Hindawi Publishing CorporationISRN CombinatoricsVolume 2013 Article ID 984549 6 pageshttpdxdoiorg1011552013984549

Research Article119896-Tuple Total Restrained Domination in Complementary Prisms

Adel P Kazemi

Department of Mathematics University of Mohaghegh Ardabili PO BOX 5619911367 Ardabil Iran

Correspondence should be addressed to Adel P Kazemi adelpkazemiyahoocom

Received 8 August 2013 Accepted 18 September 2013

Academic Editors C da Fonseca A P Godbole A V Kelarev and B Wu

Copyright copy 2013 Adel P Kazemi 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

In a graph 119866 with 120575(119866) ge 119896 ge 1 a 119896-tuple total restrained dominating set 119878 is a subset of 119881(119866) such that each vertex of 119881(119866) isadjacent to at least 119896 vertices of 119878 and also each vertex of119881(119866) minus 119878 is adjacent to at least 119896 vertices of 119881(119866) minus 119878Theminimumnumberof vertices of such sets in119866 is the 119896-tuple total restrained domination number of119866 In [119896-tuple total restrained dominationdomaticin graphs BIMS] the author initiated the study of the 119896-tuple total restrained domination number in graphs In this paper wecontinue it in the complementary prism of a graph

1 Introduction

Let 119866 be a simple graph with the vertex set 119881 = 119881(119866) andthe edge set 119864 = 119864(119866) The order |119881| and size |119864| of 119866 aredenoted by 119899 = 119899(119866) and 119898 = 119898(119866) respectively The openneighborhood and the closed neighborhood of a vertex V isin 119881

are 119873119866(V) = 119906 isin 119881 | 119906V isin 119864 and 119873119866[V] = 119873119866(V) cup

V respectively Also the open neighborhood and the closedneighborhood of a subset 119883 sube 119881(119866) are 119873119866(119883) = cupVisin119883119873119866(V)and119873119866[119883] = 119873

119866(119883) cup119883 respectively The degree of a vertex

V isin 119881 is deg(V) = |119873(V)| The minimum and maximumdegree of a graph 119866 are denoted by 120575 = 120575(119866) and Δ = Δ(119866)respectively If every vertex of119866 has degree 119896 then119866 is called119896-regular We write 119870

119899 119862119899 and 119875

119899for a complete graph a

cycle and a path of order 119899 respectively while1198701198991119899119901

denotesa complete 119901-partite graph The complement of a graph 119866 isdenoted by 119866 and is a graph with the vertex set 119881(119866) andfor every two vertices V and 119908 V119908 isin 119864(119866) if and only ifV119908 notin 119864(119866)

For each integer 119896 ge 1 the 119896-join 119866∘119896119867 of a graph 119866 to

a graph 119867 of order at least 119896 is the graph obtained from thedisjoint union of 119866 and 119867 by joining each vertex of 119866 to atleast 119896 vertices of119867 [1] Also119866∘

lowast119896119867 denotes the 119896-join119866∘

119896119867

such that each vertex of 119866 is joined to exact 119896 vertices of119867The complementary prism 119866119866 of 119866 is the graph formed

from the disjoint union119866cup119866 of119866 and119866 by adding the edgesof a perfect matching between the corresponding vertices

(same label) of 119866 and 119866 [2] For example the graph 11986251198625 isthe Petersen graph Also if 119866 = 119870

119899 the graph 119870

119899119870119899is the

corona 119870119899∘ 1198701 where the corona 119866 ∘ 119870

1of a graph 119866 is the

graph obtained from 119866 by attaching a pendant edge to eachvertex of 119866

The research of domination in graphs is an evergreen areaof graph theory Its basic concept is the dominating set Theliterature on this subject has been surveyed and detailed inthe two books by Haynes et al [3 4] And many variants ofthe dominating set were introduced and the correspondingnumerical invariants were defined for them For example the119896-tuple total dominating set is defined in [1] by Henning andKazemi which is an extension of the total dominating set (formore information see [5 6])

Definition 1 (see [1]) Let 119896 ge 1 be an integer and let 119866 be agraph with 120575(119866) ge 119896 A subset 119878 sube 119881(119866) is called a 119896-tupletotal dominating set briefly kTDS in 119866 if for each 119909 isin 119881(119866)|119873(119909)cap119878| ge 119896 Theminimum number of vertices of a 119896-tupletotal dominating set in a graph 119866 is called the 119896-tuple totaldomination number of 119866 and denoted by 120574times119896119905(119866)

A numerical invariant of a graph which is in a certainsense dual to it is the domatic number of a graphThedomaticnumber 119889(119866) and the total domatic number 119889

119905(119866) of a graph

were introduced in [7 8] respectively Sheikholeslami andVolkmann extended the last definition to the 119896-tuple totaldomatic number 119889

times119896119905(119866) in [9] and Kazemi extended it to

the star 119896-tuple total domatic number 119889lowasttimes119896119905

(119866) in [10]

2 ISRN Combinatorics

Definition 2 The 119896-tuple total domatic partition brieflykTDP of 119866 is a partition D of the vertex set of 119866 such that allclasses of D are 119896-tuple total dominating sets in 119866 The max-imum number of classes of a 119896-tuple total domatic partitionof 119866 is called the 119896-tuple total domatic number 119889

times119896119905(119866) of 119866

[9]The star 119896-tuple total domatic number 119889lowast

times119896119905(119866) of119866 is the

maximum number of classes of a kTDP of119866 such that at leastone of the 119896-tuple total dominating sets in it has cardinality120574times119896119905(119866) [10]

The author in [10] initiated the study of the 119896-tupletotal restrained domination number and the 119896-tuple total re-strained domatic number of graphs

Definition 3 (see [10]) The 119896-tuple total restrained domaticpartition briefly kTRDP of 119866 is a partition D of the vertexset of 119866 such that all classes of D are 119896-tuple total restraineddominating sets in 119866 The maximum number of classes of a119896-tuple total restrained domatic partition of 119866 is the 119896-tupletotal restrained domatic number 119889

119903

times119896119905(119866) of 119866 Similarly the

star 119896-tuple total restrained domatic number 119889119903lowast

times119896119905(119866) of 119866 is

the maximum number of classes of a kTRDP of 119866 such thatat least one of the 119896-tuple total restrained dominating sets init has cardinality 120574

119903

times119896119905(119866)

In this paper we continue our studies which is initiatedin [10] and find some sharp bounds for the 119896-tuple totalrestrained domination number of the complementary prismof a graph Also we will find the 119896-tuple total restraineddomination number of a cycle a path and a completemultipartite graph

Through this paper 119896 is a positive integer and forsimplicity we assume that119881(119866119866) is the disjoint union119881(119866)cup

119881(119866) with 119881(119866) = V | V isin 119881(119866) and 119864(119866119866) = 119864(119866) cup

119864(119866) cup VV | V isin 119881(119866) such that 119864(119866) = 119906 V | 119906V notin 119864(119866)The vertices V and V are called corresponding vertices Also fora subset 119883 sube 119881(119866) we show its corresponding subset in 119866

by 119883 Also we assume that 119881(119862119899) = 119881(119875

119899) = 119894 | 1 le 119894 le 119899

119881(119862119899) = 119881(119875

119899) = 119894 | 1 le 119894 le 119899119864(119862

119899) = 119864(119875

119899)cup1119899 = 119894119895 |

1 le 119894 le 119899 minus 1 and 119895 = 119894 + 1 cup 1119899The next known results are useful for our investigations

Proposition 4 (see Henning and Kazemi [1] 2010) Let 119901 ge 2

be an integer and let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 1198991 le 1198992 le sdot sdot sdot le 119899119901

(i) If 119896 lt 119901 then 120574times119896119905(119866) = 119896 + 1

(ii) If 119896 = 119901 and sum119896minus1

119894=1119899119894 ge 119896 then 120574times119896119905(119866) = 119896 + 2

(iii) If 2 le 119901 lt 119896 and lceil119896(119901 minus 1)rceil le 1198991le 1198992le sdot sdot sdot le 119899

119901

then 120574times119896119905

(119866) = lceil119896119901(119901 minus 1)rceil

Proposition 5 (see Kazemi [6] 2011) Let 119899 ge 4 Then

120574119905(119875119899119875119899) =

2lceil119899 minus 2

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899 minus 2

4rceil + 2 119900119905ℎ119890119903119908119894119904119890

(1)

Proposition 6 (see Kazemi [6] 2011) Let 119899 ge 4 Then

120574119905 (119862119899119862119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(2)

Proposition 7 (see Kazemi [6] 2011) If n ge 5 then120574times2119905

(119862119899119862119899) = 119899 + 2

Proposition 8 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

be acomplete119901-partite graphwith119881(119866119866) = ⋃

1le119894le119901(119883119894cup119883119894) when

for each 119894119883119894 is isomorph to119870119899

119894

If 119878 is a kTDS of 119866119866 then foreach 1 le 119894 le 119901 |119878 cap 119883

119894| ge 119896 Furthermore if |119878 cap 119883

119894| = 119896 for

some 119894 then |119878 cap 119883119894| ge 119896

Proposition 9 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

be acomplete 119901-partite graph with 1 le 1198991 le 119899

2le sdot sdot sdot le 119899

119901 Then

120574119905(119866119866) = 2119901 minus 120572 (3)

where 120572 = |119894 | 1 le 119894 le 119901 and 119899119894= 1|

Proposition 10 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph with 3 le 1198991 le 1198992 le sdot sdot sdot le 119899119901 Then120574times2119905

(119866119866) = 3119901 + 2

Proposition 11 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

be acomplete 119901-partite graph with 5 le 119896 + 2 le 1198991 le sdot sdot sdot le 119899119901 Then

120574times119896119905

(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(4)

Proposition 12 (see Kazemi [10] 2011) Let 119866 be a graph oforder 119899 in which 120575(119866) ge 119896 Then

(i) every vertex of degree at most 2119896 minus 1 of 119866 and at leastits 119896 neighbors belong to every kTRDS

(ii) 119889119903times119896119905

(119866) = 1 if 120575(119866) le 2119896 minus 1

(iii) Δ(119866) ge 2119896 if 120574119903times119896119905

(119866) lt 119899 Hence 119899 ge 2119896 + 2

(iv) 120574times119896119905(119866) le 120574119903

times119896119905(119866) and so 119889

119903

times119896119905(119866) le 119889times119896119905(119866)

Proposition 13 (see Kazemi [10] 2011) Let 119866 be a graph withminimum degree at least 119896 If 119889lowast

times119896119905(119866) ge 2 then 120574

119903

times119896119905(119866) =

120574times119896119905

(119866)

ISRN Combinatorics 3

Proposition 14 (see Kazemi [10] 2011) Let 119899 ge 119896 + 3 ge 4Then

120574119903

times119896119905(119862119899) =

119899 119894119891 119899 le 2119896 + 2

119896 + 2 119894119891 2119896 + 3 le 119899 le 3119896 + 2

119896 + 1 119894119891 119899 ge 3119896 + 3

(5)

Proposition 15 (see Kazemi [10] 2011) Let 119896 lt 119899 be positiveintegers Then

120574119903

times119896119905(119870119899) =

119899 119894119891 119899 le 2119896 + 1

119896 + 1 Otherwise(6)

Proposition 16 (see Kazemi [10] 2011) Let 119899 ge 4 Then

120574119903

119905(119862119899) =

2lceil119899

4rceil minus 1 119894119891 119899 equiv 1 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(7)

2 Some Bounds

We first give a sharp lower bound for the 119896-tuple total re-strained domination number of a regular graph

Theorem 17 Let 119896 and ℓ be integers such that 1 le 119896 minus 1 le ℓ le

2119896 minus 2 If 119866 is a ℓ-regular graph of order 119899 then

120574119903

times119896119905(119866119866) ge 119899 + 119896 (8)

with equality if and only if 119899 ge ℓ + 2119896 and 119881(119866) contains a 119896-subset 119879 such that for each vertex 119894 isin 119881(119866) |119873(119894) cap 119879| ge 119896 minus 1

and also if 119894 isin 119881(119866) minus 119879 then |119873(119894) cap (119881(119866) minus 119879)| ge 119896

Proof Let 119881(119866119866) = 119881(119866) cup 119881(119866) such that 119881(119866) = 119894 | 1 le

119894 le 119899 and 119881(119866) = 119894 | 1 le 119894 le 119899 Let 119899 ge 2119896 + ℓ and let119878 be an arbitrary kTRDS of 119866119866 Then Proposition 12(i) andthis fact that every vertex 119894 isin 119881(119866) has degree ℓ + 1 le 2119896 minus 1

imply that119881(119866) sube 119878 Let 119894 notin 119878 Then |119873(119894)cap119881(119866)cap119878| ge 119896minus1If |119873(119894) cap119881(119866)cap119878| ge 119896 then we have nothing to proveThuslet 119873(119894) cap 119881(119866) cap 119878 = 119895

119898| 1 le 119898 le 119896 minus 1 But this implies

that there exists a vertex 119905 isin 119878minus119895119898

| 1 le 119898 le 119896minus1 such thatits corresponding vertex 119905 in 119881(119866) is adjacent to some vertex119895119898 when 1 le 119898 le 119896 minus 1 Then |119878| ge 119899 + 119896 and since 119878 wasarbitrary we conclude that 120574119903

times119896119905(119866119866) ge 119899 + 119896

Obviously it can be seen that 120574119903times119896119905

(119866119866) = 119899+119896 if and onlyif 119899 ge ℓ+2119896 and119881(119866) contains a 119896-subset119879 such that for eachvertex 119894 isin 119881(119866) |119873(119894) cap 119879| ge 119896 minus 1 and also if 119894 isin 119881(119866) minus 119879then |119873(119894) cap (119881(119866) minus 119879)| ge 119896

Theorem 17 and Proposition 12(i) imply the next corol-lary

Corollary 18 Let 119896 and ℓ be integers such that 1 le 119896 minus 1 le

ℓ le 2119896 minus 2 If 119866 is a ℓ-regular graph of order 119899 le ℓ + 2119896 minus 1then 120574

119903

times119896(119866119866) = 2119899

The next theorem gives lower and upper bounds for120574119903

times119896119905(119866119866) when 119866 is an arbitrary graph

Theorem 19 If 119866 is a graph of order 119899 with 119896 le

min120575(119866) 120575(119866) then

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) le 120574

119903

times119896119905(119866119866)

le 120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

(9)

when 119896 ge 1 in the upper bound and 119896 ge 2 in the lower bound

Proof Toprove 120574119903times(119896minus1)119905

(119866)+120574119903

times(119896minus1)119905(119866) le 120574

119903

times119896119905(119866119866) let 119896 ge 2

and let119863 be a kTRDSof119866119866 Since every vertex of119881(119866) (resp119881(119866)) is adjacent to only one vertex of119881(119866) (resp119881(119866)) weconclude that119863cap119881(119866) is a (119896 minus 1) TRDS of 119866 and119863cap119881(119866)

is a (119896 minus 1) TRDS of 119866 Then

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)le|119863 cap 119881 (119866)|+

10038161003816100381610038161003816119863 cap 119881 (119866)

10038161003816100381610038161003816

= |119863| = 120574119903

times119896119905(119866119866)

(10)

Wenowprove that 120574119903times119896119905

(119866119866) le 120574119903

times119896119905(119866)+120574

119903

times119896119905(119866) when 119896 ge 1

Since for every kTRDS 119878 of 119866 and every kTRDS 1198781015840 of 119866 the

set 119878 cup 1198781015840 is a kTRDS of 119866119866 we have

120574119903

times119896119905(119866119866) le 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (11)

In Section 4 we will show that the given bounds inTheorem 19 are sharp

3 The Complementary Prism of Some Graphs

In this section we will determine 120574119903times119896119905

(119866119866)when119866 is a cyclea path or a completemultipartite graph First let119866 be a cycle

Proposition 20 Let 119899 ge 4 Then

120574119903

times2119905(119862119899119862119899) =

2119899 119894119891 119899 = 4 5

119899 + 2 119894119891 119899 ge 6(12)

Proof Corollary 18 implies that 120574119903times2119905

(119862119899119862119899) = 2119899 if 119899 = 4 5

Now let 119899 ge 6 Proposition 7 with Proposition 12(iv) implies120574119903

times2119905(119862119899119862119899) ge 119899 + 2 Since also 119881(119862119899) cup 1 4 is a 2TRDS of

119862119899119862119899 we get 120574119903

times2119905(119862119899119862119899) = 119899 + 2 when 119899 ge 6

To calculate 120574119905(119862119899119862119899) we need to prove that 119889lowast

119905(119862119899119862119899) ge

2

Proposition 21 Let 119899 ge 4 Then 119889lowast

119905(119862119899119862119899) ge 2

Proof We prove the proposition in the following four cases

Case 1 (119899 equiv 0 (mod 4)) For 119899 = 4 we set 119878 = 1 1 2 2 and1198781015840= 3 3 4 4 If 119899 gt 4 we set 119878 = 1 1 2 2 cup 5 + 4119894 6 + 4119894 |

0 le 119894 le lceil1198994rceil minus 2 and 1198781015840 = 3 3 4 4 cup 7 + 4119894 8 + 4119894 | 0 le

119894 le lceil1198994rceil minus 2

4 ISRN Combinatorics

Case 2 (119899 equiv 1 (mod 4)) For 119899 = 5 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 and for 119899 = 9 we set 119878 = 1 1 4 4 7 7 and 119878

1015840

= 2 2 5 5 8 8 If 119899 gt 9 we set 119878 = 1 1 4 4 7 7 cup 10 +

4119894 11+4119894 | 0 le 119894 le lceil1198994rceilminus4 and 1198781015840= 3 3 6 6 9 9 cup 12+

4119894 13 + 4119894 | 0 le 119894 le lceil1198994rceil minus 4

Case 3 (119899 equiv 2 (mod 4)) For 119899 = 6 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 If 119899 gt 6 we set 119878 = 1 1 4 4 cup 7+4119894 8+4119894 |

0 le 119894 le lceil1198994rceil minus 3 and 1198781015840= 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le

119894 le lceil1198994rceil minus 3

Case 4 (119899 equiv 3 (mod 4)) For 119899 = 7 we set 119878 = 1 1 4 4 6

and 1198781015840

= 2 2 5 5 7 If 119899 gt 7 we set 119878 = 1 1 4 4

119899 minus 1 cup 7 + 4119894 8 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3 and 1198781015840

=

2 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3Since in all cases 119878 and 119878

1015840 are two disjoint 120574119905(119862119899119862119899)-setswe have 119889lowast

119905(119862119899119862119899) ge 2

By Propositions 6 13 and 21 we obtain the next result

Proposition 22 Let 119899 ge 4 Then

120574119903

119905(119862119899119862119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(13)

Now we continue our work when 119866 is a path

Proposition 23 Let 119899 ge 4 Then

120574119903

119905(119875119899119875119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(14)

Proof Propositions 5 and 12(iv) imply that

120574119903

119905(119875119899119875119899) ge 120574119905 (119875119899119875119899)

=

2lceil119899 minus 2

4rceil + 1 119894119891 119899equiv0 (mod 4)

2 lceil119899 minus 2

4rceil + 2 otherwise

(15)

Let 119899 equiv 0 (mod 4) For 119899= 8 we set 119878= 1 8 3 4 5 6

and for 119899 gt 8 we set

119878 = 1 119899 minus 6 119899 minus 5 119899 119899 minus 3 119899 minus 2

cup 3 + 4119894 4 + 4119894 | 0 le 119894 le lfloor119899

4rfloor minus 3

(16)

If 119899 equiv 1 2 3 (mod4) then we set 119878= 1 119899 minus 2 119899 119899 minus

2 cup 3+4119894 4+4119894 | 0 le 119894 le lfloor1198994rfloorminus2 119878= 1 119899 cup 3+4119894 4+4119894 |

0 le 119894 le lfloor1198994rfloorminus1 and 119878= 1 119899 minus 1 119899cup3+4119894 4+4119894 | 0 le 119894 le

lfloor1198994rfloorminus1 respectively Since in each case 119878 is a TRDS of119875119899119875119899

of cardinality 120574119905(119875119899119875119899) we have completed our proof

In the next propositions we calculate 120574119903

times119896119905(119866119866) when 119866

is a complete multipartite graph

Proposition 24 If 119866 = 11987011989911198992119899119901

is a complete 119901-partitegraph then

119889119905(119866119866) = min 119899

119894| 1 le 119894 le 119901 (17)

Proof Let 119881(119866119866) = ⋃1le119894le119901

(119883119894cup 119883119894) and let ℓ = min119899

119894|

1 le 119894 le 119901 Proposition 8 implies that for every TDS119863 of119866119866|119863 cap 119883

119894| ge 1 when 1 le 119894 le 119901 Hence 119889

119905(119866119866) le ℓ Now let

11986311198632 and119863

ℓ be ℓ disjoint 2119901-sets of119881(119866119866) such that forevery 119895 = 1 2 ℓ and every 119894 = 1 2 119901 |119863119895cap119883

119894| = |119863

119895cap

119883119894| = 1 and 119909 isin 119863

119895cap 119883119894if and only if 119909 isin 119863

119895cap 119883119894 Since1198631

1198632 and 119863

ℓ are ℓ disjoint 120574119905-sets of 119866119866 by Proposition 9

we get 119889119905(119866119866) ge ℓ and so 119889119905(119866119866) = ℓ

Proposition 25 If 119866 = 1198701198991119899119901

is a complete 119901-partite graphwith 2 le 119899

1le sdot sdot sdot le 119899

119901 then 120574

119903

119905(119866119866) = 2119901

Proof Since obviously 119889119905(119866119866)= 119889

lowast

119905(119866119866) we obtain 120574

119903

119905(119866119866) =

2119901 by Propositions 9 13 and 24

Proposition 26 Let 119866 = 11987011989911198992

be a complete bipartite graphwith 4 le 119899

1le 1198992 Then 120574

119903

times2119905(119866119866) = 8

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878cap119883119894| =

|119878 cap119883119894| = 2 and 119909 isin 119878 cap119883

119894if and only if 119909 isin 119878 cap119883

119894 Since 119878 is

a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) we get 120574119903times2119905

(119866119866) = 8by Propositions 10 and 12(iv)

Proposition 27 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 5 le 1198991 le sdot sdot sdot le 119899119901 Then 120574

119903

times2119905(119866119866) = 3119901 + 2

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878 cap 119883119894| =

|119878 cap 119883119894| = 2 and 119909 isin 119878 cap 119883

119894if and only if 119909 isin 119878 cap 119883

119894 and for

119894 = 3 119901 |119878 cap 119883119894| = 3 Obviously 119878 is a 120574times2119905(119866119866)-set Itcan easily be verified that every vertex in out of 119878 is adjacentto at least two vertices in out of 119878 Hence 119878 is a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) and so 120574

119903

times2119905(119866119866) = 3119901 + 2 by

Propositions 10 and 12(iv)

Proposition28 Let119866=119870119899111989921198993

be a complete 3-partite graphwith 6 le 119899

1 le 1198992 le 1198993 Then 120574119903

times3119905(119866119866) = 16

ISRN Combinatorics 5

Proof Let 119878 be a set of vertices such that |119878 cap 1198833| = 4 and

for 119894 = 1 2 |119878 cap 119883119894| = |119878 cap 119883

119894| = 3 such that 119909 isin 119878 cap

119883119894if and only if 119909 isin 119878 cap 119883

119894 Since 119878 is a 3TRDS of 119866119866 of

cardinality 120574times3119905

(119866119866) we get 120574119903times3119905

(119866119866) = 16 by Propositions11 and 12(iv)

Proposition 29 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 7 le 2119896 + 1 le 119899

1le sdot sdot sdot le 119899

119901 Then

120574119903

times119896119905(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(18)

Proof We prove the proposition in the following three cases

Case 1 (119901 ge 119896 + 1 or 119901 = 119896 ge 4) Let 119878 be a set of verticessuch that 119878 cap 119881(119866) is a 120574

times119896119905(119866)-set and for 119894 = 1 2 119901

|119878 cap 119883119894| = 119896 + 1 Obviously 119878 is a 120574times119896119905(119866119866)-set It can easilybe verified that every vertex in out of 119878 is adjacent to at least 119896vertices in out of 119878 Hence 119878 is a kTRDS of 119866119866 of cardinality120574times119896119905

(119866119866) and so 120574119903

times119896119905(119866119866) is (119901+ 1)(119896+ 1) if 119901 ge 119896+1 and is

(119901 + 1)(119896 + 1) + 1 if 119901 = 119896 ge 4 by Propositions 11 and 12(iv)

Case 2 (119901 = 119896 = 3) This case is proved in Proposition 28when 6 = 2119896 le 119899

1le 1198992le 1198993

Case 3 (119901 lt 119896) If min2119896minus2 lceil119896119901(119901minus1)rceil = 2119896minus2 let 119878 be aset of vertices such that for 119894 = 1 2 |119878cap119883

119894| = |119878cap119883

119894| = 119896 and

119909 isin 119878 cap 119883119894 if and only if 119909 isin 119878 cap 119883119894 and for 119894 = 3 119901 |119878 cap

119883119894| = 119896 + 1 If min2119896 minus 2 lceil119896119901(119901 minus 1)rceil = lceil119896119901(119901 minus 1)rceil let 119878

be a set of vertices such that 119878 cap 119881(119866) is a 120574times119896119905

(119866)-set and for119894 = 1 2 119901 |119878 cap 119883

119894| = 119896 + 1 In all cases obviously 119878 is a

120574times119896119905

(119866119866)-set and it can easily be verified that every vertex inout of 119878 is adjacent to at least 119896 vertices in out of 119878 Hence 119878 isa kTRDS of 119866119866 of cardinality 120574times119896119905(119866119866) and so 120574

119903

times2119905(119866119866) =

119901(119896 + 1) + min2119896 minus 2 lceil119896119901(119901 minus 1)rceil by Propositions 11 and12(iv)

4 The Given Bounds in Theorem 19 Are Sharp

In this section we show that the given bounds inTheorem 19are sharp For this aim we first calculate the 119896-tuple totalrestrained domination number of a complete multipartitegraph and its complement

Proposition 30 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with 119896 + 1 le 119899

1le sdot sdot sdot le 119899

119905le 2119896 + 1 lt 119899

119905+1le

sdot sdot sdot le 119899119901 If 119901 ge 119896 + 1 then

120574119903

times119896119905(119866) = 1198991 + sdot sdot sdot + 119899119905 + (119901 minus 119905) (119896 + 1) (19)

Proof Since 119866 is the union of disjoint complete graphs 1198701198991

1198701198992

119870119899119901

Proposition 15 implies that

120574119903

times119896119905(119866) = 120574

119903

times119896119905(1198701198991

) + 120574119903

times119896119905(1198701198992

) + sdot sdot sdot 120574119903

times119896119905(119870119899119901

)

= 1198991+ sdot sdot sdot + 119899

119905+ (119901 minus 119905) (119896 + 1)

(20)

Proposition 31 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraphwith 2 le 1198991 le sdot sdot sdot le 119899119901 If119901 ge 119896+1 then 120574

119903

times119896119905(119866) = 119896+1

Proof Let 119878 be a set of vertices of cardinality 119896 + 1 such thatfor every index 119894 = 1 2 119896 + 1 |119878 cap 119883119894| = 1 Since 119878 isa kTRDS of 119866 of cardinality 120574times119896119905(119866) Propositions 12(iv) and4(i) imply that 120574119903

times119896119905(119866) = 119896 + 1

Proposition 32 Let 119866 = 11987011989911198992119899119896

be a complete 119896-partitegraph of order 119899 with 1198991 le 1198992 le sdot sdot sdot le 119899119896 If 119899minus 119899119896minus1 ge 119899minus119899119896 ge

2119896 and 119899 minus 1198991 ge sdot sdot sdot ge 119899 minus 119899119896minus2 ge 2119896 + 1 then 120574119903

times119896119905(119866) = 119896 + 2

Proof Let 119878 be a set of vertices of cardinality 119896 + 2 such that|119878cap119883119894| = 2 for 119894 = 119896minus1 119896 and |119878 cap 119883

119894| = 1 for 119894 = 1 2 119896minus

2 Since 119878 is a kTRDS of119866 of cardinality 120574times119896119905

(119866) Propositions4(ii) and 12(iv) imply that 120574119903

times119896119905(119866) = 119896 + 2

Proposition 33 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order with lceil119896(119901 minus 1)rceil le 119899

1le sdot sdot sdot le 119899

119901 and let

119901 lt 119896 If 119901 minus 1 | 119896 and for each index 119894 = 1 2 119901 we have119899 minus 119899119894ge 2119896 then 120574

119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proof Let 119878 be a set of vertices of cardinality 119896119901(119901 minus 1) suchthat for each 119894 = 1 2 119901 |119878 cap 119883

119894| = 119896(119901 minus 1) Since 119878 is

a kTRDS of 119866 of cardinality 120574times119896119905

(119866) Propositions 4(iii) and12(iv) imply that 120574119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proposition 34 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with lceil119896(119901 minus 1)rceil le 1198991 le sdot sdot sdot le 119899119901 and let119901 lt 119896 Let also 119896 = (119901 minus 1) sdot lfloor119896(119901 minus 1)rfloor + 119903 for some integer1 le 119903 le 119901 minus 2 If 119899 minus 119899119901minus119903 ge 119899 minus 119899

119901minus119903+1ge sdot sdot sdot ge 119899 minus 119899

119901ge

lceil119896119901(119901minus 1)rceil minus lceil119896(119901minus 1)rceil + 119896 and 119899minus1198991 ge sdot sdot sdot ge 119899minus 119899119901minus119903minus1 ge

lceil119896119901(119901minus1)rceilminuslceil119896(119901minus1)rceil+119896+1 then 120574119903

times119896119905(119866) = lceil119896119901(119901minus1)rceil

Proof Let 119878 be a set of vertices of cardinality 1 le 119894 le 119901minus119903minus1|119878 cap 119883

119894| = lfloor119896(119901 minus 1)rfloor and for 119901 minus 119903 le 119894 le 119901 |119878 cap 119883

119894| =

lceil119896(119901 minus 1)rceil = lfloor119896(119901 minus 1)rfloor + 1 Since 119878 is a kTRDS of 119866 ofcardinality 120574

times119896119905(119866) Propositions 4(iii) and 12(iv) imply that

120574119903

times119896119905(119866) = lceil119896119901(119901 minus 1)rceil

Comparing Propositions 29 30 31 32 33 and 34 showsthat the following two results which state the given boundsinTheorem 19 are sharp are proved

Proposition 35 For every integer 119896 ge 1 let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph of order 119899 with 2119896 + 2 le 119899

1le 1198992le

sdot sdot sdot le 119899119901 If either 119901 ge 119896 + 1 or 119901 = 119896 ge 4 and 2119896 le 119899 minus 119899

119896le

119899 minus 119899119896minus1

2119896 + 1 le 119899 minus 119899119896minus2

le sdot sdot sdot le 119899 minus 1198991 then

120574119903

times119896119905(119866119866) = 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (21)

6 ISRN Combinatorics

Proposition 36 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph and let 119896 ge 1 be an integer If 119896 le 119899

1le sdot sdot sdot le 119899

119905le

2119896 minus 1 lt 119899119905+1 le sdot sdot sdot le 119899119901and 1198991+ 1198992+ sdot sdot sdot + 119899

119905= 119901 + 1 + 119905119896 for

some integer 119905 ge 1 then

120574119903

times119896119905(119866119866) = 120574

119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) (22)

We note that there are some complete multipartite graphsthat satisfy Proposition 36 For example if 119896 ge 4 then thecomplete (119896 + 1)-partite graph 119866 = 119870

11989911198992119899119896+1

with theconditions 119899

1= 119896 + 3 119899

2= 2119896 minus 1 and 2119896 le 119899

3le sdot sdot sdot le 119899

119896+1

satisfies itAlso Propositions 14 16 and 20 by this fact that

120574119903

times2119905(119862119899) = 119899 which is obtained by Proposition 12(i) imply

that

120574119903

times2119905(119862119899119862119899) = 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899) (23)

if and only if 119899 = 5 Also they imply that

120574119903

119905(119862119899) + 120574119903

119905(119862119899) lt 120574119903

times2119905(119862119899119862119899)

lt 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899)

(24)

when 119899 ge 4 in the lower bound and 119899 ge 6 in the upper bound

5 Open Problem

Finally we finish our paper with the following problems

Problem 37 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

Problem 38 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)

References

[1] M A Henning and A P Kazemi ldquo119896-tuple total domination ingraphsrdquo Discrete Applied Mathematics vol 158 no 9 pp 1006ndash1011 2010

[2] T W Haynes M A Henning P J Slater and L C van derMerwe ldquoThe complementary product of two graphsrdquo Bulletinof the Institute of Combinatorics and Its Applications vol 51 pp21ndash30 2007

[3] TW Haynes S T Hedetniemi and P J Slater Fundamentals ofDomination in Graphs vol 208 ofMonographs and Textbooks inPure and Applied Mathematics Marcel Dekker New York NYUSA 1998

[4] TWHaynes S THedetniemi and P J Slater EdsDominationin Graphs Advanced Topics Marcel Dekker New York NYUSA 1998

[5] M A Henning and A P Kazemi ldquo119896-tuple total domination incross products of graphsrdquo Journal of Combinatorial Optimiza-tion vol 24 no 3 pp 339ndash346 2012

[6] A P Kazemi ldquo119896-tuple total domination in complementaryprismsrdquo ISRNDiscreteMathematics vol 2011 Article ID68127413 pages 2011

[7] E J Cockayne and S T Hedetniemi ldquoTowards a theory ofdomination in graphsrdquoNetworks vol 7 no 3 pp 247ndash261 1977

[8] E J Cockayne R M Dawes and S T Hedetniemi ldquoTotaldomination in graphsrdquoNetworks vol 10 no 3 pp 211ndash219 1980

[9] S M Sheikholeslami and L Volkmann ldquoThe 119896-tuple totaldomatic number of a graphrdquo submitted to Utilitas Mathemat-ica httparxivorgabs11065589

[10] A P Kazemi ldquo119896-tuple total restrained dominationdomatic ingraphsrdquo Bulletin of the IranianMathematical Sciences accepted

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 ISRN Combinatorics

Definition 2 The 119896-tuple total domatic partition brieflykTDP of 119866 is a partition D of the vertex set of 119866 such that allclasses of D are 119896-tuple total dominating sets in 119866 The max-imum number of classes of a 119896-tuple total domatic partitionof 119866 is called the 119896-tuple total domatic number 119889

times119896119905(119866) of 119866

[9]The star 119896-tuple total domatic number 119889lowast

times119896119905(119866) of119866 is the

maximum number of classes of a kTDP of119866 such that at leastone of the 119896-tuple total dominating sets in it has cardinality120574times119896119905(119866) [10]

The author in [10] initiated the study of the 119896-tupletotal restrained domination number and the 119896-tuple total re-strained domatic number of graphs

Definition 3 (see [10]) The 119896-tuple total restrained domaticpartition briefly kTRDP of 119866 is a partition D of the vertexset of 119866 such that all classes of D are 119896-tuple total restraineddominating sets in 119866 The maximum number of classes of a119896-tuple total restrained domatic partition of 119866 is the 119896-tupletotal restrained domatic number 119889

119903

times119896119905(119866) of 119866 Similarly the

star 119896-tuple total restrained domatic number 119889119903lowast

times119896119905(119866) of 119866 is

the maximum number of classes of a kTRDP of 119866 such thatat least one of the 119896-tuple total restrained dominating sets init has cardinality 120574

119903

times119896119905(119866)

In this paper we continue our studies which is initiatedin [10] and find some sharp bounds for the 119896-tuple totalrestrained domination number of the complementary prismof a graph Also we will find the 119896-tuple total restraineddomination number of a cycle a path and a completemultipartite graph

Through this paper 119896 is a positive integer and forsimplicity we assume that119881(119866119866) is the disjoint union119881(119866)cup

119881(119866) with 119881(119866) = V | V isin 119881(119866) and 119864(119866119866) = 119864(119866) cup

119864(119866) cup VV | V isin 119881(119866) such that 119864(119866) = 119906 V | 119906V notin 119864(119866)The vertices V and V are called corresponding vertices Also fora subset 119883 sube 119881(119866) we show its corresponding subset in 119866

by 119883 Also we assume that 119881(119862119899) = 119881(119875

119899) = 119894 | 1 le 119894 le 119899

119881(119862119899) = 119881(119875

119899) = 119894 | 1 le 119894 le 119899119864(119862

119899) = 119864(119875

119899)cup1119899 = 119894119895 |

1 le 119894 le 119899 minus 1 and 119895 = 119894 + 1 cup 1119899The next known results are useful for our investigations

Proposition 4 (see Henning and Kazemi [1] 2010) Let 119901 ge 2

be an integer and let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 1198991 le 1198992 le sdot sdot sdot le 119899119901

(i) If 119896 lt 119901 then 120574times119896119905(119866) = 119896 + 1

(ii) If 119896 = 119901 and sum119896minus1

119894=1119899119894 ge 119896 then 120574times119896119905(119866) = 119896 + 2

(iii) If 2 le 119901 lt 119896 and lceil119896(119901 minus 1)rceil le 1198991le 1198992le sdot sdot sdot le 119899

119901

then 120574times119896119905

(119866) = lceil119896119901(119901 minus 1)rceil

Proposition 5 (see Kazemi [6] 2011) Let 119899 ge 4 Then

120574119905(119875119899119875119899) =

2lceil119899 minus 2

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899 minus 2

4rceil + 2 119900119905ℎ119890119903119908119894119904119890

(1)

Proposition 6 (see Kazemi [6] 2011) Let 119899 ge 4 Then

120574119905 (119862119899119862119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(2)

Proposition 7 (see Kazemi [6] 2011) If n ge 5 then120574times2119905

(119862119899119862119899) = 119899 + 2

Proposition 8 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

be acomplete119901-partite graphwith119881(119866119866) = ⋃

1le119894le119901(119883119894cup119883119894) when

for each 119894119883119894 is isomorph to119870119899

119894

If 119878 is a kTDS of 119866119866 then foreach 1 le 119894 le 119901 |119878 cap 119883

119894| ge 119896 Furthermore if |119878 cap 119883

119894| = 119896 for

some 119894 then |119878 cap 119883119894| ge 119896

Proposition 9 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

be acomplete 119901-partite graph with 1 le 1198991 le 119899

2le sdot sdot sdot le 119899

119901 Then

120574119905(119866119866) = 2119901 minus 120572 (3)

where 120572 = |119894 | 1 le 119894 le 119901 and 119899119894= 1|

Proposition 10 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph with 3 le 1198991 le 1198992 le sdot sdot sdot le 119899119901 Then120574times2119905

(119866119866) = 3119901 + 2

Proposition 11 (see Kazemi [6] 2011) Let 119866 = 11987011989911198992119899119901

be acomplete 119901-partite graph with 5 le 119896 + 2 le 1198991 le sdot sdot sdot le 119899119901 Then

120574times119896119905

(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(4)

Proposition 12 (see Kazemi [10] 2011) Let 119866 be a graph oforder 119899 in which 120575(119866) ge 119896 Then

(i) every vertex of degree at most 2119896 minus 1 of 119866 and at leastits 119896 neighbors belong to every kTRDS

(ii) 119889119903times119896119905

(119866) = 1 if 120575(119866) le 2119896 minus 1

(iii) Δ(119866) ge 2119896 if 120574119903times119896119905

(119866) lt 119899 Hence 119899 ge 2119896 + 2

(iv) 120574times119896119905(119866) le 120574119903

times119896119905(119866) and so 119889

119903

times119896119905(119866) le 119889times119896119905(119866)

Proposition 13 (see Kazemi [10] 2011) Let 119866 be a graph withminimum degree at least 119896 If 119889lowast

times119896119905(119866) ge 2 then 120574

119903

times119896119905(119866) =

120574times119896119905

(119866)

ISRN Combinatorics 3

Proposition 14 (see Kazemi [10] 2011) Let 119899 ge 119896 + 3 ge 4Then

120574119903

times119896119905(119862119899) =

119899 119894119891 119899 le 2119896 + 2

119896 + 2 119894119891 2119896 + 3 le 119899 le 3119896 + 2

119896 + 1 119894119891 119899 ge 3119896 + 3

(5)

Proposition 15 (see Kazemi [10] 2011) Let 119896 lt 119899 be positiveintegers Then

120574119903

times119896119905(119870119899) =

119899 119894119891 119899 le 2119896 + 1

119896 + 1 Otherwise(6)

Proposition 16 (see Kazemi [10] 2011) Let 119899 ge 4 Then

120574119903

119905(119862119899) =

2lceil119899

4rceil minus 1 119894119891 119899 equiv 1 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(7)

2 Some Bounds

We first give a sharp lower bound for the 119896-tuple total re-strained domination number of a regular graph

Theorem 17 Let 119896 and ℓ be integers such that 1 le 119896 minus 1 le ℓ le

2119896 minus 2 If 119866 is a ℓ-regular graph of order 119899 then

120574119903

times119896119905(119866119866) ge 119899 + 119896 (8)

with equality if and only if 119899 ge ℓ + 2119896 and 119881(119866) contains a 119896-subset 119879 such that for each vertex 119894 isin 119881(119866) |119873(119894) cap 119879| ge 119896 minus 1

and also if 119894 isin 119881(119866) minus 119879 then |119873(119894) cap (119881(119866) minus 119879)| ge 119896

Proof Let 119881(119866119866) = 119881(119866) cup 119881(119866) such that 119881(119866) = 119894 | 1 le

119894 le 119899 and 119881(119866) = 119894 | 1 le 119894 le 119899 Let 119899 ge 2119896 + ℓ and let119878 be an arbitrary kTRDS of 119866119866 Then Proposition 12(i) andthis fact that every vertex 119894 isin 119881(119866) has degree ℓ + 1 le 2119896 minus 1

imply that119881(119866) sube 119878 Let 119894 notin 119878 Then |119873(119894)cap119881(119866)cap119878| ge 119896minus1If |119873(119894) cap119881(119866)cap119878| ge 119896 then we have nothing to proveThuslet 119873(119894) cap 119881(119866) cap 119878 = 119895

119898| 1 le 119898 le 119896 minus 1 But this implies

that there exists a vertex 119905 isin 119878minus119895119898

| 1 le 119898 le 119896minus1 such thatits corresponding vertex 119905 in 119881(119866) is adjacent to some vertex119895119898 when 1 le 119898 le 119896 minus 1 Then |119878| ge 119899 + 119896 and since 119878 wasarbitrary we conclude that 120574119903

times119896119905(119866119866) ge 119899 + 119896

Obviously it can be seen that 120574119903times119896119905

(119866119866) = 119899+119896 if and onlyif 119899 ge ℓ+2119896 and119881(119866) contains a 119896-subset119879 such that for eachvertex 119894 isin 119881(119866) |119873(119894) cap 119879| ge 119896 minus 1 and also if 119894 isin 119881(119866) minus 119879then |119873(119894) cap (119881(119866) minus 119879)| ge 119896

Theorem 17 and Proposition 12(i) imply the next corol-lary

Corollary 18 Let 119896 and ℓ be integers such that 1 le 119896 minus 1 le

ℓ le 2119896 minus 2 If 119866 is a ℓ-regular graph of order 119899 le ℓ + 2119896 minus 1then 120574

119903

times119896(119866119866) = 2119899

The next theorem gives lower and upper bounds for120574119903

times119896119905(119866119866) when 119866 is an arbitrary graph

Theorem 19 If 119866 is a graph of order 119899 with 119896 le

min120575(119866) 120575(119866) then

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) le 120574

119903

times119896119905(119866119866)

le 120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

(9)

when 119896 ge 1 in the upper bound and 119896 ge 2 in the lower bound

Proof Toprove 120574119903times(119896minus1)119905

(119866)+120574119903

times(119896minus1)119905(119866) le 120574

119903

times119896119905(119866119866) let 119896 ge 2

and let119863 be a kTRDSof119866119866 Since every vertex of119881(119866) (resp119881(119866)) is adjacent to only one vertex of119881(119866) (resp119881(119866)) weconclude that119863cap119881(119866) is a (119896 minus 1) TRDS of 119866 and119863cap119881(119866)

is a (119896 minus 1) TRDS of 119866 Then

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)le|119863 cap 119881 (119866)|+

10038161003816100381610038161003816119863 cap 119881 (119866)

10038161003816100381610038161003816

= |119863| = 120574119903

times119896119905(119866119866)

(10)

Wenowprove that 120574119903times119896119905

(119866119866) le 120574119903

times119896119905(119866)+120574

119903

times119896119905(119866) when 119896 ge 1

Since for every kTRDS 119878 of 119866 and every kTRDS 1198781015840 of 119866 the

set 119878 cup 1198781015840 is a kTRDS of 119866119866 we have

120574119903

times119896119905(119866119866) le 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (11)

In Section 4 we will show that the given bounds inTheorem 19 are sharp

3 The Complementary Prism of Some Graphs

In this section we will determine 120574119903times119896119905

(119866119866)when119866 is a cyclea path or a completemultipartite graph First let119866 be a cycle

Proposition 20 Let 119899 ge 4 Then

120574119903

times2119905(119862119899119862119899) =

2119899 119894119891 119899 = 4 5

119899 + 2 119894119891 119899 ge 6(12)

Proof Corollary 18 implies that 120574119903times2119905

(119862119899119862119899) = 2119899 if 119899 = 4 5

Now let 119899 ge 6 Proposition 7 with Proposition 12(iv) implies120574119903

times2119905(119862119899119862119899) ge 119899 + 2 Since also 119881(119862119899) cup 1 4 is a 2TRDS of

119862119899119862119899 we get 120574119903

times2119905(119862119899119862119899) = 119899 + 2 when 119899 ge 6

To calculate 120574119905(119862119899119862119899) we need to prove that 119889lowast

119905(119862119899119862119899) ge

2

Proposition 21 Let 119899 ge 4 Then 119889lowast

119905(119862119899119862119899) ge 2

Proof We prove the proposition in the following four cases

Case 1 (119899 equiv 0 (mod 4)) For 119899 = 4 we set 119878 = 1 1 2 2 and1198781015840= 3 3 4 4 If 119899 gt 4 we set 119878 = 1 1 2 2 cup 5 + 4119894 6 + 4119894 |

0 le 119894 le lceil1198994rceil minus 2 and 1198781015840 = 3 3 4 4 cup 7 + 4119894 8 + 4119894 | 0 le

119894 le lceil1198994rceil minus 2

4 ISRN Combinatorics

Case 2 (119899 equiv 1 (mod 4)) For 119899 = 5 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 and for 119899 = 9 we set 119878 = 1 1 4 4 7 7 and 119878

1015840

= 2 2 5 5 8 8 If 119899 gt 9 we set 119878 = 1 1 4 4 7 7 cup 10 +

4119894 11+4119894 | 0 le 119894 le lceil1198994rceilminus4 and 1198781015840= 3 3 6 6 9 9 cup 12+

4119894 13 + 4119894 | 0 le 119894 le lceil1198994rceil minus 4

Case 3 (119899 equiv 2 (mod 4)) For 119899 = 6 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 If 119899 gt 6 we set 119878 = 1 1 4 4 cup 7+4119894 8+4119894 |

0 le 119894 le lceil1198994rceil minus 3 and 1198781015840= 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le

119894 le lceil1198994rceil minus 3

Case 4 (119899 equiv 3 (mod 4)) For 119899 = 7 we set 119878 = 1 1 4 4 6

and 1198781015840

= 2 2 5 5 7 If 119899 gt 7 we set 119878 = 1 1 4 4

119899 minus 1 cup 7 + 4119894 8 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3 and 1198781015840

=

2 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3Since in all cases 119878 and 119878

1015840 are two disjoint 120574119905(119862119899119862119899)-setswe have 119889lowast

119905(119862119899119862119899) ge 2

By Propositions 6 13 and 21 we obtain the next result

Proposition 22 Let 119899 ge 4 Then

120574119903

119905(119862119899119862119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(13)

Now we continue our work when 119866 is a path

Proposition 23 Let 119899 ge 4 Then

120574119903

119905(119875119899119875119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(14)

Proof Propositions 5 and 12(iv) imply that

120574119903

119905(119875119899119875119899) ge 120574119905 (119875119899119875119899)

=

2lceil119899 minus 2

4rceil + 1 119894119891 119899equiv0 (mod 4)

2 lceil119899 minus 2

4rceil + 2 otherwise

(15)

Let 119899 equiv 0 (mod 4) For 119899= 8 we set 119878= 1 8 3 4 5 6

and for 119899 gt 8 we set

119878 = 1 119899 minus 6 119899 minus 5 119899 119899 minus 3 119899 minus 2

cup 3 + 4119894 4 + 4119894 | 0 le 119894 le lfloor119899

4rfloor minus 3

(16)

If 119899 equiv 1 2 3 (mod4) then we set 119878= 1 119899 minus 2 119899 119899 minus

2 cup 3+4119894 4+4119894 | 0 le 119894 le lfloor1198994rfloorminus2 119878= 1 119899 cup 3+4119894 4+4119894 |

0 le 119894 le lfloor1198994rfloorminus1 and 119878= 1 119899 minus 1 119899cup3+4119894 4+4119894 | 0 le 119894 le

lfloor1198994rfloorminus1 respectively Since in each case 119878 is a TRDS of119875119899119875119899

of cardinality 120574119905(119875119899119875119899) we have completed our proof

In the next propositions we calculate 120574119903

times119896119905(119866119866) when 119866

is a complete multipartite graph

Proposition 24 If 119866 = 11987011989911198992119899119901

is a complete 119901-partitegraph then

119889119905(119866119866) = min 119899

119894| 1 le 119894 le 119901 (17)

Proof Let 119881(119866119866) = ⋃1le119894le119901

(119883119894cup 119883119894) and let ℓ = min119899

119894|

1 le 119894 le 119901 Proposition 8 implies that for every TDS119863 of119866119866|119863 cap 119883

119894| ge 1 when 1 le 119894 le 119901 Hence 119889

119905(119866119866) le ℓ Now let

11986311198632 and119863

ℓ be ℓ disjoint 2119901-sets of119881(119866119866) such that forevery 119895 = 1 2 ℓ and every 119894 = 1 2 119901 |119863119895cap119883

119894| = |119863

119895cap

119883119894| = 1 and 119909 isin 119863

119895cap 119883119894if and only if 119909 isin 119863

119895cap 119883119894 Since1198631

1198632 and 119863

ℓ are ℓ disjoint 120574119905-sets of 119866119866 by Proposition 9

we get 119889119905(119866119866) ge ℓ and so 119889119905(119866119866) = ℓ

Proposition 25 If 119866 = 1198701198991119899119901

is a complete 119901-partite graphwith 2 le 119899

1le sdot sdot sdot le 119899

119901 then 120574

119903

119905(119866119866) = 2119901

Proof Since obviously 119889119905(119866119866)= 119889

lowast

119905(119866119866) we obtain 120574

119903

119905(119866119866) =

2119901 by Propositions 9 13 and 24

Proposition 26 Let 119866 = 11987011989911198992

be a complete bipartite graphwith 4 le 119899

1le 1198992 Then 120574

119903

times2119905(119866119866) = 8

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878cap119883119894| =

|119878 cap119883119894| = 2 and 119909 isin 119878 cap119883

119894if and only if 119909 isin 119878 cap119883

119894 Since 119878 is

a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) we get 120574119903times2119905

(119866119866) = 8by Propositions 10 and 12(iv)

Proposition 27 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 5 le 1198991 le sdot sdot sdot le 119899119901 Then 120574

119903

times2119905(119866119866) = 3119901 + 2

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878 cap 119883119894| =

|119878 cap 119883119894| = 2 and 119909 isin 119878 cap 119883

119894if and only if 119909 isin 119878 cap 119883

119894 and for

119894 = 3 119901 |119878 cap 119883119894| = 3 Obviously 119878 is a 120574times2119905(119866119866)-set Itcan easily be verified that every vertex in out of 119878 is adjacentto at least two vertices in out of 119878 Hence 119878 is a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) and so 120574

119903

times2119905(119866119866) = 3119901 + 2 by

Propositions 10 and 12(iv)

Proposition28 Let119866=119870119899111989921198993

be a complete 3-partite graphwith 6 le 119899

1 le 1198992 le 1198993 Then 120574119903

times3119905(119866119866) = 16

ISRN Combinatorics 5

Proof Let 119878 be a set of vertices such that |119878 cap 1198833| = 4 and

for 119894 = 1 2 |119878 cap 119883119894| = |119878 cap 119883

119894| = 3 such that 119909 isin 119878 cap

119883119894if and only if 119909 isin 119878 cap 119883

119894 Since 119878 is a 3TRDS of 119866119866 of

cardinality 120574times3119905

(119866119866) we get 120574119903times3119905

(119866119866) = 16 by Propositions11 and 12(iv)

Proposition 29 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 7 le 2119896 + 1 le 119899

1le sdot sdot sdot le 119899

119901 Then

120574119903

times119896119905(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(18)

Proof We prove the proposition in the following three cases

Case 1 (119901 ge 119896 + 1 or 119901 = 119896 ge 4) Let 119878 be a set of verticessuch that 119878 cap 119881(119866) is a 120574

times119896119905(119866)-set and for 119894 = 1 2 119901

|119878 cap 119883119894| = 119896 + 1 Obviously 119878 is a 120574times119896119905(119866119866)-set It can easilybe verified that every vertex in out of 119878 is adjacent to at least 119896vertices in out of 119878 Hence 119878 is a kTRDS of 119866119866 of cardinality120574times119896119905

(119866119866) and so 120574119903

times119896119905(119866119866) is (119901+ 1)(119896+ 1) if 119901 ge 119896+1 and is

(119901 + 1)(119896 + 1) + 1 if 119901 = 119896 ge 4 by Propositions 11 and 12(iv)

Case 2 (119901 = 119896 = 3) This case is proved in Proposition 28when 6 = 2119896 le 119899

1le 1198992le 1198993

Case 3 (119901 lt 119896) If min2119896minus2 lceil119896119901(119901minus1)rceil = 2119896minus2 let 119878 be aset of vertices such that for 119894 = 1 2 |119878cap119883

119894| = |119878cap119883

119894| = 119896 and

119909 isin 119878 cap 119883119894 if and only if 119909 isin 119878 cap 119883119894 and for 119894 = 3 119901 |119878 cap

119883119894| = 119896 + 1 If min2119896 minus 2 lceil119896119901(119901 minus 1)rceil = lceil119896119901(119901 minus 1)rceil let 119878

be a set of vertices such that 119878 cap 119881(119866) is a 120574times119896119905

(119866)-set and for119894 = 1 2 119901 |119878 cap 119883

119894| = 119896 + 1 In all cases obviously 119878 is a

120574times119896119905

(119866119866)-set and it can easily be verified that every vertex inout of 119878 is adjacent to at least 119896 vertices in out of 119878 Hence 119878 isa kTRDS of 119866119866 of cardinality 120574times119896119905(119866119866) and so 120574

119903

times2119905(119866119866) =

119901(119896 + 1) + min2119896 minus 2 lceil119896119901(119901 minus 1)rceil by Propositions 11 and12(iv)

4 The Given Bounds in Theorem 19 Are Sharp

In this section we show that the given bounds inTheorem 19are sharp For this aim we first calculate the 119896-tuple totalrestrained domination number of a complete multipartitegraph and its complement

Proposition 30 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with 119896 + 1 le 119899

1le sdot sdot sdot le 119899

119905le 2119896 + 1 lt 119899

119905+1le

sdot sdot sdot le 119899119901 If 119901 ge 119896 + 1 then

120574119903

times119896119905(119866) = 1198991 + sdot sdot sdot + 119899119905 + (119901 minus 119905) (119896 + 1) (19)

Proof Since 119866 is the union of disjoint complete graphs 1198701198991

1198701198992

119870119899119901

Proposition 15 implies that

120574119903

times119896119905(119866) = 120574

119903

times119896119905(1198701198991

) + 120574119903

times119896119905(1198701198992

) + sdot sdot sdot 120574119903

times119896119905(119870119899119901

)

= 1198991+ sdot sdot sdot + 119899

119905+ (119901 minus 119905) (119896 + 1)

(20)

Proposition 31 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraphwith 2 le 1198991 le sdot sdot sdot le 119899119901 If119901 ge 119896+1 then 120574

119903

times119896119905(119866) = 119896+1

Proof Let 119878 be a set of vertices of cardinality 119896 + 1 such thatfor every index 119894 = 1 2 119896 + 1 |119878 cap 119883119894| = 1 Since 119878 isa kTRDS of 119866 of cardinality 120574times119896119905(119866) Propositions 12(iv) and4(i) imply that 120574119903

times119896119905(119866) = 119896 + 1

Proposition 32 Let 119866 = 11987011989911198992119899119896

be a complete 119896-partitegraph of order 119899 with 1198991 le 1198992 le sdot sdot sdot le 119899119896 If 119899minus 119899119896minus1 ge 119899minus119899119896 ge

2119896 and 119899 minus 1198991 ge sdot sdot sdot ge 119899 minus 119899119896minus2 ge 2119896 + 1 then 120574119903

times119896119905(119866) = 119896 + 2

Proof Let 119878 be a set of vertices of cardinality 119896 + 2 such that|119878cap119883119894| = 2 for 119894 = 119896minus1 119896 and |119878 cap 119883

119894| = 1 for 119894 = 1 2 119896minus

2 Since 119878 is a kTRDS of119866 of cardinality 120574times119896119905

(119866) Propositions4(ii) and 12(iv) imply that 120574119903

times119896119905(119866) = 119896 + 2

Proposition 33 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order with lceil119896(119901 minus 1)rceil le 119899

1le sdot sdot sdot le 119899

119901 and let

119901 lt 119896 If 119901 minus 1 | 119896 and for each index 119894 = 1 2 119901 we have119899 minus 119899119894ge 2119896 then 120574

119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proof Let 119878 be a set of vertices of cardinality 119896119901(119901 minus 1) suchthat for each 119894 = 1 2 119901 |119878 cap 119883

119894| = 119896(119901 minus 1) Since 119878 is

a kTRDS of 119866 of cardinality 120574times119896119905

(119866) Propositions 4(iii) and12(iv) imply that 120574119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proposition 34 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with lceil119896(119901 minus 1)rceil le 1198991 le sdot sdot sdot le 119899119901 and let119901 lt 119896 Let also 119896 = (119901 minus 1) sdot lfloor119896(119901 minus 1)rfloor + 119903 for some integer1 le 119903 le 119901 minus 2 If 119899 minus 119899119901minus119903 ge 119899 minus 119899

119901minus119903+1ge sdot sdot sdot ge 119899 minus 119899

119901ge

lceil119896119901(119901minus 1)rceil minus lceil119896(119901minus 1)rceil + 119896 and 119899minus1198991 ge sdot sdot sdot ge 119899minus 119899119901minus119903minus1 ge

lceil119896119901(119901minus1)rceilminuslceil119896(119901minus1)rceil+119896+1 then 120574119903

times119896119905(119866) = lceil119896119901(119901minus1)rceil

Proof Let 119878 be a set of vertices of cardinality 1 le 119894 le 119901minus119903minus1|119878 cap 119883

119894| = lfloor119896(119901 minus 1)rfloor and for 119901 minus 119903 le 119894 le 119901 |119878 cap 119883

119894| =

lceil119896(119901 minus 1)rceil = lfloor119896(119901 minus 1)rfloor + 1 Since 119878 is a kTRDS of 119866 ofcardinality 120574

times119896119905(119866) Propositions 4(iii) and 12(iv) imply that

120574119903

times119896119905(119866) = lceil119896119901(119901 minus 1)rceil

Comparing Propositions 29 30 31 32 33 and 34 showsthat the following two results which state the given boundsinTheorem 19 are sharp are proved

Proposition 35 For every integer 119896 ge 1 let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph of order 119899 with 2119896 + 2 le 119899

1le 1198992le

sdot sdot sdot le 119899119901 If either 119901 ge 119896 + 1 or 119901 = 119896 ge 4 and 2119896 le 119899 minus 119899

119896le

119899 minus 119899119896minus1

2119896 + 1 le 119899 minus 119899119896minus2

le sdot sdot sdot le 119899 minus 1198991 then

120574119903

times119896119905(119866119866) = 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (21)

6 ISRN Combinatorics

Proposition 36 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph and let 119896 ge 1 be an integer If 119896 le 119899

1le sdot sdot sdot le 119899

119905le

2119896 minus 1 lt 119899119905+1 le sdot sdot sdot le 119899119901and 1198991+ 1198992+ sdot sdot sdot + 119899

119905= 119901 + 1 + 119905119896 for

some integer 119905 ge 1 then

120574119903

times119896119905(119866119866) = 120574

119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) (22)

We note that there are some complete multipartite graphsthat satisfy Proposition 36 For example if 119896 ge 4 then thecomplete (119896 + 1)-partite graph 119866 = 119870

11989911198992119899119896+1

with theconditions 119899

1= 119896 + 3 119899

2= 2119896 minus 1 and 2119896 le 119899

3le sdot sdot sdot le 119899

119896+1

satisfies itAlso Propositions 14 16 and 20 by this fact that

120574119903

times2119905(119862119899) = 119899 which is obtained by Proposition 12(i) imply

that

120574119903

times2119905(119862119899119862119899) = 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899) (23)

if and only if 119899 = 5 Also they imply that

120574119903

119905(119862119899) + 120574119903

119905(119862119899) lt 120574119903

times2119905(119862119899119862119899)

lt 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899)

(24)

when 119899 ge 4 in the lower bound and 119899 ge 6 in the upper bound

5 Open Problem

Finally we finish our paper with the following problems

Problem 37 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

Problem 38 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)

References

[1] M A Henning and A P Kazemi ldquo119896-tuple total domination ingraphsrdquo Discrete Applied Mathematics vol 158 no 9 pp 1006ndash1011 2010

[2] T W Haynes M A Henning P J Slater and L C van derMerwe ldquoThe complementary product of two graphsrdquo Bulletinof the Institute of Combinatorics and Its Applications vol 51 pp21ndash30 2007

[3] TW Haynes S T Hedetniemi and P J Slater Fundamentals ofDomination in Graphs vol 208 ofMonographs and Textbooks inPure and Applied Mathematics Marcel Dekker New York NYUSA 1998

[4] TWHaynes S THedetniemi and P J Slater EdsDominationin Graphs Advanced Topics Marcel Dekker New York NYUSA 1998

[5] M A Henning and A P Kazemi ldquo119896-tuple total domination incross products of graphsrdquo Journal of Combinatorial Optimiza-tion vol 24 no 3 pp 339ndash346 2012

[6] A P Kazemi ldquo119896-tuple total domination in complementaryprismsrdquo ISRNDiscreteMathematics vol 2011 Article ID68127413 pages 2011

[7] E J Cockayne and S T Hedetniemi ldquoTowards a theory ofdomination in graphsrdquoNetworks vol 7 no 3 pp 247ndash261 1977

[8] E J Cockayne R M Dawes and S T Hedetniemi ldquoTotaldomination in graphsrdquoNetworks vol 10 no 3 pp 211ndash219 1980

[9] S M Sheikholeslami and L Volkmann ldquoThe 119896-tuple totaldomatic number of a graphrdquo submitted to Utilitas Mathemat-ica httparxivorgabs11065589

[10] A P Kazemi ldquo119896-tuple total restrained dominationdomatic ingraphsrdquo Bulletin of the IranianMathematical Sciences accepted

Submit your manuscripts athttpwwwhindawicom

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Function Spaces

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Combinatorics 3

Proposition 14 (see Kazemi [10] 2011) Let 119899 ge 119896 + 3 ge 4Then

120574119903

times119896119905(119862119899) =

119899 119894119891 119899 le 2119896 + 2

119896 + 2 119894119891 2119896 + 3 le 119899 le 3119896 + 2

119896 + 1 119894119891 119899 ge 3119896 + 3

(5)

Proposition 15 (see Kazemi [10] 2011) Let 119896 lt 119899 be positiveintegers Then

120574119903

times119896119905(119870119899) =

119899 119894119891 119899 le 2119896 + 1

119896 + 1 Otherwise(6)

Proposition 16 (see Kazemi [10] 2011) Let 119899 ge 4 Then

120574119903

119905(119862119899) =

2lceil119899

4rceil minus 1 119894119891 119899 equiv 1 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(7)

2 Some Bounds

We first give a sharp lower bound for the 119896-tuple total re-strained domination number of a regular graph

Theorem 17 Let 119896 and ℓ be integers such that 1 le 119896 minus 1 le ℓ le

2119896 minus 2 If 119866 is a ℓ-regular graph of order 119899 then

120574119903

times119896119905(119866119866) ge 119899 + 119896 (8)

with equality if and only if 119899 ge ℓ + 2119896 and 119881(119866) contains a 119896-subset 119879 such that for each vertex 119894 isin 119881(119866) |119873(119894) cap 119879| ge 119896 minus 1

and also if 119894 isin 119881(119866) minus 119879 then |119873(119894) cap (119881(119866) minus 119879)| ge 119896

Proof Let 119881(119866119866) = 119881(119866) cup 119881(119866) such that 119881(119866) = 119894 | 1 le

119894 le 119899 and 119881(119866) = 119894 | 1 le 119894 le 119899 Let 119899 ge 2119896 + ℓ and let119878 be an arbitrary kTRDS of 119866119866 Then Proposition 12(i) andthis fact that every vertex 119894 isin 119881(119866) has degree ℓ + 1 le 2119896 minus 1

imply that119881(119866) sube 119878 Let 119894 notin 119878 Then |119873(119894)cap119881(119866)cap119878| ge 119896minus1If |119873(119894) cap119881(119866)cap119878| ge 119896 then we have nothing to proveThuslet 119873(119894) cap 119881(119866) cap 119878 = 119895

119898| 1 le 119898 le 119896 minus 1 But this implies

that there exists a vertex 119905 isin 119878minus119895119898

| 1 le 119898 le 119896minus1 such thatits corresponding vertex 119905 in 119881(119866) is adjacent to some vertex119895119898 when 1 le 119898 le 119896 minus 1 Then |119878| ge 119899 + 119896 and since 119878 wasarbitrary we conclude that 120574119903

times119896119905(119866119866) ge 119899 + 119896

Obviously it can be seen that 120574119903times119896119905

(119866119866) = 119899+119896 if and onlyif 119899 ge ℓ+2119896 and119881(119866) contains a 119896-subset119879 such that for eachvertex 119894 isin 119881(119866) |119873(119894) cap 119879| ge 119896 minus 1 and also if 119894 isin 119881(119866) minus 119879then |119873(119894) cap (119881(119866) minus 119879)| ge 119896

Theorem 17 and Proposition 12(i) imply the next corol-lary

Corollary 18 Let 119896 and ℓ be integers such that 1 le 119896 minus 1 le

ℓ le 2119896 minus 2 If 119866 is a ℓ-regular graph of order 119899 le ℓ + 2119896 minus 1then 120574

119903

times119896(119866119866) = 2119899

The next theorem gives lower and upper bounds for120574119903

times119896119905(119866119866) when 119866 is an arbitrary graph

Theorem 19 If 119866 is a graph of order 119899 with 119896 le

min120575(119866) 120575(119866) then

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) le 120574

119903

times119896119905(119866119866)

le 120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

(9)

when 119896 ge 1 in the upper bound and 119896 ge 2 in the lower bound

Proof Toprove 120574119903times(119896minus1)119905

(119866)+120574119903

times(119896minus1)119905(119866) le 120574

119903

times119896119905(119866119866) let 119896 ge 2

and let119863 be a kTRDSof119866119866 Since every vertex of119881(119866) (resp119881(119866)) is adjacent to only one vertex of119881(119866) (resp119881(119866)) weconclude that119863cap119881(119866) is a (119896 minus 1) TRDS of 119866 and119863cap119881(119866)

is a (119896 minus 1) TRDS of 119866 Then

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)le|119863 cap 119881 (119866)|+

10038161003816100381610038161003816119863 cap 119881 (119866)

10038161003816100381610038161003816

= |119863| = 120574119903

times119896119905(119866119866)

(10)

Wenowprove that 120574119903times119896119905

(119866119866) le 120574119903

times119896119905(119866)+120574

119903

times119896119905(119866) when 119896 ge 1

Since for every kTRDS 119878 of 119866 and every kTRDS 1198781015840 of 119866 the

set 119878 cup 1198781015840 is a kTRDS of 119866119866 we have

120574119903

times119896119905(119866119866) le 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (11)

In Section 4 we will show that the given bounds inTheorem 19 are sharp

3 The Complementary Prism of Some Graphs

In this section we will determine 120574119903times119896119905

(119866119866)when119866 is a cyclea path or a completemultipartite graph First let119866 be a cycle

Proposition 20 Let 119899 ge 4 Then

120574119903

times2119905(119862119899119862119899) =

2119899 119894119891 119899 = 4 5

119899 + 2 119894119891 119899 ge 6(12)

Proof Corollary 18 implies that 120574119903times2119905

(119862119899119862119899) = 2119899 if 119899 = 4 5

Now let 119899 ge 6 Proposition 7 with Proposition 12(iv) implies120574119903

times2119905(119862119899119862119899) ge 119899 + 2 Since also 119881(119862119899) cup 1 4 is a 2TRDS of

119862119899119862119899 we get 120574119903

times2119905(119862119899119862119899) = 119899 + 2 when 119899 ge 6

To calculate 120574119905(119862119899119862119899) we need to prove that 119889lowast

119905(119862119899119862119899) ge

2

Proposition 21 Let 119899 ge 4 Then 119889lowast

119905(119862119899119862119899) ge 2

Proof We prove the proposition in the following four cases

Case 1 (119899 equiv 0 (mod 4)) For 119899 = 4 we set 119878 = 1 1 2 2 and1198781015840= 3 3 4 4 If 119899 gt 4 we set 119878 = 1 1 2 2 cup 5 + 4119894 6 + 4119894 |

0 le 119894 le lceil1198994rceil minus 2 and 1198781015840 = 3 3 4 4 cup 7 + 4119894 8 + 4119894 | 0 le

119894 le lceil1198994rceil minus 2

4 ISRN Combinatorics

Case 2 (119899 equiv 1 (mod 4)) For 119899 = 5 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 and for 119899 = 9 we set 119878 = 1 1 4 4 7 7 and 119878

1015840

= 2 2 5 5 8 8 If 119899 gt 9 we set 119878 = 1 1 4 4 7 7 cup 10 +

4119894 11+4119894 | 0 le 119894 le lceil1198994rceilminus4 and 1198781015840= 3 3 6 6 9 9 cup 12+

4119894 13 + 4119894 | 0 le 119894 le lceil1198994rceil minus 4

Case 3 (119899 equiv 2 (mod 4)) For 119899 = 6 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 If 119899 gt 6 we set 119878 = 1 1 4 4 cup 7+4119894 8+4119894 |

0 le 119894 le lceil1198994rceil minus 3 and 1198781015840= 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le

119894 le lceil1198994rceil minus 3

Case 4 (119899 equiv 3 (mod 4)) For 119899 = 7 we set 119878 = 1 1 4 4 6

and 1198781015840

= 2 2 5 5 7 If 119899 gt 7 we set 119878 = 1 1 4 4

119899 minus 1 cup 7 + 4119894 8 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3 and 1198781015840

=

2 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3Since in all cases 119878 and 119878

1015840 are two disjoint 120574119905(119862119899119862119899)-setswe have 119889lowast

119905(119862119899119862119899) ge 2

By Propositions 6 13 and 21 we obtain the next result

Proposition 22 Let 119899 ge 4 Then

120574119903

119905(119862119899119862119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(13)

Now we continue our work when 119866 is a path

Proposition 23 Let 119899 ge 4 Then

120574119903

119905(119875119899119875119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(14)

Proof Propositions 5 and 12(iv) imply that

120574119903

119905(119875119899119875119899) ge 120574119905 (119875119899119875119899)

=

2lceil119899 minus 2

4rceil + 1 119894119891 119899equiv0 (mod 4)

2 lceil119899 minus 2

4rceil + 2 otherwise

(15)

Let 119899 equiv 0 (mod 4) For 119899= 8 we set 119878= 1 8 3 4 5 6

and for 119899 gt 8 we set

119878 = 1 119899 minus 6 119899 minus 5 119899 119899 minus 3 119899 minus 2

cup 3 + 4119894 4 + 4119894 | 0 le 119894 le lfloor119899

4rfloor minus 3

(16)

If 119899 equiv 1 2 3 (mod4) then we set 119878= 1 119899 minus 2 119899 119899 minus

2 cup 3+4119894 4+4119894 | 0 le 119894 le lfloor1198994rfloorminus2 119878= 1 119899 cup 3+4119894 4+4119894 |

0 le 119894 le lfloor1198994rfloorminus1 and 119878= 1 119899 minus 1 119899cup3+4119894 4+4119894 | 0 le 119894 le

lfloor1198994rfloorminus1 respectively Since in each case 119878 is a TRDS of119875119899119875119899

of cardinality 120574119905(119875119899119875119899) we have completed our proof

In the next propositions we calculate 120574119903

times119896119905(119866119866) when 119866

is a complete multipartite graph

Proposition 24 If 119866 = 11987011989911198992119899119901

is a complete 119901-partitegraph then

119889119905(119866119866) = min 119899

119894| 1 le 119894 le 119901 (17)

Proof Let 119881(119866119866) = ⋃1le119894le119901

(119883119894cup 119883119894) and let ℓ = min119899

119894|

1 le 119894 le 119901 Proposition 8 implies that for every TDS119863 of119866119866|119863 cap 119883

119894| ge 1 when 1 le 119894 le 119901 Hence 119889

119905(119866119866) le ℓ Now let

11986311198632 and119863

ℓ be ℓ disjoint 2119901-sets of119881(119866119866) such that forevery 119895 = 1 2 ℓ and every 119894 = 1 2 119901 |119863119895cap119883

119894| = |119863

119895cap

119883119894| = 1 and 119909 isin 119863

119895cap 119883119894if and only if 119909 isin 119863

119895cap 119883119894 Since1198631

1198632 and 119863

ℓ are ℓ disjoint 120574119905-sets of 119866119866 by Proposition 9

we get 119889119905(119866119866) ge ℓ and so 119889119905(119866119866) = ℓ

Proposition 25 If 119866 = 1198701198991119899119901

is a complete 119901-partite graphwith 2 le 119899

1le sdot sdot sdot le 119899

119901 then 120574

119903

119905(119866119866) = 2119901

Proof Since obviously 119889119905(119866119866)= 119889

lowast

119905(119866119866) we obtain 120574

119903

119905(119866119866) =

2119901 by Propositions 9 13 and 24

Proposition 26 Let 119866 = 11987011989911198992

be a complete bipartite graphwith 4 le 119899

1le 1198992 Then 120574

119903

times2119905(119866119866) = 8

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878cap119883119894| =

|119878 cap119883119894| = 2 and 119909 isin 119878 cap119883

119894if and only if 119909 isin 119878 cap119883

119894 Since 119878 is

a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) we get 120574119903times2119905

(119866119866) = 8by Propositions 10 and 12(iv)

Proposition 27 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 5 le 1198991 le sdot sdot sdot le 119899119901 Then 120574

119903

times2119905(119866119866) = 3119901 + 2

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878 cap 119883119894| =

|119878 cap 119883119894| = 2 and 119909 isin 119878 cap 119883

119894if and only if 119909 isin 119878 cap 119883

119894 and for

119894 = 3 119901 |119878 cap 119883119894| = 3 Obviously 119878 is a 120574times2119905(119866119866)-set Itcan easily be verified that every vertex in out of 119878 is adjacentto at least two vertices in out of 119878 Hence 119878 is a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) and so 120574

119903

times2119905(119866119866) = 3119901 + 2 by

Propositions 10 and 12(iv)

Proposition28 Let119866=119870119899111989921198993

be a complete 3-partite graphwith 6 le 119899

1 le 1198992 le 1198993 Then 120574119903

times3119905(119866119866) = 16

ISRN Combinatorics 5

Proof Let 119878 be a set of vertices such that |119878 cap 1198833| = 4 and

for 119894 = 1 2 |119878 cap 119883119894| = |119878 cap 119883

119894| = 3 such that 119909 isin 119878 cap

119883119894if and only if 119909 isin 119878 cap 119883

119894 Since 119878 is a 3TRDS of 119866119866 of

cardinality 120574times3119905

(119866119866) we get 120574119903times3119905

(119866119866) = 16 by Propositions11 and 12(iv)

Proposition 29 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 7 le 2119896 + 1 le 119899

1le sdot sdot sdot le 119899

119901 Then

120574119903

times119896119905(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(18)

Proof We prove the proposition in the following three cases

Case 1 (119901 ge 119896 + 1 or 119901 = 119896 ge 4) Let 119878 be a set of verticessuch that 119878 cap 119881(119866) is a 120574

times119896119905(119866)-set and for 119894 = 1 2 119901

|119878 cap 119883119894| = 119896 + 1 Obviously 119878 is a 120574times119896119905(119866119866)-set It can easilybe verified that every vertex in out of 119878 is adjacent to at least 119896vertices in out of 119878 Hence 119878 is a kTRDS of 119866119866 of cardinality120574times119896119905

(119866119866) and so 120574119903

times119896119905(119866119866) is (119901+ 1)(119896+ 1) if 119901 ge 119896+1 and is

(119901 + 1)(119896 + 1) + 1 if 119901 = 119896 ge 4 by Propositions 11 and 12(iv)

Case 2 (119901 = 119896 = 3) This case is proved in Proposition 28when 6 = 2119896 le 119899

1le 1198992le 1198993

Case 3 (119901 lt 119896) If min2119896minus2 lceil119896119901(119901minus1)rceil = 2119896minus2 let 119878 be aset of vertices such that for 119894 = 1 2 |119878cap119883

119894| = |119878cap119883

119894| = 119896 and

119909 isin 119878 cap 119883119894 if and only if 119909 isin 119878 cap 119883119894 and for 119894 = 3 119901 |119878 cap

119883119894| = 119896 + 1 If min2119896 minus 2 lceil119896119901(119901 minus 1)rceil = lceil119896119901(119901 minus 1)rceil let 119878

be a set of vertices such that 119878 cap 119881(119866) is a 120574times119896119905

(119866)-set and for119894 = 1 2 119901 |119878 cap 119883

119894| = 119896 + 1 In all cases obviously 119878 is a

120574times119896119905

(119866119866)-set and it can easily be verified that every vertex inout of 119878 is adjacent to at least 119896 vertices in out of 119878 Hence 119878 isa kTRDS of 119866119866 of cardinality 120574times119896119905(119866119866) and so 120574

119903

times2119905(119866119866) =

119901(119896 + 1) + min2119896 minus 2 lceil119896119901(119901 minus 1)rceil by Propositions 11 and12(iv)

4 The Given Bounds in Theorem 19 Are Sharp

In this section we show that the given bounds inTheorem 19are sharp For this aim we first calculate the 119896-tuple totalrestrained domination number of a complete multipartitegraph and its complement

Proposition 30 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with 119896 + 1 le 119899

1le sdot sdot sdot le 119899

119905le 2119896 + 1 lt 119899

119905+1le

sdot sdot sdot le 119899119901 If 119901 ge 119896 + 1 then

120574119903

times119896119905(119866) = 1198991 + sdot sdot sdot + 119899119905 + (119901 minus 119905) (119896 + 1) (19)

Proof Since 119866 is the union of disjoint complete graphs 1198701198991

1198701198992

119870119899119901

Proposition 15 implies that

120574119903

times119896119905(119866) = 120574

119903

times119896119905(1198701198991

) + 120574119903

times119896119905(1198701198992

) + sdot sdot sdot 120574119903

times119896119905(119870119899119901

)

= 1198991+ sdot sdot sdot + 119899

119905+ (119901 minus 119905) (119896 + 1)

(20)

Proposition 31 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraphwith 2 le 1198991 le sdot sdot sdot le 119899119901 If119901 ge 119896+1 then 120574

119903

times119896119905(119866) = 119896+1

Proof Let 119878 be a set of vertices of cardinality 119896 + 1 such thatfor every index 119894 = 1 2 119896 + 1 |119878 cap 119883119894| = 1 Since 119878 isa kTRDS of 119866 of cardinality 120574times119896119905(119866) Propositions 12(iv) and4(i) imply that 120574119903

times119896119905(119866) = 119896 + 1

Proposition 32 Let 119866 = 11987011989911198992119899119896

be a complete 119896-partitegraph of order 119899 with 1198991 le 1198992 le sdot sdot sdot le 119899119896 If 119899minus 119899119896minus1 ge 119899minus119899119896 ge

2119896 and 119899 minus 1198991 ge sdot sdot sdot ge 119899 minus 119899119896minus2 ge 2119896 + 1 then 120574119903

times119896119905(119866) = 119896 + 2

Proof Let 119878 be a set of vertices of cardinality 119896 + 2 such that|119878cap119883119894| = 2 for 119894 = 119896minus1 119896 and |119878 cap 119883

119894| = 1 for 119894 = 1 2 119896minus

2 Since 119878 is a kTRDS of119866 of cardinality 120574times119896119905

(119866) Propositions4(ii) and 12(iv) imply that 120574119903

times119896119905(119866) = 119896 + 2

Proposition 33 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order with lceil119896(119901 minus 1)rceil le 119899

1le sdot sdot sdot le 119899

119901 and let

119901 lt 119896 If 119901 minus 1 | 119896 and for each index 119894 = 1 2 119901 we have119899 minus 119899119894ge 2119896 then 120574

119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proof Let 119878 be a set of vertices of cardinality 119896119901(119901 minus 1) suchthat for each 119894 = 1 2 119901 |119878 cap 119883

119894| = 119896(119901 minus 1) Since 119878 is

a kTRDS of 119866 of cardinality 120574times119896119905

(119866) Propositions 4(iii) and12(iv) imply that 120574119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proposition 34 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with lceil119896(119901 minus 1)rceil le 1198991 le sdot sdot sdot le 119899119901 and let119901 lt 119896 Let also 119896 = (119901 minus 1) sdot lfloor119896(119901 minus 1)rfloor + 119903 for some integer1 le 119903 le 119901 minus 2 If 119899 minus 119899119901minus119903 ge 119899 minus 119899

119901minus119903+1ge sdot sdot sdot ge 119899 minus 119899

119901ge

lceil119896119901(119901minus 1)rceil minus lceil119896(119901minus 1)rceil + 119896 and 119899minus1198991 ge sdot sdot sdot ge 119899minus 119899119901minus119903minus1 ge

lceil119896119901(119901minus1)rceilminuslceil119896(119901minus1)rceil+119896+1 then 120574119903

times119896119905(119866) = lceil119896119901(119901minus1)rceil

Proof Let 119878 be a set of vertices of cardinality 1 le 119894 le 119901minus119903minus1|119878 cap 119883

119894| = lfloor119896(119901 minus 1)rfloor and for 119901 minus 119903 le 119894 le 119901 |119878 cap 119883

119894| =

lceil119896(119901 minus 1)rceil = lfloor119896(119901 minus 1)rfloor + 1 Since 119878 is a kTRDS of 119866 ofcardinality 120574

times119896119905(119866) Propositions 4(iii) and 12(iv) imply that

120574119903

times119896119905(119866) = lceil119896119901(119901 minus 1)rceil

Comparing Propositions 29 30 31 32 33 and 34 showsthat the following two results which state the given boundsinTheorem 19 are sharp are proved

Proposition 35 For every integer 119896 ge 1 let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph of order 119899 with 2119896 + 2 le 119899

1le 1198992le

sdot sdot sdot le 119899119901 If either 119901 ge 119896 + 1 or 119901 = 119896 ge 4 and 2119896 le 119899 minus 119899

119896le

119899 minus 119899119896minus1

2119896 + 1 le 119899 minus 119899119896minus2

le sdot sdot sdot le 119899 minus 1198991 then

120574119903

times119896119905(119866119866) = 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (21)

6 ISRN Combinatorics

Proposition 36 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph and let 119896 ge 1 be an integer If 119896 le 119899

1le sdot sdot sdot le 119899

119905le

2119896 minus 1 lt 119899119905+1 le sdot sdot sdot le 119899119901and 1198991+ 1198992+ sdot sdot sdot + 119899

119905= 119901 + 1 + 119905119896 for

some integer 119905 ge 1 then

120574119903

times119896119905(119866119866) = 120574

119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) (22)

We note that there are some complete multipartite graphsthat satisfy Proposition 36 For example if 119896 ge 4 then thecomplete (119896 + 1)-partite graph 119866 = 119870

11989911198992119899119896+1

with theconditions 119899

1= 119896 + 3 119899

2= 2119896 minus 1 and 2119896 le 119899

3le sdot sdot sdot le 119899

119896+1

satisfies itAlso Propositions 14 16 and 20 by this fact that

120574119903

times2119905(119862119899) = 119899 which is obtained by Proposition 12(i) imply

that

120574119903

times2119905(119862119899119862119899) = 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899) (23)

if and only if 119899 = 5 Also they imply that

120574119903

119905(119862119899) + 120574119903

119905(119862119899) lt 120574119903

times2119905(119862119899119862119899)

lt 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899)

(24)

when 119899 ge 4 in the lower bound and 119899 ge 6 in the upper bound

5 Open Problem

Finally we finish our paper with the following problems

Problem 37 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

Problem 38 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)

References

[1] M A Henning and A P Kazemi ldquo119896-tuple total domination ingraphsrdquo Discrete Applied Mathematics vol 158 no 9 pp 1006ndash1011 2010

[2] T W Haynes M A Henning P J Slater and L C van derMerwe ldquoThe complementary product of two graphsrdquo Bulletinof the Institute of Combinatorics and Its Applications vol 51 pp21ndash30 2007

[3] TW Haynes S T Hedetniemi and P J Slater Fundamentals ofDomination in Graphs vol 208 ofMonographs and Textbooks inPure and Applied Mathematics Marcel Dekker New York NYUSA 1998

[4] TWHaynes S THedetniemi and P J Slater EdsDominationin Graphs Advanced Topics Marcel Dekker New York NYUSA 1998

[5] M A Henning and A P Kazemi ldquo119896-tuple total domination incross products of graphsrdquo Journal of Combinatorial Optimiza-tion vol 24 no 3 pp 339ndash346 2012

[6] A P Kazemi ldquo119896-tuple total domination in complementaryprismsrdquo ISRNDiscreteMathematics vol 2011 Article ID68127413 pages 2011

[7] E J Cockayne and S T Hedetniemi ldquoTowards a theory ofdomination in graphsrdquoNetworks vol 7 no 3 pp 247ndash261 1977

[8] E J Cockayne R M Dawes and S T Hedetniemi ldquoTotaldomination in graphsrdquoNetworks vol 10 no 3 pp 211ndash219 1980

[9] S M Sheikholeslami and L Volkmann ldquoThe 119896-tuple totaldomatic number of a graphrdquo submitted to Utilitas Mathemat-ica httparxivorgabs11065589

[10] A P Kazemi ldquo119896-tuple total restrained dominationdomatic ingraphsrdquo Bulletin of the IranianMathematical Sciences accepted

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 ISRN Combinatorics

Case 2 (119899 equiv 1 (mod 4)) For 119899 = 5 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 and for 119899 = 9 we set 119878 = 1 1 4 4 7 7 and 119878

1015840

= 2 2 5 5 8 8 If 119899 gt 9 we set 119878 = 1 1 4 4 7 7 cup 10 +

4119894 11+4119894 | 0 le 119894 le lceil1198994rceilminus4 and 1198781015840= 3 3 6 6 9 9 cup 12+

4119894 13 + 4119894 | 0 le 119894 le lceil1198994rceil minus 4

Case 3 (119899 equiv 2 (mod 4)) For 119899 = 6 we set 119878 = 1 1 4 4 and1198781015840= 2 2 5 5 If 119899 gt 6 we set 119878 = 1 1 4 4 cup 7+4119894 8+4119894 |

0 le 119894 le lceil1198994rceil minus 3 and 1198781015840= 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le

119894 le lceil1198994rceil minus 3

Case 4 (119899 equiv 3 (mod 4)) For 119899 = 7 we set 119878 = 1 1 4 4 6

and 1198781015840

= 2 2 5 5 7 If 119899 gt 7 we set 119878 = 1 1 4 4

119899 minus 1 cup 7 + 4119894 8 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3 and 1198781015840

=

2 3 3 6 6 cup 9 + 4119894 10 + 4119894 | 0 le 119894 le lceil1198994rceil minus 3Since in all cases 119878 and 119878

1015840 are two disjoint 120574119905(119862119899119862119899)-setswe have 119889lowast

119905(119862119899119862119899) ge 2

By Propositions 6 13 and 21 we obtain the next result

Proposition 22 Let 119899 ge 4 Then

120574119903

119905(119862119899119862119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 3 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(13)

Now we continue our work when 119866 is a path

Proposition 23 Let 119899 ge 4 Then

120574119903

119905(119875119899119875119899) =

2lceil119899

4rceil + 2 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil + 1 119894119891 119899 equiv 0 (mod 4)

2 lceil119899

4rceil 119900119905ℎ119890119903119908119894119904119890

(14)

Proof Propositions 5 and 12(iv) imply that

120574119903

119905(119875119899119875119899) ge 120574119905 (119875119899119875119899)

=

2lceil119899 minus 2

4rceil + 1 119894119891 119899equiv0 (mod 4)

2 lceil119899 minus 2

4rceil + 2 otherwise

(15)

Let 119899 equiv 0 (mod 4) For 119899= 8 we set 119878= 1 8 3 4 5 6

and for 119899 gt 8 we set

119878 = 1 119899 minus 6 119899 minus 5 119899 119899 minus 3 119899 minus 2

cup 3 + 4119894 4 + 4119894 | 0 le 119894 le lfloor119899

4rfloor minus 3

(16)

If 119899 equiv 1 2 3 (mod4) then we set 119878= 1 119899 minus 2 119899 119899 minus

2 cup 3+4119894 4+4119894 | 0 le 119894 le lfloor1198994rfloorminus2 119878= 1 119899 cup 3+4119894 4+4119894 |

0 le 119894 le lfloor1198994rfloorminus1 and 119878= 1 119899 minus 1 119899cup3+4119894 4+4119894 | 0 le 119894 le

lfloor1198994rfloorminus1 respectively Since in each case 119878 is a TRDS of119875119899119875119899

of cardinality 120574119905(119875119899119875119899) we have completed our proof

In the next propositions we calculate 120574119903

times119896119905(119866119866) when 119866

is a complete multipartite graph

Proposition 24 If 119866 = 11987011989911198992119899119901

is a complete 119901-partitegraph then

119889119905(119866119866) = min 119899

119894| 1 le 119894 le 119901 (17)

Proof Let 119881(119866119866) = ⋃1le119894le119901

(119883119894cup 119883119894) and let ℓ = min119899

119894|

1 le 119894 le 119901 Proposition 8 implies that for every TDS119863 of119866119866|119863 cap 119883

119894| ge 1 when 1 le 119894 le 119901 Hence 119889

119905(119866119866) le ℓ Now let

11986311198632 and119863

ℓ be ℓ disjoint 2119901-sets of119881(119866119866) such that forevery 119895 = 1 2 ℓ and every 119894 = 1 2 119901 |119863119895cap119883

119894| = |119863

119895cap

119883119894| = 1 and 119909 isin 119863

119895cap 119883119894if and only if 119909 isin 119863

119895cap 119883119894 Since1198631

1198632 and 119863

ℓ are ℓ disjoint 120574119905-sets of 119866119866 by Proposition 9

we get 119889119905(119866119866) ge ℓ and so 119889119905(119866119866) = ℓ

Proposition 25 If 119866 = 1198701198991119899119901

is a complete 119901-partite graphwith 2 le 119899

1le sdot sdot sdot le 119899

119901 then 120574

119903

119905(119866119866) = 2119901

Proof Since obviously 119889119905(119866119866)= 119889

lowast

119905(119866119866) we obtain 120574

119903

119905(119866119866) =

2119901 by Propositions 9 13 and 24

Proposition 26 Let 119866 = 11987011989911198992

be a complete bipartite graphwith 4 le 119899

1le 1198992 Then 120574

119903

times2119905(119866119866) = 8

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878cap119883119894| =

|119878 cap119883119894| = 2 and 119909 isin 119878 cap119883

119894if and only if 119909 isin 119878 cap119883

119894 Since 119878 is

a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) we get 120574119903times2119905

(119866119866) = 8by Propositions 10 and 12(iv)

Proposition 27 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 5 le 1198991 le sdot sdot sdot le 119899119901 Then 120574

119903

times2119905(119866119866) = 3119901 + 2

Proof Let 119878 be a set of vertices such that for 119894 = 1 2 |119878 cap 119883119894| =

|119878 cap 119883119894| = 2 and 119909 isin 119878 cap 119883

119894if and only if 119909 isin 119878 cap 119883

119894 and for

119894 = 3 119901 |119878 cap 119883119894| = 3 Obviously 119878 is a 120574times2119905(119866119866)-set Itcan easily be verified that every vertex in out of 119878 is adjacentto at least two vertices in out of 119878 Hence 119878 is a 2TRDS of119866119866 of cardinality 120574times2119905(119866119866) and so 120574

119903

times2119905(119866119866) = 3119901 + 2 by

Propositions 10 and 12(iv)

Proposition28 Let119866=119870119899111989921198993

be a complete 3-partite graphwith 6 le 119899

1 le 1198992 le 1198993 Then 120574119903

times3119905(119866119866) = 16

ISRN Combinatorics 5

Proof Let 119878 be a set of vertices such that |119878 cap 1198833| = 4 and

for 119894 = 1 2 |119878 cap 119883119894| = |119878 cap 119883

119894| = 3 such that 119909 isin 119878 cap

119883119894if and only if 119909 isin 119878 cap 119883

119894 Since 119878 is a 3TRDS of 119866119866 of

cardinality 120574times3119905

(119866119866) we get 120574119903times3119905

(119866119866) = 16 by Propositions11 and 12(iv)

Proposition 29 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 7 le 2119896 + 1 le 119899

1le sdot sdot sdot le 119899

119901 Then

120574119903

times119896119905(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(18)

Proof We prove the proposition in the following three cases

Case 1 (119901 ge 119896 + 1 or 119901 = 119896 ge 4) Let 119878 be a set of verticessuch that 119878 cap 119881(119866) is a 120574

times119896119905(119866)-set and for 119894 = 1 2 119901

|119878 cap 119883119894| = 119896 + 1 Obviously 119878 is a 120574times119896119905(119866119866)-set It can easilybe verified that every vertex in out of 119878 is adjacent to at least 119896vertices in out of 119878 Hence 119878 is a kTRDS of 119866119866 of cardinality120574times119896119905

(119866119866) and so 120574119903

times119896119905(119866119866) is (119901+ 1)(119896+ 1) if 119901 ge 119896+1 and is

(119901 + 1)(119896 + 1) + 1 if 119901 = 119896 ge 4 by Propositions 11 and 12(iv)

Case 2 (119901 = 119896 = 3) This case is proved in Proposition 28when 6 = 2119896 le 119899

1le 1198992le 1198993

Case 3 (119901 lt 119896) If min2119896minus2 lceil119896119901(119901minus1)rceil = 2119896minus2 let 119878 be aset of vertices such that for 119894 = 1 2 |119878cap119883

119894| = |119878cap119883

119894| = 119896 and

119909 isin 119878 cap 119883119894 if and only if 119909 isin 119878 cap 119883119894 and for 119894 = 3 119901 |119878 cap

119883119894| = 119896 + 1 If min2119896 minus 2 lceil119896119901(119901 minus 1)rceil = lceil119896119901(119901 minus 1)rceil let 119878

be a set of vertices such that 119878 cap 119881(119866) is a 120574times119896119905

(119866)-set and for119894 = 1 2 119901 |119878 cap 119883

119894| = 119896 + 1 In all cases obviously 119878 is a

120574times119896119905

(119866119866)-set and it can easily be verified that every vertex inout of 119878 is adjacent to at least 119896 vertices in out of 119878 Hence 119878 isa kTRDS of 119866119866 of cardinality 120574times119896119905(119866119866) and so 120574

119903

times2119905(119866119866) =

119901(119896 + 1) + min2119896 minus 2 lceil119896119901(119901 minus 1)rceil by Propositions 11 and12(iv)

4 The Given Bounds in Theorem 19 Are Sharp

In this section we show that the given bounds inTheorem 19are sharp For this aim we first calculate the 119896-tuple totalrestrained domination number of a complete multipartitegraph and its complement

Proposition 30 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with 119896 + 1 le 119899

1le sdot sdot sdot le 119899

119905le 2119896 + 1 lt 119899

119905+1le

sdot sdot sdot le 119899119901 If 119901 ge 119896 + 1 then

120574119903

times119896119905(119866) = 1198991 + sdot sdot sdot + 119899119905 + (119901 minus 119905) (119896 + 1) (19)

Proof Since 119866 is the union of disjoint complete graphs 1198701198991

1198701198992

119870119899119901

Proposition 15 implies that

120574119903

times119896119905(119866) = 120574

119903

times119896119905(1198701198991

) + 120574119903

times119896119905(1198701198992

) + sdot sdot sdot 120574119903

times119896119905(119870119899119901

)

= 1198991+ sdot sdot sdot + 119899

119905+ (119901 minus 119905) (119896 + 1)

(20)

Proposition 31 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraphwith 2 le 1198991 le sdot sdot sdot le 119899119901 If119901 ge 119896+1 then 120574

119903

times119896119905(119866) = 119896+1

Proof Let 119878 be a set of vertices of cardinality 119896 + 1 such thatfor every index 119894 = 1 2 119896 + 1 |119878 cap 119883119894| = 1 Since 119878 isa kTRDS of 119866 of cardinality 120574times119896119905(119866) Propositions 12(iv) and4(i) imply that 120574119903

times119896119905(119866) = 119896 + 1

Proposition 32 Let 119866 = 11987011989911198992119899119896

be a complete 119896-partitegraph of order 119899 with 1198991 le 1198992 le sdot sdot sdot le 119899119896 If 119899minus 119899119896minus1 ge 119899minus119899119896 ge

2119896 and 119899 minus 1198991 ge sdot sdot sdot ge 119899 minus 119899119896minus2 ge 2119896 + 1 then 120574119903

times119896119905(119866) = 119896 + 2

Proof Let 119878 be a set of vertices of cardinality 119896 + 2 such that|119878cap119883119894| = 2 for 119894 = 119896minus1 119896 and |119878 cap 119883

119894| = 1 for 119894 = 1 2 119896minus

2 Since 119878 is a kTRDS of119866 of cardinality 120574times119896119905

(119866) Propositions4(ii) and 12(iv) imply that 120574119903

times119896119905(119866) = 119896 + 2

Proposition 33 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order with lceil119896(119901 minus 1)rceil le 119899

1le sdot sdot sdot le 119899

119901 and let

119901 lt 119896 If 119901 minus 1 | 119896 and for each index 119894 = 1 2 119901 we have119899 minus 119899119894ge 2119896 then 120574

119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proof Let 119878 be a set of vertices of cardinality 119896119901(119901 minus 1) suchthat for each 119894 = 1 2 119901 |119878 cap 119883

119894| = 119896(119901 minus 1) Since 119878 is

a kTRDS of 119866 of cardinality 120574times119896119905

(119866) Propositions 4(iii) and12(iv) imply that 120574119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proposition 34 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with lceil119896(119901 minus 1)rceil le 1198991 le sdot sdot sdot le 119899119901 and let119901 lt 119896 Let also 119896 = (119901 minus 1) sdot lfloor119896(119901 minus 1)rfloor + 119903 for some integer1 le 119903 le 119901 minus 2 If 119899 minus 119899119901minus119903 ge 119899 minus 119899

119901minus119903+1ge sdot sdot sdot ge 119899 minus 119899

119901ge

lceil119896119901(119901minus 1)rceil minus lceil119896(119901minus 1)rceil + 119896 and 119899minus1198991 ge sdot sdot sdot ge 119899minus 119899119901minus119903minus1 ge

lceil119896119901(119901minus1)rceilminuslceil119896(119901minus1)rceil+119896+1 then 120574119903

times119896119905(119866) = lceil119896119901(119901minus1)rceil

Proof Let 119878 be a set of vertices of cardinality 1 le 119894 le 119901minus119903minus1|119878 cap 119883

119894| = lfloor119896(119901 minus 1)rfloor and for 119901 minus 119903 le 119894 le 119901 |119878 cap 119883

119894| =

lceil119896(119901 minus 1)rceil = lfloor119896(119901 minus 1)rfloor + 1 Since 119878 is a kTRDS of 119866 ofcardinality 120574

times119896119905(119866) Propositions 4(iii) and 12(iv) imply that

120574119903

times119896119905(119866) = lceil119896119901(119901 minus 1)rceil

Comparing Propositions 29 30 31 32 33 and 34 showsthat the following two results which state the given boundsinTheorem 19 are sharp are proved

Proposition 35 For every integer 119896 ge 1 let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph of order 119899 with 2119896 + 2 le 119899

1le 1198992le

sdot sdot sdot le 119899119901 If either 119901 ge 119896 + 1 or 119901 = 119896 ge 4 and 2119896 le 119899 minus 119899

119896le

119899 minus 119899119896minus1

2119896 + 1 le 119899 minus 119899119896minus2

le sdot sdot sdot le 119899 minus 1198991 then

120574119903

times119896119905(119866119866) = 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (21)

6 ISRN Combinatorics

Proposition 36 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph and let 119896 ge 1 be an integer If 119896 le 119899

1le sdot sdot sdot le 119899

119905le

2119896 minus 1 lt 119899119905+1 le sdot sdot sdot le 119899119901and 1198991+ 1198992+ sdot sdot sdot + 119899

119905= 119901 + 1 + 119905119896 for

some integer 119905 ge 1 then

120574119903

times119896119905(119866119866) = 120574

119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) (22)

We note that there are some complete multipartite graphsthat satisfy Proposition 36 For example if 119896 ge 4 then thecomplete (119896 + 1)-partite graph 119866 = 119870

11989911198992119899119896+1

with theconditions 119899

1= 119896 + 3 119899

2= 2119896 minus 1 and 2119896 le 119899

3le sdot sdot sdot le 119899

119896+1

satisfies itAlso Propositions 14 16 and 20 by this fact that

120574119903

times2119905(119862119899) = 119899 which is obtained by Proposition 12(i) imply

that

120574119903

times2119905(119862119899119862119899) = 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899) (23)

if and only if 119899 = 5 Also they imply that

120574119903

119905(119862119899) + 120574119903

119905(119862119899) lt 120574119903

times2119905(119862119899119862119899)

lt 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899)

(24)

when 119899 ge 4 in the lower bound and 119899 ge 6 in the upper bound

5 Open Problem

Finally we finish our paper with the following problems

Problem 37 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

Problem 38 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)

References

[1] M A Henning and A P Kazemi ldquo119896-tuple total domination ingraphsrdquo Discrete Applied Mathematics vol 158 no 9 pp 1006ndash1011 2010

[2] T W Haynes M A Henning P J Slater and L C van derMerwe ldquoThe complementary product of two graphsrdquo Bulletinof the Institute of Combinatorics and Its Applications vol 51 pp21ndash30 2007

[3] TW Haynes S T Hedetniemi and P J Slater Fundamentals ofDomination in Graphs vol 208 ofMonographs and Textbooks inPure and Applied Mathematics Marcel Dekker New York NYUSA 1998

[4] TWHaynes S THedetniemi and P J Slater EdsDominationin Graphs Advanced Topics Marcel Dekker New York NYUSA 1998

[5] M A Henning and A P Kazemi ldquo119896-tuple total domination incross products of graphsrdquo Journal of Combinatorial Optimiza-tion vol 24 no 3 pp 339ndash346 2012

[6] A P Kazemi ldquo119896-tuple total domination in complementaryprismsrdquo ISRNDiscreteMathematics vol 2011 Article ID68127413 pages 2011

[7] E J Cockayne and S T Hedetniemi ldquoTowards a theory ofdomination in graphsrdquoNetworks vol 7 no 3 pp 247ndash261 1977

[8] E J Cockayne R M Dawes and S T Hedetniemi ldquoTotaldomination in graphsrdquoNetworks vol 10 no 3 pp 211ndash219 1980

[9] S M Sheikholeslami and L Volkmann ldquoThe 119896-tuple totaldomatic number of a graphrdquo submitted to Utilitas Mathemat-ica httparxivorgabs11065589

[10] A P Kazemi ldquo119896-tuple total restrained dominationdomatic ingraphsrdquo Bulletin of the IranianMathematical Sciences accepted

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Combinatorics 5

Proof Let 119878 be a set of vertices such that |119878 cap 1198833| = 4 and

for 119894 = 1 2 |119878 cap 119883119894| = |119878 cap 119883

119894| = 3 such that 119909 isin 119878 cap

119883119894if and only if 119909 isin 119878 cap 119883

119894 Since 119878 is a 3TRDS of 119866119866 of

cardinality 120574times3119905

(119866119866) we get 120574119903times3119905

(119866119866) = 16 by Propositions11 and 12(iv)

Proposition 29 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph with 7 le 2119896 + 1 le 119899

1le sdot sdot sdot le 119899

119901 Then

120574119903

times119896119905(119866119866)

=

(119901 + 1) (119896 + 1) 119894119891 119901 ge 119896 + 1

(119901 + 1) (119896 + 1) + 1 119894119891 119901 = 119896 ge 4

16 119894119891 119901 = 119896 = 3

119901 (119896 + 1) +min2119896 minus 2 lceil119896119901

119901 minus 1rceil 119894119891119901 lt 119896

(18)

Proof We prove the proposition in the following three cases

Case 1 (119901 ge 119896 + 1 or 119901 = 119896 ge 4) Let 119878 be a set of verticessuch that 119878 cap 119881(119866) is a 120574

times119896119905(119866)-set and for 119894 = 1 2 119901

|119878 cap 119883119894| = 119896 + 1 Obviously 119878 is a 120574times119896119905(119866119866)-set It can easilybe verified that every vertex in out of 119878 is adjacent to at least 119896vertices in out of 119878 Hence 119878 is a kTRDS of 119866119866 of cardinality120574times119896119905

(119866119866) and so 120574119903

times119896119905(119866119866) is (119901+ 1)(119896+ 1) if 119901 ge 119896+1 and is

(119901 + 1)(119896 + 1) + 1 if 119901 = 119896 ge 4 by Propositions 11 and 12(iv)

Case 2 (119901 = 119896 = 3) This case is proved in Proposition 28when 6 = 2119896 le 119899

1le 1198992le 1198993

Case 3 (119901 lt 119896) If min2119896minus2 lceil119896119901(119901minus1)rceil = 2119896minus2 let 119878 be aset of vertices such that for 119894 = 1 2 |119878cap119883

119894| = |119878cap119883

119894| = 119896 and

119909 isin 119878 cap 119883119894 if and only if 119909 isin 119878 cap 119883119894 and for 119894 = 3 119901 |119878 cap

119883119894| = 119896 + 1 If min2119896 minus 2 lceil119896119901(119901 minus 1)rceil = lceil119896119901(119901 minus 1)rceil let 119878

be a set of vertices such that 119878 cap 119881(119866) is a 120574times119896119905

(119866)-set and for119894 = 1 2 119901 |119878 cap 119883

119894| = 119896 + 1 In all cases obviously 119878 is a

120574times119896119905

(119866119866)-set and it can easily be verified that every vertex inout of 119878 is adjacent to at least 119896 vertices in out of 119878 Hence 119878 isa kTRDS of 119866119866 of cardinality 120574times119896119905(119866119866) and so 120574

119903

times2119905(119866119866) =

119901(119896 + 1) + min2119896 minus 2 lceil119896119901(119901 minus 1)rceil by Propositions 11 and12(iv)

4 The Given Bounds in Theorem 19 Are Sharp

In this section we show that the given bounds inTheorem 19are sharp For this aim we first calculate the 119896-tuple totalrestrained domination number of a complete multipartitegraph and its complement

Proposition 30 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with 119896 + 1 le 119899

1le sdot sdot sdot le 119899

119905le 2119896 + 1 lt 119899

119905+1le

sdot sdot sdot le 119899119901 If 119901 ge 119896 + 1 then

120574119903

times119896119905(119866) = 1198991 + sdot sdot sdot + 119899119905 + (119901 minus 119905) (119896 + 1) (19)

Proof Since 119866 is the union of disjoint complete graphs 1198701198991

1198701198992

119870119899119901

Proposition 15 implies that

120574119903

times119896119905(119866) = 120574

119903

times119896119905(1198701198991

) + 120574119903

times119896119905(1198701198992

) + sdot sdot sdot 120574119903

times119896119905(119870119899119901

)

= 1198991+ sdot sdot sdot + 119899

119905+ (119901 minus 119905) (119896 + 1)

(20)

Proposition 31 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraphwith 2 le 1198991 le sdot sdot sdot le 119899119901 If119901 ge 119896+1 then 120574

119903

times119896119905(119866) = 119896+1

Proof Let 119878 be a set of vertices of cardinality 119896 + 1 such thatfor every index 119894 = 1 2 119896 + 1 |119878 cap 119883119894| = 1 Since 119878 isa kTRDS of 119866 of cardinality 120574times119896119905(119866) Propositions 12(iv) and4(i) imply that 120574119903

times119896119905(119866) = 119896 + 1

Proposition 32 Let 119866 = 11987011989911198992119899119896

be a complete 119896-partitegraph of order 119899 with 1198991 le 1198992 le sdot sdot sdot le 119899119896 If 119899minus 119899119896minus1 ge 119899minus119899119896 ge

2119896 and 119899 minus 1198991 ge sdot sdot sdot ge 119899 minus 119899119896minus2 ge 2119896 + 1 then 120574119903

times119896119905(119866) = 119896 + 2

Proof Let 119878 be a set of vertices of cardinality 119896 + 2 such that|119878cap119883119894| = 2 for 119894 = 119896minus1 119896 and |119878 cap 119883

119894| = 1 for 119894 = 1 2 119896minus

2 Since 119878 is a kTRDS of119866 of cardinality 120574times119896119905

(119866) Propositions4(ii) and 12(iv) imply that 120574119903

times119896119905(119866) = 119896 + 2

Proposition 33 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order with lceil119896(119901 minus 1)rceil le 119899

1le sdot sdot sdot le 119899

119901 and let

119901 lt 119896 If 119901 minus 1 | 119896 and for each index 119894 = 1 2 119901 we have119899 minus 119899119894ge 2119896 then 120574

119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proof Let 119878 be a set of vertices of cardinality 119896119901(119901 minus 1) suchthat for each 119894 = 1 2 119901 |119878 cap 119883

119894| = 119896(119901 minus 1) Since 119878 is

a kTRDS of 119866 of cardinality 120574times119896119905

(119866) Propositions 4(iii) and12(iv) imply that 120574119903

times119896119905(119866) = 119896119901(119901 minus 1)

Proposition 34 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph of order 119899 with lceil119896(119901 minus 1)rceil le 1198991 le sdot sdot sdot le 119899119901 and let119901 lt 119896 Let also 119896 = (119901 minus 1) sdot lfloor119896(119901 minus 1)rfloor + 119903 for some integer1 le 119903 le 119901 minus 2 If 119899 minus 119899119901minus119903 ge 119899 minus 119899

119901minus119903+1ge sdot sdot sdot ge 119899 minus 119899

119901ge

lceil119896119901(119901minus 1)rceil minus lceil119896(119901minus 1)rceil + 119896 and 119899minus1198991 ge sdot sdot sdot ge 119899minus 119899119901minus119903minus1 ge

lceil119896119901(119901minus1)rceilminuslceil119896(119901minus1)rceil+119896+1 then 120574119903

times119896119905(119866) = lceil119896119901(119901minus1)rceil

Proof Let 119878 be a set of vertices of cardinality 1 le 119894 le 119901minus119903minus1|119878 cap 119883

119894| = lfloor119896(119901 minus 1)rfloor and for 119901 minus 119903 le 119894 le 119901 |119878 cap 119883

119894| =

lceil119896(119901 minus 1)rceil = lfloor119896(119901 minus 1)rfloor + 1 Since 119878 is a kTRDS of 119866 ofcardinality 120574

times119896119905(119866) Propositions 4(iii) and 12(iv) imply that

120574119903

times119896119905(119866) = lceil119896119901(119901 minus 1)rceil

Comparing Propositions 29 30 31 32 33 and 34 showsthat the following two results which state the given boundsinTheorem 19 are sharp are proved

Proposition 35 For every integer 119896 ge 1 let 119866 = 11987011989911198992119899119901

bea complete 119901-partite graph of order 119899 with 2119896 + 2 le 119899

1le 1198992le

sdot sdot sdot le 119899119901 If either 119901 ge 119896 + 1 or 119901 = 119896 ge 4 and 2119896 le 119899 minus 119899

119896le

119899 minus 119899119896minus1

2119896 + 1 le 119899 minus 119899119896minus2

le sdot sdot sdot le 119899 minus 1198991 then

120574119903

times119896119905(119866119866) = 120574

119903

times119896119905(119866) + 120574

119903

times119896119905(119866) (21)

6 ISRN Combinatorics

Proposition 36 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph and let 119896 ge 1 be an integer If 119896 le 119899

1le sdot sdot sdot le 119899

119905le

2119896 minus 1 lt 119899119905+1 le sdot sdot sdot le 119899119901and 1198991+ 1198992+ sdot sdot sdot + 119899

119905= 119901 + 1 + 119905119896 for

some integer 119905 ge 1 then

120574119903

times119896119905(119866119866) = 120574

119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) (22)

We note that there are some complete multipartite graphsthat satisfy Proposition 36 For example if 119896 ge 4 then thecomplete (119896 + 1)-partite graph 119866 = 119870

11989911198992119899119896+1

with theconditions 119899

1= 119896 + 3 119899

2= 2119896 minus 1 and 2119896 le 119899

3le sdot sdot sdot le 119899

119896+1

satisfies itAlso Propositions 14 16 and 20 by this fact that

120574119903

times2119905(119862119899) = 119899 which is obtained by Proposition 12(i) imply

that

120574119903

times2119905(119862119899119862119899) = 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899) (23)

if and only if 119899 = 5 Also they imply that

120574119903

119905(119862119899) + 120574119903

119905(119862119899) lt 120574119903

times2119905(119862119899119862119899)

lt 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899)

(24)

when 119899 ge 4 in the lower bound and 119899 ge 6 in the upper bound

5 Open Problem

Finally we finish our paper with the following problems

Problem 37 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

Problem 38 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)

References

[1] M A Henning and A P Kazemi ldquo119896-tuple total domination ingraphsrdquo Discrete Applied Mathematics vol 158 no 9 pp 1006ndash1011 2010

[2] T W Haynes M A Henning P J Slater and L C van derMerwe ldquoThe complementary product of two graphsrdquo Bulletinof the Institute of Combinatorics and Its Applications vol 51 pp21ndash30 2007

[3] TW Haynes S T Hedetniemi and P J Slater Fundamentals ofDomination in Graphs vol 208 ofMonographs and Textbooks inPure and Applied Mathematics Marcel Dekker New York NYUSA 1998

[4] TWHaynes S THedetniemi and P J Slater EdsDominationin Graphs Advanced Topics Marcel Dekker New York NYUSA 1998

[5] M A Henning and A P Kazemi ldquo119896-tuple total domination incross products of graphsrdquo Journal of Combinatorial Optimiza-tion vol 24 no 3 pp 339ndash346 2012

[6] A P Kazemi ldquo119896-tuple total domination in complementaryprismsrdquo ISRNDiscreteMathematics vol 2011 Article ID68127413 pages 2011

[7] E J Cockayne and S T Hedetniemi ldquoTowards a theory ofdomination in graphsrdquoNetworks vol 7 no 3 pp 247ndash261 1977

[8] E J Cockayne R M Dawes and S T Hedetniemi ldquoTotaldomination in graphsrdquoNetworks vol 10 no 3 pp 211ndash219 1980

[9] S M Sheikholeslami and L Volkmann ldquoThe 119896-tuple totaldomatic number of a graphrdquo submitted to Utilitas Mathemat-ica httparxivorgabs11065589

[10] A P Kazemi ldquo119896-tuple total restrained dominationdomatic ingraphsrdquo Bulletin of the IranianMathematical Sciences accepted

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 ISRN Combinatorics

Proposition 36 Let 119866 = 11987011989911198992119899119901

be a complete 119901-partitegraph and let 119896 ge 1 be an integer If 119896 le 119899

1le sdot sdot sdot le 119899

119905le

2119896 minus 1 lt 119899119905+1 le sdot sdot sdot le 119899119901and 1198991+ 1198992+ sdot sdot sdot + 119899

119905= 119901 + 1 + 119905119896 for

some integer 119905 ge 1 then

120574119903

times119896119905(119866119866) = 120574

119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866) (22)

We note that there are some complete multipartite graphsthat satisfy Proposition 36 For example if 119896 ge 4 then thecomplete (119896 + 1)-partite graph 119866 = 119870

11989911198992119899119896+1

with theconditions 119899

1= 119896 + 3 119899

2= 2119896 minus 1 and 2119896 le 119899

3le sdot sdot sdot le 119899

119896+1

satisfies itAlso Propositions 14 16 and 20 by this fact that

120574119903

times2119905(119862119899) = 119899 which is obtained by Proposition 12(i) imply

that

120574119903

times2119905(119862119899119862119899) = 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899) (23)

if and only if 119899 = 5 Also they imply that

120574119903

119905(119862119899) + 120574119903

119905(119862119899) lt 120574119903

times2119905(119862119899119862119899)

lt 120574119903

times2119905(119862119899) + 120574119903

times119896119905(119862119899)

(24)

when 119899 ge 4 in the lower bound and 119899 ge 6 in the upper bound

5 Open Problem

Finally we finish our paper with the following problems

Problem 37 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times119896119905(119866) + 120574

119903

times119896119905(119866)

Problem 38 Characterize graphs 119866 that satisfy 120574119903

times119896119905(119866119866) =

120574119903

times(119896minus1)119905(119866) + 120574

119903

times(119896minus1)119905(119866)

References

[1] M A Henning and A P Kazemi ldquo119896-tuple total domination ingraphsrdquo Discrete Applied Mathematics vol 158 no 9 pp 1006ndash1011 2010

[2] T W Haynes M A Henning P J Slater and L C van derMerwe ldquoThe complementary product of two graphsrdquo Bulletinof the Institute of Combinatorics and Its Applications vol 51 pp21ndash30 2007

[3] TW Haynes S T Hedetniemi and P J Slater Fundamentals ofDomination in Graphs vol 208 ofMonographs and Textbooks inPure and Applied Mathematics Marcel Dekker New York NYUSA 1998

[4] TWHaynes S THedetniemi and P J Slater EdsDominationin Graphs Advanced Topics Marcel Dekker New York NYUSA 1998

[5] M A Henning and A P Kazemi ldquo119896-tuple total domination incross products of graphsrdquo Journal of Combinatorial Optimiza-tion vol 24 no 3 pp 339ndash346 2012

[6] A P Kazemi ldquo119896-tuple total domination in complementaryprismsrdquo ISRNDiscreteMathematics vol 2011 Article ID68127413 pages 2011

[7] E J Cockayne and S T Hedetniemi ldquoTowards a theory ofdomination in graphsrdquoNetworks vol 7 no 3 pp 247ndash261 1977

[8] E J Cockayne R M Dawes and S T Hedetniemi ldquoTotaldomination in graphsrdquoNetworks vol 10 no 3 pp 211ndash219 1980

[9] S M Sheikholeslami and L Volkmann ldquoThe 119896-tuple totaldomatic number of a graphrdquo submitted to Utilitas Mathemat-ica httparxivorgabs11065589

[10] A P Kazemi ldquo119896-tuple total restrained dominationdomatic ingraphsrdquo Bulletin of the IranianMathematical Sciences accepted

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of