is a bound on the adjacent vertex distinguishing edge chromatic number

Journal of Combinatorial Theory, Series B 95 (2005) 246–256www.elsevier.com/locate/jctb

� + 300 is a bound on the adjacent vertexdistinguishing edge chromatic number

Hamed HatamiDepartment of Computer Science, University of Toronto, Ont., Canada M5S 3G4

Received 7 May 2005Available online 13 June 2005

Abstract

An adjacent vertex distinguishing edge-coloring or anavd-coloring of a simple graphG is a properedge-coloring ofG such that no pair of adjacent vertices meets the same set of colors. We prove thatevery graph with maximum degree� and with no isolated edges has anavd-coloring with at most� + 300 colors, provided that�>1020.© 2005 Elsevier Inc. All rights reserved.

MSC:05C15

Keywords:Adjacent vertex distinguishing edge coloring; Vertex distinguishing edge coloring; Adjacent strongedge coloring; Strong edge coloring

1. Introduction

We follow [10] for terminologies and notations not defined here. For every vertexv ina graphG, degG(v) or deg(v) when there is no ambiguity denotes the degree ofv in thegraphG. For every partial edge coloringc of a graphG, and every vertexv of G, let Sc(v)denote the set of the colors incident tov.

In a partial edge coloringca vertexu is calleddistinguishable, if for every vertexv adja-cent tou,Sc(u) �= Sc(v).An edge coloringc is calledadjacent vertex distinguishing or anavd-coloring if every vertex is distinguishable. Ak-avd-coloring is anavd-coloring using

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H. Hatami / Journal of Combinatorial Theory, Series B 95 (2005) 246–256 247

at mostk colors. It is clear that every graph with isolated edges does not have anyavd-coloring. Theavd-chromatic number of a graphG, the minimum number of colors inan avd-coloring ofG, is defined for every graphG without any isolated edge. Adjacentvertex distinguishing edge colorings are studied in[1,2,6,11], where different names suchasadjacent strong edge coloring [11] and1-strong edge coloring [1,6] are used torefer to anavd-coloring. Adjacent vertex distinguishing edge colorings are related to vertexdistinguishing edge colorings in which the conditionSc(u) �= Sc(v) holds for every pair ofverticesu andv, not necessarily adjacent. This concept has been studied in many papers(see for example [3–5,7]).

Another interesting problem arises when we drop the condition that the edge coloringis proper and allow the incident edges to have the same colors. The following theorem isproved by Karo´nski et al. [8].

Theorem A (Karonski et al. [8]). There exists a finite set of real numbers which can beused to weight the edges of any graph with no isolated edges so that adjacent vertices havedifferent sums of incident edge weights.

It is easy to observe that Theorem A is equivalent to the following theorem.

Theorem B (Restatement ofTheoremA). There is a finite setwhich can beused to color theedges of any graph with no isolated edges so that adjacent vertices meet different multisets(i.e. duplicate elements are allowed) of colors.

This shows that by dropping the condition of being proper from the definition ofavd-coloring, a constant number of colors would be sufficient. Obviously, when the edgecolorings are required to be proper this is not the case. The following conjecture was madein [11].

Conjecture A (Zhang et al. [11]). The avd-chromatic number of every simple connectedgraph G such thatG �= C5 (the cycle of size5) andG �= K2 is at most� + 2.

In [2], this conjecture is verified for bipartite graphs and graphs of maximum degree 3,and the following bound has been proved for general graphs.

Theorem C (Balister et al. [2]). If G is a graph with no isolated edges, then theavd-chromatic number of G is at most� +O(log �(G)).

This was the best known bound so far. We asymptotically improve it by proving thefollowing theorem.

Theorem 1. If G is a graph with no isolated edges and maximum degree� > 1020, thenthe avd-chromatic number of G is at most� + 300.

We will use the probabilistic method to prove Theorem1. The following tools of theprobabilistic method will be used several times (see [9]).

248 H. Hatami / Journal of Combinatorial Theory, Series B 95 (2005) 246–256

Theorem D (The Symmetric Lovász Local Lemma). Let A1, . . . , An be events in aprobability space, let for all i, Pr[Ai] = p and the eventAi is mutually independentof all events but at most d events. If 4pd < 1, thenPr[∧Ai] > 0.

Theorem E (The Chernoff Bound). Suppose BIN(n, p) is the sum of n independentBernoulli variables each occurring with probability p. Then for any0< t�np

Pr[|BIN(n, p)− np| > t] < 2e−t23np .

Theorem F (Talagrand’s Inequality). Let X be a non-negative random variable, notidentically0, which is determined by n independent trialsT1, . . . , Tn, and satisfying thefollowing for somec, r > 0

1. Changing the outcome of any one trial can affect X by at most c, and2. For any s and any outcome of trials, if X�s, then there is a set of at most rs trials whoseoutcomes certify thatX�s,

then for any0� t�E[x],

Pr[|X − E[x]| > t + 60c

√rE[x]

]�4e

− t2

8c2rE[X] .

Remark 1. Suppose that we have the upperboundE[X]�k, and the conditions ofTalagrand’s Inequality hold forX. Then we might applyTalagrand’s Inequality to the randomvariableY = X + k − E[X], and obtain the inequality

Pr[X > k + t + 60c

√rk]

�Pr[Y > k + t + 60c

√rk]

�4e− t2

8c2rk .

Similar arguments hold for the Chernoff bound, and also for the cases that we know alower bound forE[X] and want to apply Talagrand’s Inequality or the Chernoff bound.

We will also use the following well-known inequalities.(ab

)b�(a

b

)=(

a

a − b)

�(eab

)b, (1)

wherea andb are natural numbers.In Section 2 we present the proof of Theorem1.

2. Proof of Theorem 1

The proof is probabilistic, and consists of three steps. The first step is similarly used in[6]. LetG be a graph with no isolated edges.

• In the first step we construct a loopless multigraphG′ with multiplicity at most 2 sothat �(G′) = �(G) = �, and if there exists ap-avd-coloring forG′, thenG is alsop-avd-colorable.


• LetH be the subgraph ofG′ induced by all verticesv such thatdeg(v) < �3 . In the second

step we give a(� + 300)-edge coloringc ofG′ such that for every two adjacent verticesu, v ∈ V (G′)− V (H), Sc(u) �= Sc(v).

• In the third step we modifycby recoloring some edges inE(H), and obtain a(�+300)-avd-coloring,cfinal of G′.

2.1. First step

For every edgee = uv in G such thatdeg(u),deg(v) < �3 and none ofu andv has

any other neighbor of degree less than�3 , contract the edgee, i.e. removeeand identifyu

andv to a single vertex. Call this new multigraphG′. Obviously,�(G′) = �(G), and themultiplicity of G′ is at most 2. Note thatH, the subgraph ofG′ induced by all verticesvsuch thatdeg(v) < �

3 , does not have any isolated edges.Suppose thatG′ has ap-avd-coloring�′. Every edge ofG′ has a corresponding edge in

G. For every edgeeof G′, color its corresponding edge inG by �′(e). Now all edges ofGare colored except those edges which were contracted in the process of obtainingG′ fromG. For every such edgeuv in G, sincedeg(v) + deg(u) < �, there exist some availablecolors foruv. Coloruvwith one of these available colors. Trivially,u andv meet differentsets of colors and none of them has any other neighbor of the same degree. This impliesthat the obtained edge coloring ofG is in fact anavd-coloring.

2.2. Second step

An unused edge in a partial edge coloringcof a graphG is an edge which is not coloredby c. We will refer to the graphUc induced by all unused edges as theunused graph of c.The degree of a vertex inUc is called itsunused degree. We say that a colorx is availablefor an unused edgeuv, if x does not appear on any incident edge touv.

As it is mentioned before,H is the subgraph ofG′ induced by all verticesv such thatdeg(v) < �

3 . It is trivial thatH has no multiple edges. We start with an arbitrary� + 2 edgecoloring� of G′ (which is guaranteed to exist by Vizing’s Theorem), and then we apply atwo-phased procedure to modify�, and obtain an edge coloringc such that for every twoadjacent verticesu, v ∈ V (G′)− V (H), Sc(u) �= Sc(v). The first phase is the following.

PhaseI : 1. Uncolor each edgee ∈ E(G′)− E(H) with probability 180� .

2. For every vertexv which has unused degree greater than 290, recover the color of allunused edges which are incident to it. We say thatv is recovered.

Let �1 be the partial coloring obtained after applying PhaseI . We define the followingsets.

• UCv (uncolored) is the set of the edges which are incident tov and uncolored in PhaseI (1). Note that these edges are not necessarily unused in�1 because their colors mightbe recovered in PhaseI (2).

• R is the set of all recovered vertices.• Q is the set of all verticesv such that|UCv| < 20.


• T is the set of all verticesv ∈ V (G′)− V (H) such that there exists an edgevw ∈ UCvwherew is recovered.

• L is the set of all verticesv ∈ V (G′) − V (H) such thatdegU�1(v) < 20, whereU�1 is

the unused graph of�1.

Since we use randomness in PhaseI , intuitively after applying this phase if a vertex hasa large unused degree, then with a high probability it is distinguishable. However, there arevertices inV (G′)− V (H) which have small unused degrees which we denote them byL.Later in Lemma 4, we will show that with a positive probability the vertices inL are rareand well-distributed in the graph. In fact, we will prove that with positive probability

(a) For every vertexv ∈ V (G′)− V (H), we have|N(v) ∩ L|� �100.

(b) For every two adjacent verticesu, v ∈ V (G′)− V (H) such thatdeg(u) = deg(v) andu �∈ L, we have |S�1(u)$S�1(v)|�10, where for every two setsA and B,A$B = (A\B) ∪ (B\A).

Note thatL is a subset of the union ofR,Q, andT. The following three lemmas providethe technical details needed to prove Lemma 4.

Lemma 1. For every vertexv ∈ V (G′)− V (H), we have(a) Pr[v ∈ R]� 1

1000.(b) Pr[v ∈ Q]� 1

1000.(c) Pr[v ∈ T ]� 1

1000.

Proof. (a) Since |UCv| has Bernoulli distribution, knowing that�3 �deg(v)��, theChernoff bound implies that

Pr[v ∈ R] = Pr[|UCv| > 290]�2e−(290−180)2

3×180 � 1

1000.

(b) The worst case is whendeg(v) = �3 . Then the Chernoff bound implies that

Pr[v ∈ Q] = Pr[|UCv| < 20]�2e−(180/3−20)2

180 � 1

1000.

(c) Consider an edgevw, wherev ∈ V (G′)− V (H) andw ∈ V (G′). First, we prove anupper bound forPr[vw ∈ UCv ∧ w ∈ R]. The Chernoff bound implies that

Pr[w ∈ R|vw ∈ UCv]�2e−(290−1−180)2

3×180 � 1

106 .

Hence

Pr[vw ∈ UCv ∧ w ∈ R]� 1

106

180

�� 1

103�.

Sincev has at most� neighbors, we have

Pr[v ∈ T ]� 1

103�× �� 1

1000. �


Lemma 2. For every vertexv ∈ V (G′)− V (H),

Pr[|N(v) ∩ L| > �/100]� 1

�7 .

Proof. First notice that

|N(v) ∩ L|� |N(v) ∩ R| + |N(v) ∩ T | + |N(v) ∩Q|.Lemma1 implies that

• E [|N(v) ∩ R|] � �1000,

• E [|N(v) ∩Q|] � �1000,

and• E [|N(v) ∩ T |] � �

1000.

For edgesuv ∈ E(G′)−E(H), consider the independent Bernoulli trialsTuv where theoutcome ofTuv determines whetheruv is uncolored in PhaseI (1) or not. Next we applyTalagrand’s Inequality to prove the following three claims.

• Claim1: Pr[|N(v) ∩ R| > �/300]� 13�7 .

• Claim2: Pr[|N(v) ∩ T | > �/300]� 13�7 .

• Claim3: Pr[|N(v) ∩Q| > �/300]� 13�7 .

To see that these three claims imply Lemma2, we observe that if|N(v) ∩ L| > �100, then

either|N(v) ∩ R| > �300, |N(v) ∩ T | > �

300, or |N(v) ∩Q| > �300.

Proof of Claims 1 and 2. Changing the outcome of each trial affects|N(v) ∩ R| by atmost 2, and every assignment to trials that results|N(v) ∩ R|�k can be certified by theoutcome of 290k trials. Hence Talagrand’s Inequality implies that

Pr

[|N(v) ∩ R| > �

1000+ t + 120

√290

�

1000

]�4e−1000t2/8×4×290�.

Since changing the outcome of each trial may add or remove at most two vertices fromR,at most 2× 290 vertices may be added or removed fromT. So it affects|N(v) ∩ T | by atmost 2× 290. Also every assignment to trials that results|N(v) ∩ T |�k can be certifiedby the outcome of 290k trials. By Talagrand’s Inequality

Pr

[|N(v) ∩ T | > �

1000+ t + 34 800

√290

�

1000

]�4e−1000t2/8×4×2903�.

Substitutingt = �1000 in both inequalities completes the proof of Claims 1 and 2.�

Proof of Claim 3. To prove this claim, instead of directly applying Talagrand’s Inequalityto the random variable|N(v) ∩Q|, we apply it to the random variableXv = |N(v)\Q| =deg(v)−|N(v)∩Q|. Changing the outcome of each trial affectsXv by at most 2, and every


outcome of trials that resultsXv�k can be certified by the outcome of 20k trials. SinceXv = deg(v)− |N(v)∩Q|, and soE[Xv]�deg(v)− �

1000, Talagrand’s Inequality impliesthat

Pr[Xv < deg(v)− �

1000− t − 120

√20�

]� 4e−t2/8×4×20×(deg(v)− �

1000)

� 4e−t2/1000�.

Substitutingt = �1000 in the inequailty above shows that

Pr[Xv < deg(v)− �

300

]� 1

3�7 .

SinceXv = deg(v)− |N(v) ∩Q|, we have

Pr[|N(v) ∩Q| > �

300

]� 1

3�7 . �

In PhaseII , some of the edges that are incident to vertices inL will be uncolored. Sincethe other endpoints of these edges may lie inV (G′)− V (H) − L, to guarantee that thesevertices will remain distinguishable we need to prove a stronger condition than just showingthat they are distinguishable in�1. The following lemma is needed to prove Lemma 4(b)which can imply that with positive probability all vertices inV (G′)−V (H)−Lwill remaindistinguishable after applying PhaseII .

Lemma 3. For every two adjacent verticesu, v ∈ V (G′)−V (H)where deg(u) = deg(v),

Pr[u �∈ L ∧ (|S�1(u)$S�1(v)| < 10)] < 1

�7 .

Proof. It is sufficient to prove thatPr[(|S�1(u)$S�1(v)| < 10)|u �∈ L] < 1�7 . Since we

want to prove an upper bound forPr[(|S�1(u)$S�1(v)| < 10)|u �∈ L], we can assume thatS�(u) = S�(v) = S. (Remember that� is the edge coloring before applying PhaseI .)

Suppose thatdegU�1(u) = k andk�20. If |S�1(u)$S�1(v)| < 10, then there are at least

k− 10 colors inS−S�1(u) which are also inS−S�1(v). Since there are at most two edgesbetweenu andv, the probability of this occuring is at most(

k

k − 10

)(180

�

)k−12

<1

�7 .

So

Pr[(|S�1(u)$S�1(v)| < 10)|u �∈ L] < 1

�7 . �

Lemma 4. If we apply PhaseI to the edge coloring�, and obtain a partial edge coloring�1, then with positive probability

(a) For every vertexv ∈ V (G′)− V (H), we have|N(v) ∩ L|� �100.


(b) For every two adjacent verticesu, v ∈ V (G′)− V (H) such that deg(u) = deg(v) andu �∈ L, we have|S�1(u)$S�1(v)|�10.

Proof. In order to prove the lemma, we define the following “bad” events

• For a vertexv ∈ V (G′)− V (H), letAv be the event that|N(v) ∩ L| > �100.

• For every edgeuv such thatu, v ∈ V (G′) − V (H), deg(u) = deg(v), let Auv be theevent thatu �∈ L and|S�1(u)$S�1(v)| < 10.

It is easy to see that each eventAX is mutually independent of all eventsAY such thatall vertices inX are in a distance of at least 6 from all vertices inY. Hence each event isindependent of all events but at most 2�6 events, and by Lemmas2 and 3 each event occurswith a probability of at most1

�7 . So the Local Lemma implies this lemma.�

Phase II : 1. For every vertexu ∈ L choose 5 edgesuvi (1� i�5) such thatvi ∈ V (G′)− L, anduvi is not an unused edge uniformly at random and uncolor them.

Call this new partial edge coloring�2. For every vertexu ∈ V (G′) − V (H), let UC′u

denote the set of the edges that are incident tou and are uncolored in PhaseII .

Lemma 5. Suppose that�1 is a partial edge coloring ofG′ which satisfies Properties(a)and(b) in Lemma4. If we apply PhaseII to �1, and obtain a partial coloring�2, then withpositive probability we have

(a) For every vertexv ∈ V (G′)− V (H)− L, |UC′v|�4.

(b) For every two adjacent verticesu, v ∈ V (G′)− V (H) such that deg(u) = deg(v), wehaveS�2(u) �= S�2(v).

Proof. Suppose thatu, v ∈ V (G′) − V (H) are two adjacent vertices such thatdeg(u) =deg(v) andu �∈ L. Then since by Property (b) in Lemma 4 we know that|S�1(u)$S�1(v)|�10, (a) implies thatS�2(u) �= S�2(v). It is trivial that |UC′

u| = 5 for all verticesu ∈ L.So it is sufficient to prove (b′) instead of (b), where

(b′) For every two adjacent verticesu, v ∈ L such that deg(u) = deg(v),we haveS�2(u) �=S�2(v).

We will apply the Local Lemma to show that with positive probability (a) and(b′) hold.We define the following two types of “bad” events.

• For every vertexu ∈ V (G′)− V (H)− L, and every five edgesuv1, uv2, . . . , uv5 suchthatvi ∈ L anduvi is not an unused edge in�1, let Au,{v1,...,v5} denote the event thatuvi ∈ UC′

u for all 1� i�5. Note that sinceG′ is a multigraph,{v1, . . . , v5} is a multisetand may have duplicate elements.

• For every two adjacent verticesu, v ∈ L wheredeg(u) = deg(v), let Auv denote theevent thatS�2(u) = S�2(v).

First we should give estimates ofPr[Au,{v1,...,v5}] andPr[Auv]. For every vertexv ∈V (G′) − V (H), we havedeg(v)� �

3 , |N(v) ∩ L|� �100, and the unused degree ofv is

at most 290. So every edgeuv is uncolored in PhaseII with a probability of at most


5(�3 − �

100−290

) < 20� , and also two parallel edges are uncolored with a probability of at

most 5(�4

) × 4(�4 −1

) which is less than(20

�

)2. Sinceu �∈ L, this implies that

Pr[Au,{v1,...,v5}]�(

20

�

)5

. (2)

Suppose thatS�2(u) = S�2(v) for two adjacent verticesu, v ∈ L wheredeg(u) = deg(v).Then since for each ofuandv, five new incident edges are uncolored, we havedegU�1

(u) =degU�1

(v). SinceS�2(u) = S�2(v), all the five edges inUC′v are determined by the five

edges inUC′u, and since the edges inUC′

v are chosen independently from the edges inUC′u,

we have

Pr[Auv]� 1(�45

)�(

20

�

)5

. (3)

Construct a graphDwhose vertices are all the events of the above two types, in which twoverticesAu,{v1,...,v5} andAu′,{v′

1,...,v′5} are adjacent if and only if{v1, . . . , v5}∩{v′

1, . . . , v′5} �=

∅ (even ifu = u′), two verticesAu,{v1,...,v5} andAvw are adjacent if and only if{v1, . . . , v5}∩{v,w} �= ∅, and two verticesAvw andAv′w′ are adjacent if{v,w} ∩ {v′, w′} �= ∅. It iseasy to see that each vertex ofD is mutually independent of all the vertices that are not

adjacent to it. Since for every vertexw ∈ L, there are at most�( �

1004

)events of the form

Au,{w,v2,...,v5}, and there are at most�100 events of the formAwv, the maximum degree ofDis at most

5

(�

( �

1004

)+ �

100

)< 6�

( �

1004

)<

6�5

108 .

Then since 4(20

�

)5 6�5

108 < 1, the Local Lemma implies Lemma 5.�

Let�1 and�2 be the partial colorings that are guaranteed to exist by Lemmas 4 and 5. Themaximum unused degree of�1 is 290. So by Lemma 5(a) the maximum unused degree of�2is 294. By Vizing’s Theorem we can color the unused graphU�2 with 296 new colors, andobtain a(� + 298)-edge coloringc of G′. By Lemma 5(b) for every two adjacent verticesu, v ∈ V (G′)− V (H), we haveSc(u) �= Sc(v).

2.3. Third step

In this step we begin with the edge coloringc1 = c, and repeatedly modify it to eventuallyobtain a(� + 300)-avd-coloring,cfinal ofG′. We only recolor the edges inH, and since thedegree of every vertex inV (G′)− V (H) is more than the degree of every vertex inH, allof the vertices inV (G′)− V (H) will remain distinguishable.


Obtainck+1 from ck as in the following. Suppose that for an edgeuv ∈ H , Sck (u) =Sck (v). SinceH does not have any isolated edge, we can assume thatdegH (u)�2. Letv1, v2, . . . , vr be the neighbors ofu in H, where r = degH (u). Uncolor all edgesuv1, uv2, . . . , uvr to obtain a partial edge coloringc′k.

Lemma 6. There exist setsL(uvi) ⊆ {1, . . . ,� + 300} for 1� i�r, all of sizer + 300such that

(a) In the partial edge coloringc′k all the colors inL(uvi) are available for the edgeuvi .(b) If v′ �= u is adjacent tovi andSc′k (v

′)\Sc′k (vi) has only one element x, thenx �∈ L(uvi).

Proof. Sincedeg(u),deg(vi)� �3 , there are at least�3 + r + 300 available colors foruvi .

Furthermore, there are at most�3 colors foruvi which dissatisfy Condition (b). This implies

that the desired sets exist.�

Consider all it proper completions ofc′k such that the color of eachuvi is chosen fromL(uvi). Note that the number of these completions is at least(r + 300)(r + 299) . . .301.Pick one of them randomly and uniformly, and call itck. SinceH does not have any multipleedges, for everyvi , there are at most(r + 300)(r − 1)! different possible choices ofck forwhichSck (u) = Sck (vi). So we have

Pr

∨

1� i� rSck (u) = Sck (vi)

�

r∑i=1

Pr[Sck (u) = Sck (vi)]

�r∑i=1

(r − 1)!(r + 299) . . .301

< 1.

This implies that there exists a completionck+1 of c′k such thatu is distinguishable, andby Lemma6(b) all neighbors ofuare also distinguishable. Since we only changed the colorof the edges incident tou, the number of indistinguishable vertices is decreased. Hence, byrepeatedly applying this procedure we will eventually obtain a(� + 300)-avd-coloring ofG′.

Acknowledgments

I thank my supervisor, Michael Molloy, for reading the draft version of this paper andhis valuable comments and suggestions.

References

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is a bound on the adjacent vertex distinguishing edge chromatic number