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Research ArticleGeneral Vertex-Distinguishing Total Coloring of Graphs

Chanjuan Liu and Enqiang Zhu

School of Electronics Engineering and Computer Science Peking University Beijing 100871 China

Correspondence should be addressed to Enqiang Zhu zhuenqiang163com

Received 24 April 2014 Accepted 14 July 2014 Published 3 August 2014

Academic Editor Tak-Wah Lam

Copyright copy 2014 C Liu and E Zhu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The general vertex-distinguishing total chromatic number of a graph 119866 is the minimum integer 119896 for which the vertices and edgesof119866 are colored using 119896 colors such that any two vertices have distinct sets of colors of them and their incident edges In this paperwe figure out the exact value of this chromatic number of some special graphs and propose a conjecture on the upper bound of thischromatic number

1 Introduction

All graphs considered in this paper are simple and finite Fora graph 119866 we denote by 119881(119866) 119864(119866) Δ(119866) and 120575(119866) the setsof vertices edges maximum degree and minimum degree of119866 respectively For a vertex V of 119866 119889

119866(V) is the degree of V in

119866 For any 1198811015840 sube 119881(119866) we use 119866[1198811015840] to denote the subgraphinduced by1198811015840 For any undefined terms the reader is referredto the book [1]

The coloring problem of graphs is one of the classicalresearch areas in graph theory It has been widely appliedto various fields such as large scheduling [2] assignmentof radio frequency [3] and separating combustible chemicalcombinations [4] Due to its extensive application many newvariants of colorings have been studied [5]

Recall that a 119896-edge coloring of a graph 119866 is a mapping119891 119864(119866) rarr 119862 where 119862 is a set of 119896 colors An edgecoloring is proper if adjacent edges receive distinct colorsIn 1985 Harary and Plantholt [6] first considered point-distinguishing chromatic index which is a variant of edgecoloring After that many other variants of edge coloringwere introduced such as vertex-distinguishing proper edgecoloring [7] adjacent vertex-distinguishing edge coloring [8]and general adjacent vertex-distinguishing edge coloring [9]

A total 119896-coloring of a graph119866 is a coloring of119881(119866)cup119864(119866)using 119896 colors A total 119896-coloring is proper if no two adjacentor incident elements receive the same color The minimumnumber of colors required for a proper total coloring of 119866

is called the total chromatic number of 119866 and is denoted by120594119905(119866) Behzad [10] and Vizing [11] independently made the

conjecture that for any graph 119866

120594119905 (119866) le (119866) + 2 (1)

This is known as the total coloring conjecture (119879119862119862) and isstill unproven

Let 119891 be a total 119896-coloring of 119866 The total color set (withrespect to119891) of a vertex V isin 119881(119866) is the set denoted by119862119891(V)of colors of V and its incident edges We denote byC119891(119866) theset of total color sets of all vertices of 119866 Furthermore let 119878be a subset of 119881(119866) cup 119864(119866) we use 119862

119891(119878) to denote the set of

colors of elements of 119878Like edge coloring total coloring also has some variants

In 2005 Zhang et al [12] added a restriction to the definitionof total coloring and proposed a new type of coloring definedas follows

Definition 1 Let 119891 be a proper total 119896-coloring of a graph119866 If for all 119906 V isin 119881(119866) 119862

119891(119906) = 119862

119891(V) then 119891 is called

an adjacent vertex-distinguishing total 119896-coloring of 119866 or a119896-AVDTC of 119866 for short The minimum number 119896 for which119866 has a 119896-AVDTC is the adjacent vertex-distinguishing totalchromatic number of 119866 denoted by 120594

119886119905(119866)

Zhang et al [12] conjectured that for any graph 119866 it has

120594119886119905 (119866) le Δ (119866) + 3 (2)

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 849748 7 pageshttpdxdoiorg1011552014849748

2 Journal of Applied Mathematics

In [13ndash15] authors proved that there exists a 6-AVDTC ofgraphs with Δ = 3 which indicates conjecture (2) holdsfor such graphs For further research on adjacent vertex-distinguishing total chromatic number one may refer to [16ndash23]

For a 119896-AVDTC119891 of a graph 119866 if 119862119891(119906) = 119862

119891(V) is

required for any two distinct vertices 119906 V then 119891 is calleda vertex-distinguishing total 119896-coloring of 119866 abbreviated as119896-VDTCTheminimumnumber 119896 such that119866 has a 119896-VDTCis called the vertex-distinguishing total chromatic numberdenoted by 120594V119905(119866) [24] Zhang et al conjectured in [24] thatfor any graph 119866 it follows that

120583119905 (119866) le 120594V119905 (119866) le 120583119905 (119866) + 1 (3)

where 120583119905(119866) = min119896 | ( 119896119894+1) ge 119899119894 120575 le 119894 le Δ

In this paper we introduce a variant of vertex-distinguishing total coloring of a graph 119866 which relaxes therestriction that the coloring is proper We now present thedetailed definition as follows

Definition 2 Let 119866 be a graph and 119896 be a positive integerA total coloring 119891 of 119866 using 119896 colors is called a generalvertex-distinguishing total 119896-coloring of 119866 (or 119896-GVDTC of119866 briefly) if for all119906 V isin 119881(119866)119862119891(119906) = 119862

119891(V)Theminimum

number 119896 for which 119866 has a 119896-GVDTC is the general vertex-distinguishing total chromatic number denoted by 120594

119892V119905(119866)

It is evident that 120594119892V119905(119866) does exist for any graph 119866 In

this paper we study the general vertex-distinguishing totalcoloring of some special classes of graphs and obtain the exactvalue of the general vertex-distinguishing total chromaticnumber of these graphs Furthermore we propose a con-jecture on the upper bound of general vertex-distinguishingtotal chromatic number of a graph

2 Main Results

We first present a trivial lower bound on the general vertex-distinguishing total chromatic number of a graph

Theorem 3 Let 119866 be a graph on 119899 vertices Then

120594119892V119905 (119866) ge lceillog2 (119899 + 1)rceil (4)

Proof Let 120594119892V119905(119866) = 119896 It follows that 119899 le ( 119896

1) + ( 1198962) + sdot sdot sdot +

( 119896119896) = 2119896minus 1 so 119896 ge lceillog

2(119899 + 1)rceil

Notice that the lower bound ofTheorem 3 can be attainedin graphs such as the 119899-vertex path 119875119899 for 119899 = 1 2 7 Onecan readily check that 120594

119892V119905(1198751)= 1 and 120594

119892V119905(119875119899)= 2 for 119899 = 2 3

and 120594119892V119905(119875119899)= 3 for 119899 = 4 5 6 7

Theorem 4 Let 119866 be a graph without isolated vertices andisolated edges Then

120594119892V119905 (119866) le 1205941015840

119892V119889 (119866) (5)

Proof Suppose that 119891 is a 119896-GVDEC of 119866 For any 119906 isin 119881(119866)let 119891(119906) = 119891(119906V) where 119906V isin 119864(119866) Obviously 119891 is a119896-GVDTC of 119866

We now turn to investigating the general vertex-distinguishing total chromatic number of an 119899-vertex path

Theorem 5 Let 119875119899be a path on 119899 vertices 119899 ge 1 Then

120594119892V119905 (119875119899) =

[[[[

3radic3119899 + radic91198992 +

125

27+3radic3119899 minus radic91198992 +

125

27

]]]]

(6)

Proof Denote by 119875119899= V1V2sdot sdot sdot V119899a path 119875

119899with vertex

set V1 V2 V

119899 and edge set V

1V2 V2V3 V

119899minus1V119899 Let

120594119892V119905(119875119899) = 119896 and let 119891 be a 119896-GVDTC of 119875

119899 Let 120572

119896= ( 119896minus11)+

( 119896minus12) + ( 119896minus1

3) 120573119896= ( 1198961) + ( 1198962) + ( 1198963) 120574119896= 1 + ( 119896minus1

1) + ( 119896minus1

2)

and

119896lowast= [[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(7)

Evidently |119862119891(V119894)| le 3 and 119899 le 120573119896 (which implies 119896 ge

119896lowast) In order to prove the conclusion 119896 = 119896

lowast it sufficesto give a 119896lowast-GVDTC of 119875

119899 When 119899 le 7 it is not hard

to construct the corresponding general vertex-distinguishingtotal colorings Let 119899 ge 8 We first construct a 119896lowast-GVDTC1198911015840of 119875120573119896

recursively Note that when 119899 = 120573119896 it has 119896lowast = 119896

Procedure 1 Construct a 4-GVDTC1198914of 1198751205734

(ie 11987514) as

follows the vertices V1 V2 V

14are colored by 1 3 4 2 3 1

4 2 3 4 3 4 4 1 respectively the edges V1V2 V2V3

V3V4 V

13V14

are colored by 1 3 1 1 2 4 2 2 2 3 3 4 4respectively It is easy to see that 119891

4is a 4-GVDTC of 119875

14


based on a (119896 minus1)-GVDTC119891119896minus1 of 119875120573

119896minus1

Let 119891119896 be

119891119896(V119894+120574119896

) = 119891119896minus1

(V119894) + 1 119894 = 1 2 120572

119896minus 1

119891119896(V120572119896+120574119896

) = 119891119896minus1

(V120572119896

) = 1

119891119896 (V119894+120574119896

V119894+1+120574119896

) = 119891119896minus1 (V119894V119894+1) + 1 119894 = 1 2 120572119896 minus 1

119891119896(V1) = 1 119891

119896(V120574119896

) = 119896

(8)

V2 V3 V

120574119896minus1

are colored by

119896 minus 1 119896⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 119896 minus 2 119896 minus 1 119896⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

3 elements 2 3 119896⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119896minus1 elements (9)

when 119896 is even V1V2 V2V3 V3V4 V

120574119896

V120574119896+1

are colored by

Journal of Applied Mathematics 3

1⏟⏟⏟⏟⏟⏟⏟

1 element 119896 minus 1 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 1 119896 minus 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

3 elements 119896 minus 3 1 119896 minus 3 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

4 elements 1 119896 minus 4 1 119896 minus 4 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

5 elements

1 119896 minus (119896 minus 4) 1 119896 minus (119896 minus 4) 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119896minus3 elements 119896 minus (119896 minus 3) 1 119896 minus (119896 minus 3) 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119896minus2 elements

2 1 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119896minus2 elements 2 2⏟⏟⏟⏟⏟⏟⏟

2 elements

(10)

when 119896 is odd V1V2 V2V3 V3V4 V120574119896

V120574119896+1

are colored by

1⏟⏟⏟⏟⏟⏟⏟

1 element 119896 minus 1 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 1 119896 minus 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟



5 elements

119896 minus (119896 minus 4) 1 119896 minus (119896 minus 4) 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119896minus3 elements 1 119896 minus (119896 minus 3) 1 119896 minus (119896 minus 3) 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


1 2 1 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


2 elements

(11)

It should be pointed out that when 119895 is odd for 119895 isin

1 2 119896 minus 3 the colorsrsquo form is 119896minus 119895 1 119896 minus 119895 1 119896 minus 119895 1with totally 119895 + 1 elements and when 119895 is even for 119895 isin

1 2 119896minus3 the colorsrsquo form is 1 119896minus119895 1 119896minus119895 1 119896minus119895 1with 119895 + 1 elements in total

According to 119891119896minus1

we can see that for any 119894 119895 = 120574119896+

1 120574119896+ 2 120573

119896minus 1 and 119894 = 119895 it follows that 2 119896 is not

a total color set of vertices V119894 1 notin 119862119891119896

(V119894) and 119862119891119896

(V119894) =

119862119891119896

(V119895) In addition 1198621 1198622 119862120574minus1 are as follows 1 (1item) 1 119896 minus 1 1 119896 minus 1 119896 (2 items) 1 119896 minus 2 1 119896 minus2 119896 minus 1 1 119896 minus 2 119896 (3 items) 1 119896 minus 119895 1 119896 minus 119895 119896 minus119895 + 1 1 119896 minus 119895 119896 minus 119895 + 2 1 119896 minus 119895 119896 (119895 + 1 items) and 1 2 1 2 3 1 2 4 1 2 119896 (119896 minus 1 items) And119862119891119896

(V120574119896

) = 2 119896 So 119891119896 is a 119896-GVDTC of 119875120573119896

We now showthat 119875119899 also has a 119896-GVDTC based on a 119896-GVDTC of 119875120573

119896

for( 119896minus11) + ( 119896minus1

2) + ( 119896minus1

3) lt 119899 lt 120573

119896

Let 119903 = 120573 minus 119899 and let 119891 be a 119896-GVDTC of 119875120573119896

constructed by Procedures 1 and 2 We first delete 119903 ver-tices V

1 V2 V

119903from 119875

120573119896

Obviously the resulting graphdenoted by V

119903+1V119903+2

sdot sdot sdot V120573119896

is isomorphic to 119875119899 Let 1198911015840 be

1198911015840(V119894V119894+1) = 119891(V

119894V119894+1) for 119894 = 119903 + 1 119903 + 2 120573

119896minus 1 1198911015840(V

119894) =

119891(V119894) for 119894 = 119903 + 2 120573 and 1198911015840(V

119903+1) = 1 Then 1198911015840 is a 119896-

GVDTC of 119875119899

All the above show that the conclusion holds

According toTheorem 5 we have the same conclusion oncycles Let 119862

119899= V1V2sdot sdot sdot V119899V1be an 119899-vertex cycle with vertex

set V1 V2 V


1V2 V2V3 V

119899minus1V119899 V119899V1

Corollary 6 For any cycle 119862119899 (119899 ge 3) one has

120594119892V119905 (119862119899) =

[[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(12)Proof Let 119862119899 = V1V2 sdot sdot sdot V119899V1 and 119875119899 = 119862119899 V1V119899 let also

119896lowast= [[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(13)

and let 119891 be a 119896lowast-GVETC of 119875119899 constructed by the method

ofTheorem 5Then we can extend119891 to a 119896lowast-GVETC of119862119899by

assigning color 1 to edge V1V119899 So the conclusion holds

In the following (Theorem 7 to Theorem 9) we discussthe general vertex-distinguishing total chromatic number ofsome kinds of special trees A star 119878

119899is the complete bipartite

graph 1198701119899

(119899 ge 1) A double star 119878119898119899

is a tree containingexactly two vertices that are not leaves (which are necessarilyadjacent) A tristar 119878

119901119902119903is a tree with vertex set 119881(119878

119901119902119903) =

119906119894| 119894 = 0 1 119901cupV

119894| 119894 = 0 1 119902cup119908

119894| 119894 = 0 1 ℓ

and edge set 119864(119878119901119902119903

) = 1199060119906119894| 119894 = 1 2 119901 cup V

0V119894| 119894 =

1 2 119902 cup 1199080119908119894| 119894 = 1 2 119903 cup 119906

0V0 V01199080 where

119901 119902 119903 are positive integers

Theorem 7 For a star 119878119899(119899 ge 1) one has

120594119892V119905 (119878119899) =

2 119899 = 1 2

lceilradic8119899 + 1 minus 1

2rceil 119899 ge 3

(14)


Proof When 119899 = 1 2 the conclusion is trivial For 119899 ge 3let 120594119892V119905(119878119899) = 119896 and lceil(radic8119899 + 1 minus 1)2rceil = 119896

1015840 Since for any119906 isin 119881(119878119899) 1199060 (1199060 is the vertex with 119889119878

119899

(1199060) ge 2) |119862119891(119906)| le 2for any 119896-GVDTC119891 of 119878119899 it follows that 119899 le ( 1198961 ) + ( 1198962 ) thatis 119896 ge 119896

1015840 In order to prove 119896 = 1198961015840 we need to show that

there exists a 1198961015840-GVDTCof 119878119899 Otherwise let 119878

119899lowast be the graph

with minimum 119899lowast such that 119878

119899lowast does not have a 1198961015840-GVDTC

where 119899lowast le 119899 Let 119906 be a vertex of degree 1 in 119878119899lowast Consider the

graph 1198661015840 = 119878119899lowast minus 119906 obtained from 119878

119899lowast by deleting the vertex

119906 and its incident edge By the assumption of 119878119899lowast 1198661015840 has a

1198961015840-GVDTC denoted by 1198911015840 In addition by interchanging the

colors of some vertex and its incident edge appropriately wecan assume |1198621015840

119891(1199060)| ge 2 Since 119899lowast minus 1 lt ( 119896

1015840

1) + ( 119896

1015840

2) there

is at least one set 119886 119887 for 119886 119887 isin 1 2 1198961015840 which is not

the total color set of the vertices of 1198661015840 So on the basis of 1198911015840in 119878119899lowast we can color 119906 and its incident edge 119906119906

0by 119886 and 119887

respectively Obviously the resulting coloring is a 1198961015840-GVDTCof 119878119899lowast

For two vertices 119906 V of a graph 119866 to identify these twovertices is to replace them by a single vertex (denoted by 119906-Vin this paper) incident to all the edges which were incident in119866 to either 119906 or V The resulting graph is denoted by 119866119906 VIn what follows we denoted by [1 119896] the set of 1 2 119896

Theorem 8 Let 119878119898119899

(119898 ge 119899 ge 1) be a double star and ℓ =119898 + 119899 Then

120594119892V119905 (119878119898119899) =

3 ℓ = 2 3 4 5

4 ℓ = 6

lceilradic8ℓ + 1 minus 1

2rceil ℓ ge 7

(15)

Proof When ℓ le 6 the results are easy to be proved Whenℓ ge 7 let 119906 V be two vertices with degree more than 1 and1198661015840= 119878119898119899119906 V Evidently the graph 1198661015840 is isomorphic to the

star 119878ℓ Let 1198961015840 = lceil(radic8ℓ + 1 minus 1)2rceil Since |119862

119891(119909)| le 2 for any

119909 isin 119881(119878119898119899) 119906 V we have 120594119892V119905(119878119898119899) ge 119896

1015840By Theorem 7 1198661015840 contains a 1198961015840-GVDTC1198911015840 Evidently

|1198621198911015840(119909)| le 2 for any 119909 isin 119881(1198661015840)119906minusV and |119862

1198911015840(119906minusV)| le 1198961015840 If

|1198621198911015840(119906minusV)| le 2 thenwe can extend1198911015840 to a 1198961015840-GVDTCof 119878119898119899

by coloring vertices 119906 V and edge 119906V with any three differentcolors in [1 119896]1198911015840(119906minusV) if |1198621198911015840(119906minusV)| = ℓ

1015840ge 3 we without

loss of generality assume 1198621198911015840(119906 minus V) 1198911015840(119906 minus V) = [1 ℓ1015840] Let

119864119906(resp 119864V) be the set of edges (except edge 119906V) incident

to 119906 (resp V) in 119878119898119899

We now extend 1198911015840 to a 1198961015840-GVDTCof 119878119898119899

as follows By the fact that there remain vertices 119906 Vand edge 119906V uncolored in 119878

119898119899when 1198911015840 is restricted to 119878

119898119899

we consider the following two cases First one of 119906 V say 119906satisfies that 119862

1198911015840(119864119906) contains at most two elements Assume

1198621198911015840(119864119906) sube 1 2 we then color 119906 119906V V by 2 3 119888 respectively

where 119888 = 4 when 1198621198911015840(119864119906) = 3 4 and 119888 = 2 when

1198621198911015840(119864119906) = 3 4The resulting coloring of 119878

119898119899is also denoted

by 1198911015840 Then it follows in 119878119898119899

that |1198621198911015840(119906)| ge 3 |119862

1198911015840(V)| ge 3

and 4 notin 1198621198911015840(119906) and 4 isin 119862

1198911015840(V) So 1198911015840 is a 1198961015840-GVDTC of 119878

119898119899

Second |1198621198911015840(119864V)| ge 3 and |1198621198911015840(119864119906)| ge 3 then we will further

discuss two subcases

(1) Consider |1198621015840119891(119864119906)| = |119862

1015840

119891(119864V)| = 119896

1015840 Suppose that 119881119906

(and119881V) is the set of vertices except V (or 119906) adjacentto 119906 (and V) in 119878119898119899 Because 119891

1015840 is a 1198961015840-GVDTC of1198661015840 either 119881

119906or 119881V contains no vertices with total

color set 119894 for some 119894 isin 1 2 1198961015840 Without loss

of generality we assume that there is no vertex 119909 isin

119881119906with 119862

1015840

119891(119909) = 119894 For any vertex 119910 in 119881

119906such

that 1198621015840119891(119910) = 119894 119895 and 1198911015840(119910119906) = 119894 interchange the

two colors of 119910 and 119910119906 The resulting coloring stilldenoted by 1198911015840 satisfies that 119862

1198911015840(119864119906) does not contain

color 119894 Then we color 119906 119906V V by any three colors in1 2 119896

1015840 119894 and obtain a 1198961015840-GVDTC of 119878

119898119899

(2) Consider |1198621015840119891(119864119906)| lt 119896

1015840 or |1198621015840119891(119864V)| lt 119896

1015840 assume|1198621015840

119891(119864119906)| lt 119896

1015840 here Let 119894 notin 1198621015840119891(119864119906) Color V by 119894 and

color 119906 119906V by any two colors in [1 1198961015840] 119894 Obviouslythe resulting coloring is a 1198961015840-GVDTC of 119878119898119899

All the above show that120594119892V119905(119878119898119899) le 119896

1015840 So the conclusionholds

Theorem 9 Let 119878119901119902ℓ

be a tristar defined as above and ℓ =119901 + 119902 + 119903 Then

120594119892V119905 (119878119901119902119903) =

3 ℓ = 3 4

4 ℓ = 5 6

lceil(radic8119903 + 1 minus 1) 2rceil ℓ ge 7

(16)

Proof When ℓ = 3 4 the conclusion is easy to be checkedwhen ℓ = 5 or 6 since |119881(119878

119901119902119903)| ge 8 and ( 3

1) + ( 32) + ( 33) =

7 it follows that 120594119892V119905(119878119901119902119903) gt 3 In addition it is not hard

to give a 4-GVDTC of 119881(119878119901119902119903

) in each case of ℓ = 5 or 6so 120594119892V119905(119878119901119902119903) = 4 when ℓ ge 7 let 1198961015840 = lceil(radic8119903 + 1 minus 1)2rceil

Identify vertices 1199060and V0in 119878119901119902119903

and let 1198661015840 = 119878119901119902119903

119906 VBy Theorem 8 1198661015840 has a 1198961015840-GVDTC1198911015840 With the analogousanalysis method ofTheorem 8 we can also extend 1198911015840 to a 1198961015840-GVDTC of 119878

119901119902119903 This shows 120594

119892V119905(119878119901119902119903) le 1198961015840 On the other

hand for any 119896-GVDTC119891 of 119878119901119902119903

it has that ℓ le ( 1198961) + ( 1198962)

that is 119896 ge 1198961015840 So the result holds

In the above we construct a 119896-GVDTC of a graph 119866 byextending a 119896-GVDTC of graph 1198661015840 where 1198661015840 is the resultinggraph of identifying two vertices of degree more than 1 in 119866But thismethoddoes not alwayswork For instance the graph119866119906 V shown in Figure 1(b) has a 4-GVDTC but the graph119866 shown in Figure 1(a) does not contain any 4-GVDTC Soany 4-GVDTCof119866119906 V can not be extended to a 4-GVDTCof 119866

In the following we are devoted to the study of the generalvertex-distinguishing chromatic number of fan graph 119865119899wheel graph 119882119899 and complete graph 119870119899 Let 119866119867 be twographs such that119881(119866)cap119881(119867) = 0The join119866+119867 of119866 and119867is a graph with vertex set 119881(119866 +119867) = 119881(119866) cup119881(119867) and edgeset 119864(119866 + 119867) = 119864(119866) cup 119864(119867) cup 119906V | 119906 isin 119881(119866) V isin 119881(119867) Afan graph 119865

119899is defined as the join of a path of 119899 vertices and

an isolated vertex A wheel graph119882119899is defined as the join of

a cycle of 119899 vertices and an isolated vertex


u

(a)

u minus

(b)

Figure 1 (a) A graph 119866 (b) 119866119906 V

Theorem 10 Let 119865119899be a fan 119899 ge 2 then

120594119892V119905 (119865119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(17)

where ( 119896minus11)+( 119896minus12)+( 119896minus13)+( 119896minus14) lt 119899 le ( 119896

1)+( 1198962)+( 1198963)+( 1198964)

Proof Let 119881(119865119899) = V

119894| 119894 = 0 1 119899 and 119864(119865

119899) =

V1V2 V2V3 V

119899minus1V119899 cup V0V119894| 119894 = 1 2 119899 When 119899 le 14

the conclusion is easy to show We now consider the case of119899 ge 15 (which implies 119896 ge 5) Since ( 119896minus1

1) + ( 119896minus1

2) + ( 119896minus1

3) +

( 119896minus14) lt 119899 and |119862

119891(V119894)| le 4 for 119894 isin [1 119899] we can easily deduce

that 120594119892V119905(119865119899) ge 119896 So it suffices to show that 119865

119899contains a 119896-

GVDTC In particular we prove that119865119899contains a 119896-GVDTC

such that the total color set of V0contains at least 5 elements

By induction on 119899 When 119899 = 15 it is not hard to constructsuch a 119896(ge 5)-GVDTC of119865119899 Suppose that for any1198651198991015840 119899

1015840lt 119899

there exists a 119896-GVDTC of 1198651198991015840 Consider the fan graph 119865119899minus1and let 119891 be a 119896-GVDTC of 119865119899minus1 Anyway we can assumethat for any color 119909 isin [1 119896] there is an edge V119894V119894+1 for some119894 isin [1 119899minus1] such that 119891(V119894V119894+1) = 119909 (If not we can permutate119909 and 119891(V119894V119894+1))

Note that for any edge V119894V119894+1 of 119865119899minus1 119894 isin [1 119896] if wereplace this edge by a vertex 119906 and connect this vertex to V0V119894 and V

119894+1 then the resulting graph denoted by 119865V

119894119906V119894+1

119899minus1 is

isomorphic to119865119899Wewill use this to construct a 119896-GVDTCof

119865119899based on119891 It is obvious that there remain only 4 uncolored

elements V119894119906 119906V119894+1 119906V0 and 119906 in 119865V

119894119906V119894+1

119899minus1 if we restrict 119891 to

119865V119894119906V119894+1

119899minus1 We need to consider the following 2 cases

Case 1 There exist colors 119909 119910 119911 isin [1 119896] such that 119909 119910 119911 notinC119891(119865119899minus1) Let V

119894V119894+1

be the edge with 119891(V119894V119894+1) = 119909 119894 isin

[1 119899 minus 1] In 119865V119894119906V119894+1

119899minus1 let 119891(119906V119894) = 119891(119906V119894+1) = 119909 119891(119906V0) = 119910

and 119891(119906) = 119911 and the resulting coloring is still denoted by119891 Evidently in 119865V

119894119906V119894+1

119899minus1 119862119891(119906) = 119909 119910 119911 meanwhile 119862119891(V119894)

and 119862119891(V119894+1) are the same as those in 119865119899minus1 and |119862119891(V0)| ge 5So 119891 a 119896-GVDTC of 119865V

119894119906V119894+1

119899minus1

Case 2 Four different colors 119909 119910 119911 119908 isin [1 119896] such that119909 119910 119911 119908 notin C119891(119865119899minus1) Select an edge V119894V119894+1 119894 isin [1 119899 minus 1]for which 119891(V119894V119894+1) = 119909 Since 119891 is a 119896-GVDTC of 119865119899minus1119862119891(V119894) 119862119891(V119894+1) contains at least one element (here weassume |119862

119891(V119894)| ge |119862

119891(V119894+1)|) say 119888 Obviously 119888 = 119909

119888 isin 119862119891(V119894) and 119888 notin 119862

119891(V119894+1) in 119865

119899minus1 We can permutate

the colors so that 119888 isin 119910 119909 119908 and 119891(V119894) = 119888 in 119865

119899minus1 say

119888 = 119910 Then in 119865V119894119906V119894+1

119899minus1 erase the color of vertex V

119894and

recolor it by color 119909 and let 119891(119906V119894) = 119910 119891(119906V

119894+1) = 119909

119891(119906V0) = 119911 and 119891(119906) = 119908 Obviously in 119865V

119894119906V119894+1

119899minus1 it follows

that 119862119891(119906) = 119909 119910 119911 119908 119862

119891(V119894) and 119862

119891(V119894+1) are the same as

those in 119865119899minus1

and |119862119891(V0)| ge 5 So 119891 a 119896-GVDTC of 119865V

119894119906V119894+1

119899minus1

All of the above show that 119865119899has a 119896-GVDTC

Theorem 11 Let119882119899be a wheel graph 119899 ge 2 then

120594119892V119905 (119882119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(18)


1)+( 1198962)+( 1198963)+( 1198964)

We omit the proof forTheorem 11 since it is analogous tothat of Theorem 10

Theorem 12 For a complete graph 119870119899 119899 ge 1 one has

120594119892V119905 (119870119899) = 1 + lceillog2119899rceil (19)

Proof When 119899 lt 10 the conclusion is easy to show So weassume 119899 ge 10

Denote by119881(119870119899) = V

1 V2 V

119899 and 1 2 119896 the set

of 119896 colors For integer ℓ = lceil1198992rceil we construct a ℓ-GVDTC1198911015840of 119870119899as follows let 1198911015840(V

119894) = 119894 for any 119894 isin [1 ℓ] 1198911015840(V

119894V119895) = 1

for 119894 = 119895 119894 isin [1 ℓ] and 119895 isin [1 119899] 119891(V119894) = 119894 minus ℓ + 2 for

119894 isin [ℓ + 1 119899 minus 2] 1198911015840(V119894V119895) = 2 for 119894 = 119895 119894 isin [ℓ + 1 119899 minus 2] and

119895 isin [ℓ + 1 119899] 1198911015840(V119899minus1

V119899) = 3 1198911015840(V

119899minus1) = 4 1198911015840(V

119899) = 5 One

can readily check that 1198621015840119891(V1) = 1 1198621015840

119891(V119894) = 1 119894 for 119894 =

1 2 ℓ1198621015840119891(V119894) = 1 2 119894minusℓ+2 for 119894 = ℓ+1 ℓ+2 119899minus2

1198621015840

119891(V119899minus1) = 1 2 3 4 and 1198621015840

119891(V119899minus1) = 1 2 3 5 Thus 1198911015840 is

a ℓ-GVDTC of119870119899 which shows 120594119892V119905(119870119899) le ℓSuppose that 120594119892V119905(119870119899) = 119896 and 119891 is a 119896-GVDTC of 119870119899

Since for any two vertices V119894 V119895 (119894 = 119895 isin [1 119896]) 119862119891(V119894) cap119862119891(V119895) = 0 one can see that there is at most one vertex whosetotal color set contains only one color If there is a vertex Vwith |119862(V)| = 1 without loss of generality assume 119862(V) = 1then the total color set of each vertex contains color 1 whichindicates

119899 le 1 + (119896 minus 1

1) + (

119896 minus 1

2) + sdot sdot sdot + (

119896 minus 1

119896 minus 1) = 2

119896minus1 (20)

If there is no vertex whose total color set contains only onecolor then for each V

119894it has |119862(V

119894)| ge 2 Since for any vertex

V119895= V119894 |119862(V119894)cap119862(V

119895)| ge 1 it follows that119862(V) and [1 119896]119862(V)

can not be two total color sets with respect to 119891 This implies119899 le 2119896minus1 So 119896 ge 1 + lceillog

2119899rceil


To prove 119896 = 1 + lceillog2119899rceil we need to show that 119870

119899has a

(1+lceillog2119899rceil)-GVDTC In particular we show that any119870

119899has

a (1 + lceillog2119899rceil)-GVDTC such that for each color 119888 isin [1 1 +

lceillog2119899rceil] there is at least one vertex in 119870

119899being colored by 119888

We prove this by induction on 119899 When 119899 = 10 5-GVDTC is the 1198911015840 defined above for 119899 = 10 ℓ = 5 where1198621015840

119891(V1) = 1 Consider the graph 119870

119899minus V119899obtained from 119870

119899

by deleting vertex V119899 and its incident edges Obviously119870119899minusV119899is isomorphic to119870119899minus1 By the induction hypothesis119870119899minus1 hasa (1 + lceillog

2119899 minus 1rceil)-GVDTC say 119891 such that for each color

119888 isin [1 1 + lceillog2119899 minus 1rceil] there is at least one vertex of 119870

119899minus1

being colored by 119888 Since 119899 le 2119896minus1 there must be some set

119862 (denoted by 1198881 1198882 119888

ℓ) notin C

119891(119870119899minus1) We consider the

following two cases

(1) Consider lceillog2119899 minus 1rceil = lceillog

2119899rceil By the induction

hypothesis each color 119888119894 isin [1 1+ lceillog2119899minus1rceil] appearsat a vertex Without loss of generality assume 119891(V119894) =119888119894 for 119894 = 1 2 ℓ Then 119891 is extended to a (1 +lceillog2119899 minus 1rceil)-GVDTC of 119870119899 via coloring V119899 by one

of the colors in [1198881 1198882 119888ℓ] coloring V119899V119894 for 119894 =1 2 ℓ by 119888119894 and coloring V119899V119895 for 119895 = ℓ+1 119899minus1by one of the colors in 119862119891(V119895) (V119895 isin 119881(119870119899minus1))


2119899rceil minus 1 Then on the basis

of 119891 we only need to color V119899and all of its incident

edges in119870119899by color 1 + lceillog

2119899 minus 1rceil

One can readily check that the resulting coloring of119870119899in

the above two cases is (1 + lceillog2119899rceil)-GVDTCs of119870

119899such that

each color in [1 1+ lceillog2119899rceil] appears at a vertex of119870

119899 Hence

119870119899 has a (1+lceillog2119899rceil)-GVDTC and the conclusion holds

In the following we present a trivial upper bound of thegeneral vertex-distinguishing total chromatic number of thejoin graph of two graphs

Theorem 13 Suppose119866119867 are two simple graphs and119866cap119867 =

0 Then

120594119892V119905 (119866 + 119867) le 120594119892V119905 (119866) + 120594119892V119905 (119867) (21)

Proof Let 119881(119866)=119906119894| 119894 = 1 2 119898 and 119881(119867) = V

119894| 119894 =

1 2 119899 Suppose that 1198911is a 120594119892V119905(119866)-GVDTC of 119866 and 119891

2

is a 120594119892V119905(119867)-GVDTC of119867 where the sets of colors of 119891

1and

1198912are 1198621and 119862

2(1198621cap 1198622= 0) respectively

Define 119891 as 119891(119906119894V119895) = 119891

1(119906119894) (or 119891

2(V119895)) 119894 =

1 2 119898 119895 = 1 2 119899Combining colorings 119891

1 1198912 119891 we can obtain a (120594

119892V119905(119866)+

120594119892V119905(119867))-GVDTC of 119866 + 119867

3 Remarks

Based on the above results we propose two conjectures asfollows

Conjecture 14 Let 119866 be a graph without isolated verticesThen

120594119892V119905 (119866) le lceil119899

2rceil (22)

Conjecture 15 Let119866 be a connected graph on 119899 verticesThen

120594119892V119905 (119866) le 1 + lceillog2119899rceil (23)

Note that if Conjecture 15 is true then Conjecture 14 istrue On the other hand if Conjecture 14 is true then theupper bound cannot be improved For instance the graph 119866contains exactly three1198702 components It is easy to show that120594119892V119905(119866) = 3

In addition there is a very interesting observation aboutthe general vertex-distinguishing total chromatic number

Observation 1 Let 119867 be a subgraph of a graph 119866 Then itpossibly follows that

120594119892V119905 (119867) gt 120594119892V119905 (119866) (24)

As an illustration of this observationwe consider the path11987515

and the fan graph 11986514 11987515

is a subgraph of 11986514 while

120594119892V119905(11987515)(= 5) gt 120594119892V119905(11986514)(= 4)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported by the National Natural ScienceFoundation of China (Grant nos 61127005 and 61309015) andtheNational Basic Research Programof China (973 Program)(no 2010CB328103)

References

[1] J A Bondy and U S R Murty GraphTheory Springer 2008[2] F T Leighton ldquoA graph coloring algorithm for large scheduling

problemsrdquo Journal of Research of National Bureau of Standardsvol 84 no 6 pp 489ndash506 1979

[3] D de Werra and Y Gay ldquoChromatic scheduling and frequencyassignmentrdquo Discrete Applied Mathematics vol 49 no 1ndash3 pp165ndash174 1994

[4] G Jonathan and Y Jay GraphTheory and Its Applications CRCpress New York NY USA 1999

[5] F Harary ldquoConditional colorability in graphsrdquo in Graphs andApplications (Boulder Colo 1982) pp 127ndash136 John Wiley ampSons New York NY USA 1985

[6] F Harary and M Plantholt ldquoThe point-distinguishing chro-matic indexrdquo in Graphs and Applications F Harary and J SMaybee Eds pp 147ndash162 Wiley-Interscience New York NYUSA 1985

[7] A C Burris and R H Schelp ldquoVertex-distinguishing properedge-coloringsrdquo Journal of Graph Theory vol 26 no 2 pp 73ndash82 1997

[8] Z Zhang L Z Liu and J F Wang ldquoAdjacent strong edgecoloring of graphsrdquo Applied Mathematics Letters vol 15 no 5pp 623ndash626 2002

[9] E Gyori M Hornak C Palmer and M Wozniak ldquoGeneralneighbour-distinguishing index of a graphrdquoDiscreteMathemat-ics vol 308 no 5-6 pp 827ndash831 2008


[10] M Behzad Graphs and Their Chromatic Numbers MichiganState University 1965

[11] V G Vizing ldquoSome unsolved problems in Graph TheoryrdquoRussian Mathematical Surveys vol 23 pp 125ndash141 1968

[12] Z Zhang X E Chen J Li B Yao X Lu and J Wang ldquoOnadjacent-vertex-distinguishing total coloring of graphsrdquo Sciencein China Series A Mathematics vol 48 no 3 pp 289ndash2992005

[13] X Chen ldquoOn the adjacent vertex distinguishing total coloringnumbers of graphs with Δ = 3rdquo Discrete Mathematics vol 308no 17 pp 4003ndash4007 2008

[14] H Wang ldquoOn the adjacent vertex-distinguishing total chro-matic numbers of the graphs with 998779 = 3rdquo Journal ofCombinatorial Optimization vol 14 no 1 pp 87ndash109 2007

[15] J Hulgan ldquoConcise proofs for adjacent vertex-distinguishingtotal coloringsrdquoDiscrete Mathematics vol 309 no 8 pp 2548ndash2550 2009

[16] D Huang W Wang and C Yan ldquoA note on the adjacent vertexdistinguishing total chromatic number of graphsrdquo DiscreteMathematics vol 312 no 24 pp 3544ndash3546 2012

[17] M Chen and X Guo ldquoAdjacent vertex-distinguishing edge andtotal chromatic numbers of hypercubesrdquo Information ProcessingLetters vol 109 no 12 pp 599ndash602 2009

[18] Y Wang and W Wang ldquoAdjacent vertex distinguishing totalcolorings of outerplanar graphsrdquo Journal of CombinatorialOptimization vol 19 no 2 pp 123ndash133 2010

[19] V Pedrotti and C P de Mello ldquoAdjacent-vertex-distinguishingtotal coloring of indifference graphsrdquo Matematica Contem-poranea vol 39 pp 101ndash110 2010

[20] W Wang and D Huang ldquoThe adjacent vertex distinguishingtotal coloring of planar graphsrdquo Journal of CombinatorialOptimization vol 27 no 2 pp 379ndash396 2014

[21] X Chen and Z Zhang ldquoAVDTC numbers of generalized Halingraphs with maximum degree at least 6rdquo Acta MathematicaeApplicatae Sinica vol 24 no 1 pp 55ndash58 2008


[23] A Papaioannou and C Raftopoulou ldquoOn the AVDTC of 4-regular graphsrdquoDiscrete Mathematics vol 330 pp 20ndash40 2014

[24] Z Zhang P X Qiu B G Xu J Li X Chen and B Yao ldquoVertex-distinguishing total coloring of graphsrdquo Ars Combinatoria vol87 pp 33ndash45 2008

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


In [13ndash15] authors proved that there exists a 6-AVDTC ofgraphs with Δ = 3 which indicates conjecture (2) holdsfor such graphs For further research on adjacent vertex-distinguishing total chromatic number one may refer to [16ndash23]

For a 119896-AVDTC119891 of a graph 119866 if 119862119891(119906) = 119862

119891(V) is

required for any two distinct vertices 119906 V then 119891 is calleda vertex-distinguishing total 119896-coloring of 119866 abbreviated as119896-VDTCTheminimumnumber 119896 such that119866 has a 119896-VDTCis called the vertex-distinguishing total chromatic numberdenoted by 120594V119905(119866) [24] Zhang et al conjectured in [24] thatfor any graph 119866 it follows that

120583119905 (119866) le 120594V119905 (119866) le 120583119905 (119866) + 1 (3)

where 120583119905(119866) = min119896 | ( 119896119894+1) ge 119899119894 120575 le 119894 le Δ

In this paper we introduce a variant of vertex-distinguishing total coloring of a graph 119866 which relaxes therestriction that the coloring is proper We now present thedetailed definition as follows

Definition 2 Let 119866 be a graph and 119896 be a positive integerA total coloring 119891 of 119866 using 119896 colors is called a generalvertex-distinguishing total 119896-coloring of 119866 (or 119896-GVDTC of119866 briefly) if for all119906 V isin 119881(119866)119862119891(119906) = 119862

119891(V)Theminimum

number 119896 for which 119866 has a 119896-GVDTC is the general vertex-distinguishing total chromatic number denoted by 120594

119892V119905(119866)

It is evident that 120594119892V119905(119866) does exist for any graph 119866 In

this paper we study the general vertex-distinguishing totalcoloring of some special classes of graphs and obtain the exactvalue of the general vertex-distinguishing total chromaticnumber of these graphs Furthermore we propose a con-jecture on the upper bound of general vertex-distinguishingtotal chromatic number of a graph

2 Main Results

We first present a trivial lower bound on the general vertex-distinguishing total chromatic number of a graph

Theorem 3 Let 119866 be a graph on 119899 vertices Then

120594119892V119905 (119866) ge lceillog2 (119899 + 1)rceil (4)

Proof Let 120594119892V119905(119866) = 119896 It follows that 119899 le ( 119896

1) + ( 1198962) + sdot sdot sdot +

( 119896119896) = 2119896minus 1 so 119896 ge lceillog

2(119899 + 1)rceil

Notice that the lower bound ofTheorem 3 can be attainedin graphs such as the 119899-vertex path 119875119899 for 119899 = 1 2 7 Onecan readily check that 120594

119892V119905(1198751)= 1 and 120594

119892V119905(119875119899)= 2 for 119899 = 2 3

and 120594119892V119905(119875119899)= 3 for 119899 = 4 5 6 7

Theorem 4 Let 119866 be a graph without isolated vertices andisolated edges Then

120594119892V119905 (119866) le 1205941015840

119892V119889 (119866) (5)

Proof Suppose that 119891 is a 119896-GVDEC of 119866 For any 119906 isin 119881(119866)let 119891(119906) = 119891(119906V) where 119906V isin 119864(119866) Obviously 119891 is a119896-GVDTC of 119866

We now turn to investigating the general vertex-distinguishing total chromatic number of an 119899-vertex path

Theorem 5 Let 119875119899be a path on 119899 vertices 119899 ge 1 Then

120594119892V119905 (119875119899) =

[[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(6)

Proof Denote by 119875119899= V1V2sdot sdot sdot V119899a path 119875

119899with vertex

set V1 V2 V


1V2 V2V3 V

119899minus1V119899 Let

120594119892V119905(119875119899) = 119896 and let 119891 be a 119896-GVDTC of 119875

119899 Let 120572

119896= ( 119896minus11)+

( 119896minus12) + ( 119896minus1

3) 120573119896= ( 1198961) + ( 1198962) + ( 1198963) 120574119896= 1 + ( 119896minus1

1) + ( 119896minus1

2)

and

119896lowast= [[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(7)

Evidently |119862119891(V119894)| le 3 and 119899 le 120573119896 (which implies 119896 ge

119896lowast) In order to prove the conclusion 119896 = 119896

lowast it sufficesto give a 119896lowast-GVDTC of 119875

119899 When 119899 le 7 it is not hard

to construct the corresponding general vertex-distinguishingtotal colorings Let 119899 ge 8 We first construct a 119896lowast-GVDTC1198911015840of 119875120573119896

recursively Note that when 119899 = 120573119896 it has 119896lowast = 119896


(ie 11987514) as

follows the vertices V1 V2 V

14are colored by 1 3 4 2 3 1

4 2 3 4 3 4 4 1 respectively the edges V1V2 V2V3

V3V4 V

13V14

are colored by 1 3 1 1 2 4 2 2 2 3 3 4 4respectively It is easy to see that 119891

4is a 4-GVDTC of 119875

14


based on a (119896 minus1)-GVDTC119891119896minus1 of 119875120573

119896minus1

Let 119891119896 be

119891119896(V119894+120574119896

) = 119891119896minus1

(V119894) + 1 119894 = 1 2 120572

119896minus 1

119891119896(V120572119896+120574119896

) = 119891119896minus1

(V120572119896

) = 1

119891119896 (V119894+120574119896

V119894+1+120574119896

) = 119891119896minus1 (V119894V119894+1) + 1 119894 = 1 2 120572119896 minus 1

119891119896(V1) = 1 119891

119896(V120574119896

) = 119896

(8)

V2 V3 V

120574119896minus1

are colored by

119896 minus 1 119896⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 119896 minus 2 119896 minus 1 119896⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

3 elements 2 3 119896⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119896minus1 elements (9)

when 119896 is even V1V2 V2V3 V3V4 V

120574119896

V120574119896+1

are colored by


1⏟⏟⏟⏟⏟⏟⏟

1 element 119896 minus 1 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 1 119896 minus 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟



5 elements




2 1 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


2 elements

(10)


V120574119896+1

are colored by

1⏟⏟⏟⏟⏟⏟⏟

1 element 119896 minus 1 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 1 119896 minus 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟



5 elements




1 2 1 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


2 elements

(11)






1 120574119896+ 2 120573



(V119894) and 119862119891119896

(V119894) =

119862119891119896


(V120574119896

) = 2 119896 So 119891119896 is a 119896-GVDTC of 119875120573119896


119896

for( 119896minus11) + ( 119896minus1

2) + ( 119896minus1

3) lt 119899 lt 120573

119896



1 V2 V

119903from 119875

120573119896


119903+1V119903+2



1198911015840(V119894V119894+1) = 119891(V

119894V119894+1) for 119894 = 119903 + 1 119903 + 2 120573

119896minus 1 1198911015840(V

119894) =

119891(V119894) for 119894 = 119903 + 2 120573 and 1198911015840(V

119903+1) = 1 Then 1198911015840 is a 119896-

GVDTC of 119875119899




set V1 V2 V


1V2 V2V3 V

119899minus1V119899 V119899V1


120594119892V119905 (119862119899) =

[[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]


119896lowast= [[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(13)






graph 1198701119899

(119899 ge 1) A double star 119878119898119899



119901119902119903) =

119906119894| 119894 = 0 1 119901cupV

119894| 119894 = 0 1 119902cup119908

119894| 119894 = 0 1 ℓ

and edge set 119864(119878119901119902119903

) = 1199060119906119894| 119894 = 1 2 119901 cup V

0V119894| 119894 =

1 2 119902 cup 1199080119908119894| 119894 = 1 2 119903 cup 119906

0V0 V01199080 where



120594119892V119905 (119878119899) =

2 119899 = 1 2


2rceil 119899 ge 3

(14)




119899














1015840

1) + ( 119896

1015840

2) there



0by 119886 and 119887



Theorem 8 Let 119878119898119899


120594119892V119905 (119878119898119899) =

3 ℓ = 2 3 4 5

4 ℓ = 6


2rceil ℓ ge 7

(15)



119891(119909)| le 2 for any

119909 isin 119881(119878119898119899) 119906 V we have 120594119892V119905(119878119898119899) ge 119896



1198911015840(119906minusV)| le 1198961015840 If






to 119906 (resp V) in 119878119898119899




119898119899








that |1198621198911015840(119906)| ge 3 |119862

1198911015840(V)| ge 3


1198911015840(V) So 1198911015840 is a 1198961015840-GVDTC of 119878

119898119899



(1) Consider |1198621015840119891(119864119906)| = |119862

1015840

119891(119864V)| = 119896

1015840 Suppose that 119881119906






119881119906with 119862

1015840

119891(119909) = 119894 For any vertex 119910 in 119881

119906such



1198911015840(119864119906) does not contain



119898119899

(2) Consider |1198621015840119891(119864119906)| lt 119896

1015840 or |1198621015840119891(119864V)| lt 119896

1015840 assume|1198621015840

119891(119864119906)| lt 119896





Theorem 9 Let 119878119901119902ℓ


120594119892V119905 (119878119901119902119903) =

3 ℓ = 3 4

4 ℓ = 5 6


(16)


119901119902119903)| ge 8 and ( 3

1) + ( 32) + ( 33) =


to give a 4-GVDTC of 119881(119878119901119902119903



and let 1198661015840 = 119878119901119902119903


119901119902119903 This shows 120594

119892V119905(119878119901119902119903) le 1198961015840 On the other


it has that ℓ le ( 1198961) + ( 1198962)








u

(a)

u minus

(b)



120594119892V119905 (119865119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(17)


1)+( 1198962)+( 1198963)+( 1198964)

Proof Let 119881(119865119899) = V

119894| 119894 = 0 1 119899 and 119864(119865

119899) =

V1V2 V2V3 V



1) + ( 119896minus1

2) + ( 119896minus1

3) +

( 119896minus14) lt 119899 and |119862



119899contains a 119896-




1015840lt 119899




119894119906V119894+1

119899minus1 is




119894119906V119894+1


119865V119894119906V119894+1



119894V119894+1


[1 119899 minus 1] In 119865V119894119906V119894+1

119899minus1 let 119891(119906V119894) = 119891(119906V119894+1) = 119909 119891(119906V0) = 119910


119894119906V119894+1



119894119906V119894+1

119899minus1


119891(V119894)| ge |119862

119891(V119894+1)|) say 119888 Obviously 119888 = 119909

119888 isin 119862119891(V119894) and 119888 notin 119862

119891(V119894+1) in 119865



119899minus1 say

119888 = 119910 Then in 119865V119894119906V119894+1


119894and


119894+1) = 119909


119894119906V119894+1


that 119862119891(119906) = 119909 119910 119911 119908 119862

119891(V119894) and 119862

119891(V119894+1) are the same as



119894119906V119894+1

119899minus1



120594119892V119905 (119882119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(18)


1)+( 1198962)+( 1198963)+( 1198964)



120594119892V119905 (119870119899) = 1 + lceillog2119899rceil (19)


Denote by119881(119870119899) = V

1 V2 V

119899 and 1 2 119896 the set


119894) = 119894 for any 119894 isin [1 ℓ] 1198911015840(V

119894V119895) = 1



119895 isin [ℓ + 1 119899] 1198911015840(V119899minus1

V119899) = 3 1198911015840(V

119899minus1) = 4 1198911015840(V

119899) = 5 One


119891(V119894) = 1 119894 for 119894 =


1198621015840

119891(V119899minus1) = 1 2 3 4 and 1198621015840

119891(V119899minus1) = 1 2 3 5 Thus 1198911015840 is



119899 le 1 + (119896 minus 1

1) + (

119896 minus 1


119896 minus 1

119896 minus 1) = 2

119896minus1 (20)


119894it has |119862(V


V119895= V119894 |119862(V119894)cap119862(V



2119899rceil



119899has a


119899has







119899




119899minus1


119862 (denoted by 1198881 1198882 119888

ℓ) notin C


following two cases












119899such that


119899 Hence




0 Then

120594119892V119905 (119866 + 119867) le 120594119892V119905 (119866) + 120594119892V119905 (119867) (21)

Proof Let 119881(119866)=119906119894| 119894 = 1 2 119898 and 119881(119867) = V

119894| 119894 =


2


1and

1198912are 1198621and 119862


Define 119891 as 119891(119906119894V119895) = 119891

1(119906119894) (or 119891

2(V119895)) 119894 =


1 1198912 119891 we can obtain a (120594

119892V119905(119866)+

120594119892V119905(119867))-GVDTC of 119866 + 119867

3 Remarks



120594119892V119905 (119866) le lceil119899

2rceil (22)






120594119892V119905 (119867) gt 120594119892V119905 (119866) (24)




120594119892V119905(11987515)(= 5) gt 120594119892V119905(11986514)(= 4)



Acknowledgments


References

































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1⏟⏟⏟⏟⏟⏟⏟

1 element 119896 minus 1 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 1 119896 minus 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟



5 elements




2 1 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


2 elements

(10)


V120574119896+1

are colored by

1⏟⏟⏟⏟⏟⏟⏟

1 element 119896 minus 1 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

2 elements 1 119896 minus 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟



5 elements




1 2 1 2 1⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


2 elements

(11)






1 120574119896+ 2 120573



(V119894) and 119862119891119896

(V119894) =

119862119891119896


(V120574119896

) = 2 119896 So 119891119896 is a 119896-GVDTC of 119875120573119896


119896

for( 119896minus11) + ( 119896minus1

2) + ( 119896minus1

3) lt 119899 lt 120573

119896



1 V2 V

119903from 119875

120573119896


119903+1V119903+2



1198911015840(V119894V119894+1) = 119891(V

119894V119894+1) for 119894 = 119903 + 1 119903 + 2 120573

119896minus 1 1198911015840(V

119894) =

119891(V119894) for 119894 = 119903 + 2 120573 and 1198911015840(V

119903+1) = 1 Then 1198911015840 is a 119896-

GVDTC of 119875119899




set V1 V2 V


1V2 V2V3 V

119899minus1V119899 V119899V1


120594119892V119905 (119862119899) =

[[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]


119896lowast= [[[[

3radic3119899 + radic91198992 +

125


125

27

]]]]

(13)






graph 1198701119899

(119899 ge 1) A double star 119878119898119899



119901119902119903) =

119906119894| 119894 = 0 1 119901cupV

119894| 119894 = 0 1 119902cup119908

119894| 119894 = 0 1 ℓ

and edge set 119864(119878119901119902119903

) = 1199060119906119894| 119894 = 1 2 119901 cup V

0V119894| 119894 =

1 2 119902 cup 1199080119908119894| 119894 = 1 2 119903 cup 119906

0V0 V01199080 where



120594119892V119905 (119878119899) =

2 119899 = 1 2


2rceil 119899 ge 3

(14)




119899














1015840

1) + ( 119896

1015840

2) there



0by 119886 and 119887



Theorem 8 Let 119878119898119899


120594119892V119905 (119878119898119899) =

3 ℓ = 2 3 4 5

4 ℓ = 6


2rceil ℓ ge 7

(15)



119891(119909)| le 2 for any

119909 isin 119881(119878119898119899) 119906 V we have 120594119892V119905(119878119898119899) ge 119896



1198911015840(119906minusV)| le 1198961015840 If






to 119906 (resp V) in 119878119898119899




119898119899








that |1198621198911015840(119906)| ge 3 |119862

1198911015840(V)| ge 3


1198911015840(V) So 1198911015840 is a 1198961015840-GVDTC of 119878

119898119899



(1) Consider |1198621015840119891(119864119906)| = |119862

1015840

119891(119864V)| = 119896

1015840 Suppose that 119881119906






119881119906with 119862

1015840

119891(119909) = 119894 For any vertex 119910 in 119881

119906such



1198911015840(119864119906) does not contain



119898119899

(2) Consider |1198621015840119891(119864119906)| lt 119896

1015840 or |1198621015840119891(119864V)| lt 119896

1015840 assume|1198621015840

119891(119864119906)| lt 119896





Theorem 9 Let 119878119901119902ℓ


120594119892V119905 (119878119901119902119903) =

3 ℓ = 3 4

4 ℓ = 5 6


(16)


119901119902119903)| ge 8 and ( 3

1) + ( 32) + ( 33) =


to give a 4-GVDTC of 119881(119878119901119902119903



and let 1198661015840 = 119878119901119902119903


119901119902119903 This shows 120594

119892V119905(119878119901119902119903) le 1198961015840 On the other


it has that ℓ le ( 1198961) + ( 1198962)








u

(a)

u minus

(b)



120594119892V119905 (119865119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(17)


1)+( 1198962)+( 1198963)+( 1198964)

Proof Let 119881(119865119899) = V

119894| 119894 = 0 1 119899 and 119864(119865

119899) =

V1V2 V2V3 V



1) + ( 119896minus1

2) + ( 119896minus1

3) +

( 119896minus14) lt 119899 and |119862



119899contains a 119896-




1015840lt 119899




119894119906V119894+1

119899minus1 is




119894119906V119894+1


119865V119894119906V119894+1



119894V119894+1


[1 119899 minus 1] In 119865V119894119906V119894+1

119899minus1 let 119891(119906V119894) = 119891(119906V119894+1) = 119909 119891(119906V0) = 119910


119894119906V119894+1



119894119906V119894+1

119899minus1


119891(V119894)| ge |119862

119891(V119894+1)|) say 119888 Obviously 119888 = 119909

119888 isin 119862119891(V119894) and 119888 notin 119862

119891(V119894+1) in 119865



119899minus1 say

119888 = 119910 Then in 119865V119894119906V119894+1


119894and


119894+1) = 119909


119894119906V119894+1


that 119862119891(119906) = 119909 119910 119911 119908 119862

119891(V119894) and 119862

119891(V119894+1) are the same as



119894119906V119894+1

119899minus1



120594119892V119905 (119882119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(18)


1)+( 1198962)+( 1198963)+( 1198964)



120594119892V119905 (119870119899) = 1 + lceillog2119899rceil (19)


Denote by119881(119870119899) = V

1 V2 V

119899 and 1 2 119896 the set


119894) = 119894 for any 119894 isin [1 ℓ] 1198911015840(V

119894V119895) = 1



119895 isin [ℓ + 1 119899] 1198911015840(V119899minus1

V119899) = 3 1198911015840(V

119899minus1) = 4 1198911015840(V

119899) = 5 One


119891(V119894) = 1 119894 for 119894 =


1198621015840

119891(V119899minus1) = 1 2 3 4 and 1198621015840

119891(V119899minus1) = 1 2 3 5 Thus 1198911015840 is



119899 le 1 + (119896 minus 1

1) + (

119896 minus 1


119896 minus 1

119896 minus 1) = 2

119896minus1 (20)


119894it has |119862(V


V119895= V119894 |119862(V119894)cap119862(V



2119899rceil



119899has a


119899has







119899




119899minus1


119862 (denoted by 1198881 1198882 119888

ℓ) notin C


following two cases












119899such that


119899 Hence




0 Then

120594119892V119905 (119866 + 119867) le 120594119892V119905 (119866) + 120594119892V119905 (119867) (21)

Proof Let 119881(119866)=119906119894| 119894 = 1 2 119898 and 119881(119867) = V

119894| 119894 =


2


1and

1198912are 1198621and 119862


Define 119891 as 119891(119906119894V119895) = 119891

1(119906119894) (or 119891

2(V119895)) 119894 =


1 1198912 119891 we can obtain a (120594

119892V119905(119866)+

120594119892V119905(119867))-GVDTC of 119866 + 119867

3 Remarks



120594119892V119905 (119866) le lceil119899

2rceil (22)






120594119892V119905 (119867) gt 120594119892V119905 (119866) (24)




120594119892V119905(11987515)(= 5) gt 120594119892V119905(11986514)(= 4)



Acknowledgments


References

































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119899














1015840

1) + ( 119896

1015840

2) there



0by 119886 and 119887



Theorem 8 Let 119878119898119899


120594119892V119905 (119878119898119899) =

3 ℓ = 2 3 4 5

4 ℓ = 6


2rceil ℓ ge 7

(15)



119891(119909)| le 2 for any

119909 isin 119881(119878119898119899) 119906 V we have 120594119892V119905(119878119898119899) ge 119896



1198911015840(119906minusV)| le 1198961015840 If






to 119906 (resp V) in 119878119898119899




119898119899








that |1198621198911015840(119906)| ge 3 |119862

1198911015840(V)| ge 3


1198911015840(V) So 1198911015840 is a 1198961015840-GVDTC of 119878

119898119899



(1) Consider |1198621015840119891(119864119906)| = |119862

1015840

119891(119864V)| = 119896

1015840 Suppose that 119881119906






119881119906with 119862

1015840

119891(119909) = 119894 For any vertex 119910 in 119881

119906such



1198911015840(119864119906) does not contain



119898119899

(2) Consider |1198621015840119891(119864119906)| lt 119896

1015840 or |1198621015840119891(119864V)| lt 119896

1015840 assume|1198621015840

119891(119864119906)| lt 119896





Theorem 9 Let 119878119901119902ℓ


120594119892V119905 (119878119901119902119903) =

3 ℓ = 3 4

4 ℓ = 5 6


(16)


119901119902119903)| ge 8 and ( 3

1) + ( 32) + ( 33) =


to give a 4-GVDTC of 119881(119878119901119902119903



and let 1198661015840 = 119878119901119902119903


119901119902119903 This shows 120594

119892V119905(119878119901119902119903) le 1198961015840 On the other


it has that ℓ le ( 1198961) + ( 1198962)








u

(a)

u minus

(b)



120594119892V119905 (119865119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(17)


1)+( 1198962)+( 1198963)+( 1198964)

Proof Let 119881(119865119899) = V

119894| 119894 = 0 1 119899 and 119864(119865

119899) =

V1V2 V2V3 V



1) + ( 119896minus1

2) + ( 119896minus1

3) +

( 119896minus14) lt 119899 and |119862



119899contains a 119896-




1015840lt 119899




119894119906V119894+1

119899minus1 is




119894119906V119894+1


119865V119894119906V119894+1



119894V119894+1


[1 119899 minus 1] In 119865V119894119906V119894+1

119899minus1 let 119891(119906V119894) = 119891(119906V119894+1) = 119909 119891(119906V0) = 119910


119894119906V119894+1



119894119906V119894+1

119899minus1


119891(V119894)| ge |119862

119891(V119894+1)|) say 119888 Obviously 119888 = 119909

119888 isin 119862119891(V119894) and 119888 notin 119862

119891(V119894+1) in 119865



119899minus1 say

119888 = 119910 Then in 119865V119894119906V119894+1


119894and


119894+1) = 119909


119894119906V119894+1


that 119862119891(119906) = 119909 119910 119911 119908 119862

119891(V119894) and 119862

119891(V119894+1) are the same as



119894119906V119894+1

119899minus1



120594119892V119905 (119882119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(18)


1)+( 1198962)+( 1198963)+( 1198964)



120594119892V119905 (119870119899) = 1 + lceillog2119899rceil (19)


Denote by119881(119870119899) = V

1 V2 V

119899 and 1 2 119896 the set


119894) = 119894 for any 119894 isin [1 ℓ] 1198911015840(V

119894V119895) = 1



119895 isin [ℓ + 1 119899] 1198911015840(V119899minus1

V119899) = 3 1198911015840(V

119899minus1) = 4 1198911015840(V

119899) = 5 One


119891(V119894) = 1 119894 for 119894 =


1198621015840

119891(V119899minus1) = 1 2 3 4 and 1198621015840

119891(V119899minus1) = 1 2 3 5 Thus 1198911015840 is



119899 le 1 + (119896 minus 1

1) + (

119896 minus 1


119896 minus 1

119896 minus 1) = 2

119896minus1 (20)


119894it has |119862(V


V119895= V119894 |119862(V119894)cap119862(V



2119899rceil



119899has a


119899has







119899




119899minus1


119862 (denoted by 1198881 1198882 119888

ℓ) notin C


following two cases












119899such that


119899 Hence




0 Then

120594119892V119905 (119866 + 119867) le 120594119892V119905 (119866) + 120594119892V119905 (119867) (21)

Proof Let 119881(119866)=119906119894| 119894 = 1 2 119898 and 119881(119867) = V

119894| 119894 =


2


1and

1198912are 1198621and 119862


Define 119891 as 119891(119906119894V119895) = 119891

1(119906119894) (or 119891

2(V119895)) 119894 =


1 1198912 119891 we can obtain a (120594

119892V119905(119866)+

120594119892V119905(119867))-GVDTC of 119866 + 119867

3 Remarks



120594119892V119905 (119866) le lceil119899

2rceil (22)






120594119892V119905 (119867) gt 120594119892V119905 (119866) (24)




120594119892V119905(11987515)(= 5) gt 120594119892V119905(11986514)(= 4)



Acknowledgments


References

































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










u

(a)

u minus

(b)



120594119892V119905 (119865119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(17)


1)+( 1198962)+( 1198963)+( 1198964)

Proof Let 119881(119865119899) = V

119894| 119894 = 0 1 119899 and 119864(119865

119899) =

V1V2 V2V3 V



1) + ( 119896minus1

2) + ( 119896minus1

3) +

( 119896minus14) lt 119899 and |119862



119899contains a 119896-




1015840lt 119899




119894119906V119894+1

119899minus1 is




119894119906V119894+1


119865V119894119906V119894+1



119894V119894+1


[1 119899 minus 1] In 119865V119894119906V119894+1

119899minus1 let 119891(119906V119894) = 119891(119906V119894+1) = 119909 119891(119906V0) = 119910


119894119906V119894+1



119894119906V119894+1

119899minus1


119891(V119894)| ge |119862

119891(V119894+1)|) say 119888 Obviously 119888 = 119909

119888 isin 119862119891(V119894) and 119888 notin 119862

119891(V119894+1) in 119865



119899minus1 say

119888 = 119910 Then in 119865V119894119906V119894+1


119894and


119894+1) = 119909


119894119906V119894+1


that 119862119891(119906) = 119909 119910 119911 119908 119862

119891(V119894) and 119862

119891(V119894+1) are the same as



119894119906V119894+1

119899minus1



120594119892V119905 (119882119899) =

3 119899 = 2 3 6

4 119899 = 7 8 14

119896 119899 ge 15

(18)


1)+( 1198962)+( 1198963)+( 1198964)



120594119892V119905 (119870119899) = 1 + lceillog2119899rceil (19)


Denote by119881(119870119899) = V

1 V2 V

119899 and 1 2 119896 the set


119894) = 119894 for any 119894 isin [1 ℓ] 1198911015840(V

119894V119895) = 1



119895 isin [ℓ + 1 119899] 1198911015840(V119899minus1

V119899) = 3 1198911015840(V

119899minus1) = 4 1198911015840(V

119899) = 5 One


119891(V119894) = 1 119894 for 119894 =


1198621015840

119891(V119899minus1) = 1 2 3 4 and 1198621015840

119891(V119899minus1) = 1 2 3 5 Thus 1198911015840 is



119899 le 1 + (119896 minus 1

1) + (

119896 minus 1


119896 minus 1

119896 minus 1) = 2

119896minus1 (20)


119894it has |119862(V


V119895= V119894 |119862(V119894)cap119862(V



2119899rceil



119899has a


119899has







119899




119899minus1


119862 (denoted by 1198881 1198882 119888

ℓ) notin C


following two cases












119899such that


119899 Hence




0 Then

120594119892V119905 (119866 + 119867) le 120594119892V119905 (119866) + 120594119892V119905 (119867) (21)

Proof Let 119881(119866)=119906119894| 119894 = 1 2 119898 and 119881(119867) = V

119894| 119894 =


2


1and

1198912are 1198621and 119862


Define 119891 as 119891(119906119894V119895) = 119891

1(119906119894) (or 119891

2(V119895)) 119894 =


1 1198912 119891 we can obtain a (120594

119892V119905(119866)+

120594119892V119905(119867))-GVDTC of 119866 + 119867

3 Remarks



120594119892V119905 (119866) le lceil119899

2rceil (22)






120594119892V119905 (119867) gt 120594119892V119905 (119866) (24)




120594119892V119905(11987515)(= 5) gt 120594119892V119905(11986514)(= 4)



Acknowledgments


References

































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119899has a


119899has







119899




119899minus1


119862 (denoted by 1198881 1198882 119888

ℓ) notin C


following two cases












119899such that


119899 Hence




0 Then

120594119892V119905 (119866 + 119867) le 120594119892V119905 (119866) + 120594119892V119905 (119867) (21)

Proof Let 119881(119866)=119906119894| 119894 = 1 2 119898 and 119881(119867) = V

119894| 119894 =


2


1and

1198912are 1198621and 119862


Define 119891 as 119891(119906119894V119895) = 119891

1(119906119894) (or 119891

2(V119895)) 119894 =


1 1198912 119891 we can obtain a (120594

119892V119905(119866)+

120594119892V119905(119867))-GVDTC of 119866 + 119867

3 Remarks



120594119892V119905 (119866) le lceil119899

2rceil (22)






120594119892V119905 (119867) gt 120594119892V119905 (119866) (24)




120594119892V119905(11987515)(= 5) gt 120594119892V119905(11986514)(= 4)



Acknowledgments


References

































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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