α2,··· ,α⌈ℓ(P2)/2⌉,β⌊ℓ(P2)/2⌋,...,β2,β1. Otherwise, by Lemma 1, there exists an

optimal (Hi−1,x2y2)-ear P2= Py2z2such that P1and P2have only one continuous

common segment containing z2, where z2is the other foot of P2. Thus we can

color P2 symmetrically with colors α1,α2,··· ,α⌈ℓ(P2)/2⌉,β⌊ℓ(P2)/2⌋,...,β2,β1 by

preserving the coloring of P1. We can pick the vertices of X and color optimal

Hi-ears until that for any x ∈ X, at least two incident edges of x are colored.

Since for any leg e of Hi−1, ℓ(e) ≤ m by Lemma 3, we use at most m coloring in

the above coloring process.

Let Hibe the graph obtained from Hi−1by adding vertices in V (G)\V (Hi−1),

which has at least two new colored incident edges, and adding new colored edges.

Clearly, Ni[u] ⊆ V (Hi). It is suffices to show that Hiis rainbow connected. Let

x and y be two distinct vertices in Hi. If x,y ∈ V (Hi−1), then there exists a

rainbow path between x and y by inductive hypothesis. If exactly one of x and y

belongs to V (Hi−1), say x. Let P be a symmetrical colored Hi−1-ear containing

y and y′be a foot of P. There exists a rainbow path Q between x and y′in Hi−1

by inductive hypothesis. Thus, xQy′Py is a rainbow path between x and y in Hi.

Suppose none of x and y belongs to Hi−1. Let P and Q be symmetrical

colored Hi−1-ear containing x and y, respectively.Furthermore, let x′,x′′be

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the foot of P and y′,y′′be the foot of Q. Without loss of generality, assume

that P is colored from x′to x′′by α1,α2,··· ,α⌈ℓ(P)/2⌉,β⌊ℓ(P)/2⌋,...,β2,β1in that

order, and Q is colored from y′to y′′by α1,α2,··· ,α⌈ℓ(Q)/2⌉,β⌊ℓ(Q)/2⌋,...,β2,β1

in that order. If ℓ(x′Px) ≤ ℓ(y′Qy).

and y′′in Hi−1. Then xPx′Ry′′Qy is a rainbow path between x and y in Hi.

Otherwise, ℓ(x′Px) > ℓ(y′Qy). Let R be a rainbow path between y′and x′′in

Hi−1. Then yPy′Rx′′Qx is a rainbow path between x and y in Hi. Thus, there

exists a rainbow path between any two distinct vertices in Hi, that is, Hi is

(Σi

Let R be a rainbow path between x′

j=1min{2(rad(G) − j) + 1,η(G)})-rainbow connected.

?

The following optimal example is from [1].

u = xr

xr−1

x2

x1

x0= v

Pr

P2

P1

Figure 3. Graph Hr,η(G). Every Piis a path between xi

and xi−1of length ℓ(Pi) = min{2i,η(G) − 1}.

Example 2. For any r ≥ 1 and 3 ≤ η(G) ≤ 2r + 1, we first construct the

graph Hr,η(G)as in Figure 3. Clearly, Hr,η(G)is a bridgeless graph with radius

rad(G) = ecc(u) = r and every edge of Hr,η(G)is contained in a cycle of length

at most η(G).

Let m =?r

mr+1, and V (Hj) = {xj: x ∈ V (Hr,η(G)) and E(Hj) = {xjyj|xy ∈ E(Hr,η(G))}.

Identify the vertex ujas a new vertex u. The resulting graph is denoted by G.

It is easy to check that G is a bridgeless graph with radius rad(G) = ecc(u) = r

and every edge of Hr,η(G)is contained in a cycle of length at most η(G). Thus,

rc(G) ≤ Σr

and any k-edge coloring of G, every r-length Puvj path can be colored in at most kr

different ways. By the Pigeonhole Principle, there exist p ?= q, 1 ≤ p < q ≤ mr+1

such that c(xp

vpto vq. For every 1 ≤ i ≤ r, xp

belong to P. Thus, ℓ(P) ≥ Σr

any rainbow path between vpand vq. Hence, rc(G) =?r

i=1min{2i+1,η(G)} and let Hjbe a copy of Hr,η(G), where 1 ≤ j ≤

i=1min{2i+1,η(G)} by Theorem 4. On the other hand, for any k < m

i−1xp

i) = c(xq

i−1xq

i) for 1 ≤ i ≤ r. Consider any rainbow path P from

i−1xp

i=1min{2i+1,η(G)} = m, and there does not exist

ibelongs to P if and only if xq

i−1xq

idoes not

i=1min{2i + 1,η(G)}.

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The following example shows that our result is better than that of Theorem 3.

Example 3. Let r ≥ 3,k ≥ 2r be two integers, and Wk= Ck∨ K1be an wheel,

where V (Ck) = {u1,u2,...,uk} and V (K1) = {u}. Let H be the graph obtained

from Wkby inserting r−1 vertices between every edge uui, 1 ≤ i ≤ k. For every

edge e = xy of H, add a new vertex veand new edges vex,vey. Denote by G the

resulting graph. It is easy to check that rad(G) = r, diam(G) = 2r, η(G) = 3

and ζ(G) = 2r − 1. By Theorem 2, we have rc(G) ≤?r

r2+ 2r − 2. But, by Theorem 7 we have rc(G) ≤ 3r. On the other hand,

rc(G) ≥ 2r since diam(G) = 2r.

i=1min{2i + 1,ζ(G)} ≤

The remaining results are similar to those in Section 3.

Theorem 9. Let G be a plane graph. If the length of the boundary of every face

is at most k, then rc(G) ≤ k rad(G).

Corollary 4. Let G be a maximal plane (resp. outerplane) graph. Then rc(G) ≤

3rad(G).

Theorem 10. Let G be a bridgeless edge-transitive graph. Then rc(G) ≤ rad(G)g(G),

where g(G) is the girth of G.

Theorem 11. Let G = (V1∪ V2,E) be a bipartite graph with |V1| = n and

|V2| = m. If d(x) ≥ k > ⌈m/2⌉ for any x ∈ V1, d(y) ≥ r > ⌈n/2⌉ for any y ∈ V2,

then rc(G) ≤ 12.

Remark 4. The degree condition is optimal. Let m,n be two even numbers with

n,m ≥ 2. Since Kn/2,m/2∪Kn/2,m/2is disconnected, rc(Kn/2,m/2∪Kn/2,m/2) = ∞.

Corollary 5. Let G = (V1∪ V2,E) be a k-regular bipartite graph with |V1| =

|V2| = n. If k > ⌈n/2⌉, then rc(G) ≤ 12.

The following theorem holds for general graphs.

Theorem 12. Let G be a graph.

(i) If there exists an integer k ≥ 2 such that |Nk(u)| > n/2−1 for every vertex

u in G, then rc(G) ≤ 4k2+ 2k.

(ii) If δ(G) > n/2, then rc(G) ≤ 6.

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Remark 5. The above condition is almost optimal since Kn/2∪ Kn/2is discon-

nected for even n.

Corollary 6. Let G be a graph with minimum degree δ(G) and girth g(G). If

there exists an integer k such that k < g(G)/2 and δ(G)(δ(G)−1)k−1> n/2−1,

then then rc(G) ≤ 4k2+ 2k.

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