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ISBN 978-1-59973-146-9

VOLUME 1, 2011

MATHEMATICAL COMBINATORICS

(INTERNATIONAL BOOK SERIES)

Edited By Linfan MAO

THE MADIS OF CHINESE ACADEMY OF SCIENCES

March, 2011



Vol.1, 2011 ISBN 978-1-59973-146-9

Mathematical Combinatorics(International Book Series)

Edited By Linfan MAO

The Madis of Chinese Academy of Sciences

March, 2011



Aims and Scope: The Mathematical Combinatorics (International Book Series)(ISBN 978-1-59973-146-9 ) is a fully refereed international book series, sponsored by the MADIS of Chinese Academy of Sciences and published in USA quarterly comprising 100-150 pagesapprox. per volume, which publishes original research papers and survey articles in all as-

pects of Smarandache multi-spaces, Smarandache geometries, mathematical combinatorics,non-euclidean geometry and topology and their applications to other sciences. Topics in detailto be covered are:

Smarandache multi-spaces with applications to other sciences, such as those of algebraicmulti-systems, multi-metric spaces, · · ·, etc.. Smarandache geometries;

Differential Geometry; Geometry on manifolds;Topological graphs; Algebraic graphs; Random graphs; Combinatorial maps; Graph and

map enumeration; Combinatorial designs; Combinatorial enumeration;Low Dimensional Topology; Differential Topology; Topology of Manifolds;Geometrical aspects of Mathematical Physics and Relations with Manifold Topology;

Applications of Smarandache multi-spaces to theoretical physics; Applications of Combi-natorics to mathematics and theoretical physics;Mathematical theory on gravitational elds; Mathematical theory on parallel universes;Other applications of Smarandache multi-space and combinatorics.

Generally, papers on mathematics with its applications not including in above topics arealso welcome.

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Editorial Board

Editor-in-Chief Linfan MAOChinese Academy of Mathematics and SystemScience, P.R.ChinaEmail: [email protected]

Editors

S.BhattacharyaDeakin UniversityGeelong Campus at Waurn Ponds

AustraliaEmail: [email protected]

An ChangFuzhou University, P.R.ChinaEmail: [email protected]

Junliang CaiBeijing Normal University, P.R.ChinaEmail: [email protected]

Yanxun Chang

Beijing Jiaotong University, P.R.ChinaEmail: [email protected]

Shaofei DuCapital Normal University, P.R.ChinaEmail: [email protected]

Florentin Popescu and Marian PopescuUniversity of CraiovaCraiova, Romania

Xiaodong Hu

Chinese Academy of Mathematics and SystemScience, P.R.ChinaEmail: [email protected]

Yuanqiu HuangHunan Normal University, P.R.ChinaEmail: [email protected]

H.IseriManseld University, USAEmail: [email protected]

M.KhoshnevisanSchool of Accounting and Finance,Griffith University, Australia

Xueliang LiNankai University, P.R.ChinaEmail: [email protected]

Han RenEast China Normal University, P.R.ChinaEmail: [email protected]

W.B.Vasantha Kandasamy

Indian Institute of Technology, IndiaEmail: [email protected]

Mingyao XuPeking University, P.R.ChinaEmail: [email protected]

Guiying YanChinese Academy of Mathematics and SystemScience, P.R.ChinaEmail: [email protected]

Y. ZhangDepartment of Computer ScienceGeorgia State University, Atlanta, USA



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By Thomas Edison, an American inventor.



Math.Combin.Book Ser. Vol.1 (2011), 01-19

Lucas Graceful Labeling for Some Graphs

M.A.Perumal 1 , S.Navaneethakrishnan 2 and A.Nagarajan 2

1. Department of Mathematics, National Engineering College,

K.R.Nagar, Kovilpatti, Tamil Nadu, India

2. Department of Mathematics, V.O.C College, Thoothukudi, Tamil Nadu, India.

Email : [email protected], [email protected], [email protected]

Abstract : A Smarandache-Fibonacci triple is a sequence S (n), n ≥ 0 such thatS (n) = S (n − 1) + S (n − 2), where S (n) is the Smarandache function for integers

n ≥ 0. Clearly, it is a generalization of Fibonacci sequence and Lucas sequence . LetG be a ( p, q )-graph and S (n)|n ≥ 0 a Smarandache-Fibonacci triple. An bijectionf : V (G) → S (0) , S (1) , S (2) , . . . , S (q ) is said to be a super Smarandache-Fibonacci grace-

ful graph if the induced edge labeling f ∗(uv) = |f (u) − f (v)| is a bijection onto the set

S (1) , S (2) , . . . , S (q ). Particularly, if S (n), n ≥ 0 is just the Lucas sequence, such a label-ing f : V (G) → l0 , l1 , l2 , · · · , la (a ǫ N ) is said to be Lucas graceful labeling if the inducededge labeling f 1 (uv) = |f (u) −f (v)| is a bijection on to the set l1 , l2 , · · · , lq . Then G iscalled Lucas graceful graph if it admits Lucas graceful labeling. Also an injective functionf : V (G) → l0 , l1 , l2 , · · · , lq is said to be strong Lucas graceful labeling if the induced edgelabeling f 1 (uv ) = |f (u) −f (v)| is a bijection onto the set l1 , l2 ,...,l q. G is called strongLucas graceful graph if it admits strong Lucas graceful labeling. In this paper, we show

that some graphs namely P n , P +n −e, S m,n , F m @P n , C m @P n , K 1 ,n ⊙2P m , C 3 @2P n andC n @K 1 , 2 admit Lucas graceful labeling and some graphs namely K 1 ,n and F n admit strongLucas graceful labeling.

Key Words : Smarandache-Fibonacci triple, super Smarandache-Fibonacci graceful graph,Lucas graceful labeling, strong Lucas graceful labeling.

AMS(2010) : 05C78

§1. Introduction

By a graph, we mean a nite undirected graph without loops or multiple edges. A path of length n is denoted by P n . A cycle of length n is denoted by C n .G+ is a graph obtained fromthe graph G by attaching a pendant vertex to each vertex of G. The concept of graceful labelingwas introduced by Rosa [3] in 1967.

A function f is a graceful labeling of a graph G with q edges if f is an injection from

1 Received November 11, 2010. Accepted February 10, 2011.



2 M.A.Perumal, S.Navaneethakrishnan and A.Nagarajan

the vertices of G to the set 1, 2, 3, · · · , q such that when each edge uv is assigned the la-bel |f (u) −f (v)|, the resulting edge labels are distinct. The notion of Fibonacci gracefullabeling was introduced by K.M.Kathiresan and S.Amutha [4]. We call a function, a Fi-bonacci graceful labeling of a graph G with q edges if f is an injection from the vertices of

G to the set 0, 1, 2,...,F q, where F q is the q th

Fibonacci number of the Fibonacci seriesF 1 = 1 , F 2 = 2 , F 3 = 3 , F 4 = 5 , ..., and each edge uv is assigned the label |f (u) −f (v)|. Basedon the above concepts we dene the following.

Let G be a ( p, q ) -graph. An injective function f : V (G) → l0 , l1 , l2 , · · · , la , (a ǫ N ),is said to be Lucas graceful labeling if an induced edge labeling f 1(uv) = |f (u) − f (v)| is abijection onto the set l1 , l2 , · · · , lq with the assumption of l0 = 0 , l1 = 1 , l2 = 3 , l3 = 4 , l4 =7, l5 = 11 , · · · ,. Then G is called Lucas graceful graph if it admits Lucas graceful labeling. Alsoan injective function f : V (G) → l0 , l1 , l2 , · · · , lq is said to be strong Lucas graceful labeling if the induced edge labeling f 1(uv) = |f (u)−f (v)| is a bijection onto the set l1 , l2 , · · · , lq. ThenG is called strong Lucas graceful graph if it admits strong Lucas graceful labeling. In this paper,we show that some graphs namely P n , P +n

−e, S m,n , F m @P n , C m @P n , K 1,n

⊙

2P m , C 3@2P nand C n @K 1,2 admit Lucas graceful labeling and some graphs namely K 1,n and F n admit strongLucas graceful labeling. Generally, let S (n), n ≥ 0 with S (n) = S (n − 1) + S (n − 2) be aSmarandache-Fibonacci triple , where S (n) is the Smarandache function for integers n ≥ 0. Anbijection f : V (G) → S (0) , S (1) , S (2) , . . . , S (q ) is said to be a super Smarandache-Fibonacci graceful graph if the induced edge labeling f ∗(uv) = |f (u) − f (v)| is a bijection onto the set

S (1), S (2), · · · , S (q ).

§2. Lucas graceful graphs

In this section, we show that some well known graphs are Lucas graceful graphs.

Denition 2.1 Let G be a ( p, q ) -graph. An injective function f : V (G) → l0 , l1 , l2 , · · · , la , ,(a ǫ N ) is said to be Lucas graceful labeling if an induced edge labeling f 1(uv) = |f (u) −f (v)| is a bijection onto the set l1 , l2 , · · · , lq with the assumption of l0 = 0 , l1 = 1 , l2 = 3 , l3 = 4 , l4 =7, l5 = 11 , · · · ,. Then G is called Lucas graceful graph if it admits Lucas graceful labeling.

Theorem 2.2 The path P n is a Lucas graceful graph.

Proof Let P n be a path of length n having ( n + 1) vertices namely v1 , v2 , v3 , · · · , vn , vn +1 .Now, |V (P n )| = n + 1 and |E (P n )| = n. Dene f : V (P n ) → l0 , l1 , l2 , · · · , la , , a ǫ N byf (u i ) = li +1 , 1 ≤ i ≤ n. Next, we claim that the edge labels are distinct. Let

E = f 1(vi vi+1 ) : 1 ≤ i ≤ n = |f (vi ) −f (vi +1 )| : 1 ≤ i ≤ n= |f (v1) −f (v2)| , |f (v2) −f (v3)| , · · · , |f (vn ) −f (vn +1 )| , = |l2 −l3| , |l3 −l4| , · · · , |ln +1 −ln +2 | = l1 , l2 , · · · , ln .

So, the edges of P n receive the distinct labels. Therefore, f is a Lucas graceful labeling.Hence, the path P n is a Lucas graceful graph.



Lucas Graceful Labeling for Some Graphs 3

Example 2.3 The graph P 6 admits Lucas graceful Labeling, such as those shown in Fig. 1 following.

l2 l3 l4 l5 l6 l7 l8

l1 l2 l3 l4 l5 l6

Fig.1

Theorem 2.4 P +n −e, (n ≥ 3) is a Lucas graceful graph.

Proof Let G = P +n −e with V (G) = u1, u 2 , · · · , un +1 v2 , v3 , · · · , vn +1 be the vertexset of G. So, |V (G)| = 2 n + 1 and |E (G)| = 2 n. Dene f : V (G) → l0 , l1 , l2 , · · · , la , ,a ǫ N,by

f (u i ) = l2i−1 , 1 ≤ i ≤ n + 1 and f (vj ) = l2( j −1) , 2 ≤ j ≤ n + 1 .

We claim that the edge labels are distinct. Let

E 1 = f 1(u i u i+1 ) : 1 ≤ i ≤ n = |f (u i ) −f (u i+1 )| : 1 ≤ i ≤ n= |f (u1) −f (u2)|, |f (u2) −f (u3)|, · · · , |f (un ) −f (un +1 )|= |l1 −l3|, |l3 −l5|, · · · , |l2n −1 −l2n +1 | = l2, l4 , · · · , l2n ,

E 2 = f 1(u i vj ) : 2 ≤ i, j ≤ n= |f (u2) −f (v2)|, |f (u3) −f (v3)|, · · · , |f (un +1 ) −f (vn +1 )|= |l3 −l2|, |l5 −l4|, · · · , |l2n +1 −l2n | = l1 , l3 , · · · , l2n −1.

Now, E = E 1 ∪E 2 = l1 , l3 , · · · , l2n −1 , l2n . So, the edges of G receive the distinct labels.Therefore, f is a Lucas graceful labeling. Hence, P +n −e, (n ≥ 3) is a Lucas graceful graph.

Example 2.5 The graph P +8 −e admits Lucas graceful labeling, such as thsoe shown in Fig.2.

l1 l3 l5 l7 l9 l11 l13 l15 l17

l2 l4 l6 l8 l10 l12 l14 l16

l1 l3 l5 l7 l9 l11 l13 l15

l2 l4 l6 l8 l10 l12 l14 l16

Fig.2

Denition 2.6([2]) Denote by S m,n such a star with n spokes in which each spoke is a path of length m.

Theorem 2.7 The graph S m,n is a Lucas graceful graph when m is odd and n ≡ 1, 2(mod 3).




Proof Let G = S m,n and let V (G) = u ij : 1 ≤ i ≤ m and 1 ≤ j ≤ n be the vertex set of

S m,n . Then |V (G)| = mn + 1 and |E (G)| = mn . Dene f : V (G) → l0 , l1 , l2 , · · · , la , ,a ǫ N by

f (u0) = l0 for i = 1 , 2, · · · , m −2 and i ≡ 1(mod 2);

f u ij = ln ( i−1)+2 j −1 , 1 ≤ j ≤ n for i = 1 , 2, · · · , m −1 and i ≡ 0(mod 2);f u i

j = lni +2 −2j , 1 ≤ j ≤ n and for s = 1 , 2, · · · , n3

,

f umj = ln (m −1)+2( j +1) −3s , 3s −2 ≤ j ≤ 3s.


E 1 =m

i =1i≡1( mod 2)

f 1 u0 ui1 =

m

i =1i≡1( mod 2)

f (u0) −f u i1

=m

i =1i

≡1( mod 2)

l0 −ln ( i−1)+1 =m

i =1i

≡1( mod 2)

ln ( i−1)+1

= l1 , l2n +1 , l4n +1 , · · · , ln (m −1)+1 ,

E 2 =m −1

i =1i≡1( mod 2)

f 1 u0 ui1 =

m −1

i =1i≡1( mod 2)

f (u0) −f u i1

=m −1

i =1

i≡1( mod 2)

|l0 −lni | =m −1

i =1

i≡1( mod 2)

lni = l2n , l4n , · · · , ln (m −1)

E 3 =m −2

i =1

i≡1( mod 2)

f 1 u ij u i

j +1 : 1 ≤ j ≤ n −1

=m −2

i =1i≡1( mod 2)

f u ij −f u i

j +1 : 1 ≤ j ≤ n −1

=m −2

i =1i≡1( mod 2)

ln ( i−1)+2 j −1 −ln ( i−1)+2 j +1 : 1 ≤ j ≤ n −1

=m −2

i =1i≡1( mod 2)

ln ( i

−1)+2 j : 1

≤ j

≤ n

−1

=m −2

i =1i≡1( mod 2)

ln ( i−1)+2 , ln ( i−1)+4 , · · · , : ln ( i−1)+2( n −1)

= l2 , l2n +2 ,...,l n (m −3)+2 ∪ l4 , l2n +4 , · · · , ln (m −3)+4 ∪ · · ·∪ l2n −2 , l4n −2 ,...,l n (m −3)+2 n −2 ,




E 4 =m −2

i =1

i≡1( mod 2)

f 1 u ij ui

j +1 : 1 ≤ j ≤ n −1

=m −2

i =1i≡1( mod 2)

f u ij

−f u i

j +1 : 1

≤ j

≤ n

−1

=m −2

i =1i≡1( mod 2)

|lni −2j +2 −lni −2j | : 1 ≤ j ≤ n −1

=m −2

i =1

i≡1( mod 2)

lni −2j +1 : 1 ≤ j ≤ n −1

=m −2

i =1

i≡1( mod 2)

lni −1 , lni −3 , · · · , lni −(2 n −3)

= l2n −1 , l2n −3 , · · · , l3 , l4n −1 , l4n −3 , · · · , l2n +3 , ln (m −1) −1, · · · , ln (m −1)−(2 n −3) .

For n ≡ 1(mod 3), let

E 5 =

n − 13

s =1f 1 um

j umj +1 : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1f um

j −f umj +1 : 3s −2 ≤ j ≤ 3s −1

=

n − 1

3

s =1ln (m −1)+2 j −3s +2 −ln (m −1)+2 j −3s +4 : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1ln (m −1)+2 j −3s +2 : 3s −2 ≤ j ≤ 3s −1 =

n − 13

s =1ln (m −1)+3 s−1 , ln (m −1)+3 s +1

= ln (m −1)+2 , ln (m −1)+4 , ln (m −1)+5 , ln (m −1)+7 , · · · , ln m −2 , lmn .

We nd the edge labeling between the end vertex of sth loop and the starting vertex of (s + 1) th loop and s = 1 , 2, · · · ,

n −13

. Let

E 6 =

n − 13

s =1f 1 um

3s um3s +1 =

n − 13

s =1f (um

3s ) −f um3s +1

= |f (um3 ) −f (um

4 )| , |f (um6 ) −f (um

7 )| , |f (um9 ) −f (um

10 )| , · · · , f umn −1 −f (um

n ) = ln (m −1)+5 −ln (m −1)+4 , ln (m −1)+8 −ln (m −1)+7 , · · · , ln (m −1)+ n +1 −ln (m −1)+ n

= ln (m −1)+3 , ln (m −1)+6 , · · · , ln (m −1)+ n −1 = ln (m −1)+3 , ln (m −1)+6 , · · · , lnm −1 .





E ′5 =

n − 13

s =1f 1 um

j umj +1 : 3s −2 ≤ j ≤ 3s −1

=

n − 1

3

s =1f um

j −f umj +1 : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1ln (m −1)+2 j −3s +2 −ln (m −1)+2 j −3s +4 : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1ln (m −1)+2 j −3s +3 : 3s −2 ≤ j ≤ 3s −1 =

n − 13

s =1ln (m −1)+3 s−1 , ln (m −1)+3 s +1

= ln (m −1)+2 , ln (m −1)+4 , ln (m −1)+5 , ln (m −1)+7 , · · · , ln (m −1)+ n −2 , ln (m −1)+ n .

We determine the edge labeling between the end vertex of s th loop and the starting vertexof (s + 1) th loop and s = 1 , 2, 3, ...,

n

−1

3 .

Let E ′

6 =

n − 13

s =1f 1 um

3s um3s +1 =

n − 13

s =1f (um

3s ) −f m3s +1

= |f (um3 ) −f (um

4 )| , |f (um6 ) −f (um

7 )| , |f (um9 ) −f (um

10 )| , · · · , f umn −1 −f (um

n )

= ln (m −1)+5 −ln (m −1)+4 , ln (m −1)+8 −ln (m −1)+7 , · · · , ln (m −1)+ n +1 −ln (m −1)+ n

= ln (m −1)+3 , ln (m −1)+6 , · · · , lnm −1 .

Now,E =6

i =1E i if n ≡ 1(mod 3) and E =

6

i =1E i E

′

5 E ′

6 if n ≡ 2(mod 3). So the

edges of S m,n (when m is odd and n

≡ 1, 2(mod 3)), receive the distinct labels. Therefore, f is

a Lucas graceful labeling. Hence, S m,n is a Lucas graceful graph if m is odd, n ≡ 1, 2(mod 3).

Example 2.8 The graphs S 5,4 and S 5,5 admit Lucas graceful labeling, such as those shown inFig.3 and Fig 4.

l1 l3 l5 l7

l8 l6 l7 l2

l9 l11 l13 l15

l16 l14 l12 l10

l17 l19 l21 l20

l2 l4 l6

l7 l5 l3

l10 l12 l14

l15 l13 l11

l18 l20 l19

l1

l8

l9

l16

l17

l0

Fig.3






For s = 1 , 2, 3, · · · , n −1

3 and n ≡ 1(mod 3), let

E 4 =

n − 13

s =1f 1 (u j , u j +1 ) : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1|f (u j ) −f (u j +1 )| : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1|l2m +2 j +3 −3s −l2m +2 j +5 −3s | : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1(l2m +2 j +4 −3s : 3s −2 ≤ j ≤ 3s −1)

= l2m +2 j −2 : 4 ≤ j ≤ 5 l2m +2 j −5 : 7 ≤ j ≤ 8 · · ·

l2m +2 j −n +4

: n

−3

≤ j

≤ n

−2

= l2m +6 , l2m +8 ∪l2m +9 , l2m +11 · · · l2m + n −2 , l2m + n = l2m +6 , l2m +8 , l2m +9 , l2m +11 , · · · , l2m + n −2 , l2m + n

We nd the edge labeling between the end vertex of sth loop and the starting vertex of

(s + 1) th loop and s = 1 , 2, 3, · · · , n −1

3 , n ≡ 1(mod 3). Let

E 5 =

n − 13

s =1f 1 (u j uj +1 ) : j = 3 s =

n − 13

s =1|f (u j ) −f (u j +1 )| : j = 3 s

=

n − 13

s =1 |l2m +2 j +3 −3s −l2m +2 j +5 −3s | : j = 3 s= |l2m +2 j −l2m +2 j −1| : j = 3∪|l2m +2 j −3 −l2m +2 j −4| : j = 6 · · ·

|l2m +2 j −l2m +2 j −1 | : j = n −1= l2m +2 j −2 : j = 3∪l2m +2 j −5 : j = 6∪, · · · ,∪l2m +2 j −n +3 : j = n −1= l2m +4 , l2m +7 , · · · , l2m + n +1 .

For s = 1 , 2, 3, · · · , n −2

3 and n ≡ 2(mod 3), let

E ′

4 =

n − 23

s =1 f 1(u j uj +1 ) : 3s −2 ≤ j ≤ 3s −1=

n − 23

s =1|f (u j ) −f (u j +1 )| : 3s −2 ≤ j ≤ 3s −1

=

n − 23

s =1|l2m +2 j +3 −3s −l2m +2 j +5 −3s | : 3s −2 ≤ j ≤ 3s −1




=

n − 23

s =1(l2m +2 j +4 −3s : 3s −2 ≤ j ≤ 3s −1)

= l2m +2 j −2 : 4 ≤ j ≤ 5 l2m +2 j −5 : 7 ≤ j ≤ 8 · · ·l2m +2 j −n +4 : n −3 ≤ j ≤ n −2

= l2m +6 , l2m +8 l2m +9 , l2m +11 · · · l2m + n −2 , l2m + n = l2m +6 , l2m +8 ,2m +9 , l2m +11 , · · · , l2m + n −2 , l2m + n

We determine the edge labeling between the end vertex of s th loop and the starting vertexof (s + 1) th loop and s = 1 , 2, 3, ...,

n −23

, n ≡ 2(mod 3). Let

E ′

5 =

n − 23

s =1f 1 (u j , u j +1 ) : j = 3 s

=

n − 23

s =1 |f (u j ) −f (u j +1 )| : j = 3 s =

n − 23

s =1 |l2m +2 j +3 −3s −l2m +2 j +5 −3s | : j = 3 s= |l2m +2 j −l2m +2 j −1 | : j = 3 |l2m +2 j −3 −l2m +2 j −4 | : j = 6 · · ·

|l2m +2 j −n +4 −l2m +2 j −n +5 | : j = n −1= l2m +2 j −2 : j = 3∪l2m +2 j −5 : j = 6 · · · l2m +2 j −(n −3) : j = n −1

= l2m +4 , l2m +7 ,...,l 2m + n +1 .

Now, E =5

i=1E i if n ≡ 1(mod 3) and E =

5

i=1E i E

′

4 E ′

5 if n ≡ 2(mod 3). So, the

edges of F m @P n (whenn ≡ 1, 2(mod 3)) are the distinct labels. Therefore, f is a Lucas graceful

labeling. Hence, G = F m @P n (if n ≡ 1, 2(mod 3)) is a Lucas graceful labeling.

Example 2.11 The graph F 5 @P 4 admits a Lucas graceful labeling shown in Fig.5.

l1 l3 l5 l7 l9 l11

l0 l12 l14 l16 l15

l2 l4 l6 l8 l10

l12 l13 l15 l14

l1l3

l5 l7l9

l11

Fig.5

Denition 2.12 ([2]) The Graph G = C m @P n consists of a cycle C m and a path of P n of length n which is attached with any one vertex of C m .




Theorem 2.13 The graph C m @P n is a Lucas graceful graph when m ≡ 0(mod 3) and n =1, 2(mod 3).

Proof Let G = C m @P n and let u1 , u2 , · · · , um be the vertices of a cycle C m and v1 , v2 , · · · , vn , vn +1

be the vertices of a path P n which is attached with the vertex ( u1 = v1) of C m . Let V (G) =

u1 = v1∪u2 , u3 , · · · , um ∪v2 , v3 ,...,v n , vn +1 be the vertex set of G. So, |V (G)| = m + nand |E (G)| = m + n. Dene f : V (G) → l0 , l1 , · · · , la , a ǫ N by f (u1) = f (v1) = l0 ; f (u i ) =l2i−3s , 3s −1 ≤ j ≤ 3s + 1 for s = 1 , 2, 3, · · · ,

m3

, i = 2 , 3, · · · , m; f (vj ) = lm +2 j −3r , 3r −1 ≤ j ≤ 3r + 1 for r = 1 , 2, · · · ,

n + 13

and j = 2 , 3, · · · , n + 1.We claim that the edge labels are distinct. Let

E 1 = f 1 (u1 u2) = |f (u1) −f (u2)| = ( |l0 −l1|) = l1,

E 2 =

m3

s =1 f 1 (u i ui +1 ) : 3s

−1

≤ i

≤ 3s and u m +1 = u1

=

m3

s =1f 1 (u i ) −f (u i+1 ) : 3s −1 ≤ i ≤ 3s and u m +1 = u1

= |f (u2) −f (u3)| , |f (u3) −f (u4)| ,..., |f (um ) −f (um +1 )|= |l1 −l3| , |l3 −l5| , |l4 −l6| , |l6 −l8| , · · · , |lm −l0|= l2 , l4 , l5 , l7 , · · · , lm

We determine the edge labeling between the end vertex of s th loop and the starting vertexof (s + 1) th loop and s = 1 , 2,...,

m3 −1. Let

E 3 =

m

3 −1

s =1f 1(u3s +1 u3s +2 ) =

m

3 −1

s =1|f (u3s +1 ) −f (u3s +2 )|

= |f (u4) −f (u5)| , |f (u7) −f (u8)| , · · · , |f (um −2) −f (um −1)|= |l5 −l4| , |l8 −l7| , · · · , |lm −1 −lm −2|= l3 , l6 , · · · , lm −3,

E 4 = f 1(v1 v2) = |f (v1) −f (v2)| = |l0 −lm +4 −3||l0 −lm +4 −3|= |l0 −lm +1 | = |l0 −lm +1 | = lm +1 .

For n

≡ 1(mod 3), let

E 5 =

n − 13

r =1f 1(vj vj +1 ) : 3r −1 ≤ j ≤ 3r

=

n − 13

r =1|f (vj ) −f (vj +1 )| : 3r −1 ≤ j ≤ 3r




= |f (v2) −f (v3)| , |f (v3) −f (v4)| , · · · , |f (vn −1) −f (vn )|= |lm +4 −3 −lm +6 −3| , |lm +6 −3 −lm +8 −3| , |lm +10 −6 −lm +12 −6| , |lm +12 −6 −lm +14 −6| ,

· · · , |lm +2 n −2−n +1 −lm +2 n −n +1 |=

|lm +1

−lm +3

|,

|lm +3

−lm +5

|,

|lm +4

−lm +6

|,

|lm +6

−lm +8

|,

· · · ,

|lm + n

−1

−lm + n +1

|= lm +2 , lm +4 , lm +5 , lm +7 , · · · , lm + n .

We calculate the edge labeling between the end vertex of r th loop and the starting vertexof (r + 1) th loop and r = 1 , 2, · · · ,

n −13

. Let

E 6 =

n − 13

r =1f 1(v3r +1 v3r +2 ) =

n − 13

r =1|f (v3r +1 ) −f (v3r +2 )|

= |f (v4 ) −f (v5 )| , |f (v7) −f (v8)| , · · · , |f (vn −2) −f (vn −1)|= |lm +8 −3 −lm +10 −6| , |lm +14 −6 −lm +16 −9| , · · · , |lm +2 n −4−n +2 −lm +2 n −2−n +1 |= |lm +5 −lm +4 | , |lm +8 −lm +7 | , · · · , |lm + n −2 −lm + n |= lm +3 , lm +6 , lm +9 , · · · , lm + n −1


E ′

5 =

n − 13

r =1f 1(vj vj +1 ) : 3r −1 ≤ j ≤ 3r =

n − 13

r =1|f (vj ) −f (vj +1 )| : 3r −1 ≤ j ≤ 3r

= |f (v2 ) −f (v3)| , |f (v3) −f (v4 )| , · · · , |f (vn −1) −f (vn )|= |lm +4 −3 −lm +6 −3| , |lm +6 −3 −lm +8 −3| , |lm +10 −6 −lm +12 −6| , |lm +12 −6 −lm +14 −6| ,

· · · , |lm +2 n −2−2n +1 −lm +2 n −n +1 |= lm +2 , lm +4 , lm +5 , lm +7 ,...,l m + n .

We nd the edge labeling between the end vertex of r th loop and the starting vertex of

(r + 1) th loop and r = 1 , 2, · · · , n −2

3 . Let

E ′

6 =

n − 23

r =1f 1(v3r +1 v3r +2 ) =

n − 23

r =1|f (v3r +1 ) −f (v3r +2 )|

= |f (v4 ) −f (v5 )| , |f (v7) −f (v8)| , · · · , |f (vn −2) −f (vn −1)|= |lm +8 −3 −lm +10 −6| , |lm +14 −6 −lm +16 −9| , · · · , |lm +2 n −4−n +2 −lm +2 n −2−n +1 |= |lm +5 −lm +4 | , |lm +8 −lm +7 | , · · · , |lm + n −2 −lm + n |= lm +3 , lm +6 , lm +9 , · · · , lm + n −1

Now, E =6

i=1E i if n ≡ 1(mod 3) and E =

4

i=1E i E

′

5 E ′

6 if n ≡ 2(mod 3). So,

the edges of G receive the distinct labels. Therefore, f is a Lucas graceful labeling. Hence,G = C m @P n is a Lucas graceful graph when m ≡ 0(mod 3) and n ≡ 1, 2(mod 3).








Now, E =4

i =1E i = l1 , l2 ,...,l (2 m +1) n . So, the edge labels of G are distinct. Therefore,

f is a Lucas graceful labeling. Hence, G = K 1,n ⊙2P m is a Lucas graceful labeling.

Example 2.17 The graph K 1,4

⊙2P 4 admits Lucas graceful labeling, such as those shown in

Fig.7.

l3 l5 l7 l9

l4 l6 l8 l10

l12

l14

l16

l18

l13

l15

l17

l19

l21l23l25l27

l28 l26 l24 l22

l30

l32

l34

l36

l31

l33

l35

l37

l2

l11

l20

l29

l4 l6 l8

l5 l7 l9

l13

l15

l17

l14

l16

l18

l22l24l26

l27 l25 l23

l31

l33

l35 l36

l34

l32

l1

l3

l10l12

l19

l21

l28 l30

l2l11

l20

l29

l0

Fig.7

Theorem 2.18 The graph C 3 @2P n is Lucas graceful graph when n ≡ 1(mod 3).

Proof Let G = C 3@2P n with V (G) = wi : 1 ≤ i ≤ 3∪u i : 1 ≤ i ≤ n∪vi : 1 ≤ i ≤ nand the vertices w2 and w3 of C 3 are identied with v1 and u1 of two paths of length nrespectively. Let E (G) = wi wi +1 : 1 ≤ i ≤ 2∪u i u i +1 , vi vi+1 : 1 ≤ i ≤ n be the edge set of G. So, |V (G)| = 2 n + 3 and |E (G)| = 2 n + 3. Dene f : V (G) → l0 , l1 , l2 , · · · , la , a ǫ N by f (w1) = ln +4 ; f (u i ) = ln +3 −i , 1 ≤ i ≤ n + 1; f (vj ) = ln +4+2 j −3s , 3s − 2 ≤ j ≤ 3s for

s = 1 , 2,..., n −1

3 and f (vj ) = ln +4+2 j −3s 3s −2 ≤ j ≤ 3s −1 for s =

n −13

+ 1.





E 1 =n

i=1f 1(u i u i +1 =

n

i=1|f (u i ) −f (u i +1 )|

=n

i=1|ln +3 −i −ln +3 −i−1| =

n

i=1|ln +3 −i −ln +2 −i |

=n

i=1ln +1 −i = ln , ln −1 , · · · , l1,

E 2 = f 1(u1w1), f 1(w1 v1 ), f 1(v1 u1)= |f (u1) −f (w1 )| , |f (w1 −f (v1)| , |f (v1 ) −f (u1)|= |ln +2 −ln +4 | , |ln +4 −ln +3 | , |ln +3 −ln +2 | = ln +3 , ln +2 , ln +1 .

For s = 1 , 2, · · · , n −1

3 , let

E 3 =

n − 1

3

s =1f 1 (vj vj +1 ) : 3s −2 ≤ j ≤ 3s −1

=

n − 13

s =1|f (vj ) −f (vj +1 )| : 3s −2 ≤ j ≤ 3s −1

= |f (v1 ) −f (v2 )| , |f (v2) −f (v3)|∪|f (v4) −f (v5)| , |f (v5) −f (v6)|· · · |f (vn −3) −f (vn −2)| , |f (vn −2 ) −f (vn −1)|

= |ln +3 −ln +5 | , |ln +5 −ln +7 |∪|ln +6 −ln +8 | , |ln +8 −ln +10 |· · · |l2n −1 −l2n +1 | , |l2n +1 −l2n +3 |

= ln +4 , ln +6 ln +7 , ln +9 · · · l2n , l2n +2 .

We nd the edge labeling between the end vertex of sth loop and the starting vertex of (s + 1) th loop and 1 ≤ s ≤

n −13

. Let

E 4 = f 1(vj vj +1 ) : j = 3 s = |f (vj ) −f (vj +1 )| : j = 3 s= |f (v3 ) −f (v4 )| , |f (v6) −f (v7)| , · · · , |f (vn −1) −f (vn )|= |ln +7 −ln +6 | , |ln +10 −ln +9 | , · · · , |l2n +3 −l2n +2 | = l5 , l8 , · · · , l2n +1 .

For s = n −1

3 + 1, let

E 5 = f 1(vj v( j +1 ) : j = 3 s −2 = |f (vj ) −f (vj +1 )| : j = n= |f (vn ) −f (vn +1 )| = |ln +4+2 n −n −2 −ln +4+2 n +2 −n −2|= |l2n +2 −l2n +4 | = l2n +3 .

Now, E =5

s =1E i = l1 , l2 ,...,l 2n +3 . So, the edge labels of G are distinct. Therefore, f is

a Lucas graceful labeling. Hence, G = C 3 @2P n is a Lucas graceful graph if n ≡ 1(mod 3).




Example 2.19 The graph C 3 @2P 4 admits Lucas graceful labeling shown in Fig.8.

v1 v2 v3 v4 v5

l7 l9 l11 l10 l12

l8 l10 l9 l11

l5

l6 l5 l4 l3 l2u1 u2 u3 u4 u5l4 l3 l2 l1

w3

w2

w1

l6

l8

l7

Fig.8

Theorem 2.20 The graph C n @K 1,2 is a Lucas graceful graph if n

≡ 1(mod 3).

Proof Let G = C n @K 1,2 with V (G) = u i : 1 ≤ i ≤ n ∪ v1 , v2, E (G) = u i u i +1 :1 ≤ i ≤ n −1∪un u1 , un vn , un v2. So, |V (G)| = n +2 and |E (G)| = n +2. Dene f : V (G) →l0 , l1 , l2 ,...,l a , a ǫ N by f (u1) = 0 , f (v1 ) = ln , f (v2) = ln +3 ; f (u i ) = l2i−3s , 3s −1 ≤ i ≤3s + 1 for s = 1 , 2,...,

n −43

and f (u i ) = l2i−3s , 3s −1 ≤ i ≤ 3s for s = n −1

3 . We claim that

the edge labels are distinct. Let

E 1 = f 1 (u1u2) , f 1(un v1), f 1(un v2), f 1(un u1)= |f (u1) −f (u2)| , |f (un ) −f (v1)| , |f (un ) −f (v2)| , |f (un ) −f (v1)|=

|l0

−l1

|,

|ln +1

−ln

|,

|ln +1

−ln +3

|,

|ln +1

−l0

|= l1 , ln −1 , ln +2 , ln +1 ,

E 2 =

n − 43

s =1f 1(u i u i+1 ) : 3s −1 ≤ i ≤ 3s

=

n − 43

s =1|f (u i ) −f (u i+1 )| : 3s −1 ≤ i ≤ 3s

= |f (u2) −f (u3)| , |f (u3) −f (u4)| |f (u5) −f (u6)| , |f (u6) −f (u7)|

· · · |f (un

−5)

−f (un

−4)

|,

|f (un

−4)

−f (un

−3)

|= |l1 −l3| , |l3 −l5| |l4 −l6| , |l6 −l8|· · · |ln −6 −ln −4| , |ln −5 −ln −2|

= l2 , l4 l5 , l7 · · · ln −5 , ln −3 = l2 , l4 , l5 , l7 , · · · , ln −5 , ln −3We determine the edge labeling between the end vertex of s th loop and the starting vertex






7, l5 = 11 , · · · ,. Then G is called strong Lucas graceful graph if it admits strong Lucas graceful labeling.

Theorem 3.2 The graph K 1,n is a strong Lucas graceful graph.

Proof Let G = K 1,n and V = V 1 ∪ V 2 be the bipartition of K 1,n with V 1 = u1 andV 2 = u1 , u 2 ,...,u n . Then, |V (G)| = n +1 and |E (G)| = n. Dene f : V (G) → l0 , l1 , l2 ,...,l n by f (u0) = l0 , f (u1) = l1 , 1 ≤ i ≤ n. We claim that the edge labels are distinct. Notice that

E = f 1(u0u1) : 1 ≤ i ≤ n = f (u0) −f (u1) : 1 ≤ i ≤ n= |f (u0) −f (u1)| , |f (u0) −f (u2)| , ..., |f (u0) −f (un )|= |l0 −l1| , |l0 −l2| ,..., |l0 −ln | = l1 , l2 ,...,l n

So, the edges of G receive the distinct labels. Therefore, f is a strong Lucas graceful labeling.Hence, K 1 , n the path is a strong Lucas graceful graph.

Example 3.3 The graph K 1,9 admits strong Lucas graceful labeling shown in Fig.10.l0

l1 l2 l3 l4 l5 l6 l7l8

l9

l1l2

l3 l4 l5 l6

l7 l8

l9

Fig.10

Denition 3.4([2]) Let u1 , u2 ,...,u n , un +1 be the vertices of a path and u0 be a vertex which is attached with u1 , u 2 ,...,u n , un +1 . Then the resulting graph is called Fan and is denoted by F n = P n + K 1 .

Theorem 3.5 The graph F n = P n + K 1 is a Lucas graceful graph.

Proof Let G = F n and u1 , u2 ,...,u n , un +1 be the vertices of a path P n with the centralvertex u0 joined with u1 , u2 ,...,u n , u n +1 . Clearly, |V (G)| = n + 2 and |E (G)| = 2 n + 1. Dene

f : V (G) → l0 , l1 , l2 ,...,l 2n +1 by f (u0) = l0 and f (u i ) = l2i−1 , 1 ≤ i ≤ n + 1. We claim thatthe edge labels are distinct.Calculation shows that

E 1 = f 1(u i u i+1 ) : 1 ≤ i ≤ n = |f ( u i ) −f (u i +1 )| : 1 ≤ i ≤ n= |f (u1) −f (u2)|, |f (u2) −f (u3)|,..., |f (un ) −f (un +1 )|= |l1 −l3|, |l3 −l5|, ..., |l2n −1 −l2n +1 | = l2, l4 ,...,l 2n ,




E 2 = f 1(u0u i ) : 1 ≤ i ≤ n + 1 = |f (u0) −f (u i )| : 1 ≤ i ≤ n + 1 = |f (u0) −f (u1)|, |f (u0) −f (u2)|, ..., |f (u0) −f (un +1 )|= |l0 −l1|, |l0 −l3|,..., |l0 −l2n +1 | = l1 , l3 ,...,l 2n +1 .

Whence, E = E 1 ∪E 2 = l1 , l2 ,...,l 2n , l2n +1 . Thus the edges of F n receive the distinct labels.Therefore, f is a Lucas graceful labeling. Consequently, F n = P n + K 1 is a Lucas gracefulgraph.

Example 3.6 The graph F 7 = P 7 + K 1 admits Lucas graceful graph shown in Fig.11.

l1

l3

l5

l7

l9

l11

l13

l15

l0

l2

l4

l6

l8

l10

l12

l14

l1l3

l5

l7

l9

l11

l13l15

Fig.11

References

[1] David M.Burton, Elementary Number Theory ( Sixth Edition), Tata McGraw - Hill Edition,Tenth reprint 2010.

[2] G.A.Gallian, A Dynamic Survey of Graph Labeling, The Electronic Journal of Combina-torics , 16(2009) # DS 6, pp 219.

[3] A.Rosa, On certain valuations of the vertices of a graph, Theory of Graphs International Symposium , Rome, 1966.

[4] K.M.Kathiresan and S.Amutha, Fibonacci Graceful Graphs , Ph.D. thesis of Madurai Ka-

maraj University, October 2006.




Sequences on Graphs with Symmetries

Linfan Mao

Chinese Academy of Mathematics and System Science, Beijing, 10080, P.R.China

Beijing Institute of Civil Engineering and Architecture, Beijing, 100044, P.R.China

E-mail: [email protected]

Abstract : An interesting symmetry on multiplication of numbers found byProf.Smarandache recently. By considering integers or elements in groups on graphs, weextend this symmetry on graphs and nd geometrical symmetries. For extending further,Smarandache’s or combinatorial systems are also discussed in this paper, particularly, theCC conjecture presented by myself six years ago, which enables one to construct more sym-metrical systems in mathematical sciences.

Key Words : Smarandache sequence, labeling, Smarandache beauty, graph, group,Smarandache system, combinatorial system, CC conjecture.

AMS(2010) : 05C21, 05E18

§1. Sequences

Let Z + be the set of non-negative integers and Γ a group. We consider sequences

i(n)

|n

∈Z +

and gn ∈ Γ|n ∈Z + in this paper. There are many interesting sequences appeared in literature.For example, the sequences presented by Prof.Smarandache in references [2], [13] and [15]following:

(1) Consecutive sequence

1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, · · ·;(2) Digital sequence

1, 11, 111, 1111 , 11111 , 11111 , 1111111 , 11111111 , · · ·(3) Circular sequence

1, 12, 21, 123, 231, 312, 1234, 2341, 3412, 4123, · · ·;(4) Symmetric sequence

1, 11, 121, 1221, 12321, 123321, 1234321, 12344321, 123454321, 1234554321, · · ·;(5) Divisor product sequence

1 Reported at The 7th Conference on Number Theory and Smarandache’s Notion, Xian, P.R.China 2 Received December 18, 2010. Accepted February 18, 2011.



Sequences on Graphs with Symmetries 21

1, 2, 3, 8, 5, 36, 7, 64, 27, 100, 11, 1728, 13, 196, 225, 1024, 17, 5832, 19, · · ·;(6) Cube-free sieve

2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25, 26, 28, 29, 30, · · ·.He also found three nice symmetries for these integer sequences recently.

First Symmetry

1 ×8 + 1 = 9

12 ×8 + 2 = 98

123 ×8 + 3 = 987

1234 ×8 + 4 = 9876

12345 ×8 + 5 = 98765

123456 ×8 + 6 = 9876541234567 ×8 + 7 = 9876543

12345678 ×8 + 8 = 98765432

123456789 ×8 + 9 = 987654321

Second Symmetry

1 ×9 + 2 = 11

12 ×9 + 3 = 111

123 ×9 + 4 = 11111234 ×9 + 5 = 11111

12345 ×9 + 6 = 111111

123456 ×9 + 7 = 1111111

1234567 ×9 + 8 = 11111111

12345678 ×9 + 9 = 111111111

123456789 ×9 + 10 = 1111111111

Third Symmetry

1 ×1 = 1

11 ×11 = 121

111 ×111 = 12321

1111 ×1111 = 1234321

11111 ×11111 = 12345431



22 Linfan Mao

111111 ×111111 = 12345654321

1111111 ×1111111 = 1234567654321

11111111 ×11111111 = 13456787654321111111111 ×111111111 = 12345678987654321

Notice that a Smarandache sequence is not closed under operation, but a group is, whichenables one to get symmetric gure in geometry. Whence, we also consider labelings on graphsG by that elements of groups in this paper.

§2. Graphs with Labelings

A graph G is an ordered 3-tuple ( V (G), E (G); I (G)), where V (G), E (G) are nite sets, called

vertex and edge set respectively, V (G) = ∅ and I (G) : E (G) → V (G) × V (G). Usually, thecardinality |V (G)| is called the order and |E (G)| the size of a graph G.A graph H = ( V 1 , E 1 ; I 1) is a subgraph of a graph G = ( V, E ; I ) if V 1 ⊆ V , E 1 ⊆ E and

I 1 : E 1 → V 1 ×V 1 , denoted by H ⊂ G.

Example 2.1 A graph G is shown in Fig.2.1, where, V (G) = v1 , v2 , v3 , v4, E (G) =

e1 , e2 , e3 , e4 , e5 , e6 , e7 , e8 , e9 , e10and I (ei ) = ( vi , vi ), 1 ≤ i ≤ 4; I (e5) = ( v1 , v2) = ( v2 , v1), I (e8)= ( v3 , v4 ) = ( v4 , v3), I (e6) = I (e7) = ( v2 , v3 ) = ( v3 , v2 ), I (e8) = I (e9) = ( v4 , v1) = ( v1 , v4).

v1 v2

v3v4

e1 e2

e3e4

e5

e6e7

e8

e9 e10

Fig. 2.1

An automorphism of a graph G is a 1

−1 mapping θ : V (G)

→ V (G) such that

θ(u, v ) = ( θ(u), θ(v)) ∈ E (G)

holds for ∀(u, v ) ∈ E (G). All such automorphisms of G form a group under compositionoperation, denoted by Aut G. A graph G is vertex-transitive if AutG is transitive on V (G).

A graph family F P is the set of graphs whose each element possesses a graph property P .Some well-known graph families are listed following.




Walk. A walk of a graph G is an alternating sequence of vertices and edges u1 , e1 , u 2 , e2 ,

· · · , en , u n 1 with ei = ( u i , u i+1 ) for 1 ≤ i ≤ n.

Path and Circuit. A walk such that all the vertices are distinct and a circuit or a cycle issuch a walk u1 , e1 , u 2 , e2 ,

· · · , en , u n 1 with u1 = un and distinct vertices. A graph G = ( V, E ; I )

is connected if there is a path connecting any two vertices in this graph.

Tree. A tree is a connected graph without cycles.

n-Partite Graph. A graph G is n -partite for an integer n ≥ 1, if it is possible to partitionV (G) into n subsets V 1 , V 2 , · · · , V n such that every edge joints a vertex of V i to a vertex of V j , j = i, 1 ≤ i, j ≤ n. A complete n-partite graph G is such an n-partite graph with edgesuv ∈ E (G) for ∀u ∈ V i and v ∈ V j for 1 ≤ i, j ≤ n, denoted by K ( p1 , p2 , · · · , pn ) if |V i | = pi forintegers 1 ≤ i ≤ n . Particularly, if |V i | = 1 for integers 1 ≤ i ≤ n , such a complete n-partitegraph is called complete graph and denoted by K n .

K (4, 4) K 6

Fig. 2.2

Two operations of graphs used in this paper are dened as follows:

Cartesian Product. A Cartesian product G1

× G2 of graphs G1 with G2 is dened by

V (G1 ×G2) = V (G1 ) ×V (G2 ) and two vertices ( u1 , u2) and ( v1 , v2 ) of G1 ×G2 are adjacent if and only if either u1 = v1 and ( u2 , v2 ) ∈ E (G2) or u2 = v2 and (u1 , v1) ∈ E (G1 ).

The graph K 2 ×P 6 is shown in Fig.2 .3 following.

u

v

1 2 3 4 5K 2

6

P 6

K 2 ×P 6

u1 u2 u3 u4 u5 u6

v1 v2 v3 v4 v5 v6

Fig. 2.3



24 Linfan Mao

Union. The union G∪H of graphs G and H is a graph ( V (G∪H ), E (G∪H ), I (G∪H )) with

V (G∪H ) = V (G)∪V (H ), E (G∪H ) = E (G)∪E (H ) and I (G∪H ) = I (G)∪I (H ).

Labeling. Now let G be a graph and N

⊂ Z+ . A labeling of G is a mapping lG : V (G)

∪E (G) → N with each labeling on an edge ( u, v ) is induced by a ruler r(lG (u), lG (v)) withadditional conditions.

Classical Labeling Rulers. The following rulers are usually found in literature.

Ruler R1. r(lG (u), lG (v)) = |lG (u) −lG (v)|.

5 4 3 2 15 0 4 1 3 2 01

2

31

4

32

4

Fig. 2.4

Such a labeling lG is called to be a graceful labeling of G if lG (V (G)) ⊂ 0, 1, 2, · · · , |V (G)|and lG (E (G)) = 1, 2, · · · , |E (G)|. For example, the graceful labelings of P 6 and S 1.4 are shownin Fig.2 .4.

Graceful Tree Conjecture (A.Rose, 1966) Any tree is graceful.

There are hundreds papers on this conjecture. But it is opened until today.

Ruler R2. r(lG (u), lG (v)) = lG (u) + lG (v).Such a labeling lG on a graph G with q edges is called to be harmonious on G if lG (V (G)) ⊂

Z (mod q ) such that the resulting edge labels lG (E (G)) = 1, 2, · · · , |E (G)| by the inducedlabeling lG (u, v ) = lG (u) + lG (v) (mod q ) for ∀(u, v ) ∈ E (G). For example, ta harmoniouslabeling of P 6 are shown in Fig.2 .5 following.

2 1 0 5 4 33 1 5 4 2

Fig. 2.5

Update results on classical labeling on graphs can be found in a survey paper [4] of Gallian.

Smarandachely Labeling Rulers. There are many new labelings on graphs appeared inInternational J.Math.Combin. in recent years. Such as those shown in the following.

Ruler R3. A Smarandachely k-constrained labeling of a graph G(V, E ) is a bijective mappingf : V ∪E → 1, 2,.., |V |+ |E | with the additional conditions that |f (u) −f (v)| ≥ k whenever




uv ∈ E , |f (u) −f (uv)| ≥ k and |f (uv) −f (vw)| ≥ k whenever u = w, for an integer k ≥ 2.A graph G which admits a such labeling is called a Smarandachely k-constrained total graph,abbreviated as k −CT G. An example for k = 5 on P 7 is shown in Fig.2 .6.

11 1 7 13 3 9 15 56 12 2 8 14 4 10

Fig.2 .6

The minimum positive integer n such that the graph G ∪K n is a k − CT G is called k-constrained number of the graph G and denoted by t k (G), the corresponding labeling is calleda minimum k-constrained total labeling of G. Update results for tk (G) in [3] and [12] are asfollows:

(1) t2(P n )=

2 if n = 2 ,

1 if n = 3 ,

0 else.(2) t2(C n ) = 0 if n ≥ 4 and t 2(C 3) = 2.(3) t2(K n ) = 0 if n ≥ 4.

(4) t2(K (m, n ))=

2 if n = 1 and m = 1 ,

1 if n = 1 and m ≥ 2,

0 else.

(5) tk (P n )=

0 if k ≤ k0 ,

2(k −k0) −1 if k > k 0 and 2n ≡ 0(mod 3),

2(k

−k0) if k > k 0 and 2n

≡ 1 or 2(mod 3).

(6) tk (C n ) =

0 if k ≤ k0 ,

2(k −k0) if k > k 0 and 2n ≡ 0 (mod 3),

3(k −k0) if k > k 0 and 2n ≡ 1 or 2(mod 3),where k0 = ⌊2n −1

3 ⌋. More results on tk (G) cam be found in references.

Ruler R4. Let G be a graph and f : V (G) → 1, 2, 3, · · · , |V | + |E (G)| be an injection.For each edge e = uv and an integer m ≥ 2, the induced Smarandachely edge m-labeling f ∗S isdened by

f ∗S (e) =f (u) + f (v)

m.

Then f is called a Smarandachely super m-mean labeling if f (V (G)) ∪ f ∗(e) : e ∈ E (G) =

1, 2, 3, · · · , |V | + |E (G)|. A graph that admits a Smarandachely super mean m-labeling iscalled Smarandachely super m-mean graph. Particularly, if m = 2, we know that

f ∗(e) =

f (u) + f (v)2

if f (u) + f (v) is even;f (u) + f (v) + 1

2 if f (u) + f (v) is odd.



26 Linfan Mao

A Smarandache super 2-mean labeling on P 26 is shown in Fig.2 .7.

1 2 3 5 7 8 9 11 13 14 15

4 6 10 12

Fig. 2.7

Now we have know graphs P n , C n , K n ,K(2,n) , (n ≥ 4), K (1, n ) for 1 ≤ n ≤ 4, C m ×P nfor n ≥ 1, m = 3 , 5 have Smarandachely super 2-mean labeling. More results on Smarandachelysuper m-mean labeling of graphs can be found in references in [1], [11], [17] and [18].

§3. Smarandache Sequences on Symmetric Graphs

Let lS G : V (G) → 1, 11, 111, 1111 , 11111 , 111111 , 1111111 , 11111111 , 111111111 be a vertex la-

beling of a graph G with edge labeling lS G (u, v ) induced by lS

G (u)lS G (v) for (u, v ) ∈ E (G) such that

lS G (E (G)) = 1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321,

12345678987654321, i.e., lS G (V (G)∪E (G)) contains all numbers appeared in the Smarandachely

third symmetry. Denote all graphs with lS G labeling by L S . Then it is easily nd a graph with

a labeling lS G in Fig.3.1 following.

1 111 11

111 111

1111 111111111 11111111111 111111

1111111 111111111111111 11111111

111111111 111111111

1121

123211234321

12345432112345654321

1234567654321123456787654321

12345678987654321

Fig. 3.1

Generally, we know the following result.

Theorem 3.1 Let G ∈ L S . Then G =n

i=1H i for an integer n ≥ 9, where each H i is a

connected graph. Furthermore, if G is vertex-transitive graph, then G = nH for an integer n ≥ 9, where H is a vertex-transitive graph.

Proof Let C (i) be the connected component with a label i for a vertex u, where i ∈1, 11, 111, 1111 , 11111 , 111111 , 1111111 , 11111111 , 111111111 . Then all vertices v in C (i) mustbe with label lS

G (v) = i. Otherwise, if there is a vertex v with lS G (v) = j ∈ 1, 11, 111, 1111 , 11111 ,

111111 , 1111111 , 11111111 , 111111111 \i, let P (u, v ) be a path connecting vertices u and v.




Then there must be an edge ( x, y ) on P (u, v ) such that lS G (x) = i, lS

G (y) = j . By denition,i × j ∈ lS

G (E (G)), a contradiction. So there are at least 9 components in G.Now if G is vertex-transitive, we are easily know that each connected component C (i) must

be vertex-transitive and all components are isomorphic.

The smallest graph in L S v is the graph 9 K 2 shown in Fig.3 .1. It should be noted that eachgraph in L S

v is not connected. For nding a connected one, we construct a graph Qk followingon the digital sequence

1, 11, 111, 1111 , 11111 , · · ·, 11 · · ·1 k

.

byV (Qk ) = 1, 11, · · · , 11 · · ·1 k

1′, 11′, · · · , 11 · · ·1′

k,

E (Qk ) = (1, 11 · · ·1

k

), (x, x ′), (x, y )|x, y ∈ V (Q) differ in precisely one 1.

Now label x ∈ V (Q) by lG (x) = lG (x′) = x and (u, v ) ∈ E (Q) by lG (u)lG (v). Then we havethe following result for the graph Qk .

Theorem 3.2 For any integer m ≥ 3, the graph Qm is a connected vertex-transitive graph of order 2m with edge labels

lG (E (Q)) = 1, 11, 121, 1221, 12321, 123321, 1234321, 12344321, 12345431, · · ·,i.e., the Smarandache symmetric sequence.

Proof Clearly,

Qm is connected. We prove it is a vertex-transitive graph. For simplicity,

denote 11 · · ·1

i

, 11 · · ·1′

i

by i and i ′, respectively. Then V (Qm ) = 1, 2, · · · , m. We dene an

operation + on V (Qk ) by

k + l = 11 · · ·1 k+ l(mod k)

and k′ + l′ = k + l′, k′′ = k

for integers 1 ≤ k, l ≤ m. Then an element i naturally induces a mapping

i∗ : x → x + i, for x ∈ V (Qm ).

It should be noted that i∗ is an automorphism of Qm because tuples x and y differ in preciselyone 1 if and only if x + i and y + i differ in precisely one 1 by denition. On the other hand,

the mapping τ : x → x′ for ∀x ∈ is clearly an automorphism of Qm . Whence,

G = τ , i∗ | 1 ≤ i ≤ m Aut Qm ,

which acts transitively on V (Q) because ( y −x)∗(x) = y for x, y ∈ V (Qm ) and τ : x → x′.Calculation shows easily that

lG (E (Qm )) = 1, 11, 121, 1221, 12321, 123321, 1234321, 12344321, 12345431, · · ·,







30 Linfan Mao

Corollary 4.4 |N

Q m,n,k[x]| = mk for ∀x ∈ 1Γ , x1 , · · · , xn and integers m,n,k ≥ 1.

1Γ

1Γ

1Γ

1Γ

1Γ

1Γ

x1

x1

x1

x1

x1

x1

x2

x2

x2

x2

x2

x2

x3

x3

x3

x3

x3

x3

x4

x4

x4

x4

x4

x4

Fig. 4.1

§5. Speculation

It should be noted that the essence we have done is a combinatorial notion, i.e., combining math-ematical systems on that of graphs. Recently, Sridevi et al. consider the Fibonacci sequenceon graphs in [16]. Let G be a graph and F 0 , F 1 , F 2 , · · · , F q , · · · be the Fibonacci sequence,where F q is the q th Fibonacci number. An injective labeling lG : V (G) → F 0 , F 1 , F 2 , · · · , F qis called to be super Fibonacci graceful if the induced edge labeling by lG (u, v ) = |lG (u) −lG (v)|is a bijection onto the set

F 1 , F 2 ,

· · · , F q

with initial values F 0 = F 1 = 1. They proved a

few graphs, such as those of C n ⊕P m , C n ⊕K 1,m have super Fibonacci labelings in [18]. Forexample, a super Fibonacci labeling of C 6 ⊕P 6 is shown in Fig.5 .1.

F 0

F 7F 9

F 11

F 10 F 12

F 6 F 4 F 5 F 3 F 1 F 2F 1F 2F 4F 3F 5F 6

F 7

F 8

F 10

F 9

F 11

F 12

Fig. 5.1

All of these are not just one mathematical system. In fact, they are applications of Smaran-dache multi-space and CC conjecture for developing modern mathematics, which appeals oneto nd combinatorial structures for classical mathematical systems, i.e., the following problem.

Problem 5.1 Construct classical mathematical systems combinatorially and characterize them.




For example, classical algebraic systems, such as those of groups, rings and elds by combina-torial principle.

Generally, a Smarandache multi-space is dened by the following.

Denition 5.2([6],[14]) For an integer m ≥ 2, let (Σ 1 ;R1), (Σ 2 ;R2), · · ·, (Σ m ;Rm ) be mmathematical systems different two by two. A Smarandache multi-space is a pair (Σ; R) with

Σ =m

i =1

Σ i , and R =m

i=1Ri .

Denition 5.3([10]) A combinatorial system C G is a union of mathematical systems (Σ 1 ;R1),(Σ 2 ;R2), · · ·, (Σ m ;Rm ) for an integer m , i.e.,

C G = (m

i=1

Σ i ;m

i=1Ri )

with an underlying connected graph structure G, where V (G) = Σ 1 , Σ 2 , · · · , Σ m , E (G) = (Σ i , Σ j ) | Σ i Σ j = ∅, 1 ≤ i, j ≤ m.

We have known a few Smarandache multi-spaces in classical mathematics. For examples,these rings and elds are group multi-space, and topological groups, topological rings andtopological elds are typical multi-space are both groups, rings, or elds and topological spaces.Usually, if m ≥ 3, a Smarandache multi-space must be underlying a combinatorial structure G.Whence, it becomes a combinatorial space in that case. I have presented the CC conjecturefor developing modern mathematical science in 2005 [5], then formally reported it at The 2th Conference on Graph Theory and Combinatorics of China (2006, Tianjing, China)([7]-[10]).

CC Conjecture (Mao, 2005) Any mathematical system (Σ; R) is a combinatorial system C G (lij , 1 ≤ i, j ≤ m).

This conjecture is not just an open problem, but more likes a deeply thought, which opensa entirely way for advancing the modern mathematical sciences. In fact, it indeed means acombinatorial notion on mathematical objects following for researchers.

(1) There is a combinatorial structure and nite rules for a classical mathematical system,which means one can make combinatorialization for all classical mathematical subjects.

(2) One can generalizes a classical mathematical system by this combinatorial notion suchthat it is a particular case in this generalization.

(3) One can make one combination of different branches in mathematics and nd newresults after then.

(4) One can understand our WORLD by this combinatorial notion, establish combinatorialmodels for it and then nd its behavior, for example,

what is true colors of the Universe, for instance its dimension?

and · · ·. For its application to geometry and physics, the reader is refereed to references [5]-[10],particularly, the book [10] of mine.



32 Linfan Mao

References

[1] S.Avadayappan and R.Vasuki, New families of mean graphs, International J.Math. Com-bin. Vol.2 (2010), 68-80.

[2] D.Deleanu, A Dictionary of Smarandache Mathematics , Buxton University Press, London& New York, 2006.

[3] P.Devadas Rao, B. Sooryanarayana and M. Jayalakshmi, Smarandachely k-ConstrainedNumber of Paths and Cycles, International J.Math. Combin. Vol.3 (2009), 48-60.

[4] J.A.Gallian, A dynamic survey of graph labeling, The Electronic J.Combinatorics , # DS6,16(2009), 1-219.

[5] Linfan Mao, Automorphism Groups of Maps, Surfaces and Smarandache Geometries , Amer-ican Research Press, 2005.

[6] Linfan Mao, Smarandache Multi-Space Theory , Hexis, Phoenix, USA, 2006.[7] Linfan Mao, Selected Papers on Mathematical Combinatorics , World Academic Union,

2006.

[8] Linfan Mao, Combinatorial speculation and combinatorial conjecture for mathematics,International J.Math. Combin. Vol.1(2007), No.1, 1-19.

[9] Linfan Mao, An introduction to Smarandache multi-spaces and mathematical combina-torics, Scientia Magna , Vol.3, No.1(2007), 54-80.

[10] Linfan Mao, Combinatorial Geometry with Applications to Field Theory , InforQuest, USA,2009.

[11] A.Nagarajan, A.Nellai Murugan and S.Navaneetha Krishnan, On near mean graphs, In-ternational J.Math. Combin. Vol.4 (2010), 94-99.

[12] ShreedharK, B.Sooryanarayana and RaghunathP, Smarandachely k-Constrained labelingof Graphs, International J.Math. Combin. Vol.1 (2009), 50-60.

[13] F.Smarandache, Only Problems, Not Solutions! Xiquan Publishing House, Chicago, 1990.[14] F.Smarandache, Mixed noneuclidean geometries, eprint arXiv: math/0010119 , 10/2000.[15] F.Smarandache, Sequences of Numbers Involved in Unsolved Problems , Hexis, Phoenix,

Arizona, 2006.[16] R.Sridevi, S.Navaneethakrishnan and K.Nagarajan, Super Fibonacci graceful labeling, In-

ternational J.Math. Combin. Vol.3(2010), 22-40.[17] R. Vasuki and A.Nagarajan, Some results on super mean graphs, International J.Math.

Combin. Vol.3 (2009), 82-96.[18] R. Vasuki and A.. Nagarajan, Some results on super mean graphs, International J.Math.

Combin. Vol.3 (2009), 82-96.




Supermagic Coverings of Some Simple Graphs

P.Jeyanthi

Department of Mathematics, Govindammal Aditanar College for Women, Tiruchendur-628 215

P.SelvagopalDepartment of Mathematics,Cape Institute of Technology,

Levengipuram, Tirunelveli Dist.-627 114

E-mail: [email protected], [email protected]

Abstract : A simple graph G = ( V, E ) admits an H -covering if every edge in E belongs to

a subgraph of G isomorphic to H . We say that G is Smarandachely pair s, l H -magic if there is a total labeling f : V ∪E → 1, 2, 3, · · · , |V | + |E | such that there are subgraphsH 1 = ( V 1 , E 1 ) and H 2 = ( V 2 , E 2 ) of G isomorphic to H , the sum v ∈ V 1

f (v) + e∈ E 1

f (e) = s

and v ∈ V 2f (v)+ e∈ E 2

f (e) = l. Particularly, if s = l, such a Smarandachely pair s, l H -magic

is called H -magic and if f (V ) = 1, 2, · · · , |V |, G is said to be a H -supermagic. In thispaper we show that edge amalgamation of a nite collection of graphs isomorphic to any2-connected simple graph H is H -supermagic.

Key Words : H -covering, Smarandachely pair s, l H -magic, H -magic, H -supermagic.

AMS(2010) : 05C78

§1. Introduction

The concept of H -magic graphs was introduced in [3]. An edge-covering of a graph G is a familyof different subgraphs H 1 , H 2 , . . . , H k such that each edge of E belongs to at least one of thesubgraphs H i , 1 ≤ i ≤ k. Then, it is said that G admits an ( H 1 , H 2 , . . . , H k ) - edge covering.If every H i is isomorphic to a given graph H , then we say that G admits an H -covering.

Suppose that G = ( V, E ) admits an H -covering. We say that a bijective function f :V ∪E → 1, 2, 3, · · · , |V |+ |E | is an H -magic labeling of G if there is a positive integer m(f ),which we call magic sum, such that for each subgraph H ′ = ( V ′, E ′) of G isomorphic to H ,we have, f (H ′) =

v∈V ′ f (v) +

e∈E ′ f (e) = m(f ). In this case we say that the graph G

is H -magic. When f (V ) = 1, 2, |V |, we say that G is H -supermagic and we denote itssupermagic-sum by s(f ).

We use the following notations. For any two integers n < m , we denote by [ n, m ], the setof all consecutive integers from n to m. For any set I ⊂ N we write, I = x∈I

x and for any

integers k, I + k = x + k : x ∈I. Thus k +[ n, m ] is the set of consecutive integers from k+ n to

1 Received December 29, 2010. Accepted February 20, 2011.



34 P.Jeyanthi and P.Selvagopal

k+ m. It can be easily veried that (I + k) = I + k|I|. If P = X 1 , X 2 , · · · , X n is a partitionof a set X of integers with the same cardinality then we say P is an n-equipartition of X . Alsowe denote the set of subsets sums of the parts of P by P = X 1, X 2 , · · · , X n .Finally,given a graph G = ( V, E ) and a total labeling f on it we denote by f (G) =

f (V ) + f (E ).

§2. Preliminary Results

In this section we give some lemmas which are used to prove the main results in Section 3.

Lemma 2.1 Let h and k be two positive integers and h is odd. Then there exists a k-

equipartition P = X 1, X 2 , · · · , X k of X = [1, hk ] such that X r = (h −1)(hk + k + 1)

2 + r

for 1 ≤ r ≤ k. Thus, P is a set of consecutive integers given by P = (h −1)(hk + k + 1)

2 +

[1, k].

Proof Let us arrange the set of integers X = [1, hk ] in a h ×k matrix A as given below.

A =

1 2 · · · k −1 k

n + 1 n + 2 · · · 2k −1 2k

2n + 1 2 n + 2 · · · 3k −1 3k...

......

......

(h −1)k + 1 (h −1)k + 2 · · · hk −1 hkh×k

That is, A = ( a i,j )h ×k where a i,j = ( i −1)k + j for 1 ≤ i ≤ h and 1 ≤ j ≤ k. For 1 ≤ r ≤ k,dene X r = a i,r / 1 ≤ i ≤ h+1

2 ∪a i,k −r +1 / h +32 ≤ i ≤ h. Then

X r =

h +12

i=1

a i,r +h

i= h +32

a i,k −r +1

=

h +12

i=1

(i −1)k + rh

i= h +32

(i −1)k + k −r + 1

= h2k + h −k −1

2 + r

= (h −1)(hk + k + 1)

2 + r for 1 ≤ r ≤ k.

Hence, P = (h −1)(hk + k + 1)2

+ [1, k].

Example 2.2 Let h = 9, k = 6 and X = [1, 54]. Then the partition subsets are X 1 =

1, 7, 13, 19, 25, 36, 42, 48, 54, X 2 = 2, 8, 14, 20, 26, 35, 41, 47, 53, X 3 = 3, 9, 15, 21, 27, 34,40, 46, 52, X 4 = 4, 10, 16, 22, 28, 33, 39, 45, 51, X 5 = 5, 11, 17, 23, 29, 32, 38, 44, 50and X 6 =

6, 12, 18, 24, 30, 31, 37, 43, 49. X r = (h −1)(hk + k + 1)

2 + r = 244 + r for 1 ≤ r ≤ 6.






Example 2.4 Let h = 6, k = 5 and X = [1, 30]. Y 1 = 1, 6, 11, 20, 25, Y 2 = 2, 7, 12, 19, 24,Y 3 = 3, 8, 13, 18, 23, Y 4 = 4, 9, 14, 17, 22and Y 5 = 5, 10, 15, 16, 21. By denition the parti-tion subsets are, X r = Y σ (r ) ∪(h −1)k + π(r ) for 1 ≤ r ≤ 5. X 1 = 2, 7, 12, 19, 24, 27, X 2 =

1, 6, 11, 20, 25, 29, X 3 = 5, 10, 15, 16, 21, 26X 4 = 4, 9, 14, 17, 22, 28X 5 = 3, 8, 13, 18, 23, 30,

Now, X r = (h −1)(hk + k + 1)2 + r = 90 + r for 1 ≤ r ≤ 5.

Lemma 2.5 If h is even, then there exists a k-equipartition P = X 1 , X 2 , · · · , X k of X =

[1, hk ] such that X r = h(hk + 1)

2 for 1 ≤ r ≤ k. Thus, the subsets sum are equal and is

equal to h(hk + 1)

2 .

Proof Let us arrange the set of integers X = 1, 2, 3, · · · , hkin a h ×k matrix A as givenbelow.

A =

1 2 · · · k −1 k

n + 1 n + 2 · · · 2k −1 2k2n + 1 2 n + 2 · · · 3k −1 3k

......

......

...

(h −1)k + 1 (h −1)k + 2 · · · hk −1 hkh×k

That is, A = ( a i,j )h ×k where a i,j = ( i −1)k + j for 1 ≤ i ≤ h and 1 ≤ j ≤ k. For 1 ≤ r ≤ k,dene X r = a i,r / 1 ≤ i ≤ h

2 ∪a i,k −r +1 / h2 + 1 ≤ i ≤ h −1. Then

X r =

h2

i =1

a i,r +h

i = h2 +1

a i,k −r +1

=h2

i =1(i −1)k + r+

h

i= h2 +1

(i −1)k + k −r + 1 = h(hk + 1)

2

Thus, the subsets sum are equal and is equal to h(hk + 1)

2 .

Example 2.6 Let h = 6, k = 5 and X = [1, 30]. Then the partition subsets are X 1 =

1, 6, 11, 20, 25, 30, X 2 = 2, 7, 12, 19, 24, 29, X 3 = 3, 8, 13, 18, 23, 28, X 4 = 4, 9, 14, 17,

22, 27 and X 5 = 5, 10, 15, 16, 21, 26. Now, X r = h(hk + 1)

2 = 93 for 1 ≤ r ≤ 5.

Lemma 2.7 Let h and k be two even positive integers and h

≥ 4. If X = [1 , hk +1]

−k

2 + 1

,

there exists a k-equipartition P = X 1 , X 2 , · · · , X kof X such that X r = h2k + 3 h −k −2

2 + r

for 1 ≤ r ≤ k. Thus P is a set of consecutive integers h2 k + 3 h −k −2

2 + [1, k].

Proof First we prove this lemma for h = 2 and we generalize for any even integer h ≥ 4.

Case 1: h = 2.



Supermagic Coverings of Some Simple Graphs 37

X = [1 , 2k + 1] − k2 + 1. For 1 ≤ r ≤ k, dene

X r = k2 + 1 −r, k + 1 + 2 r for 1 ≤ r ≤ k

2

3k2 + 2 −r, 2r for k

2 + 1 ≤ r ≤ k.

Hence, X r = 3k2 + 2 + r for 1 ≤ r ≤ k.

Case 2: h ≥ 4

Let Y = [1 , 2k + 1] −k2

+ 1 and Z = [2k + 2 , hk + 1]. Then X = Y ∪Z . By Case 1, thereexists a k-equipartition P1 = Y 1 , Y 2 , · · · , Y k of Y such that

Y r = 3k

2 + 2 + r for 1 ≤ r ≤ k (1)

Since h −2 is even, by Lemma 2.5, there exists a k-equipartition

P ′2 = Z ′1 , Z ′2 , · · · , Z ′k of [1, (h −2)k] such that Z ′r = (h −2)(hk −2k + 1)

2 for 1 ≤ r ≤ k.

Adding 2 k+1 to [1 , (h−2)k], we get a k-equipartition P 2 = Z 1 , Z 2 , · · · , Z kof Z = [2k+2 , hk +

1] such that Z r = ( h −2)(2k + 1) + (h −2)(hk −2k + 1)2 for 1 ≤ r ≤ k. Let X r = Y r ∪Z rfor 1 ≤ r ≤ k. Then,

X r = Y r ∪ Z r

= h2k + 3 h −k −2

2 + r for 1 ≤ r ≤ k.

Hence, P is a set of consecutive integers h2k + 3 h −k −2

2 + [1, k].

Example 2.8 Let h = 6, k = 6 and X = [1, 37] − 4.Then the partition subsets are X 1 =

3, 9, 14, 20, 31, 37, X 2 = 2, 11, 15, 21, 30, 36, X 3 = 1, 13, 16, 22, 29, 35, X 4 = 7, 8, 17, 23,28, 34

, X 5 =

6, 10, 18, 24, 27, 33

and X 6 =

5, 12, 19, 25, 26, 32

. Now,

X r = h2k + 3 h −k −2

2 + r = 113 + r

for 1 ≤ r ≤ 6.

Lemma 2.9 Let h and k be two even positive integers. If X = [1, hk + 2] − 1, k2

+ 2 , there

exists a k-equipartition P = X 1 , X 2 , · · · , X k of X such that X r = h2k + 5 h −k −2

2 + r for

1 ≤ r ≤ k. Thus P is a set of consecutive integers h2k + 5 h −k −2

2 + [1, k].

Proof First we prove this lemma for h = 2 and we generalize for any even integer h ≥ 4.

Case 1: h = 2

X = [1 , 2k + 2] − 1, k2

+ 2. For 1 ≤ r ≤ k, dene

X r = k2

+ 1 −r, k + 2 + 2 r for 1 ≤ r ≤ k2

,

3k2

+ 3 −r, 2r + 1 for k

2 + 1 ≤ r ≤ k.




Hence, X r = 3k

2 + 4 + r for 1 ≤ r ≤ k.

Case 2: h ≥ 4

Let Y = [1, 2k + 2] − 1, k2

+ 2 and Z = [2k + 3 , hk + 2]. Then X = Y ∪Z . By Case 1,

there exists a k-equipartition P 1 = Y 1 , Y 2 , · · · , Y k of Y such that

Y r = 3k

2 + 4 + r for 1 ≤ r ≤ k (2)

Since h −2 is even, by Lemma 2.5, there exists a k-equipartition

P ′2 = Z ′1 , Z ′2 , · · · , Z ′k of [1, (h −2)k] such that Z ′r = (h −2)(hk −2k + 1)

2 for 1 ≤ r ≤ k.

Adding 2 k+2 to [1 , (h−2)k], we get a k-equipartition P 2 = Z 1 , Z 2 , · · · , Z kof Z = [2k+3 , hk +

2] such that Z r = ( h −2)(2k + 2) + (h −2)(hk −2k + 1)

2 for 1 ≤ r ≤ k. Let X r = Y r ∪Z r

for 1 ≤ r ≤ k. Then,

X r = Y r

∪Z r

= h2k + 5 h −k −2

2 + r for 1 ≤ r ≤ k.

Hence, P is a set of consecutive integers h2k + 5 h −k −2

2 + [1, k].

Example 2.10 Let h = 6, k = 6 and X = [1, 38] − 1, 5.Then the partition subsetsare X 1 = 4, 10, 15, 21, 32, 38, X 2 = 3, 12, 16, 22, 31, 37, X 3 = 2, 14, 17, 23, 30, 36, X 4 =

8, 9, 18, 24, 29, 35, X 5 = 7, 11, 19, 25, 28, 34 and X 6 = 6, 13, 20, 26, 27, 33. Now, X r =h2k + 5 h −k −2

2 + r = 119 + r for 1 ≤ r ≤ 6.

§3. Main Results

Denition 3.1(Edge amalgamation of a nite collection of graphs, [1]) For any nite collection (G i , u i vi ) of graphs Gi , each with a xed edge ui vi , Carlson [1] dened the edge amalgamation

E dgeamal (G i , u i vi ) as the graph obtained by taking the union of all the Gi ’s and identifying their xed edges.

Denition 3.2( Generalized Book) If all the Gi ’s are cycles then E dgeamal (G i , u i vi ) is called a generalized book.

Theorem 3.3 Let H be a 2-connected ( p, q ) simple graph. Then the edge amalgamation

E dgeamal

(H i , u i vi )

of any nite collection

H i , u i vi

of graphs H i , each with a xed edge

u i vi isomorphic to H is H -supermagic for all values of p and q .

Proof Let H i , u i vi be a collection of n graphs H i , each with a xed edge ui vi andisomorphic to a 2-connected simple graph H .Let G = E dgeamal (H i , u i vi ) with vertex set V and edge set E . Note that |V | = n( p−2)+2and |E | = n(q − 1) + 1. Let H i = ( V i , E i ) for 1 ≤ i ≤ n. Label the common edge of G ase = w1 w2 . Let V ′i = V i −w1 , w2 and E ′i = E −e for 1 ≤ i ≤ n.




Case 1: n is odd

Subcase (i): p is even and q is odd

Since p −2 and q −1 are even by Lemma 2.5 there exists n -equipartitions P′1 = X ′1 , X ′2 ,

· · · , X ′n of [1, ( p−2)n] and P ′2 = Y ′1 , Y ′2 , · · · , Y ′n of [1, (q −1)n] such thatX ′i =

( p −2)( pn −2n + 1)2

, Y ′i = (q −1)(qn −n + 1)

2 .

Add 2 to each element of the set [1 , ( p−2)n] and ( p−2)n+3 to each element of the set [1 , (q −1)n].We get n-equipartitions P1 = X 1 , X 2 , · · · , X n of [3, pn −2n + 3] and P2 = Y 1 , Y 2 , · · · , Y n of [ pn −2n + 4 , ( p + q −3)n + 3] such that

X i = ( p−2)2+ ( p−2)( pn −2n + 1)

2 , Y i = ( q −1)( pn −2n +3)+

(q −1)(qn −n + 1)2

.

Dene a total labeling f : V ∪E → [1, ( p + q −3)n + 3] as follows:

f (w1) = 1 and f (w2 ) = 2 .

f (e) = pn

−2n + 3 .

f (V ′i ) = X i for 1 ≤ i ≤ n.

f (E ′i ) = Y n −i+1 for 1 ≤ i ≤ n.

Then for 1 ≤ i ≤ n,

f (H i ) = f (w1 ) + f (w2) + f (e) + f (V ′i ) + f (E ′i )

= f (w1 ) + f (w2) + f (e) + X ′i + Y ′n −i +1

= n( p + q )2 + p + q + 5( n −1)

2 −(n −1)(2 p + 3 q )

= constant.

Since H i ∼= H for 1 ≤ i ≤ n, G is H -supermagic.Subcase (ii): p is odd and q is even

Since p−2 and q −1 are odd, by Lemma 2.1 there exists n-equipartitions P ′1 = X ′1 , X ′2 , · · · ,X ′n of [1, ( p−2)n] and P′2 = Y ′1 , Y ′2 , · · · , Y ′n of [1, (q −1)n] such that

X ′i = ( p −3)( pn −n + 1)

2 + i, Y ′i =

(q −2)(qn + 1)2

+ i

for 1 ≤ i ≤ n . Add 2 to each element of the set [1 , ( p−2)n] and ( p −2)n + 3 to each elementof the set [1, (q −1)n]. We get n -equipartitions P1 = X 1 , X 2 , · · · , X n of [3, pn −2n + 3] andP 2 = Y 1 , Y 2 , · · · , Y n of [ pn −2n + 4 , ( p + q −3)n + 3] such that

X i = ( p

−2)2+

( p−3)( pn −n + 1)

2 + i, Y i = ( q

−1)( pn

−2n +3)+

(q −2)(nq + 1)

2 + i

for 1 ≤ i ≤ n. Dene a total labeling f : V ∪E → [1, ( p + q −3)n + 3] as follows:

f (w1) = 1 and f (w2 ) = 2 .

f (e) = pn −2n + 3 .

f (V ′i ) = X i for 1 ≤ i ≤ n.

f (E ′i ) = Y n −i+1 for 1 ≤ i ≤ n.






= f (w1 ) + f (w2) + f (e) + X ′i + Y ′n −i +1

= n( p + q )2 + p + q + 5( n −1)2 −(n −1)(2 p + 3 q )

= constant.

Since H i ∼= H for 1 ≤ i ≤ n, G is H -supermagic.

Subcase (iii): p and q are odd

Since p −2 is odd, by Lemma 2.1 there exists an n-equipartition P′1 = X ′1, X ′2 , · · · , X ′n of [1, ( p−2)n] such that X ′i =

( p −3)( pn −n + 1)2

+ i for 1 ≤ i ≤ n. Since q −1 is even andn is odd, by Lemma 2.3 there exists an n-equipartition P′2 = Y ′1 , Y ′2 , · · · , Y ′n of [1, (q −1)n]

such that Y ′i = (q −2)(qn + 1)

2 + i for 1

≤ i

≤ n. Add 2 to each element of the set

[1, ( p −2)n] and ( p −2)n + 3 to each element of the set [1 , (q −1)n]. We get n-equipartitionsP 1 = X 1 , X 2 , · · · , X n of [3, pn−2n+3] and P 2 = Y 1 , Y 2 , · · · , Y n of [ pn−2n+4 , ( p+ q −3)n+3]such that

X i = ( p −2)2 + ( p −3)(np −n + 1)

2 + i,

Y i = ( q −1)( pn −2n + 3) + (q −2)(qn + 1)

2 + i


f (w1) = 1 and f (w2 ) = 2 .

f (e) = pn

−2n + 3 .

f (V ′i ) = X i for 1 ≤ i ≤ n.

f (E ′i ) = Y n −i+1 for 1 ≤ i ≤ n.



= f (w1 ) + f (w2) + f (e) + X ′i + Y ′n −i +1

= n( p + q )2 + p + q + 5( n −1)

2 −(n −1)(2 p + 3 q )

= constant.


Subcase (iv): p and q are even

Since p − 2 is even and n is odd, by Lemma 2.3 there exists an n-equipartition P′1 =

X ′1 , X ′2 , · · · , X ′n of [1, ( p − 2)n] such that X ′i = ( p−3)( pn −n + 1)

2 + i for 1 ≤ i ≤ n.

Since q − 1 is odd, by Lemma 2.1 there exists an n-equipartition P′2 = Y ′1 , Y ′2 , · · · , Y ′n of




[1, (q −1)n] such that Y ′i = (q −2)(qn + 1)

2 + i for 1 ≤ i ≤ n. Add 2 to each element of the

set [1, ( p−2)n] and ( p−2)n + 3 to each element of the set [1 , (q −1)n]. We get n-equipartitionsP 1 = X 1 , X 2 , · · · , X n of [3, pn−2n+3] and P 2 = Y 1 , Y 2 , · · · , Y n of [ pn−2n+4 , ( p+ q −3)n+3]such that

X i = ( p−2)2+ ( p−3)( pn −n + 1)

2 + i, Y i = ( q −1)( pn −2n +3)+

(q −2)(qn + 1)2

+ i


f (w1) = 1 and f (w2 ) = 2 .

f (e) = pn −2n + 3 .

f (V ′i ) = X i for 1 ≤ i ≤ n.

f (E ′i ) = Y n −i+1 for 1 ≤ i ≤ n.



= f (w1 ) + f (w2) + f (e) + X ′i + Y ′n −i +1

= n( p + q )2 + p + q + 5( n −1)

2 −(n −1)(2 p + 3 q )

= constant.


Case 2: n is even

Subcase (i): p is even and q is odd

The argument in Subcase(i) of Case (1) is independent of the nature of n. Hence we getG is H -supermagic.

Subcase (ii): p is odd and q is even

The argument in Subcase(ii) of Case (1) is independent of the nature of n . Hence we getG is H -supermagic.

Subcase (iii): p and q are odd

Since p −2 is odd, by Lemma 2.1 there exists an n-equipartition P′1 = X ′1, X ′2 , · · · , X ′n of [1, ( p − 2)n] such that X ′i =

( p −3)( pn −n + 1)2

+ i for 1 ≤ i ≤ n. Since q − 1 and nare even, by Lemma 2.7 there exists an n-equipartition P ′2 =

Y ′1 , Y ′2 ,

· · · , Y ′n

of [1, (q

−1)n +

1]− n2 + 1 such that Y ′i = (q −1)2n + 3( q −1) −n −22

+ i for 1 ≤ i ≤ n . Add 2 to eachelement of the set [1 , ( p −2)n] and ( p − 2)n + 2 to each element of the set [1 , (q − 1)n]. Weget n-equipartitions P1 = X 1 , X 2 , · · · , X n of [3, pn − 2n + 3] and P2 = Y 1 , Y 2 , · · · , Y n of [ pn −2n + 3 , ( p + q −3)n + 3] − ( p−2)n +

n2

+ 3 such that

X i = ( p −2)2 + ( p−3)( pn −n + 1)

2 + i,




Y i = ( q −1)( pn −2n + 2) + (q −1)2n + 3( q −1) −n −2

2 + i


f (w1) = 1 and f (w2 ) = 2 .

f (e) = ( p −2)n + n2

+ 3 .

f (V ′i ) = X i for 1 ≤ i ≤ n.

f (E ′i ) = Y n −i+1 for 1 ≤ i ≤ n.



= f (w1 ) + f (w2) + f (e) + X ′i + Y ′n −i+1

= n( p + q )2 + p + q

2 −(n −1)(2 p + 3 q −3)

= constant.


Subcase (iv): p and q are even

Since p −2 and n are even, by Lemma 2.9 there exists an n-equipartition P1 = X 1 , X 2 ,

· · · , X n of [1, ( p−2)n + 2] −1, n2

+ 2 such that X i = ( p−2)2n + 5( p−2) −n −2

2 + i for

1 ≤ i ≤ n.Since q −1 is odd, by Lemma 2.1 there exists an n -equipartition P ′2 = Y ′1 , Y ′2 , · · · , Y ′n of

[1, (q −1)n] and

Y ′i = (q−2)( qn +1)

2 + i for 1 ≤ i ≤ n. Add ( p−2)n + 3 to each element of theset [1, (q

−1)n]. We get an n-equipartition P 2 =

Y 1 , Y 2 ,

· · · , Y n

of [ pn

−2n +4 , ( p+ q

−3)n +3]

such that Y i = ( q −1)( pn −2n + 3) + (q−2)( qn +1)2 + i for 1 ≤ i ≤ n . Dene a total labeling

f : V ∪E → [1, ( p + q −3)n + 3] as follows:

f (w1) = 1 and f (w2 ) = n2

+ 2 .

f (e) = pn −2n + 3 .

f (V ′i ) = X i for 1 ≤ i ≤ n.

f (E ′i ) = Y n −i+1 for 1 ≤ i ≤ n.


f (H i ) = f (w1 ) + f (w2) + f (e) + f (V ′i ) + f (E ′i )= f (w1 ) + f (w2) + f (e) + X ′i + Y ′n −i+1

= n( p + q )2 + p + q

2 −(n −1)(2 p + 3 q −3)

= constant.





Hence, the edge amalgamation E dgeamal (H i , u i vi ) of any nite collection H i , u i vi of graphs H i , each with a xed edge ui vi and isomorphic to H is H -supermagic for all values of p and q .

Illustration 3.4 Let H 1 ,H 2 ,H 3 ,H 4 and H 5 be ve graphs isomorphic to the wheel W 4 =C 4 + K 1 and their xed edges given by dotted lines. Then the Edge amalgamation graph

E dgeamal (H i , u i vi ) of the given collection is W 4 -supermagic with supermagic sum 303.

1

2

1

2

1

2

1

25

10

15

7

8

17 16

9

6

4

11

14

12

3

21

3136

18 18

18

18

26

41

51

46

2225

3235

42

45 52

23

24

3334

4443

53

1928

2938

48

3949

40

5047

20

27

3037

13

Fig. 1

Illustration 3.5 Let H 1 ,H 2 ,H 3 and H 4 be four graphs isomorphic to H and their xed edgesgiven by dotted lines. Then the Edge amalgamation graph

E dgeamal

(H i , u i vi )

of the given

collection is H -supermagic with supermagic sum 300.

1 419

40 47

39

32183

31

24

38

3025

33

2

10 12

17

9

14 4

15

15

20

2728

35

3619

43

44

1

4

6 19

21

34 29

26

7

16

42

45

42

13

23

22

4146

Fig. 2




Denition 3.6(Book with m-gon pages) Let n and m be any positive integers with n ≥ 1 and m ≥ 3 . Then, n copies of the cycle C m with an edge in common is called a book with n m-gon pages. That is, if Gi,u i vi is a collection of n copies of the cycle C m each with a xed edge u i vi then E dgeamal (G i , u i vi ) is called a book with n m-gon pages.

A book with 3 pentagon pages is given below.

Fig. 3

Corollary 3.7 Books with n m-gon pages are C m -supermagic for every positive integers n ≥ 1and m ≥ 3.

Illustration 3.8 C 5 -supermagic covering of a book with 3 hexagon pages is given below. Thesupermagic sum is 167.

4

7

10

14

5

8

9

12

3

6

1113

16 21

22

2728

17 20

23

2629

18 19

24

2530

2

151

Fig. 4

Theorem 3.9 Let H i = K 1,k with vertex set V (H i ) = vi , vir : 1 ≤ r ≤ k and the edge set E (H i ) = vi vir : 1 ≤ r ≤ k where 1 ≤ i ≤ k and G be a graph obtained by joining a new vertex w with v11 , v21 ,

· · · , vk1 . Then G is K 1,k -supermagic.

Proof Let V i = vi , vir : 1 ≤ r ≤ k and E i = V i vjr : 1 ≤ r ≤ k for 1 ≤ i ≤ k.Then the vertex and edge set of G = ( V, E ) are given by V = ∪ki=1 V i ∪v and E = ∪ki=1 E i ∪vv1 , vv2 , · · · , vvk. Also |V | = k2 + k + 1 and |E | = k2 + k. Let V k+1 = w, v1 , v2 , · · · , vkand E k+1 = wv1 , wv2 , · · · , wvk and H k+1 = ( V k+1 , E k+1 ) be the graph with vertex set V k+1

and edge set E k+1 . Note that any edge of E belongs to at least one of the subgraphs H i for1 ≤ i ≤ k + 1. Since H i ∼= K 1,k for 1 ≤ i ≤ k + 1, G admits a K 1,k -covering.




Case 1: k is odd

Since k + 1 is even, by Lemma 2.3, there exists a k-equipartition P = X 1 , X 2 , · · · , X k of X = [1 , (k + 1) k] such that

X i = k(k + 1)2

2 + i for 1 ≤ i ≤ k (3)

It can be easily veried from the denition of X r in Lemma 2.3 thatk + 1

2 −1 k + σ(r ) ∈ X r

for 1 ≤ r ≤ k, where σ denotes the permutation of 1, 2, · · · , k given by

σ(r ) =k −2r + 1

2 for 1 ≤ r ≤

k −12

3k −2r + 12

for k + 1

2 ≤ r ≤ k.

Construct X k+1 = k + 1

2 −1 k + σ(r ) : 1 ≤ r ≤ k∪k2 + k + 1.

X k+1 =k

r =1

k + 12 −1 k + σ(r ) + k2 + k + 1

= k2(k −1)

2 +

k(k + 1)2

+ k2 + k + 1

= k(k + 1) 2

2 + k + 1 (4)

From (1) and (2) we have

X i = k(k + 1) 2

2 + i for 1 ≤ i ≤ k + 1 (5)

As k is odd, by Lemma 1, there exists a k + 1-equipartition Q ′ =

Y ′1 , Y ′2 ,

· · · , Y ′k+1

of the

set Y = [1 , k(k + 1)] such that Y ′i = (k −1)[(k + 1) 2 + 1]2

+ i for 1 ≤ i ≤ k + 1.Adding k2 + k + 1 to [1, k(k + 1)], we get a k + 1-equipartition Q = Y 1 , Y 2 , · · · , Y k+1 of

the set Y = [ k2 + k + 2 , 2k2 + 2 k + 1] such that

Y i = k(k2 + k + 1) + (k −1)[(k + 1) 2 + 1]

2 + i for 1 ≤ i ≤ k + 1 (6)

Dene a total labeling f : V ∪E → [1, 2k2 + 2 k + 1] as follows:

(i) f (w) = k2 + k + 1.

(ii) f (V i ) = X i with f (vi1 ) =k + 1

2 −1 k + σ(r ) for 1 ≤ i ≤ k + 1.

(iii) f (E i ) = Y k+2 −i for 1 ≤ i ≤ k + 1.

Then for 1 ≤ i ≤ k + 1,

f (H i ) = f (V i ) + f (E i ) = X i + Y k+2 −i

= 4k3 + 5 k2 + 5 k + 2

2 ,




which is a constant. Since H i ∼= K 1,k for 1 ≤ i ≤ k + 1, G is K 1,k -supermagic.

Case 2: k is even

Since k + 1 is odd, by Lemma 1,there exists a k-equipartition P = X 1 , X 2 , · · · , X k of

X = [1 , (k + 1) k] such that

X i = k(k + 1) 2

2 + i for 1 ≤ i ≤ k (7)

It can be easily veried from the denition of X r in Lemma 2.3 thatk + 2

2 −1 k + r ∈ X r for

1 ≤ r ≤ k2

, andk2 −1 k + r ∈ X r for

k2

+ 1 ≤ r ≤ k. Construct X k+1 = k + 2

2 −1 k + r :

1 ≤ r ≤ k2∪

k2 −1 k + r :

k2

+ 1 ≤ r ≤ k∪k2 + k + 1.

X k+1 =

k2

r =1

k + 2

2 −1 k + r +

k

k2 +1

k

2 −1 k + r + k2 + k + 1

= k2(k −1)

2 +

k(k + 1)2

+ k2 + k + 1

= k(k + 1) 2

2 + k + 1 (8)

From (5) and (6) we have

X i = k(k + 1) 2

2 + i for 1 ≤ i ≤ k + 1 (9)

As k is even, by Lemma 2.3, there exists a k + 1-equipartition Q ′ = Y ′1 , Y ′2 , · · · , Y ′k+1 of the set Y = [1, k(k + 1)] such that

Y ′i = (k−1)[( k+1) 2 +1]

2 + i for 1 ≤ i ≤ k + 1. Adding

k2

+ k + 1 to [1, k(k + 1)], we get a k + 1-equipartition Q = Y 1 , Y 2 , · · · , Y k+1 of the setY = [k2 + k + 2 , 2k2 + 2 k + 1] such that

Y i = k(k2 + k + 1) + (k −1)[(k + 1) 2 + 1]

2 + i for 1 ≤ i ≤ k + 1 (10)

Dene a total labeling f : V ∪E → [1, 2k2 + 2 k + 1] as follows:

(i) f (w) = k2 + k + 1.

(ii) f (V i ) = X i with f (vi1 ) =k + 2

2 −1 k + r for 1 ≤ i ≤ k2

and f (vi1 ) =k2 −1 k + r

for k2

+ 1 ≤ i ≤ k.

(iii) f (E i ) = Y k+2 −i for 1 ≤ i ≤ k + 1.Then for 1 ≤ i ≤ k + 1,

f (H i ) = f (V i ) + f (E i )

= X i + Y k+2 −i

= 4k3 + 5 k2 + 5 k + 2

2 ,




which is a constant. Since H i ∼= K 1,k for 1 ≤ i ≤ k + 1, G is K 1,k -supermagic. Thus, in both

the cases G is K 1,k -supermagic with supermagic sum s(f ) = 4k3 + 5 k2 + 5 k + 2

2 .

Illustration 3.10

11 12

13

14

1531

1

6

20

2529 27

19 24

27

3

8

18

2330

4

9

17

22

28

5

10

16

21

26

3738

4950

61

36

3948

5160

3540

47

52

59 34

41

4653

58

33

42 45

54

5732

4344

55

56

Fig.1. G - is K 1 , 5 -supermagic with supermagic sum 236 .

19

20 21

16

1718

431

713

30

36 42

2

8

14

29 35

413

9

15

28

34

40

4

10 22

27

3339

5

11

23

2632

38

6 12

2425

31

37

5461

6875

85

47

4855

62

6774

83

49

5663

66

7381

50

57 64

65

79 72

44 5158

7178

84

45

52

5970

77

82

46

53 60

69

7680

Fig.2. G - is K 1 , 6 -supermagic with supermagic sum 538 .




References

[1] K.Carlson, Generalized books and Cm-snakes are prime graphs, Ars Combin . 80(2006)215-221.

[2] J.A.Gallian, A Dynamic Survey of Graph labeling(DS6), The Electronic Journal of Com-binatorics , 5(2005).

[3] A.Gutierrez, A.Llado, Magic coverings, J. Combin. Math. Combin. Comput. , 55(2005),43-56.

[4] P.Selvagopal, P.Jeyanthi, On Ck-supermagic graphs, International Journal of Mathematics and Computer Science , 3(2008), No. 1, 25-30.




Elementary Abelian Regular Coverings of Cube

Furong Wang

(Beijing Wuzi University, Beijing 101149, P.R. China)

Lin Zhang(Capital University of Economics and Business, Beijing 100070, P. R. China)

E-mail: [email protected], [email protected]

Abstract : For a give nite connected graph Γ, a group H of automorphisms of Γ and a nitegroup A, a natural question can be raised as follows: Find all the connected regular coveringsof Γ having A as its covering transformation group, on which each automorphism in H can

be lifted. In this paper, we classify all the connected regular covering graphs of the cubesatisfying the following two properties: (1) the covering transformation group is isomorphicto the elementary Abelian p-groups; (2) the group of bre-preserving automorphisms actsedge-transitively.

Key Words : Connected graph, graph covering, cube, Smarandachely covering, regularcovering.

AMS(2010) :

§1. Introduction

All graphs considered in this paper are nite, undirected and simple. For a graph Γ, we useV (Γ), E (Γ), A(Γ) and Aut(Γ) to denote its vertex set, edge set, arc set and full automorphismgroup, respectively. For any v ∈ V (Γ) , by N (v) we denote the neighborhood of v in Γ. For anarc (u, v ) ∈ A(Γ) , we denote the corresponding undirected edge by uv.

A graph Γ is called a covering of the graph Γ with projection p : Γ → Γ if there is asurjection p : V (Γ) → V (Γ) such that p|N ( v) : N (v) → N (v) is a bijection for any vertexv ∈ V (Γ) and v ∈ p−1(v). The graph Γ is called the covering graph and Γ is the base graph .A p : Γ → Γ is called to be a Smarandachely covering of Γ if there exist u, v ∈ V (Γ) such that

| p−1(u)| = | p−1(v)|. Conversely, if | p−1(v)| = n for each v ∈ V (Γ), then such a covering p issaid to be n -fold . Each p−1(v) is called a bre of

Γ. An automorphism of

Γ which maps a bre

to a bre is said to be bre-preserving. The group K of all automorphisms of

Γ which x each

of the bres setwise is called the covering transformation group . A covering p : Γ → Γ is saidto be regular (simply, A-covering ) if there is a subgroup A of K acting regularly on each bre.Moreover, if Γ is connected, then A = K.

A purely combinatorial description of a covering was introduced through a voltage graph

1 Supported by NNSF(10971144) and BNSF(1092010).2 Received December 23, 2010. Accepted February 23, 2011.



50 Furong Wang and Lin Zhang

by Gross and Tucker [4,5] and also a very similar idea was appeared in Biggs’ monograph [1,2].Let A be a nite group. An ( ordinary ) voltage assignment (or, A-voltage assignment ) of Γ isa function φ : A(Γ) → A with the property that φ(u, v ) = φ(v, u )−1 for each (u, v ) ∈ A(Γ) .The values of φ are called voltages , and A is called the voltage group. The graph Γ ×φ A

derived from φ is dened by V (Γ ×φ A) = V (Γ) ×A and E (Γ ×φ A) = ((u, g ), (v, φ(u, v )g))(u, v ) ∈ E (Γ) , g ∈ A. Clearly, the graph Γ ×φ A is a covering of the graph Γ with the rstcoordinate projection p : Γ×φ A → Γ, which is called the natural projection . For each u ∈ V (Γ) ,

(u, g ) g ∈ A is a bre of Γ ×φ A. Moreover, by dening ( u, g ′)g := ( u, g ′g) for any g ∈ Aand ( u, g ′) ∈ V (Γ ×φ A), A can be identied with a bre-preserving automorphism subgroup of Aut(Γ ×φ A) acting regularly on each bre. Therefore, p can be viewed as a A-covering . Givena spanning tree T of the graph Γ , a voltage assignment φ is called T -reduced if the voltageson the tree arcs are identity. Gross and Tucker ([4]) showed that every regular covering of agraph Γ can be derived from an ordinary T -reduced voltage assignment φ with respect to anarbitrary xed spanning tree T of Γ.

An automorphism α of Γ can be lifted to an automorphism ˜ α of a covering graph

Γ if

pα = αp , where p is the covering projection from Γ to Γ . We say a subgroup of H of Aut(Γ)can be lifted if each element of H can be lifted.

For a given nite connected graph Γ , a group H of automorphisms of Γ and a nite groupA, a natural question can be raised as follows: Find all the connected regular coverings of Γhaving A as its covering transformation group, on which each automorphism in H can be lifted.In [3], Du, Kawk and Xu investigate the regular coverings with A = Z n p , an elementary Abeliangroup and get some new matrix-theoretical characterizations for an automorphism of the basegraph to be lifted, and as one of the applications, they gave a classication of all connectedregular coverings of the Petersen graph with the covering transformation group Z n p , whosebre-preserving automorphism subgroup acts arc-transitively.

In this paper, we use the same method to classify all the connected regular covering graphsof the cube satisfying the following two properties: (1) the covering transformation group isisomorphic to the elementary Abelian p-group; (2) the group of bre-preserving automorphismsacts edge-transitively.

The cube is identied with the graph Γ as shown in Figure (a). Fix a spanning tree T inΓ as shown in Figure (b). Let V 1 = 2, 6, 4, 7, 3, 5. Then the induced subgraph Γ( V 1) is a lineas shown in Figure (c).

1765

432

8 1765

432

8

Figure (a): the graph Γ; (b) a spanning tree T of Γ.

2 6 4 7 3 5

Figure (c): the induced subgraph Γ( V 1 )



Elementary Abelian Regular Coverings of Cube 51

First, we introduce ve families of covering graphs Γ ×φ Z n p of cube Γ by giving a T -reducedvoltage assignment φ. Since φ is T -reduced, we only need to give the voltages on the cotree arcs(see Figure (c)). Let t denote the transposition of a matrix.

(1) X (2, 1):=Γ

×φ Z 2 , where φ26 = φ47 = φ35 = 1 and φ46 = φ37 = 0,

X ( p, 1):=Γ ×φ Z p, where p = 3 or p ≡ 1(mod 6), φ26 = φ37 = 1, φ46 = φ35 = 1+ √ −32 and

φ47 = 0.

(2) X ( p, 2):=Γ ×φ Z 2 p , where φ26 = φ37 = (0 , 1), φ46 = φ35 = (1 , 0) and φ47 = (0 , 0).

(3) X ( p, 3):=Γ ×φ Z 3 p , where ((φ26 ) t , (φ47 ) t , (φ35 ) t ) = I 3×3 , φ46 = (0 , 1, −1) and φ37 =(−1, 1, 0).

(4) X ( p, 4):=Γ ×φ Z 4 p , where p = 3 or p ≡ 1(mod 6), ((φ26 )t , (φ46 )t , (φ47 ) t , (φ37 )t ) = I 4×4

and φ35 = (1 −√ −3

2 , −1,

1 + √ −32

, 1−√ −3

2 ).

(5) X ( p, 5):=Γ ×φ Z 5

p , where (( φ26 )t, (φ46 )

t, (φ47 )

t, (φ37 )

t, (φ35 )

t) = I 5×5 .

Now we state the main theorem of this paper.

Theorem 1.1 Let Γ be a connected regular covering of the cube Γ whose covering transformation group is isomorphic to Z n p and whose bre-preserving automorphism subgroup G acts edge-transitively on Γ. Then, Γ is isomorphic to one of the graphs in (1) -(5) listed above. Moreover, for the graphs X (2, 1), X (3, 1), X ( p, 2), X ( p, 3), X (3, 4) and X ( p, 5), Aut(Γ) can be lifted, and so they are 2-arc-transitive; and for the graphs X ( p, 1) and X ( p, 4) for p ≡ 1(mod 6) , the subgroup isomorphic to A4×Z 2 can be lifted but AutΓ cannot, and so they are arc-transitive, in particular, all these ve families of graphs are vertex transitive.

§2. Algorithm for the Lifting

In this section, we present the algorithm given by Du, Kwak and Xu [3], which deals withthe lifting problem for regular coverings of a graph Γ whose covering transformation group iselementary Abelian.

Throughout this section, let Γ be a connected graph and let Γ = Γ ×φ Z n p be a connectedregular covering of the Γ . The voltage group Z n p will be identied with the additive group of the n-dimensional vector space V (n, p ) over the nite eld GF ( p). Since Γ is connected, thenumber β (Γ) = |E (Γ) | − |V (Γ) |+ 1 is equal to the number of independent cycles in Γ and it is

referred to as the Betti number of Γ.Let V (Γ) = 0, 1, . . . , |V (Γ) | −1. For any arc ( i, j ) ∈ A(Γ) , by φi,j we denote the voltageon the arc, which is identied with a row vector in V (n, p ). An arc ( i, j ) ∈ A(Γ) is calledpositive (resp. negative ) if i < j (resp. i > j ). For each subset F in E (Γ), we denote theset of its arcs, positive arcs and negative arcs by A(F ), A+ (F ) and A−(F ), respectively, sothat A(F ) = A+ (F )∪A−(F ). In particular, if F = E (Γ) , we prefer to use A(Γ), A+ (Γ) andA−(Γ) to denote A(F ), A+ (F ) and A−(F ), respectively. Fix a spanning tree T in the graph






Now we state the algorithm for solving lifting problem for the connected regular coveringsof a graph Γ whose covering transformation group is elementary Abelian.

(1st) Choose a xed spanning tree T in Γ and write down the arcs in A+ (E 0), A+ (E 1 ) andA+ (E 2) in a certain order so that Φ 0 = 0, Φ1 = I n

×n and Φ2 = M.

(2nd) Calculate the incidence matrix P for the fundamental cycles of Γ with respect to T .

(3rd) Assume that the voltage generating matrix M = ( a ij )m ×n , where the entries aij areunknowns. Let ∆ = (( −M, I m ×m )P, −M, I m ×m ), whose columns are indexed by thearcs in A+ (E 0 ), A+ (E 1), A+ (E 2) according to the given order. We call the matrix ∆the discriminant matrix for a lift of φ. For convenience, we write ∆ 0 = ( −M, I m ×m )P ,∆ 1 = −M and ∆ 2 = I m ×m , so that ∆ = (∆ 0 , ∆ 1 , ∆ 2), as a block matrix.

(4th) Let ∆ = ( · · · , ci,j , · · ·), where ci,j is the column indexed by ( i, j ) ∈ A+ (Γ) . For a givenσ ∈ Aut(Γ) , let cσ

i,j = ci σ − 1 ,j σ − 1 , where we assume that c i,j = −cj,i for any arc ( i, j ). Let∆ σ = ( · · · , cσ

i,j , · · ·) for any ( i, j ) ∈ A+ (Γ) , and let (∆ σ )0 , (∆ σ )1 and (∆ σ )2 denote the

rst, the second and the third blocks of the matrix ∆ σ respectively, as before. Then onecan say that

σ can be lifted ⇐⇒ (∆ σ )1 + (∆ σ )2 M = 0 ⇐⇒ ∆ σ1 + ∆ σ

2 M = 0. (5)

Proof of Theorem 1.1

Let E 0 = E (T ) and E = E (Γ) \ E 0 . Give an ordering for the arcs in A+ (E 0) and A+ (E ) asfollows:

A+ (E 0) =

15, 16, 17, 25, 28, 38, 48

,

A+ (E ) = 26, 46, 47, 37, 35.

Give fundamental cycles in Γ as follows:

C (2, 6; T ) = (2 , 6, 1, 5, 2),

C (4, 6; T ) = (4 , 6, 1, 5, 2, 8, 4),

C (4, 7; T ) = (4 , 7, 1, 5, 2, 8, 4),

C (3, 7; T ) = (3 , 7, 1, 5, 2, 8, 3),

C (3, 5; T ) = (3 , 5, 2, 8, 3).

It is well-known that Aut(Γ)

∼= S 4

×Z 2 . Take four automorphisms of Γ as follows: α =

(243)(567), β = (14)(23)(58)(67) , γ = (18)(27)(36)(45) and δ = (23)(67).It is easy to check that M = < α,β > is subgroup of Aut(Γ) isomorphic to A4 , N = <

M,γ > is subgroup of Aut(Γ) isomorphic to A4×Z 2 , and < N,δ > = Aut(Γ). Thus we have:

(1) M can be lifted if and only if α and β can be lifted;(2) N can be lifted if and only if α , β and γ can be lifted;(3) Aut(Γ) can be lifted if and only if α , β , γ and δ can be lifted.




Since β (Γ) = 5 , we get n 5. If n = 5 , then Γ is nothing but X ( p, 5) and by [6, Theorem2.11], Aut(Γ) can be lifted. So, in what follows, we assume n < 5 and divide our proof intofour cases for each n with 1 n 4.

3.1 The Case of n = 1Suppose that n = 1 . Then K = Z p = V (1, p). Since the element (18)(25)(36)(47) of Aut(Γ)maps 2, 6, 4, 7, 3, 5 to 5, 3, 7, 4, 6, 2, respectively, without loss of any generality, we have thefollowing three essentially different cases for the set E 1 and E 2 :

(1) E 1 = 26 and E 2 = 46, 47, 37, 35;

(2) E 1 = 46 and E 2 = 26, 47, 37, 35;

(3) E 1 = 47 and E 2 = 26, 46, 37, 35.

Case (1): E 1 = 26 and E 2 = 46, 47, 37, 35.

In this case, the incidence matrix is:

P =

(1,5) (1,6) (1,7) (2,5) (2,8) (3,8) (4,8)

(2,6) 1 −1 0 −1 0 0 0(4,6) 1 −1 0 −1 1 0 −1(4,7) 1 0 −1 −1 1 0 −1(3,7) 1 0 −1 −1 1 −1 0(3,5) 0 0 0 −1 1 −1 0

.

Let D = ( D0 , D 1 , D 2) be the discriminant matrix with D0 = ( −M, I 4×4)P , D1 = ( −M )and D2 = ( I 4×4 ) with a voltage generating matrix M = ( a,b,c,d)t . A direct computation givesthat D 0 = ( −M, I 4×4)P is equal to

D 0 =

(1, 5) (1, 6) (1, 7) (2 , 5) (2 , 8) (3, 8) (4 , 8)

−a + 1 a −1 0 a −1 1 0 −1

−b + 1 b −1 b−1 1 0 −1

−c + 1 c −1 c −1 1 −1 0

−d d 0 d −1 1 −1 0

.

For an automorphism σ of Γ can be lifted if and only if

∆ σ1 + ∆ σ

2 M = ( c26 )σ + ( c46 c47 c37 c35 )σ M = 0. (6)

Inserting α = (243)(567) to (6), we have

0 = Dα

1 + Dα

2 M = ( c26 )α

+ ( c46 c47 c37 c35 )α

M = ( c35 ) + ( c25 c26 c46 c47 )M

=

a2 −a −ab + c

ab −a −b2 + d

ac −a −bc

ad −a −bd + 1

= ( xi,j ).




Inserting β = (14)(23)(58)(67) to (6), we have

0 = Dβ1 + D β

2 M = ( c26 )β + ( c46 c47 c37 c35 )β M

= ( c37 ) + ( c17 c16 c26 c28 )M

=

ab −b−ac + d

−a + b2 −bc + d

−a + bc−c2 + d + 1

bd−cd + d

= ( yij ).

Now assume that α and β can be lifted. By y41 = 0, we distinguish two cases: (1) d = 0;(2) b−c + 1 = 0. If (1) happens, by x41 = 0 and x11 = 0, we have a = 1 and b = c. But itdoesn’t satisfy y21 = 0. If (2) happens, by y11 = 0 , we have −a −b + d = 0. By x21 = 0, wedistinguish two subcases: (i) a −b + 1 = 0; (ii) b = 0. If (i) happens, by ( x ij ) = 0, we have

that either a = c = 0, b = d = 1 and p = 2 or a = d = 2, b = 0, c = 1 and p = 3. If (ii)happens, we have c = 1 and a = d. By x11 = 0, we have a2 −a + 1 = 0. Thus the solution of (x ij ) = ( yij ) = 0 are: a = c = 0, b = d = 1 and p = 2; or a = d = 1±√ −3

2 , b = 0, c = 1 and p = 3 or p ≡ 1(mod 6). Or equivalently,

M 1 = (0 , 1, 0, 1) t , p = 2;

M 21 = (1 + √ −3

2 , 0, 1,

1 + √ −32

)t , p = 3 or p ≡ 1(mod 6);

M 22 = ( 1 −√ −32

, 0, 1, 1 −√ −32

)t , p = 3 or p ≡ 1(mod 6) .

By X (2, 1), X ( p, 1) and X ′( p, 1), we denote the covering graphs determined by M 1 , M 21

and M 22 , respectively. In particular, X ( p, 1) and X ′( p, 1) are the same one if p = 3.

Inserting γ = (18)(27)(36)(45) to (6), we have

0 = Dγ 1 + D γ

2 M = ( c26 )γ + ( c46 c47 c37 c35 )γ M

= ( c73 ) + ( c53 c52 c62 c64 )M

=

b−ab + ac −d

b−b2 + bcb−bc + c2 −1

−a + b−bd + cd

.

Clearly, the matrices M 1 , M 21 and M 22 satisfy this equation and γ can be lifted, and sothe group N can be lifted.




Inserting δ = (23)(67) to (6), we have

0 = Dδ1 + D δ

2 M = ( c26 )δ + ( c46 c47 c37 c35 )δ M

= ( c37 ) + ( c47 c46 c26 c25 )M

=

b−ac + ad −da −bc + bd−d

−c2 + cd −d + 1

−cd + d2 −d

.

Clearly, the matrix M 1 satisfy this equation and δ can be lifted. But the matrices M 21 andM 22 satisfy this equation and δ can be lifted when p = 3. So, for graphs X (2, 1) and X (3, 1),Aut(Γ) can be lifted.

Finally, we show that for p ≡ 1(mod 6), the graphs X ( p, 1) and X ′( p, 1) are isomor-phic as graphs. Let V := V (X ( p, 1)) = V (X ′( p, 1)) = (i; x) | 1 ≤ i ≤ 8, x ∈ GF ( p). Let

R = ( −1+ √

−3

2 ), and let ζ = (18)(25)(36)(47) ∈Aut(Γ). Dene a permutation Υ on V by(i; x)Υ = ( iζ ; (x)R). A direct checking shows that Υ is isomorphism from X ( p, 1) to X ′( p, 1).

Cases (2) and (3): By a computation similar to case (1), we can get the graph X ( p, 1) incase (2) and the graph X (2, 1) in the case (3).

3.2 The case of n = 2 , 3, and 4

In this subsection, the case n = 2 , 3, and 4 will be described briey.

Case n = 2 . In the case, K = Z 2 p = V (2, p). As before, without loss we may assume one of the

following happens:

(1) E 1 = 46, 37 and E 2 = 26, 47, 35;

(2) E 1 = 26, 46 and E 2 = 47, 37, 35;

(3) E 1 = 26, 47 and E 2 = 46, 37, 35;

(4) E 1 = 26, 37 and E 2 = 46, 47, 35;

(5) E 1 = 26, 35 and E 2 = 46, 47, 37;

(6) E 1 =

46, 47

and E 2 =

26, 37, 35

.

For the case of (1), let

M =

a1 b1

a2 b2

a3 b3

.




be a voltage generating matrix. Then a computation similar to case n = 1 gives that Aut(Γ)can be lifted if and only if a1 = a2 = b2 = b3 and b1 = a3 = 1 . Accordingly, we can get thegraph X ( p, 2).

In the Cases of (2), (3), (4), (5) and (6), the group M can not be lifted.

Case n = 3 . In this case, K = Z 3 p = V (3, p). Without any loss of generality, we may assumeone of the following happens:

(1) E 1 = 26, 47, 35 and E 2 = 46, 37;

(2) E 1 = 26, 46, 47 and E 2 = 37, 35;

(3) E 1 = 26, 46, 37 and E 2 = 47, 35;

(4) E 1 = 26, 46, 35 and E 2 = 47, 37;

(5) E 1 = 46, 47, 37 and E 2 = 26, 35;

(6) E 1 = 26, 47, 37 and E 2 = 46, 35.

For the case of (1), one can show that Aut(Γ) can be lifted if and only if the voltagegeneration matrix

M = 0 1 −1

−1 1 0.

Thus, we get the graph X ( p, 3).In the case (2), (3), (4), (5) and (6), the group M can not be lifted.

Case n = 4 . In this case, K = Z 4 p = V (4, p). Without any loss of generality, we may divide ourdiscussion into mutually exclusive three cases as followings:

Case (1): E 1 = 26, 46, 47, 37 and E 2 = 35.Let D = ( D0 , D 1 , D 2) be the discriminant matrix with D0 = ( −M, I 1×1)P , D1 = ( −M )

and D 2 = ( I 1×1) with a voltage generating matrix M = ( a,b,c,d).The group Aut(Γ) can be lifted if and only if

M 1 = (1 −√ −3

2 ,−1,

1 + √ −32

, 1 −√ −3

2 ), p = 3 or p ≡ 1(mod 6);

M 2 = (1 + √ −3

2 ,−1,

1 −√ −32

, 1 + √ −3

2 ), p = 3 or p ≡ 1(mod 6).

By X ( p, 4) and X ′( p, 4), we denote the covering graphs determined by M 1 and M 2 , respec-

tively. In particular, X ( p, 4) and X ′( p, 4) are the same one if p = 3.

For p ≡ 1(mod 6), the graphs X ( p, 4) and X ′( p, 4) are isomorphic as graphs. Let V :=




V (X ( p, 4)) = V (X ′( p, 4)) = (i; x,y,z,w ) | 1 ≤ i ≤ 8, x,y, z,w ∈ GF ( p). Let

R =

c2 −1 1 −c2 c2 −1

0 0 0

−1

0 0 −1 0

0 −1 0 0

.

And let and let ζ = (18)(25)(36)(47) ∈Aut(Γ). Dene a permutation Υ on V by (i; x,y,z,w )Υ =

(iζ ; (x,y,z,w )R). A direct checking shows that Υ is isomorphism from X ( p, 4) to X ′( p, 4).

Case (2): E 1 = 26, 46, 47, 35 and E 2 = 37, and φ26 , φ46 , φ47 φ37, φ46 , φ47 , φ37 φ35 are

linear dependant, hence letting

M = (0 ,b,c, 0).

Now, one can show that M cannot be lifted.By a similar computation, in case (3) , E 1 = 26, 46, 37, 35and E 2 = 47, and the group

M can not be lifted.

Combining subsection 3.1 and 3.2, we nish the proof of Theorem 1 .1.

References

[1] N.L. Biggs, Algebraic graph theory , Cambridge University Press, Cambridge, 1974.[2] N.L. Biggs, Homological coverings of graphs, J. London Math. Soc. 30(1984), 1–14.[3] S.F. Du, J.H. Kwak and M.Y. Xu, Linear criteria for liftings of automorphisms of the

elementary abelian regular coverings, to appear in Linear Algebra and Its applications ,2003.

[4] J.L. Gross and T.W. Tucker, Generating all graph coverings by permutation voltage as-signments, Discrete Math. 18(1977) 273–283.

[5] J.L. Gross and T.W. Tucker, Topological Graph Theory , Wiley-Interscience, New York,1987.




Super Fibonacci Graceful Labeling of

Some Special Class of Graphs

R.Sridevi 1 , S.Navaneethakrishnan 2 and K.Nagarajan 1

1. Department of Mathematics, Sri S.R.N.M.College, Sattur - 626 203, Tamil Nadu, India

2. Department of Mathematics, V.O.C.College, Tuticorin - 628 008, Tamil Nadu, India

E-mail: r.sridevi [email protected], [email protected], k nagarajan [email protected]

Abstract : A Smarandache-Fibonacci Triple is a sequence S (n), n ≥ 0 such that S (n) =S (n −1) + S (n −2), where S (n) is the Smarandache function for integers n ≥ 0. Certainly,

it is a generalization of Fibonacci sequence. A Fibonacci graceful labeling and a super Fi-bonacci graceful labeling on graphs were introduced by Kathiresan and Amutha in 2006.Generally, let G be a ( p, q )-graph and S (n)|n ≥ 0 a Smarandache-Fibonacci Triple. An bijection f : V (G) → S (0) , S (1) , S (2) , . . . , S (q ) is said to be a super Smarandache-Fibonaccigraceful graph if the induced edge labeling f ∗ (uv) = |f (u) −f (v)| is a bijection onto the set

S (1) , S (2) , . . . , S (q ). Particularly, if S (n), n ≥ 0 is just the Fibonacci sequence F i , i ≥ 0,such a graph is called a super Fibonacci graceful graph. In this paper, we show that somespecial class of graphs namely F tn , C tn and S tm,n are super bonacci graceful graphs.

Key Words : Smarandache-Fibonacci triple, graceful labeling, Fibonacci graceful labeling,super Smarandache-Fibonacci graceful graph, super Fibonacci graceful graph.

AMS(2010) : 05C78

§1. Introduction

By a graph, we mean a nite undirected graph without loops or multiple edges. A path of length n is denoted by P n +1 . A cycle of length n is denoted by C n . G+ is a graph obtainedfrom the graph G by attaching pendant vertex to each vertex of G. Graph labelings, wherethe vertices are assigned certain values subject to some conditions, have often motivated bypractical problems. In the last ve decades enormous work has been done on this subject [1].

The concept of graceful labeling was rst introduced by Rosa [6] in 1967. A function f isa graceful labeling of a graph G with q edges if f is an injection from the vertices of G to theset 0, 1, 2, . . . , q such that when each edge uv is assigned the label |f (u) −f (v)|, the resultingedge labels are distinct.

The notion of Fibonacci graceful labeling and Super Fibonacci graceful labeling were in-troduced by Kathiresan and Amutha [5]. We call a function f , a bonacci graceful labeling of agraph G with q edges if f is an injection from the vertices of G to the set 0, 1, 2, . . . , F q, where

1 Received November 1, 2010. Accepted February 25, 2011.





Super Fibonacci Graceful Labeling of Some Special Class of Graphs 61

Let E 2 = f ∗(u j1u j +1

1 ) : 1 ≤ j ≤ n −1. Then

E 2 = |f (u j1) −f (u j +1

1 )| : 1 ≤ j ≤ n −1= |f (u1

1) −f (u21)|, |f (u2

1) −f (u31)|, . . . , |f (un −2

1 ) −f (un −11 )|, |f (un −1

1 ) −f (un1 )|

= |F 1 −F 3|, |F 3 −F 5|, . . . , |F 2n −5 −F 2n −3|, |F 2n −3 −F 2n −1 |= F 2 , F 4 , . . . , F 2n −4 , F 2n −2.

For i = 2, we know that

E 3 = f ∗(u0u j2) : 1 ≤ j ≤ n = |f (u0) −f (u j

2)| : 1 ≤ j ≤ n= |f (u0) −f (u1

2)|, |f (u0) −f (u22)|, . . . , |f (u0) −f (un −1

2 )|, |f (u0) −f (un2 )|

= |F 0 −F 2n |, |F 0 −F 2n +2 |, . . . , |F 0 −F 4n −4 |, |F 0 −F 4n −2 |= F 2n , F 2n +2 , . . . , F 4n −4 , F 4n −2,

E 4 = f ∗(uj2u

j +12 ) : 1 ≤ j ≤ n −1= |f (u j

2) −f (u j +12 )| : 1 ≤ j ≤ n −1

= |f (u12) −f (u2

2)|, |f (u22) −f (u3

2)|, . . . , |f (un −22 ) −f (un −1

2 )|, |f (un −12 ) −f (un

2 )|= |F 2n −F 2n +2 |, |F 2n +2 −F 2n +4 |, . . . , |F 4n −6 −F 4n −4|, |F 4n −4 −F 4n −2 |= F 2n +1 , F 2n +3 , . . . , F 4n −5 , F 4n −3.

For i = 3, let E 5 = f ∗(u0u j3) : 1 ≤ j ≤ n. Then

E 5 = |f (u0) −f (u j3)| : 1 ≤ j ≤ n

= |f (u0) −f (u13)|, |f (u0) −f (u2

3)|, . . . , |f (u0) −f (un −13 )|, |f (u0) −f (un

3 )|= |F 0 −F 4n −1 |, |F 0 −F 4n +1 |, . . . , |F 0 −F 6n −5|, |F 0 −F 6n −3|= F 4n −1 , F 4n +1 , . . . , F 6n −5 , F 6n −3.

Let E 6 = f ∗(u j3u j +1

3 ) : 1 ≤ j ≤ n −1. Then

E 6 = |f (u j3) −f (u j +1

3 )| : 1 ≤ j ≤ n −1= |f (u1

3) −f (u23)|, |f (u2

3) −f (u33)|, . . . , |f (un −2

3 ) −f (un −13 )|, |f (un −1

3 ) −f (un3 )|

= |F 4n −1 −F 4n +1 |, |F 4n +1 −F 4n +3 |, . . . , |F 6n −7 −F 6n −5|, |F 6n −5 −F 6n −3 |= F 4n , F 4n +2 , . . . , F 6n −6 , F 6n −4

· · · · · · · · · · · · · · · ,

Now, for i = t −1, let E t−1 = f ∗(u0u jt−1) : 1 ≤ j ≤ n. Then

E t−1 = |f (u0) −f (u jt−1)| : 1 ≤ j ≤ n

= |f (u0) −f (u1t−1)|, |f (u0) −f (u2

t−1)|, . . . , |f (u0) −f (un −1t−1 )|, |f (u0) −f (un

t−1)|= |F 0 −F 2nt −4n −t +3 |, |F 0 −F 2nt −4n −t +5 |, . . . , |F 0 −F 2nt −2n −t−1 |, |F 0 −F 2nt −2n −t +1 |= F 2nt −4n −t +3 , F 2nt −4n −t +5 , . . . , F 2nt −2n −t−1 , F 2nt −2n −t +1 .



62 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan

Let E t−1 = f ∗(u jt−1u j +1

t−1 ) : 1 ≤ j ≤ n −1. Then

E t−1 = |f (u jt−1) −f (u j +1

t−1 )| : 1 ≤ j ≤ n −1= |f (u1

t−1) −f (u2t−1)|, |f (u2

t−1) −f (u3t−1)|,

. . . , |f (un

−2

t−1 ) −f (un

−1

t−1 )|, |f (un

−1

t−1 ) −f (unt−1)|= |F 2nt −4n −t +3 −F 2nt −4n −t +5 |, |F 2nt −4n −t +5 −F 2nt −4n −t +7 |,

. . . , |F 2nt −2n −t−3 −F 2nt −2n −t−1 |, |F 2nt −2n −t−1 −F 2nt −2n −t +1 |= F 2nt −4n −t +4 , F 2nt −4n −t +6 , . . . , F 2nt −2n −t−2 , F 2nt −2n −t .

F 1

F 8F 10F 12F 14

F 19

F 21

F 22 F 26 F 28

F 0

F 24F 23 F 25 F 27

F 22 F 24 F 28

F 1

F 3

F 5

F 8

F 26

F 7

F 10F 12F 14

F 15

F 17

F 19

F 21

F 3

F 7

F 5

F 2

F 4

F 6

F 9F 11F 13

F 17

F 15

F 16

F 18

F 20

Fig.1

F 44 :

For i = t, let E t =

f ∗(u0u j

t) : 1

≤ j

≤ n

. Then

E t = |f (u0) −f (u jt )| : 1 ≤ j ≤ n

= |f (u0) −f (u1t )|, |f (u0) −f (u2

t )|, . . . , |f (u0) −f (un −1t )|,

|f (u0) −f (unt )|

= |F 0 −F 2nt −2n −t +2 |, |F 0 −F 2nt −2n −t +4 |, . . . , |F 0 −F 2nt −t−2 |, |F 0 −F 2nt −t |= F 2nt −2n −t +2 , F 2nt −2n −t +4 , . . . , F 2nt −t−2 , F 2nt −t .






of the rst loop. Let E 3 = f ∗(u(n −1)( i−1)+2 u(n −1)( i−1)+3 ) : 1 ≤ i ≤ t. Then

E 3 = |f (u(n −1)( i−1)+2 ) −f (u(n −1)( i−1)+3 )| : 1 ≤ i ≤ t= |f (u2) −f (u3)|, |f (un +1 ) −f (un +2 )|, |f (u2n ) −f (u2n +1 )|, . . . ,

|f (unt −2n −t +4 ) −f (unt −2n −t +5 )|, |f (unt −n −t +3 ) −f (unt −n −t +4 )|= |F nt −2 −F nt −1|, |F nt −n −2 −F nt −n −1|, |F nt −2n −2 −F nt −2n −1|,

. . . , |F 2n −2 −F 2n −1|, |F n −2 −F n −1|= F nt −3 , F nt −n −3 , F nt −2n −3 , . . . , F 2n −3 , F n −3.

Now, for s = 1, let E 4 =t

i=1 f ∗(u (n −1)( i−1)+ j u(n −1)( i−1)+ j +1 ) : 3 ≤ j ≤ 4. Then

E 4 = ∪ti=1 |f (u(n −1)( i−1)+ j ) −f (u (n −1)( i−1)+ j +1 )| : 3 ≤ j ≤ 4= |f (u3) −f (u4)|, |f (u4) −f (u5)|∪|f (un +2 ) −f (un +3 )|, |f (un +3 ) −f (un +4 )|∪, . . . ,

∪|f (unt −2n −t +5 ) −f (unt −2n −t +6 )|, |f (unt −2n −t +6 ) −f (unt −2n −t +7 )|∪|f (unt −n −t +4 ) −f (unt −n −t +5 )|, |f (unt −n −t +5 ) −f (unt −n −t +6 )|

= |F nt −1 −F nt −3|, |F nt −3 −F nt −5|∪|F nt −n −1 −F nt −n −3|, |F nt −n −3 −F nt −n −5|∪, . . . ,∪|F 2n −1 −F 2n −3|, |F 2n −3 −F 2n −5|

∪|F n

−1

−F n

−3

|,

|F n

−3

−F n

−5

|= F nt −2 , F nt −4∪F nt −n −2 , F nt −n −4∪, . . . ,

∪F 2n −2 , F 2n −4∪F n −2 , F n −4

For the edge labeling between the end vertex ( u(n −1)( i−1)+5 ) of the rst loop and startingvertex ( u (n −1)( i−1)+6 ) of the second loop, calculation shows that

E 14 = |f (u(n −1)( i−1)+5 ) −f (u (n −1)( i−1)+6 )| : 1 ≤ i ≤ t=

|f (u5)

−f (u6)

|,

|f (un +4 )

−f (un +5 )

|, . . . ,

|f (unt −2n −t +7 ) −f (unt −2n −t +8 )|, |f (unt −n −t +6 ) −f (unt −n −t +7 )|= |F nt −5 −F nt −4|, |F nt −n −5 −F nt −n −4|, . . . , |F 2n −5 −F 2n −4 |, |F n −5 −F n −4|= F nt −6 , F nt −n −6 , . . . , F 2n −6 , F n −6

For s = 2, let E 5 = ∪ti =1 f ∗(u(n −1)( i−1)+ j u(n −1)( i−1)+ j +1 ) : 6 ≤ j ≤ 7. Then




E 5 = ∪ti =1 |f (u (n −1)( i−1)+ j ) −f (u(n −1)( i−1)+ j +1 )| : 6 ≤ j ≤ 7= |f (u6) −f (u7)|, |f (u7) −f (u8)|∪|f (un +5 ) −f (un +6 )|, |f (un +6 ) −f (un +7 )|∪, . . . ,

∪|f (unt −2n −t +8 ) −f (unt −2n −t +9 )|, |f (unt −2n −t +9 ) −f (unt −2n −t +10 )|∪|f (unt −n −t +7 ) −f (unt −n −t +8 )|, |f (unt −n −t +8 ) −f (unt −n −t +9 )|

= |F nt −4 −F nt −6|, |F nt −6 −F nt −8|∪|F nt −n −4 −F nt −n −6|, |F nt −n −6 −F nt −n −8|∪, . . . ,∪|F 2n −4 −F 2n −6|, |F 2n −6 −F 2n −8|∪|F n −4 −F n −6 |, |F n −6 −F n −8|

= F nt −5 , F nt −7∪F nt −n −5 , F nt −n −7∪, . . . ,

∪F 2n −5 , F 2n −7

∪F n −5 , F n −7

Similarly, for nding the edge labeling between the end vertex ( u(n −1)( i−1)+8 ) of the secondloop and starting vertex ( u(n −1)( i−1)+9 ) of the third loop, calculation shows that

E 15 = |f (u (n −1)( i−1)+8 ) −f (u(n −1)( i−1)+9 )| : 1 ≤ i ≤ t= |f (u8) −f (u9)|, |f (un +7 ) −f (un +8 )|, . . . ,

|f (unt −2n −t +10 ) −f (unt −2n −t +11 )|, |f (unt −n −t +9 ) −f (unt −n −t +10 )|= |F nt −8 −F nt −7 |, |F nt −n −8 −F nt −n −7|, . . . , |F 2n −8 −F 2n −7 |,

|F n −8 −F n −7|= F nt −9 , F nt −n −9 , . . . , F 2n −9 , F n −9,

· · · · · · · · · · · · ,

For s = n −3

3 −1, let E n − 33 −1 = ∪ti=1 f ∗(u (n −1)( i−1)+ j u(n −1)( i−1)+ j +1 ) : n −6 ≤ j ≤ n −5.

Then

E n − 33 −1 = ∪ti=1 |f (u(n −1)( i−1)+ j ) −f (u (n −1)( i−1)+ j +1 )| : n −6 ≤ j ≤ n −5

= |f (un −6) −f (un −5)|, |f (un −5) −f (un −4)|∪|f (u2n −7) −f (u2n −6)|, |f (u2n −6) −f (u2n −5)|∪, . . . ,

∪|f (unt −n −t−4) −f (unt −n −t−3)|, |f (unt −n −t−3) −f (unt −n −t−2)|∪|f (unt −t−5) −f (unt −t−4)|, |f (unt −t−4) −f (unt −t−3)|= |F nt −n +8 −F nt −n +6 |, |F nt −n +6 −F nt −n +4 |∪|F nt −2n +8 −F nt −2n +6 |,|F nt −2n +6 −F nt −2n +4 |∪, . . . ,∪|F n +8 −F n +6 |, |F n +6 −F n +4 |∪|F 8 −F 6|, |F 6 −F 4 |

= F nt −n +7 , F nt −n +5 ∪F nt −2n +7 , F nt −2n +5 ∪, . . . ,∪F n +7 , F n +5 ∪F 7 , F 5




We calculate the edge labeling between the end vertex ( u (n −1)( i−1)+ n −4) of the ( n −33 −1) th

loop and starting vertex ( u(n −1)( i−1)+ n −3) of the ( n −33 )rd loop as follows.

E 1n − 33 −1 = |f (u(n −1)( i−1)+ n −4) −f (u (n −1)( i−1)+ n −3)| : 1 ≤ i ≤ t

= |f (un −4) −f (un −3)|, |f (u2n −5) −f (u2n −4)|, . . . ,

|f (unt −n −t +2 ) −f (unt −n −t−1)|, |f (unt −t−3) −f (unt −t−2)|= |F nt −n +6 −F nt −n +5 |, |F nt −2n +4 −F nt −2n +5 |, . . . ,

|F n +4 −F n +5 |, |F 4 −F 5|= F nt −n +4 , F nt −2n +3 , . . . , F n +3 , F 3

For s = n −3

3 , let E n − 3

3=

t

i =1 f ∗(u (n −1)( i−1)+ j u(n −1)( i−1)+ j +1 ) : n − 3 ≤ j ≤ n − 2.

Then

E n − 3

3=

∪t

i=1 |f (u

(n −1)( i−1)+ j)

−f (u

(n −1)( i−1)+ j +1)

| : n

−3

≤ j

≤ n

−2

= |f (un −3) −f (un −2)|, |f (un −2) −f (un −1)|∪|f (u2n −4) −f (u2n −3)|, |f (u2n −3) −f (u2n −2)|∪, . . . ,

∪|f (unt −n −t−1) −f (unt −n −t )|, |f (unt −n −t ) −f (unt −n −t +1 )|∪|f (unt −t−2) −f (unt −t−1)|, |f (unt −t−1) −f (unt −t )|

= |F nt −n +5 −F nt −n +3 |, |F nt −n +3 −F nt −n +1 |∪|F nt −2n +5 −F nt −2n +3 |, |F nt −2n +3 −F nt −2n +1 |∪, . . . ,

∪|F n +5 −F n +3 |, |F n +3 −F n +1 |∪|F 5 −F 3 |, |F 3 −F 1|= F nt −n +4 , F nt −n +2 ∪F nt −2n +4 , F nt −2n +2 ∪, . . . ,

∪F n +4 , F n +2 ∪F 4 , F 2Calculation shows the edge labeling between the end vertex ( u(n −1)( i−1)+ n −1) of the ( n −3

3 )rd

loop and the vertex u0 are

E ∗ = |f (u(n −1)( i−1)+ n −1) −f (u0)| : 1 ≤ i ≤ t= |f (un −1) −f (u0)|, |f (u2n −2) −f (u0)|, . . . ,

|f (unt −n −t +1 ) −f (u0)|, |f (unt −t ) −f (u0)|= |F nt −n +1 −F 0|, |F nt −2n +1 −F 0|, . . . , |F n +1 −F 0 |, |F 1 −F 0|= F nt −n +1 , F nt −2n +1 , . . . , F n +1 , F 1.

Therefore,

E = ( E 1 ∪E 2∪, . . . ,∪E n − 33

)∪(E 14 ∪E 15∪, . . . ,∪E 1n − 33 −1)∪E ∗

= F 1 , F 2 , . . . , F nt Thus, the edge labels are distinct. Therefore, C tn admits a super bonacci graceful labeling.






For the edge labeling between the vertex vi2 and starting vertex vi

3 of the rst loop, letE 3 = f ∗(vi

2vi3) : 1 ≤ i ≤ mt. Calculation shows that

E 3 = |f (vi

2) −f (vi

3)| : 1 ≤ i ≤ mt= |f (v12 ) −f (v1

3 )|, |f (v22 ) −f (v2

3 )|, . . . , |f (vmt −12 ) −f (vmt −1

3 )|,|f (vmt

2 ) −f (vmt3 )|

= |F mnt −2 −F mnt −1|, |F mnt −n −2 −F mnt −n −1 |, . . . , |F 2n −2 −F 2n −1|,|F n −2 −F n −1 |

= F mnt −3 , F mnt −n −3 , . . . , F 2n −3 , F n −3.

For s = 1, let E 4 = ∪mti =1 f ∗(vi

j vij +1 ) : 3 ≤ j ≤ 4. Then

E 4 = ∪mti=1 |f (vi

j ) −f (vij +1 )| : 3 ≤ j ≤ 4

= |f (v13 ) −f (v1

4 )|, |f (v14 ) −f (v1

5 )|∪|f (v23 ) −f (v2

4 )|, |f (v24 ) −f (v2

5 )|∪, . . . ,∪|f (vmt −1

3 ) −f (vmt −14 )|, |f (vmt −1

4 ) −f (vmt −15 )|

∪|f (vmt3 ) −f (vmt

4 )|, |f (vmt4 ) −f (vmt

5 )|= |F mnt −1 −F mnt −3|, |F mnt −3 −F mnt −5|∪|F mnt −n −1 −F mnt −n −3|, |F mnt −n −3 −F mnt −n −5|∪, . . . ,∪|F 2n −1 −F 2n −3 |, |F 2n −3 −F 2n −5|∪|F n −1 −F n −3|, |F n −3 −F n −5|

=

F mnt

−2 , F mnt

−4

∪F mnt

−n

−2 , F mnt

−n

−4

∪, . . . ,

∪F 2n

−2 , F 2n

−4

∪F n −2 , F n −4.

We nd the edge labeling between the vertex vi5 of the rst loop and starting vertex vi

6 of the second loop. Let E 14 = f ∗(vi

5vi6) : 1 ≤ i ≤ mt. Then

E 14 = |f (vi5) −f (vi

6)| : 1 ≤ i ≤ mt= |f (v1

5 ) −f (v16 )|, |f (v2

5 ) −f (v26 )|, . . . , |f (vmt −1

5 ) −f (vmt −16 )|,

|f (vmt5 ) −f (vmt

6 )|= |F mnt −5 −F mnt −4|, |F mnt −n −5 −F mnt −n −4|, . . . , |F 2n −5 −F 2n −4 |,

|F n −5 −F n −4|= F mnt −6 , F mnt −n −6 , . . . , F 2n −6 , F n −6.

For s = 2, let E 5 = ∪mti =1 f ∗(vi

j vij +1 ) : 6 ≤ j ≤ 7. Then




E 5 = ∪mti=1 |f (vi

j ) −f (vij +1 )| : 6 ≤ j ≤ 7

= |f (v16 ) −f (v1

7 )|, |f (v17 ) −f (v1

8 )|∪|f (v26 ) −f (v2

7 )|, |f (v27 ) −f (v2

8 )|∪, . . . , . . . , |f (vmt −16 ) −f (vmt −17 )|, |f (vmt −17 ) −f (vmt −18 )|∪|f (vmt

6 ) −f (vmt7 )|, |f (vmt

7 ) −f (vmt8 )|

= |F mnt −4 −F mnt −6 |, |F mnt −6 −F mnt −8 |∪|F mnt −n −4 −F mnt −n −6|,|F mnt −n −6 −F mnt −n −8|∪, . . . ,∪|F 2n −4 −F 2n −6|, |F 2n −6 −F 2n −8|∪|F n −4 −F n −6|, |F n −6 −F n −8 |

= F mnt −5 , F mnt −7∪F mnt −n −5 , F mnt −n −7∪, . . . ,∪F 2n −5 , F 2n −7∪F n −5 , F n −7.

Let E 15 = f ∗(vi8vi

9) : 1 ≤ i ≤ mt . Calculation shows that the edge labeling between the

vertex vi8 of the second loop and starting vertex v

i9 of the third loop are

E 15 = |f (vi8) −f (vi

9 )| : 1 ≤ i ≤ mt= |f (v1

8 ) −f (v19 )|, |f (v2

8 ) −f (v29 )|, . . . , |f (vmt −1

8 ) −f (vmt −19 )|,

|f (vmt8 ) −f (vmt

9 )|= |F mnt −8 −F mnt −7 |, |F mnt −n −8 −F mnt −n −7|, . . . ,

|F 2n −8 −F 2n −7|, |F n −8 −F n −7 |= F mnt −9 , F mnt −n −9 , . . . , F 2n −9 , F n −9.

For s = n −4

3 −1, let E n − 4

3

−1 =

∪mti=1

f ∗(vi

j vij +1 ) : n

−7

≤ j

≤ n

−6

. Then

E n − 43 −1 = ∪mt

i=1 |f (vij ) −f (vi

j +1 )| : n −7 ≤ j ≤ n −6= |f (v1

n −7) −f (v1n −6)|, |f (v1

n −6) −f (v1n −5)|

∪|f (v2n −7) −f (v2

n −6)|, |f (v2n −6) −f (v2

n −5 )|∪, . . . ,∪|f (vmt −1

n −7 ) −f (vmt −1n −6 )|, |f (vmt −1

n −6 ) −f (vmt −1n −5 )|

∪|f (vmtn −7) −f (vmt

n −6)|, |f (vmtn −6) −f (vmt

n −5 )|= |F mnt −n +9 −F mnt −n +7 |, |F mnt −n +7 −F mnt −n +5 |∪|F mnt −2n +9 −F mnt −2n +7 |, |F mnt −2n +7 −F mnt −2n +5 |∪, . . . ,∪|F n +9 −F n +7 |, |F n +7 −F n +5 |∪|F 9 −F 7|, |F 7 −F 5|

= F mnt −n +8 , F mnt −n +6 ∪F mnt −2n +8 , F mnt −2n +6 ∪, . . . ,

∪F n +8 , F n +6 ∪F 8 , F 6.

Similarly, for the edge labeling between the end vertex vin −5 of the ( n −4

3 −1)th loop andstarting vertex vi

n −4 of the ( n −43 )rd loop, let E 1n − 4

3 −1 = f ∗(vin −5 vi

n −4) : 1 ≤ i ≤ mt. Calcula-




tion shows that

E 1n − 43 −1 = |f (vi

n −5) −f (vin −4)| : 1 ≤ i ≤ mt

= |f (v1n −5) −f (v1

n −4)|, |f (v2n −5) −f (v2

n −4)|, . . . ,

|f (vmt

−1

n −5 ) −f (vmt

−1

n −4 )|, |f (vmtn −5 ) −f (v

mtn −4)|= |F mnt −n +5 −F mnt −n +6 |, |F mnt −2n +5 −F mnt −2n +6 |, . . . ,

|F n +5 −F n +6 |, |F 5 −F 6|= F mnt −n +4 , F mnt −2n +4 , . . . , F n +4 , F 4.

Now for s = n −4

3 , let E n − 4

3=

mt

i =1 f ∗(vij vi

j +1 ) : n −4 ≤ j ≤ n −3. Then

E n − 43

= ∪mti=1 |f (vi

j ) −f (vij +1 )| : n −4 ≤ j ≤ n −3

= |f (v1n −4) −f (v1

n −3)|, |f (v1n −3 ) −f (v1

n −2)|∪|f (v2

n −4) −f (v2n −3 )|, |f (v2

n −3) −f (v2n −2)|∪

, . . . , |f (vmt

−1

n −4 ) −f (vmt

−1

n −3 )|, |f (vmt

−1

n −4 ) −f (vmt

−1

n −3 )|∪|f (vmt

n −4) −f (vmtn −3 )|, |f (vmt

n −3) −f (vmtn −2)|

= |F mnt −n +6 −F mnt −n +4 |, |F mnt −n +4 −F mnt −n +2 |∪|F mnt −2n +6 −F mnt −2n +4 |, |F mnt −2n +4 −F mnt −2n +2 |∪, . . . ,∪|F n +6 −F n +4 |, |F n +4 −F n +2 |∪|F 6 −F 4 |, |F 4 −F 2 |

= F mnt −n +5 , F mnt −n +3 ∪F mnt −2n +5 , F mnt −2n +3 ∪, . . . ,

∪F n +5 , F n +3 ∪F 5 , F 3.

We nd the edge labeling between the end vertex vin −2 of the ( n −4

3 )rd loop and the vertex

vin −1 . Let E ∗1 = f ∗(vin −2 vin −1) : 1 ≤ i ≤ mt. Then

E ∗1 = |f (vin −2) −f (vi

n −1)| : 1 ≤ i ≤ mt= |f (v1

n −2) −f (v1n −1)|, |f (v2

n −2 ) −f (v2n −1 )|, . . . , |f (vmt −1

n −2 ) −f (vmt −1n −1 )|,

|f (vmtn −2) −f (vmt

n −1)|= |F mnt −n +2 −F mnt −n +3 |, |F mnt −2n +2 −F mnt −2n +3 |, . . . ,

|F n +2 −F n +3 |, |F 2 −F 3 |= F mnt −n +1 , F mnt −2n +1 , . . . , F n +1 , F 1.

Let E ∗2 = f ∗(vin −1vi

n ) : 1 ≤ i ≤ mt. Then

E ∗2 = |f (vin −1) −f (v

in )| : 1 ≤ i ≤ mt= |f (v1

n −1) −f (v1n )|, |f (v2

n −1) −f (v2n )|, . . . ,

|f (vmt −1n −1 ) −f (vmt −1

n )|, |f (vmtn −1) −f (vmt

n )|= |F mnt −n +3 −F mnt −n +1 |, |F mnt −2n +3 −F mnt −2n +1 |, . . . ,

|F n +3 −F n +1 |, |F 3 −F 1|= F mnt −n +2 , F mnt −2n +2 , . . . , F n +2 , F 2.




Therefore,

E = ( E 1 ∪E 2∪, . . . ,∪E n − 43

)∪(E 14 ∪E 15∪, . . . ,∪E 1n − 43 −1)∪E ∗1 ∪E ∗2

= F 1 , F 2 , . . . , F mnt Thus, S tm,n admits a super bonacci graceful labeling.

For example the super bonacci graceful labeling of S 23,7 is shown in Fig.3.

F 42 F 40F 41 F 41F 39 F 39F 40 F 37F 38 F 38F 36 F 36F 37

F 35

F 35

F 33F 34 F 34F 32 F 32F 33 F 30F 31 F 31F 29 F 29F 30

F 28

F 28

F 26F 27 F 27F 25 F 25F 26 F 23F 24 F 24F 22 F 22F 23

F 21

F 21

F 19

F 20

F 20

F 18

F 18

F 19

F 16

F 17

F 17

F 15

F 15

F 16

F 14

F 14

F 12

F 13

F 13

F 11

F 11

F 12

F 9

F 10

F 10

F 8

F 8F 9

F 7

F 7

F 5F 6

F 6

F 4

F 4F 5

F 2F 3

F 3F 1

F 1F 2

F 0

F 42

Fig.3

Theorem 2.9 The complete graph K n is a super bonacci graceful graph if n ≤ 3.

Proof Let v0 , v1 , . . . , v n −1 be the vertex set of K n . Then vi (0 ≤ i ≤ n −1) is adjacentto all other vertices of K n . Let v0 and v1 be labeled as F 0 and F q respectively. Then v2 mustbe given F q−1 or F q−2 so that the edge v1 v2 will receive a bonacci number F q−2 or F q−1 .Therefore, the edges will receive the distinct labeling. Suppose not, Let v0 and v1 be labeled asF 1 and F q or F 0 and F q−2 respectively. Then v2 must be given F q−1 or F q−2 so that the edgesv0v2 and v1v2 will receive the same edge label F q−2 , which is a contradiction by our denition.Hence, K n is super bonacci graceful graph if n ≤ 3.




References

[1] G.J.Gallian, A Dynamic survey of graph labeling, The electronic Journal of Combinotorics ,16(2009), #DS6, PP219.

[2] Henry Ibstedt, Surng on the Ocean of Numbers a Few Smarandache Notions and Similar Topics , Ethus University Press, Vail 1997.

[3] Ian Anderson, A First Course in Combinatorial Mathematics , Claridon Press-Oxford,28(1989) 3-13.

[4] K.Kathiresan, Two classes of graceful graphs, Ars Combin. , 55(2000) 129-132.[5] K.M.Kathiresan and S.Amutha, Fibonacci Graceful Graphs , Ph.D.Thesis, Madurai Kama-

raj University, October 2006.[6] A.Rosa, On certain valuations of the vertices of a graph, Theory of Graphs (International

Symposium, Rome), July (1966).[7] R.Sridevi, S.Navaneethakrishnan and K.Nagarajan, Super Fibonacci graceful graphs, In-

ternational Journal of Mathematical Combinatorics , Vol.3(2010) 22-40.





74 Yanpei Liu

0

1

2

34

a

b

c

d

α

β γ

λδ

σ

0

1

2

3

4

2

3

43

4

4

ab

cd

α

β

γ δ

λ

σ

(a) T (b) T Fig.1

§2. Tree-Travels

Let C = C (V ; e) be the tree travel obtained from the boundary of T with 0 as the startingvertex. Apparently, the travel as a edge sequence C = C (e) provides a double covering of G = ( V, E ), denoted by

C (V ; e) = 0 P 0,i 1 vi 1 P i 1 ,i 2 vi 2 P i 2 ,i ′

1vi ′

1P i ′

1 ,i ′

2vi ′

2P i ′

2 ,2ǫ0 (1)

where ǫ = |E |.For a vertex-edge sequence Q as a tree-travel, denote by [ Q]eg the edge sequence induced

from Q missing vertices, then C eg = [C (V ; e)]eg is a polyhegon( i.e. , a polyhedron with only oneface).

Example 1 From T in Fig.1(b), obtain the thee-travel

C (V ; e) = 0 P 0,80P 8,14 0P 14 ,18 0P 18 ,20 0

where v0 = v8 = v14 = v18 = v20 = 0 and

P 0,8 = a1α2α−11β 3β −11γ 4γ −11a−1 ;

P 8,14 = b2δ 3δ −12λ4λ−12b−1;

P 14 ,18 = c3σ4σ−13c−1;

P 18 ,20 = d4d−1 .

For natural number i, if avi a−1 is a segment in C , then a is called a reective edge andthen vi , the reective vertex of a.



Surface Embeddability of Graphs via Tree-travels 75

Because of nothing important for articulate vertices(1-valent vertices) and 2-valent verticesin an embedding, we are allowed to restrict ourselves only discussing graphs with neither 1-valent nor 2-valent vertices without loss of generality. From vertices of all greater than or equalto 3, we are allowed only to consider all reective edges as on the cotree.

If vi 1 and vi 2 are both reective vertices in (1), their reective edges are adjacent in G andvi ′

1= vi 1 and vi ′

2= vi 2 , [P v i 1 ,i 2

]eg ∩[P v i ′

1 ,i ′

2]eg = ∅, but neither vi ′

1nor vi ′

2is a reective vertex,

then the transformation from C to

v i 1 ,v i 2C (V ; e) = 0 P 0,i 1 vi 1 P i ′

1 ,i ′

2vi 2 P i 2 ,i ′

1vi ′

1P i 1 ,i 2 vi ′

2P i ′

2 ,00. (2)

is called an operation of interchange segments for vi 1 , vi 2 .

Example 2 In C = C (V ; e)) of Example 1, v2 = 2 and v4 = 3 are two reective vertices, theirreective edges α and β , v9 = 2 and v1 5 = 3. For interchange segments once on C , we have

2,3C = 0 P 0,22P 9,15 3P 4,92P 2,43P 15 ,20 0 (= C 1).

where

P 0,2 = a1α (= P 1;0 ,2);

P 9,15 = δ 3δ −12λ4λ−12b−10c3 (= P 1;2 ,8 );

P 4,9 = β −11γ 4γ −11a−10b2 (= P 1;8 ,13 );

P 2,4 = α−11β (= P 1;13 ,15 );

P 15 ,20 = σ4σ−13c−10d4d−1 (= P 1;15 ,20 ).

Lemma 1 Polyhegon v i ,v j C eg is orientable if, and only if, C eg is orientable and the genusof

v i

1,v i

2C eg is exactly 1 greater than that of C eg .

Proof Because of the invariant of orientability for -operation on a polyhegon, the rststatement is true.

In order to prove the second statement, assume cotree edges α and β are reective edgesat vertices, respectively, vi 1 and vi 2 . Because of

C eg = Aαα −1 Bββ −1 CDE

where

Aα = [P 0,i 1 ]eg ; α−1 Bβ = [P i 1 ,i 2 ]eg ;

β −1

C = [P i 2 ,i′

1 ]eg ; D = [P i′

1 ,i′

2 ]eg ;E = [P i ′

2 ,i ǫ ]eg ,

we have

v i 1 ,v i 2C eg = AαDβ −1 Cα −1 BβE

∼top ABCDEαβα −1 β −1 , (Theorem 3.3.3 in [5])

= C eg αβα −1β −1 (Transform 1, in §3.1 of [5]).



76 Yanpei Liu

Therefore, the second statement is true.

If interchange segments can be done on C successively for k times, then C is called a k-tree travel . Since one reective edge is reduced for each interchange of segments on C and C has atmost m = ⌊β/ 2⌋ reective edges, we have 0 k m where β = β (G) is the Betti number(or

corank) of G. When k = m, C is also called normal .For a k-tree travel C k (V ; e, e−1) of G, graph Gk is dened as

Gk = T [E ref E T −k

j =1ej , e′j ] (3)

where T is a spanning tree, [ X ] represents the edge induced subgraph by edge subset X , ande ∈ E ref , e ∈ E T , ej , e′j are, respectively, reective edge, cotree edge, pair of reective edgesfor interchange segments.

Example 3 On C 1 in Example 2, v1;3 = 3 and v1;5 = 4 are two reective vertices, v1;8 = 3and v1;10 = 4. By doing interchange segments on C 1 , obtain

3,4C 1 = 0 P 1;0 ,10 3P 1;17 ,19 4P 1;12 ,15 3P 1;10 .12 4P 1;19 ,20 0 (= C 2 )

where

P 1;0 ,10 = a1α2b−10c3β −11γ 4γ −11a−10b2δ (= P 2;0 ,10 );

P 1;17 ,19 = c−10d(= P 2;10 ,12 );

P 1;12 .17 = α−1 2α−11β 3σ4σ−1(= P 2;12 ,17 );

P 1;10 ,12 = δ −12λ(= P 2;17 ,19 );

P 1;19 ,20 = d−1(= P 2;19 ,20 ).

Because of [P 2;6 ,16 ]eg ∩ [P 2;12 ,19 ]eg = ∅ for v2;12 = 4 and v2;19 = 4, only v2;6 = 4 andv2;16 = 4 with their reective edges γ and σ are allowed for doing interchange segments on C 2 .The protracted tree T in Fig.1(b) provides a 2-tree travel C , and then a 1-tree travel as well.

However, if interchange segments are done for pairs of cotree edges as β, γ , δ, λ and

α, σ in this order, it is known that C is also a 3-tree travel.On C of Example 1, the reective vertices of cotree edges β and γ are, respectively, v4 = 3

and v6 = 4, choose 4 ′ = 15 and 6 ′ = 19, we have

4,6C = 0 P 1;0 ,43P 1;4 ,84P 1;8 ,17 3P 1;17 ,19 4P 1;19 ,20 0(= C 1)

where

P 1;0 ,4 = P 0,4 ; P 1;4 ,8 = P 15 ,19 ; P 1;8 ,17 = P 6,15 ;

P 1;17 ,19 = P 4,6 ; P 1;19 ,20 = P 19 ,20 .

On C 1 , subindices of the reective vertices for reective edges δ and λ are 5 and 8, choose5′ = 17 and 8 ′ = 19, nd

5,8C 1 = 0 P 2;0 ,53P 2;5 ,74P 2;7 ,16 3P 2;16 ,19 4P 2;19 ,20 0(= C 2)




where

P 2;0 ,12 = P 1;0 ,12 ; P 2;12 ,14 = P 1;17 ,19 ; P 2;14 ,17 = P 1;14 ,17 ;

P 2;17 ,19 = P 1;12 ,14 ; P 2;19 ,20 = P 1;19 ,20 .

On C 2 , subindices of the reective vertices for reective edges α and σ are 2 and 5, choose2′ = 18 and 5 ′ = 19, nd

5,8C 2 = 0 P 3;0 ,23P 3;2 ,34P 3;3 ,16 3P 3;16 ,19 4P 3;19 ,20 0(= C 3)

where

P 3;0 ,2 = P 2;0 ,2 ; P 2;2 ,3 = P 2;18 ,19 ; P 3;3 ,16 = P 2;5 ,18 ;

P 3;16 ,19 = P 2;2 ,5 ; P 3;19 ,20 = P 2;19 ,20 .

Because of β (K 5 ) = 6, m = 3 = ⌊β/ 2⌋. Thus, the tree-travel C is normal.This example tells us the problem of determining the maximum orientable genus of a graph

can be transformed into that of determining a k-tree travel of a graph with k maximum as shownin [4].

Lemma 2 Among all k-tree travel of a graph G, the maximum of k is the maximum orientablegenus γ max (G) of G.

Proof In order to prove this lemma, the following two facts have to be known(both of them can be done via the nite recursion principle in §1.3 of [5]!).

Fact 1 In a connected graph G considered, there exists a spanning tree such that anypair of cotree edges whose fundamental circuits with vertex in common are adjacent in G.

Fact 2 For a spanning tree T with Fact 1, there exists an orientation such that on theprotracted tree T , no two articulate subvertices(articulate vertices of T ) with odd out-degreeof cotree have a path in the cotree.

Because of that if two cotree edges for a tree are with their fundamental circuits withoutvertex in common then they for any other tree are with their fundamental circuits withoutvertex in common as well, Fact 1 enables us to nd a spanning tree with number of pairs of adjacent cotree edges as much as possible and Fact 2 enables us to nd an orientation suchthat the number of times for dong interchange segments successively as much as possible. FromLemma 1, the lemma can be done.

§3. Tree-Travel Theorems

The purpose of what follows is for characterizing the embeddability of a graph on a surface of genus not necessary to be zero via k-tree travels.

Theorem 1 A graph G can be embedded into an orientable surface of genus k if, and only if,there exists a k-tree travel C k (V ; e) such that Gk is planar.



78 Yanpei Liu

Proof Necessity. Let µ(G) be an embedding of G on an orientable surface of genus k.From Lemma 2, µ(G) has a spanning tree T with its edge subsets E 0 , |E 0| = β (G) −2k, suchthat G = G − E 0 is with exactly one face. By successively doing the inverse of interchangesegments for k times, a k-tree travel is obtained on G. Let K be consisted of the k pairs of

cotree edge subsets. Thus, from Operation 2 in §3.3 of [5], Gk = G−K = G −K + E 0 is planar.Sufficiency. Because of G with a k-tree travel C k (V ; e), Let K be consisted of the k pairs of

cotree edge subsets in successively doing interchange segments for k times. Since Gk = G −K is planar, By successively doing the inverse of interchange segments for k times on C k (V ; e) inits planar embedding, an embedding of G on an orientable surface of genus k is obtained.

Example 4 In Example 1, for G = K 5, C is a 1-tree travel for the pair of cotree edges α andβ . And, G1 = K 5 − α, β is planar. Its planar embedding is

[4σ−13c−10d4]eg = ( σ−1c−1d);

[4d−10a1γ 4]eg = ( d−1 aγ );

[3σ4λ−12δ 3]eg = ( σλ −1 δ ); [0c3δ −12b−10]eg = ( cδ −1b−1);[2λ4γ −11a−10b2]eg = ( λγ −1a−1b).

By recovering α, β to G and then doing interchange segments once on C , obtain C 1 . FromC 1 on the basis of a planar embedding of G1 , an embedding of G on an orientable surface of genus 1(the torus) is produced as

[4σ−13c−10d4]eg = ( σ−1c−1d); [4d−10a1γ 4]eg = ( d−1aγ );

[3σ4λ−12δ 3β −11a−10b2α−11β 3]eg = ( σλ −1 δβ −1 a−1b2α−1β );

[0c3δ −12b−10]eg = ( cδ −1b−1); [2λ4γ −11α2]eg = ( λγ −1α).

Similarly, we further discuss on nonorientable case. Let G = ( V, E ), T a spanning tree,and

C (V ; e) = 0 P 0,i vi P i,j vj P j, 2ǫ0 (4)

is the travel obtained from 0 along the boundary of protracted tree T . If vi is a reective vertexand vj = vi , then

ξ C (V ; e) = 0 P 0,i vi P −1i,j vj P j, 2ǫ0 (5)

is called what is obtained by doing a reverse segment for the reective vertex vi on C (V ; e).If reverse segment can be done for successively k times on C , then C is called a k-tree

travel . Because of one reective edge reduced for each reverse segment and at most β reectiveedges on C , we have 0 k β where β = β (G) is the Betti number of G(or corank). Whenk = β , C (or G) is called twist normal .

Lemma 3 A connected graph is twist normal if, and only if, the graph is not a tree.

Proof Because of trees no cotree edge themselves, the reverse segment can not be done,this leads to the necessity. Conversely, because of a graph not a tree, the graph has to be witha circuit, a tree-travel has at least one reective edge. Because of no effect to other reective




edges after doing reverse segment once for a reective edge, reverse segment can always bedone for successively β = β (G) times, and hence this tree-travel is twist normal. Therefore,sufficiency holds.

Lemma 4 Let C be obtained by doing reverse segment at least once on a tree-travel of agraph. Then the polyhegon [ i C ]eg is nonorientable and its genus

g([ξ C ]eg ) =2g(C ) + 1 , when C orientable;

g(C ) + 1 , when C nonorientable . (6)

Proof Although a tree-travel is orientable with genus 0 itself, after the rst time of doingthe reverse segment on what are obtained the nonorientability is always kept unchanged. Thisleads to the rst conclusion. Assume C eg is orientable with genus g(C )(in fact, only g(C ) = 0will be used!). Because of

[

i C ]eg = AξB −1ξC

where [P 0,i ]eg = Aξ , [P i,j ]eg = ξ −1 B and [P j,ǫ ]eg = C , From (3.1.2) in [5]

[i C ]eg ∼top ABCξξ.

Noticing that from Operation 0 in §3.3 of [5], C rseg ∼top ABC , Lemma 3.1.1 in [5] leads to

g([ξ C ]eg ) = 2 g([C ]eg ) + 1 = 2 g(C ) + 1 .

Assume C eg is nonorientable with genus g(C ). Because of

C eg = Aξξ −1BC ∼top ABC,

g([ξ C ]eg ) = g(C ) + 1. Thus, this implies the second conclusion.

As a matter of fact, only reverse segment is enough on a tree-travel for determining thenonorientable maximum genus of a graph.

Lemma 5 Any connected graph, except only for trees, has its Betti number as the nonori-entable maximum genus.

Proof From Lemmas 3-4, the conclusion can soon be done.

For a k-tree travel C k (V ; e) on G, the graph Gk is dened as

Gk = T [E ref

−

k

j =1 ej

] (7)

where T is a spanning tree, [ X ] the induced graph of edge subset X , and e ∈ E ref and ej , e′j ,respectively, a reective edge and that used for reverse segment.

Theorem 2 A graph G can be embedded into a nonorientable surface of genus k if, and only if, G has a k-tree travel C k (V ; e) such that Gk is planar.



80 Yanpei Liu

Proof From Lemma 3, for k, 1 k β (G), any connected graph G but tree has a k-treetravel.

Necessity. Because of G embeddable on a nonorientable surface S k of genus k, let

µ(G)

be an embedding of G on S k . From Lemma 5,

µ(G) has a spanning tree T with cotree edge

set E 0 , |E 0| = β (G) −k, such that G = G −E 0 has exactly one face. By doing the inverse of reverse segment for k times, a k-tree travel of G is obtained. Let K be a set consisted of the kcotree edges. From Operation 2 in §3.3 of [5], Gk = G −K = G −K + E 0 is planar.

Sufficiency. Because of G with a k-tree travel C k (V ; e), let K be the set of k cotree edgesused for successively dong reverse segment. Since Gk = G −K is planar, by successively doingreverse segment for k times on C k (V ; e) in a planar embedding of Gk , an embedding of G on anonorientable surface S k of genus k is then extracted.

Example 5 On K 3,3 , take a spanning tree T , as shown in Fig.2(a) by bold lines. In (b), givena protracted tree T of T . From T , get a tree-travel

C = 0 P 0,11 2P 11 ,15 2P 15 ,00 (= C 0)

where v0 = v18 and

P 0,11 = c4δ 5δ −14γ 3γ −14c−10d2e3β 1β −13e−1;

P 11 ,15 = d−1 0a1b5α;

P 15 ,0 = α−15b−11a−1 .

Because of v15 = 2 as the reective vertex of cotree edge α and v11 = v15 ,

3C 0 = 0 P 1;0 ,11 2P 1;11 ,15 2P 1;15 ,00 (= C 1)

where

P 1;0 ,11 = P 0,11 = c4δ 5δ −14γ 3γ −14c−10d2e3β 1β −13e−1;

P 1;11 ,15 = P −111 ,15 = α−15b−11a−10d;

P 1;15 ,0 = P 15 ,0 = α−15b−11a−1.

Since G1 = K 3,3 −α is planar, from C 0 we have its planar embedding

f 1 = [5P 16 ,00P 0,20]eg = ( b−1a−1cδ );

f 2 = [3P 4,83]eg = ( γ −1c−1de);

f 3 = [1P 13 ,14 5P 2,43P 8,91]eg = ( δ −1γβb);

f 4 = [1P 9,13 1]eg = ( β −1 e−1d−1 a).

By doing reverse segment on C 0 , get C 1 . On this basis, an embedding of K 3,3 on the projectiveplane( i.e. , nonorientable surface S

1 of genus 1) is obtained as

f 1 = [5P 1;16 ,00P 1;0 ,20]eg = f 1 = ( b−1a−1cδ );

f 2 = [3P 1;4 ,83]eg = f 2 = ( γ −1c−1de);

f 3 = [1P 1;9 ,11 2P 1;11 ,13 1]eg = be−1e−1α−1b−1);

f 4 = [0P 1;14 ,15 2P 1;15 ,16 5P 1;2 ,43P 1;8 ,91P 1;13 ,14 0]eg

= ( dα−1δ −1γβa −1).




01

2

34

5 α

β

γ

δ

a

bc

d

e

0

1 2

345

25

31

a

b

c d

e

αδ

γ

β

(a) T (b) T Fig.2

§4. Research Notes

A. For the embeddability of a graph on the torus, double torus etc or in general orientablesurfaces of genus small, more efficient characterizations are still necessary to be further con-templated on the basis of Theorem 1.B. For the embeddability of a graph on the projective plane(1-crosscap), Klein bottle(2-crosscap), 3-crosscap etc or in general nonorientable surfaces of genus small, more efficient

characterizations are also necessary to be further contemplated on the basis of Theorem 2.C. Tree-travels can be extended to deal with all problems related to embedings of a graph onsurfaces as joint trees in a constructive way.

References

[1] Archdeacon, D., A Kuratowski theorem for the projective plane, J. Comb. Theory , 5(1981), 243–246.

[2] Liu, Yi. and Liu, Y.P., A characterization of the embeddability of graphs on the surfaceof given genus, Chin. Ann. Math. , 17B(1996), 457–462.

[3] Liu, Y.P., The nonorientable maximum genus of a graph(in Chinese with English abstract),Scintia Sinica , Special Issue on Math. I(1979), 191–201.

[4] Liu, Y.P., The maximum orientable genus of a graph, Scientia Sinica , Special Issue onMath. II(1979), 41–55.

[5] Liu, Y.P., Topological Theory on Graphs , USTC Press, 2008.




Edge Maximal C 3 and C 5-Edge Disjoint Free Graphs

M.S.A. Bataineh 1 and M.M.M. Jaradat 1,2

1 Department of Mathematics, Yarmouk University, Irbid-Jordan

2 Department of Mathematics, Statistics and Physics, Qatar University, Doha-Qatar

E-mail: [email protected]; [email protected]

Abstract : For a positive integer k, let G(n; E 2k +1 ) be the class of graphs on n vertices con-taining no two 2 k + 1-edge disjoint cycles. Let f (n ; E 2k +1 ) = max E (G) : G ∈ G(n; E 2k +1 ).In this paper we determine f (n ; E 2k +1 ) and characterize the edge maximal members in

G(n; E 2k +1 ) for k = 1 and 2.Key Words : Extremal graphs; Edge disjoint: Cycles, Smarandache-Tur´ an graph.

AMS(2010) :

§1. Introduction

For our purposes a graph G is nite, undirected and has no loops or multiple edges. We denotethe vertex set of G by V (G) and edge set of G by E (G). The cardinalities of these sets aredenoted by ν (G) and E (G), respectively. The cycle on n vertices is denoted by C n . Let G bea graph and u ∈ V (G). The degree of u in G, denoted by dG (u), is the number of edges of G

incident to u. The neighbor set of u in G is a subgraph H of G, denoted by N H (u), consists of the vertices of H adjacent to u; observe that dG (u) = |N H (u)|. For a proper subgraph H of Gwe write G[V (H )] and G-V (H ) simply as G[H ] and G −H respectively.

Let G1 and G2 be graphs. The union G1 ∪G2 of G1 and G2 is a graph with vertex setV (G1 ) ∪ V (G2 ) and edge set E (G1) ∪ E (G2). G1 and G2 are vertex disjoint if and only if V (G1 ) ∩V (G2) = ∅; G1 and G2 are edge disjoint if E (G1 ) ∩E (G2) = ∅. If G1 and G2 arevertex disjoint, we denote their union by G1 + G2 . The intersection G1 ∩G2 of graphs G1 andG2 is dened similarly, but in this case we need to assume that V (G1 ) ∩ V (G2 ) = ∅. The join G ∨ H of two disjoint graphs G and H is the graph obtained from G1 + G2 by joiningeach vertex of G to each vertex of H . For a vertex disjoint subgraphs H 1 and H 2 of G we letE (H 1 , H 2) =

xy

∈ E (G) : x

∈ V (H 1), y

∈ V (H 2)

and

E (H 1 , H 2) =

|E (H 1 , H 2)

|.

Let F 1 , F 2 be two graph families and n be a positive integer. Let G(n;F 1 , F 2) be aSmarandache-Tur´ an graph family consisting of graphs being F 1-free but containing a subgraphisomorphic to a graph in F 2 on n vertices. Dene

f (n;F 1 , F 2) = max |E (G)| : G ∈ G(n;F 1 , F 2).

1 Received August 25, 2010. Accepted February 25, 2011.



Edge Maximal C 3 and C 5 -Edge Disjoint Free Graphs 83

The problem of determining f (n ;F 1 , F 2) is called the Smarandache-Tur´ an-type extremal prob-lem. It is well known that in case F 2 = ∅ or F 2 = edge the problem is called the Tur´ an-typeextremal problem and abbreviated by f (n;F 1) and the class by G(n;F 1). In this paper weconsider the Tur´ an-type extremal problem with the odd edge disjoint cycles being the forbid-

den subgraph. Since a bipartite graph contains no odd cycles, we only consider non-bipartitegraphs. For convenience, in the case when F 1 consists of only one member C r , where r is anodd integer, we write

G(n; r ) = G(n;F 1), f (n; r ) = f (n;F 1).

An important problem in extremal graph theory is that of determining the values of the functionf (n;F 1). Further, characterize the extremal graphs G(n;F 1) where f (n ;F 1) is attained. For agiven r , the edge maximal graphs of G(n; r ) have been studied by a number of authors [1, 2, 3,6, 7, 8, 9, 11].

Let G(n; E 2k+1 ) denote the class of graphs on n vertices containing no two (2 k + 1)- edgedisjoint cycles. Let

f (n; E 2k+1 ) = max E (G) : G ∈ G(n; E 2k+1 ).

In this paper we determine f (n; E 2k+1 ) and characterize the edge maximal members in fork = 1 and 2. Now, we state a number of results, which we use to prove our main results.

Lemma 1.1 ( Bondy and Murty, [4]) Let G be a graph on n vertices. If E (G) > n 2 / 4, then Gcontains a cycle of length r for each 3 ≤ r ≤ ⌊(n + 3) / 2⌋.

Theorem 1.2 (Brandt, [5]) Let G be a non-bipartite graph with n vertices and more than (n −1)2/ 4 + 1 edges. Then G contains all cycles of length between 3 and the length of the

longest cycle.Let G∗(n) denote the class of graphs obtained by adding a triangle, two vertices of which

are new, to the complete bipartite graph K ⌊(n −2) / 2⌋,⌈(n −2) / 2⌉

. For an example of G∗(n), seeFigure 1.Theorem 1.3 ( Jia, [10]) Let G ∈ G(n; 5), n ≥ 10 . Then

E (G) ≤ (n −2)2/ 4 + 3 .

Furthermore, equality holds if and only if G ∈ G∗(n).

In this paper we determine f (n, E 2k+1 ) and characterize the edge maximal members in

G(n, E 2k+1 ) for k = 1 , 2, which is the rst step toward solving the problem for each positive

integer k.

§2. Edge-Maximal C 3-Disjoint Free Graphs

In this section we determine f (n, E 3) and characterize the edge maximal members in G(n, E 3).We begin with some constructions. Let Ω( G) denote to the class of graphs obtained by adding



84 M.S.A. Bataineh and M.M.M. Jaradat

Figure 1: The gure represent a member of G∗(n).

an edge to the complete bipartite graph K ⌊n/ 2⌋,⌈n/ 2⌉

. Figure 2 displays a member of Ω( G).Observe that Ω( G) ⊆ G(n, E 2k+1 ) and every graph in Ω( G) contains n 2/ 4 + 1 edges. Thus,we have established that

f (n; E 2k+1 ) ≥ n 2 / 4 + 1 (1)

We now establish that equality (1) holds for k = 1. Further we characterize the edge maximalmembers in G(n ; E 3 ).

Figure 2: The gure represents a member of Ω( G).

In the following theorem we determine edge maximum members in G(n; E 3).




Theorem 2.1 Let G ∈ G(n; E 3). For n ≥ 8,

f (n; E 3) ≤ n2 / 4 + 1 .

Furthermore, equality holds if and only if G ∈ Ω(G).

Proof Let G ∈ G(n, E 3 ). If G contains no cycle of length three, then by Lemma 1.1,

E (G) ≤ n2 / 4 . Thus, E (G) < n2 / 4 + 1 . So, we need to consider the case when G has cyclesof length 3. Let xyzx be a cycle of length 3 in G. Let H = G −xy,xz,yz . Observe that H cannot have cycles of length 3 as otherwise G would have two edge disjoint cycles of length 3.To this end we consider the following two cases.

Case1: H is not a bipartite graph. Since H contains no cycles of length 3, by Theorem 1.2,

E (G) ≤ (n −1)2/ 4 + 1 .

Now,

E (H ) = E (H ) + 3

≤(n −1)2

4+ 4 <

n2

4+ 1 ,

for n ≥ 8.

Case 2: H is a bipartite graph. Let X and Y be the bipartition of V (H ). Thus, E (H ) ≤ |X ||Y |.Observe |X | + |Y | = n. The maximum of the above is when |X | = ⌊n/ 2⌋ and |Y | = ⌈n/ 2⌉.Thus, E (G) ≤ n2 / 4 . Now we divide our work into two subcases.

Subcase 2.1. All the edges xy,xz, and yz are edges in on of X and Y , say in X . Observethat for any two vertices of Y , say u and v, we have that E (x,y,z , u, v) ≤ 5 as otherwiseG would have two edge disjoint cycles of length 3. Thus,

E (

x,y,z

, Y )

≤ 2

|Y

|+ 1. Now,

E (G) = E (X −x,y,z , Y ) + E (x,y,z , Y ) + E (x,y,z )

≤ (|X | −3)|Y |+ 2 |Y |+ 1 + 3

≤ |X ||Y | − |Y |+ 4 ≤ (|X | −1)|Y |+ 4 .

Observe |X | + |Y | = n . The maximum of the above equation is when |X | = ⌊(n + 1) / 2⌋ and

|Y | = ⌈(n −1)/ 2⌉, Thus,

E (G) ≤(n −1)2

4+ 4 <

n2

4+ 1.

Subcase 2.2. At most one of xy,xz and yz is an edge in X and Y . So,

E (G) = E (H ) + 1 ≤n2

4+ 1.

This completes the proof of the theorem.

We now characterize the extremal graphs. Through the proof, we notice that the only timewe have equality is in case when G obtained by adding an edge to the complete bipartite graphK ⌊n/ 2⌋,⌈n/ 2⌉

. This gives rise to the class Ω( G).



86 M.S.A. Bataineh and M.M.M. Jaradat

§3. Edge-Maximal C 5-Disjoint Free Graphs

In this section we determine f (n; E 2k+1 ) and characterize the edge maximal members in G(n; E 2k+1 )for k = 2. We now establish that equality (1) holds for k = 2. Further we characterize the

edge maximal members in G(n; E 5). To do that we employ the same method as in the abovetheorem

Theorem 3.1 Let G ∈ G(n, E 5). For n ≥ 9,

f (n; E 5) ≤ n2 / 4 + 1 .

Furthermore, equality holds if and only if G ∈ Ω(G).

Proof Let G ∈ G(n; E 5 ). If G does not have a cycle of length 5 , then by Lemma 1.1,

E (G) ≤ n2 / 4 . Thus, E (G) < n2 / 4 + 1. Hence, we consider the case when G has cyclesof length 5. Assume x1 x2 . . . x 5 x1 be a cycle of length 5 in G. As in the proof of the above

theorem, we consider H = G − e1 = x1x2 , e2 = x2x3 , . . . , e 5 = x5x1. Observe that H cannothave 5-cycles as otherwise G would have two 5 - edges disjoint cycles. Now, we consider twocases.

Case 1: H is not a bipartite graph. Then, by Theorem 1.3, we have

E (H ) ≤ (n −2)2 / 4 + 1 .

Now,

E (G) = E (H ) + 5

≤(n −2)2

4+ 8 <

n2

4 −n + 9 ,

for n ≥ 9, we have

E (G) <n2

4+ 1 .

Case 2: H is a bipartite graph. Let X and Y be the bipartition of V (H ). Thus, E (G) ≤|X ||Y |. Observe |X | + |Y | = n. The maximum of the above is when |X | = ⌊n/ 2⌋ and |Y | =⌈n/ 2⌉. Thus, E (G) < n 2

4 . Now, we consider the following subcases.

Subcase 2.1. One of X and Y contains two edges. Observe that those two edges must beconsecutive, say e1 and e2 . Let z be a vertex in X . If |N Y (x1 )∩N Y (x2) ∩N Y (x3 )∩N Y (z)| ≥ 4,then G contains two 5 edges disjoint cycles. Thus,

E (x1 , x2 , x3 , z, Y ) ≤ 3|Y |+ 3 .

So,

E (G) = E (X −x1 , x2 , x3 , z, Y ) + E (x1 , x2 , x3 , z, Y ) + E (X ) + E (Y )

≤ (|X | −4)|Y |+ 3 |Y |+ 3 + 2 + 3

≤ |X ||Y | − |Y |+ 8 ≤ (|X | −1)|Y |+ 8




Observe |X | + |Y | = n. The maximum of the above equation is when |Y | = n −12 and

|X | −1 = n −12 . Thus,

E (G) ≤(n −1)2

4+ 8

For n ≥ 9, we have

E (G) <n2

4+ 1 .

Subcase 2.2. E (X ) = 1 and E (Y ) = 0 or E (X ) = 0 and E (Y ) = 1. Then

E (G) ≤ E (H ) + 1 ≤n2

4+ 1

This completes the proof of the theorem.

We now characterize the extremal graphs. Through the proof, we notice that the only timewe have equality is in case when G obtained by adding an edge to the complete bipartite graph.This gives rise to the class Ω( G).

References

[1] Bataineh, M. (2007), Some Extremal problems in graph theory , Ph.D Thesis, Curtin Uni-versity of Technology, Australia.

[2] Bondy, J. (1971a), Large cycle in graphs, Discrete Mathematics 1, 121-132.[3] Bondy, J. (1971b), Pancyclic graphs, J. Combin. Theory Ser. B 11, 80-84.[4] Bondy J. & Murty U. (1976), Graph Theory with Applications , The MacMillan Press,

London.[5] Brandt, S. (1997), A sufficient condition for all short cycles, Discrete Applied Mathematics

79, 63-66.[6] Caccetta, L. (1976), A problem in extremal graph theory, Ars Combinatoria 2, 33-56.[7] Caccetta, L. & Jia, R. (1997), Edge maximal non-bipartite hamiltonian graphs without

cycles of length 5, Technical Report .14/97,School of Mathematics and Statistics, CurtinUniversity of Technology, Australia.

[8] Caccetta, L. & Jia, R. (2002), Edge maximal non-bipartite graphs without odd cycles of prescribed length, Graphs and Combinatorics , 18, 75-92.

[9] Furedi, Z. (1996), On the number of edges of quadrilateral-free graphs, J. Combin. Theory ,Series B 68, 1-6.

[10] Jia, R. (1998), Some extremal problems in graph theory , Ph.D Thesis, Curtin University of Technology, Australia.

[11] Tur an, P. (1941), On a problem in graph theory, Mat. Fiz. Lapok 48, 436-452.




A Note on Admissible Mannheim Curves in Galilean Space G 3

S.Ersoy, M.Akyigit AND M.Tosun

Department of Mathematics, Faculty of Arts and Sciences,

Sakarya University, Sakarya, 54187 Turkey

E-mail: [email protected], [email protected], [email protected]

Abstract : In this paper, the denition of the Mannheim partner curves in Galilean spaceG3 is given. The relationship between the curvatures and the torsions of the Mannheimpartner curves with respect to each other are obtained.

Key Words : Galilean space, Mannheim curve, admissible curve.

AMS(2010) : 53A35, 51M30

§1. Introduction

The theory of curves is a fundamental structure of differential geometry. An increasing interestof the theory curves makes a development of special curves to be examined. A way for clas-sication and characterization of curves is the relationship between the Frenet vectors of thecurves. The well known of such curve is Bertrand curve which is characterized as a kind of corresponding relation between the two curves. A Bertrand curve is dened as a spatial curve

which shares its principal normals with another spatial curve (called Bertrand mate). Anotherkind of associated curve has been called Mannheim curve and Mannheim partner curve. Aspace curve whose principal normal is the binormal of another curve is called Mannheim curve.The notion of Mannheim curve was discovered by A. Mannheim in 1878. The articles concern-ing Mannheim curves are rather few. In [1], R. Blum studied a remarkable class of Mannheimcurves. O. Tigano obtained general Mannheim curves in Euclidean 3-space in [2]. Recently,H.Liu and F. Wang studied the Mannheim partner curves in Euclidean 3-space and Minkowski3-space and obtained the necessary and sufficient conditions for the Mannheim partner curvesin [3]. On the other hand, a range of new types of geometries have been invented and devel-oped in the last two centuries. They can be introduced in a variety of manners. One possibleway is through projective manner, where one can express metric properties through projectiverelations. For this purpose a xed conic in innity is taken and all metric relations with respectto the absolute. This approach is due to A. Cayley and F. Klein who noticed that due tothe nature of the absolute, various geometries are possible. Among this geometries, there arealso Galilean and pseudo-Galilean geometries. They are very important in physics. Galileanspace-time plays the same role in non relativistic physics. The Geometry of the Galilean space

1 Received October 12, 2010. Accepted February 26, 2011.



A Note an Admissible Mannheim Curves in Galilean Space G 3 89

G3 has treated in detail in R¨ oschel’s habilitiation, [4]. Furthermore, Kamenarovic and Sipusstudied about Galilean space, [5]-[6]. The properties of the curves in the Galilean space arestudied [7]-[9]. The aim of this paper is to study the Mannheim partner curves in Galileanspace G3 and give some characterizations for this curves.

§2. Preliminaries

The Galilean space G3 is a Cayley-Klein space equipped with the projective metric of signature(0,0,+,+), as in [10]. The absolute gure of the Galilean Geometry consist of an ordered triple

w,f,I , where w is the ideal (absolute) plane, f is the line (absolute line) in w and I is thexed elliptic involution of points of f , [5]. In the non-homogeneous coordinates the similaritygroup H 8 has the form

x = a11 + a12 x

y = a21 + a22 x + a23 y cosϕ + a23 z sin ϕ

z = a31 + a32 x −a23 y sin ϕ + a23 z cosϕ

(2.1)

where a ij and ϕ are real numbers, [7].In what follows the coefficients a12 and a23 will play the special role. In particular, for

a12 = a23 = 1, (2.1) denes the group B6 ⊂ H 8 of isometries of Galilean space G3 .In G3 there are four classes of lines:

i) (proper) non-isotropic lines- they don’t meet the absolute line f .ii) (proper) isotropic lines- lines that don’t belong to the plane w but meet the absolute

line f .iii) un-proper non-isotropic lines-all lines of w but f .

iv) the absolute line f .Planes x =constant are Euclidean and so is the plane w. Other planes are isotropic, [6].The Galilean scalar product can be written as

< u 1 , u2 > =x1x2 , if x 1 = 0 ∨ x2 = 0

y1y2 + z1z2 , if x 1 = 0 ∧ x2 = 0

where u1 = ( x1 , y1 , z1) and u2 = ( x2 , y2 , z2). It leaves invariant the Galilean norm of the vectoru = ( x,y,z ) dened by

u =x , x = 0

y2 + z2 , x = 0 .

Let α be a curve given in the coordinate form

(α : I → G3 , I ⊂R

t → α(t) = ( x(t), y(t), z(t)) (2.2)

where x(t), y(t), z(t) ∈ C 3 and t is a real interval. If x′(t) = 0, then the curve α is calledadmissible curve.



90 S. Ersoy, M. Akyigit, M. Tosun

Let α be an admissible curve in G3 parameterized by arc length s and given by

α(s) = ( s, y (s), z(s))

where the curvature κ(s) and the torsion τ (s) are

κ(s) = y′′2 (s) + z′′2(s) and τ (s) = det[α ′ (s) , α ′′ (s) , α ′′′ (s)]

κ2(s) , (2.3)

respectively. The associated moving Frenet frame is

T (s) = α ′(s) = (1 , y′(s), z ′(s))

N (s) = 1κ (s ) α ′(s) = 1

κ (s ) (0, y′′(s), z ′′(s))

B (s) = 1κ (s ) (0, −z′′(s), y′′(s)) .

(2.4)

Here T, N and B are called the tangent vector, principal normal vector and binormal vectorelds of the curve α, respectively. Then for the curve α, the following Frenet equations are givenby

T ′(s) = κ(s)N (s)

N ′(s) = τ (s)B (s)

B ′(s) = −τ (s)N (s)

(2.5)

where T,N ,B are mutually orthogonal vectors, [6].

§3. Admissible Mannheim Curves in Galilean Space G3

In this section, we dened the admissible Mannheim curve and gave some theorems related tothese curves in G3 .

Denition 3.2 Let α and α∗ be an admissible curves with the Frenet frames along T,N,B and T ∗, N ∗, B∗, respectively. The curvature and torsion of α and α∗, respectively, κ(s), τ (s)and κ∗(s), τ ∗(s) never vanish for all s ∈ I in G3 . If the principal normal vector eld N of αcoincidence with the binormal vector eld B∗ of α∗ at the corresponding points of the admissible curves α and α∗. Then α is called an admissible Mannheim curve and α∗ is an admissible

Mannheim mate of α . Thus, for all s ∈ I α∗(s) = α(s) + λ(s)N (s). (3.1)

The mate of an admissible Mannheim curve is denoted by ( α, α∗), (see Figure 1).




Figure 1. The admissible Mannheim partner curves

Theorem 3.1 Let (α, α∗) be a mate of admissible Mannheim pair in G3 . Then function λ is constant.

Proof Let α be an admissible Mannheim curve in G3 and α∗ be an admissible Mannheimmate of α . Let the pair of α(s) and α∗(s) be corresponding points of α and α∗. Then the curveα∗(s) is given by (3.1). Differentiating (3.1) with respect to s and using Frenet equations,

T ∗ds∗ds

= T + λ ′N + λτB (3.2)

is obtained.Here and here after prime denotes the derivative with respect to s . Since N is coincident

with B∗ in the same direction, we have

λ ′(s) = 0 , (3.3)

that is, λ is constant. This theorem proves that the distance between the curve α and itsMannheim mate α∗ is constant at the corresponding points of them. It is notable that if α∗is an admissible Mannheim mate of α. Then α is also Mannheim mate of α∗ because therelationship obtained in theorem between a curve and its Mannheim mate is reciprocal one.

Theorem 3.2 Let α be an admissible curve with arc length parameter s. The curve α is an admissible Mannheim curve if and only if the torsion τ of α is constant.

Proof Let (α, α∗) be a mate of an admissible Mannheim curves, then there exists therelation

T (s) = cos θT ∗(s) + sin θN ∗(s)

B (s) = −sin θT ∗(s) + cos θN ∗(s)(3.4)

and

T ∗(s) = cos θT (s) −sin θB (s)

N ∗(s) = sin θT (s) + cos θB (s)(3.5)



92 S. Ersoy, M. Akyigit, M. Tosun

where θ is the angle between T and T ∗ at the corresponding points of α(s) and α∗(s), (seeFigure 1).

By differentiating (3.5) with respect to s, we get

τ ∗B∗ds ∗

ds = d(sin θ)

ds T + sin θκN + cos θτN + d(cos θ)

ds B. (3.6)

Since the principal normal vector eld N of the curve α and the binormal vector eld B∗ of itsMannheim mate curve, then it can be seen that θ is a constant angle.

If the equations (3.2) and (3.5) is considered, then

λτ cot θ = 1 (3.7)

is obtained. According to Theorem 3.1 and constant angle θ, u = λ cot θ is constant. Thenfrom equation (3.7), τ = 1

u is constant, too. Hence the proof is completed.

Theorem 3.3(Schell’s Theorem) Let (α, α∗) be a mate of an admissible Mannheim curves with torsions τ and τ ∗, respectively. The product of torsions τ and τ ∗ is constant at the corresponding points α(s) and α∗(s).

Proof Since α is an admissible Mannheim mate of α∗, then equation (3.1) also can begiven by

α = α∗ −λB∗. (3.8)

By taking differentiation of last equation and using equation (3.4),

τ ∗ = 1λ tan θ (3.9)

can be given. By the helps of (3.7), the equation below is easily obtained;

ττ ∗ = tan 2 θλ 2 = constant (3.10)

This completes the proof.

Theorem 3.4 Let (α, α∗) be an admissible Mannheim mate with curvatures κ, κ∗ and torsions τ, τ ∗ of α and α∗, respectively. Then their curvatures and torsions satisfy the following relations

i) κ∗ = − dθds ∗

ii ) κ = τ ∗ds ∗

ds sin θ

iii ) τ = −τ ∗ds ∗

ds cosθ.

Proof i) Let us consideration equation (3.4), then we have

< T,T ∗ > = cos θ. (3.11)

By differentiating last equation with respect to s∗ and using the Frenet equations of α andα∗, we reach

< κ (s)N (s) dsds ∗ , T ∗(s) > + < T (s), κ∗(s)N ∗(s) > = −sin θ dθ

ds ∗ . (3.12)




Since the principal normal N of α and binormal B∗ of α∗ are linearly dependent. Byconsidering equations (3.4) and (3.12), we reach

κ∗(s) = − dθds ∗ . (3.13)

If we take into consideration < T, B ∗ > , < B ,B ∗ > scalar products and (2.5), (3.4), (3.5)equations, then we can easily prove ii) and iii) items of the theorem, respectively.The relations given in ii) and iii) of the last theorem, we obtain

κτ

= −tan θ = constant.

References

[1] R. Blum, A remarkable class of Mannheim curves, Canad. Math. Bull. 9 (1966), 223–228.[2] O. Tigano, Sulla determinazione delle curve di Mannheim, Matematiche Catania 3 (1948),

25–29.[3] H. Liu and F. Wang, Mannheim partner curves in 3-space, Journal of Geometry , 88 (2008),

120–126.[4] O. Roschel, Die Geometrie des Galileischen Raumes , Habilitationsschrift, Leoben, (1984).[5] Z. M. Sipus, Ruled Weingarten surfaces in the Galilean space, Periodica Mathematica

Hungarica , 56 (2) (2008), 213–225.[6] I. Kamenarovic, Existence theorems for ruled surfaces in the Galilean space G3 , Rad HAZU

Math. 456 (10) (1991), 183–196.[7] B. J. Pavkovic and I. Kamenarovic, The equiform differential geometry of curves in the

Galilean space G3 , Glasnik Matematicki , 22 (42) (1987), 449–457.[8] A. O. Ogrenmis, M. Bektas and M. Erg¨ ut, The characterizations for helices in the Galilean

space G3 , Int. J. of Pure and Appl. Math. , 16 (1) (2004), 31–36.[9] A. O. Ogrenmis, M. Erg¨ut and M. Bektas, On the helices in the Galilean space G3 , Iranian

J. of Sci. Tec, Trans. A., 31 (2) (2007), 177–181.[10] E. Molnar, The projective interpretation of the eight 3-dimensional homogeneous geome-

tries, Beitrage zur Algebra und Geometrie Contributions to Algebra and Geometry 38 (2)(1997), 261–288.





The Number of Spanning Trees in GeneralizedComplete Multipartite Graphs 95

edges, respectively. The degree of a vertex v in a graph G is the number of edges incident withv and is denoted by d(v) = dG (v).

For simple graphs Gi = V i (G), E i (G) with vertex set V i = vi1 , vi2 , · · · , vin i , the emptygraphs of order n i are denoted by N n i = V i , φ , i = 1 , 2, · · · , k. A complete k-partite graph

K k,n = K n 1 ,n 2 ,···,n k = V 1 , V 2 , · · · , V k , E is said to be one spanned by the empty graph setN = N n 1 , N n 2 , · · · , N n k , denoted by K N

n 1 ,n 2 ,···,n k or K N k,n where n = n1 , n 2 , · · · , nk.

In generally, for the graph set G = Gn 1 , Gn 2 , · · · , Gn k , the graph

K G k,n = K N

k,n ∪Gn 1 ∪Gn 2 ∪ · · ·∪Gn k (1)

is said to be a generalized complete k-partite graph spanned by the graph set G .For all graph theoretic terminology not described here we refer to [1]. Let G be a connected

graph and E 1 , E 2 ⊂ E (E ) with E 1 = E 2 . The Smarandache (E 1 , E 2)-number tS (E 1 , E 2 ) of treesis the number of such spanning trees T in G with E (T )∩E 1 = ∅but E (T )∩E 2 = ∅. Particularly,if E 1 = E (G) and E 2 = ∅, i.e., such number is just the number of labeled spanning trees of agraph G, denoted by t(G). For a few special families of graphs there exist simple formulas thatmake it much easier to calculate and determine the number of corresponding spanning trees.One of the rst such results is due to Cayley [2] who showed in 1889 that complete graph on nvertices, K n , has nn −2 spanning trees. That is

t(K n ) = nn −2 for n 2. (2)

Another result is due to Sedlacek [3] who derived in 1970 a formula for the wheel on n + 1vertices, W n +1 , which is formed from a cycle C n on n vertices by adding a new vertex adjacentto every vertex of C n . That is

t(W n +1 ) = (3 + √ 5

2 )n + (

3 −√ 52

)n

−2 for n 3. (3)

Sedlacek [4] also later derived a formula for the number of spanning trees in a M¨ obiusladder, M n , is formed from a cycle C 2n on 2n vertices ladder v1 , v2 , · · · , v2n by adding edgesvi vn + i for every vertex vi , where i n. That is

t(M n ) = n2

[(2 + √ 3)n + (2 −√ 3)n + 2] for n 2. (4)

In 1985, Baron et al [5] derived the formula for the number of spanning trees in a squareof cycle, C 2n , which is expressed as follows.

t(C 2n ) = nF −n, n 5, (5)

where F −n is the n , th Fibonacci Number. Similar results can also be found in [6].The next result is due to Boesch and Bogdanowicz [7] who derived in 1987 a formula for

the prism on 2 n vertices, Rn , which is formed from two disjoint cycles C n with vertex setV (C n ) = v1 , v2 , · · · , vn and C ′n with vertex set V (C ′n ) = v′1 , v′2 , · · · , v′n by adding all edgesof the form vi v′i . That is

t(Rn ) = n2

[(2 + √ 3)n + (2 −√ 3)n −2] for n 3. (6)



96 Junliang Cai and Xiaoli Liu

In 2007, Bleller and Saccoman [8] derived a formula for a threshold graph on n vertices,T n , which is formed from that for all pairs of vertices u and v in T n , N (u) − v ⊂ N (v) − uwhenever d(u) d(v). That is

t(T n ) =k

i=1dn i

i

s−1

j = k+2(dj + 1) n j (dk+1 + l)n k +1 −1(ds + 1) n s −1 , (7)

where d(T n ) = ( d(n 1 )1 , d(n 2 )

2 , · · · , d(n s )s ) is the degree sequence of T n , di < d i +1 for i = 1 , 2, · · · , s−

1, k = ⌊s−12 ⌋ and s ≡ l(mod 2).

In 2008, Bogdanowicz [9] also derived a formula for an n-fan on n +1 vertices, F n +1 , whichis formed from a n-path P n by adding an additional vertex adjacent to every vertex of P n . Thatis

t(F n +1 ) = 2

5 −3√ 5[(3 −√ 5

2 )n +1 −(

3 + √ 52

)n −1] for n 2. (8)

In this paper it is proved that: Let K k,n be a complete k-partite graph of order n where

the vector n = n1 , n 2 , · · · , nk and n = n1 + n2 + · · ·+ nk for 1 k n, then the number of spanning trees in K k,n is

t(K k, n ) = nk−2k

i=1

(n −n i )n i −1 . (9)

Moreover, let K F k,n be a generalized complete k-partite graph of order n spanned by the

fan set F = F n 1 , F n 2 , · · · , F n k where n = n1 , n 2 , · · · , nk and n = n1 + n2 + · · ·+ nk for1 k n, then the number of spanning trees in K F

k,n is

t(K F k,n ) = n2k−2

k

i=1

α n i −1i −β n i −1

iα i

−β i

(10)

where α i = ( di + d2i −4)/ 2 and β i = ( di − d2

i −4)/ 2,di = n −n i + 3.In particular, from (9) we can obtain K 1,n = K cn with t(K cn ) = 0 and K n,n = K n with

t(K n ) = nn −2 which is just the Cayley’s formula (2). From (10) we can also obtain K F 1,n = F n

with t(K F 1,n ) = ( α n −1 −β n −1)/ √ 5 where α = 1

2 (3+ √ 5) and β = 12 (3−√ 5) which is the formula

(8), too.

§2. Some Lemmas

In order to calculate the number of spanning trees of G, we rst denote by A(G), or A =

(a ij )n ×n , the adjacency matrix of G, which has the rows and columns corresponding to thevertices, and entries aij = 1 if there is an edge between vertices vi and vj in V (G), aij = 0otherwise.

In addition, let D(G) represent the diagonal matrix of the degrees of the vertices of G. Wedenote by H (G), or H = ( h ij )n ×n , the Laplacian matrix (also known as the nodal admittancematrix ) D(G) − A(G) of G. From H (G) = D(G) − A(G), we can see that hii = d(vi ) andh ij = −a ij if i = j .




A well-known result states the relationship between the number of spanning trees of agraph and the eigenvalues of its nodal admittance matrix:

Lemma 2.1([10]) The value of t(G), the number of spanning trees of a graph G, is related to

the eigenvalues λ1 (G), λ 2 (G), · · · , λn (G) of its nodal admittance matrix H = H (G) as follows:

t(G) = 1n

n

i =2

λ i (G), 0 = λ1 (G) λ2 (G) · · · λn (G). (11)

A classic result of Kirchhoff can be used to determine the number of spanning trees for G.Next, we state the well-known theorem of Kirchhoff:

Lemma 2.2(Kirchhoff’s Matrix-Tree Theorem, [11]) All cofactors of H are equal and their common value is the number of spanning trees.

Lemma 2.3 Let b > 2 be a constant, then the determinant of order m

1 1 1 · · · 1

−1 b −1 0...

0 . . . . . . . . . 0

... 0 −1 b −1

0 · · · 0 −1 b−1m ×m

= αm −β m

α −β , (12)

where α = ( b + √ b2 −4)/ 2, β = ( b−√ b2 −4)/ 2.

Proof Let am stand for the determinant in (12) as above, by expending the determinant

according to the rst column we then have

am =

1 1 1 · · · 1

−1 b −1 0...

0 . . . . . . . . . 0

... 0 −1 b −1

0 · · · 0 −1 b−1m ×m

= cm −1 + am −1 , (13)

where

cm =

b −1 0 · · · 0

−1 b −1 0 ...

0 . . . . . . . . . 0

... 0 −1 b −1

0 · · · 0 −1 b−1m ×m

= bcm −1 −cm −2 (14)

in which c0 = 1, c1 = b−1 and c2 = b2 −b−1.




Constructing a function as follows:

f (x) =m 0

cm xm , (15)

thenf (x) = 1 + ( b−1)x + m 2

cm xm

= 1 + ( b−1)x + m 2(bcm −1 −cm −2)xm

= 1 + ( b−1)x + bx(f (x) −1) −x2 f (x),

i.e.

f (x) = 1−x

x2 −bx + 1 =

m 0

α m +1 + β m

1 + α xm (16)

where α = ( b + √ b2 −4)/ 2 and β = ( b −√ b2 −4)/ 2 which is just the two solutions of theequation x2

−bx + 1 = 0.

Thus, from (11.4) and (11.5) we have

cm = αm +1 + β m

1 + α . (17)

Since

a3 =

1 1 1

−1 b −1

0 −1 b−1

= b2 −1 = c2 + c1 + c0 , (18)

from (11.2),(11.6) and (11.7) we can obtain that

am =m −1

k=0

ck =m −1

k=0

α k+1 + β k1 + α

= αm −β mα −β

.

§3 Complete Multipartite Graphs

For a complete k-partite graph K k, n of order n where the vector n = n1 , n 2 , · · ·, nk andn = n1 + n2 + · · ·+ nk for 1 k n with n1 n 2 · · · n k . Let V (K k,n ) = V 1∪V 2∪ · · ·∪V kbe the k partitions of the graph K k, n such that |V i | = n i for i = 1 , 2, · · · , k. It is clear that forany vertex vi ∈ V i , d(vi ) = n −n i for i = 1 , 2, · · · , k. So the degree sequence of vertices in K k,n

d(K k, n ) = (( n −n1)n 1 , (n −n2)n 2 , · · · , (n −n k )n k ). (19)

Now, let E i be an identity matrix of order i, I i×j = (1) i×j a total module matrix of orderi × j and Oi×j = (0) i×j a total zero matrix of order i × j . Then the diagonal block matrix of the degrees of the vertices of K k,n corresponding to (12) is

D (K k,n ) = ( D ij )k×k , (20)




where D ii = ( n −n i )E n i and D ij = On i ×n j when i = j for 1 i, j k. The adjacency matrixof K k,n corresponding to (19) is

A(K k, n ) = ( Aij )k×k , (21)

where A ii = On i and A ij = I n i

×n j when i

= j for 1 i, j k.

Thus, we have from (13) and (14) the Laplacian block matrix (or the nodal admittancematrix ) of K k, n

H (K k,n ) = D(K k,n ) −A(K k,n ) = ( H ij )k×k , (22)

where H ii = ( n −n i )E n i and H ij = −I n i ×n j when i = j for 1 i, j k.

Theorem 3.1 Let K k,n be a complete k-partite graph of order n where the vector n =

n1 , n 2 , · · · , nk and n = n1 + n 2 + · · · + nk for 1 k n with n1 n2 · · · nk .Then the number of spanning trees of K k, n is

t(K k,n ) = nk−2k

i=1

(n

−n i )n i −1 . (23)

Proof According to Lemma 2.1, we need to determine all eigenvalues of nodal admittancematrix H (K k, n ) of K k, n . From (15) we can get easily the characteristic polynomial of H (K k, n )as follows:

|H (K k,n ) −λE | =

(n −n1 −λ)E n 1 −I n 1 ×n 2 · · · −I n 1 ×n k

−I n 2 ×n 1 (n −n2 −λ)E n 2 · · · −I n 2 ×n k

· · · · · · · · · · · ·−I n k ×n 1 −I n k ×n 2 · · · (n −nk −λ)E n k n ×n

.

(24)

Since the summations of entries in every column in |H (K k,n ) −λE | all are −λ, by addingthe entries of all rows other than the rst row to the rst row in |H (K k, n ) −λE |, all entries of the rst row then become −λ. Thus, the determinant becomes

|H (K k,n ) −λE | =

−λ

H +n 1 ∗ · · · ∗

−I n 2

×n 1 (n

−n2

−λ)E n 2

· · · −I n 2

×n

k

· · · · · · · · · · · ·−I n k ×n 1 −I n k ×n 2 · · · (n −nk −λ)E n k n ×n

,

(25)

where

H +n 1=

1 I 1×(n 1 −1)

O(n 1 −1)×1 (n −n1 −λ)E (n 1 −1)n 1 ×n 1

(26)




and the stars ”*” in (18) stand for the matrices with all entries in the rst row being 1. Byadding the entries in the rst row to the rows from row n1 + 1 to row n in the determinant(18), it then becomes

|H (K k,n ) −λE | =

−λ

H +n 1 ∗ · · · ∗On 2 ×n 1 (n −n 2 −λ)E n 2 + I n 2 · · · On 2 ×n k

· · · · · · · · · · · ·On k ×n 1 On k ×n 2 · · · (n −nk −λ)E n k + I n k n ×n

,

i.e.

|H (K k,n ) −λE | = −λ|H +n 1 |1<i k

|(n −n i −λ)E n i + I n i |. (27)

From (19) we have

|H +n 1 | =1 I 1×n 1

On 1 ×1 (n −n1 −λ)E n 1

= ( n −n1 −λ)n 1 −1 . (28)

For 1 < i k we have

|(n −n i −λ)E n i + I n i | =n −λ (n −λ)I 1×(n i −1)

I (n i −1)×1 (n −n i −λ)E n i −1 + I n i −1

i.e.

|(n −n i −λ)E n i + I n i | = ( n −λ) 1 I 1×(n i −1)

O(n i −1) ×1 (n −n i −λ)E n i −1

= ( n −λ)(n −n i −λ)n i −1 .

(29)

By substituting (21),(22) to (20) we have

|H (K k,n ) −λE | = −λ(n −λ)k−1k

i=1

(n −n i −λ)n i −1 . (30)

So, from (23) we derive all eigenvalues of H (K k,n ) as follows: λ1 = 0 and

(n i −1)-multiple roots λi +1 (K k, n ) = n −n i , i = 1 , 2, · · · , k;

(k −1)-multiple root λn (K k, n ) = n.(31)

By substituting (24) to (11) we have

t(K k, n ) = 1n

n

i =2

λ i (K k, n ) = nk−2k

i =1

(n −n i )n i −1 . (32)




This is just the theorem.

Corollary 3.2 (Cayley’s formula) The total number of spanning trees of complete graph K nis

t(K n ) = nn

−2

. (33)

Proof Let k = n, then n1 = n2 = · · · = nn = 1 and K n,n = K n . By substituting it to (25)we then have

t(K n ) = t(K n, n ) = nn −2n

i =1

(n −1)0 = nn −2 .

§3. Generalized Complete Multipartite Graphs

For a generalized complete k-partite graph K F k,n of order n spanned by the fan set F =

F n 1 , F n 2 , · · · , F n k where n = n1 + n 2 + · · ·+ n k with n 1 n2 · · · nk for 1 k n. It isclear that the degree sequence of vertices of the fan F n i in K F

k,n

d(F n i ) = ( n −n i + 2 , (n −n i + 3) n i −3 , n −n i + 2 , n −1) (34)

for i = 1 , 2, · · · , k. So the degree sequence of vertices of K F k,n is

d(K F k,n ) = ( d(F n 1 ), d(F n 2 ), · · · , d(F n k )) . (35)

Now, the diagonal block matrix of the degrees of the vertices of K F k,n corresponding to (28)

isD (K F

k,n ) = ( D ij )k

×k , (36)

where from (27) we have

D ii = diag n −n i + 2 , n −n i + 3 , · · · , n −n i + 3 , n −n i + 2 , n −1 (37)

and D ij = On i ×n j when i = j for 1 i, j k.The adjacency matrix of K F

k,n corresponding to (28) is

A(K F k,n ) = ( Aij )k×k , (38)

where

Aii =

0 1 1

1 0 . . . 1. . . . . . . . .

.... . . 0 1

1 1 · · · 1 0n i ×n i

(39)

and A ij = I n i ×n j when i = j for 1 i, j k.




Thus, we have from (29) ∼ (32) the Laplacian block matrix (or the nodal admittancematrix ) of K F

k,n

H (K F k,n ) = D(K F

k,n ) −A(K F k,n ) = ( H ij )k×k , (40)

where

H ii = D ii −Aii =

di −1 −1 −1

−1 di −1 −1. . . . . . . . .

...

−1 di −1

−1 di −1 −1

−1 −1 · · · −1 n −1n i ×n i

(41)

with di = n −n i + 3 and H ij = −I n i ×n j when i = j for 1 i, j k.

Theorem 4.1 Let K F k,n be a generalized complete k-partite graph of order n spanned by the

fan set F = F n 1 , F n 2 , · · · , F n k where n = n1 , n 2 , · · · , nk and n = n 1 + n2 + · · ·+ nk for 1 k n , then the number of spanning trees in K F

k,n is

t(K F

k,n ) = n2k

−2

k

i=1

α n i −1i

−β n i −1

i

α i −β i (42)

where α i = ( di + d2i −4)/ 2 and β i = ( di − d2

i −4)/ 2,di = n −n i + 3 .

Proof According to the Kirchhoff’s Matrix-Tree theorem, all cofactors of H (K F k,n ) are

equal to the number of spanning trees t(K F k,n ). Let H ∗(K F

k,n ) be the cofactor of H (K F k,n )

corresponding to the entry hnn of H (K F k,n ), then

t(K F k,n ) = |H ∗(K F

k,n )| =

H 11 · · · −I n 1 ×n i · ·· −I n 1 ×(n k −1)

· · · · · · · · · · · · · · ·−I n i ×n 1 · · · H ii · ·· −I n i ×(n k −1)

· · · · · · · · · · · · · · ·−I (n k −1)×n 1 · · · −I (n k −1)×n i · · · H ∗kk

(43)




where

H ∗kk =

dk −1 −1

−1 dk −1. . . . . . . . .

−1 dk −1

−1 dk −1(n k −1)×(n k −1)

. (44)

Since the summations of entries in every column in |H ∗(K F k,n )| all are 1 by adding the

entries of all rows other than the rst row to the rst row in |H ∗(K F k,n )| the entries of the rst

row become 1. By adding the entries in the rst row to the rows from row n1 + 1 to row n −1in the last determinant, it becomes from (36) and (37)

|H ∗(K F k,n )| =

H ∗n 1 · · · ∗ · · · ∗

· · · · · · · · · · · · ·· ·On i ×n 1 · · · H ii + I n i · · · ∗· · · · · · · · · · · · ·· ·

O(n k −1)×n 1 · · · O(n k −1)×n i · · · H ∗n k −1

,

i.e.

t(K F k,n ) = |H ∗(K F

k,n )| = |H ∗n 1 | ·1<i<k

|H ii + I n i | · |H ∗n k −1|, (45)

where

|H ∗n 1 | =

1 1 1 · · · 1 1

−1 d1 −1 −1. . . . . . . . . ...

−1 d1 −1 −1

−1 d1 −1 −1

−1 −1 · · · −1 −1 n −1n 1 ×n 1

,

by adding the entries in the rst row to the correspondence entries of the other row in |H ∗n 1 |we then have

|H ∗n 1 | =

1 1 1 · · · 1 1

0 d1 + 1 0 1 ... 0

1 . . . . . . . . . 1

...... 1 0 d1 + 1 0 0

1 · · · 1 0 d1 0

0 0 · · · 0 0 nn 1 ×n 1

,




i.e.

|H ∗n 1

| = n

1 1 1 · · · 1

0 d1 + 1 0 1...

1

. . . . . . . . . 1... 1 0 d1 + 1 0

1 · · · 1 0 d1 (n 1 −1)×(n 1 −1)

; (46)

It is clear that

|H ii + I n i | =

di 0 1 · · · 1 0

0 di + 1 0 1...

1 . . . . . . . . . 1

...... 1 0 di + 1 0

1 · · · 1 0 di 00 · · · 0 n

n i ×n i

,

i.e.

|H ii + I n i | = n

di 0 1 · · · 1

0 di + 1 0 1...

1 . . . . . . . . . 1

... 1 0 di + 1 0

1 · · · 1 0 di (n i −1)×(n i −1)

.

Since the summations of all entries in every column in the last determinant, we have

|H ii + I n i | = n2

1 1 1 · · · 1

0 di + 1 0 1...

1 . . . . . . . . . 1

... 1 0 di + 1 0

1 · · · 1 0 di (n i −1)×(n i −1)

; (47)

and

|H ∗n k −1 | =

dk 0 1 · · · 1

0 dk + 1 0 1 ...

1 . . . . . . . . . 1

... 1 0 dk + 1 0

1 · · · 1 0 dk (n k −1)×(n k −1)

,

Similarly, we have




|H ∗n

k −1

| = n

1 1 1 · · · 1

0 dk + 1 0 1...

1

. . . . . . . . . 1... 1 0 dk + 1 0

1 · · · 1 0 dk(n k −1) ×(n k −1)

. (48)

By substituting (39),(40) and (41) to (38), it becomes

t(K F k,n ) = n2k−2

1 i k

1 1 1 · · · 1

0 di + 1 0 1...

1 . . . . . . . . . 1

.

.. 1 0 di + 1 01 · · · 1 0 di

(n i −1)×(n i −1)

,

or

t(K F k,n ) = n2k−2

1 i k

1 1 1 · · · 1

−1 di −1 0...

0 . . . . . . . . . 0

... 0 −1 di −1

0 · · · 0 −1 di −1 (n i −1)×(n i −1)

. (49)

Thus, from (11.1) and (42) we have

t(K F k,n ) = n2k−2

1 i k

α n i −1i −β n i −1

iα i −β i

(50)

where α i = ( di + d2i −4)/ 2, β i = ( di − d2

i −4)/ 2 and di = n −n i + 3.This is just the theorem.

Corollary 4.2 The total number of spanning trees of a fan graph F n is

t(F n ) = 1√ 5(α n −1 −β n −1) (51)

where α = (3 + √ 5)/ 2 and β = (3 −√ 5)/ 2.

Proof Let k = 1, then n 1 = n and K F 1,n = F n . Substituting this fact into (35), this result

is followed.




References

[1] J.A.Bondy and U.S.R.Murty, Graph Theory with Applications . The Macmilan Press LTD,1976.[2] G.A.Cayley, A Theorem on Trees. Quart. J. Math. ,23 ,(1889),276-387.[3] J.Sedlacek, Lucas Numbers in Graphs Theory. In: Mqthematics(Geometry and Graph

Theory)(Chech.). Univ.Karlova, Prague , (1970),111-115.[4] J.Sedlacek, On the Skeleton of a graph or Digraph. In: Combinatorial Structures and

Their Applications(R.Guy,M.Hanani,N.Saver and J.Schonheim,eds.),Gordon and Breach,New york(1970),387-391.

[5] G.Baron,F.Boesch,H.Prodinger,R.Tichy and J,Wang, The Number of spanning trees in theSquare of Cycle. The Fibonacci Quart. , 23 ,(1985),258-264.

[6] G.J.F.Wang and C.S.Yang, On the Number of Spanning Trees of Circulant Graphs. Inter.J. Comput.Math. ,16 ,(1984),229-241.

[7] G.F.T.Boesch and Z.R.Bogdanowicz, The Number of Spanning Trees in a Prism. Inter. J.Comput.Math. ,21 ,(1987),229-243.

[8] S.K.Blelleer and J.T.Saccoman, A Correction in the Formula for the Number of SpanningTrees in threshold graphs. Australasian of Combinatorics ,37 ,(2007),205-213.

[9] Z.R.Bogdanowicz, Formulas for the Number of Spanning Trees in a Fan. Applied Mathe-matical Sciences ,2,(2008),781-786.

[10] V.M.Chelnokov and A.K.Kelmans, A Certain Polynomial of a Graph and Graphs with anExtremal Number of Trees. J. Combin. Theory, Ser. B , 16 ,(1974),197-214.

[11] G.G.Kirhhoff, Uber die Au osung der Gleichungen, auf welche man bei der Untersuchungder linearen Verteilung galvanischer Strme gefhrt wird. Ann Phys. Chem. , 72,(1847),497-508.





108 P.Siva Kota Reddy, B.Prashanth and Kavita. S.Permi

set has been partitioned into positive and negative edges. He called a cycle positive if it had aneven number of negative edges, and he called a signed graph balanced if every cycle is positive.Then he gave both necessary and sufficient conditions for balance.

Since then, mathematicians have written numerous papers on the topic of signed graphs.

Many of these papers demonstrate the connection between signed graphs and different subjects:circuit design (Barahona [3], coding theory (Sol´ e and Zaslavsky [20]), physics (Toulouse [22])and social psychology (Abelson and Rosenberg [1]). While these subjects seem unrelated,balance plays an important role in each of these elds.

Four years after Harary’s paper, Abelson and Rosenberg [1], wrote a paper in which theydiscuss algebraic methods to detect balance in a signed graphs. It was one of the rst papersto propose a measure of imbalance, the “complexity” (which Harary called the “line index of balance”). Abelson and Rosenberg introduced an operation that changes a signed graph whilepreserving balance and they proved that this does not change the line index of imbalance. Formore new notions on signed graphs refer the papers ([7]-[11], [13]-[19]).

A marking of S is a function µ : V (G)

→ + ,

−; A signed graph S together with a

marking µ is denoted by S µ . Given a signed graph S one can easily dene a marking µ of S asfollows: For any vertex v ∈ V (S ),

µ(v) =uv∈E (S )

σ(uv),

the marking µ of S is called canonical marking of S . In a signed graph S = ( G, σ ), for anyA ⊆ E (G) the sign σ(A) is the product of the signs on the edges of A.

The following characterization of balanced signed graphs is well known.

Proposition 1(E. Sampathkumar, [9]) A signed graph S = ( G, σ ) is balanced if, and only if,

there exists a marking µ of its vertices such that each edge uv in S satises σ(uv) = µ(u)µ(v).Let S = ( G, σ ) be a signed graph. Consider the marking µ on vertices of S dened

as follows: each vertex v ∈ V , µ(v) is the product of the signs on the edges incident at v.Complement of S is a signed graph S = ( G, σ ′), where for any edge e = uv ∈ G, σ′(uv) =µ(u)µ(v). Clearly, S as dened here is a balanced signed graph due to Proposition 1.

The idea of switching a signed graph was introduced in [1]in connection with structuralanalysis of social behavior and also its deeper mathematical aspects, signicance and connec-tions may be found in [24].

Switching S with respect to a marking µ is the operation of changing the sign of every edgeof S to its opposite whenever its end vertices are of opposite signs (See also R. Rangarajan and

P. S. K. Reddy [8]). The signed graph obtained in this way is denoted by S µ (S ) and is calledµ-switched signed graph or just switched signed graph . Two signed graphs S 1 = ( G, σ ) andS 2 = ( G′, σ ′) are said to be isomorphic , written as S 1 ∼= S 2 if there exists a graph isomorphismf : G → G′ (that is a bijection f : V (G) → V (G′) such that if uv is an edge in G thenf (u)f (v) is an edge in G′) such that for any edge e ∈ G, σ(e) = σ′(f (e)). Further a signedgraph S 1 = ( G, σ ) switches to a signed graph S 2 = ( G′, σ ′) (or that S 1 and S 2 are switching equivalent ) written S 1 ∼ S 2 , whenever there exists a marking µ of S 1 such that S µ (S 1) ∼= S 2 .



A Note on Antipodal Signed Graphs 109

Note that S 1 ∼ S 2 implies that G ∼= G′, since the denition of switching does not involve changeof adjacencies in the underlying graphs of the respective signed graphs.

Two signed graphs S 1 = ( G, σ ) and S 2 = ( G′, σ ′) are said to be weakly isomorphic (see[21]) or cycle isomorphic (see [23]) if there exists an isomorphism φ : G → G′ such that the

sign of every cycle Z in S 1 equals to the sign of φ(Z ) in S 2 . The following result is well known(See [23]).

Proposition 2(T. Zaslavsky, [23]) Two signed graphs S 1 and S 2 with the same underlying graph are switching equivalent if, and only if, they are cycle isomorphic.

§2. Antipodal Signed Graphs

Singleton [11] has introduced the concept of antipodal graph of a graph G as the graph A(G)having the same vertex set as that of G and two vertices are adjacent if they are at a distanceof diam (G) in G.

Motivated by the existing denition of complement of a signed graph, we extend the notionof antipodal graphs to signed graphs as follows: The antipodal signed graph A(S ) of a signedgraph S = ( G, σ ) is a signed graph whose underlying graph is A(G) and sign of any edge uv isA(S ) is µ(u)µ(v), where µ is the canonical marking of S . Further, a signed graph S = ( G, σ ) iscalled antipodal signed graph, if S ∼= A(S ′) for some signed graph S ′. The following resultindicates the limitations of the notion A(S ) as introduced above, since the entire class of unbalanced signed graphs is forbidden to be antipodal signed graphs.

Proposition 3 For any signed graph S = ( G, σ ), its antipodal signed graph A(S ) is balanced.

Proof Since sign of any edge uv in A(S ) is µ(u)µ(v), where µ is the canonical marking of

S , by Proposition 1, A(S ) is balanced.

For any positive integer k, the k th iterated antipodal signed graph A(S ) of S is dened asfollows:

A0(S ) = S , Ak (S ) = A(Ak−1 (S ))

Corollary 4 For any signed graph S = ( G, σ ) and any positive integer k, Ak (S ) is balanced.

In [2], the authors characterized those graphs that are isomorphic to their antipodal graphs.

Proposition 5(Aravamudhan and Rajendran, [2]) For a graph G = ( V, E ), G ∼= A(G) if, and only if, G is complete.

We now characterize the signed graphs that are switching equivalent to their antipodalsigned graphs.

Proposition 6 For any signed graph S = ( G, σ ), S ∼ A(S ) if, and only if, G = K p and S is balanced signed graph.



110 P.Siva Kota Reddy, B.Prashanth and Kavita. S.Permi

Proof Suppose S ∼ A(S ). This implies, G ∼= A(G) and hence G is K p. Now, if S is anysigned graph with underlying graph as K p, Proposition 3 implies that A(S ) is balanced andhence if S is unbalanced and its A(S ) being balanced can not be switching equivalent to S inaccordance with Proposition 2. Therefore, S must be balanced.

Conversely, suppose that S is a balanced signed graph and G is K p. Then, since A(S ) isbalanced as per Proposition 3 and since G ∼= A(G), this follows from Proposition 2 again.

Proposition 7 For any two signed graphs S and S ′ with the same underlying graph, their antipodal signed graphs are switching equivalent.

Proposition 8(Aravamudhan and Rajendran, [2]) For a graph G = ( V, E ), G ∼= A(G) if, and only if, i). G is diameter 2 or ii). G is disconnected and the components of G are complete graphs.

In view of the above, we have the following result for signed graphs:

Proposition 9 For any signed graph S = ( G, σ ), S ∼ A(S ) if, and only if, G satises conditions of Proposition 8.

Proof Suppose that A(S ) ∼ S . Then clearly we have A(G) ∼= G and hence G satisesconditions of Proposition 8.

Conversely, suppose that G satises conditions of Proposition 8. Then G ∼= A(G) by Propo-sition 8. Now, if S is a signed graph with underlying graph satises conditions of Proposition8, by denition of complementary signed graph and Proposition 3, S and A(S ) are balancedand hence, the result follows from Proposition 2.

The notion of negation η(S ) of a given signed graph S dened in [6] as follows: η(S ) has the same underlying graph as that of S with the sign of each edge opposite to that given to it in S . However, this denition does not say anything about what to do with nonadjacent pairs of vertices in S while applying the unary operator η(.) of taking the negation of S .

Propositions 6 and 9 provides easy solutions to two other signed graph switching equiva-lence relations, which are given in the following results.

Corollary 10 For any signed graph S = ( G, σ ), S ∼ A(η(S )) .

Corollary 11 For any signed graph S = ( G, σ ), S ∼ A(η(S )) .

Problem 12 Characterize signed graphs for which i) η(S ) ∼ A(S ) or ii) η(S ) ∼ A(S ).

For a signed graph S = ( G, σ ), the A(S ) is balanced (Proposition 3). We now examine,the conditions under which negation η(S ) of A(S ) is balanced.

Proposition 13 Let S = ( G, σ ) be a signed graph. If A(G) is bipartite then η(A(S )) is balanced.

Proof Since, by Proposition 3, A(S ) is balanced, if each cycle C in A(S ) contains even



A Note on Antipodal Signed Graphs 111

number of negative edges. Also, since A(G) is bipartite, all cycles have even length; thus, thenumber of positive edges on any cycle C in A(S ) is also even. Hence η(A(S )) is balanced.

§3. Characterization of Antipodal Signed Graphs

The following result characterize signed graphs which are antipodal signed graphs.

Proposition 14 A signed graph S = ( G, σ ) is an antipodal signed graph if, and only if, S is balanced signed graph and its underlying graph G is an antipodal graph.

Proof Suppose that S is balanced and G is a A(G). Then there exists a graph H suchthat A(H ) ∼= G. Since S is balanced, by Proposition 1, there exists a marking µ of G suchthat each edge uv in S satises σ(uv) = µ(u)µ(v). Now consider the signed graph S ′ = ( H, σ ′),where for any edge e in H , σ′(e) is the marking of the corresponding vertex in G. Then clearly,A(S ′) ∼= S . Hence S is an antipodal signed graph.

Conversely, suppose that S = ( G, σ ) is an antipodal signed graph. Then there exists asigned graph S ′ = ( H, σ ′) such that A(S ′) ∼= S . Hence G is the A(G) of H and by Proposition3, S is balanced.

Acknowledgement

The authors would like to thank Sri. B. Premnath Reddy, Chairman, Acharya Institutes, forhis constant support and encouragement for research and development.

References

[1] R.P.Abelson and M. J.Rosenberg, Symbolic psychologic:A model of attitudinal cognition,Behav. Sci. , 3 (1958), 1-13.

[2] R.Arvamudhan and B.Rajendran, On antipodal graphs, Discrete Math. , 58(1984), 193-195.[3] F.Barahona, M.Grotschel, M.Junger, and G.Reinelt, An application of combinatorial opti-

mization to statistical physics and circuit layout design, Operations Research , 36 (3) (1988),493-513.

[4] F.Harary, Graph Theory , Addison-Wesley Publishing Co., 1969.[5] F.Harary, On the notion of balance of a signed graph, Michigan Math. J. , 2(1953), 143-146.[6] F.Harary, Structural duality, Behav. Sci. , 2(4) (1957), 255-265.[7] R.Rangarajan, M. S.Subramanya and P.S.K.Reddy, The H -Line Signed Graph of a Signed

Graph, International J. Math. Combin. , 2 (2010), 37-43.[8] R.Rangarajan and P. S. K.Reddy, The edge C 4 signed graph of a signed graph, Southeast Asain Bull. Math. , 34 (6) (2010), 1077-1082.

[9] E.Sampathkumar, Point signed and line signed graphs, Nat. Acad. Sci. Letters , 7(3)(1984), 91-93.

[10] E.Sampathkumar, P. S. K.Reddy and M. S.Subramanya, Directionally n-signed graphs,Ramanujan Math. Soc., Lecture Notes Series (Proc. Int. Conf. ICDM 2008) , 13 (2010),






Group Connectivity

of 1-Edge Deletable IM-Extendable Graphs

Keke Wang, Rongxia Hao

(Department of Mathematics, Beijing Jiaotong University, Beijing 100044, P.R. China)

Jianbing Liu(Department of Mathematics, Northwest Normal University, Lanzhou 730070, P.R. China)


Abstract : A graph G is called a k-edges deletable IM-extendable graph, if G

−F is IM-

extendable for every F ⊆ E (G) with |F | = k. Denoted by ∧g (G) the group connectivity of a graph G. In this paper, ∧g (G) = 3 is gotten if G is a 4-regular claw-free 1-edge deletableIM-extendable graph.

Key Words : Graph, multi-group connectivity, group connectivity, induced matching.

AMS(2010) : 05C75, 05C45

§1. Introduction and Lemmas

In 1950s, Tutte introduced the theory of nowhere-zero ows as a tool to investigate the coloring

problem of maps, together with his most fascinating conjectures on nowhere-zero ows. Thesehave been extended by Jaeger, Linial, Payan and Tarsil in 1992 to group connectivity, the

generalized form of nowhere-zero ows. Let G be an undirected graph and A = (m

i =1Ai ;+ i , 1 ≤

i ≤ m) be an Abelian multi-group. Let A∗ denote the set of non-zero elements of A. A functionb : V (G) → A is called an A-valued zero-sum function of G if v∈V (G ) b(v) = 0 + i , 1 ≤ i ≤ min G. The set of all A-valued zero-sum function on G is denoted by Z (G, A). We dene:F (G, A) = f : E (G) → A and F ∗(G, A) = f : E (G) → A∗. Let G1 be an orientation of a graph G. If an edge e ∈ E (G) is directed from a vertex u to a vertex v, then let tail( e) = uand head( e) = v. For a vertex v ∈ V (G), let E −(v) = e ∈ E (G1) : v = tail (e), andE + (v) = e ∈ E (G1) : v = head (e). Given a function f ∈ F (G,

A), dene ∂f : V (G) → A

by ∂ f (v) = e∈E + (v) f (e) − e∈E − (v) f (e). A graph G is A-connected if G has an orientation G1

such that for every function b ∈ Z (G, A), there is a function f ∈ F ∗(G1 , A) such that b = ∂f .Let A be the family of graphs that are A-connected. The multi-group connectivity of G isdened as: ∧g (G)=min k | if A is an Abelian group with |A| ≥ k, then G ∈ A . Particularly,

1 Supported by NNSFC No.10871021.2 Received December 22, 2010. Accepted March 3, 2011.



114 Keke Wang, Rongxia Hao and Jianbing Liu

if m = 1, i.e., A = ( A, +) an Abelian group, such connectivity is called group connectivity .Let v be a vertex of G, denote N (v) by : N (v) = u ∈ V (G) − v : uv ∈ E (G). Let u

be a vertex of G, denote N 2(u) = N (N (u))\(N (u)∪ u). Graph G is called claw-free , if itdoesn’t contain k1,3 as an induced subgraph. Let C kn denote the graph with V (C kn ) = V (C n ),

E (C kn ) = uv : u, v ∈ V (C n ) and d cn (u, v ) ≤ k, where dcn (u, v ) is a distance between u and vin C n . Let G be a graph. A triangle-path in G is a sequence of distinct triangles T 1 T 2 · ·· T min G such that for 1 ≤ i ≤ m −1, the following formula (*) holds:

|E (T i ) ∩E (T i+1 )| = 1 and E (T i ) ∩E (T j ) = Ω if j > i + 1 . (*)

Furthermore, if m ≥ 3 and (*) holds for all i, 1 ≤ i ≤ m, with the additionally taken modm, then the sequence is called a triangle-cycle . The number m is the length of the triangle-path(triangle-cycle). A connected graph G is triangularly connected if for any distinct e, e1 ∈E (G), which are not parallel, there is a triangle-path T 1T 2 · · · T m such that e ∈ E (T 1 ) ande1 ∈ E (T m ).

Let G be a connected graph, V (G) and E (G) denote its sets of vertices and edges,

respectively. For S ⊆ V (G), let E (S ) = uv ∈ E (G),u ,v ∈ S . For M ⊆ E (G), letV (M ) = u ∈ V (G): there is v ∈ V (G) such that uv ∈ M . A set of edges M ⊆ E (G)is called a matching of G if they are independent in G, and no two of them share a commonend vertex. A matching is called perfect if it covers all vertices of G. A matching M is calledinduced matching if E (V (M )) = M . G is called induced matching extendable if every inducedmatching M of G is contained in a perfect matching of G. For simplicity, induced matchingextendable will often be abbreviated as IM-extendable .

Notations undened in this paper will follow [1]. In this paper, we give some properties of the 4-regular claw-free 1-edge deletable IM-extendable and prove that its group connectivity is3.

Lemma 1.1.([1]) A graph G has a perfect matching if and only if for every S ⊆ V (G), o(G−S ) ≤|S |, Where o(H ) is the number of odd components of H .

For group connectivity, some conclusions have reached. For example, complete graphs,complete bipartite graphs and triangularly connected graphs etc in [2,3,4,5,6]. For more resultsabout IM-extendable graphs, one can see references [7,8,9,10].

A k-circuit is a circuit of k vertices. A wheel W k is the graph obtained from a k-circuitby adding a new vertex, called the center of the wheel, which is joined to every vertex of thek-circuit. W k is an odd(even) wheel if k is odd(even). For a technical reason, a single edge isregarded as 1-circuit, and thus W 1 is a triangle, called the trivial wheel.

Lemma 1.2([6]) (1)W 2n ∈ Z 3 .(2) Let G ∼= W 2n +1 , b ∈ Z (G, Z 3). T hen there exists a (Z 3 , b)-NZF f ∈ F ∗(G, Z 3) if and onlyif b = 0.

Lemma 1.3([4]) Let G be a connected graph with n vertices and m edges . Then ∧g (G)=2 if and only if n = 1 ( and so G has m loops ).

Lemma 1.4([3]) Let H ≤ G be Z k -connected. If G/H is Z k -connected, then so is G.



Group Connectivity of 1-Edge Deletable IM-Extendable Graphs 115

§2. Main Results

Lemma 2.1 C 26 is 4-regular claw -free 1-edge deletable connected IM-extendable graph .

Proof Obviously, C 26

is 4-regular and claw-free. The following we will prove C 26

is 1-edgedeletable IM-extendable graph. Supposing the vertices of C 26 denoted by vi , 1 ≤ i ≤ 6, along aclockwise. Supposing M is an induced matching of an induced graph G[N (u)]. Since G[N (u)]has four vertices, so |M | ≤ 2. If |M | = 2, u is an isolated vertex of G −V (M ), this conictwith that G is 1-edge deletable IM-extendable graph, thus, |M | ≤ 1. Since E (C 26 ) = E 1 ∪E 2where E 1 = E (C 6 ), E 2 = E \E 1 , the following discussions are divided into two cases.

Case 1 Deleting one edge in E 1 . Without loss of generality, suppose deleting edge v1v2 .From the structure of C 26 , if M is an induced matching of C 26 , then |M | < 2. If let M = v2 v3

be an induced matching of G − v1v2, it extended to a perfect matching v2v3 , v4v5 , v6v1.Otherwise, let M = v2v4 be an induced matching of G − v1v2, it extended to a perfectmatching

v2v4 , v3v5 , v6 v1

.

Case2 Deleting one edge in E 2 . Without loss of generality, let M = v2v3 be an inducedmatching of G − v1v3, it extended to a perfect matching v2v3 , v4 v5 , v6v1. Otherwise,let M = v3v5 be an induced matching of G − v1v3 , it extended to a perfect matching

v3v5 , v2v4 , v6 v1. So C 26 is a 4-regular claw-free 1-edge deletable connected IM-extendablegraph.

Lemma 2.2 Let G be a 4-regular claw -free 1-edge deletable connected IM-extendable graph ,then G = C 26 .

Proof For a given u ∈ V (G), since G is 4-regular, let N (u) = x1 , x2 , x3 , x4, N 2(u) =N (N (u))

\(N (u)

∪u

), f (u) =

|E (N (u))

|, g(N (u)) =

|N 2(u)

|.

Since G is 4-regular and claw-free, we have 2 ≤ f (u) ≤ 6. If f (u) = 6, G is isomorphicto K 5 , there is no perfect matching in K 5 which is a contradiction with 1-edge deletable IM-extendable graph. If f (u) = 5, G[N (u)] is isomorphic to K 4 −e. Without loss of generality, lete = x2x4 and y1 ∈ N (x2) \ (u∪N (u)). Since G is 4-regular, there exists z ∈ N (y1) \ N (u).Let M = y1z,ux 1 be an induced matching of G − x2 x3. It could not extend to a perfectmatching of G − x2x3 which is a contradiction. Next we discuss: 2 ≤ f (u) ≤ 4. There arethree cases according to the value of f (u).

Case 1 f (u) = 2.

Let x1 , x2 , x3 , x4 be neighbor vertices of u, we have three different subcases: Subcase(a)

x1x2 , x2x3 ∈ E (G); Subcase(b) x1 x2 , x3x4 ∈ E (G); Subcase(c) x1x2 , x1x3 ∈ E (G).For subcases (a), the induced subgraph of G by vertices u, x 1 , x3 , x4 consists of an inducedsubgraph k1,3 which contradict with the assumption. For subcase (b), M = x1x2 , x3 x4 isan induced matching of G −ux 2, however, u could not included in vertex set of any perfectmatching of G − ux 2 which contradict with the assumption. For subcase (c), the inducedsubgraph of G by vertices u, x 2 , x3 , x4 consists of k1,3 , which contradicts with the assumption.Therefore f (u) = 2.




Case 2 f (u) = 3.

G[N (u)] is isomorphic to P 4 or K 3∪K 1 or K 1,3 , where P n is the path with n vertices.If G[N (u)] is isomorphic to P 4 . Supposing x1 x2 , x2 x3 , x3x4 ∈ E (G), then M = x1x2 , x3x4

is an induced matching of G

− x2x3

, but it does not extended to a perfect matching of

G − x2x3.If G[N (u)] is isomorphic to k1,3 , obviously G consists of k1,3 as its induced subgraph,

contradiction.If G[N (u)] is isomorphic to K 3 ∪ K 1 , supposing x1 x2 , x2 x3 , x1x3 ∈ E (G), then 3 ≤

g(N (u)) ≤ 6, there are four subcases according to the value of g(N (u)).

Subcase 1. If g(N (u)) = 3, let N 2(u) = y1 , y2 , y3 then yi x4 ∈ E (G) (i = 1 , 2, 3). SinceG is claw-free and 4-regular, then y1y2 , y1y3 , y2y3 ∈ E (G) and each vertex of xi is adjacentto only one vertices of yj (where i, j ∈ 1, 2, 3 are distinct). Without loss of generality, letx1 , x2 , x3 are adjacent to y1 , y2 , y3 respectively. Let M = x1x3 , x4 y2 be an induced matchingof G

−ux 2

. It does not extended to a perfect matching of G

−ux 2

which is a contradiction.

Subcase 2. If g(N (u)) = 4, let yi ∈ N 2(u), i = 1 , 2, 3, 4. Since d(x4) = 4, there is three of yi (1 ≤ i ≤ 4) adjacent to x4 , without loss of generality, supposing yi x4 ∈ E (G) (i = 2 , 3, 4).Because G is claw-free, one have y2y3 , y3y4 , y2y4 ∈ E (G). Obviously y1 is adjacent to at leastone vertex of x i (i=1,2,3).

If y1 is adjacent to each of xi (i=1,2,3), that is x1 y1 , x2y1 , x3 y1 ∈ E (G). If y1y4 ∈ E (G),let M = x1y1 , x4 y4 be an induced matching of G − x2x3 which could not extended to aperfect matching of G −x2x3. If y1y4 ∈ E (G), let M = x1y1 , x4y3 be an induced matchingof G − x2x3 which could not extended to a perfect matching of G − x2x3.

If y1 is adjacent to two vertices of x i (i=1,2,3), without loss of generality, let x1y1 , x2 y1 ∈E (G) and x3 y1

∈ E (G), x3 is adjacent to one of yi (which i

∈ 2, 3, 4

). If x3 y3

∈ E (G), let

M = x2x3 , x4y3 be an induced matching of G −y2y4 which could not extended to a perfectmatching of G − y2y4. If x3y3 ∈ E (G), let M = x2x3 , x4y4 be an induced matching of G − ux 1 which could not extended to a perfect matching of G − ux 1.

If y1 is adjacent to only one of xi (i=1,2,3), without loss of generality, let x1 y1 ∈ E (G).Supposing x2y2 , x3 y3 ∈ E (G), let M = x1x2 , x4 y3 be an induced matching of G − ux 3which could not extended to a perfect matching of G −ux 3.

Subcase 3. If g(N (u)) = 5, let N 2(u) = y1 , y2 , y3 , y4 , y5 and supposing y3x4 , y4x4 , y5x4 ∈E (G). Because G is claw-free, y3y4 , y4y5 , y5y3 ∈ E (G). Since no more than two vertices of xi

(i = 1 , 2, 3) is adjacent to yj ( j = 1 , 2), without loss of generality, let x1y1 , x2y2 ∈ E (G).If x3 is adjacent to y1 or y2 , supposing x3 y1

∈ E (G). Let M =

x1x3 , x4 y3

be an induced

matching of G − ux 2. It does not extended to a perfect matching of G − ux 2 which is acontradiction.

If x3 yi ∈ E (G) (i=1,2). Without loss of generality, supposing x3 y3 ∈ E (G). Let M =

x1x3 , x4y5 be an induced matching of G − ux 2 which could not extended to a perfectmatching of G −ux 2.

Subcase 4. If g(N (u)) = 6, without loss of generality, supposing yi x4 ∈ E (G) (i = 4 , 5, 6).



Group Connectivity of 1-Edge Deletable IM-Extendable Graphs 117

Because each vertex of xi (i=1,2,3) has degree 3 in N (u), each of xi (i = 1 , 2, 3) is adjacentto only one of yj ( j = 1 , 2, 3), without loss of generality, let xi yi ∈ E (G) (i = 1 , 2, 3). LetM = x1x3 , x4y4 be an induced matching of G −ux 2 which could not extended to a perfectmatching of G −ux 2. So f (u) = 3.

Case 3 If f (u) = 4, G[N (u)] is isomorphic to C 4 or K 1,3 + e.

If G[N (u)] is isomorphic to C 4 . Since there are only four edges between N (u) and N 2(u),then 1 ≤ g(N (u)) ≤ 4.

(3.1) If g(N (u)) = 1. Supposing v ∈ N 2(u), there exists x i v ∈ E (G) (i = 1 , 2, 3, 4), it isisomorphic to C 26 .

(3.2) If g(N (u)) = 2. Supposing y1 , y2 ∈ N 2(u). There are two subcases. Subcases 1,there are two vertices adjacent to y1 , two vertices adjacent to y2 . Subcases 2, there are threevertices adjacent to y1 , only one vertex adjacent to y2 .

Subcase 1. Supposing x1y1 , x2y1 , x3 y2 , x4y2

∈ E (G), there exists z, satisfying y1z

∈ E (G).

Let M = x3x4 , y1z be an induced matching of G −x1x2 which could not extended to aperfect matching of G −x1x2.

Subcase 2. Supposing x1y1 , x2y1 , x3y1 , x4y2 ∈ E (G). However, x4 , x3 , x1 , y2 induced a K 1,3

which is a contradiction.

(3.3) If g(N (u)) = 3. Supposing y1 , y2 , y3 ∈ N 2(u). Obviously, only two vertices of xi

(i=1,2,3,4) are adjacent to one of yi (i=1,2,3). Without loss of generality, let xi yi , x4y3 ∈ E (G)(i = 1 , 2, 3). x2 , x1 , y2 , x3 induced a K 1,3 which is a contradiction.

(3.4) If g(N (u)) = 4. Supposing y1 , y2 , y3 , y4 ∈ N 2(u). Without loss of generality, letx i yi ∈ E (G) (i = 1 , 2, 3, 4). x2 , x1 , y1 , x4 induced a K 1,3 subgraph. If G[N (u)] is isomorphic to

C 4 , it does not extended to a perfect matching. If G[N (u)] is isomorphic to K 1,3 + e, supposingx1x2 , x1x3 , x1x4 , x3 x4 ∈ E (G), since there are only 4 edges between N (u) and N 2(u), one have2 ≤ g(N (u)) ≤ 4. There are three subcases according to the value of g(N (u)).

Subcase 1. If g(N (u)) = 2. Supposing N 2(u) = y1 , y2, then x2 y1 , x2 y2 ∈ E (G). BecauseG is claw-free, y1y2 ∈ E (G). If both of x3 , x4 are adjacent to y1 , let M = ux 1 , y1y2 be aninduced matching of G−x3x4 which could not extended to a perfect matching of G−x3x4.If x3 y1 , x4y2 ∈ E (G), let M = ux 1 , y1y2 be an induced matching of G −x3 x4 which couldnot extended to a perfect matching of G −x3 x4.

Subcase 2. If g(N (u)) = 3, supposing N 2(u) = y1 , y2 , y3. Without loss of generality, letx2y1 , x2y2

∈ E (G). Because G is claw-free, y1y2

∈ E (G). If both of x3 , x4 are adjacent to

y3 , let M = y1y2 , x3 x4 be an induced matching of G − x1x3. However, not all vertices of x1 , x2 , u could be included in a perfect matching vertices of G−x1x3 which is a contradiction.If x3y3 , x4y2 ∈ E (G), let M = x1x3 , y1y2 be an induced matching of G − x1x2. However,not all vertices of x2 , x4 , u could be included in a vertex set of perfect matching of G −x1x1which is a contradiction.

Subcase 3. If g(N (u)) = 4, Supposing N 2(u) = y1 , y2 , y3 , y4. Without loss of generality,




let x2y1 , x2 y2 , x3y3 , x4 y4 ∈ E (G). Because G is claw-free, y1y2 ∈ E (G). Let M = x1x4 , y1y2 ∈E (G) be an induced matching of G − ux 2 which could not extended to a perfect matchingof G − ux 2. The lemma 2.2 is proved.

Lemma 2.3 If n ≥ 5, then C 2n is Z 3-connected .

Proof By the denition of C kn , for n ≥ 5, there exists a subgraph of C 2n isomorphic to W 2k .By lemma 1.2, W 2k is Z 3-connected. Since C 2n is triangularly connected, by contracting W 2k inC 2n and Lemma 1.4, we have C 2n is Z 3-connected.

Theorem 2.4 T he group connectivity of 4-regular claw -free 1-edge deletable IM-extendablegraph is 3.

Proof By applying lemma 2.1 and lemma2.2, 4-regular claw-free 1-edge deletable IM-extendable graph is C 26 . From Lemma 2.3, the group connectivity of 4-regular claw-free 1-edgedeletable IM-extendable graph is not more than 3. By lemma 1.3 we conclude the group con-

nectivity of 4-regular claw-free 1-edge deletable IM-extendable graph is more than 2. Therefore,the group connectivity of 4-regular claw-free 1-edge deletable IM-extendable graph is 3. Thistheorem is proved.

References

[1] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications , American Elsevier, NewYork, 1976.

[2] Jing-jing Chen, Elaine Eschen and Hong-Jian Lai, Group connectivity of certain graphs,Ars Combinatoria , 141-158(2008).

[3] M. DeVos, R. Xu and G. Yu, Nowhere-zero Z 3-ows through Z 3-connectivity, Discrete Math. 306(2006), 26-30.

[4] Genghua Fan, Hongjian Lai, Rui Xu, Cun-Quan Zhang, Chuixiang Zhou, Nowhere-zero3-ows in triangularly connected graphs, Journal of Combin. Theory , Series B98(2008)1325-1336.

[5] F. Jaeger, N. Linial, C. Payan and M. Tarsi, Group connectivity of graphs-a nonhomoge-neous analogue of nowhere-zero ow properties, J.Combin.Theory Ser.B 56(1992) 165-182.

[6] H.-J. Lai, Group connectivity of 3-edge-connected chordal graphs, Graphs and Combina-torics , 16(2000), 165-176.

[7] Y. Liu, J.-J. Yuan and S.-Y. Wang, Degree conditions of IM-extendable graphs, Appl.Math. JCU , 15B 1 (2000) 1-6.

[8] Q. Wang and J.-J. Yuan, 4-regular claw-free IM-extendable graphs, Discrete Math. , 294(2005), 303-309.

[9] Q. Wang and J.-J. Yuan, Degree sum conditions of induced matching extendable graphs,Zhengzhou Univ. 32 (1) (2000), 19-21.

[10] J.-J. Yuan, Induced matching extendable graph, J. Graph Theory , 28 (1998) 203-313.




A Note On Line Graphs

P. Siva Kota Reddy, Kavita. S. Permi and B. Prashanth

Department of Mathematics, Acharya Institute of Technology

Soladevanahalli, Bangalore-560 090, India


Abstract : In this note we dene two generalizations of the line graph and obtain someresults. Also, we mark some open problems.

Key Words : Smarandachely ( k, r ) line graph, ( k, r )-graphs, line graphs

AMS(2010) : 05C75, 05C45

For standard terminology and notion in graph theory we refer the reader to Harary [2]; thenon-standard will be given in this paper as and when required. We treat only nite simplegraphs without self loops and isolates.

The line graph L(G) of a graph G is dened to have as its vertices the edges of G, with twobeing adjacent if the corresponding edges share a vertex in G. Line graphs have a rich history.The name line graph was rst used by Harary and Norman [3] in 1960. But line graphs werethe subject of investigation as far back as 1932 in Whitney s paper [7], where he studied edgeisomorphism and showed that for connected graphs, edge-isomorphism implies isomorphismexcept for K 3 and K 1,3 . The rst characterization (partition into complete subgraphs) was

given by Krausz [5]. Instead, we refer the interested reader to a somewhat older but still anexcellent survey on line graphs and line digraphs by Hemminger and Beineke [4]. An excellentbook by Prisner [6] describes many interesting generalizations of line graphs. In this note wegeneralize the line graph L(G) of G as follows:

Let G = ( V, E ) be a graph of order p ≥ 3, k and r be integers with 1 ≤ r < k ≤ p. LetU = S 1 , S 2 ,...,S n be the set of all distinct connected acyclic subgraphs of G of order k andU ′ = T 1 , T 2 ,...,T m be the set of all distinct connected subgraphs of G with size k.

The vertex (k, r )-graph Lv(k,r ) (G), where 1 ≤ r < k ≤ p is the graph has the vertex set

U where two vertices S i and S j , i = j are adjacent if, and only if, S i ∩ S j has a connectedsubgraph of order r .

The edge (k, r )-graph Le

(k,r )(G), where 0

≤ r < k

≤ q is the graph has the vertex set U ′

where two vertices T i and T j , i = j are adjacent if, and only if, T i ∩T j has a connected subgraphof size r .

The Smarandachely (k, r ) line graph LS (k,r ) (G) of a graph G is such a graph with vertex

set U ′ and two vertices T i and T j , i = j are adjacent if and only if, T i ∩T j has a connectedsubgraph with order or size r . Clearly, Lv

(k,r ) (G) ≤ LS (k,r ) (G) and Le

(k,r ) (G) ≤ LS (k,r ) (G). In

1 Received December 22, 2010. Accepted March 3, 2011.



120 P. Siva Kota Reddy, Kavita. S. Permi and B. Prashanth

Figure 1, we depicted L(G), Lv(k,r ) (G) and Le

(k,r ) (G) for the graph G.

Figure.1One can easily verify that: L3(K 1,3) = K 3 , L3(C n ) = C n , for n ≥ 3, L3(P n ) = P n −1 and

L3(K 4) = L(K 4 −e) = K 4 .For any positive integer k, the kth iterated line graph L(G) of G is dened as follows:

L0(G) = G, Lk (G) = L(Lk−1(G)) .A graph G is a (3, 2)-graph if there exists a graph H such that L(3 ,2) (H ) ∼= G. First we

prove the following result:

Proposition 1 For any graph G, L(3 ,2) ∼= L2(G).

Proof First we show that L(3 ,2)( G ) and L2(G) have the same number of vertices. Let S

be a vertex in L(3 ,2) (G). Then S corresponds to a subgraph of order 3 in G. Say, S consistsof two adjacent edges ab and bc. Then corresponding to S we have an edge in L(G) with endvertices ab and bc, and corresponding this edge, we have a vertex say abc in L2(G). Similarly,we can show that very vertex in L2(G) corresponds to a connected subgraph of order 3. Thisproves that L (3 ,2) (G) and L2(G) have the same number of vertices. Now, let S 1 and S 2 be twoadjacent edges in L (3 ,2) (G). Then S 1 and S 2 correspond to two connected subgraphs of order3 each, having a common edge. These in turn will give two adjacent edges, say e(S 1) and e(S 2)in L(G) and this will give an edge in L2(G) with end vertices e(S 1) and e(S 2). This proves theresult.

In general one can establish the following result.

Proposition 2 Let G be a graph of order p and 2 ≥ r < k ≤ p. If (G) ≤ 2 then L(k,k −1) (G) ∼=Lk−1(G).

Note that this is true only when (G) ≤ 2. For example, we nd Ln −1(n,n −1) (K 1,n ) = K n =

L(K 1,n ).

Proposition 3 The star K 1,3 is not a 3-line graph.



A Note On Line Graphs 121

Proof Let K 1,3 be a 3-line graph. Then there exists a graph G such that L(K 1,3) = G.Since K 1,3 has four vertices, G should have exactly four connected subgraphs each of orderthree. All connected graphs having exactly four induced subgraphs are as follows: i) C 4 , ii) K 4 ,iii) K 4 −e and iv) P 6 . None of these graphs have K 1,3 as its 3-line graph.

Clearly, any graph having K 1,3 as an induced subgraph is not a 3-line graph. Hence, wehave

Corollary 4 K 1,n , n ≥ 3 is not a 3-line graph.

It is not true in general that the line graph L(G) of a graph G is a subgraph of L3(G). Forexample, L3(K 1,4) does not contain K 4 , the line graph of K 1,4 as a subgraph.

Problem 5 Characterize 3-line graphs.

A graph G is a self 3-line graph , if it is isomorphic to its 3-line graph.

Problem 6 Characterize self 3-line graphs.

Proposition 7 Let L3(G) be the 3-line graph of a graph G of order p ≥ 3. The degree of a vertex s in L3(G) is denoted by degs and is dened as follows:

Let S be the subgraph of G corresponding to the vertex s in L3(G). For an edge x = uv in S , let d(x) = ( degG u + degG v) −(degS u + degS v), where degG u and degS u are the degrees of uin G and S respectively. Then d(s) =

x∈S

d(x).

Proof Consider an edge uv in S . Suppose y = uz is an edge of G at u which is not in S .Then y belongs to a connected subgraph S 1 of cardinality three containing the edge uv which

is distinct form S . Since S and S 1 have common edge, ss 1 is an edge in L3(G), where s1 is thevertex in L3(G) corresponding to the subgraph S 1 in G. Similarly, for any edge y1 = vz1 at vin G which is not in S , we have an edge ss 2 in L3(G). This implies that corresponding to theedge x = uv in S , we have (degG u −degS u) + ( degG v −degS v) edges in L3(G), and hence theresult follows.

Acknowledgement

The authors are grateful to Sri. B. Premnath Reddy, Chairman, Acharya Institutes, for hisconstant support and encouragement for research and development.

References

[1] F.Barahona, M.Grotschel, M.Junger, and G.Reinelt, An application of combinatorial opti-mization to statistical physics and circuit layout design, Operations Research , 36 (3) (1988),493-513.

[2] F. Harary, Graph Theory , Addison-Wesley Publishing Co., 1969.



122 P. Siva Kota Reddy, Kavita. S. Permi and B. Prashanth

[3] F.Harary and R. Z. Norman, Some properties of line digraphs, Rend. Circ. Mat. Palermo ,(2) 9 (1960), 161-168.

[4] R. L.Hemminger and L. W. Beineke, Line graphs and line digraphs, Selected Topics in Graph Theory (W. B. Lowell and R. J. Wilson, eds.), Academic Press, New York, 1978,

pp. 271-305.[5] J.Krausz, D emonstration nouvelle d u ne th´ eoreme de Whitney sur les r´eseaux, Mat. Fiz.

Lapok , 50 (1943), 75-85 (Hungarian).[6] E. Prisner, Graph Dynamics , Pitman Research Notes in Mathematics Series, vol. 338,

Longman, Harlow, 1995.[7] H. Whitney, Congruent graphs and the connectivity of graphs, Amer. J. Math. , 54 (1932),

150-168.




One-Mother Vertex Graphs

F. Salama

(Mathematics Department, Faculty of Science, Tanta University, Tanta, Egypt)


Abstract : In this paper we will dene a new type of graph. The idea of this denitionis based on when we illustrate the cardiovascular system by a graph we nd that not allvertices have the same important so we dene this new graph and call it 1- mother vertexgraph.

Key Words : Smarandache mother-father graph, 1-mother graph, matrices

AMS(2010) : 97K30, 94C15

§1. Introduction

Unlike other areas in mathematics, graph theory traces its beginning to denite time and place:the problem of the seven bridges of K¨ onigsberg, which was solved in 1736 by Leonhard Euler.And in 1752 we nd Euler’s Theorem for planer graph. However, after this development, littlewas accomplished in this area for almost a century [4].

here are many physical systems whose performance depends not only on the characteristicsof the components but also on the relative locations of the elements. An obvious example is an

electrical network. One simple way of displaying a structure of a system is to draw a diagramconsisting of points called vertices and line segments called edges which connect these vertices sothat such vertices and edges indicate components and relationships between these components.Such a diagram is called linear graph. A graph G is a triple consisting of a vertex-set V(G), anedge-set E(G) and a relation that associated with each edge two vertices called its endpoints.

§2. Denitions and Background

Denition 2.1 An abstract graph G is a diagram consisting of nite non empty set of elements called vertices denoted by V(G) together with a set of unordered pairs of these elements called

edges denoted by E(G). The set of vertices of the graph G is called the vertex-set of G and the list of the edges is called the edge-list of G [1,5,9,10].

Denition 2.2 An oriented abstract graph is a pair (V,E) where V is nite non empty set of vertices and E is a set of ordered pairs of distinct elements of E with the property that if (v,w)∈E then (w,v) /∈E where the element (v,w) denote the edge from v to w [4,5].

1 Received December 22, 2010. Accepted March 3, 2011.



124 F. Salama

Denition 2.3 An empty graph is a graph with no vertices and no edges [5].

Denition 2.4 A null graph is a graph containing no edges [9,10].

Denition 2.5 A multiple edges dened as two or more edges joining the same pair of vertices [1,8,9,10].

Denition 2.6 A loop is an edge joining a vertex to itself [1,8,9,10].

Denition 2.7 A simple graph is a graph with no loops or multiple edges [9].

Denition 2.8 A multiple graph is a graph with al lows multiple edges and loops [1,8,9,10].

Denition 2.9 A complete graph is a graph in which every two distinct vertices are joined by exactly one edge [5,6,9,10].

Denition 2.10 A connected graph is a graph that in one piece, where as one which splits in to several pieces is disconnected [9].

Denition 2.11 Given a graph G, a graph H is called a subgraph of G if the vertices of H are vertices of G and the edges of H are edges of G [5,6,8].

Denition 2.12 Let v and w be two vertices of a graph. If v and w are joined by an edges,then v and w are said to be adjacent. Also, v and w are said to be incident with e then e is said to be incident with v and w [10].

Denition 2.13 Let G be a graph without loops, with n-vertices labeled 1, 2, 3,...,n . The

adjacency matrix A(G) is the n × n matrix in which the entry in row i and column j is the number of edges joining the vertices i and j [10].

Denition 2.14 Let G be a graph without loops, with n- vertices labeled 1,2,3,. . . ., n and m edges labeled 1,2,3,. . . .,m. The incidence matrix I(G) is the n ×m matrix in which the entry in row i and column j is l if vertex i is incident with edge j and 0 otherwise [10].

§3. Main Results

In this article, we will dene new types of graphs as follows:

Denition 3.1 A Smarandache mother-father graph is a graph G in which there are vertices u1

m , u2m , · · · , un

m , v1m , v2

m , · · · , vnm in G with a partition of V 1 , v2 , · · · , V n of V (G) such that vi

m

is important than vi1 , vi

1 is important than vi2 · · ·, and vi

j is important than vij +1 , · · ·, important

than uim for ∀1 ≤ i ≤ n, j 1 , we call vi

m , u im , 1 ≤ i ≤ n mother vertices and father vertices.

Particularly, if n = 1 and there are no father vertices in a graph G, we call such a graph G1-mother graph, seeing Figure 1.



One-Mother Vertex Graphs 125

vm v1 v2 v3 v4

Figure 1Now we will classify the 1-mother vertex graph with respect to the number of the family

which contacts with the mother vertex as follows:

Denition 3.2 A 1-mother vertex graph with n families of vertices is a graph Gm which it’s vertex-set has the form V vm ; v1

1 , v12 , v1

3 , · · · ; v21 , v2

2 , v23 , · · · ; · · · ; vn

1 , vn2 , vn

3 , · · · , where vi1 , vi

2 , vi3 , · · ·

is the i-th family, seeing Figure 2.

vm v1 v2 v3 v4

(a) 1 family (b) 2 families

vmv1v2v3v4 v1 v2 v3 v4

(c) 3 families

vm v11 v1

2

v21

v22

v31

v32

(d) n families

vmv11v1

2

v21

v22

v41

v32

v31

v42

vn1

vn2

Figure 2

Denition 3.3 Any edge has vm as a vertex is called a mother edge.

In Figure 2, there is a one mother edge in (a), two mother edges in (b), three mother edgesin (c) and n mother edges in (d).

Note 1) The families of vertices in a 1-mother vertex graph with n families not necessary havethe same number of vertices , seeing Figure 3.

1 family 3 families 2 familiesvm v1 vm v1

1v21v2

2v23

v31

v32

vm v11 v1

2v21v2

2v23

Figure 3

2) The following graph is not 1-mother vertex graph, seeing Figure 4.



126 F. Salama

v

Figure 4

Denition 3.4 An empty 1-mother vertex graph is an 1-mother vertex graph with no vertices and no edges.

Denition 3.5 A simple 1-mother vertex graph is an 1-mother vertex graph with no loops and no multiple edges, seeing Figure 3.

Denition 3.6 A multiple 1-mother vertex graph is an 1-mother vertex graph allows multiple edges and loops, seeing Figure 5.

1 family 3 families 2 familiesvm v1 vm v1

1v21v2

2v23

v31

v32

vm v11 v1

2v21v2

2v23

Figure 5

Denition 3.7 A connected 1-mother vertex graph is 1-mother vertex graph that in one piece and the one which splits into several pieces is disconnected,seeing Figure 6.

vm v11 v1

2v21v2

2v23

(a) A connected1-mother vertex graph with 2 families

vm v21 v2

2v31v3

2v33

(a) A connected1-mother vertex graph with 2 families

v11

Figure 6

Note The following graph is not disconnected 1-mother vertex graph and also is not 1-mothervertex graph.

vm v11 v1

2v21v2

2v23

Figure 7




Denition 3.8 A graph H im is said to be main supgraph of Gnm ,where n, i ∈ z+ and i ≤ n ,

if V (H im ) ⊆ V (Gnm ) , E (H im ) ⊆ E (Gn

m ) and vm ∈ V (H im ).

Proposition 3.1 The main supgraph H im of Gnm is 1-mother vertex graph.

Denition 3.9 A graph H is a supgraph of Gnm if V (H ) ⊆ V (Gn

m ) , E (H ) ⊆ E (Gnm ) and

vm /∈ V (H im ).

Proposition 3.2 A supgraph H of Gnm is not 1-mother vertex graph.

Example 3.1 As shown in Figure 8, H 2m is a main supgraph of G2m and H is a supgraph of

G2m .

vm v21v2

1v22v2

3 v22v2

3vm v11 v1

2v21v2

2v23 G2

m H 2m H

Figure 8

Denition 3.10 An oriented 1-mother vertex graph is a pair (V, E) where V is nite non empty set of vertices and E is a set of ordered pairs of distinct elements of E with the property that if (v, w) ∈ E , then (w, v) /∈ E , where the element (v, w) denote the edge from v to w, seeing Figure 9.

vm v11 v1

2v21v2

2v23

G2m

Figure 9

Denition 3.11 Let Gnm be a 1-mother vertex graph, with n-families of vertices. The adjacency

matrix A( Gnm ) is the (n+1)x(n+1) matrix in which the entry in row i and column j is matrix

its elements are the number of edges joining the families i and j.

Denition 3.12 Let Gnm be a 1-mother vertex graph, with n-families. The incidence matrix I(

Gnm ) is the (n+1) x n matrix in which the entry in row i and column j is matrix its elements is

l if vertex in family i incident with edge in family j and 0 otherwise.

Example 3.2 The adjacency matrix and the incidence matrix of a 1-mother vertex graph G2m



128 F. Salama

as shown in Figure 9 are given by

A(G2m ) =

0 1 0 1 0 0

1 1 1 0 0 0

0 1 0 0 0 0

1 0 0 0 2 0

0 0 0 2 0 1

0 0 0 0 1 0

I (G2m ) =

1 11 0 0 0 0

0 1 1 0 0 0

1 0 0 1 0 0

0 0 0 1 1 0

0 0 0 1 1 0

0 0 0 0 1 1

where the symbol 1 1 in the matrix in the row 1 and column 2 of the incidence matrix meansthat there exists a loop at the vertex v1

1 with the edge e11 .

Theorem A A complete 1-mother vertex graph is not dened.

Proof Let there exist a complete 1-mother vertex graph. Then this mean that every twodistinct vertices are joined which is contradict with the denition of the 1-mother vertex graph.Hence the complete 1-mother vertex graph is not dene.

New we will dene the union of any 1-mother vertex graphs as follows:

Denition 3.13 The union of Gsm and Gv

m , denoted Gsm∪ Gv

m is the graph with vertex set V 1∪V 2 and edge set E 1 ∪E 2 .

Proposition 3.3 The union of any 1-mother vertex graphs is 1-mother vertex graph if vm ∈V 1∩V 2 .

Proof Let we have two 1-mother vertex graphs, the union of these graphs has one of twotypes.

1) If vm ∈V1 ∩ V 2 , i.e. the new graph has one mother vertex, then the new graph is1-mother vertex graph, seeing Figure 10.a.

2) If vm /∈V1 ∩ V 2 , i.e. the new graph has more than one mother vertex, then the newgraph is not 1-mother vertex graph, seeing Figure 10-b.

vm v11v21v22v23H 2m

G2m

v21v2

2v23 vm v1

1 v12

unionv4

1v42v4

3 v31 v3

2

v23

v22

v2

1

v11

vm

H 2m ∪G2m

Figure 10-a




vm v11v2

1v22v2

3H 2m

G2m

v2

1v2

2v2

3 vm v1

1 v1

2

unionvm v1

1v21v2

2v23 v2

3 v22 v2

1 vm v11 v1

2

H 2m ∪G2m

Figure 10-b

The intersection of 1-mother vertex graphs will be dened as follows:

Denition 3.14 The intersection of Gsm and Gv

m , denoted Gsm ∩Gv

m is the graph with vertex set V 1 ∩V 2 and edge set E 1 ∩E 2 .

Proposition 3.4 The intersection of any number of 1-mother vertex graphs is 1-mother vertex graph if vm ∈ V 1 ∩V 2 or V 1 ∩V 2 = φ and E 1 ∩E 2 .

Proof Let we have n number of 1-mother vertex graphs, the intersection of these graphshas one of two types.

1) If vm ∈ V 1 ∩V 2 · · ·∩V n , i.e. the new graph has one mother vertex, then the new graphis 1-mother vertex graph, seeing Figure 11-a.

2) If vm /∈ V 1 ∩V 2 · · ·∩V n = φ, i.e. the new graph is the empty 1-mother vertex graph.

3) If vm /∈ V 1 ∩V 2 · · ·∩V n = φ, i.e. the new graph has more than one mother vertex, thenthe new graph is not 1-mother vertex graph, see Figure 11-b.

vm v11v2

1v22v2

3

vm v11 v1

2v21v2

2v23

H 2m

G2m

Intersection

vm v11

H 2m ∩G2m

Figure 11-a



130 F. Salama

vm v11v2

1v22v2

3

vm v11 v1

2v21v2

2v23

H 2m

G2m

Intersection

H 2m ∩G2m

v21v2

2v23

Figure 11-b

In this section we will dene special types of 1-mother vertex graphs.

Denition 3.15 A spider mother graph S nm is 1-mother vertex graph has the form as shown in Figure 12.

vm

v11

v12

v13

v21

v22

v23

v31

v32

v33

vm v11

v12

v13

v14

v21

v22

v23

v24

v31

v32

v33

v34

v41

v42

v43

v44

(a) Spider mother graph with 3 families (b) Spider mother graph with 4 families

Figure 12

Note The least number of families which the spider graph has is three.

Denition 3.16 A tree mother graph T nm is 1-mother vertex graph has the form as shown in Figure 13.

T 1m T 2m T 3m T 4m T 5m

Figure 13




Example 3.3 The adjacency matrix of the spider graph as shown in Figure 12-a are given by

0 1 0 0 1 0 0 1 0 0

1 0 1 0 1 0 0 1 0 0

0 1 0 1 0 1 0 0 1 0

0 0 1 0 0 0 1 0 0 1

1 1 0 0 0 1 0 1 0 0

0 0 1 0 1 0 1 0 1 0

0 0 0 1 0 1 0 0 0 1

1 1 0 0 1 0 0 0 1 0

0 0 1 0 0 1 0 1 0 1

0 0 0 1 0 0 1 0 1 0

the existence of the unit matrix in column I and raw j means that the family in the column Iand the family in the raw j have the relation between the vertices which have the same order.

Denition 3.17 An orbit mother graph is 1-mother vertex graph is a 1-mother vertex graph containing no edges and the elements in the same family have the same distance from the mother vertex, seeing Figure 14.

vm

v11

v12

v13

v21 v2

2

v31

v32

Figure 14

§4. Applications

(1) The solar system is orbit mother graph.

(2) If we illustrate the nervous system by using the graph we nd that the nervous system is1-mother vertex graph, seeing Figure 16.





Perfect understanding will sometimes almost extinguish pleasure.

By A.E.Housman, a British scholar and poet



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Books

[4]Linfan Mao, Combinatorial Geometry with Applications to Field Theory , InfoQuest Press,2009. [12]W.S. Massey, Algebraic topology: an introduction , Springer-Verlag, New York 1977.

Research papers

[6]Linfan Mao, Combinatorial speculation and combinatorial conjecture for mathematics, In-ternational J.Math. Combin. , Vol.1, 1-19(2007).[9]Kavita Srivastava, On singular H-closed extensions, Proc. Amer. Math. Soc. (to appear).

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