(College of Mathematics and Information Science,Northwest Normal University, Lanzhou,730070)

Vertex-distinguishing IE-total Colorings of Cycles and Wheels

CHEN XIANG-EN,HE WEN-YU,LI ZE-PENG AND YAO BING

(College of Mathematics and Information Science,Northwest Normal University, Lanzhou,730070)

Communicated by Du Xian-kun

Let G be a simple graph.An IE-total coloring f of G refers to a coloring of the vertices and edges of G so that no two adjacent vertices receive the same color. Let C(u)be the set of colors of vertex u and edges incident to u under f.For an IE-total coloring f of G using k colors,if C(u)C(v)for any two di ff erent vertices u and v of V(G),thenfis called a k-vertex-distinguishing IE-total-coloring of G, or a k-VDIET coloring of G for short.The minimum number of colors required for a VDIET coloring of G is denoted by(G),and is called the VDIET chromatic number of G.We get the VDIET chromatic numbers of cycles and wheels,and propose related conjectures in this paper.

graph,IE-total coloring,vertex-distinguishing IE-total coloring,vertexdistinguishing IE-total chromatic number

1 Introduction and Preliminaries

For an edge coloring(proper or not)of a graph G and a vertex v of G,denote by S(v)the set of colors used to color the edges incident to v.

A proper edge coloring of a graph G is said to be vertex-distinguishing if each pair of vertices is incident to a di ff erent set of colors.In other words,S(u)S(v)whenever uv. A graph G has a vertex-distinguishing proper edge coloring if and only if it has no more than one isolated vertex and no isolated edges.Such a graph is referred to as a vdec-graph. The minimum number of colors required for a vertex-distinguishing proper edge coloringof a vdec-graph G is denoted by(G).The concept of vertex-distinguishing proper edge coloring has been considered in several papers(see[1–7]).

A general edge coloring(not necessarily proper)of a graph G is called vertex-distinguishing if S(u)≠S(v)is required for any two distinct vertices u and v.The point-distinguishing chromatic index of a vdec-graph G,denoted by χ0(G),refers to the minimum number of colors required for a vertex-distinguishing general edge coloring of G.This parameter was introduced by Harary and Plantholt in[8].In spite of the fact that the structure of complete bipartite graph is simple,it seems that the problem of determining χ0(Km,n)is not easy, especially in the case m=n,as documented by papers of Horˇn′ak and Sot′ak[9–10],Zagaglia Salvi[11–12]and Horˇn′ak and Zagaglia Salvi[13].

For a total coloring(proper or not)f of G and a vertex v of G,denote by Cf(v),or simply C(v)if no confusion arises,the set of colors used to color the vertex v as well as the edges incident to v.Let(v)be the complementary set of C(v)in the set of all colors we used.Obviously|C(v)|≤dG(v)+1 and the equality holds if the total coloring is proper.

For a proper total coloring,if C(u)C(v),i.e.,(u)(v)for any two distinct vertices u and v,then the coloring is called vertex-distinguishing(proper)total coloring and the minimum number of colors required for a vertex-distinguishing(proper)total coloring is denoted by χvt(G).This concept has considered in[14–15].In[15],the authors give the following conjecture.

From[15]we know that the above conjecture is valid for complete graphs,complete bipartite graphs,path and cycle,etc.

In this paper we propose a kind of vertex-distinguishing general total coloring.The relationship of this coloring and vertex-distinguishing proper total coloring is similar to the relationship of vertex-distinguishing general edge coloring and vertex-distinguishing proper edge coloring.

When we de fi ne a proper total coloring of a graph G,we need three conditions for a total coloring which are listed as follows∶

Condition(v)∶No two adjacent vertices receive the same color;

Condition(e)∶No two adjacent edges receive the same color;

Condition(i)∶No edge receives the same color as one of its endpoints.

If we only consider the total coloring of the graph G such that the Condition(v)is satis fi ed,then such a coloring is called an IE-total coloring of the graph G.

If f is an IE-total coloring of the graph G using k colors and for all u,v∈V(G),uv, we have C(u)C(v),then f is called a k-vertex-distinguishing IE-total coloring,or a k-VDIET coloring.The minimum number k for which G has a k-VDIET coloring is called the vertex-distinguishing IE-total chromatic number(or VDIET chromatic number)of the graph G and is denoted by

The following proposition is obviously true.

Proof.For a graph G,let nibe the number of the vertices of degree i,δ≤i≤Δ.Suppose that

In Section 2 we consider the VDIET colorings for cycles and paths.The discussions of the VDIET colorings for wheels and fans are long,so we put them into the Section 3.In Section 4 we give two conjectures.

2 Vertex Distinguishing IE-total Chromatic Numbers of Cycle and Path

Theorem 2.1LetCnbe a cycle of ordern(n≥3).Then

Proof.When n=3,4,5,6,the results are obviously true.Let Cn=u1u2u3···unu1.In this proof when we want to give a k-VDIET coloring fnof Cnwith all colors 1,2,···,k,we need only to give(fn(ui?1ui),fn(ui),fn(uiui+1))or fn(ui?1ui)fn(ui)fn(uiui+1)for each i=1,2,···,n,i.e.,

such that

and

If n=7,then

Assume that C7has a 3-VDIET coloring g.Then there are three vertices such that their color sets are{1},{2}and{3},respectively.But no two such vertices are adjacent.So without loss of generality we assume that C(u1)={1},C(u3)={2},C(u5)={3}.As no two adjacent vertices receive the same color,the colors that u2and u4are assigned under g are 3 and 1,respectively.Thereby u2and u4have the same color set,which is a contradiction.So there does not exist a VDIET coloring of C7using 3 colors.

Obviously,

If

i.e.,8≤n≤14,then

Let

If

i.e.,15≤n≤25,then

Let

where“+”denotes the concatenation of sequence.Obviously,when 15≤n≤25,fnis a 5-VDIET coloring of Cn,so the result holds.

We now want to apply the same technique to give fnfor

To do so,we de fi ne Ckat fi rst.Let

If l≡0(mod 4),let

If l≡1(mod 4),let

If l≡2(mod 4),let

If l≡3(mod 4),let

For l=6,7,···,k?2,we do the above work step by step.Finally we obtain.The last term ofis(k?1)1k or k1(k?1).We change the last term(k?1)1k to(k?1)k1 and k1(k?1)to k(k?1)1.And the resulting sequence is denoted by Ck.

Now we determine fnwhen

Suppose

Let

Theorem 2.2For a pathPnof ordern(≥8),if

Proof.If

Actually,based on the k-VDIET coloring fnof Cn=u1u2u3···unu1de fi ned in the above theorem,if we delete the edge which connects two vertices with color sets{4,2}and{4,3}, i.e.,the corresponding terms are 424 and 434,then we can obtain Pnand its k-VDIET coloring when

3 Vertex Distinguishing IE-total Chromatic Numbers of Wheel and Fan

Proof.Let

In this proof,when we want to give a k-VDIET coloring fnof Wn,we always appoint that the colors which we use are 1,2,···,k,the color of u0is k,and then it suffices to give (fn(ui?1ui),fn(u0ui),fn(ui),fn(uiui+1))or fn(ui?1ui)fn(u0ui)fn(ui)fn(uiui+1)for each i=1,2,···,n,equivalently to give

such that fn(ui)fn(ui+1),i=1,2,···,n,and

If n=4,then

Case 1.|C(u0)|=1.Then C(u0)={3},and each C(ui),i=1,2,3,4,is one of{1,2,3}, {2,3},{1,3}.Three subsets can not distinguish 4 vertices.It is a contradiction.

Case 2.|C(u0)|=2.We may suppose C(u0)={3,2}.Then each C(ui),i=1,2,3,4,is one of{1,2,3},{1,2},{1,3},{2}.Let C(u1)={2}.Then the colors of u2and u4are only 1.Thus{C(u2),C(u4)}={{1,2,3},{1,2}}.So C(u3)={1,3}and the color of u3is 2 for ensuring the vertex coloring proper.This is a contradiction.

Case 3.|C(u0)|=3.Of course C(u0)={1,2,3}.Then each C(ui),i=1,2,3,4,is one of{1,2},{2,3},{1,3},{1},{2}.

(a)If{C(u1),C(u2),C(u3),C(u4)}={{1,2},{1,3},{1},{2}},then two vertices which has color sets{1}and{2}are not adjacent.But these two vertices must have the same color.This is a contradiction.

(b)If{C(u1),C(u2),C(u3),C(u4)}={{1,2},{1,3},{2,3},{1}},without loss of generality,we assume C(u1)={1}.Then 1∈C(u2)∩C(u4).Thus C(u3)={2,3},whichyields g(u3)=2.But g(u1)=1,so the color that u2is assigned under g is one of g(u1), g(u3)and g(u0).This is a contradiction.

(c)If{C(u1),C(u2),C(u3),C(u4)}={{1,2},{2,3},{1},{2}},then similar to(a)we can get a contradiction.

(d)If{C(u1),C(u2),C(u3),C(u4)}={{1,2},{1,3},{2,3},{2}},then similar to(b) we can get a contradiction.

(e)If{C(u1),C(u2),C(u3),C(u4)}={{1,3},{2,3},{1},{2}},then similar to(a)we can get a contradiction.

Thus there does not exist a VDIET coloring of W4using 3 colors,so≥4.We can prove=4 by giving a 4-VDIET coloring f4of W4(note that f4(u0)=4)as follows∶f4=(1132,2214,4323,3411).

As the(proper)vertex chromatic number of W5is χ(W5)=4,χievt(W5)≥4.We can show that=4 by giving a 4-VDIET coloring f5of W5as follows∶

Note that

Case 1.|C(u0)|=1.Then C(u0)={3},and each C(ui),i=1,2,···,6,is one of {1,2,3},{2,3},{1,3}.This is a contradiction.

Case 2.|C(u0)|=2.We may suppose C(u0)={3,1}.Then each C(ui),i=1,2,···,6, is one of{1,2,3},{1,2},{2,3},{1}.This is a contradiction.

Case 3.|C(u0)|=3.Then C(u0)={1,2,3},and each C(ui),i=1,2,···,6,is one of{1,2},{2,3},{1,3},{1},{2}.Five subsets do not distinguish six vertices.This is a contradiction.

f13=(1132,2112,2222,2114,4224,4114,4334,4223,3333,3113,3223,3431,1111). It is easy to see that fnis a 4-VDIET coloring of Wnand(Wn)=4 when 7≤n≤13.

Case 1.|C(u0)|=1.Then C(u0)={4},and each C(ui),i=1,2,···,14,is one of {1,4},{2,4},{3,4},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}.This is a contradiction.

Case 2.|C(u0)|=2.We may suppose C(u0)={1,4}.Then each C(ui),i=1,2,···,14, is one of{1},{1,2},{1,3},{2,4},{3,4},{1,3,4},{1,2,4},{1,2,3},{2,3,4},{1,2,3,4}. This is a contradiction.

Case 3.|C(u0)|=3.We may suppose C(u0)={1,2,4}.Then each C(ui),i= 1,2,···,14,is one of{1},{2},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,3,4}, {2,3,4},{1,2,3,4}.This is a contradiction.

Case 4.|C(u0)|=4,that is,C(u0)={1,2,3,4}.Then each C(ui),i=1,2,···,14,is one of{1},{2},{3},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4}, {2,3,4}.This is a contradiction.

f14=(1132,2112,2222,2114,4444,4224,4114,4334,4223,3333,3113,3223,3431,1111).

(I)We prove that what wheel has(k?1)-VDIET coloring with colors 1,2,···,k?1 for k≥6.

Suppose that Wnhas a(k?1)-VDIET coloring g.Let|C(u0)|=i.Denote by mithe number of subsets of{1,2,···,k?1}which can be acted as the color set of some uj(1≤j≤n).Then

This yields that if

(II)Next we prove that k colors can color Wn,so that the coloring is IE-total coloring and is vertex distinguishing when

Assume that we have constructed all fiwith

such that

(1)The fi rst term of fiis 1132;

(2)When

and fn(u0)=l,the fi rst term abcd and the last term a1b1c1d1of fnsatisfy that d1=a=1, c1?=c and if there are at most 3 di ff erent numbers in{a1,b1,c1,d1}then c1=d1=1;

(3)fqhas used up all 4-combinations,3-combinations,2-combinations and 1-combination of{1,2,···,k?1}except for 1-combination{k?1}.Under fq,C(u0)={1,2,···,k?1}.

Let

and let Cibe the sequence de fi ned in the proof of Theorem 2.1 for i≥7.We change each term abc of Cs,4≤s≤k?1,into akbc,and the resulted sequence is denoted byLetObviously,Dkhasterms.

(A)Let fq+1be the sequence formed by inserting the term(k?1)(k?1)(k?1)(k?1) into the place between(k?1)33(k?1)and(k?1)22(k?1)from fq.

(B)Let

Let

If the(r?1)-th term of Dkis akb(k?1),then we can obtain fq+r+1by adding the fi rst r?1 terms of Dkto fq+1and then adding(k?1)k(k?2)1.

If the(r?1)-th term of Dkis(k?1)kab(but not(k?1)k1s,s≤k?2),then we can obtain fq+r+1by adding the fi rst r?1 terms of Dkto fq+1and then adding bk(k?1)1.

Let

If the(r?1)-th term of Bkis abbk,then we can obtainby adding the fi rst r?1 terms of Bktoand then adding the term k(k?1)11.

If the(r?1)-th term of Bkis of the form kaab(≠k11s,i.e.,a≠1),then we can obtainby adding the fi rst r?1 terms of Bktoand then adding the term bk11.

Theorem 3.2LetFnbe a fan of ordern+1.If3≤ n≤ 4,then(Fn)=3;if5≤n≤13,then

thenχievt(Fn)=k.

Proof.Let

In this proof when we want to give a k-VDIET coloring gnof Fnwith the set of all colors {1,2,···,k},we always appoint that the color of v0is k and then it suffices to give the desired sequence

Assume that F5has a 3-VDIET coloring h and h(v0)=3.

If|C(v0)|=1,then 3∈C(vi),i=0,1,···,5.But the number of the subsets of{1,2,3} which contain 3 is 4.This is a contradiction.

If|C(v0)|≥2,then{3}is not the color set of any vertex.Since the vertex coloring is proper,we may suppose h(v1)=h(v3)=h(v5)=1 and h(v2)=h(v4)=2.As{C(v0), C(v1),···,C(v5)}={{1},{2},{1,2},{1,3},{2,3},{1,2,3}},and C(v3)≠{1},without loss of generality we assume that C(v1)={1}and C(v4)={2}.This yields that 2∈C(vi), i=0,2,3,4,5.This is a contradiction for we just have 4 subsets of{1,2,3}which contain 2.

Similarly to proving that W6has no 3-VDIET coloring in the proof of Theorem 3.1 we can show that F6has no 3-VDIET coloring.And

is a 4-VDIET coloring of F6.So=4.

Similarly to the proof that W14has no 4-VDIET coloring in the proof of Theorem 3.1, we can show that F14have no 4-VDIET coloring.And

g14=(132,2112,2222,2114,4444,4224,4114,4334,4223,3333,3113,3223,3431,111). So

If we delete the fi rst number of the fi rst term of fnand also delete the last number of the last term of fn,then the resulting sequence gnis a 4 or 5-VDIET coloring of Fnwhen 7≤n≤13 or 15≤n≤29.Thus=4 if 7≤n≤13,and=5 if 15≤n≤29. Su

ppose that

and n≥30.

(I)We can show that Fnhas no(k?1)-VDIET coloring.The process is the same as(I) in the proof of Theorem 3.1.So≥k.

(II)If we delete the fi rst number of the fi rst term of fnand also delete the last number of the last term of fn,then the resulting sequence gnis a k-VDIET coloring of Fn.

and n≥30.

The proof is completed.

Theorem 3.3LetKnbe the complete graph of ordern(n≥3),then(Kn)=n.

Proof.Any two vertices in Knmust receive di ff erent colors under arbitrary VDIET coloring.Therefore(Kn)≥n.Of course,we are able to show that(Kn)=n by giving a VDIET coloring of Knusing n colors 1,2,···,n as follows.Assign colors 1,2,···,n to vertices v1,v2,···,vnof Knrespectively,and then let all edges receive the same color 1. The proof is completed.

4 Conjectures

From the results obtained in this paper,we know that for any graph G discussed in this paper but Kn(n≥5),we have χievt(G)=ξ(G)or ξ(G)+1.So we propose the following conjectures.

Conjecture 4.1For a simple graphG,if(proper vertex coloring)chromatic numberχ(G)≤4,we have=ξ(G)orξ(G)+1.

Conjecture 4.2For a simple graphG,we have≤max{ξ(G)+1,χ(G)}.

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tion:05C15

A

1674-5647(2014)03-0222-15

10.13447/j.1674-5647.2014.03.04

Received date:Oct.18,2011.

Foundation item:The NSF(61163037,61163054)of China and the Scienti fi c Research Project(nwnu-kjcxgc-03-61)of Northwest Normal University.

E-mail address:chenxe@nwnu.edu.cn(Chen X E).