MATHEMATICAL COMBINATORICS

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1 ISSN VOLUME, 6 INTERNATIONAL JOURNAL OF MATHEMATICAL COMBINATORICS EDITED BY THE MADIS OF CHINESE ACADEMY OF SCIENCES AND ACADEMY OF MATHEMATICAL COMBINATORICS & APPLICATIONS, USA June, 6

2 Vol., 6 ISSN International Journal of Mathematical Combinatorics Edited By The Madis of Chinese Academy of Sciences and Academy of Mathematical Combinatorics & Applications, USA June, 6

3 Aims and Scope: The International J.Mathematical Combinatorics (ISSN ) is a fully refereed international journal, sponsored by the MADIS of Chinese Academy of Sciences and published in USA quarterly comprising -5 pages approx. per volume, which publishes original research papers and survey articles in all aspects of Smarandache multi-spaces, Smarandache geometries, mathematical combinatorics, non-euclidean geometry and topology and their applications to other sciences. Topics in detail to be covered are: Smarandache multi-spaces with applications to other sciences, such as those of algebraic multi-systems, multi-metric spaces,, etc.. Smarandache geometries; Topological graphs; Algebraic graphs; Random graphs; Combinatorial maps; Graph and map enumeration; Combinatorial designs; Combinatorial enumeration; Differential Geometry; Geometry on manifolds; 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 Combinatorics to mathematics and theoretical physics; Mathematical theory on gravitational fields; 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 are also welcome. It is also available from the below international databases: Serials Group/Editorial Department of EBSCO Publishing Estes St. Ipswich, MA 938-6, USA Tel.: (978) , Ext. 6 Fax: (978) and Gale Directory of Publications and Broadcast Media, Gale, a part of Cengage Learning 75 Drake Rd. Farmington Hills, MI , USA Tel.: (48) , ext. 36; GALE Fax: (48) Indexing and Reviews: Mathematical Reviews (USA), Zentralblatt Math (Germany), Referativnyi Zhurnal (Russia), Mathematika (Russia), Directory of Open Access (DoAJ), International Statistical Institute (ISI), International Scientific Indexing (ISI, impact factor.659), Institute for Scientific Information (PA, USA), Library of Congress Subject Headings (USA). Subscription A subscription can be ordered by an directly to Linfan Mao The Editor-in-Chief of International Journal of Mathematical Combinatorics Chinese Academy of Mathematics and System Science Beijing, 9, P.R.China maolinfan@63.com Price: US$48.

4 Editorial Board (4th) Editor-in-Chief Linfan MAO Chinese Academy of Mathematics and System Science, P.R.China and Academy of Mathematical Combinatorics & Applications, USA Deputy Editor-in-Chief Guohua Song Beijing University of Civil Engineering and Architecture, P.R.China Editors Arindam Bhattacharyya Jadavpur University, India Said Broumi Hassan II University Mohammedia Hay El Baraka Ben M sik Casablanca B.P.795 Morocco Junliang Cai Beijing Normal University, P.R.China caijunliang@bnu.edu.cn Yanxun Chang Beijing Jiaotong University, P.R.China yxchang@center.njtu.edu.cn Shaofei Du Capital Normal University, P.R.China dushf@mail.cnu.edu.cn Xiaodong Hu Chinese Academy of Mathematics and System Science, P.R.China xdhu@amss.ac.cn Yuanqiu Huang Hunan Normal University, P.R.China hyqq@public.cs.hn.cn H.Iseri Mansfield University, USA hiseri@mnsfld.edu Xueliang Li Nankai University, P.R.China lxl@nankai.edu.cn Guodong Liu Huizhou University lgd@hzu.edu.cn W.B.Vasantha Kandasamy Indian Institute of Technology, India vasantha@iitm.ac.in Ion Patrascu Fratii Buzesti National College Craiova Romania Han Ren East China Normal University, P.R.China hren@math.ecnu.edu.cn Jingan Cui Ovidiu-Ilie Sandru Beijing University of Civil Engineering and Politechnica University of Bucharest Architecture, P.R.China Romania cuijingan@bucea.edu.cn

5 ii International Journal of Mathematical Combinatorics Mingyao Xu Peking University, P.R.China Guiying Yan Chinese Academy of Mathematics and System Science, P.R.China Y. Zhang Department of Computer Science Georgia State University, Atlanta, USA Famous Words: If you would go up high, then use your own legs! Do not let yourselves carried aloft; do not seat yourselves on other people s backs and heads. By F.W.Nietzsche, a German philosopher.

6 International J.Math. Combin. Vol.(6), -5 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space Süha Yılmaz (Dokuz Eylül University, Buca Educational Faculty, 355, Buca-Izmir, Turkey) Ümit Ziya Savcı (Celal Bayar University, Department of Mathematics Education, 459, Manisa-Turkey) Abstract: In this paper, we investigate Smarandache curves according to type- Bishop frame in Euclidean 3- space and we give some differential geometric properties of Smarandache curves. Also, some characterizations of Smarandache breadth curves in Euclidean 3- space are presented. Besides, we illustrate examples of our results. Key Words: Smarandache curves, Bishop frame, curves of constant breadth. AMS(): 53A5, 53B5, 53B3.. Introduction A regular curve in Euclidean 3-space, whose position vector is composed by Frenet frame vectors on another regular curve, is called a Smarandache curve. M. Turgut and S. Yılmaz have defined a special case of such curves and call it Smarandache TB curves in the space E 4 []. Moreover, special Smarandache curves have been investigated by some differential geometric [6]. A.T.Ali has introduced some special Smarandache curves in the Euclidean space []. Special Smarandache curves according to Sabban frame have been studied by [5]. Besides, It has been determined some special Smarandache curves E 3 by []. Curves of constant breadth were introduced by L.Euler [3]. We investigate position vector of curves and some characterizations case of constant breadth according to type- Bishop frame in E 3.. Preliminaries The Euclidean 3-space E 3 proved with the standard flat metric given by <, >= dx + dx + dx 3 Received November 6, 5, Accepted May 6, 6.

7 Süha Yılmaz and Ümit Ziya Savcı where (x, x, x 3 ) is rectangular coordinate system of E 3. Recall that, the norm of an arbitrary vector a E 3 given by a = < a, a >. ϕ is called a unit speed curve if velocity vector υ of ϕ satisfied υ = The Bishop frame or parallel transport frame is alternative approach to defining a moving frame that is well defined even when the curve has vanishing second derivative. One can express parallel transport of orthonormal frame along a curve simply by parallel transporting each component of the frame [8]. The type- Bishop frame is expressed as ξ ξ B ε ξ = ε. ξ (.) ε ε B In order to investigate type- Bishop frame relation with Serret-Frenet frame, first we B = τn = ε ξ + ε ξ (.) Taking the norm of both sides, we have κ(s) = dθ(s) ds, τ(s) = ε + ε (.3) Moreover, we may express ε (s) = τ cosθ(s), ε (s) = τ sin θ(s) (.4) By this way, we conclude θ(s) = Arc tan ε. The frame {ξ, ξ, B} is properly oriented, ε and τ and θ(s) = s κ(s)ds are polar coordinates for the curve α(s). We write the tangent vector according to frame {ξ, ξ, B} as and differentiate with respect to s T = sinθ(s)ξ cosθ(s)ξ T = κn = θ (s)(cosθ(s)ξ + sin θ(s)ξ ) + sinθ(s)ξ cosθ(s)ξ (.5) Substituting ξ = ε B and ξ = ε B in equation (.5) we have κn = θ (s)(cos θ(s)ξ + sinθ(s)ξ ) In the above equation let us take θ (s) = κ(s). So we immediately arrive at N = cosθ(s)ξ + sin θ(s)ξ

8 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space 3 Considering the obtained equations, the relation matrix between Serret-Frenet and the type- Bishop frame can be expressed T sin θ(s) cos θ(s) ξ N = cos θ(s) sin θ(s). ξ (.6) B B 3. Smarandache Curves According to Type- Bishop Frame in E 3 Let α = α(s) be a unit speed regular curve in E 3 and denote by {ξ α, ξ α, B α } the moving Bishop frame along the curve α. The following Bishop formulae is given by ξ α = εα Bα, ξ α = εα Bα, B α = ε α ξα + εα ξα 3. ξ ξ -Smarandache Curves Definition 3. Let α = α(s) be a unit speed regular curve in E 3 and {ξ α, ξα, Bα } be its moving Bishop frame. ξ ξ -Smarandache curves can be defined by β(s ) = (ξ α + ξ α ) (3.) Now, we can investigate Bishop invariants of ξ ξ -Smarandache curves according to α = α(s). Differentiating (3..) with respect to s, we get where β = dβ ds ds ds = (ε α + ε α )B α T β ds ds = (ε α + εα )Bα (3.) The tangent vector of curve β can be written as follow; Differentiating (3.4) with respect to s, we obtain ds ds = (ε α + ε α ) (3.3) T β = B α = (ε α ξ α + ε α ξ α ) (3.4) dt β ds ds ds = εα ξ α + ε α ξ α (3.5)

9 4 Süha Yılmaz and Ümit Ziya Savcı Substituting (3.3) in (3.5), we get T β = ε α + (ε α εα ξα + εα ξα ) Then, the curvature and principal normal vector field of curve β are respectively, T β N β = = κ β = ε α + εα (ε α ) + (ε α ) (ε α ) + (ε α ) (ε α ξα + εα ξα ) On the other hand, we express B β = det (ε α ) + (ε α ) ξ α ξ α B α ε α ε α. So, the binormal vector of curve β is B β = (ε α ) + (ε α ) (ε α ξα εα ξα ) We differentiate (3.) with respect to s in order to calculate the torsion of curve β β = {[(ε α ) + ε α εα ]ξα + [ε α εα + (εα ) ]ξ α + [ ε α + ε α ]}Bα ] and similarly where β = (δ ξ α + δ ξ α + δ 3B α ) δ = 3ε α ε α +ε α ε α +ε α ε α - (ε α ) 3 -(ε α ) ε α δ = ε α εα +εα ε α +3εα ε α -εα (εα ) -(ε α )3 δ 3 = ε α + ε α The torsion of curve β is τ β = ε α +εα 4 [(ε α ) +(ε α ) ] {[(εα + ε α )(ε α ε α + (ε α ) ]δ [(ε α + ε α )((ε α ) + ε α ε α )]δ } 3. ξ B-Smarandache Curves Definition 3. Let α = α(s) be a unit speed regular curve in E 3 and {ξ α, ξα, Bα } be its moving

10 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space 5 Bishop frame. ξ B-Smarandache curves can be defined by β(s ) = (ξ α + Bα ) (3.6) Now, we can investigate Bishop invariants of ξ B-Smarandache curves according to α = α(s). Differentiating (3.6) with respect to s, we get where β = dβ ds ds ds = (ε α B α + ε α ξ α + ε α ξ α ) T β ds ds = ( ε α B α + ε α ξ α + ε α ξ α ) ds ds = (ε α ) + (ε α ) The tangent vector of curve β can be written as follow; (3.7) (3.8) T β = (ε α ) + (ε α ) (ε α ξα + εα ξα εα Bα ) (3.9) Differentiating (3.9) with respect to s, we obtain dt β ds ds ds = [ 3 (ε α ) +(ε α )] (µ ξ α +µ ξ α +µ 3 B α ) (3.) where µ = ε α ε α ε α + ε α (ε α ) µ = (ε α ) ε α -ε α ε α ε α + (ε α ) ε α -(ε α ) 3 ε α -ε α (ε α ) 3 µ 3 = ε α ε α ε α (ε α ) 4 + (ε α ) (ε α ) ε α (ε α ) Substituting (3.8) in (3.), we have T β = [ (ε α ) + (ε α )] (µ ξ α + µ ξ α + µ 3B α ) Then, the first curvature and principal normal vector field of curve β are respectively T β =κ β = µ [(ε α ) +(ε α ) ] + µ + µ 3 N β = µ + µ + µ 3 (µ ξ α + µ ξ α + µ 3 B α )

11 6 Süha Yılmaz and Ümit Ziya Savcı On the other hand, we get B β = µ +µ +µ 3 (ε α ) + (ε α ) [(µ ε α +µ 3ε α )ξα (µ ξ α +µ 3 ξ α )ξ α +(µ ε α -µ ε α )B α ] We differentiate (3.7) with respect to s in order to calculate the torsion of curve β β = {[- (ε α ) + ε α ]ξα + [ ε α ε α + ε α (ε α ) ]ξ α ε α B α } and similarly where β = (Γ ξ α + Γ ξ α + Γ 3B α ) Γ = -6ε α ε α + ε α + (ε α ) 3 Γ = - ε α ε α -ε α ε α + ε α -ε α ε α +ε α (ε α ) - ε α ε α + (ε α ) 3 Γ 3 = - ε α The torsion of curve β is τ β = - [ (εα ) + (ε α ) ] 4 4 (µ +µ +µ 3 ) {[(-εα ε α - ε α +(ε α ) )Γ -((ε α ) ε α )Γ +(-ε α ε α - ε α + (ε α ) )Γ 3 ]ε α -[( ε α (ε α ) )Γ 3 + ε α Γ ]ε α } 3.3 ξ B-Smarandache Curves Definition 3.3 Let α = α(s) be a unit speed regular curve in E 3 and {ξ α, ξ α, B α } be its moving Bishop frame. ξ B-Smarandache curves can be defined by β(s ) = (ξ α + B α ) (3.) Now, we can investigate Bishop invariants of ξ B-Smarandache curves according to α = α(s). Differentiating (3.) with respect to s, we get β = dβ ds ds ds = ( εα B α + ε α ξ α + ε α ξ α ) T β ds ds = (εα ξ α + ε α ξ α ε α B α ) (3.)

12 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space 7 where ds ds = (ε α ) + (ε α ) The tangent vector of curve β can be written as follow; (3.3) T β = εα ξα + εα ξα εα Bα (ε α ) + (ε α ) (3.4) Differentiating (3.4) with respect to s, we obtain dt β ds ds ds = [ 3 (ε α ) +(ε α )] (η ξ α +η ξ α +η 3 B α ) (3.5) where η = ( ε α (ε α ) -ε α ε α ) η = (ε α ) ε α + (ε α ) ε α -ε α ε α ε α η 3 = (ε α ) ε α + (εα )3 -(ε α )4 - (ε α )4-3 (ε α ) (ε α ) Substituting (3.3) in (3.5), we have T β = [ (ε α ) + (ε α )] (η ξ α + η ξ α + η 3 B α ) Then, the first curvature and principal normal vector field of curve β are respectively T β η =κ β = + η + η 3 [(ε α ) + (ε α )] N β = η + η + (η ξ η α + η ξ α + η 3B α ) 3 On the other hand, we express B β = η +η +η 3 (ε α ) + (ε α ) [(η ε α +η 3ε α )ξα (η ξ α +η 3 ξ α )ξ α +(η ε α -η ε α )B α ] We differentiate (3.) with respect to s in order to calculate the torsion of curve β β = {[ε α ξα + ε α (εα ) ]ξ α + [ ε α (εα ) ]ξ α ε α Bα }

13 8 Süha Yılmaz and Ümit Ziya Savcı and similarly where β = (η ξ α + η ξ α + η 3B α ) η = - ε α ε α 5 ε α ε α + ε α +(ε α ) ε α + (ε α ) 3 The torsion of curve β is η = -4 ε α ε α + ε α + ε α η 3 = - ε α τ β = - [(εα ) + (ε α ) ] 4 4 (η +η +η 3 ) {[ ε α η + (ε α (ε α ) )η 3 ]ε α +[ (ε α ) η +(ε α ε α - ε α +(ε α ) )η +(-ε α ε α + ε α )η 3 ]ε α } 3.4 ξ ξ B-Smarandache Curves Definition 3.4 Let α = α(s) be a unit speed regular curve in E 3 and {ξ α, ξα, Bα } be its moving Bishop frame. ξ α ξ B-Smarandache curves can be defined by β(s ) = 3 (ξ α + ξα + Bα ) (3.6) Now, we can investigate Bishop invariants of ξ α ξ B-Smarandache curves according to α = α(s). Differentiating (3.6) with respect to s, we get where β = dβ ds ds ds = 3 [(ε α + εα )Bα ε α ξα εα ξα ] T β ds ds = 3 [(ε α + εα )Bα ε α ξα εα ξα )] ds ds = [(ε α ) + ε α εα + (εα ) ] 3 The tangent vector of curve β can be written as follow; T β = εα ξ α + ε α ξ α (ε α + ε α )B α [(ε α ) + ε α εα + (εα ) ] (3.7) (3.8) (3.9)

14 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space 9 Differentiating (3.9) with respect to s, we get dt β ds ds ds = (λ ξ α+λ ξ α+λ 3B α ) [ 3 (ε α ) + ε α εα + (εα )] (3.) where λ = [ ε α - (εα ) -ε α εα ]u(s)-εα [εα ε α + ε α εα +εα εα + ε α εα ] λ = [ ε α - (εα ) -ε α εα ]u(s)-εα [ ε α + εα ε α + εα ε α ] λ 3 = [-ε α - ε α ]u(s)+εα [εα ε α + 3ε α εα +εα ε α + ε α εα ] + ε α [ ε α (εα ) + (ε α ) ] Substituting (3.8) in (3.), we have T β = 3(λ ξ α +λ ξ α +λ 3 B α ) 4 [(ε α ) + ε α εα + (εα )] Then, the first curvature and principal normal vector field of curve β are respectively T β N β = 3 λ =κ β = + λ + λ 3 4 [(ε α ) + ε α εα + (εα )] λ + λ + λ 3 (λ ξ α + λ ξ α + λ 3B α ) (3.) On the other hand, we express B β = [(ε α ) +ε α εα +(εα ) ] λ det +λ +λ 3 ξ α ξ α B α ε α ε α -(ε α +ε α ) λ λ λ 3 So, the binormal vector field of curve β is B β = [(ε α ) +ε α εα + (εα ) ] λ {[(ε α + ε α )λ +λ +λ 3 ε α λ 3]ξ α +[ εα λ 3-(ε α + εα )]ξα +[εα λ -ε α λ ]B α } We differentiate (3.) with respect to s in order to calculate the torsion of curve β β = - 3 {[ (ε α ) +ε α ξ α - ε α ]ξ α +[ (ε α ) +ε α ε α - ε α ]ξ α + [ε α + ε α ]B α }

15 Süha Yılmaz and Ümit Ziya Savcı and similarly where β = - 3 (σ ξ α + σ ξ α + σ 3B α ) η = 4 ε α ε α +3 ε α ε α - ε α -(ε α ) 3 - (ε α ) ε α η = 5 ε α ε α + ε α ε α +ε α ε α - ε α -(ε α ) 3 -ε α (ε α ) η 3 = ε α +ε α The torsion of curve β is τ β = - 6[(εα ) +ε α εα +(εα ) ] 9 {[( (ε α 3 λ ) +ε α ε α - ε α )σ +(- ε α - (ε α ) -ε α ε α )σ +λ +λ 3 +( (ε α ) +ε α εα - ε α )σ 3]ε α +[- ε α - ε α +(εα ) +ε α εα )σ +(- (ε α ) -ε α ε α + ε α )σ + ( (ε α ) +ε α ε α - ε α )σ 3 ]ε α }. 4. Smarandache Breadth Curves According to Type- Bishop Frame in E 3 A regular curve with more than breadths in Euclidean 3-space is called Smarandache breadth curve. Let α = α(s) be a Smarandache breadth curve. Moreover, let us suppose α = α(s) simple closed space-like curve in the space E 3. These curves will be denoted by (C). The normal plane at every point P on the curve meets the curve at a single point Q other than P. We call the point Q the opposite point P. We consider a curve in the class Γ as in having parallel tangents ξ and ξ opposite directions at opposite points α and α of the curves. A simple closed curve having parallel tangents in opposite directions at opposite points can be represented with respect to type- Bishop frame by the equation α (s) = α(s) + λξ + ϕξ + ηb (4.) where λ(s), ϕ(s) and η(s) are arbitrary functions also α and α are opposite points. Differentiating both sides of (4.) and considering type- Bishop equations, we have dα ds ds =ξ ds = (dλ ds + ηε + )ξ + ( dϕ ds + ηε )ξ + ( λε ϕε + dη ds )B (4.)

16 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space Since ξ = ξ rewriting (4.) we have dλ ds = dϕ ds = ϕε ηε ds ds dη ds = λε + ϕε (4.3) If we call θ as the angle between the tangent of the curve (C) at point α(s) with a given direction and consider dθ = κ, we have (4.3) as follow: ds dλ dθ = ηε κ f(θ) dϕ dθ = ϕε κ (4.4) dη dθ = λε κ + ϕε κ where f(θ) = δ + δ, δ = κ, δ = κ denote the radius of curvature at α and α respectively. And using system (4.4), we have the following differential equation with respect to λ as d 3 λ dθ 3 [ κ d ε dθ (ε λ κ )]d dθ +[ ε κ -ε κ - d dθ ( κ ) d ε dθ (ε κ ) κ d ε dθ (ε κ )]dλ dθ + [ε d κ dθ (ε κ ) ε ε κ ]λ+ +[- κ ε - κ ε d dθ (ε κ )]d f dθ [ κ ε + κ ε d dθ (ε κ )]df dθ (4.5) [ ε ε κ +ε ε + d dθ ( κ ε ) d dθ (ε κ )+ κ ε d dθ (ε κ )]f(θ) = Equation (4.5) is characterization for α. If the distance between opposite points of (C) and (C ) is constant, then we can write that Hence, we write Considering system (4.4) we obtain α α = λ + ϕ + η = l = constant (4.6) λ dλ dθ + ϕdϕ dθ + ηdη dθ = (4.7) λ f(θ) = (4.8)

17 Süha Yılmaz and Ümit Ziya Savcı We write λ = or f(θ) =. Thus, we shall study in the following subcases. Case. λ =. Then we obtain and θ η = κ ε f(θ)dθ, ϕ = θ θ ( η ε κ dθ)ε κ dθ (4.9) d f dθ df dθ [(τ κ ) sin3 θ cosθ τ cosθ]f = (4.) κ General solution of (4.) depends on character of τ. Due to this, we distinguish following κ subcases. Subcase. f(θ) =. then we obtain λ = θ ϕ = θ η = θ η ε κ dθ η ε κ dθ λ ε κ dθ + θ ϕ ε κ dθ (4.) Case. Let us suppose that λ, ϕ,η and λ, ϕ, η constant. Thus the equation (4.4) we obtain ε κ = and ε κ =. Moreover, the equation (4.5) has the form d3 λ dθ 3 = The solution (4.) is λ = L θ + L θ + L 3 where L, L and L 3 real numbers. And therefore we write the position vector ant the curvature α = α + A ξ + A ξ + A 3 B where A = λ, A = ϕ and A 3 = η real numbers. And the distance between the opposite points of (C) and (C ) is α α = A + A + A 3 = constant 5. Examples In this section, we show two examples of Smarandache curves according to Bishop frame in E 3. Example 5. First, let us consider a unit speed curve of E 3 by β(s)= ( 5 36 sin(9s) 9 85 sin(5s), 5 36 cos(9s) cos(5s), 5 36 sin(8s))

18 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space 3 Fig. The curve β = β(s) See the curve β(s) in Fig.. One can calculate its Serret-Frenet apparatus as the following T = ( 5 34 cos9s cos5s, 34 sin 9s sin5s, 7 cos8s) N = ( csc8s(sin 9s sin 5s), 34 csc 8s(cos9s cos5s), 7 ) B = ( 34 (5 sin9s 9 sin 5s), 34 (5 cos9s + 9 cos5s), 5 7 sin 8s) κ = 5 sin8s and τ = 5 cos8s In order to compare our main results with Smarandache curves according to Serret-Frenet frame, we first plot classical Smarandache curve of β Fig.. Now we focus on the type- Bishop trihedral. In order to form the transformation matrix (.6), let us express θ(s) = Since, we can write the transformation matrix s 5 sin(8s)ds = 5 8 cos(8s) T sin( 5 8 cos8s) cos(5 8 cos8s) ξ N = cos( 5 8 cos8s) sin(5 8 cos8s). ξ B B

19 4 Süha Yılmaz and Ümit Ziya Savcı Fig. ξ ξ Smarandache curve By the method of Cramer, one can obtain type- Bishop frame of β as follows ξ = (sin θ( 5 34 cos9s cos5s) + 34 cosθ csc8s(sin 9s sin5s), sin θ( 5 34 sin 9s sin 5s) 34 cosθ csc 8s(cos9s cos5s), 5 7 sin θ cos8s cosθ) ξ = ( cosθ( 5 34 cos9s cos5s) + 34 sin θ csc 8s(sin9s sin 5s), - cosθ( 5 34 sin 9s sin 5s) 34 sin θ csc8s(cos9s cos5s), 5 7 cosθ cos8s sinθ) B = ( 34 (5 sin9s 9 sin5s), 34 (5 cos9s + 9 cos5s), 5 7 sin 8s) where θ = 5 8 cos(8s). So, we have Smarandache curves according to type- Bishop frame of the unit speed curve β = α(s), see Fig.-4 and Fig.5. Fig.3 ξ B Smarandache curve

20 Smarandache Curves and Applications According to Type- Bishop Frame in Euclidean 3-Space 5 Fig.4 ξ B Smarandache curve Fig.5 ξ ξ B Smarandache curve References [] A. Mağden and Ö. Köse, On the curves of constant breadth in E4 space, Tr. J. of Mathematics, (997), [] A.T. Ali, Special Smarandache curves in The Euclidean space, Math. Cobin. Bookser, (), [3] L. Euler, De curvis trangularibus, Acta Acad Petropol, (778), 3-3. [4] L.R. Bishop, There is more than one way to frame a curve, Amer. Math. Monthly, 8 (975), [5] K. Taşköprü, M.Tosun, Smarandache curves on S, Bol. Soc. Paran. Mat., 3() (4), [6] M. Çetin, Y.Tuncer and M. K. Karacan, Smarandache curves according to Bishop frame in Euclidean 3-space, Gen. Math. Notes, (4), [7] M.K. Karacan, B.Bükcü, An alternative moving frame for tubular surface around the spacelike curve with a spacelike binormal in Minkowski 3-space, Math. Morav., (7), [8] M. Özdemir, A.A Ergin, Parallel frames of non-lightlike curves, Missouri J. Math. Sci., () (8), -. [9] M.Petroviç-Torgasev and E. Nesoviç, Some characterizations of the space-like, the timelike and the null curves on the pseudohyperbolic space H in E3, Kragujevac J. Math., (), 7-8. [] M.Turgut, S. Yılmaz, Smarandache curves in Minkowski space-time, International J. Math. Combin., 3 (8), [] M.Turgut, Smarandache breadth pseudo null curves in Minkowski space-time, International J. Math. Combin., (9), [] N.Bayrak, Ö.Bekt as and S. Yüce, Special Smarandache curves in E3, Mathematical Phy. Sics. and [3] S. Yılmaz, M. Turgut, A new version of Bishop frame and an application to spherical images, Journal of Mathematical Analysis and Applications, 37() (),

21 International J.Math. Combin. Vol.(6), 6-8 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 Evren ERGÜN (Ondokuz Mayıs University, Çarşamba Chamber of Commerce Vocational School, Samsun, Turkey) Mustafa ÇALIŞKAN (Gazi University, Faculty of Sciences, Department of Mathematics, Ankara Turkey) eergun@omu.edu.tr, mustafacaliskan@gazi.edu.tr Abstract: In this study,firstly, the darboux vector W of the natural lift α of a curve α are calculated in terms of those of α in R 3. Secondly, we obtained striction lines and distribution parameters of ruled surface pair generated by by Darboux vectors of the curve α and its natural lift α. Finally, for α and α those notions are compared with each other. Key Words: Lift, ruled surface, striction line, distribution parameter. AMS(): 5B,53A5,53A4.. Introduction and Preliminary Notes The concepts of the natural lift curve and geodesic sprays have first been given by Thorpe in [7]. Thorpe proved the natural lift α of the curve α is an integral curve of the geodesic spray iff α is an geodesic on M. Çalışkan at al. studied the natural lift curves of the spherical indicatries of tangent, principal normal, binormal vectors and fixed centrode of a curve in [6]. They gave some interesting results about the original curve, depending on the assumption that the natural lift curve should be the integral curve of the geodesic spray on the tangent bundle T ( S ). Some properties of M-vector field Z defined on a hypersurface M of M were studied by Agashe in []. M-integral curve of Z and M-geodesic spray are defined by Çalışkan and Sivridağ. They gave the main theorem: The natural lift α of the curve α (in M) is an M-integral curve of the geodesic spray Z iff α is an M-geodesic in [8]. Bilici et al. have proposed the natural lift curves and the geodesic sprays for the spherical indicatrices of the the involute evolute curve couple in Euclidean 3-space. They gave some interesting results about the evolute curve, depending on the assumption that the natural lift curve of the spherical indicatrices of the involute should be the integral curve on the tangent bundle T ( S ) in [6]. Then Bilici applied this problem to involutes of a timelike curve in Minkowski 3-space (see [7]). Ergün and Çalışkan defined the concepts of the natural lift curve and geodesic spray in Minkowski 3-space in []. The analogue of the theorem of Thorpe was given in Minkowski 3-space by Ergün and Çalışkan in Received June 6, 5, Accepted May 8, 6.

22 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 7 []. Çalışkan and Ergün defined M-vector field Z, M-geodesic spray, M-integral curve of Z, M- geodesic in [9]. The analogue of the theorem of Sivridağ and Çalışkan was given in Minkowski 3-space by Ergün and Çalışkan in []. Walrave characterized the curve with constant curvature in Minkowski 3-space in [6]. In differential geometry, especially the theory of space curve, the Darboux vector is the areal velocity vector of the Frenet frame of a spacere curve. It is named after Gaston Darboux who discovered it. In term of the Frenet-Serret apparatus, the darboux vector W can be expressed as W = τt + κb, details are given in Lambert et al. in [3]. Let Minkowski 3-space R 3 be the vector space R3 equipped with the Lorentzian inner product g given by g (X, X) = x + x + x 3 where X = (x, x, x 3 ) R 3. A vector X = (x, x, x 3 ) R 3 is said to be timelike if g (X, X) <, spacelike if g (X, X) > and lightlike (or null) if g (X, X) =. Similarly, an arbitrary curve α = α (t) in R 3 where t is a pseudo-arclength parameter, can locally be timelike, spacelike or null (lightlike), if all of its velocity vectors α (t) are respectively timelike, spacelike or null (lightlike), for every t I R. A lightlike vector X is said to be positive (resp. negative) if and only if x > (resp.x < ) and a timelike vector X is said to be positive (resp. negative) if and only if x > (resp. x < ).The norm of a vector X is defined by [4] X IL = g (X, X). We denote by {T (t),n (t), B (t)} the moving Frenet frame along the curve α. Then T, N and B are the tangent, the principal normal and the binormal vector of the curve α, respectively. Let α be a unit speed timelike space curve with curvature κ and torsion τ. Let Frenet vector fields of α be {T, N, B}. In this trihedron, T is timelike vector field, N and B are spacelike vector fields.for this vectors, we can write T N = B, N B = T, B T = N, where is the Lorentzian cross product, [4]. in space R 3. Then, Frenet formulas are given by T = κn, N = κt + τb, B = τn, [6]. The Frenet instantaneous rotation vector for the timelike curve is given by W = τt + κb. Let α be a unit speed spacelike space curve with a spacelike binormal. In this trihedron, we assume that T and B are spacelike vector fields and N is a timelike vector field. In this situation, T N = B, N B = T, B T = N, Then, Frenet formulas are given by T = κn, N = κt + τb, B = τn, [6]. The Frenet instantaneous rotation vector for the spacelike space curve with a spacelike binormal is given by W = τt κb.

23 8 Evren ERGÜN and Mustafa ÇALIŞKAN Let α be a unit speed spacelike space curve with a timelike binormal. In this trihedron, we assume that T and N are spacelike vector fields and B is a timelike vector field.in this situation, Then, Frenet formulas are given by, T N = B, N B = T, B T = N, T = κn, N = κt + τb, B = τn, [6]. The Frenet instantaneous rotation vector for the spacelike space curve with a timelike binormal is given by W = τ T + κb. Lemma.([5]) Let Xand Y be nonzero Lorentz orthogonal vectors in R 3. If X is timelike, then Y is spacelike. Lemma.([5]) Let X and Y be pozitive (negative ) timelike vectors in R 3. Then g (X, Y ) X Y whit equality if and only if X and Y are linearly dependent. Lemma.3([5]) () Let X and Y be pozitive (negative) timelike vectors in R 3. By the Lemma, there is unique nonnegative real number ϕ(x, Y ) such that g (X, Y ) = X Y coshϕ(x, Y ), the Lorentzian timelike angle between X and Y is defined to be ϕ(x, Y ). () Let X and Y be spacelike vektors in R 3 that span a spacelike vector subspace. Then we have g (X, Y ) X Y. Hence, there is a unique real number ϕ(x, Y ) between and π such that g (X, Y ) = X Y cosϕ(x, Y ), the Lorentzian spacelike angle between X and Y is defined to be ϕ(x, Y ). (3) Let X and Y be spacelike vectors in R 3 that span a timelike vector subspace. Then we have g (X, Y ) > X Y. Hence, there is a unique pozitive real number ϕ(x, Y ) between and π such that g (X, Y ) = X Y coshϕ(x, Y ), the Lorentzian timelike angle between X and Y is defined to be ϕ(x, Y ). (4) Let X be a spacelike vector and Y be a pozitive timelike vector in R 3. Then there is

24 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 9 a unique nonnegative reel number ϕ(x, Y ) such that g (X, Y ) = X Y sinhϕ(x, Y ), the Lorentzian timelike angle between X and Y is defined to be ϕ(x, Y ). For the curve α with a timelike tanget, let θ be a Lorentzian timelike angle between the spacelike binormal unit B and the Frenet instantaneous rotation vector W. a) If κ > τ, then W is a spacelike vector. In this situation, from Lemma 3 (3) we can write κ = W coshθ, τ = W sinhθ W = g (W, W) = κ τ and C = W W = sinhθt + coshθb, where C is unit vector of direction W. b) If κ < τ, then W is a timelike vector. In this situation, from Lemma 3 (4) we can write κ = W sinhθ, τ = W coshθ W = g (W, W) = ( κ τ ) and C = coshθt + sinhθb. For the curve α with a timelike principal normal, let θ be an angle between the B and the W, if B and W spacelike vectors that span a spacelike vektor subspace then by the Lemma 3 () we can write κ = W cosθ, τ = W sinθ W = g (W, W) = κ + τ and C = sin θt cosθb. For the curve α with a timelike binormal, let θ be a Lorentzian timelike angle between the B and the W. a) If κ < τ, then W is a spacelike vector. In this situation, from Lemma 3 (4) we can write κ = W sinhθ, τ = W coshθ W = g (W, W) = τ κ and C = coshθt + sinhθb. (b) If κ > τ, then W is a timelike vector. In this situation, from Lemma 3 () we have κ = W coshθ, τ = W sinhθ W = g (W, W) = ( τ κ ) and C = sinhθt + coshθb. From [], we know that if α be a unit speed timelike space curve, then the natural lift α of α is a spacelike space curve; if α be a unit speed spacelike space curve with a spacelike binormal, then the natural lift α of α is a timelike space curve; if α be a unit speed spacelike space curve with a timelike binormal, then the natural lift α of α is a spacelike space curve. If α be a unit speed timelike space curve and α be the natural lift of α, then from [] we know

25 Evren ERGÜN and Mustafa ÇALIŞKAN that T (s) = N (s), N (s) = κ (s) W T (s) τ (s) W τ (s) κ (s) B (s), B (s) = T (s) W W B (s), and if α be a unit speed spacelike space curve with a spacelike binormal and α be the natural lift of α, then T (s) = N (s), N (s) = κ (s) W T (s) + τ (s) W τ (s) κ (s) B (s), B (s) = T (s) W W B (s), and if α be a unit speed spacelike space curve with a timelike binormal and α be the natural lift of α, then T (s) = N (s), N (s) = κ (s) W T (s) τ (s) W τ (s) κ (s) B (s), B (s) = T (s) + W W B (s). Definition.([4]) Let M be a hypersurface in R 3 and let α : I M be a parametrized curve. α is called an integral curve of X if d (α (s)) = X (α (s)) (for all s I), ds where X is a smooth tangent vector field on M. We have TM = T P M = χ (M), P M where T P M is the tangent space of M at P and χ (M) is the space of vector fields on M. Definition.([5]) A parameterized curve α : I M, α : I TM given by α (s) = ( ) α (s),α (s) = α (s) α(s) is called the natural lift of α on TM. Thus, we can write where D is the Levi-Civita connection on R 3. dα ds = d ) (α (s) α(s) = D ds α (s) α (s) A ruled surface is generated by a one-parameter family of straight lines and it possesses a parametric representation X (s, v) = α (s) + ve (s), where α (s) represents a space curve which is called the base curve and e is a unit vector representing the direction of a straight line. The striction point on a ruled surface X is the foot of the common normal between two consecutive generators (or ruling ). The set of striction points defines the striction curve given

26 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 as β (s) = α (s) g (α, e ) e (s) []. g (e, e ) The distribution parameter of the ruled surface X is defined by ([]) P e = det (α, e, e ) e and the ruled surface is developable if and only if P e =.. Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 In this section the darboux vector W of the natural lift α of a curve α are calculated in terms of those of α in R 3. We obtained striction lines and distribution parameters of ruled surface pair generated by Darboux vectors of the curve α and its natural lift α. Let α be a unit speed timelike space curve. Then the natural lift α of α is a spacelike space curve. Proposition. Let α be a unit speed timelike space curve and the natural lift α of the curve α be a space curve with curvature κ and torsion τ. Then κ(s) = W (s)τ (s) + κ (s)τ (s), τ (s) = κ κ (s) κ (s) W. Proposition. Let α be a unit speed timelike space curve and the natural lift α of the curve α be a space curve. (i) If the natural lift α is a unit speed spacelike space curve with a spacelike binormal, then ( ) W = τ (s) κ (s) T κ (s)τ (s) + κ (s)τ (s) κ (s) W N + B (ii) If the natural lift α is a unit speed spacelike space curve with a timelike binormal, then ( ) W = τ (s) κ (s) T + κ (s)τ (s) + κ (s)τ (s) κ (s) W N + B Let X and X be two ruled surfaces which is given by X (s, v) = α (s) + vc (s), X (s, v) = α (s) + vc (s) The striction curves of X and X are given byβ (s) = α (s) λc (s) and β (s) = α (s)

27 Evren ERGÜN and Mustafa ÇALIŞKAN µc (s),respectively. The distribution parameters of the ruled surfaces X and X are defined by P C = det (α, C, C ) C and P C = ) det (α, C, C C. Proposition.3 If the natural lift α is a unit speed spacelike space curve with a timelike binormal, then λ = τ (s) [κ (s)] [τ (s)] W, W κ (s)σ (s) µ = [ ] κ (s)τ(s)+κ(s)τ, (s) + κ (s)σ (s) κ(s) + [σ (s)] + [σ (s)τ (s)] ( ) σ (s)τ (s) κ κ (s) (s)τ(s)+κ(s)τ (s) + κ (s)σ (s) κ(s) P C =, P C = [ κ (s)τ(s)+κ(s)τ (s) + κ (s)σ (s)] κ(s) + [σ (s)] + [σ (s)τ (s)] where σ (s) = κ (s)τ(s)+κ(s)τ (s) κ(s) W. Proposition.4 If the natural lift α is a unit speed spacelike space curve with a spacelike binormal, then τ (s) λ = [κ (s)] [τ W, (s)] µ = W κ (s)σ [ κ (s)τ(s)+κ(s)τ (s) κ (s)σ (s) κ(s) (s) ], + [ σ (s)] + [ σ (s)τ (s)] ( ) σ (s)τ (s) κ κ (s) (s)τ(s)+κ(s)τ (s) κ (s)σ (s) κ(s) P C =, P C = [ κ (s)τ(s)+κ(s)τ (s) κ (s)σ (s)] κ(s) + [ σ (s)] + [ σ (s)τ (s)] where σ (s) = κ (s)τ(s)+κ(s)τ (s) κ(s) W. Let α be a unit speed spacelike space curve with a spacelike binormal. Then the natural lift α of α is a timelike space curve. Proposition.5 Let α be a unit speed spacelike space curve with a spacelike binormal and the natural lift α of the curve α be a space curve with curvature κ and torsion τ. Then. κ(s) = W κ (s)τ (s) + κ (s)τ (s), τ (s) = κ (s) κ (s) W.

28 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 3 Proposition.6 Let α be a unit speed spacelike space curve with a spacelike binormal and the natural lift α of the curve α be a space curve, then. ( ) W = τ (s) κ (s) T + κ (s)τ (s) + κ (s)τ (s) κ (s) W N + B. Let X and X be two ruled surfaces which is given by X (s, v) = α (s) + vc (s), X (s, v) = α (s) + vc (s) Proposition.7 The striction curves of X and X are given by β (s) = α (s) λc (s) and β (s) = α (s) µc (s),respectively. The distribution parameters of the ruled surfaces X and X are defined by P C = det (α,c,c ) C and P C = det (α,c,c ) C. Then we have τ (s) λ = [κ (s)] + [τ W, (s)] ( ) τ (s) + σ (s) κ (s) W µ = [ ] κ (s)τ(s)+κ(s)τ, (s) κ(s) [τ (s) + σ (s)] + [σ (s)τ (s)] ( ) κ(s)τ κ (s) (s) κ (s)τ(s) + κ (s)σ (s) + σ (s)τ (s) κ(s) P C =, P C = [ κ (s)τ(s)+κ(s)τ (s) + κ (s)σ (s)] κ(s) [τ (s) + σ (s)] + [σ (s)τ (s)] where σ (s) = κ (s)τ(s)+κ(s)τ (s) κ(s) W. Let α be a unit speed spacelike space curve with a timelike binormal. Then the natural lift α of α is a spacelike space curve. Proposition.8 Let α be a unit speed spacelike space curve with a timelike binormal and the natural lift α of the curve α be a space curve with curvature κ and torsion τ.then κ(s) = W κ (s)τ (s) κ (s)τ (s), τ (s) = κ (s) κ (s) W. Proposition.9 Let α be a unit speed spacelike space curve with a timelike binormal and the natural lift α of the curve α be a space curve. (i) If the natural lift α is a unit speed spacelike space curve with a spacelike binormal, then ( ) W = τ (s) κ (s) T + κ (s)τ (s) κ (s)τ (s) κ (s) W N B.

29 4 Evren ERGÜN and Mustafa ÇALIŞKAN (ii) If the natural lift α is a unit speed spacelike space curve with a timelike binormal, then ( ) W = τ (s) κ (s) T + κ (s)τ (s) + κ (s)τ (s) κ (s) W N + B. Let X and X be two ruled surfaces which is given by X (s, v) = α (s) + vc (s), X (s, v) = α (s) + vc (s) The striction curves of X and X are given by β (s) = α (s) λc (s) and β (s) = α (s) µc (s),respectively. The distribution parameters of the ruled surfaces X and X are defined by P C = det (α, C, C ) C and P C = ) det (α, C, C C. Proposition. If the natural lift α is a unit speed spacelike space curve with a timelike binormal, then τ (s) λ = [τ (s)] [κ W, (s)] ( ) τ (s) + σ (s) κ (s) W µ = [ ] κ (s)τ(s)+κ(s)τ, (s) κ(s) + [τ (s) + σ (s)] [σ (s)τ (s)] ( ) κ(s)τ κ (s) (s) κ (s)τ(s) κ (s)σ (s) + σ (s)τ (s) κ(s) P C =, P C = [ κ (s)τ(s)+κ(s)τ (s) κ (s)σ (s)] κ(s) + [τ (s) + σ (s)] [σ (s)τ (s)] where σ (s) = κ (s)τ(s)+κ(s)τ (s) κ(s) W. Proposition. If the natural lift α is a unit speed spacelike space curve with a spacelike binormal, then τ (s) λ = [τ (s)] [κ W, (s)] ( ) τ (s) + σ (s) κ (s) W µ = [ ] κ (s)τ(s)+κ(s)τ, (s) κ(s) + [ τ (s) + σ (s)] [σ (s)τ (s)] ( ) κ (s) κ(s)τ (s) κ (s)τ(s) + κ (s)σ (s) σ (s)τ (s) κ(s) P C =, P C = [ κ (s)τ(s)+κ(s)τ (s) + κ (s)σ (s)] κ(s) + [ τ (s) σ (s)] [ σ (s)τ (s)] where σ (s) = κ (s)τ(s)+κ(s)τ (s) κ(s) W.

30 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R 3 5 Example. Let α (s) = timelike circular helix with ( 3 T (s) = N (s) = B (s) = C (s) = (,, ). ( 3 3 s, 3 cos( 3s ), 3 sin ( 3s )) be a unit speed (timelike curve) 3 ( ) 3, 3 sin 3s, ( ( ), cos 3s, sin ( 3 3, 3 X (s, t) = ) 3 ( ) 3 cos 3s, ( )) 3s, 3 sin ( 3s ), 3 3 cos ( 3s ) ( 3 3 s + t, 3 cos ( 3s ), 3 sin ( 3s ) ) ) and κ =, τ =, and X (s, t) = ( 3 + t 3 ( ) ) 3 ( ), sin 3s, 3 cos 3s + t. 3.eps Figure Example. Let α (s) = helix with T (s) = N (s) = ( cosh ( ) s ( sinh ( cosh, s, sinh ( ( s ), )) s, ( s ),, sinh be a unit speed spacelike hyperbolic ( )) s cosh ( s )),

31 6 Evren ERGÜN and Mustafa ÇALIŞKAN B (s) = C (s) = ( sinh ( sinh ( s ), ( s ),, cosh, ( )) s cosh ( s )),, and κ =, τ = X (s, t) = ( ( ) ( ) ( ) ( )) s s s s s cosh + t sinh,, sinh + t cosh and ( ( ) s X (s, t) = sinh, + t, ( )) s cosh.eps Figure ( 3 Example.3 Let α (s) = 3 s, 3 cos( 3s ), 3 sin ( 3s )) be a unit speed (spacelike curve with timelike binormal) spacelike circular helix with T (s) = N (s) = B (s) = C (s) = (,, ) ( 3 3 ( ) 3, 3 sin 3s, 3 ( ( ) ( )), cos 3s, sin 3s, ( 3 3 ( ) 3, 3 sin 3s, X (s, t) = ) ( ) 3 cos 3s 3 3 cos ( 3s ) ( 3 3 s + t, 3 cos ( 3s ), 3 sin ( 3s ) ) ), and κ =, τ = and ( 3 X (s, t) = 3 s t, ) 3 ( ) 3 3 sin 3s, 3 ( ) 3 cos 3s t 3

32 Ruled Surface Pair Generated by Darboux Vectors of a Curve and Its Natural Lift in R eps Figure 3 References [] Agashe N. S., Curves associated with an M-vector field on a hypersurfacem of a Riemmanian manifold M, Tensor N.S., 8 (974), 7-. [] Akutagawa K., Nishikawa S.,The Gauss Map and Spacelike Surfacewith Prescribed Mean Curvature in Minkowski3-Space, Töhoko Math. J.,4, 67-8,(99) [3] A.Turgut and H.H. Hacısalihoğlu, Spacelike Ruled Surfaces in the Minkowski Space, Commun.Fac.Sci.Univ.Ank.Series, Vol.46.No.,(997),83-9. [4] A.Turgut and H.H. Hacısalihoğlu, On the Distribution Parameter of Timelike Ruled Surfaces in the Minkowski Space, Far. East J. Math.Sci., 3(38),(997). [5] A.Turgut and H.H. Hacısalihoğlu, Timelike Ruled Surfaces in the Minkowski Space II, Turkish Journal of Math., Vol.. No.,(998), [6] Bilici M., Çalışkan M. and Aydemir İ., The natural lift curves and the geodesic sprays for the spherical indicatrices of the pair of evolute-involute curves, International Journal of Applied Mathematics, Vol.,No.4(),45-4, [7] Bilici, M., Natural lift curves and the geodesic sprays for the spherical indicatrices of the involutes of a timelike curve in Minkowski 3-space, International Journal of the Physical Sciences, 6(),, [8] Çalışkan, M., Sivridağ, A.İ., Hacısalihoğlu H.H, Some Characterizations for the natural lift curves and the geodesic spray, Communications, Fac. Sci.Univ. Ankara Ser. A Math., 33 (984), Num. 8,35-4 [9] Çalışkan M., Ergün E., On the M-integral curves and M-geodesic sprays in Minkowski 3- space, International Journal of Contemp. Math. Sciences, Vol.6, No.39, (), [] Ergün E., Çalışkan M., On geodesic sprays in Minkowski 3-space, International Journal of

33 8 Evren ERGÜN and Mustafa ÇALIŞKAN Contemp. Math. Sciences, Vol.6, No.39(), [] E.Ergün, M.Çalışkan, Ruled surface pair generated by a curve and its natural lift In R 3, Pure Mathematical Sciences, Vol., No.(),75-8 [] E.Ergün, M.Çalışkan, On natural lift of a curve, Pure Mathematical Sciences, Vol., No.(), [3] Lambert MS, Mariam TT, Susan FH, Darboux Vector, VDMPublishing House,. [4] O Neill B., Semi-Riemannian Geometry with Applications to Relativity, Academic Press, New York, 983. [5] Ratcliffe J.G., Foundations of Hyperbolic Manifolds, Springer-Verlag,New York, Inc., New York, 994. [6] Sivridağ A.İ, Çalışkan M., On the M-integral curves and M-geodesic sprays, Erc.Uni. Fen Bil. Derg. 7, (99), [7] Thorpe J.A., Elementary Topics In Differential Geometry, Springer-Verlag,New York, Heidelberg-Berlin, 979. [8] Walrave J., Curves and Surfaces in Minkowski Space K. U., Leuven Faculteit, Der Wetenschappen, 995.

34 International J.Math. Combin. Vol.(6), 9-4 Tutte Polynomial of Generalized Flower Graphs Nithya Sai Narayana (N.E.S. Ratnam College of Arts, Science and Commerce, Bhandup, Mumbai-478, India) narayana Abstract: The book graph denoted by B n, is the Cartesian Product S n+ P where S n+ is a star graph with n vertices of degree and one vertex of degree n and P is the path graph of vertices. Let X n,p denote the generalized form of Book graph where a family of p cycles which are n in number, is merged at a common edge. The generalized flower graph is obtained by merging t copies of X n,p with a base cycle C t of length t at the common edges. The resultant structure looks like flower with petals. In this paper we discuss some properties satisfied by Tutte polynomial of this special graph and the related graphs. Key Words: Tutte polynomial, recurrence relation, flower graph. AMS(): 5C3, 5C99, 68R5.. Introduction and Preliminaries Tutte polynomial is a polynomial in two variables x, y with remarkable properties and it can be defined for a graph, matrix and more generally for matroids. Tutte polynomial is closely associated with many graphical invariants and in fact the following are the special cases of Tutte polynomial along particular curves of (x, y) plane. () The chromatic and flow polynomial of a graph; () The partition function of a Q-state Pott s model; (3) The Jone s polynomial of an alternating knot; (4) The weight enumerator of a linear code over GF(q); (5) The all terminal reliability probability of a network; (6) The number of spanning trees,number of forests, number of connected spanning subgraphs, the dimension of bicycle space and so on. Tutte polynomial is widely studied for the reason that it provides structural information about the graph. Definition. (i) Let G = (V, E) be an undirected connected multi-graph. The Tutte polyno- Received November 3, 5, Accepted May, 6.

35 3 Nithya Sai Narayana mial of the graph G is given by T(G, x, y) = if E(G) = φ; = xt(g.e, x, y) if e E and e is a cut edge; = yt(g e, x, y) if e E and e is a loop; = T(G e, x, y) + T(G.e, x, y) if e E and e is neither a loop nor a cut edge. (ii) If G is a disconnected graph with connected components G, G,, G t with t, then the Tutte Polynomial of G denoted by T(G, x, y) is defined as T(G, x, y) = t T(G i, x, y). Tutte polynomial of some of standard graphs are given below. Theorem. Let T n be a tree on n vertices and let C n be a cycle on n vertices then () T(T n, x, y) = x n ; () T(C n, x, y) = y + n x i. i= Theorem.3 Let G be a bi connected graph. Let u, v be two vertices in G such that u, v are joined by a path P s of length s where degree of each vertex in P s is two except possibly for u, v then T(G) = ( + x + x + x s )T(G P s ) + T(G.P s ). Proof Let e, e, e s be the s edges in the path P s, then T(G) = T(G e ) + T(G.e ) = x s T(G P s ) + x s T(G P s ) + x s s T(G P s ) + T(G.P s ) (G is bi connected, e t is not a bridge in G G G G t ) = ( + x + x + x s )T(G P s ) + T(G.P s ). ¾ We study the Tutte polynomial of generalized Book graph. Cartesian product of two graphs G, G denoted by G G is a graph with V (G G ) = V (G ) V (G ) and two vertices (u, v ), (u, v ) of G G are adjacent if and only if either u = u and (v, v ) is an edge in G or v = v and (u, u ) is an edge of G. The book graph denoted by B n, or simplyb n is the Cartesian Product S n+ P where S n+ is a star graph with n vertices of degree and one vertex of degree n and P is the path graph of vertices. It can be observed that book graphs are planar. Some book graphs and their planar representation are given below. i= Figure Book graph B 4, B 5 and their Planar representation

36 Tutte Polynomial of Generalized Flower Graphs 3 We make a generalization of this graph. Through out this section T(G, x, y) = T(G) denotes the Tutte polynomial for the graph G. We make use of the following notation. X n,p denote a graph with n number of p-cycles with a common edge e = xy and let Y n,p = X n,p e and Z n,p = X n,p.e. Note that Z n,p is actually a graph with n number of p cycles with a common vertex. Figure X 3,5, X 5,3,X,6 Thus B n = X n,4 is a particular case of the graph we have defined which we call as generalized book graph. We first arrive at some recurrence relation satisfied by these graphs. Before we prove the relations satisfied by these graphs we will prove some preliminary results. Notations and Conventions.4 () Let G and G be two disjoint graphs each of them having a unique identified vertex. The graph obtained by merging an identified vertex of G to an identified vertex of G is denoted by G G. () Let G, G be two disjoint graphs each of them having two designated vertices namely x, y and x, y. The graph obtained by merging the identified vertex x with x and the vertex y with y is denoted by G G. (3) Let G be a graph with an identified vertex v. The graph obtained by taking n copies of G and joining all the copies at the identified vertex v is denoted by G (n). (4) Let G, G be two disjoint graphs each with two identified vertices x, y and x, y respectively. Let xy E(G) and x y E(G ). The graph obtained by merging the two vertices x, x and y, y and the edges xy and x y to a single edge is denoted by G G. Proposition.5 G be a graph which can be expressed as G = H P l where P l is a tree of order l +, then T(G) = x l T(H). [ ] Proposition.6 T(C p (n) ) = T(C p ) n = y + p n x k. k= Proof Induction on n. For n =, T(C p ) = y+ p k= vertex. Assume that the result is true for n. Let G = C (n) p x k by Theorem.. Let v be the identified. Note that C (n) p = C (n ) p C p.

37 3 Nithya Sai Narayana Let e be any edge adjacent to v. By recurrence relation T(G) = T(G e) + T(G.e) = T(C (n ) p P p ) + T(C (n ) p C p ) = x p T(C p (n ) ) + T(C p (n ) C p ) = x p T(C p (n ) ) + T(C p (n ) P p ) + T(C p (n ) C p ) = x p T(C (n ) p ) + x p T(C p (n ) ) + T(C p (n ) C p ) by using Proposition.5 = (x p + x p + + x)t(c p (n ) ) + T(C p (n ) C ) = (x p + x p + + x)t(c (n ) p = (x p + x p + + x + y)t(c p (n ) ) [ ] p p n [ ] p n = {y + x k } y + x k = y + x k. k= k= ) + yt(c p (n ) ) as C is a loop k= ¾. Tutte Polynomial of Generalized Book Graph Theorem. Let X n,p denote a graph with n number of p-cycles with a common edge e = xy and let Y n,p = X n,p e then, X n,p and Y n,p satisfy the following recurrence relations (i) T(X n,p ) = T(Y n,p ) + T(C p ) n ; (ii) T(Y n,p ) = [ p ] x k T(Y n,p ) + T(X n,p ) for n with T(Y,p ) = x p. k= Proof (i) e is neither a loop nor a cut edge and hence using recurrence relation of Tutte polynomial T(X n,p ) = T(X n,p e) + T(X n,p.e) using Proposition.6. This proves (). = T(Y n,p ) + T(C (n) p ) = T(Y n,p) + T(C p ) n (ii) Clearly, Y,p = a path of length p and hence T(Y,p ) = x p. We prove this result by induction on n. For n =, T(Y,p ) = T(Y,p e ) + T(Y,p.e ) where e is any edge of Y,p adjacent to x

38 Tutte Polynomial of Generalized Flower Graphs 33 other than e. T(Y,p ) = x p T(Y,p ) + T(Y,p P p ) = x p T(Y,p ) + x p 3 T(Y,p ) + T(Y,p P p 3 ) = x p T(Y,p ) + x p 3 T(Y,p ) + + xt(y,p ) + T(Y,p P ) = [ p ] x k T(Y,p ) + T(X,p ), k= which proves the result for n =. Assume that the result is true for a graph Y n,p. Consider Y n,p and let e is any edge of Y n,p adjacent to x other than e. Then, T(Y n,p ) = T(Y n,p e ) + T(Y n,p.e ) = x p T(Y n,p ) + T(Y n,p P p ) = x p T(Y n,p ) + x p 3 T(Y n,p ) + T(Y n,p P p 3 ) = x p T(Y n,p ) + x p 3 T(Y n,p ) + + xt(y n,p ) + T(Y n,p P ) = which proves the result for n. [ p x ]T(Y k n,p ) + T(X n,p ), k= ¾ [ n ] Theorem. (i) T(Y n,p ) = b n x p + b k (b + y ) n k n. k= [ n ] (ii) T(X n,p ) = b n x p + b k (b + y ) n k k= for n ; + (b + y ) n, where b = p k= x k for Proof By Theorem., T(Y n,p ) = = = [ p ] x k T(Y n,p ) + T(X n,p ) k= [ p ] x k T(Y n,p ) + T(Y n,p ) + T(C p ) n k= [ p ] x k T(Y n,p ) + T(C p ) n k= = bt(y n,p ) + T(C p ) n. Note that T(C p ) = x p + x p x + y = b + y. We solve the recurrence

39 34 Nithya Sai Narayana relation, T(Y,p ) = x p, T(Y,p ) = bt(y,p ) + T(C p ) = bx p + (y + b ) = b x p + T(Y 3,p ) = bt(y,p ) + T(C p ) [ ] b k (b + y ) k, k= = b x p + bt(c p ) + (y + b ) = b x p + b(y + b ) + (y + b ) [ 3 ] = b 3 x p + b k (b + y ) 3 k, k= T(Y 4,p ) = bt(y 3,p ) + T(C p ) 3 = b 3 x p + b T(C p ) + bt(c p ) + T(C p ) 3 = b 3 x p + b (b + y ) + b(b + y ) + (b + y ) 3 [ 4 ] = b 4 x p + b k (b + y ) 4 k. Assume that by induction T(Y n,p ) = b n x p + [ n 3 k= ] b k (b + y ) n k, k= T(Y n,p ) = bt(y n,p ) + T(C p ) n { [ n 3 ]} = b b n x p + b k (b + y ) n k + (b + y ) n k= = b n x p + b[(b + y ) n + b(b + y ) n b n 3 ] + (b + y ) n = b n x p + (b + y ) n + b(b + y ) n + + b n (b + y ) [ n ] = b n x p + b k (b + y ) n k. This completes the proof of (i). k= T(X n,p ) = T(Y n,p ) + (b + y ) n = b n x p + which completes proof of (ii). [ n ] b k (b + y ) n k + (b + y ) n, k= ¾

40 Tutte Polynomial of Generalized Flower Graphs 35 Remark.3 For n =, T(X n,p ) = T(C p ) = x p + x p + + x + yx p + (b + y ), which matches with the Theorem. An equivalent representation of Tutte polynomial for generalized book graph is the following. [ n ] Theorem.4 T(X n,p ) = xb n + b k (y + b ) n k y k= Proof By Theorem., T(X n,p ) = b n x p + [ n ] b k (b + y ) n k + (b + y ) n < k= Hence, T(X n,p ) = xb n b n + b n + Now consider b = x p + x p x + = xp x x p = b(x ) + = bx b + x p b n = (bx b + )b n = xb n b n + b n. [ n ] b k (b + y ) n k + (b + y ) n k= = xb n b n + b n + (b + y ) n + b(b + y ) n + xb n + +b n (b + y ) + (b + y ) n. ( ) [ n b k (y + b ) ]y n k k= = xb n + y [ (y + b ) n + b(y + b ) n + + b n ] = xb n + (y + b (b )) [ (y + b ) n + b(y + b ) n + + b n ] = xb n + (y + b ) [ (y + b ) n + b(y + b ) n + + b n ] (b ) [ (y + b ) n + b(y + b ) n + + b n ] = xb n + [ (y b ) n + b(y b + ) n + + b n (y + b ) ] b( [ (y + b ) n + b(y + b ) n + + b n ] + [ (y + b ) n + b(y + b ) n + + b n ]

41 36 Nithya Sai Narayana = xb n + (y + b ) n + b(y + b ) n + + (y + b )b n b(y + b ) n b (y + b ) n b n (b + y ) b n +(y + b ) n + b(y + b ) n + + b n (y + b ) + b n = xb n + (y + b ) n b n + (y + b ) n + b(y + b ) n + +b n (y + b ) + b n. ( ) From (*) and (**) we get T(X n,p ) = xb n + [ n b k (y + b ) ]y. n k ¾ k= 3. The Generalized Flower Graph The generalized flower graph is obtained by merging X n,p at each of the edge of a basic cycle of length t. We define the generalized complete flower graph and generalized Flower graph with k petals. Definition 3. (i) A graph in which i copies of X n,p is taken and is merged with any of the i out of t edges of the base cycle C t of length t where, i t is called a generalized flower graph with i petals and is denoted by G (i) n,p,t. (ii) A graph obtained by taking a base cycle C t of length t and t copies of X n,p and merging the two graphs at the common edge of X n,p with each of the edge of the basic cycle C t is referred to as Generalized Flower Graph or Generalized Complete Flower graph and is denoted by G n,p,t. In fact G n,p,t = G (t) n,p,t. (iii) The graph obtained by taking i copies of X n,p with each of the cycle containing a designated edge and joining the i copies at the end vertices of the designated edges is denoted by H (i) n,p. Figure 3 Generalized complete flower graph G,4,5 and flower graph G (3),4,6 with 3 petals and generalized flower graph with one petal Theorem 3. Let X n,p has the common edge e. Let G () n,p,t be the generalized flower graph with

42 Tutte Polynomial of Generalized Flower Graphs 37 one petal. Then, T(G () n,p,t) = ( + x + x + + x t )T(X n,p ) + y(y + b ) n. Proof Let e be any edge on C t other than e. By deletion contraction formula we get T(G n,p,t ) = x t T(X n,p ) + T(G n,p,t ) = x t T(X n,p ) + x t 3 T(X n,p ) + + xt(x n,p ) + T(G n,p, ) = (x + x + + x t )T(X n,p ) + T(X n,p ) + yt(cp n ) = ( + x + x + + x t )T(X n,p ) + y(y + b ) n, which completes the proof. ¾ Corollary 3.3 T(G n,p,p ) = b n x n + n b k (b + y ) n k + (y + b ) n+ k= Proof By Theorem.5 taking p = t we get T(G n,p,p ) = ( + x + x + + x p )T(X n,p ) + y(y + b ) n { [ n ] } = b b n x p + b k (b + y ) n k + (b + y ) n + y(y + b ) n = b n x p + = b n x p + = b n x p + = b n x p + [ n k= ] b k+ (b + y ) n k + b(b + y ) n + y(y + b ) n k= [ n ] b k+ (b + y ) n k + (b + y)(b + y ) n k= [ n ] b k+ (b + y ) n k + (b + y + )(y + b ) n k= [ n ] b k+ (b + y ) n k + (b + y ) n + (y + b ) n+ k= = b n x p + b(b + y ) n + b (b + y ) n + + b n (b + y ) +(b + y ) n + (b + y ) n+ [ n ] = b n x p + b k+ (b + y ) n k + (y + b ) n+. k= Lemma 3.4 Let u, v be two vertices of a graph G which are joined by n disjoint paths of length p, namely P, P, P n such that degree of each of vertices in P, P, P n other than u, v is in G and removal of these n paths does not disconnect u and v, then T(G) = b n T(G ) + [ n b k (y + b ) ]T(G n k ), k=

43 38 Nithya Sai Narayana where G is obtained from G by removing the n disjoint paths of length p between u, v and identifying u, v and G is the graph is obtained by removing the n disjoint paths of length p from G. Hence Proof Using Theorem.3, T(G) = bt(g P ) + T(G.P ). But T(G.P ) = T(C n p G ) = (y + b ) n T(G ). T(G) = bt(g P ) + (y + b ) n T(G ) = b [ bt(g P P ) + (y + b ) n T(G ) ] + (y + b ) n T(G ) = b T(G P P ) + [ b(y + b ) n + (y + b ) n ] T(G ) = b T(G P P P 3 ) + [ b (y + b ) n 3 + b(y + b ) n +(y + b ) n ] T(G ) = b n T(G P P P n ) + [ b n (y + b ) + b n (y + b ) + = b n T(G ) + which completes the proof. +b(y + b ) n + (y + b ) n ] T(G ) [ n ] b k (y + b ) n k T(G ), k= ¾ From the above theorem we get another method of proving Theorem.4. [ n ] Corollary 3.5 T(X n,p ) = xb n + b k (y + b ) n k y. k= Proof Applying Lemma 3.5 to X n,p we get G = K and G = a single loop so that T(G ) = x, T(G ) = y and we obtain the result. ¾ Theorem 3.6 Let H (i) n,p denote a graph obtained by taking i copies of X n,p and joining it at a common vertex in succession. Then, H (i) n,p = [T(X n,p )] i. Proof Let e = uv be the common edge of the i th copy of X n,p. If G is the graph obtained by removing the n distinct paths of length p, then the resultant graph is a graph obtained by joining the i copies of X n,p with edge e = uv at the vertex u and hence T(G ) = xt(h (i ) n,p ). If G is obtained by removing n disjoint paths of length p between u, v and identifying u and v then, T(G ) = yt(h (i ) n,p ). Thus using Lemma 3.4

44 Tutte Polynomial of Generalized Flower Graphs 39 u v u v u Figure 4 H (i) n,p graph G and G segregation T(H (i) n,p ) = = xbn T(H n,p (i ) ) + y { xb n + y [ n k= = [T(X n,p )] T(H (i ) n,p ) [ n ] b k (y + b ) n k T(H n,p (i ) ) k= = [T(X n,p )] T(H (i ) n,p ), ]} b k (y + b ) n k T(H n,p (i ) ) [T(X n,p )] i T(H () n,p) = [T(X n,p )] i T(X n,p ) = [ T(X n,p ) i]. ¾ Corollary 3.7 Let G denote a graph obtained by taking i copies of X n,p and t copies of K and joining it in succession in any order then, T(G) = x t T(H (i) n,p) = x t [T(X n,p )] i. Theorem 3.8 Let G n,p,t denote a graph obtained by taking t copies of X n,p and taking product with C t in succession, then T(G n,p,t ) = (b n + αy) k (b n x + αy) t k b n t k= +(b n + αy) t αy(b + y ) n + (b n + αy) t y(b + y ) n, where α = [ n b k (y + b ) ]. n k k=

45 4 Nithya Sai Narayana Proof Let G be the graph obtained by removing n distinct paths of length p between the two vertices which are end points of common edge e of any copy of X n,p on the cycle C t and let G be the graph obtained by removing n distinct paths of length p as described for G and identifying the two end vertices of e in C t. Then by Lemma 3.4. Figure 5 G n,p,t graph G and G segregation T(G n,p,t ) = b n T(G ) + αt(g ), [ n where α = b k (y + b ) ]. n k Also using deletion contraction formula of Tutte polyno- k= mial T(G ) = yt(g n,p,t ) and T(G ) = T(H (t ) n,p ) + T(G n,p,t ) = T(X n,p ) t + T(G n,p,t ) from Theorem 3.6. But by Corollary 3.5, T(X n,p ) = xb n + [ n ] b k (y + b ) n k y = b n x + αy k= Thus, T(G ) = (b n x + αy) t + T(G n,p,t ) and T(G n,p,t ) = b n (b n x + αy) t + b n T(G n,p,t ) + αyt(g n,p,t ) = b n (b n x + αy) t + (b n + αy)t(g n,p,t ) = b n (b n x + αy) t + (b n + αy) [ b n (b n x + αy) t + (b n + αy)t(g n,p,t ) ] = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) + (b n + αy) T(G n,p,t ) = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) +(b n + αy) [ b n (b n x + αy) t 3 + (b n + αy)t(g n,p,t 3 ) ] = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) +b n (b n x + αy) t 3 (b n + αy) + (b n + αy) 3 T(G n,p,t 3 )

46 Tutte Polynomial of Generalized Flower Graphs 4 = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) +b n (b n x + αy) t 3 (b n + αy) + + (b n + αy) t T(G n,p,t (t ) ) = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) + b n (b n x + αy) t 3 (b n + αy) + + (b n + αy) t T(G n,p,) ). But T(G n,p, ) = b n T(X n,p ) + αy(b + y ) n + y(b + y ) n = b n (b n x + αy) + αy(b + y ) n + y(b + y ) n. Thus T(G n,p,t ) = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) +b n (b n x + αy) t 3 (b n + αy) + + b n (b n x + αy) (b n + αy) t 3 +(b n + αy) t [ b n (b n x + αy) + αy(b + y ) n + y(b + y ) n] = b n (b n x + αy) t + b n (b n x + αy) t (b n + αy) +b n (b n x + αy) t 3 (b n + αy) + +b n (b n x + αy) (b n + αy) t 3 + b n (b n x + αy)(b n + αy) t +(b n + αy) t αy(b + y ) n + (b n + αy) t y(b + y ) n t = b n (b n + αy) k (b n x + αy) t k k= +(b n + αy) t αy(b + y ) n + (b n + αy) t y(b + y ) n, which completes the proof. ¾ Theorem 3.9 T(G (i) n,p,t) = (+x+ +x t i )(b n x+αy) i +b n i (b n +αy) k (b n x+αy) i k + (b n + αy) i αy(b + y ) n + (b n + αy) i y(b + y ) n. Proof In G (i) n,p,t there are t i sides of C t without petals. Let e be any side of G (i) n,p,t without petal. Clearly e is neither a loop nor a bridge. Applying deletion and contraction formula G (i) n,p,t = x t i T(H (i) n,p ) + T(G(i) n,p,t ) = x t i T(H (i) n,p) + x t i T(H (i) n,p) + T(G (i) n,p,t ) k= = (x t i + x t i + + x t i (t i) )T(H (i) n,p ) + T(G(i) n,p,t (t i) ) = ( + x + + x t i )T(H (i) n,p + T(G (i) n,p,i ) = ( + x + + x t i )T(X i n,p ) + T(G n,p,i) i = ( + x + + x t i )(b n x + αy) i + b n (b n + αy) k (b n x + αy) i k k= +(b n + αy) i αy(b + y ) n + (b n + αy) i y(b + y ) n.

47 4 Nithya Sai Narayana 4. Conclusion Tutte Polynomial has been an open topic for research for mathematicians for the last 3 years. It is a two variable polynomial which reduces to many graph polynomials associated with the graph. It gives various information about the graph like the number of spanning trees, number of cyclic orientations not resulting in oriented cycles and colorability of graphs. In this research paper, Tutte polynomial of many specialized graphs have been studied in detail. The properties and Tutte polynomials of, generalized Book graph, Generalized Book graph with petals, Complete Generalized Book graph have been arrived at which in turn reveal various other related information by substituting appropriate values for the two variables. References [] Gary Haggard, David J. Pearce, and Gordon Royle Computing Tutte Polynomials, files/preprints/ ni94.pdf. [] Ira M. Gessel, Bruce E. Sagan, The Tutte polynomial of a graph, depth-first search, and simplicial complex partitions, Electronic J. Combin. (Foata Festschrift Volume), 3, (996) Volume 69, Issue 3, pages , March. [3] Joanna A. Ellis-Monaghan, Criel Merino, Graph Polynomials and Their Applications I: The Tutte Polynomial, Structural Analysis of Complex Networks,, pp 9-55, org/pdf/83.379v.pdf. Ser. B 84 (), pages [4] Martin Trinks, From spanning forests to edge subsets, ARS Mathematica Contemporanea, 7(4), pages 4-5. [5] A. R. Nizami, M. Munir, Ammara Usman, Saima Jabeen, M. Athar, Some specializations and Evaluations of Tutte Polynomial of a family of graphs, Asian Journal of Natural & Applied Sciences, Vol., No., June 3. [6] Mphako-Banda.E, Tutte Polynomials of Flower Graphs, Bulletin of the Iranian Mathematical Society, Vol. 35 No. (9), pp [7] Tutte W.T., Chromatic polynomials, in Selected Topics in Graph Theory(III), Academic Press, 988, p [8] Tutte W.T., A contribution to the theory of chromatic polynomials, Can. J. Math., 6, pages 8-9 (954)

48 International J.Math. Combin. Vol.(6), 43-5 Entire Equitable Dominating Graph B.Basavanagoud, V.R.Kulli and Vijay V. Teli 3 Department of Mathematics, Karnatak University, Dharwad , India Department of Mathematics, Gulbarga University, Gulbarga , India 3 Department of Mathematics, KLS s, Vishwanathrao Deshpande Rural, Institute of Technology, Haliyal , India b.basavanagoud@gmail.com, vrkulli@gmail.com, vijayteli@gmail.com Abstract: The entire equitable dominating graph EE qd(g) of a graph G with vertex set V S, where S is the collection of all minimal equitable dominating sets of G and two vertices u, v V S are adjacent if u, v are not disjoint minimal equitable dominating sets in S or u, v D, where D is the minimal equitable dominating set in S or u V and v is a minimal equitable dominating set in S containing u. In this paper, we initiate a study of this new graph valued function and also established necessary and sufficient conditions for EE qd(g) to be connected and complete. Other properties of EE qd(g) are also obtained. Key Words: Dominating set, equitable dominating set, entire equitable dominating graph, Smarandachely dominating set. AMS(): 5C69(7).. Introduction All graphs considered here are finite, undirected with no loops and multiple edges. We denote by p the order(i.e number of vertices) and by q the size (i.e number of edges) of such a graph G. Any undefined term and notation in this paper may be found in Harary [5]. A set of vertices which covers all the edges of a graph G is called vertex cover for G. The smallest number of vertices in any vertex cover for G is called its vertex covering number and is denoted by α (G) or α. A set of vertices in G is independent if no two of them are adjacent. The largest number of vertices in such a set is called the vertex independence number of G and is denoted by β (G) or β. The connectivity κ = κ(g) of a graph G is the minimum number of vertices whose removal results a disconnected or trivial graph. Analogously the edgeconnectivity λ = λ(g) is the minimum number of edges whose removal results a disconnected or trivial graph. The diameter of a connected graph is the maximum distance between two vertices in G and is denoted by diam(g). If G and H are graphs with the property that the identification of any vertex of G with an arbitrary vertex of H results in a unique graph (up to Supported by UGC-SAP DRS-II New Delhi, India: for -5 and the University Grants Commission, New Delhi, India. No.F.4-/6(BSR)/7-/7(BSR) dated: th June,. Received April 8, 5, Accepted May 3, 6.

49 44 B.Basavanagoud, V.R.Kulli and Vijay V. Teli isomorphism), then we write as G H for this graph. A subset D of V is called a dominating set of G if every vertex in V D is adjacent to at least one vertex in D. The domination number γ(g) of G is the minimum cardinality taken over all minimal dominating sets of G. (See Ore []). A subset D of V is called an equitable dominating set if for every v V D, there exists a vertex u D such that uv E(G) and deg(u) deg(v). The minimum cardinality of such a dominating set is called the equitable domination number of G and is denoted by γ e (G). For more details about graph valued functions, domination number and their related parameters we refer [-4, 6 -, ]. The opposite of equitable dominating set is the Smarandachely dominating set with deg(u) deg(v) for uv E(G). The purpose of this paper is to introduce a new graph valued function in the field of domination theory in graphs.. Entire Equitable Dominating Graph Definition. The entire equitable dominating graph EE q D(G) of a graph G with vertex set V S, where S is the collection of all minimal equitable dominating sets of G and two vertices u, v V S adjacent if u, v are not disjoint minimal equitable dominating sets in S or u, v D, where D is the minimal equitable dominating set in S or u V and v is a minimal equitable dominating set in S containing u. In Fig., a graph G and its entire equitable dominating graph EE q D(G) are shown. Here D = {, 3}, D = {, 4}, D 3 = {, 3} and D 4 = {, 4} are minimal equitable dominating sets of G. D G: 3 4 EE q D(G): 3 D D 3 4 Fig. D 4 3. Preliminary Results The following will be useful in the proof of our results. Theorem 3.([5]) For any nontrivial graph G, α + β = p = α + β. Theorem 3.([5]) A connected graph G is Eulerian if and only if every vertex of G has even degree.

50 Entire Equitable Dominating Graph Results First we obtain a necessary and sufficient condition on a graph G such that the entire equitable dominating graph EE q D(G) is connected. Theorem 4. For any graph G with at least three vertices, the entire equitable dominating graph EE q D(G) is connected if and only if (G) < p. Proof Let (G) < p and u, v be any two vertices in G. We consider the following cases: Case. If u and v are adjacent vertices in G, then there exist two not disjoint minimal equitable dominating sets D and D containing u and v respectively. Therefore by the definition., u and v are adjacent in EE q D(G). Case. Suppose there exist two vertices u D and v D such that u and v are not adjacent in G. Then there exists a minimal equitable dominating set D 3 containing both u and v and by definition., D and D are connected in EE q D(G). Conversely, suppose EE q D(G) is connected. Suppose (G) = p and u is a vertex of degree p. Then the degree of u in EE q D(G) is minimum. If every vertex of G has degree p, then every vertex of G forms a minimal equitable dominating set. Therefore EE q D(G) has at least two components, a contradiction. Thus (G) < p. ¾ Proposition 4. EE q D(G) = pk if and only if G = K p ; p. Proof Suppose G = K p ; p. Then clearly each vertex of G will form a minimal equitable dominating set. Hence by definition., EE q D(G) = pk. Conversely, suppose EE q D(G) = pk and G K p. Then there exists at least one minimal equitable dominating set D containing two vertices of G. Then D will form C 3 in EE q D(G), a contradiction. Hence G = K p ; p. ¾ Theorem 4. For any graph G, EE q D(G) is either connected or it has at least one component which is K. Proof If (G) < p, then by Theorem 4., EE q D(G) is connected. If G is complete graph K p ; p and by Proposition 4., then each component of EE q D(G) is K. Next, we must prove that δ(g) < (G) = p. Let v, v,, v n be the set of vertices in G such that deg(v i ) = p, then it is clear that {v i } forms a minimal equitable dominating set and which forms a component isomorphic to K. Hence EE q D(G) has at least one component which is K. ¾ In the next theorem, we characterize the graphs G for which EE q D(G) is complete. Theorem 4.3 EE q D(G) = K p+ if and only if G is K,p ; p 3. Proof Suppose G = K,p ; p 3. Then there exists a minimal equitable dominating set D

51 46 B.Basavanagoud, V.R.Kulli and Vijay V. Teli contains all the vertices of G i.e D = {u, v, v, v 3,, v p } = p+. Hence EE q D(G) = K p+. Conversely, EE q D(G) = K p+, then we prove that G is K,p ; p 3. Let us suppose that, G K,p ; p 3. Then there exists a minimal equitable dominating set D of cardinality is maximum p i.e D = {v, v, v 3,, v p } = p, a contradiction. Therefore G must be K,p ; p 3. ¾ Theorem 4.4 Let G be a nontrivial connected graph of order p and size q. The entire equitable dominating graph is a graph with order p and size p if and only if G = K p ; p. Proof Let G be a complete graph with p, then by Proposition 4., G = K p ; p. Conversely, suppose EE q D(G) be a (p, p) graph. Then pk is the only graph with order p and size q. ¾ In the next results, we obtain the bounds on the order and size of EE q D(G). Theorem 4.5 For any graph G, p p p(p ) +, where p denotes the number of vertices in EE q D(G). Further, the lower bound is attained if and only if G is either P 4 or K p ; p and upper bound is attained if and only if G is K 3 K, K 3 K or C 4 K. Proof The lower bound follows from the fact that the twice the number of vertices in G and the upper bound follows that the maximum number of edges in G. Suppose the lower bound is attained. Then every vertex of G forms a minimal equitable dominating set or every vertex of G is in exactly two minimal equitable dominating sets. This implies that the necessary condition. Conversely, suppose G is P 4 or K p ; p. Then by definition of entire equitable dominating graph, V (EE q D(G)) = p. If the upper bound is attained. Then G must be one of the following graphs are K 3 K, K 3 K or C 4 K. If G = K 3 K, then every vertex of G is in exactly two minimal equitable dominating sets hence V (EE q D(G)) = p(p ) + = pq +. Suppose G = K 3 K. Then the pendant vertex of G is in all the minimal equitable dominating sets and forms (p ) minimal equitable dominating sets. Therefore the upper bound holds. Now if G is C 4 K. Then every equitable dominating sets contains an isolated vertex and they are not disjoint sets and by definition.. Therefore upper bound holds. Conversely, suppose G is one of the following graphs K 3 K, K 3 K or C 4 K. Then it is obvious that V (EE q D(G)) = p(p ) +. ¾ Theorem 4.6 For any graph G, p q p(p+) +, where q denotes the number of edges in EE q D(G). Further, the lower bound is attained if and only G = K p and the upper bound is attained if and only if G is K 3 K. Proof The proof follows from Theorem 4.5. ¾

52 Entire Equitable Dominating Graph 47 In the next result, we find the diameter of EE q D(G). Theorem 4.7 Let G be any graph with (G) < p, then diam(ee q D(G)), where diam(g) is the diameter of G. Proof Let G be any graph with (G) < p, then by Theorem 4., EE q D(G) is connected. Let u, v be any arbitrary vertices in EE q D(G). We consider the following cases. Case. Suppose u, v V, u and v are nonadjacent in G. Then there exists a minimal equitable dominating set containing u and v and by definition., d EEqD(G)(u, v) =. If u and v are adjacent in G and there is no minimal equitable dominating set containing u and v, then there exists another vertex w V which is not adjacent to both u and v. Let D and D be two minimal equitable dominating sets containing (u, w) and (w, v) respectively. This implies that d EEqD(G)(u, v) =. Case. Suppose u V and v S. Then v = D is a minimal equitable dominating set of G. If u S, then u and v are adjacent in EE q D(G). Otherwise, there exists another vertex w D. This implies that d EEqD(G)(u, v) d EEqD(G)(u, w) + d EEqD(G)(w, v) =. Case 3. Suppose u, v S. Then u D and v D are two minimal equitable dominating sets of G and by Definition., d EEqD(G)(u, v) =. ¾ We now characterize graphs G for which SE q D(G) = EE q DG. A semientire equitable dominating graph SE q D(G) of a graph G is the graph with vertex set V S and two vertices u, v V S adjacent if u, v D, where D is a minimal equitable dominating set or u V and v = D is a minimal equitable dominating set containing u ([]). Proposition 4.([3]) The semientire equitable dominating graph SE q D(G) is pk if and only if G = K p ; p. Remark 4.([3]) For any graph G, SE q D(G) is a subgraph of EE q D(G). Theorem 4.8 For any graph G, SE q D(G) EE q D(G). Further, equality G, SE q D(G) = EE q D(G) if and only if G has exactly one minimal equitable dominating set containing all vertices of G. Proof By Remark 4., SE q D(G) EE q D(G). Suppose SE q D(G) = EE q D(G). Then by Theorem 4.3, D is the only minimal equitable dominating set contains all the vertices of G. Therefore G must be K,n ; n 3. The converse is obvious. In the next results, we discuss about α and β of EE q D(G). Theorem 4.9 For any graph G with no isolated vertices, () α (EE q D(G)) = S +, where S is the collection of all minimal equitable dominating ¾

53 48 B.Basavanagoud, V.R.Kulli and Vijay V. Teli sets of G; () β (EE q D(G)) = γ(g). Proof (i) Let G be graph of order p. Let S = {s, s, s i } be the set of all minimal equitable dominating sets. Then by definition. and Theorem??. Therefore the minimum number of vertices in EE q D(G) which covers all the edges. Hence α (EE q D(G)) = S +. (ii) By definition of EE q D(G), for any vertex v i ; i p of EE q D(G) are not adjacent. Hence these vertices forms a maximum independent set of EE q D(G). Hence (ii) follows. ¾ In the next two results, we prove the vertex connectivity and edge- connectivity of EE q D(G). Theorem 4. For any graph G, κ(ee q D(G)) = min{min(deg EEqD(G) i p v i ), min j n S j }, where S j s is the collection of all minimal equitable dominating sets of G. Proof Let G be any graph with order p and size q. We consider the following cases. Case. Let u v i (EE qd(g)) for some i, having the minimum degree among all v i in EE q D(G). If the degree of u is less than any other vertex in EE q D(G), then by deleting the vertices which are adjacent to u, results a disconnected graph. Case. Let v S j for some j, having the minimum degree among all S j s in EE q D(G). If degree of v is less than any other vertex in EE q D(G), then by deleting all the vertices which are adjacent to v. This results the graph is disconnected. Hence the result follows. ¾ Theorem 4. For any graph G, λ(ee q D(G)) = min{min(deg EEqD(G) i p v i ), min j n S j }, where S j s is the collection of all minimal equitable dominating sets of G. Proof Let G be any (p, q) graph. We consider two cases. Case. Let u v i (EE qd(g)), having minimum degree among all v i in EE qd(g). If the degree of u is less than any other vertex in EE q D(G), then by deleting those edges of EE q D(G) which are incident with u, results a disconnected graph. Case. Let v S j, having the minimum degree among all vertices of S j. If degree of v is less than any other vertex in EE q D(G), then by deleting those edges which are adjacent to v, results in a disconnected. Hence the result follows. ¾ Next, we prove the necessary and sufficient condition for EE q D(G) to be Eulerian. Theorem 4. For any graph G, EE q D(G) is Eulerian if and only if one of the following conditions are satisfied: () There exists a vertex u V is in all minimal equitable dominating sets and cardinality of every minimal equitable dominating set D of G is even; () If v V is a vertex of odd degree, then it is in odd number of minimal equitable dominating sets, otherwise it is in even number of minimal equitable dominating sets of G. Proof Suppose < p and by Theorem 4., EE q D(G) is connected. Suppose EE q D(G)

54 Entire Equitable Dominating Graph 49 is Eulerian. on the contrary if condition (i) is not satisfied, then there exists a minimal equitable dominating set contains odd number of vertices and does not contains a vertex of odd degree, a contradiction. Therefore by Theorem 3., EE q D(G) is Eulerian. Hence condition () holds. Suppose () does not hold. Then there exists v V of even degree which is in odd number of minimal equitable dominating sets, a contradiction. Hence (ii) hold. Conversely, suppose the conditions () and () are satisfied. Then every vertex of EE q D(G) has even degree and hence EE q D(G) is Eulerian. ¾ 5. Domination in EE q D(G) We calculate the domination number of EE q D(G) of some standard class of graphs. Theorem 5. For any graph G with no isolated vertices. () If G = K p ; p, then γ(ee q D(K p ) = p; () If G = K,p ; p 3, then γ(ee q D(K,p ) = ; (3) If G = C p, p 3, then γ(ee q D(C p ) =. Theorem 5. For any graph G, γ(ee q D(G)) =, if and only if G is K,p ; p 3. Proof If G is K,p ; p 3, then there exists a minimal equitable dominating set D contains all the vertices of G and by Theorem??, it is clear that, EE q D(G) is complete. Hence γ(ee q D(G)) =. Conversely, suppose γ(ee q D(G)) = and G K,p ; p 3. Then there exists a minimal dominating set D in EE q D(G) of cardinality greater than or equal to, a contradiction. Therefore G must be K,p ; p 3. ¾ We conclude this paper by exploring one open problem on EE q D(G). Problem. Give necessary and sufficient condition for a given graph G is entire equitable dominating graph of some graph. References [] B.Basavanagoud, V.R.Kulli and Vijay V.Teli, Equitable Dominating Graph, Int. J. of Mathematical Science & Engineering Applications, 9 (), (5), 9-4. [] B.Basavanagoud, V.R.Kulli and Vijay V.Teli, Equitable Total Domination in Graphs, Journal of Computer and Mathematical Sciences, 5 (), (4), [3] B.Basavanagoud, V.R.Kulli and Vijay V.Teli, Semientire Equitable Dominating Graph, Intenational Journal of Mathematical Combinatorics, 3 (4), [4] B.Basavanagoud, V.R.Kulli and Vijay V.Teli, Equitable Total Minimal Dominating Graph, International Research Journal of Pure Algebra, 3 (), (3), [5] F.Harary, Graph Theory, Addison-Wesley, Reading, Mass, (969).

55 5 B.Basavanagoud, V.R.Kulli and Vijay V. Teli [6] T.W.Haynes, S.T.Hedetniemi and P. J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, Inc., New York, (998). [7] T.W.Haynes, S.T.Hedetniemi and P.J.Slater, Domination in Graphs - Advanced Topics, Marcel Dekker, Inc., New York, (998). [8] V.R.Kulli, Theory of Domination in Graphs, Vishwa International Publications, Gulbarga, India (). [9] V.R.Kulli, Advances in Domination Theory I, ed. V.R.Kulli, Vishwa International Publications, Gulbarga, India (). [] V.R.Kulli, Advances in Domination Theory II, ed. V.R.Kulli, Vishwa International Publications, Gulbarga, India (3). [] V.R.Kulli, On Entire Equitable Dominating Transformation Graphs and Fuzzy Transformation Graphs, Journal of Computer and Mathematical Sciences, 6(8), (5) [] O.Ore, Theory of Graphs, Amer. Math. Soc. Colloq. Publ., 38, Providence, (96). [3] V.Swaminathan and K.M.Dharmalingam, Degree Equitable Domination on Graphs, Kragujevak Journal of Mathematics, 35(), 9-97 ().

56 International J.Math. Combin. Vol.(6), 5-64 Radio Mean Number of Certain Graphs R.Ponraj and S.Sathish Narayanan (Department of Mathematics, Sri Paramakalyani College, Alwarkurichi-674, India) Abstract: A radio mean labeling of a connected graph G is a one to one map f from the vertex set V (G) to the set of natural numbers N such that for each distinct vertices u and v of G, d(u, v) + f(u)+f(v) + diam (G). The radio mean number of f, rmn(f), is the maximum number assigned to any vertex of G. The radio mean number of G, rmn(g) is the minimum value of rmn(f) taken over all radio mean labeling f of G. In this paper we find the radio mean number of Jelly fish, subdivision of jelly fish, book with n pentagonal pages and K,n : m. Key Words: Radio mean number, subdivision of a graph, complete bipartite graph. AMS(): 5C78.. Introduction For standard terminology and notion we follow Harary [6] and Gallian [4]. Unless or otherwise mentioned, G = (V (G), E(G)) is a simple, finite, connected and undirected graph. A graph labeling is an assignment of integers to the vertices, or edges, or both, subject to certain conditios. Graph labeling used for several areas of science and few of them are communication network, coding theory, database management ete. In particular, radio labeling applied for channel assignment problem. The concept of radio labeling was introduced by Chatrand et al. [] in. Also in [, 3], radio number of several graphs were found. Motivated by the above labeling, Ponraj et al. [7] introduced the notion of radio mean labeling of G. A radio mean labeling is a one to one mapping f from V (G) to N satisfying the condition f (u) + f (v) d (u, v) + + diam (G) (.) for every u, v V (G). The span of a labeling f is the maximum integer that f maps to a vertex of Graph G. The radio mean number of G, rmn (G) is the lowest span taken over all radio mean labelings of the graph G. The condition. is called radio mean condition. In [7, 8, 9], they have found the radio mean number of some graphs like three diameter graphs, lotus inside a circle, gear graph, Helms, Sunflower graphs, subdivision of complete bipartite, corona of complete graph with path, one point union of cycle C 6 and wheel related graphs. In Received October 6, 5, Accepted May 5, 6.

57 5 R.Ponraj and S.Sathish Narayanan this article we find the radio mean number of Jelly fish, subdivision of jelly fish, book with n pentogonal pages and K,n : m. We write d(u, v) for the distance between the vertices u and v in G. The maximum distance between any pair of vertices is called the diameter of G and denoted by diam(g). Let x be any real number. Then x stands for smallest integer greater than or equal to x.. Main Results First we look into the Jelly fish graphs. Jelly fish graphs J (m, n) obtained from a cycle C 4 : uvxyu by joining x and y with an edge and appending m pendent edges to u and n pendent edges to v. Theorem. The radio mean number of a jelly fish graph J(m, n) is m + n + 4. Proof Let V (J(m, n)) = {u, v, x, y} {u i, v j : i m; j n} and E(J(m, n)) = {uy, yv, vx, xu, xy} {uu i, vv j : i m; j n}. It is clear that diam(j(m, n)) = 4. The vertex labeling of J(, ), J(, ) given in Figure shows that their radio mean numbers are 6, 7 respectively Figure Assume m and n 3. We define a vertex labeling f as follows. Assign the label to u. Then put the label to v, 3 to v and so on. In this sequence v n received the label n +. Then assign the label n + to u, n + 3 to u 3 and so on. Clearly label of u m is m + n. Then assign the labels m + n + 3, m + n +, m + n +, m + n + 4 respectively to the vertices u, v, x, y. Now we check the radio mean condition f (u) + f (v) d (u, v) + + diam (J(m, n)) for all u, v V (J(m, n)). It is easy to verify that the vertices u, v, x, y are mutually satisfies the radio mean condition. Case. Check the pair (u, u i ). f (u) + f (ui ) m + n d (u, u i )

58 Radio Mean Number of Certain Graphs 53 Case. Consider the pair (u, u i ). Case 3. Check the pair (x, u i ). Case 4. Verify the pair (x, v j ). f (u) + f (vj ) m + n d (u, v j ) f (x) + f (ui ) m + n + + d (x, u i ) f (x) + f (vj ) m + n + + d (x, v j ) Case 5. Consider the pair (y, u i ). Case 6. Check the pair (y, v j ). Case 7. Check the pair (v, v j ). Case 8. Verify the pair (v, u i ). f (y) + f (ui ) m + n d (y, u i ) f (y) + f (vj ) m + n d (y, v j ) f (v) + f (vj ) m + n + + d (v, v j ) f (v) + f (ui ) m + n + + d (v, u i ) Case 9. Consider the pair (u i, v j ). f (ui ) + f (v j ) + d (u i, v j ) Hence rmn(j(m, n)) = m + n + 4. Now, we find the radio mean number of subdivision of jelly fish graph. If x = uv is an edge of G and w is not a vertex of of G, then x is subdivided when it is replaced by the lines uw and wv. If every edges of G is subdivided, the resulting graph is the subdivision graph S(G). Theorem. For a subdivision of graph J m,n, 6 if m = n = rmn (S (J m,n )) = m + n + otherwise ¾

59 54 R.Ponraj and S.Sathish Narayanan Proof Let V (S(J m,n )) = {z i : i 9} {u i, u i : i m} {v j, v j : j n} and E(S(J m,n )) = {z i z i+ : i 7} {z 8 z, z 7 z 9, z 9 z 3 } {z u i, u i u i, z 5v j, v j v j : i m, j n}. Clearly diam(s(j m,n )) = 8. Case. m = n =. In this case should be a label of the vertex u or u or v or v. If not, is a label of any one of the remaining vertices, say x, and suppose is a label of any other vertex, say x. Then a contradiction. f(x) + f(x d(x, x ) + ) , Subcase. u receives the label. Then should be a label of v otherwise we get a contradiction as previously. For satisfying the radio mean condition, 3 should be a label of a vertex which is at least at a distance 6 from the vertex v and 7 from u, such a vertex doesn t exists. Therefore, in this case, rmn(s(j, )) 4. Subcase. u receives the label. Then should be a label of either v or v. Otherwise as in subcase a, the radio mean condition is not satisfied. If v or v receives the label then 3 can not be a label of any of the remaining vertices. Suppose 3 is a label of any other vertices, say x, then f(u d(u, x) + ) + f(x) 8. or or f(v d(v, x) + ) + f(x) 8. f(v ) + f(x) d(v, x) + 8, a contradiction. Thus, here also rmn(s(j, )) 4. By symmetry, the same case arises when v or v receives the label. Therefore in all the cases rmn(s(j,)) 4. Now we will try to label the vertices of S(J, ) with the property that the sum of the distance between the any pair of vertices and the mean value of labels of that pair of vertices exceeds the integer 9. We drop the label from the set of integers {,,, 3} and add a new label 4. Thus the labels are {, 3,, 4}. Suppose l, m, n are any three vertices of S(J, ) with their respective labels are, 3, 4. Then d(l, m) 6, d(l, n) 6 and d(m, n) 5. It is clear that, such type of vertices l, m, n doesn t exists. So rmn(s(j, )) 5. Now consider the labels from the set {3, 4,, 5}. Since the vertices with labels 3 and 4 are at least at a distance 5, any one of the vertices with these label should be a pendent vertex and the other is either z 6 or z 9 or z 4. Now suppose either 3 or 4 is a label of z 6 or z 4 then 5 can not be a label of any of the rest vertices. So 3 or

60 Radio Mean Number of Certain Graphs 55 4 should be a label of z 9. Suppose 3, 4 are the labels of u, z 9 then 5 should be the label of v. Then 6 can not be a label of any of the remaining vertices. The same problem arises when 4, 3 are the labels of u, z 9. By symmetry, if we assign the label 3 or 4 to the vertex v, then 6 can not be a label of any other vertices as discussed above. Hence rmn(s(j, )) 6. Consider the labeling given in Figure Figure From Figure, it is clear that rmn(s(j, )) 6. Hence rmn(s(j, )) = 6. Case. m, n. Subcase. m + n 4. As discussed in case, clearly it is not possible to label the vertices of S(J m,n ) from the sets {,,, m+n+9} and {,,, m+n+}. That is rmn(s(j m,n )) m+n+. The following Figure 3 shows that rmn(s(j m,n )) m + n + where m + n Figure 3 Hence rmn(s(j m,n )) = m + n + for m + n 4 and m, n. Subcase. m + n > 4. Define an injective map f : V (S(J m,n )) {,,, m + n + } by f(v ) = 3, f(v ) =

61 56 R.Ponraj and S.Sathish Narayanan m + n +, f(u i ) = i + 3, i m f(v i ) = m + + i, i n f(v n i+ ) = m + n + + i, i n f(u m i+ ) = m + n + + i, i m f(z 3 ) = m + n + 3, f(z ) = m + n + 4, f(z ) = m + n + 5, f(z 8 ) = m + n + 6, f(z 7 ) = m + n + 7, f(z 6 ) = m + n + 8, f(z 5 ) = m + n + 9, f(z 4 ) = m + n + and f(z 9 ) = m + n +. Now we check the radio mean condition that for every pair of vertices u, v V (S(J m,n )). Case. Consider the pair (z i, z j ). f(u) + f(v) d(u, v) + 9. f(zi ) + f(z j ) m + n m + n + 4 d(z i, z j ) Case. Check the pair (u i, u i ). d(u i, u i ) + f(ui ) + f(u i ) + Case 3. Check the pair (u i, u j), i j. m + n f(u d(u i, u j) + i ) + f(u j ) 4 + m + n Case 4. Examine the pair (u i, u j ). f(ui ) + f(u j ) m + n + + m + n + 3 d(u i, u j ) + +. Case 5. Verify the pair (u i, u j ). Case 6. Check the pair (u i, z j). f(u d(u i, u j ) + i ) + f(u j ) f(u d(u i, z j ) + i ) + f(z j ) m + n Case 7. Examine the pair (u i, z j ). f(ui ) + f(z j ) m + n m + n + d(u i, z j ) + +.

62 Radio Mean Number of Certain Graphs 57 Case 8. Verify the pair (u i, v j ). f(ui ) + f(v j ) m + n + + m + n + 3 d(u i, v j ) Case 9. Consider the pair (u i, v j ). d(u i, v j ) + f(ui ) + f(v j ) 7 + Case. Examine the pair (u i, v j ). m + n f(u d(v i, v j) i ) + f(v j + ) Case. Verify the pair (u i, v j), i j. f(u d(u i, v j) + i ) + f(v j ) 4 + m + n Case. Check the pair (v i, v j ), i j. f(vi ) + f(v j ) m + n m + n + 4 d(v i, v j ) + +. Case 3. Verify the pair (v i, v j ), i j. In this case, obviously m. f(v d(v i, v j ) + i ) + f(v j ) 4 + Case 4. Consider the pair (v i, v i ). Case 5. Check the pair (v i, v j ). d(v i, v i ) + f(vi ) + f(v i ) m m + n f(vi d(v i, v j) ) + f(v j + ) 3 + m + n Case 6. Verify the pair (v i, z j ), i j. f(vi ) + f(z j ) m + n m + n + 3 d(v i, z j ) + +. Case 7. Check the pair (v i, z j), i j. f(v d(v i, z j) + i ) + f(z j ) 3 + m + n

63 58 R.Ponraj and S.Sathish Narayanan Hence rmn(s(j m,n )) m + n + where m + n > 4. As in argument in case, rmn(s(j m,n )) m + n + for this case also. Hence rmn(s(j m,n )) = m + n + when m + n > 4. ¾ Next investigation is about book with n pentogonal pages. n copies of the cylce C 5 with one edge common is called book with n pentogonal pages. Theorem.3 The radio mean number of a book with n pentogonal pages, BP n, is 3n +. Proof Let V (BP n ) = {u, v} {u i, v i, w i : i n} and E(BP n ) {u, v} {uu i, u i w i, w i v i, v i v : i n}. Note that if n = diam(bp n ) = 4 otherwise For n =,, the labeling given in Figure 4 satisfies the radio mean condition Figure 4 For n 3, define an injective map f : V (BP n ) {,,, 3n + } by f(w i ) = i, f(v n i+ ) = n + i, f(u n i+ ) = n + i f(u) = 3n +, and f(v) = 3n + i n i n i n Now we check the condition f(u) + f(v) d(u, v) + 5 for every pair of vertices u, v V (BP n ). Case. Check the pair (u i, w i ). f(ui ) + f(w i ) 3n + d(u i, w i ) Case. Verify the pair (v i, w i ). f(vi ) + f(w i ) n + d(v i, w i )

64 Radio Mean Number of Certain Graphs 59 Case 3. Examine the pair (u i, v i ). f(ui ) + f(v i ) n + + n + d(u i, v i ) Case 4. Consider the pair (w i, u j ). f(wi ) + f(u j ) + n + d(w i, u j ) Case 5. Consider the pair (v i, w j ). Case 6. Verify the pair (u i, v j ). f(vi ) + f(w j ) n + + d(v i, w j ) f(ui ) + f(v j ) n + + n + d(u i, v j ) Case 7. Check the pair (w i, w j ). Case 8. Examine the pair (u i, u j ). f(wi ) + f(w j ) + d(w i, w j ) f(ui ) + f(u j ) n + + n + d(u i, u j ) + +. Case 9. Consider the pair (v i, v j ). Since f(vi ) + f(v j ) n + + n + d(v i, v j ) f(u)+f(x) = 3n++f(x) 6, the pair (u, x) for every x V (BP n ) satisfy the radio mean condition. Similarly the pair (v, y) for every y V (BP n ) also satisfy the condition. Hence rmn(bp n ) 3n +. Since f is injective, rmn(bp n ) = 3n +. ¾ The following result is used for the next theorem. Theorem.4([7]) Let G be a (p, q)-connected graph with diameter =. Then rmn(g) = p. Let K,n : m denotes the graph obtained by taking m disjoint copies of K,n and joining a new vertex to the centers of the m copies of K,n. Let V ( K,n : m ) = {v} {v i : i m} {u i j : i m, j n} and E( K,n : m ) = {vv i : i m} {v i u i j : i m, j n}.

65 6 R.Ponraj and S.Sathish Narayanan Theorem.5 For integers m, n, 6 if m =, n = rmn( K,n : m ) = mn + m + otherwise Proof First we observe that Case. m =. if m = diam( K,n : m ) = 4 otherwise In this case K,n : = K,n+, which is a -diameter graph and hence by Theorem.4, rmn( K,n : ) = n +. Case. m =. Subcase. n =. Since and are labels of the vertices which are at least at a distance 3, either or is a label of a pendent vertex. Assume that the label of u is. Then is a label of either v or u. Then 3 can not be a label of the remaining vertices. Similarly we can show that if is a label of u, is a label of either v or u and then 3 can not be a label of the remaining vertices. Hence rmn( K, : ) 6. Obviously, Figure 5 shows that rmn( K, : ) 6. Hence rmn( K, : ) = 6. Subcase. n Figure 5 Define an injective function f : V ( K,n : ) {,,, n + 3} by f(u ) =, f(v) = n +, f(v ) = n + 3, f(v ) = n +, f(u i ) = i +, i n f(u i ) = n + i, n. We now check whether the labeling f is a required labeling. It is easy to check that the pairs (v, v ), (v, v) and (v, v) satisfy the radio mean condition. Subcase. Check the pair (u i, u j ), i j. f(u d(u i,u i ) + f(u j j) + )

66 Radio Mean Number of Certain Graphs 6 Subcase. Verify the pair (u i, u j ). f(u d(u i, u j ) + i ) + f(u j ) 4 + Subcase 3. Consider the pair (u i, u j ), i j f(u d(u i, u i ) + f(u j j) + ) + n Subcase 4. Examine the pair (u i, v ). f(u d(u i, v ) + i ) + f(v ) + n Subcase 5. Check the pair (u i, v). f(u d(u i, v) + i ) + f(v) + Subcase 6. Consider the pair (u i, v ). + n + 5. f(u d(u i, v ) + i ) + f(v ) + n Subcase 7. Verify the pair (u i, v ). f(u d(u i, v ) + i ) + f(v ) + n Subcase 8. Check the pair (u i, v). f(u d(u i, v) + i ) + f(v) + Subcase 9. Examine the pair (u i, v ). + n + 6. f(u d(u i, v ) + i ) + f(v ) + n Therefore, rmn( K,n : ) n + 3. But rmn( K,n : ) n + 3 and hence rmn( K,n : ) = n + 3. Case 3. m 3. For n =, m = 3, Figure 6 shows that rmn( K, : 3 ) = 7. Now we consider the cases n, m = 3 and n, m 4. Define a function f : V ( K,n : m ) {,,, mn + m + }

67 6 R.Ponraj and S.Sathish Narayanan by f(v) = mn +, f(u i j ) = (j )m + i, i m, j n f(v i ) = mn + + i, i m Figure 6 We show that f is a valid radio mean labeling. Subcase. Check the pair (u j i, uj k ). d(u j i, uj k ) + f(u j i ) + f(uj k ) + + m + 5. Subcase. Consider the pair (u j i, ur k ), j r. d(u j i, ur k ) + f(u j i ) + f(ur k ) Subcase 3. Verify the pair (v i, v j ). f(vi ) + f(v j ) mn + + mn + 3 d(v i, v j ) Subcase 4. Examine the pair (v, u j i ). d(v, u j i ) + Subcase 5. Verify the pair (v, v i ). f(v) + f(u j i ) + mn f(v) + f(vi ) mn + + mn + d(v, v i ) Subcase 6. Check the pair (v i, u i j ). For n and m = 3, f(vi d(v i, u i ) + f(u i j j) + ) mn

68 Radio Mean Number of Certain Graphs 63 If n and m 4 then, f(vi d(v i, u i ) + f(u i j j) + ) mn Subcase 7. Consider the pair (v i, u k j ), i k. d(v i, u k j ) + f(vi ) + f(u k j ) 3 + mn Hence rmn( K,n : m ) = mn + m +. ¾ Example.6 A radio mean labeling of K,5 : 4 is given in Figure Figure 7 References [] Chartrand, Gray and Erwin, David and Zhang, Ping and Harary, Frank, Radio labeling of graphs, Bull. Inst. Combin. Appl., 33(), [] Daphne Der-Fen Liu, Radio number for trees. Discrete Math., 38(7)(8), [3] Daphne Der-Fen Liu and Xuding Zhu. Multilevel distance labelings for paths and cycles. SIAM J. Discrete Math., 9(3)(5), 6-6. [4] J.A.Gallian, A Dynamic survey of graph labeling, The Electronic Journal of Combinatorics, 9 () #Ds6. [5] Gutman, Distance of thorny graphs, Publ. Inst. Math. (Beograd), 63 (998), [6] F.Harary, Graph Theory, Addision Wesley, New Delhi (969). [7] R.Ponraj, S.Sathish Narayanan and R.Kala, Radio mean labeling of a graph, AKCE International Journal of Graphs and Combinatorics, (5), 4-8.

69 64 R.Ponraj and S.Sathish Narayanan [8] R.Ponraj, S.Sathish Narayanan and R. Kala, On radio mean number of some graphs, International journal of Mathematical Combinatorics, 3(4), [9] R.Ponraj, S.Sathish Narayanan and R. Kala, Radio mean number of some wheel related graphs, Jordan Journal of Mathematics and Statistics (JJMS), 7(4)(4), [] R.Ponraj, and S.Sathish Narayanan, Radio mean number of some subdivision graphs, Jordan Journal of Mathematics and Statistics (JJMS), 9()(6),

70 International J.Math. Combin. Vol.(6), 65-8 Some Results on Vertex Version and Edge Versions of Modified Schultz Index Mahdieh Azari (Department of Mathematics, Kazerun Branch, Islamic Azad University, P. O. Box: , Kazerun, Iran) Abstract: Let G and G be two simple connected graphs with disjoint vertex sets V (G ) and V (G ), respectively. For given vertices a V (G ) and a V (G ), a splice of G and G by vertices a and a is defined by identifying the vertices a and a in the union of G and G and a link of G and G by vertices a and a is obtained by joining a and a by an edge in the union of these graphs. The modified Schultz index of a simple connected graph G is defined as the sum of the terms d(u G)d(v G)d(u, v G) over all unordered pairs {u, v} of vertices in G, where d(u G) and d(u, v G) denote the degree of the vertex u of G and the distance between the vertices u and v of G, respectively. In this paper, explicit formulas for computing the vertex and edge-modified Schultz indices of splice and link of graphs are presented. Key Words: Distance, vertex-degree, modified Schultz index, edge-modified Schultz index, splice, link. AMS(): 5C7, 5C, 5C76.. Introduction In this paper, we consider connected finite graphs without any loops or multiple edges. A topological index Top(G) of a graph G is a real number with the property that for every graph H isomorphic to G, T op(h) = T op(g). There are numerous topological indices that have been found to be useful in chemical documentation, isomer discrimination, quantitative structure-property relationships (QSPR), quantitative structure-activity relationships (QSAR), and pharmaceutical drug design [7, ]. The Wiener index is the first reported distance-based topological index which was introduced in 947 by Wiener [9, ] who used it for modeling the shape of organic molecules and for calculating several of their physico-chemical properties. The Wiener index of a graph G is defined as the sum of distances between all pairs of vertices of G, W(G) = d(u, v G), {u,v} V (G) where d(u, v G) denotes the distance between the vertices u and v of G which is defined as the Received May 4, 5, Accepted May 6, 6.

71 66 Mahdieh Azari length of any shortest path in G connecting them. The molecular topological index or Schultz index [7] was introduced by Harry Schultz in 989. The Schultz index of a graph G is defined as S(G) = {u,v} V (G) where d(u G) denotes the degree of the vertex u of G. [ d(u G) + d(v G) ] d(u, v G), The vertex version of the modified Schultz index [9]of a graph G was introduced by Ivan Gutman in 994 as S (G) = d(u G)d(v G)d(u, v G). {u,v} V (G) The modified Schultz index is also known as Gutman index. The edge versions of the modified Schultz index [5] were introduced by Khormali et al. in. Two possible distances between the edges e = uv and f = zt of a graph G can be considered. The first distance is denoted by d (e, f G) and defined as d (e, f G) + e f, d (e, f G) = e = f, where d (e, f G) = min{d(u, z G), d(u, t G), d(v, z G), d(v, t G)}. It is easy to see that, d (e, f G) = d(e, f L(G)), where L(G) is the line graph of G. The second distance is denoted by d 4 (e, f G) and defined as d (e, f G) e f, d 4 (e, f G) = e = f, where d (e, f G) = max{d(u, z G), d(u, t G), d(v, z G), d(v, t G)}. Related to the distances d and d 4, two edge versions of the modified Schultz index can be defined. The first and second edge-modified Schultz indices of G are denoted by (W ) e (G) and (W ) e4 (G), respectively and defined as (W ) ei (G) = {e,f} E(G) d(e G)d(f G)d i (e, f G), i {, 4}, where d(e G) denotes the degree of the edge e in G which is the degree of the vertex e in the line graph L(G). For more information on the edge-modified Schultz indices, see [4]. In this paper, we compute the vertex version and edge versions of the modified-schultz index for splice and link of graphs. Readers interested in more information on computing topological indices of splice and link of graphs can be referred to [ C 6, 8,, 3, 6, 8].. Definitions and Preliminaries Let G be a graph with vertex set V (G) and edge set E(G). We denote by V (e) the set of two

72 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 67 end-vertices of the edge e of G. For u V (G) and e = ab E(G), we define D (u, e G) = min{d(u, a G), d(u, b G)}, D (u, e G) = max{d(u, a G), d(u, b G)}. Note that, D (u, e G) is a nonnegative integer and D (u, e G) = if and only if u V (e). Also, D (u, e G) is a positive integer and D (u, e G) = if and only if u V (e) or u and the end vertices of e form a 3-cycle in G. For u V (G), let N(u G) denote the set of all first neighbors of u in G. We denote by δ(u G), the sum of degrees of all neighbors of u in G, i.e., δ(u G) = v N(u G) d(v G). We denote by M (G), the first Zagreb index [] of G which is defined as M (G) = u V (G) d(u G). The first Zagreb index can also be expressed as a sum over edges of G, M (G) = uv E(G) [ d(u G) + d(v G) ]. Let e be an edge of G with V (e) = {a, b}. It is easy to see that, d(e G) = d(a G) + d(b G). Therefore, d(e G) = M (G) E(G). Also for u V (G), we have {e,f} E(G);u V (e) V (f) e E(G) e E(G);u V (e) d(e G) = d(u G) ( d(u G) ) + δ(u G), [ d(e G) + d(f G) ] = ( d(u G) )[ d(u G)(d(u G) ) + δ(u G) ]. 3. Results and Discussion In this section, we compute the vertex version and edge versions of the modified Schultz index for splice and link of graphs. 3. Splice Let G and G be two graphs with disjoint vertex sets V (G ) and V (G ) and edge sets E(G ) and E(G ), respectively. For given vertices a V (G ) and a V (G ), a splice [8] of G and G by vertices a and a is denoted by (G.G )(a, a ) and defined by identifying the vertices

73 68 Mahdieh Azari a and a in the union of G and G. We denote by n i, m i and α i, the order and size of the graph G i and the degree of a i in G i, respectively, where i {, }. It is easy to see that, V ( (G.G )(a, a ) ) = n + n and E ( (G.G )(a, a ) ) = m + m. In the following lemmas, the degree of an arbitrary vertex of (G.G )(a, a ) and the distance between two arbitrary vertices of this graph are computed. The results follow easily from the definition of the splice of graphs, so their proofs are omitted. Lemma 3. Let u V ( (G.G )(a, a ) ). Then d ( u (G.G )(a, a ) ) d(u G ) u V (G ) \ {a }, = d(u G ) u V (G ) \ {a }, α + α u = a or u = a. Lemma 3. Let u, v V ( (G.G )(a, a ) ). Then d ( u, v (G.G )(a, a ) ) d(u, v G ) u, v V (G ), = d(u, v G ) u, v V (G ), d(u, a G ) + d(a, v G ) u V (G ), v V (G ). In the following theorem, the modified Schultz index of (G.G )(a, a ) is computed. Theorem 3.3 The modified Schultz index of G = (G.G )(a, a ) is given by S (G) =S (G ) + S (G ) + m d(u G )d(u, a G ) + m u V (G )\{a } u V (G )\{a } d(u G )d(u, a G ). Proof We partition the sum in the formula of S (G) into three sums as follows. The first sum S consists of contributions to S (G) of pairs of vertices from G. Using Lemmas 3. and 3., we obtain S = d(u G )d(v G )d(u, v G ) {u,v} V (G )\{a } + ( ) α + α d(u G )d(u, a G ) u V (G )\{a } =S (G ) + α d(u G )d(u, a G ). u V (G )\{a } The second sum S consists of contributions to S (G) of pairs of vertices from G. Similar

74 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 69 to the previous case, we obtain S = S (G ) + α u V (G )\{a } d(u G )d(u, a G ). The third sum S 3 is taken over all pairs {u, v} of vertices in G such that u V (G ) \ {a } and v V (G ) \ {a }. Using Lemmas 3. and 3., we obtain S 3 = u V (G )\{a } v V (G )\{a } = ( m α ) u V (G )\{a } + ( m α ) u V (G )\{a } d(u G )d(v G ) [ d(u, a G ) + d(a, v G ) ] d(u G )d(u, a G ) d(u G )d(u, a G ). The formula of S (G) is obtained by adding S, S and S 3 and simplifying the resulting expression. ¾ In the following lemmas, the degree of an arbitrary edge of (G.G )(a, a ) and the distances d and d 4 between two arbitrary edges of this graph are computed. The results are direct consequences of Lemmas 3. and 3., respectively, so their proofs are omitted. Lemma 3.4 Let e E ( (G.G )(a, a ) ). Then d(e G ) d ( e (G.G )(a, a ) ) d(e G ) + α = d(e G ) d(e G ) + α e E(G ), a / V (e), e E(G ), a V (e), e E(G ), a / V (e), e E(G ), a V (e). Lemma 3.5 Let G = (G.G )(a, a ) and e, f E ( G). Then d (e, f G ) e, f E(G ), (i) d (e, f G) = d (e, f G ) e, f E(G ), D (a, e G ) + D (a, f G ) + e E(G ), f E(G ), d 4 (e, f G ) e, f E(G ), (ii) d 4 (e, f G) = d 4 (e, f G ) e, f E(G ), D (a, e G ) + D (a, f G ) e E(G ), f E(G ). In the following theorem, the first and second edge-modified Schultz indices of (G.G )(a, a ) are computed. Theorem 3.6 The first and second edge-modified Schultz indices of G = (G.G )(a, a ) are

75 7 Mahdieh Azari given by (i) (W ) e (G) = (W ) e (G ) + (W ) e (G ) + α d(f G )d (e, f G ) e,f E(G );a V (e)\v (f) + α e,f E(G );a V (e)\v (f) + ( M (G ) m + α α ) d(f G )d (e, f G ) e E(G ) + ( M (G ) m + α α ) f E(G ) d(e G )D (a, e G ) d(f G )D (a, f G ) + ( )( ) M (G ) m M (G ) m + α (α )δ(a G ) + α (α )δ(a G ) [ + α α M (G ) + M (G ) + ( ) 7( ) α + α α + α (m + m ) ]. (ii) (W ) e4 (G) = (W ) e4 (G ) + (W ) e4 (G ) + α d(f G )d 4 (e, f G ) e,f E(G );a V (e)\v (f) + α e,f E(G );a V (e)\v (f) + ( M (G ) m + α α ) d(f G )d 4 (e, f G ) e E(G ) + ( M (G ) m + α α ) + α {u,v} N(a G ) + α {u,v} N(a G ) + α ( α + α 4 ) f E(G ) d(e G )D (a, e G ) d(f G )D (a, f G ) [ d(u G ) + d(v G ) ] d(u, v G ) [ d(u G ) + d(v G ) ] d(u, v G ) {u,v} N(a G ) + α ( α + α 4 ) {u,v} N(a G ) d(u, v G ) d(u, v G ) + α α [ M (G ) + M (G ) (m + m α α ) ]. Proof We prove the first part of the theorem. The second part can be proved by a similar method. At first, we partition the sum in the formula of (W ) e (G) into three sums as follows. The first sum S consists of contributions to (W ) e (G) of pairs of edges from G. In order to compute S, we partition it into three sums S, S and S 3 as follows. The sum S is equal to S = {e,f} E(G );a / V (e) V (f) d(e G)d(f G)d (e, f G).

76 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 7 Using Lemmas 3.4 and 3.5, we obtain S = The sum S is equal to S = {e,f} E(G );a / V (e) V (f) {e,f} E(G );a V (e) V (f) d(e G )d(f G )d (e, f G ). d(e G)d(f G)d (e, f G). Note that, for every pairs {e, f} of edges in G such that a V (e) V (f), d (e, f G ) =. Now, using Lemmas 3.4 and 3.5, we obtain S = = The sum S 3 is equal to {e,f} E(G );a V (e) V (f) {e,f} E(G );a V (e) V (f) ( d(e G ) + α )( d(f G ) + α ) d(e G )d(f G ) + α (α ) [ α (α ) + δ(a G ) ] + α S 3 = e,f E(G );a V (e)\v (f) Using Lemmas 3.4 and 3.5, we obtain S 3 = = e,f E(G );a V (e)\v (f) e,f E(G );a V (e)\v (f) + α e,f E(G );a V (e)\v (f) By adding S, S and S 3, we obtain S =(W ) e (G ) + α + α ( α e,f E(G );a V (e)\v (f) ( α ). d(e G)d(f G)d (e, f G). ( d(e G ) + α ) d(f G )d (e, f G ) d(e G )d(f G )d (e, f G ) d(f G )d (e, f G ). ) + α (α ) [ α (α ) + δ(a G ) ] d(f G )d (e, f G ). The second sum S consists of contributions to (W ) e (G) of pairs of edges from G.

77 7 Mahdieh Azari Using the same argument as in the computation of S, we obtain S =(W ) e (G ) + α + α ( α e,f E(G );a V (e)\v (f) ) + α (α ) [ α (α ) + δ(a G ) ] d(f G )d (e, f G ). The third sum S 3 is taken over all pairs {e, f} of edges in G such that e E(G ) and f E(G ). In order to compute S 3, we partition it into four sums S 3, S 3, S 33 and S 34 as follows. The sum S 3 is equal to S 3 = d(e G)d(f G)d (e, f G). e E(G );a / V (e) f E(G );a / V (f) Using Lemmas 3.4 and 3.5, we obtain S 3 = d(e G )d(f G ) [ D (a, e G ) + D (a, f G ) + ] e E(G );a / V (e) f E(G );a / V (f) = [ M (G ) m α (α ) δ(a G ) ] d(e G )D (a, e G ) e E(G ) + [ M (G ) m α (α ) δ(a G ) ] + [ M (G ) m α (α ) δ(a G ) ] f E(G ) d(f G )D (a, f G ) [ M (G ) m α (α ) δ(a G ) ]. The sum S 3 is equal to S 3 = d(e G)d(f G)d (e, f G). e E(G );a / V (e) f E(G );a V (f) Using Lemmas 3.4 and 3.5, we obtain S 3 = d(e G ) ( )[ d(f G ) + α D (a, e G ) + ] e E(G );a / V (e) f E(G );a V (f) = [ ][ α (α ) + δ(a G ) + α α d(e G )D (a, e G ) e E(G ) + M (G ) m α (α ) δ(a G ) ]. The sum S 33 is equal to S 33 = d(e G)d(f G)d (e, f G). e E(G );a V (e) f E(G );a / V (f)

78 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 73 Using the same argument as in the computation of S 3, we obtain S 33 = [ ][ α (α ) + δ(a G ) + α α d(f G )D (a, f G ) f E(G ) + M (G ) m α (α ) δ(a G ) ]. The sum S 34 is equal to S 34 = d(e G)d(f G)d (e, f G). e E(G );a V (e) f E(G );a V (f) Using Lemmas 3.4 and 3.5, we obtain ( )( ) S 34 = d(e G ) + α d(f G ) + α e E(G );a V (e) f E(G );a V (f) = [ α (α ) + δ(a G ) + α α ][ α (α ) + δ(a G ) + α α ]. By adding S 3, S 3, S 33 and S 34, we obtain S 3 = ( ) M (G ) m + α α d(e G )D (a, e G ) e E(G ) + ( ) M (G ) m + α α d(f G )D (a, f G ) f E(G ) + ( M (G ) m )( M (G ) m ) + α α [ M (G ) m + M (G ) m + α α ]. The formula of (W ) e (G) is obtained by adding S, S and S 3 and simplifying the resulting expression. ¾ 3. Link Let G and G be two graphs with disjoint vertex sets V (G ) and V (G ) and edge sets E(G ) and E(G ), respectively. For vertices a V (G ) and a V (G ), a link [8] of G and G by vertices a and a is denoted by (G G )(a, a ) and obtained by joining a and a by an edge in the union of these graphs. We denote by n i, m i and α i, the order and size of the graph G i and the degree of a i in G i, respectively, where i {, }. It is easy to see that, V ( (G G )(a, a ) ) = n + n and E ( (G G )(a, a ) ) = m + m +. In the following lemmas, the degree of an arbitrary vertex of (G G )(a, a ) and the distance between two arbitrary vertices of this graph are computed. The results follow easily from the definition of the link of graphs, so their proofs are omitted.

79 74 Mahdieh Azari Lemma 3.7 Let u V ( (G G )(a, a ) ). Then d(u G ) u V (G ) \ {a }, d ( u (G G )(a, a ) ) d(u G ) u V (G ) \ {a }, = α + u = a, α + u = a. Lemma 3.8 Let u, v V ( (G G )(a, a ) ). Then d ( u, v (G G )(a, a ) ) d(u, v G ) u, v V (G ), = d(u, v G ) u, v V (G ), d(u, a G ) + d(a, v G ) + u V (G ), v V (G ). In the following theorem, the modified Schultz index of (G G )(a, a ) is computed. Theorem 3.9 The modified Schultz index of G = (G G )(a, a ) is given by S (G) =S (G ) + S (G ) + (m + ) d(u G )d(u, a G ) + (m + ) u V (G )\{a } u V (G )\{a } d(u G )d(u, a G ) + (m + m + m m ) +. Proof We partition the sum in the formula of S (G) into three sums as follows. The first sum S consists of contributions to S (G) of pairs of vertices from G. By Lemmas 3.7 and 3.8, we obtain S = d(u G )d(v G )d(u, v G ) {u,v} V (G )\{a } + ( α + ) d(u G )d(u, a G ) u V (G )\{a } =S (G ) + d(u G )d(u, a G ). u V (G )\{a } The second sum S consists of contributions to S (G) of pairs of vertices from G. Similar to the previous case, we obtain S = S (G ) + d(u G )d(u, a G ). u V (G )\{a } The third sum S 3 is taken over all pairs {u, v} of vertices in G such that u V (G ) and v V (G ). In order to compute S 3, we partition it into four sums S 3, S 3, S 33 and S 34 as

80 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 75 follows. The sum S 3 is taken over all pairs {u, v} of vertices in G such that u V (G ) \ {a } and v = a. Using Lemmas 3.7 and 3.8, we obtain S 3 = ( α + ) u V (G )\{a } = ( α + )[ u V (G )\{a } d(u G ) [ d(u, a G ) + ] d(u G )d(u, a G ) + m α ]. The sum S 3 is taken over all pairs {u, v} of vertices in G such that u V (G ) \ {a } and v = a. Similar to the previous case, we obtain S 3 = ( α + )[ u V (G )\{a } d(u G )d(u, a G ) + m α ]. The sum S 33 is taken over all pairs {u, v} of vertices in G such that u V (G ) \ {a } and v V (G ) \ {a }. By Lemmas 3.7 and 3.8, we obtain S 33 = u V (G )\{a } v V (G )\{a } = ( m α ) u V (G )\{a } + ( m α ) v V (G )\{a } d(u G )d(v G ) [ d(u, a G ) + d(a, v G ) + ] d(u G )d(u, a G ) d(v G )d(a, v G ) + ( m α )( m α ). The sum S 34 is equal to S 34 = d(a G)d(a G)d(a, a G) = ( α + )( α + ). By adding S 3, S 3, S 33 and S 34, we obtain S 3 = ( m + ) d(u G )d(u, a G ) u V (G )\{a } + ( m + ) d(v G )d(a, v G ) v V (G )\{a } + (m + m + m m ) +. Now, the formula of S (G) is obtained by adding S, S and S 3 and simplifying the resulting expression. ¾ In the following lemmas, the degree of an arbitrary edge of (G G )(a, a ) and the distances d and d 4 between two arbitrary edges of this graph are computed. The results are direct consequences of Lemmas 3.7 and 3.8, respectively, so their proofs are omitted.

81 76 Mahdieh Azari Lemma 3. Let e E ( (G G )(a, a ) ). Then d(e G ) e E(G ), a / V (e), d(e G ) + e E(G ), a V (e), d ( e (G G )(a, a ) ) = d(e G ) e E(G ), a / V (e), d(e G ) + e E(G ), a V (e), α + α e = a a. Lemma 3. Let G = (G G )(a, a ) and e, f E(G). Then d (e, f G ) e, f E(G ), d ( (e, f G ) e, f E(G ), (i) d e, f G) = D (a, e G ) + e E(G ), f = a a, D (a, e G ) + e E(G ), f = a a, D (a, e G ) + D (a, f G ) + e E(G ), f E(G ), d 4 (e, f G ) e, f E(G ), d ( 4 (e, f G ) e, f E(G ), (ii) d 4 e, f G) = D (a, e G ) + e E(G ), f = a a, D (a, e G ) + e E(G ), f = a a, D (a, e G ) + D (a, f G ) + e E(G ), f E(G ). In the following theorem, the first and second edge-modified Schultz indices of (G G )(a, a ) are computed. Theorem 3. The first and second edge-modified Schultz indices of G = (G G )(a, a ) are given by (i) (W ) e (G) = (W ) e (G ) + (W ) e (G ) + d(f G )d (e, f G ) e,f E(G );a V (e)\v (f) + d(f G )d (e, f G ) e,f E(G );a V (e)\v (f) + ( ) M (G ) m + α + α d(e G )D (a, e G ) e E(G ) + ( ) M (G ) m + α + α d(f G )D (a, f G ) f E(G ) + ( )( ) M (G ) m M (G ) m + (α )δ(a G ) + (α )δ(a G ) + (3α + α ) ( ) M (G ) m + (3α + α ) ( ) M (G ) m ( ) ( ) + α 3 + α 3 α α + 4α α 3 3.

82 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 77 (ii) (W ) e4 (G) = (W ) e4 (G ) + (W ) e4 (G ) + + e,f E(G );a V (e)\v (f) d(f G )d 4 (e, f G ) + ( M (G ) m + α + α ) e E(G ) + ( M (G ) m + α + α ) + + {u,v} N(a G ) {u,v} N(a G ) + (α 3) f E(G ) [ d(u G ) + d(v G ) ] d(u, v G ) [ d(u G ) + d(v G ) ] d(u, v G ) {u,v} N(a G ) d(u, v G ) + (α 3) e,f E(G );a V (e)\v (f) d(e G )D (a, e G ) d(f G )D (a, f G ) {u,v} N(a G ) + ( M (G ) m )( M (G ) m ) + (3α + α ) ( M (G ) m ) + (3α + α ) ( M (G ) m ) + (α + α ) + 3α α. d(f G )d 4 (e, f G ) d(u, v G ) Proof We prove the second part of the theorem. The first part can be proved by a similar method. At first, we partition the sum in the formula of (W ) e4 (G) into three sums as follows. The first sum S consists of contributions to (W ) e4 (G) of pairs of edges from G. In order to compute S, we partition it into three sums S, S and S 3 as follows. The sum S is equal to S = d(e G)d(f G)d 4 (e, f G). {e,f} E(G );a / V (e) V (f) Using Lemmas 3. and 3., we obtain S = d(e G )d(f G )d 4 (e, f G ). {e,f} E(G );a / V (e) V (f) The sum S is equal to S = d(e G)d(f G)d 4 (e, f G). {e,f} E(G );a V (e) V (f) Using Lemmas 3. and 3., we obtain

83 78 Mahdieh Azari S = = {e,f} E(G );a V (e) V (f) {e,f} E(G );a V (e) V (f) + {u,v} N(a G ) + (α 3) The sum S 3 is equal to S 3 = ( d(e G ) + )( d(f G ) + ) d 4 (e, f G ) d(e G )d(f G )d 4 (e, f G ) [ d(u G ) + d(v G ) ] d(u, v G ) {u,v} N(a G ) e,f E(G );a V (e)\v (f) Using Lemmas 3. and 3., we obtain S 3 = = e,f E(G );a V (e)\v (f) e,f E(G );a V (e)\v (f) + e,f E(G );a V (e)\v (f) By adding S, S and S 3, we obtain S =(W ) e4 (G ) + + (α 3) {u,v} N(a G ) {u,v} N(a G ) d(u, v G ). d(e G)d(f G)d 4 (e, f G). ( d(e G ) + ) d(f G )d 4 (e, f G ) d(e G )d(f G )d 4 (e, f G ) d(f G )d 4 (e, f G ). [ d(u G ) + d(v G ) ] d(u, v G ) d(u, v G ) + e,f E(G );a V (e)\v (f) d(f G )d 4 (e, f G ). The second sum S consists of contributions to (W ) e4 (G) of pairs of edges from G. Using the same argument as in the computation of S, we obtain S =(W ) e4 (G ) + + (α 3) {u,v} N(a G ) {u,v} N(a G ) [ d(u G ) + d(v G ) ] d(u, v G ) d(u, v G ) + e,f E(G );a V (e)\v (f) d(f G )d 4 (e, f G ). The third sum S 3 is taken over all pairs {e, f} of edges in G such that e E(G ) and f = a a. In order to compute S 3, we partition it into two sums S 3 and S 3 as follows. The sum S 3 is equal to S 3 = e E(G );a / V (e),f=a a d(e G)d(f G)d 4 (e, f G).

84 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 79 Using Lemmas 3. and 3., we obtain S 3 = d(e G )(α + α ) [ D (a, e G ) + ] e E(G );a / V (e) =(α + α ) [ d(e G )D (a, e G ) + M (G ) m α (α ) δ(a G ) ]. e E(G ) The sum S 3 is equal to S 3 = e E(G );a V (e),f=a a d(e G)d(f G)d 4 (e, f G). Using Lemmas 3. and 3., we obtain ( S 3 = d(e G ) + ) (α + α ) = (α + α ) [ ] α (α ) + δ(a G ) + α. e E(G );a V (e) By adding S 3 and S 3, we obtain S 3 = (α + α ) [ ] d(e G )D (a, e G ) + M (G ) m + α. e E(G ) The fourth sum S 4 is taken over all pairs {e, f} of edges in G such that e E(G ) and f = a a. Using the same argument as in the computation of S 3, we obtain S 4 = (α + α ) [ e E(G ) d(e G )D (a, e G ) + M (G ) m + α ]. The fifth sum S 5 is taken over all pairs {e, f} of edges in G such that e E(G ) and f E(G ). In order to compute S 5, we partition it into four sums S 5, S 5, S 53 and S 54 as follows. The sum S 5 is equal to S 5 = e E(G );a / V (e) f E(G );a / V (f) d(e G)d(f G)d 4 (e, f G). Using Lemmas 3. and 3., we obtain S 5 = d(e G )d(f G ) [ D (a, e G ) + D (a, f G ) + ] e E(G );a / V (e) f E(G );a / V (f) = [ M (G ) m α (α ) δ(a G ) ] [ d(e G )D (a, e G ) α (α ) δ(a G ) ] e E(G ) + [ M (G ) m α (α ) δ(a G ) ]

85 8 Mahdieh Azari [ d(f G )D (a, f G ) α (α ) δ(a G ) ] f E(G ) + [ M (G ) m α (α ) δ(a G ) ] [ M (G ) m α (α ) δ(a G ) ]. The sum S 5 is equal to S 5 = d(e G)d(f G)d 4 (e, f G). e E(G );a / V (e) f E(G );a V (f) Using Lemmas 3. and 3., we obtain S 5 = d(e G ) ( d(f G ) + )[ D (a, e G ) + ] e E(G );a / V (e) f E(G );a V (f) = [ α (α ) + δ(a G ) ][ d(e G )D (a, e G ) α (α ) δ(a G ) e E(G ) + ( M (G ) m α (α ) δ(a G ) )]. The sum S 53 is equal to S 53 = d(e G)d(f G)d 4 (e, f G). e E(G );a V (e) f E(G );a / V (f) Using the same argument as in the computation of S 5, we obtain S 53 = [ α (α ) + δ(a G ) ][ d(f G )D (a, f G ) α (α ) δ(a G ) f E(G ) + ( M (G ) m α (α ) δ(a G ) )]. The sum S 54 is equal to S 54 = d(e G)d(f G)d 4 (e, f G). e E(G );a V (e) f E(G );a V (f) Using Lemmas 3. and 3., we obtain ( S 54 =3 d(e G ) + )(d(f G ) + ) e E(G );a V (e) f E(G );a V (f) =3 [ α (α ) + δ(a G ) ][ α (α ) + δ(a G ) ]. By adding S 5, S 5, S 53 and S 54, we obtain S 5 = ( ) M (G ) m + α d(e G )D (a, e G ) e E(G ) + ( ) M (G ) m + α d(f G )D (a, f G ) f E(G )

86 Some Results on Vertex Version and Edge Versions of Modified Schultz Index 8 + ( M (G ) m )( M (G ) m ) + α ( M (G ) m ) + α ( M (G ) m ) + 3α α. The formula of (W ) e4 (G) is obtained by adding S, S, S 3, S 4 and S 5 and simplifying the resulting expression. ¾ References [] A.R.Ashrafi, A.Hamzeh and S.Hosseinzadeh, Calculation of some topological indices of splices and links of graphs, J. Appl. Math. Inf., 9 (), [] M.Azari, Sharp lower bounds on the Narumi-Katayama index of graph operations, Appl. Math. Comput., 39C (4), [3] M.Azari, A note on vertex-edge Wiener indices, Iranian J. Math. Chem., 7() (6), 7. [4] M.Azari and F.Falahati-Nezhad, On vertex-degree-based invariants of link of graphs with application in dendrimers, J. Comput. Theor. Nanosci., () (5), [5] M.Azari and A.Iranmanesh, Edge-Wiener type invariants of splices and links of graphs, Politehn. Univ. Bucharest Sci. Bull. Ser. A Appl. Math. Phys., 77(3) (5), [6] M.Azari, A.Iranmanesh and I.Gutman, Zagreb indices of bridge and chain graphs, MATCH Commun. Math. Comput. Chem., 7 (3), [7] M.V.Diudea, QSPR/QSAR Studies by Molecular Descriptors, NOVA, New York,. [8] T.Došlić, Splices, links and their degree-weighted Wiener polynomials, Graph Theory Notes N. Y., 48 (5), [9] I.Gutman, Selected properties of the Schultz molecular topological index, J. Chem. Inf. Comput. Sci., 34 (994), [] I.Gutman and O.E.Polansky, Mathematical Concepts in Organic Chemistry, Springer, Berlin, 986. [] I.Gutman and N.Trinajstić, Graph theory and molecular orbitals, Total π-electron energy of alternant hydrocarbons, Chem. Phys. Lett., 7 (97), [] M.A.Hosseinzadeh, A.Iranmanesh and T.Došlić, On the Narumi-Katayama index of composite graphs, Croat. Chem. Acta, 86(4) (3), [3] A.Iranmanesh and M.Azari, The first and second Zagreb indices of several interesting classes of chemical graphs and nanostructures, In: M. V. Putz, O. Ori (Eds.), Exotic Properties of Carbon Nanomatter, Springer Netherlands, Dordrecht, vol. 8, 5, pp [4] A.Iranmanesh, O.Khormali and A.Ahmadi, Generalized edge-schultz indices of some graphs, MATCH Commun. Math. Comput. Chem., 65 (), 93. [5] O.Khormali, A.Iranmanesh, I.Gutman and A.Ahmadi, Generalized Schultz index and its edge versions, MATCH Commun. Math. Comput. Chem., 64 (), [6] M.Mogharrab and I.Gutman, Bridge graphs and their topological indices, MATCH Commun. Math. Comput. Chem., 69 (3),

87 8 Mahdieh Azari [7] H.P.Schultz, Topological organic chemistry. Graph theory and topological indices of alkanes, J. Chem. Inf. Comput. Sci., 9 (989), 7 8. [8] R.Sharafdini and I. Gutman, Splice graphs and their topological indices, Kragujevac J. Sci., 35 (3), [9] H.Wiener, Structural determination of paraffin boiling points, J. Am. Chem. Soc., 69 (947), 7. [] H.Wiener, Correlation of heats of isomerization and differences in heats of vaporization of isomers among the paraffin hydrocarbons, J. Am. Chem. Soc., 69 (947),

88 International J.Math. Combin. Vol.(6), 83-9 Folding of Cayley Graphs M.R.Zeen El-Deen (Department of Mathematics, Faculty of Science, Suez University, Suez, Egypt) Abstract: The aim of this paper is to discuss the folding of Cayley graphs of finite group.we prove that, for any finite group G, G = n and H is a subgroup of G. Then Cayley graph Γ = Cay(G, S) of G with respect to S = H \{ G} can be folded into a complete graph K r where r = H. Hence every Cayley graph Γ = Cay(G, S) of valency n can not be folded. Also every Cayley graph Γ = Cay(G, S) of valency one can be folded and Γ = Cay(G, S), where S is generating set, every elements in it is self inverse and S = G, can be folded to an edge. Theorems governing these types of foldings are achieved. Key Words: Cayley graph, folding, graph folding. AMS(): 54C5, 54A5.. Introduction It was Robertson, S.A. [7] who in 977 introduced the idea of folding on manifolds. Following this first paper there has been huge progress in the folding theory. All are focusing on topology and manifolds.many other studies on the folding of different types of manifolds introduced by many others [5], [6], [8]. Also a graph folding has been introduced by E. El-Kholy [4]. But EL-Ghoul in [3], turns this idea to algebras branch by giving a definition of the folding of abstract rings and studying its properties. Zeen El-Deen in [9] introduced the folding of groups and studying its properties. Some applications on the folding of a manifold into itself was introduced by P. Di.Francesco []. Graph Theory began with Leonhard Euler in his study of the Bridges of Königsburg problem. Since Euler solved this very first problem in Graph Theory, the field has exploded, becoming one of the most important areas of applied mathematics we currently study. Generally speaking, Graph Theory is a branch of Combinatorics but it is closely connected to Applied Mathematics, Topology and Computer Science. There are frequent occasions for which graphs with a lot of symmetry are required. One such family of graphs is constructed using groups. The study of graphs of groups is innovative because through this description one can immediately look at the graph and deduce many properties of this group. Cayley graphs are an example where graphs theory can be applied to groups. These graphs are useful for studying the structure of groups and the relationships Received September 4, 5, Accepted May 8, 6.

89 84 M.R.Zeen El-Deen between elements with respect to subsets of these groups (for example, generating sets, inverse closed sets,, etc.). Cayley (878) used graphs to draw picture of groups as we will see in the following definitions.. Definitions and Notations. We will start putting down some definitions which are needed in this paper. We begin with a short review of some basic definitions and properties of graphs. A graph Γ consists of a set of elements called vertices V (Γ), and a set of unordered pairs of these elements, called edges E(Γ). We will write (x, y) for directed edge, and xy or {x, y} for an undirected edge. we will only deal with simple graphs; that is, graph with no loops and no multiple edges and we will define all graphs to have a nonempty vertex set. A graph with no edges, but at least one vertex, is called empty graph. It is important to note that a graph may have many different geometric representation, but we just use these as a visualization tools and focus on V (Γ) and E(Γ) for our analysis. A graph is said to be connected if every pair of vertices has a path connecting them. Otherwise the graph is disconnected. The valency of a vertex is the number of edges with the vertex as an end point.if all the vertices of a graph have the same valency then it called a regular graph. A graph is complete if every vertex is connected to every other vertex, and we denote the complete graph on n vertices by K n. A graph is said to be bipartite if its vertex set can be partitioned into two sets, V and V, such that there are no edges of the form {x, y} where x, y V or x, y V. The complete bipartite graph K m,n is a bipartite graph with vertex set V V, such that V and V have size m and n respectively, and edge set {{x, y}, x V, y V }. A clique of a graph is its maximal complete subgraph. Definition. Let S be a subset of a finite group G. The Cayley digraph Γ = Cay(G, S) of G with respect to S is the directed graph given as follows. The vertices of Γ = Cay(G, S) are the elements of the group G. There is an arc between two vertices g and h if and only if g h S. In other words, for every vertex g G and element s S, there is an arc from g to gs. Notice that if the identity of G is in S, then there is a loop at every vertex, while if / S, the digraph has no loops. For convenience we will assume that the latter case holds; it makes no difference to the results. Also notice that since S is a set, it contains no multiple entries and hence there are no multiple arcs. Definition. A Cayley digraph can be consider to be a Cayley graph if whenever S is closed inverse, that is ; if s S, we also have s S, since in this case every arc is a part of a digon, and we can replace a digons with undirected edges Definition.3 A non empty subset S of a group G is called a Cayley subset if S = S and G / S.

90 Folding of Cayley Graphs 85 It should be noted that, the Cayley graph depends very much on the given Cayley subset as well as on the group. Also Cayley graph Γ = Cay(G, S) has valency S and that Γ = Cay(G, S) is connected if and only if S is generating set for G i.e., S = G. The complement S of Cayley subset S with respect to G = G \ { G } is also a Cayley subset.because if x S then x / S and since S is a Cayley subset then x / S.Hence x S,i.e., S is a Cayley subset. It is clear that The Γ = Cay(G, S) and Γ = Cay(G, S) have the same vertex set as G, where vertex g and h are adjacent in Γ = Cay(G, S) if and only if they are not adjacent in Γ = Cay(G, S). Definition.4([4]) A graph map f : Γ Γ between two graphs Γ and Γ is a graph folding if and only if f maps vertices to vertices and edges to edges, i.e., if, () for each v V (Γ ), f(v) is a vertex in V (Γ ); () for each e E(Γ ), f(e) is an edge in E(Γ ). Note that if the vertices of an edge e = (u, v ) E(Γ ) are mapped to the same vertex, then the edge e will collapse to this vertex and hence we cannot get a graph folding. In the case of a graph folding f the set of singularities, f, consists of vertices only. The graph folding is non trivial iff f φ. In this case the no. V (f(γ )) no. V (Γ ), also no. E(f(Γ )) no. E(Γ ). 3. Folding of Cayley Graphs In this section we will discuss the folding of Cayley graph Γ = Cay(G, S) to a subgraph Γ of it. We notice that not every Cayley graph Γ = Cay(G, S) can be folded into a subgraph of it, for example, Let G = S 3 be the Symmetric group of order 6, G = {(), ( ), ( 3), ( 3), ( 3 ), ( 3 )} and let S = { ( ), ( 3), ( 3 ) } since S is generating set and S = 3, so Γ = Cay(S 3, S ) is connected graph of valency 3. This graph cannot be folded into the induced subgraph Γ which shown in Figure. Because the vertex ( 3) is adjacent with the vertices ( 3 ), ( ) and ( 3) then it can not mapped by any folding to these vertices. Also the vertex ( 3) can not mapped to the vertex ( ) because the edge {( 3), ( 3)} has no image in Γ. Finally the vertex ( 3) can not mapped to the vertex ( 3 ) because the edge {( ), ( 3)} has no image in Γ. ( ) () ( ) () (3) (3) (3) (3) (3) (3) Γ (3) Γ Figure Γ = Cay(G, S) can not be folded into Γ

91 86 M.R.Zeen El-Deen It is known that, up to isomorphism, for any finite group G, there is Cayley graph of G of valency zero. which is a trivial graph and this graph has a trivial foldings since there are no edges, then any graph map on the vertices is a graph folding. Theorem 3. Let G be a finite group and H is a subgroup of G. Then Cayley graph Γ = Cay(G, S) of G with respect to S = H \ { G } can be folded and the end of these foldings is a complete subgraph (clique) K r where r = H i.e., the map φ : Γ = Cay(G, S) K r is graph folding. Proof Let G be a finite group and H is a subgroup of G. Define the set S to be the subgroup H with the identity removed S = H \ { G }, so S is closed inverse. Then we can define Cayley graph Γ = Cay(H, S) as follows, from the definition, there exist an edge between { G } and every element x S. Also for all x, y S there exist an edge between them, because, since S is inverse closed, then x, y S, xy H and yx H = xy or yx = xy S or yx S Then Cayley graph of H with respect to S is a complete graph K H. Also we can define Cayley graph Γ = Cay(G, S) such that {g, h} be an edge of Γ = Cay(G, S) if gh S and hence gh an element of H. From the properties of cosets gh H implies that Hg = Hh. This means that, two vertices are adjacent if and only if they are in the same cosets.thus Γ = Cay(G, S) is a graph depicting or describing the cosets of H in G. Since the number of the cosets of H in G is the index m = G : H, thus there are m components in Γ = Cay(G, S) each of which is a clique of size H i.e., K H. Then Γ = Cay(G, S) = {H, Hx, Hx,..., Hx m } where x i / H, i =,,, m We can define a graph maps φ i : Γ = Cay(G, S) Γ = Cay(H, S) by φ i : V (Hx i ) V (H), where, φ i (ax i ) = a for all a H. These are graph foldings since any edge in Hx i be in the form e = {ax i, bx i } where a, b H will mapped under φ into the edge e = {a, b} in H. Then φ i preserves the edges between vertices. The end of these foldings is the folding φ : Γ = Cay(G, S) K r where K r is a complete subgraph (clique) and r = H. ¾ Example 3. Let G = D 8 = { α, β α 4 = β = (αβ) = ; αβα = β } be the dihedral group of order 8, G = {, α, α, α 3, β, αβ, α β, α 3 β }. () Let H = α = {, α, α, α 3 } be a subgroup of G. Since H is closed inverse but not generating set of G and G : H =.So there exist two cosets of H in G i.e., H, Hβ = {β, αβ, α β, α 3 β }. Let S = H \ {}, then Γ = Cay(D 8, S) has two disjoint component { H, Hβ } each of which is a clique of size r = H = 4, see Figure.

92 Folding of Cayley Graphs 87 β α αβ Φ ¹ α α α β α α 3 α 3 β α 3 Figure Folding of Cayley graph Γ = Cay(D 8, {α, α, α 3 } ) Then, the map φ : Γ = Cay(G, S) Γ = Cay(H, S) = K 4 defined by φ : V (Hβ) V (H), where φ (α i β) = α i, i =,,, 3 where α i H This is a graph folding since it preserves the edges between vertices and the limit of the foldings is a clique of order 4. () Let H = β = {, β} be a subgroup of G, H is closed inverse but not generating set of G. Since G : H = 4, there exist four cosets of H in G i.e., { H, αh, α H, α 3 H }.Let S = H \ {}, then Γ = Cay(D 8, S) has four disjoint component { H, αh, α H, α 3 H } each of which is a clique of size r =, see Figure 3. α α α 3 Φ ¹ β αβ α β α 3 β β Figure 3 Folding of Cayley graph Γ = Cay(D 8, {β } ) Then the maps φ i : Γ = Cay(G, S) Γ = Cay(H, S) = K defined by φ i : V (α i H) V (H) where φ i (α i a) = a, i =,,, 3 where a H are graph foldings and the end of these foldings is a clique of order. Theorem 3. For any finite group G of order n, G = n. Every Cayley graph Γ = Cay(G, S) of valency n can be folded to a clique of order n. Proof Let G be a finite group of order n, G = {, a, a,, a n }, let S be a Cayley subset of G, since the identity element / S and valency of Cayley graph Γ = Cay(G, S)

93 88 M.R.Zeen El-Deen is the valency of S, if Γ has valency n so S = n. From the definition of Cayley graph, the identity element is adjacent to all the elements in S and since Γ = Cay(G, S) has no loop, then there exists exactly one element y G, y / S not adjacent with the identity element. this means that the two elements which is not in S is not adjacent and they adjacent to all elements in S. For any element a i S, since a i = n, {, a i } E(Γ) and {y, a i } E(Γ) then there exist one element b S such that {b, a i } / E(Γ) and a i must adjacent to all other elements in S. Then, the vertices of Γ = Cay(G, S) can be partitioned into n sets {A, A,, A n } each set has two vertices which are not adjacent, for example A = {, y }, y / S and A i = {a k, a r }, a k, a r S such that a k and a r not adjacent and each elements in A i are adjacent to all elements in A j, i j, i, j =,,, n. Then we have a complete n partite graph K,,,, so we can define n foldings on Γ as follows φ k : Γ Γ defined by φ k (x ) = y, if x, y A i, i, k =,,, n are graph foldings and the end of these foldings is a clique of order n. ¾ Example 3. Let G = D.3 = { α, β α 3 = β = (αβ) = ; αβα = β } be the dihedral group of order 6, G = {, α, α, β, αβ, α β }, and S = { α, α, β, αβ } be a Cayley subset of G. The Cayley graph Γ = Cay(D 6, S ) is shown in Figure 4. α β α β αβ Φ ¹ α αβ α α α α α β β Figure 4 Folding of Cayley graph Γ = Cay(D 6, { α, α, β, αβ } ) The vertices of the graph Γ = Cay(D 6, S ) can divides into n = 3 sets, each set has two elements which are not adjacent in Γ = Cay(D 6, S ).So we have A = {, α β }, A = {α, β }, A 3 = {α, αβ }. Then we can define three foldings φ ( α β) =, φ ( β) = α and φ 3 ( αβ) = α. The composition of these foldings is the map φ : V (Γ = Cay(D 6, S )) V (Γ = Cay(D 6, S )) defined by if x A φ (x) = α if x A α if x A 3 Since the image of any edge of E (Γ) will be the edge, then φ is a graph folding. Proposition 3. For any finite group G, G = n. Every Cayley graph Γ = Cay(G, S) of valency n can not be folded.

94 Folding of Cayley Graphs 89 Proof Let G be a finite group G = n, let S be a Cayley subset of G, since valency of Cayley graph Γ = Cay(G, S) is the valency of S, if Γ has valency n so S = n. Then S = G \ { G } is a generating and closed inverse which implies that Cayley graph Γ = Cay(G, S) is connected. Let H = S {} = G, so H is a subgroup of G and the index G : H =. Then we have one component which is a clique of order S = n i.e., Γ = Cay(G, S) = K n, so every vertex of Γ is adjacent to all other vertices. Then we can not define any folding on Γ since mapping any vertex to another will collapse the edge between them. ¾ Proposition 3. For any finite group G. Every Cayley graph Γ = Cay(G, S) of valency one can be folded. Proof Let G be a finite group G = n. Up to isomorphism, there is one Cayley graph Γ = Cay(G, S) of valency one. This graph Γ = Cay(G, S) is disconnected graph n consists of disconnected components each component is an edge between two vertices. Since if S = {a }, a G we have two cases (i) If a is self inverse, a = a, then H = {, a } is subgroup of G. Then from Theorem 3. Γ = Cay(G, S) consists of disjoint components {H, Hx i }, x i / H, i =,,..., n, where H is a clique of order two, i.e., there is an edge between the two vertices on Hx i = {x i, ax i }. so each Hx i is a clique of order two. This graph Γ = Cay(G, S) can be folded into H. (ii) If a is not self inverse a a, then H is not subgroup of G. Let S = H \ { G } = {a }, then Γ = Cay(G, S) consists of an edge between elements of H = {, a }. For any other vertex x G, then x G so there exists only one vertex y G such that xy = a H, which implies an edge between x and y. Then the graph Γ = Cay(G, S) is disconnected graph consists of disconnected components each component is an edge between n two vertices, if G = n then Γ consists of disconnecting edges, which can be folded into one component.. ¾ 4. Folding Cayley Graph of Non-Abelian Group In this section we will discuss the folding of Cayley graph of finite non-abelian groups. Theorem 4. For any finite non-abelian group G. Every Cayley graph of G with respect to a Cayley subset S, Γ = Cay(G, S), where S is generating set, every elements in it is self inverse and S = G, can be folded to an edge. Proof Let G be a finite non-abelian group, G = n and S G such that S is generating set, every elements in it is self inverse and S = G. Since the valency of Cayley graph Γ = Cay(G, S) is equal to the valency of S then Γ = S = G. If x and y S then x = x S and y = y S but there is no edge between x and y in Γ = Cay(G, S), since if there exist an edge between them this must implies that x y S and x y = x y, but x y / G because (x y) = y x = y x xy. So to have a graph Γ = Cay(G, S) of valency n, every element in S must connected to every element

95 9 M.R.Zeen El-Deen in G S. Then we have a complete bipartite graph Γ = Cay(G, S) = K n, n. Let G = {a, a,, a n, b, b,, b n } = V (Γ) and let S = {a, a,, a n }, then each vertex of S is joined to each vertex of (G S) by exactly one edge, thus E (Γ) = { (a, b ), (a, b ),, (a, b n ), (a, b ), (a, b ),, (a, b n ),, (a n, b ), (a n, b ),, (a n, b n ) } Now, let φ : V ( Γ) V (Γ) be a graph map defined by a if x S φ (x) = b if x G S Thus the image of any edge of E (Γ) will be the edge ( a, b ), then φ is a graph folding.¾ Example 4. Let G = S 3 be the Symmetric group of order 6, G = {(), ( ), ( 3), ( 3), ( 3 ), ( 3 )} and let S = { ( ), ( 3), ( 3) } since S is generating set, every elements in it is self inverse and S = G = 3, so Γ = Cay(S 3, S ) has valency 3 and the vertices of Γ can be partitioned into two sets S and G S such that there are no edges between any two vertices on the same set, see Figure 5. Then Γ = K 3,3 which can be folded by the function ( ) if x S φ (x) = ( ) if x G S () ( ) (3) () (3) (3) () Φ ¹ (3) (3) (3) ( ) (3) (3) ( ) Figure 5 Folding of Cayley graph Γ = Cay(S 3, {( ), ( 3), ( 3) } ) References [] C.Adiga, Some Properties of Cayley Graphs on Symmetric Groups S n, International Journal of Algebra, Vol. 6, No.7 (), [] P.DL-Francessco, Folding and Coloring problem in mathematics and physics, Bulletin of the American Mathematics Society, Vol. 37, No.3 (), [3] M.El-Ghoul, S.I.Nada and R.M.Abo Elanin, On the folding of rings, International Journal of Algebra, Vol. 3, No., (9), [4] E.El-Kholy and A.El-Esawy, Graph folding of some special graphs, Journal of Mathematics and Statistics, Sci. Pub. Vol. I, No.(5), 66-7.

96 Folding of Cayley Graphs 9 [5] E.M.El-Kholy, M.Basher, and M.R.Zeen El-Deen, Perfect folding of the plane, Soochow J. Math., Vol.3, No.4 (6), [6] H.R.Farran, E.EL-Kholy and S.A.Robertson, Folding a surface to a polygon, Geometriae Dedicata, Vol 63. (996), [7] S.A.Robertson, Isometric folding of Riemannian manifolds, Proc. Roy. Soc. Edinburgh, Vol.77 (977), [8] M.R.Zeen El-Deen, Relation between chromatic number and graph folding, Int.J.of Comp. and Applied Math., Vol.5, No.(), [9] M.R.Zeen El-Deen, On the folding of groups, J. Math. Comput. Sci., Vol, No. 5 (), 8-3.

97 International J.Math. Combin. Vol.(6), 9- The Merrifield-Simmons Indices of Triangle-Trees with k Pendant-Triangles Xuezheng Lv, Zixu Yan and Erling Wei (Department of Mathematics, Renmin University of China, Beijing 87, P.R. China) werling@ruc.edu.cn Abstract: Triangle-trees are a kind of graphs derived from Koch networks. The Merrifield- Simmons index of a graph is the total number of the independent sets of the graph. We prove that Pk,n k is the triangle-tree with maximal Merrifield-Simmons index among all the triangle-trees with n triangles and k pendant triangles. Key Words: Triangle-tree; Merrifield-Simmons index; pendant-triangle. AMS(): 5C5.. Introduction The Koch networks (see [], [3]) are derived from the Koch fractals (see [4], [9]) and are constructed iteratively. Let K m,g (m is a natural number) denote the Koch network after g iterations. Then, the family of Koch networks can be generated in the following way: initially (g = ), K m, consists of a triangle with three nodes labeled respectively by x, y, z, which have the highest degree among all nodes in the networks. For g, K m,g is obtained from K m,g by performing the following operation. For each of the three nodes in every existing triangle in K m,g, we add m groups of nodes. Each node group contains two nodes, both of which and their mother node are connected to one another forming a new triangle. In other words, to get K m,g from K m,g, we can replace each triangle in K m,g by a connected cluster on the right-hand side of the arrow in Fig.. Note that a Koch network does not have any cycle except for the triangles, we can call such a graph a triangle-tree. Definition. Let Tn (n is a natural number) denote a triangle-tree with n triangles. The family of triangle-trees can be generated in the following way: initially n =, T consists of a triangle with three vertices labeled respectively by x, y, z. For n, Tn is obtained from T n by adding a pair of new vertices u, v, both of them are joined to a vertex of Tn and the edge uv is also added to form a new triangle. In other words, to get Tn from Tn, we add a new triangle to Tn by identifying a vertex of the new triangle with a vertex of T n. Supported by National Natural Science Foundation of China, No Received June 4, 5, Accepted May, 6.

98 The Merrifield-Simmons Indices of Triangle-Trees with k Pendant-Triangles 93 m = = m = Fig. v v 4 v 6 v n v n 3 n n v v 3 v 5 v 7 v n 3 v n v n+ v 3 v 4 v v n+ n v 3. v 5 v 6 v 7 v n P n Fig. S n Obviously Koch networks are all triangle-trees. Suppose T is a triangle-tree, is a triangle of T, if there are two vertices with degree two in, we call the triangle a pendant triangle of T. The triangle-path Pn (see Fig.) is the only triangle-tree with only two pendant triangles and the triangle-star Sn (see Fig.) is the only triangle-tree with n pendant triangles. For any two triangles and of T, if and have a common vertex, we say and are adjacent, and the distance between and is, denoted by d(, ) =. If and do not have a common vertex, there is only one triangle-path between them. If the triangle-path between and contains d triangles, we say the distance between and is d, denoted by d(, ) = d. The diameter of a triangle-tree is denoted by d, defined as d (Tn ) = max{d(, ), are two triangles of Tn }. Throughout this paper G = (V, E) is a finite simple undirected graph with vertex set V = V (G) and edge set E = E(G). The neighborhood of a vertex v V is the set N G (v) = {w : w V, vw E}, d G (v) = N G (v), and N G [v] = N G (v) {v}. For S V, we use G S for the subgraph induced by V (G) \ S, G[S] for the subgraph of G induced by S and N S (v) = {w : w S, vw E(G)}. For F E(G), we use G F for the subgraph of G obtained by deleting F. Let G be a graph on n vertices. Two vertices of G are said to be independent if they are not adjacent in G. A k-independent set of G is a set of k-mutually independent vertices. Denote by f k (G) the number of the k-independent sets of G. For convenience, we regard the

99 94 Xuezheng Lv, Zixu Yan and Erling Wei empty vertex set as an independent set. Then f (G) = for any graph G. Let α(g) denote the cardinality of a maximal independent set of G. The Merrifield-Sommons index was introduced by Prodinger and Tichy in 98, which is defined by i(g) = α(g) s= f s (G), although it is called Fibonacci number of a graph in [8]. It is one of the most popular topological indices in chemistry, which was extensively studied in monograph [7]. Now there have been many papers studying the Merrifield-Simmons index. In [8], Prodinger and Tichy showed that, for trees with order n, the star has the maximal Merrifield-Simmons index and the path has the minimal Merrifield-Simmons index. In [6], Li et al characterized the tree with the maximal Merrifield-Simmons index among the trees with given diameter. In [], Yu and Lv characterized the trees with maximal Merrifield-Simmons indices, among the trees with given pendant vertices. For more results on Merrifield-Simmons index, see [-3], [5] and []. Due to the similarity of triangle-trees and ordinary trees, it is very interesting to study the Merrifield-Simmons indices of triangle-trees. It is easy verify that, among all the triangle-trees with n triangles, Sn is the triangle-tree with maximal Merrifield-Simmons index and Pn is the triangle-tree with minimal Merrifield-Simmons index. As noting this result is similar to the result of ordinary trees, we consider all the triangle-trees with n triangles and k pendant triangles. It is very interesting to find that Pk,n k (as shown in Fig. 3) is the triangle-tree with maximal Merrifield-Simmos index among all such triangle-trees, and this result is also similar to the result of ordinary trees. k k+ k+ n n n k P k,n k Fig.3. Lemmas and Results We first introduce the following lemma, which is obvious and well-known. Lemma. For a graph G, we have () i(g) = i(g v) + i(g N[v]) for any v V (G); () i(g) = i(g e) i(g N[e]) for any e E(G); (3) If G = G G, then i(g) = i(g )i(g ). Using the above lemma, we can derive some recursion formulas on the Merrifield-Simmons

100 The Merrifield-Simmons Indices of Triangle-Trees with k Pendant-Triangles 95 index of the triangle-path P n. Denote a n = i(p n ). It is easy to see that a = 4, a =, a 3 = 4. Let Q n = P n v, where v is one of the vertices with degree two of the pendant-triangle of P n (as shown in Fig ) and b n = i(q n ). It is easy to see that b = 3, b = 7, b 3 = 7. Let R n = Q n v n+, where v n+ is one of the vertices with degree two of another pendant-triangle of P n (as shown in Fig.). It is easy to see that c =, c = 5, c 3 =. By Lemma., we know So we have a n = b n + b n, b n = a n + b n = c n + c n, c n = b n + c n. b n+ = b n + b n, a n+ = a n + a n, c n+ = c n + c n. Let Pk = k be a path of a triangle-tree T, where i = v i v i v i+. If d T (v ) 6, d T (v k+ ) 6, d T (v i ) = ( i k) and d T (v i+ ) = 4 ( i k ), we call Pk an internal triangle-path of T. If the triangle = v v v 3 is a pendant triangle of T, d T (v k+ ) 6, d T (v i ) = ( i k) and d T (v i+ ) = 4 ( i k ), we call P a pendant triangle-path of T. Let s(t ) be the number of vertices in T with degree not less than 6 and p(t ) be the number of pendant triangle-paths in T with length not less than. 3 s w u v G G T G u 3 s w u v G Operation I 3 s wu u v G T G T Fig.4 Denote Tn,k (3 k n ) be the set of all triangle-trees with n triangles and k pendant

101 96 Xuezheng Lv, Zixu Yan and Erling Wei triangles. In the following, we shall define two kinds of operations of T Tn,k and show that these two kinds of operations make the Merrifield-Simmons indices of the triangle-tree increase strictly. If T Tn,k, T = P k,n k and p(t ), then T can be seen as the triangle-trees T or T as shown in Fig.4, where s (s ) is a pendant path of T with s triangles, G and G are two subtriangle-trees of T and V (G ) 3, V (G ) 3. If T is obtained from T or T by Operation I (as shown in Fig.4), it is easy to see that T T n,k. Now we show that operation I makes the Merrifield-Simmons indices of the triangle-trees increase strictly. Lemma. If T is obtained from T or T by operation I, then i(t ) > i(t ) and i(t ) > i(t ). Proof Let N G [v] = V, N G [u] = V in T, N G [u ] = V in T and N G [w] = V 3 in T. If s 3, by Lemma., we have i(t ) = i(t v) + i(t N T [v]) = i(g v)(i(g u)b s + i(g V )b s ) + i(g V )i(g u)b s, i(t ) = i(t v) + i(t N T [v]) = i(g v)(i(g u )a s + i(g V )b s) + i(g V )i(g u )b s, i(t ) = i(t v) + i(t N T [v]) = i(g v)(3i(g w)c s + i(g V 3 )c s ) +i(g V )(3i(G w)c s + i(g V 3 )c s ). Obviously, i(g w) = i(g u ) = i(g u) and i(g V 3 ) = i(g V ) = i(g V ), so we have i(t ) i(t ) = i(g v)i(g u)(3c s b s ) + i(g v)i(g V )(c s b s ) +i(g V )i(g u)(3c s b s ) + i(g V )i(g V )c s = i(g v)i(g u)c s i(g v)i(g V )c s i(g V )i(g u)c s + i(g V )i(g V )c s = c s (i(g v) i(g V ))(i(g u) i(g V )). Since s 3, c s >, i(g v) i(g V ) > and i(g u) i(g V ) >, we know i(t ) i(t ) > when s 3.

102 The Merrifield-Simmons Indices of Triangle-Trees with k Pendant-Triangles 97 Similarly, i(t ) i(t ) = i(g v)i(g u )(3c s a s ) + i(g v)i(g V )(c s b s ) +i(g V )i(g u )(3c s b s ) + i(g V )i(g V )c s = i(g v)i(g u )(c s + c s ) + i(g v)i(g V )( c s c s ) +i(g V )i(g u )( c s ) + i(g V )i(g V )c s = c s (i(g v) i(g V ))(i(g u ) i(g V )) +c s i(g v)(i(g u ) i(g V )) >. Therefore, i(t ) i(t ) > when s 3. If s =, similarly, we have i(t ) i(t ) = i(g v)i(g u)(3c b ) + i(g v)i(g V )(c b ) +i(g V )i(g u)(3c b ) + i(g V )i(g V ) = (i(g v) i(g V ))(i(g u) i(g V )). i(t ) i(t ) = i(g v)i(g u )(3c a ) + i(g v)i(g V )(c b ) +i(g V )i(g u )(3c b ) + i(g V )i(g V ) = (i(g v) i(g V ))(i(g u ) i(g V )) +4i(G v)(i(g u ) i(g V )) >. Therefore, i(t ) i(t ) > and i(t ) i(t ) > when s =. ¾ From Lemma., we can immediately get the following result. Lemma.3 Let T T n,k (3 k n ), T = P k,n k and p(t ). () If s(t ) =, we can finally get a triangle-tree T by operation I with i(t ) > i(t ), and p(t ) = ; it is easy to see that T = P k,n k ; () If s(t ) >, we can finally get a triangle-tree T by operation I with i(t ) > i(t ) and p(t ) =. If T Tn,k (3 k n ), T = P n,k and p(t ) =, then we can find two pendant triangles and of T such that d(, ) = d (T ). Suppose = uu u and = vv v, where u, u, v, v are the vertices with degree and d(u) 6, d(v) 6. Then the triangle-tree can be seen as the triangle-tree T shown in Fig 5, where,,, s are pendant triangles with common vertex u,,,, t are pendant triangles with common ( s vertex v, G is the subgraph of T induced by V (T ) \ V ( i ) t V ( i ). ) i= i=

103 98 Xuezheng Lv, Zixu Yan and Erling Wei x 3... s u x G T y y v 3 Operation II t s u t u x u x x u x G T G T Delta y v v y v y v t y s Fig.5 Note that if d(, ) = 3, then x = y; if d(, ) 4, then V (G ) 5. T is a triangle-tree got from T by moving the pendant triangles, 3,, t from v to u, and T is a triangle-tree got from T by moving the pendant triangles, 3,, s from u to v. We say both of T and T are obtained from T by Operation II. It is easy to see that T, T Γ n,k p(t ) = p(t ) = and s(t ) = s(t ) = s(t ). Lemma.4 If T and T are obtained from T by Operation II, then either i(t ) > i(t ) or i(t ) > i(t ). Proof If d(, ) 3, then N G (u) = {x, x } and N G [v] = {y, y }. Note that if d(, ) = 3, then x = y. By Lemma., we have i(t ) = i(t u) + i(t N T [u]) = 3 s (3 t i(g ) + i(g {y, y })) + 3 t i(g {x, x }}) + i(g {x, x, y, y }), i(t ) = i(t u) + i(t N T [u]) = 3 s+t (3i(G ) + i(g {y, y })) + 3i(G {x, x }) + i(g {x, x, y, y }), i(t ) = i(t u) + i(t N T [u]) = 3(3 s+t i(g ) + i(g {y, y })) + 3 s+t i(g {x, x }) + i(g {x, x, y, y }). It is easy to see that i(t ) i(t ) = 3(3 t )(3 s i(g {y, y }) i(g {x, x })),

104 The Merrifield-Simmons Indices of Triangle-Trees with k Pendant-Triangles 99 i(t ) i(t ) = 3(3 s )(3 t i(g {x, x }) i(g {y, y })). Note that s, t. If i(t ) i(t ), we have 3 s i(g {y, y }) i(g {x, x }). Then we have i(t ) i(t ) 3(3 s )(3 s 3 t )i(g {y, y }) >. If d(, ) =, we have T = T. Suppose N G (u) = {v, w}, N G (v) = {u, w}, then i(t ) i(t ) = 3(3 t )(3 s )i(g w) >. Therefore, if T and T are obtained from T by operation II, then either i(t ) > i(t ) or i(t ) > i(t ). ¾ Theorem.5 Let T T n,k. Then i(t ) 3 k b n k+ +b n k, the equality holds if and only if T = P k,n k. Proof By Lemma., it is easy to see that i(p k,n k ) = 3k b n k+ + b n k. Since T n, = {P n } and P n = P n,, T n,n = {S n } and S n = P,n, we may assume 3 k n. It is sufficient to show that i(t ) < i(p k,n k ) for any T T n,k and T = P k,n k. For T Tn,k (3 k n ) and T = P k,n k, we know s(t ) n k, we shall show i(t ) i(pk,n k ) by induction on s(t ). When s(t ) =, since T = P k,n k, we have p(t ). By () of Lemma.3, we have i(t ) < i(pk,n k ). Suppose the result holds for any triangle-tree T with s(t ) = s. Let s(t ) = s. If p(t ), by () of Lemma.3, we can get a triangle-tree T Tn,k such that p(t ) =, s(t ) = s and i(t ) > i(t ). Then by Lemma.4, we can get a triangle-tree T Tn,k from T such that p(t ) =, s(t ) = s and i(t ) > i(t ). By the induction hypothesis, we have i(t ) < i(t ) < i(t ) < i(p k,n k ). Therefore, if T Tn,k, then i(t ) 3 k b n k+ + b n k = i(pk,n k ) and the equality holds if and only if T = Pk,n k. ¾ Lemma.6 For 3 k n, i(p n k+,k ) > i(p n k+3,k 3 ). Proof By Lemma., it is easy to see that i(p n k+,k ) = 3 k b n k+ + b n k, i(p n k+3,k 3 ) = 3k b n k+ + b n k+.

105 Xuezheng Lv, Zixu Yan and Erling Wei Since b n+ = b n + b n, we have i(p k,n k) i(p n k+3,k 3) = 3 k b n k+ + b n k 3 k b n k+ + b n k+ = (3 k )(b n k+ b n k ) >. Hence i(pn k+,k ) > i(p n k+3,k 3 ) for 3 k n. ¾ From Theorem.5 and Lemma.6, we can immediately get the following result. Corollary.7 Let T be a triangle-tree with n + vertices and n triangles. Then () i(t ) 3 n + and the equality holds if and only if T = S n ; () If T = S n, then i(t ) 7 3 n +3 and the equality holds if and only if T = P 3,n 3. References [] I.Gutman, Extremal Hexagonal Chains, J. Math. Chem. (993) 97-. [] I.Gutman, B. Furtula, D. Vidovic and H. Hosoya, A concealed property of the topological index, Z. Bull. Chem. Soc. Jpn. 77(4), [3] I.Gutman and S. Wagner, Maximal and Minimal of the Hosoya index and the Merrifield- Simmons index: A survey of results and techniques, Acta. Appl. Math. (), [4] A.Lakhtakia, V.K. Varadan, R. Messier and V.V. Varadan, Generalisations and randaomisation of the plane Koch curve, J. Phys. A (987), no., [5] X.L.Li, Z.M.Li and L.S.Wang, The inverse problem for some topological indices in combinatorial chemistry, J. Comput. Bio. (3), [6] X.L.Li, H.X.Zhao and I.Gutman, On the Merrifield-Simmons index of trees, MATCH Commun. Math. Comput. Chem. 54(5) [7] R.E.Merrifield and H.E. Simmons, Topological Methods in Chemistry (Wiley, New York, 989). [8] H.Prodinger and R.F.Tichy, Fibonacci numbers of graphs, Fibonacci Quart. (98) 6-. [9] J.E.Schneider, A generalization of the von Koch curve, Math. Mag. 38(965), [] B.Wu, Z.Zhang and G.Chen, Properties and applications of Laplacian spectra for Koch networks, J. Phy. A: Math. Theor. 45() no., 5, pp. [] A. Yu and X.Lv, The Merrifield-Simmons indices and Hosoya indices of trees with k- pendant vertices, Journal of Mathematical Chemistry, 4(7), [] A.Yu and F.Tian, A kind of graphs with minimal Hosoya indices and maximal Merrifield- Simmons indices, MATCH Commun. Comput. Chem. 55(6) 3-8. [3] Z.Zhang, S.Gao, L.Chen, S.Zhou, H.Zhang and J.Guan, Mapping Koch curves into scalefree small-world networks, J. Phy. A: Math. Theor. 39() 395, 6 pp.

106 International J.Math. Combin. Vol.(6), -8 On Linear Codes Over a Non-Chain Ring Abdullah Dertli, Yasemin Cengellenmis, Senol Eren. Ondokuz Mayıs University, Faculty of Arts and Sciences, Mathematics Department, Samsun, Turkey. Trakya University, Faculty of Arts and Sciences, Mathematics Department, Edirne, Turkey abdullah.dertli@gmail.com, seren@omu.edu.tr, ycengellenmis@gmail.com Abstract: In this paper, we study skew cyclic and quasi cyclic codes over the ring S = F + uf + vf where u = u, v = v, uv = vu =. We investigate the structural properties of them. Using a Gray map on S we obtain the MacWilliams identities for codes over S. The relationships between Symmetrized, Lee and Hamming weight enumerator are determined. Key Words: Line code, chain ring, non-chain ring, Gray map, MacWilliams identity. AMS(): 94B5.. Introduction Since the revelation in 994 [], there are a lot of works on codes over finite rings. The structure of certain type of codes over many finite rings are determined such as cyclic, quasicyclic. Recently, it is introduced the class of skew codes which are generalized the notion cyclic, quasi-cyclic in [5,6,,4]. In [], T. Abualrub, P. Seneviratre studied skew cyclic codes over F + vf, v = v. In [], T. Abualrub, A. Ghrayeb, N. Aydın, I. Siap introduced skew quasi-cyclic codes. They obtained several new codes with Hamming distance exceeding the distance of the previously best known linear codes with comparable parameters. In [4], they investigated the structures of skew cyclic and skew quasi-cyclic of arbitrary length over Galois rings. They shown that the skew cyclic codes are equivalent to either cyclic and quasi-cyclic codes over Galois rings. Moreover, they gave a necessary and sufficient condition for skew cyclic codes over Galois rings to be free. Jian Gao, L.Shen, F. W. Fu studied a class of generalized quasi cyclic codes called skew generalized quasi-cyclic codes. They gave the Chinese Remainder Theorem over the skew polynomial ring which lead to a canonical decomposition of skew generalized quasi-cyclic codes. Moreover, they focused on -generator skew generalized quasi-cyclic code in [7]. J.Gao also presented skew cyclic codes over F p + vf p in [8]. The MacWilliams identity supplies the relationship between the weight enumerator of a linear code and that of its dual code []. The distribution of weights for a linear code is important for its performance analysis such as linear programming bound, error correcting Received August, 5, Accepted May, 6.

107 Abdullah Dertli, Yasemin Cengellenmis, Senol Eren capabilities, etc. There are a lot of work about the MacWilliams identities in [3,9,5]. This paper is organized as follows. In section, we give some basic knowledges about the finite ring S. In section 3, we define a new Gray map from S to F 3, Lee weights of elements of S and Lee distance in the linear codes over S. In section 4, we define a new non trivial automorphism and we introduce skew codes over S. In section 5, we obtain the MacWilliams identities and give an example.. Preliminaries Let S be the ring F + uf + vf where u = u, v = v, uv = vu = and F = {, }, a finite commutative ring with 8 elements. S is semi local ring with three maximal ideals and a principal ideal ring. It is not finite chain ring. The ideals are follows; I = {}, I = S I u = {, u}, I v = {, v}, I +u+v = {, + u + v} I u+v = {, u, v, u + v}, I +u = {, v, + u, + u + v} I +v = {, u, + v, + u + v} A linear code C over S length n is a S submodule of S n. An element of C is called a codeword. For any x = (x, x,, x n ), y = (y, y,, y n ) the inner product is defined as n x.y = x i y i i= If x.y = then x and y are said to be orthogonal. Let C be linear code of length n over S, the dual code of C C = {x : y C, x.y = }, which is also a linear code over S of length n. A code C is self orthogonal if C C and self dual if C = C. A cyclic code C over S is a linear code with the property that if c = (c, c,, c n ) C then σ (C) = (c n, c,, c n ) C. A subset C of S n is a linear cyclic code of length n iff it is polynomial representation is an ideal of S [x] / x n. Let C be code over F of length n and ć = (ć, ć,, ć n ) be a codeword of C. The Hamming weight of ć is defined as w H (ć) = n i= w H (ć i ) where w H (ć i ) = if ć i = and w H (ć i ) = if ć i =. Hamming distance of C is defined as d H (C) = mind H (c, ć), where for any ć C, c ć and d H (c, ć) is Hamming distance between two codewords with d H (c, ć) = w H (c ć). Let a F 3n with a = (a, a,, a 3n ) = ( a () a () a ()), a (i) F n for i =,,. Let ϕ be a map from F 3n to F 3n given by ϕ(a) = ( σ ( a ()) ( ) ( σ a () )) σ a () where σ is a cyclic

108 On Linear Codes Over a Non-Chain Ring 3 shift from F n to F n given by σ ( a (i)) = ((a (i,n ) ), (a (i,) ), (a (i,) ),, (a (i,n ) )) for every a (i) = (a (i,),, a (i,n ) ) where a (i,j) F, j n. A code of length 3n over F is said to be quasi cyclic code of index 3 if ϕ(c) = C. 3. Gray Map Let x = a + ub + uc be an element of S where a, b, c F. We define Gray map Ψ from S to F 3 by Ψ : S F 3 Ψ (a + ub + vc) = (a, a + b, a + c) The Lee weight of elements of S are defined w L (a + ub + vc) = w H (a, a + b, a + c) where w H denotes the ordinary Hamming weight for binary codes. Hence, there is one element whose weight is, there are u, v, + u + v elements whose weights are, there are + u, + v, u + v elements whose weights are, there is one element whose weight are 3. Let C be a linear code over S of length n. For any codeword c = (c,, c n ) the Lee weight of c is defined as w L (c) = n i= w L (c i ) and the Lee distance of C is defined as d L (C) = min d L (c, ć), where for any ć C, c ć and d L (c, ć) is Lee distance between two codewords with d L (c, ć) = w L (c ć). Gray map Ψ can be extended to map from S n to F 3n. Theorem 3. The Gray map Ψ is a weight preserving map from (S n,lee weight) to (F 3n, Ham ming weight). Moreover it is an isometry from S n to F 3n. Theorem 3. If C is an [n, k, d L ] linear codes over S then Ψ (C) is a [3n, k, d H ] linear codes over F, where d H = d L. 4. Skew Codes over S We are interested in studying skew codes using the ring S = F +uf +vf where u = u, v = v, uv = vu =.. We define non-trivial ring automorphism θ on the ring S by θ (a + ub + vc) = a + vb + uc for all a + ub + vc S. The ring S[x, θ] = {a + a x + + a n x n : a i S, n N} is called a skew polynomial ring. This ring is a non-commutative ring. The addition in the ring S[x, θ] is the usual polynomial addition and multiplication is defined using the rule, (ax i )(bx j ) = aθ i (b)x i+j. Note that θ (a) = a for all a S. This implies that θ is a ring automorphism of order. Definition 4. A subset C of S n is called a skew cyclic code of length n if C satisfies the following conditions, (i) C is a submodule of S n ; (ii) If c = (c, c,, c n ) C, then σ θ (c) = ( θ(c n ), θ(c ),, θ(c n ) ) C.

109 4 Abdullah Dertli, Yasemin Cengellenmis, Senol Eren Let (f(x)+(x n )) be an element in the set S n = S [x, θ] /(x n ) and let r(x) S [x, θ]. Define multiplication from left as follows: for any r(x) S [x, θ]. r(x)(f(x) + (x n )) = r(x)f(x) + (x n ) Theorem 4. S n is a left S [x, θ]-module where multiplication defined as in above. Theorem 4.3 A code C in S n is a skew cyclic code if and only if C is a left S [x, θ]-submodule of the left S [x, θ]-module S n. Theorem 4.4 Let C be a skew cyclic code in S n and let f(x) be a polynomial in C of minimal degree. If f(x) is monic polynomial, then C = (f(x)) where f(x) is a right divisor of (x n ). Theorem 4.5 Let n be odd and C be a skew cyclic code of length n. Then C is equivalent to cyclic code of length n over S. Proof Since n is odd, gcd(, n) =. Hence there exist integers b, c such that b + nc =. So b = nc = + zn where z >. Let a(x) = a + a x + + a n x n be a codeword in C. Note that x b a(x) = θ b (a )x +zn + θ b (a )x +zn + + θ b (a n )x n+zn = a n + a x + + a n x n C. Thus C is a cyclic code of length n. ¾ Corollary 4.6 Let n be odd. Then the number of distinct skew cyclic codes of length n over S is equal to the number of ideals in S [x] /(x n ) because of Theorem 7. If x n = r i= psi i (x) where p i (x) are irreducible polynomials over F. Then the number of distinct skew cyclic codes of length n over S is r i= (s i + ). Example 4.7 Let n = 5 and g(x) = x 4 + x 3 + x + x +. Then g(x) generates a skew cyclic codes of length 5. This code is equivalent to a cyclic code of length 5. Since x 5 = (x + )(x + x + )(x 4 + x + )(x 4 + x 3 + )(x 4 + x 3 + x + x + ), it follows that there are 8 skew cyclic code of length 5. Definition 4.8 A subset C of S n is called a skew quasi-cyclic code of length n if C satisfies the following conditions: (i) C is a submodule of S n ; (ii) If e = (e,,, e,l, e,,, e,l,, e s,,, e s,l ) C, then τ θ,s,l (e) = (θ(e s, ),, θ(e s,l ), θ(e, ),, θ(e,l ), θ(e s, ),, θ(e s,l )) C. We note that x s is a two sided ideal in S [x, θ] if m s where m = is the order of θ. So S [x, θ] /(x s ) is well defined. The ring M l s = (S [x, θ] /(xs )) l is a left M s = S [x, θ] /(x s ) module by the following multiplication on the left f(x)(g (x),, g l (x)) = (f(x)g (x),, f(x)g l (x)). If the map γ is defined by γ : S n M l s

110 On Linear Codes Over a Non-Chain Ring 5 (e,,, e,l, e,,, e,l,, e s,,, e s,l ) (c (x),, c l (x)) such that e j (x) = s i= e i,jx i M l s where j =,,, l then the map γ gives a one to one correspondence S n and the ring M l s. Theorem 4.9 A subset C of S n is a skew quasi-cyclic code of length n = sl and index l if and only if γ(c) is a left S s -submodule of M l s. 5. MacWilliams Identities Let the elements of S be represented as S = {f, f,, f 8 } = {,, u, v, + u, + v, u + v, + u + v} where the order of elements is fixed. Definition 5. Define χ : S C by χ(a+ub+vc) = ( ) a+b+c. χ is a non-trivial character of each non-zero ideal I of S. Hence we have a I χ(a) =. Lemma 5. Let C be a linear code over S of length n. Then for any m S n,,if m / C χ(c.m) = C,if m C c C Theorem 5.3([]) Let C be a linear code over S of length n and f(c) = m S χ(c.m)f(m). n Then m C f(m) = f(c). C c C Let A is a 8 8 matrix.a matrix defined by A(i, j) = χ(f i f j ). The matrix A is given as follows A = Definition 5.4 Let C be a linear code of length n over S, then Lee C (x, y) = c C x3n wl(c) y wl(c) can be called as the Lee weight enumerator of C and Ham C (x, y) = c C xn wh(c) y wh(c) can be called as the Hamming weight enumerator of C. Besides, Swe C (x, y, z, w) = c C x n(c) y n(c) z n(c) w n3(c) is the symmetric weight enumerator where n i (c) denote the number of elements of c with Lee

111 6 Abdullah Dertli, Yasemin Cengellenmis, Senol Eren weight,,, 3, respectively. Definition 5.5 The complete weight enumerator of a linear code C over R is defined as cwe C (x, x,, x 8 ) = c C xn f (c) x n f 8 (c) 8 where n fi (c) is the number of appearances of f i in the vector c. The complete weight enumerator gives us a lot of information about the code, such as the size of the code, the minimum weight of the code and the weight enumerator of the code for any weight function. We can define the symmetrized weight enumerator as follows. Definition 5.6 Let C be a linear code of length n over S. Then define the symmetrized weight enumerator of C as Swe C (x, y, z, w) = cwe C (x, w, y, y, z, z, z, y) Here x represents the elements that have weight (the element), y represents the elements with weight (the elements u, v, +u+v), z represents the elements with weight (the elements + u, + v, u + v), w represents the elements with weight 3 (the element ). Theorem 5.7 Let C be a linear code of length n over S and let C be its dual. Then cwe C (x, x,, x 8 ) = C cwe C(A.(x x x 8 ) ) where () denotes the transpose. Theorem 5.8 Let C be a linear code of length n over S and let C be its dual. Then Swe C (x, y, z, w) = C Swe C(x + w + 3y + 3z, x w 3y + 3z, x w + y z, x + w y z). Proof The proof follows simply from calculating the matrix product A.(x w y y z z z y) where () denotes the transpose. ¾ Theorem 5.9 Let C be a linear code of length n over S. Then, (i) Lee C (x, y) = Swe C (x 3, x y, y x, y 3 ); (ii) Lee C (x, y) = C Lee C(x + y, x y). Proof (i) Let w L (c) = n (c)+n (c)+3n 3 (c) where n i (c) denote the number of elements of c with Lee weight,,, 3, respectively. Since n = n (c) + n (c) + n (c) + n 3 (c), 3n w L (c) = 3n (c) + n (c) + n (c). From the definition, Lee C (x, y) = c C x 3n wl(c) y wl(c) = c C x 3n(c)+n(c)+n(c) y n(c)+n(c)+3n3(c) = c C x 3n(c) (x y) n(c) (y x) n(c) y 3n3(c) = Swe C (x 3, x y, y x, y 3 )

112 On Linear Codes Over a Non-Chain Ring 7 (ii) From Theorems 5.7 and 5.8, Lee C (x, y) = C Swe C(x 3 + 3x y + 3y x + y 3, x 3 y 3 3x y + 3xy, x 3 y 3 + x y xy, x 3 + y 3 x y xy ) = C Swe C((x + y) 3, (x + y) (x y), (x y) (x + y), (x y) 3 ) = C Lee C(x + y, x y). Theorem 5. Let C be a linear code of length n over S. Then we have (i) Ham C (x, y) = C Ham C(x + 7y, x y); (ii) Ham C (x, y) = Swe C (x, y, y, y). Proof (i) It is straightforward from [3]. (ii) The Hamming weight w H (c) is defined as w H (c) = n (c) + n (c) + n (c) + n 3 (c). Ham C (x, y) = x n wh(c) y wh(c) = x n(c) y n(c)+n(c)+n3(c) c C c C = Swe C (x, y, y, y). ¾ Example 5. Let C = {(, ), (v, v)} be a linear code of length over S. The Lee weight enumerator is Lee C (x, y) = x 6 + x 4 y ; the Hamming enumerator is Ham C (x, y) = x + y. Lee weight enumerator of C is Lee C (x, y) = x 6 + 4x 5 y + 7x 4 y + 8x 3 y 3 + 7x y 4 + 5xy 5 + y 6 ; Hamming weight enumerator of C is Ham C (x, y) = x + 6xy + 5y. ¾ References [] T.Abualrub, P.Seneviratne, On θ-cyclic codes over F +vf, Australasian Journal of Com., 54(), 5-6. [] T.Abualrub, A.Ghrayeb, N.Aydın, I.Siap, On the construction of skew quasi-cyclic codes, IEEE Transsactions on Information Theory, Vol 56, No.5(), 8-9. [3] R.K.Bandi, M.Bhaintwal, Codes over Z 4 + vz 4, IEEE, (4), [4] M.Bhaintwal, Skew quasi-cyclic codes over Galois rings, Des. Codes Cryptogr., DOI..7/ s [5] D.Boucher, W.Geiselmann, F.Ulmer, Skew cyclic codes, Appl. Algebra. Eng.Commun Comput., Vol.8, No.4(7) [6] D.Boucher, F.Ulmer, Coding with skew polynomial rings, Journal of Symbolic Computation, 44(9), [7] J.Gao, L.Shen, F.W.Fu, Skew generalized quasi-cyclic codes over finite fields, arxiv:39, 6v. [8] J.Gao, Skew cyclic codes overf p + vf p, J. Appl. Math. & Informatics, 3(3),

113 8 Abdullah Dertli, Yasemin Cengellenmis, Senol Eren [9] J.Gao, Y.Gao, Some Results on Linear Codes over Z 4 + vz 4, arxiv: v, 4. [] A.R.Hammons, V.Kumar, A.R.Calderbank, N.J.A.Sloane, P. Sole, The Z 4 -linearity of Kerdock, Preparata, Goethals and related codes, IEEE Trans. Inf. Theory, 4 (994) [] F.J.MacWilliams and N.J.A.Sloane, The Theory of Error Correcting Codes, North-Holland, Amsterdam, The Netherlands, 977. [] I.Siap, T.Abualrub, N.Aydın, P.Seneviratne, Skew cyclic codes of arbitrary length, Int. Journal of Information and Coding Theory,. [3] J.Wood, Duality for modules over finite rings and applications to coding theory, Amer. J. Math., (993), [4] M.Wu, Skew cyclic and quasi-cyclic codes of arbitrary length over Galois rings, International Journal of Algebra, Vol.7, No.7(3), [5] B.Yildiz, S.Karadeniz, Linear codes over Z 4 +uz 4, MacWilliams identities, projections and formally self-dual codes, Finite Fields and Their Applications, 7(4), 4-4.

114 International J.Math. Combin. Vol.(6), 9- Nonsplit Geodetic Number of a Graph Tejaswini K M, Venkanagouda M Goudar Sri Gauthama Research Centre, (Affiliated to Kuvempu University, Shimoga) Department of Mathematics, Sri Siddhartha Institute of Technology, Tumkur-57 5, Karnataka, India Venkatesh Department of mathematics, Kuvempu University, Shimoga, India tejaswini.ssit@gmail.com, vmgouda@gmail.com vensprema@gmail.com Abstract: Let G be a graph. If u, v V (G), a u v geodesic of G is the shortest path between u and v. The closed interval I[u, v] consists of all vertices lying in some u - v geodesic of G. For S V (G) the set I[S] is the union of all sets I[u, v] for u, v S. A set S is a geodetic set of G if I(S) = V (G). The cardinality of a minimum geodetic set of G is the geodetic number of G, denoted by g(g). In this paper, we study the nonsplit geodetic number of a graph g ns(g). The set S V (G) is a nonsplit geodetic set in G if S is a geodetic set and V (G) S is connected, nonsplit geodetic number g ns(g) of G is the minimum cardinality of a nonsplit geodetic set of G. We investigate the relationship between nonsplit geodetic number and geodetic number. We also obtain the nonsplit geodetic number in the cartesian product of graphs. Key Words: Cartesian products, distance, edge covering number, Smarandachely k- geodetic set, geodetic number, vertex covering number. AMS(): 5C5, 5C.. Introduction As usual n = V and m = E denote the number of vertices and edges of a graph G respectively. The graphs considered here are finite, undirected,simple and connected. The distance d(u, v) between two vertices u and v in a connected graph G is the length of a shortest u v path in G. It is well known that this distance is a metric on the vertex set V (G). For a vertex v of G, the eccentricity e (v) is the distance between v and a vertex farthest from v. The minimum eccentricity among the vertices of G is radius, rad G and the maximum eccentricity is the diameter, diam G. A u v path of length d(u, v) is called a u v geodesic. We define I[u, v] to the set (interval) of all vertices lying on some u v geodesic of G and for a nonempty subset S of V (G), I[S] = u,v S I[u, v]. A set S of vertices of G is called a geodetic set in G if I[S] = V (G), and a geodetic set of minimum cardinality is a minimum geodetic set, and generally, if there is a k-subset T of V (G) such that I(S) T = V (G), where k < G S, then S is called a Smarandachely k-geodetic set of G. The cardinality of a minimum geodetic set in G is called Received December 3, 4, Accepted December 6, 5.

115 Tejaswini K M, Venkanagouda M Goudar and Venkatesh the geodetic number and is denoted by g (G). The concept of geodetic number of a graph was introduced in [, 4, 7], further studied in [, 3], and the split geodetic number of a graph was introduced in []. It was shown in [7] that determining the geodetic number of a graph is an NP-hard problem. A set of vertices S in a graph G is a nonsplit geodetic set if S is a geodetic set and the subgraph G[V S] induced by V (G) S is connected. The minimum cardinality of a nonsplit geodetic set, denoted g ns (G), is called the nonsplit geodetic number of G. x u v y w Figure. Consider the graph G of Figure.. For the vertices u and y in G d (u, y) = 3 and every vertex of G lies on an u y geodesic in G. Thus S = {u, y} is the geodetic set of G and so g (G). Here the induced subgraph V (G) S is connected. So that S is a minimum nonsplit geodetic set of G. Therefore nonsplit geodetic number g ns (G) =. A vertex v is an extreme vertex in a graph G, if the subgraph induced by its neighbours is complete. A vertex cover in a graph G is a set of vertices that covers all edges of G. The minimum number of vertices in a vertex cover of G is the vertex covering number α (G) of G. An edge cover of a graph G without isolated vertices is a set of edges of G that covers all the vertices of G. The edge covering number α (G) of a graph G is the minimum cardinality of an edge cover of G. For any undefined term in this paper, see [, 6]. Preliminary Notes We need the following results to prove our results. Theorem. Every geodetic set of a graph contains its extreme vertices. Theorem. For any tree T with k pendant vertices, g(t) = k. Theorem.3 For any graph G of order n, α (G) + β (G) = n.

116 Nonsplit Geodetic Number of a Graph Theorem.4 For cycle C n of order n 3, if n even, g(c n ) = 3 if n odd. Theorem.5 If G is a nontrivial connected graph, then g(g) g(g K ). 3. Nonsplit Geodetic Number Theorem 3. For cycle C n of order n 3, g ns (C n ) = n n + if n is even, + if n is odd. Proof Suppose C n be cycle with n 3, we have the following Case. Let n be even. Consider C p = {v, v,, v p, v } be a cycle with p vertices. Then v p+ is the antipodal vertex of v. Suppose S = {v, v p+ } be the geodetic set of G. It is clear that V (G) S is not connected. Thus S is not a nonsplit geodetic set. But S = {v, v,, v p+ } is a nonsplit geodetic set of G. So that g ns (G) (p + ). If S is any set of vertices of G with S < S then S contains at most p-elements. Hence V (G) S is not connected. This follows that g ns (G) = p + = n +. Case. Let n be odd. Consider C p+ = {v, v,, v p+, v } be a cycle with p+ vertices. Then v p+ and v p+ are the antipodal vertices of v. Now consider S = {v, v p+, v p+ } be the geodetic set of G and it is clear that V (G) S is not connected. Thus S is not a nonsplit geodetic set. But S = {v, v,, v p+, v p+ } is a nonsplit geodetic set of G so that g ns (G) p +. If S is any set of vertices of G with S < S then S contains at most p-elements. Hence V (G) S is not connected. This follows that g ns (G) = p + = n +. ¾ Theorem 3. For any nontrivial tree T with k-pendant-vertices, then g ns (T) = k. Proof Let S = {v, v,, v k } be the set containing pendant vertices of a tree T. By Theorem., g(t) S. On the other hand, for an internal vertex v of T there exist pendant vertices x,y of T such that v lies on the unique x-y geodesic in T. Thus, v I [S] and I [S] = V (T).Then g(t) S. Thus S itself a minimum geodetic set of T.Therefore g(t) = S = k and V S is connected. Hence g ns (T) = k. ¾ Theorem 3.3 For any integers r, s,g ns (K r,s ) = r + s. Proof Let G =K r,s, such that U = {u, u,, u r }, W = {w, w,, w s } are the partite

117 Tejaswini K M, Venkanagouda M Goudar and Venkatesh sets of G, where r s and also V = U W. Consider S = U W x for any x W. Every w k W, k s lies on u i u j geodesic for i j r, so that S is a geodetic set of G. Since V (G) S is connected and hence S itself a nonsplit geodetic set of G. Let S be any set of vertices such that S < S. If S is not a subset of U then V (G) S is not connected and so S is not a nonsplit geodetic set of G. If S is not a subset of W x, again S is not a nonsplit geodetic set of G, by a similar argument.if S = U then S is a geodetic set but V (G) S is not connected, so S is not nonsplit geodetic set. If S = W x then S is not a nonsplit geodetic set of G. From the above argument, it is clear that S is a minimum nonsplit geodetic set of G. Hence g ns (Kr, s) = S = r + s. ¾ Theorem 3.4 If G is a star then g ns (G) = n. Proof Let V (G) = {v, v,, v n, v n } and let S = {v, v,, v n } be the set of pendant vertices of G and is the geodetic set of G. Clearly, the subgraph induced by V (G) S = v n is connected. Hence S = {v, v,, v n } is a minimum nonsplit geodetic set of G. Therefore g ns (G) = n. ¾ Theorem 3.5 For any nontrivial connected graph G different from star of order n and diameter d, g ns (G) n d +. Proof Let u and v be the vertices of G for which d(u, v) = d and let u = v, v,, v d = v be a u v path of length d. Now S = V (G) {v, v,, v d } then I[S] = V [G] and consequently g ns (G) S n d +. ¾ Theorem 3.6 For any tree T, g ns (T) + g(t) < m. Proof Suppose S = {v, v, v 3,, v k } be the set of all pendant vertices in T, forms a minimal geodetic set of I [S] = V (T). Further {u, u, u 3,, u l } V (G) S is the set of internal vertices in T. Then V (G) S forms a minimal non split geodetic set of T, it follows that S + S < m. Hence g ns (T) + g(t) < m. ¾ Theorem 3.7 For any graph G of order n, g ns (G) g s (G), where G is not a cycle.. Proof Let G be any graph with n vertices. Consider a nonsplit geodetic set S = {v, v,, v k } of a graph G. Since V (G) S is connected, the set S is not a split geodetic set of G. Now, we consider a set S = S {a, b} for any a, b V (G) such that V (G) S is disconnected. Therefore S is the split geodetic set of G with minimum cardinality. Thus S < S. Clearly g ns (G) g s (G). ¾ Theorem 3.8 Let G be a cycle of order n then g s (G) g ns (G). Proof Let G be a cycle of order n, we discuss the following cases. Case. Suppose n is even. Let S = {v i, v j } be the split geodetic set of G where v i, v j are the two antipodal vertices of G. The v i v j geodesic includes all the vertices of G and V (G) S

118 Nonsplit Geodetic Number of a Graph 3 is disconnected. But S = {v i, v i+,, v j } is a nonsplit geodetic set of G and the induced subgraph V (G) S is connected. Thus S S. Clearly g s (G) g ns (G). Case. Suppose n is odd. Let S = {v i, v j, v k } be the split geodetic set of G. By the Theorem.4, no two vertices of S form a non split geodetic set and V (G) S is disconnected. But S = {v i, v i+,, v j, v k } is a nonsplit geodetic set of G and the induced subgraph V (G) S is connected. Thus S S. Clearly g s (G) g ns (G). ¾ Theorem 3.9 For the wheel W n = k + C n (n 5), g ns (W n ) = n if n is even n if n is odd Proof Let W n = K +C n and let V (W n )={x, u, u,, u n }, where deg(x) = n > 3 and deg(u i ) = 3 for each i {,,, n }. We discuss the following cases. Case. Let n be even. Consider geodesic P : {u, u, u 3 }, Q : {u 3, u 4, u 5 },, R : {u n, u n, u n+, }. It is clear that the vertices u, u 4, u n lies on the geodesic P, Q and R. Also u, u 3, u 5,, u n, u n+ is a minimum nonsplit geodetic set such that V (G) S is connected and it has n vertices. Hence g ns(w n ) = n. Case. Let n be odd. Consider geodesic P : {u, u, u 3 }, Q : {u 3, u 4, u 5 },, R : {u n, u n, u n+, }. It is clear that the vertices u, u 4,, u n lies on the geodesic P, Q and R. Also u, u 3, u 5,, u n, u n+ is a minimum nonsplit geodetic set such that V (G) S is connected and it has n vertices. Hence g ns (W n ) = n. ¾ Theorem 3. Let G be a graph such that both G and G are connected then g ns (G)+g ns (G) n(n 3) +. Proof Since both G and G are connected, we have (G) (G) < n.thus β (G), β (G). Hence, g ns n g ns (G) (n ) n + g ns (G) m n +. Similarly, g ns (G) m n +. Thus, g ns (G) + g ns (G) (m + (m)) n + g ns (G) + g ns (G) n(n ) n + Theorem 3. For any nontrivial tree T, g ns (T) α (T). g ns (G) + g ns (G) n n n + g ns (G) + g ns (G) n 3n + g ns (G) + g ns (G) n(n 3) +. ¾

119 4 Tejaswini K M, Venkanagouda M Goudar and Venkatesh Proof Let S be a minimum cover set of vertices in T. Then S has at least one vertex and every vertex in S is adjacent to some vertices in V (G) S. This implies that S is a nonsplit geodetic set of G. Thus g ns (T) α (T). ¾ Theorem 3. For any nontrivial tree T with m edges, g ns (T) m α(t) +,where α (T) is an edge covering number. Proof Suppose S = {e, e,, e i } be the set of all end edges in T and J E(T) S be the minimal set of edges such that S J = α (T). By the Theorem. S is the minimal geodetic set of G. Also it follows that V (G) S is connected. Clearly, g ns (T) E(T) S J α (T) + g ns(t) m +. ¾ Theorem 3.3 For a cycle C n of order n, g ns (G) = α (C n ) +. Proof Consider a cycle C n of order n. We discuss the following cases. Case. Suppose that n is even and α (C n ) is the vertex covering number of C n. We have by Theorem 3., g ns (G) = n + and also for an even cycle, vertex covering number α (C n ) = n. Hence, g ns (G) = n + = α (C n ) +. Case. Suppose that n is odd and α (C n ) is the vertex covering number of C n. We have by Theorem 3., g ns (G) = n + and also for an odd cycle, vertex covering number α (C n ) = n +. Hence, g ns(g) = n + g ns(g) = α (C n ) +. ¾ Theorem 3.4 If is a connected noncomplete graph G of order n,g ns (n κ(g)) +, where κ(g) is vertex connectivity. Proof Let κ(g) = k. Since G is connected but not complete κ(g) n. Let U = {u, u,, u k } be a minimum cut set of G, let G, G,, G r (r ) be the components of G U and let W = V (G) (U ) then every vertex u i ( i k) is adjacent to at least one vertex of G j for every (i j r). Therefore, every vertex u i belongs to a W geodesic path. Thus g ns (G) = W (V (G) U) + (n κ(g)) +. ¾ 4. Corona Graph Let G and H be two graphs and let n be the order of G. The corona product G H is defined as the graph obtained from G and H by taking one copy of G and n copies of H and then joining by an edge, all the vertices from the i th -copy of H with the i th -vertex of G. Theorem 4. Let G be a connected graph of order n and H be any graph of order m then g ns (G H) = nm.

120 Nonsplit Geodetic Number of a Graph 5 Proof Let S be a nonsplit geodetic set in G H, v i V (G), i n and u j V (H), j m. For each v i there is a copy Hv i which contains u j vertices. Clearly V (Hu j ) S is a geodetic set of G H and V (G) S is connected. Further every w k (G H) lies on the geodesic path in S. Therefore S is the minimum nonsplit geodetic set. Thus, S = g ns (G H) = nm. ¾ 5. Adding a Pendant Vertex An edge e = (u, v) of a graph G with deg(u) = and deg(v) > is called an pendant edge and u an pendant vertex. Theorem 5. Let G be the graph obtained by adding an pendant edge (u, v) to a cycle G = C n of order n > 3, with u G and v / G, then g ns (G if n is even ) = 3 if n is odd Proof Let {u, u, u 3,, u n, u } be a cycle with n vertices. Let G be the graph obtained from G = C n by adding an pendant edge (u, v) such that u G and v / G. We discuss the following cases. Case. For G = C n, let S = {v, u i } be a non split geodetic set of G, where v is the pendant vertex of G and diam(g ) = v u i path, clearly I[S] = V [G ]. Also for all x, y V (G ) S, V (G ) S is connected. Hence, g ns (G ) =. Case. For G = C n+, let S = {v, a, b} be a non split geodetic set of G, where v is the pendant vertex of G and a, b G such that d(v, a) = d(v, b). Thus I[S] = V [G ] and V (G ) S is connected. Hence, g ns (G ) = 3. ¾ Theorem 5. Let G be the graph obtained by adding a pendant vertex (u i, v i )for i =,, 3,, n to each vertex of G = C n such that u G,v i / G, then g ns (G ) = k. Proof Let G = C n = {u, u, u 3,, u n, u } be a cycle with n vertices. Let G be the graph obtained by adding an pendant vertex {u i, v i }, i =,, 3,, n to each vertex of G such that u i G and v i / G. Let S = {v, v, v 3,, v k } be a non split geodetic set of G. Clearly I[X] V (G ). Also, x, y V (G ) S with V (G ) S connected. Thus, g ns (G ) = k. ¾ Theorem 5.3 Let G be the graph obtained by adding k pendant vertices {(u, v ),, (u, v k )} to a cycle G = C n of order n > 3, with u G and {v, v,, v k } / G. Then g ns (G k + ) = k + if n is even if n is odd Proof Consider {u, u, u 3,, u n, u } be a cycle with n vertices. Let G be the graph

121 6 Tejaswini K M, Venkanagouda M Goudar and Venkatesh obtained from G = C n by adding k pendant edges {u i v, u i v,, u i v k } such that u i a single vertex of G and {v, v, v 3,, v k } does not belongs to G. We discuss the following cases. Case. Let G = C n. Consider X = {v, v, v 3,, v k } u i, for any vertex u i of G. Now S = {X} be a non split geodetic set, such that {v, v, v 3,, v k } are the pendant vertices of G and u j is the antipodal vertex of u i in G. Thus I[X] = V [G ]. Consider P = {v, v, v 3,, v k } as a set of pendant vertices such that P < S is not a non split geodetic set i.e for some vertex u j V G,u j / I[P]. If P = X, then P is not nonsplit geodetic set. Thus S is a minimum non split geodetic set of G and V (G ) S is connected. Thus, g ns (G ) = k +. Case. Let G = C n+. Consider S = {v, v, v 3,, v k, a, b} be a non split geodetic set, where {v, v,, v k } / G are k pendant vertices of G not in G and a, b G such that d(u, a) = d(u, b). Thus I[S] = V [G ]. Also x, y V (G ) S it follows that V (G ) S is connected. Therefore, g ns (G ) = k +. ¾ 6. Cartesian Products The cartesian product of the graphs H and H written as H H, is the graph with vertex set V (H ) V (H ), two vertices u, u and v, v being adjacent in H H iff either u = v and (u, v ) E(H ), or u = v and (u, v ) E(H ). Theorem 6. Let K and G = C n be the graphs then if n is even g ns (K G) = 3 if n > 5 is odd 4 if n=3 Proof Consider G = C n, let K G be graphs formed from two copies G and G of G. Let V = {v, v,, v n } be the vertices of G and W = {w, w,, w n } be the vertices of G and U = V W. We consider the following cases. Case. Let n be even. Consider S = {v i, w j } be the non split geodetic of K G such that v i to w j path is equal to diam(k G) which includes all the vertices of K G. Hence U S is connected. Therefore, g ns [K G] =. Case. Let n be odd. Consider S = {v i, w j, v k } be the non split geodetic set of K G such that v i to w j path is equal to diam(k G) and is equal to w j v k path and also v i w j w j v k path includes all the vertices of K G. Hence U S is connected. Therefore, g ns [K G] = 3. Case 3. For n = 3, let S = {v i, w j, v k } be the geodetic set of K G, that is v i w j is equal to diam(k G) and is equal to w j v k and also I[S] = U(K G). But U S is not connected. Let S = S {v n } = {v i, w j, v k, v n } be the non split geodetic set of K G. Hence, U S is connected. Therefore, g ns [K G] = 4. ¾ Theorem 6. For any complete graph K n of order n, g ns [K K n ] = n +.

122 Nonsplit Geodetic Number of a Graph 7 Proof Consider K K n be graph formed from two copies of G and G of G. Now, let us prove the result by mathematical induction, For n =, g ns [k K ] = 3, since K K = C 4 by Theorem 3. we have g ns [C 4 ] = 3 the result is true. Let us assume that the result is true for n=m,that is g ns [K K m ] = m +. Now, we shall prove the result for n = m+. Let S = {v, v, v 3,, v m+ } be the nonsplit geodetic set formed from some elements in G and the elements which are not corresponds to elements in G of K K m+. Clearly I[S] = V (K K n ). Let P be any set of vertices such that P < S. Suppose P = {v, v, v 3,, v m } which is not non split geodetic set, because I[P] V [K K m+ ]. So S itself a minimum geodetic set of K K m+. Hence, g ns [K K m+ ] = m + +. Thus, g ns (K K n ) = n +. ¾ Theorem 6.3 For any complete graph of order n 3, g ns (K n K n ) = n. Proof We shall prove the result by mathematical induction, For n 3, let us assume that the result is true for n = m, that is g ns (K m K m ) = m. Now, we shall prove the result for n = m +. Let S = {v, v, v 3,, v m+ } be the non split geodetic set formed from some elements in G and the elements which are not corresponds to elements in G of K m+ K m+. Clearly I[S] = V (K K n ). Now, consider P be any set of vertices such that P < S. Suppose P = {v, v, v 3,, v m } which is not non split geodetic set, because I[P] V (K m+ K m+ ). So S itself a minimum geodetic set of K m+ K m+. Hence, g ns (K m+ K m+ ) = m +. Thus g ns (K n K n ) = n. ¾ Theorem 6.4 Let G and H be graphs then g ns (G H) max{g(g), g(h)}.equality holds when G,H are complete graphs and n 3. Proof If S is a minimum geodetic set in G H then we have I[S] = a,b S I[a, b] = a,b S I[a, b ] I[a, b ] ( a,b SI[a, b ]) ( a,b SI[a, b ]) = I[S ] I[S ],V (G H) = I[S] I[S ] I[S ]. Therefore S and S geodetic set in G, H respectively, so g ns (G H) = S max{ s, s } max{g(g), g(h)}, proving the lower bound. Consider complete graphs G, H with vertex sets V (G) = {u, u,, u p } and V (H) = {v, v,, v q } where without loss of generality p q. Then g(g) = p and g(h) = q. Let S = {(u, v ), (u, v ),, (u q, v q ), (u q+, v q ), (u q+, v q ),, (u p, v q )}. It is straight forward to verify that S is a non split geodetic set for G H. Hence, g ns (G H) S p = max{g(g), g(h)} g ns (G H), so equality holds. ¾ Theorem 6.5 Let G = T and H = K be the graphs with g(g) = p g(h) = q then g ns (G H) pq q. Proof Set X = G H. Let S = {g, g,..., g p } and T = {h, h,, h q } be the geodetic sets of G and H respectively, and U = {(S T)/ p,q i,j= {(g i, h j )}}. We claim that I X [U] = V (X). Let (g, h) be an arbitrary vertex of X.Then there exists indices i and i such that g I G [g i, g i ] and there are indices j and j such that h I H [h j, h j ]. Since p, q we may assume that i = i and j = j. Indeed,if say g = g i then i to be an

123 8 Tejaswini K M, Venkanagouda M Goudar and Venkatesh arbitrary index from {,,, p} different from i. Set B = {(g i, h j ), (g i, h j ), (g i, h j ), (g i, h j )}. Suppose that one of the vertices from B is not in U. We may without loss of generality assume (g i, h j ) inu. This means that i = j. Therefore i j and i j. Then we infer that (g, h) I X [(g i, h j ), (g i, h j )]. Otherwise, all vertices from B are in U, then (g, h) I X [(g i, h j ), (g i, h j )]. Hence, g ns [G H] pq q. ¾ Theorem 6.6 Let K and T be the graphs then g ns (K T) = g ns (T). Proof Consider a tree T. Let K T be a graph formed from two copies T and T of T and S be a minimum non split geodetic set of K T. Now, we define S to be the union of those vertices of S in T and the vertices of T corresponding to vertices of T belonging to S. Let v V (T ) lies on some x y geodesic,where x, y S.Since S is a non split geodetic set by Theorem 3., i.e., g ns (T) = k at least one of x and y belongs to V. If both x, y V then x, y S. Hence, we may assume that x V, y V. If y corresponds to x then v = x S. Hence, we assume that y corresponds to y S,where y x. Since d(x, y) = d(x, y ) + and the vertex v lies on an x y geodesic in K G. Hence, v lies on x y geodesic in G that is g ns (G) g ns (K G). Conversely, let S contains a vertex with the property that every vertex of T lies on x w geodesic T for some w S. Let S consists of x together with those vertices of T corresponding to those S {x}. Thus, S = S. We show that S is a non split geodetic set of K T. Hence g ns (K T) g ns (T). Thus, g ns (K T) = g ns (T). ¾ Theorem 6.7 Let K and G = P n be the two graphs, ifn 3 g ns (K G) = 3 ifn = Proof Consider a trivial graph K as a connected graph. Let G and G be the two copies of G and also V (G ) = {a, a,, a n }, V (G ) = {b, b,, b n }. Let S = {a, b n } be the non split geodetic set of K G and also d(a, b n ) = diam(a, b n ). Thus, V S = {a, a 3,, a n, b, b,, b n } is the induced subgraph and it is connected. Hence g ns [K G] =. Similarly,the result is obvious for n = that is g ns [K G] = 3. ¾ 7. Block Graphs A block graph has a subgraph G of G(not a null graph) such that G is non separable and if G is any other graph of G, then G G = G or G G is separable. For any graph G a complete subgraph of G is called clique of G. The number of vertices in a largest clique of G is called the clique number of G and denoted by ω(g). Theorem 7. For any block graph G, g ns (G) = n c i where n be the number of vertices and c i be the number of cut vertices.

124 Nonsplit Geodetic Number of a Graph 9 Proof Let V = {v, v,, v n } be the number of vertices of G. Consider S be the geodetic set of G and V (G) S is connected. Thus S itself a nonsplit geodetic set of G. Since every geodetic set does not contain any cut vertices. Hence, g ns (G) = n c i. ¾ Theorem 7. For any block graph G, g ns (G) ω(g) + c i where ω(g) be the clique number and c i be the number of cut vertices. Proof Let V = {v, v,, v n } be the number of vertices of G. In a block graph, every geodetic set is a nonsplit geodetic set. Consider S be the geodetic set of G and V (G) S is connected. Thus S itself a nonsplit geodetic set of G. By the definition, the number of vertices in a largest clique of G is ω(g) and also every geodetic set does not contain any cut vertices of G. It follows that g ns (G) ω(g) + c i. ¾ Theorem 7.3 For any block graph G, g ns (G) = α (G) + where α (G) be the vertex covering number. Proof Let G be a block graph of order n. Now, we prove the result by mathematical induction. For c i =, the vertex covering number of G is α (G) = n c i α (G) = n α (G) + = n, by Theorem 7., we have g ns (G) = n c i g ns (G) = n. Therefore, g ns (G) = α (G) +. Thus the result is result is true for c i =. Let us assume that the result is true for c i = m that is g ns (G) = α (G) +. Now, we shall prove the result for c i = m +, where m+ is the number of cut vertices. Let S = {v, v,, v n } be the minimum nonsplit geodetic set of G. Since every geodetic set does not contain any cut vertex, by Theorem 7. we have g ns (G) = n m. Therefore, α (G) = n c i α (G) = (n m ) α (G) + = n m. Thus, g ns (G) = α (G) +. ¾ References [] F.Buckley, F.Harary, Distance in Graphs, Addison-Wesley, Redwood City,CA, 99. [] F.Buckley, F.Harary, L V.Quintas, Extremal results on the geodetic number of a graph, Scientia A, Redwood City,CA, (988), 7-6. [3] G.Chartrand, F.Harary and P.Zhang, On the geodetic number of a graph, Networks, 39(988), -6. [4] G.Chartrand, E.M.Palmer and P.Zhang, The geodetic number of a graph, A Survey, Congr.Numer., 56()

125 Tejaswini K M, Venkanagouda M Goudar and Venkatesh [5] G.Chartrand and P.Zhang, The forcing geodetic number of a graph, Discuss.Math, Graph Theory, 9(999), [6] F.Buckley, F.Harary, Graph Theory, Addison-Wesley, 969. [7] F.Harary, E.Loukakis, C.Tsouros, The geodetic number of a graph, Math.Comput.Modelling, 7()(993), [8] P.A.Ostrand, Graphs with specified radius and diameter, Discrete Math., 4(973), [9] A.P.Santhakumaran,P.Titus and J.John, On the connected geodetic number of a graph, Congr.Numer., (in press). [] Venkanagouda M Goudar, Ashalatha K S and Venkatesh, Split geodetic number of a graph, Advances and Applications in Discrete Mathematics, Vol.3,(4), 9-.

126 International J.Math. Combin. Vol.(6), -3 k-difference cordial labeling of graphs R.Ponraj, M.Maria Adaickalam and R.Kala 3.Department of Mathematics, Sri Paramakalyani College, Alwarkurichi-674, India.Department of Economics and Statistics, Government of Tamilnadu, Chennai-6 6, India 3.Department of Mathematics, Manonmaniam Sundaranar University, Tirunelveli-67, India Abstract: In this paper we introduce new graph labeling called k-difference cordial labeling. Let G be a (p,q) graph and k be an integer, k V (G). Let f : V (G) {,,, k} be a map. For each edge uv, assign the label f(u) f(v). f is called a k- difference cordial labeling of G if v f (i) v f (j) and e f () e f () where v f (x) denote the number of vertices labelled with x, e f () and e f () respectively denote the number of edges labelled with and not labelled with. A graph with a k-difference cordial labeling is called a k-difference cordial graph. In this paper we investigate k-difference cordial labeling behavior of star, m copies of star and we prove that every graph is a subgraph of a connected k-difference cordial graph. Also we investigate 3-difference cordial labeling behavior of some graphs. Key Words: Path, complete graph, complete bipartite graph, star, k-difference cordial labeling, Smarandachely k-difference cordial labeling. AMS(): 5C78.. Introduction All graphs in this paper are finite and simple. The graph labeling is applied in several areas of sciences and few of them are coding theory, astronomy, circuit design etc. For more details refer Gallian []. Let G, G respectively be (p, q ), (p, q ) graphs. The corona of G with G, G G is the graph obtained by taking one copy of G and p copies of G and joining the i th vertex of G with an edge to every vertex in the i th copy of G. The subdivision graph S (G) of a graph G is obtained by replacing each edge uv by a path uwv. The union of two graphs G and G is the graph G G with V (G G ) = V (G ) V (G ) and E (G G ) = E (G ) E (G ). In [], Cahit introduced the concept of cordial labeling of graphs. Recently Ponraj et al. [4], introduced difference cordial labeling of graphs. In this way we introduce k-difference cordial labeling of graphs. Also in this paper we investigate the k-difference cordial labeling behavior of star, m copies of star etc. x denote the smallest integer less than or equal to x. Terms and results not here follows from Harary [3]. Received June 4, 5, Accepted May 6, 6.

127 R.Ponraj, M.Maria Adaickalam and R.Kala. k-difference Cordial Labeling Definition. Let G be a (p, q) graph and k be an integer k V (G). Let f : V (G) {,,, k} be a function. For each edge uv, assign the label f(u) f(v). f is called a k-difference cordial labeling of G if v f (i) v f (j) and e f () e f (), and Smarandachely k-difference cordial labeling if v f (i) v f (j) > or e f () e f () >, where v f (x) denote the number of vertices labelled with x, e f () and e f () respectively denote the number of edges labelled with and not labelled with. A graph with a k-difference cordial labeling or Smarandachely k-difference cordial labeling is called a k-difference cordial graph or Smarandachely k-difference cordial graph, respectively. Remark. () p-difference cordial labeling is simply a difference cordial labeling; () -difference cordial labeling is a cordial labeling. Theorem.3 Every graph is a subgraph of a connected k-difference cordial graph. Proof Let G be (p, q) graph. Take k copies of graph K p. Let G i be the i th copy of K p. Take k copies of the K ( p ) and the ith copies of the K ( p ) is denoted by G i. Let V (G i) = {u j i : j k, i p}. Let V (G i ) = {vj i : j k, i p}. The vertex and edge set of super graph G of G is as follows: Let V (G ) = k V (G i ) k V (G i ) {w i : i k} {w}. E(G ) = k i= i= i= E(G i ) {u j vj i : i ( ) p, j k } {u k w, wvi k : i ( p ) } {u j p w j : j k} {u j uj+ : j k } {w w }. Assign the label i to the vertices of G i, i k. Then assign the label i + to the vertices of G i, i k. Assign the label to the vertices of G k. Then assign to the vertex w. Finally assign the label i to the vertex w i, i k. Clearly v f (i) = p + ( p ) +, i =, 3,...,k, v f () = p + ( ) p + and ef () = k ( ) p + k, ef () = k ( p ) + k +. Therefore G is a k-difference cordial graph. Theorem.4 If k is even, then k-copies of star K,p is k-difference cordial. Proof Let G i be the i th copy of the star K,p. Let V (G i ) = {u j, v j i : j k, i p} and E(G i ) = {u j v j i : j k, i p}. Assign the label i to the vertex u j, j k. Assign the label i + to the pendent vertices of G i, i k. Assign the label k i + to the pendent vertices of G k +i, i k. Finally assign the label to all the pendent vertices of the star G k. Clearly, v f (i) = p +, i k, e f () = e f () = kp. Therefore f is a k-difference cordial labeling of k-copies of the star K,p. Theorem.5 If n (mod k) and k 6, then the star K,n is not k-difference cordial. Proof Let n = kt. Suppose f is a k-difference cordial labeling of K,n. Without loss of generality, we assume that the label of central vertex is r, r k. Clearly v f (i) = t, ¾ ¾

128 k-difference cordial labeling of graphs 3 i n and i r, v f (r) = t +. Then e f () t and e f () (k )t. Now e f () (k )t t (k 4)t, which is a contradiction. Thus f is not a k-difference cordial. ¾ Next we investigate 3-difference cordial behavior of some graph Difference Cordial Graphs First we investigate the path. Theorem 3. Any path is 3-difference cordial. Proof Let u u...u n be the path P n. The proof is divided into cases following. Case. n (mod 6). Let n = 6t. Assign the label, 3,,, 3, to the first consecutive 6 vertices of the path P n. Then assign the label, 3,,, 3, to the next 6 consecutive vertices. Then assign the label, 3,,, 3, to the next six vertices and assign the label, 3,,, 3, to the next six vertices. Then continue this process until we reach the vertex u n. Case. n (mod 6). This implies n (mod 6). Assign the label to the vertices of u i, i n as in case. If u n receive the label, then assign the label to the vertex u n ; if u n receive the label, then assign the label to the vertex u n. Case 3. n (mod 6). Therefore n (mod 6). As in case, assign the label to the vertices u i, i n. Next assign the label 3 to u n. Case 4. n 3 (mod 6). This forces n (mod 6). Assign the label to the vertices u, u,... u n as in case 3. Assign the label or to u n according as the vertex u n receive the label or. Case 5. n 4 (mod 6). This implies n 3 (mod 6). As in case 4, assign the label to the vertices u, u,, u n. Assign the label or to the vertex u n according as the vertex u n receive the label to. Case 6. n 5 (mod 6). This implies n 4 (mod 6). Assign the label to the vertices u, u,, u n as in Case 5. Next assign the label 3 to u n. ¾ Example 3. A 3-difference cordial labeling of the path P 9 is given in Figure Figure

129 4 R.Ponraj, M.Maria Adaickalam and R.Kala Corollary 3.3 If n, 3 (mod 4), then the cycle C n is 3-difference cordial. Proof The vertex labeling of the path given in Theorem 3. is also a 3-difference cordial labeling of the cycle C n. ¾ Theorem 3.4 The star K,n is 3-difference cordial iff n {,, 3, 4, 5, 6, 7, 9}. Proof Let V (K,n ) = {u, u i : i n} and E(K,n ) = {uu i : i n}. Our proof is divided into cases following. Case. n {,, 3, 4, 5, 6, 7, 9}. Assign the label to u. The label of u i is given in Table. n\u i u u u 3 u 4 u 5 u 6 u 7 u 8 u Case. n / {,, 3, 4, 5, 6, 7, 9} Table Let f(u) = x where x {,, 3}. To get the edge label, the pendent vertices receive the label either x or x +. Subcase. n = 3t. Subcase a. x = or x = 3. When x =, e f () = t or t + according as the pendent vertices receives t s or (t+) s. Therefore e f () = t or t. Thus e f () e f () = t >, t > 4 a contradiction. When x = 3, e f () = t or t + according as the pendent vertices receives t s or (t+) s. Therefore e f () = t or t. Thus e f () e f () = t or t. Therefore, e f () e f () >, a contradiction. Subcase b. x =. In this case, e f () = t or t + according as pendent vertices receives t s or (t-) s. Therefore e f () = t or t. e f () e f () = t or t + as t > 3. Therefore, e f () e f () >, a contradiction. Subcase. n = 3t +.

130 k-difference cordial labeling of graphs 5 Subcase a. x = or 3. Then e f () = t or t + according as pendent vertices receives t s or (t+) s. Therefore e f () = t + or t. e f () e f () = t + or t as t > 3. Therefore, e f () e f () > 3, a contradiction. Subcase b. x =. In this case e f () = t or t + according as pendent vertices receives t s and t s 3 and t s and (t+3) s 3. Therefore e f () = t + or t. e f () e f () = t or t as t > 3. Therefore, e f () e f () >, a contradiction. Subcase 3. n = 3t +. Subcase 3a. x = or 3. This implies e f () = t + and e f () = t +. e f () e f () = t as t > 3. Therefore, e f () e f () >, a contradiction. Subcase 3b. x =. This implies e f () = t + and e f () = t. e f () e f () = t + as t >. Therefore, e f () e f () >, a contradiction. Thus K,n is 3-difference cordial iff n {,, 3, 4, 5, 6, 7, 9}. ¾ Next, we research the complete graph. Theorem 3.5 The complete graph K n is 3-difference cordial if and only if n {,, 3, 4, 6, 7, 9, }. Proof Let u i, i n be the vertices of K n. The 3-difference cordial labeling of K n, n {,, 3, 4, 6, 7, 9, } is given in Table. n\u i u u u 3 u 4 u 5 u 6 u 7 u 8 u 9 u Table Assume n / {,, 3, 4, 6, 7, 9, }. Suppose f is a 3-difference cordial labeling of K n. Case. n (mod 3). Let n = 3t, t > 3. Then v f () = v f () = v f () = t. This implies e f () = ( t ) + ( t ) + ( t ) +

131 6 R.Ponraj, M.Maria Adaickalam and R.Kala t = 5t 3t. Therefore e f () = t + t = t. e f () e f () = 5t 3t t > as t > 3, a contradiction. Case. n (mod 3). Let n = 3t +, t > 3. Subcase. v f () = t +. Therefore v f () = v f (3) = t. This forces e f () = ( ) ( t+ + t ) ( + t ) + t(t + ) = (5t + t). e f () = t(t + ) + t = t + t. Then e f () e f () = (5t + t) (t + t) > as t > 3, a contradiction. Subcase. v f (3) = t +. Similar to Subcase. Subcase 3. v f () = t +. Therefore v f () = v f (3) = t. In this case e f () = 5t +t and e f () = t(t + ) + t(t + ) = t + t. This implies e f () e f () = 5t +t (t + t) > as t > 3, a contradiction. Case 3. n (mod 3). Let n = 3t +, t. Subcase. v f () = t. Therefore v f () = v f (3) = t +. This gives e f () = ( t ) + ( t+ ) + ( t+ ) + t(t + ) = 5t +3t and e f () = t(t+)+(t+) = t +3t+. This implies e f () e f () = 5t +3t (t +3t+) > as t, a contradiction. Subcase. v f (3) = t. Similar to Subcase. Subcase 3. v f () = t. Therefore v f () = v f (3) = t +. In this case e f () = ( ) ( t+ + t+ ) ( + t ) + (t + )(t + ) = 5t +5t+ and e f () = t(t + ) + t(t + ) = t + t. This implies e f () e f () = 5t +5t+ (t + t) > as t, a contradiction. ¾ Theorem 3.6 If m is even, the complete bipartite graph K m,n (m n) is 3-difference cordial. Proof Let V (K m,n ) = {u i, v j : i m, j n} and E(K m,n ) = {u i v j : i

132 k-difference cordial labeling of graphs 7 m, j n}. Define a map f : V (K m,n ) {,, 3} by f(u i ) =, i m f(u m +i ) =, i m f(v i ) = 3, i m+n 3 f(v m+n 3 +i ) =, i m+n 3 m if m + n, (mod 3) i m+n 3 m if m + n (mod 3) f(v m+n 3 m +i) =, i n m+n 3 + m + if m + n, (mod 3) f(v m+n 3 m+i) =, i n m+n 3 + m if m + n (mod 3) Since e f () = e f () = mn, f is a 3-difference cordial labeling of K m,n. ¾ Example 3.7 A 3-difference cordial labeling of K 5,8 is given in Figure Next, we research some corona of graphs. Figure Theorem 3.8 The comb P n K is 3-difference cordial. Proof Let P n be the path u u... u n. Let V (P n K ) = V (P n ) {v i : i n} and E(P n K ) = E(P n ) {u i v i : i n}. Case. n (mod 6). Define a map f : V (G) {,, 3} by f(u 6i 5 ) = f(u 6i ) =, i n 6 f(u 6i 4 ) = f(u 6i ) = 3, i n 6 f(u 6i 3 ) = f(u 6i ) =, i n 6. In this case, e f () = n and e f () = n. Case. n (mod 6). Assign the label to the vertices u i, v i ( i n ) as in case. Then assign the labels, to the vertices u n, v n respectively. In this case, e f () = n, e f () = n.

133 8 R.Ponraj, M.Maria Adaickalam and R.Kala Case 3. n (mod 6). As in Case, assign the label to the vertices u i, v i ( i n ). Then assign the labels 3, 3 to the vertices u n, v n respectively. In this case, e f () = n, e f () = n. Case 4. n 3 (mod 6). Assign the label to the vertices u i, v i ( i n ) as in case 3. Then assign the labels, to the vertices u n, v n respectively. In this case, e f () = n, e f () = n. Case 5. n 4 (mod 6). As in Case 4, assign the label to the vertices u i, v i ( i n ). Then assign the labels, 3 to the vertices u n, v n respectively. In this case, e f () = n, e f () = n. Case 6. n 5 (mod 6). Assign the label to the vertices u i, v i ( i n ) as in case 5. Then assign the labels 3, to the vertices u n, v n respectively. In this case, e f () = n, e f () = n. Therefore P n K is 3-difference cordial. ¾ Theorem 3.9 P n K is 3-difference cordial. Proof Let P n be the path u u u n. Let V (P n K ) = V (P n ) {v i, w i : i n} and E(P n K ) = E(P n ) {u i v i, u i w i : i n}. Case. n is even. Define a map f : V (P n K ) {,, 3} as follows: f(u i ) =, i n f(u i ) =, i n f(v i ) =, i n f(v i ) =, i n f(w i ) = 3, i n. In this case, v f () = v f () = v f (3) = n, e f () = 3n and e f() = 3n. Case. n is odd. Define a map f : V (P n K ) {,, 3} by f(u ) =, f(u ) =, f(u 3 ) = 3, f(v ) = f(v 3 ) =, f(w ) = f(w ) = 3, f(v ) = f(w 3 ) =, f(u i+ ) =, i n 3 f(u i+3 ) =, i n 3 f(v i+ ) =, i n 3 f(v i+3 ) =, i n 3 f(w i+3 ) = 3, i n 3.

134 k-difference cordial labeling of graphs 9 Clearly, v f () = v f () = v f (3) = n, e f () = e f () = 3n. ¾ Next we research on quadrilateral snakes. Theorem 3. The quadrilateral snakes Q n is 3-difference cordial. Proof Let P n be the path u u u n. Let V (Q n ) = V (P n ) {v i, w i : i n } and E(Q n ) = E(P n ) {u i v i, v i w i, w i u i+ : i n }. Note that V (Q n ) = 3n and E(Q n ) = 4n 4. Assign the label to the path vertices u i, i n. Then assign the labels, 3 to the vertices v i, w i i n respectively. Since v f () = n, v f () = v f (3) = n, e f () = e f () = n, f is a 3-difference cordial labeling. ¾ The next investigation is about graphs B n,n, S(K,n ), S(B n,n ). Theorem 3. The bistar B n,n is 3-difference cordial. Proof Let V (B n,n ) = {u, v, u i, v i : i n} and E(B n,n ) = {uv, uu i, vv i : i n}. Clearly B n,n has n + vertices and n + edges. Case. n (mod 3). Assign the label, to the vertices u and v respectively. Then assign the label to the vertices u i, v i ( i n 3 ). Assign the label to the vertices u n 3 +i, v n 3 +i ( i n 3 ). Finally assign the label 3 to the vertices u n +i, 3 vn 3 +i ( i n 3 ). In this case e f() = n + and e f () = n. Case. n (mod 3). Assign the labels to the vertices u, v, u i, v i ( i n ) as in Case. Then assign the label 3, to the vertices u n, v n respectively. In this case e f () = n and e f () = n +. Case 3. n (mod 3). As in Case, assign the label to the vertices u, v, u i, v i ( i n ). Finally assign, 3 to the vertices u n, v n respectively. In this case e f () = n and e f () = n +. Hence the star B n,n is 3-difference cordial. ¾ Theorem 3. The graph S(K,n ) is 3-difference cordial. Proof Let V (S(K,n )) = {u, u i, v i : i n} and E(S(K,n )) = {uu i, u i v i : i n}. Clearly S(K,n ) has n + vertices and n edges. Case. n (mod 3). Define a map f : V (S(K,n )) {,, 3} as follows: f(u) =, f(u i ) f(u t+i ) f(v i ) =, i t =, i t = 3, i t f(v t+i ) =, i t.

135 3 R.Ponraj, M.Maria Adaickalam and R.Kala Case. n (mod 3). As in Case, assign the label to the vertices u, u i, v i ( i n ). Then assign the label, 3 to the vertices u n, v n respectively. Case 3. n (mod 3). As in Case, assign the label to the vertices u, u i, v i ( i n ). Then assign the label, to the vertices u n, v n respectively. f is a 3-difference cordial labeling follows from the following Table 3. Values of n v f () v f () v f (3) e f () e f () n = 3t t t + t 3t 3t n = 3t + t + t + t + 3t + 3t + n = 3t + t + t + t + 3t + 3t + Table 3 Theorem 3.3 S(B n,n ) is 3-difference cordial. Proof Let V (S(B n,n )) = {u, w, v, u i, w i, v i, z i : i n} and E(S(B n,n )) = {uw, wv, uu i, u i w i, vv i, v i z i : i n}. Clearly S(B n,n ) has 4n + 3 vertices and 4n + edges. Case. n (mod 3). Define a map f : V (S(B n,n )) {,, 3} by f(u) =, f(w) = 3, f(v) =, f(w i ) = f(v i ) = f(z i ) =, i n, i n 3, i n f(u i ) =, i n 3 f(u n 3 +i ) =, i n 3 f(u n 3 +i ) = 3, i n 3. Case. n (mod 3). As in Case, assign the label to the vertices u, w, v, u i, v i, w i, z i ( i n ). Then assign the label,,, 3 to the vertices u n, w n, v n, z n respectively. Case 3. n (mod 3). As in Case, assign the label to the vertices u, w, v, u i, v i, w i, z i ( i n ). Then assign the label,,, 3 to the vertices u n, w n, v n, z n respectively. f is a 3-difference cordial labeling follows from the following Table 4.

136 k-difference cordial labeling of graphs 3 Values of n v f () v f () v f (3) e f () e f () n (mod 3) 4n+3 3 n (mod 3) 4n+5 3 n (mod 3) 4n+4 3 4n+3 3 4n+ 3 4n+4 3 4n+3 3 4n+ 3 4n+ 3 4n+ 4n+ 4n+ 4n+ 4n+ 4n+ Table 4 Finally we investigate cycles C (t) 4. Theorem 3.4 C (t) 4 is 3-difference cordial. Proof Let u be the vertices of C (t) 4 and i th cycle of C (t) 4 be uu i ui ui 3u. Define a map f from the vertex set of C (t) 4 to the set {,, 3} by f(u) =, f(u i ) = 3, i t, f(u i ) =, i t, f(u i 3 ) =, i t. Clearly v f() = t +, v f () = v f (3) = t and e f () = e f () = t. Hence f is 3-difference cordial. ¾ References [] I.Cahit, Cordial Graphs: A weaker version of Graceful and Harmonious graphs, Ars Combin., 3 (987), -7. [] J.A.Gallian, A Dynamic survey of graph labeling, The Electronic Journal of Combinatorics, 7 (4) #Ds6. [3] F.Harary, Graph Theory, Addision wesley, New Delhi (969). [4] R.Ponraj, S.Sathish Narayanan and R.Kala, Difference cordial labeling of graphs, Global Journal of Mathematical Sciences: Theory and Practical, 5(3),

137 International J.Math. Combin. Vol.(6), 3-38 Traversability and Covering Invariants of Token Graphs Keerthi G. Mirajkar and Priyanka Y. B (Department of Mathematics, Karnatak Arts College, Dharwad - 58, India) keerthi.mirajkar@gmail.com, priyankaybpriya@gmail.com Abstract: Let F k (G), k, G be the token graph of a connected graph G. In this paper, we investigate the Eulerian and Hamiltonian property of token graphs and obtain the covering invariants for complete graph of token graph. Key Words: Token graph, α, α, β, β, symmetric difference. AMS(): 5C45, 5C7.. Introduction All graphs considered here are simple, connected,undirected graphs. Let G be a graph with vertex set V(G)={v, v, v 3,, v n } and edge set E(G) = {e, e, e 3,, e m }. We refer the reader to Harary [3]. R.Fabila-Monroy and et.al. introduced a model in which, k indistinguished tokens move from vertex to vertex along the edges of a graph. This idea is formalized as follows, for a graph G and integer k, we define F k (G) to be the graph with vertex set ( ) V (G) k, where two vertices A and B of F k (G) are adjacent whenever their symmetric difference A B is a pair {a, b} such that a A, b B and ab E(G). Thus the vertices of F k (G) correspond to configurations of k- indistinguishable tokens placed at distinct vertices of G, where two configurations are adjacent whenever one configuration can be reached from the other by moving one token along an edge from its current position to an unoccupied vertex. The F k (G) is called the k-token graph of G. Many problems in mathematics and computer science are modeled by moving objects on the vertices of a graph according to certain prescribed rules. In graph pebbeling, a pebbling step consists of removing two pebbels from a vertex and placing one pebble on an adjacent vertex; [4] and [5] for surveys. Related pebbling games have been used to study rigidity [6,7], motion planning[,9], and as models of computation[]. In the chip firing game, a vertex v fires by distributing one chip to each of its neighbors(assuming the number of chips at v is at least its degree). This model has connections with matroid, the Tutte polynomial, and mathematical physics [8]. Inspired by this we investigate the some more properties like traversability and covering invariants of token graphs. Received September 6, 5, Accepted May 8, 6.

138 Traversability and Covering Invariants of Token Graphs 33 Remark.([]) Let G be a graph and F k (G) be the token graph of G with k n, V (F k (G)) = ( ) n k, E(Fk (G)) = ( n k ) E(G). Remark.([]) Two vertices A and B are adjacent in F k (G) if and only if V (G) \ A and V (G) \ B are adjacent in F n (G), F k (G) = F n k (G), with only one token, the token graph is isomorphic to G. Thus, F (G) = G. Remark.3 Degree of vertices in F k (G) is deg(v Fk (G)) = n deg G (V i ) i= (sum of pairs of vertices v i ; i k of G which are the elements of V ) if v i and v j are two vertices in G, then in F k (G) v i v j =, if v i is adjacent to v j in G. =, if v i is not adjacent to v j in G. Remark.4 If degree of all the vertices in a graph G is even or even regular then by the Remark 3 degree of all the vertices in F k (G) is even, irrespective of tokens being odd or even. Remark.5 If degree of all the vertices in a graph G is odd or odd regular then by Remark 3, degree of all the vertices in F k (G) is even, only when k is even token. Remark.6 If G contains both even and odd degree vertices then the vertices in F k (G) are also of odd and even degree, irrespective of tokens being odd or even.. Traversability of Token Graphs In this section we obtain the traversability properties of token graphs. Theorem. Let G be a connected graph. Then F k (G) is Eulerian if and only if it satisfies either of the following conditions. (i) Every vertex in G is of even degree; (ii) Every vertex in G is of odd degree and k-is even. Proof Let F k (G) be a token graph of graph G. Assume F k (G) is Eulerian, that is each vertices in F k (G) is of even degree. By the Remark.3, we have, d(v F k (G)) = deg(u) + deg(v)- (sum of pair of adjacent elements of G in V of F k (G)). Depending upon the degree, we consider the following cases. Case. Suppose deg(u)+deg(v) is odd, then by Remark 3, d(v F k (G)) is odd, a contradiction. Thus condition (i) is satisfied. Case. Suppose deg(u)+deg(v) is even, where u and v are odd or odd regular with odd tokens

139 34 Keerthi G. Mirajkar and Priyanka Y. B then by Remarks.3 and.5. F k (G) is non-eulerian, a contradiction. Thus the condition (ii) is satisfied. If G is Eulerian, that is it contains even degree of vertices. Then by the Remark 4, F k (G) is Eulerian. That is it contains even degree vertices. The converse follows from Remarks.4 and.5. ¾ Corollary. If G be Eulerian graph, then F n (G) is also Eulerian. Proof Let G be Eulerian graph. Then by the Remark., F n (G) is Eulerian. ¾ Lemma.3 If G is hamiltonian, then F n (G) is also hamiltonian. Proof Suppose G is hamiltonian, by the Remark., we know that G = F (G) and F k (G) = F n (G). If k = then, F (G) = F n (G) Therefore, G = F (G), G = F n (G). Thus, F n (G) is also hamiltonian. ¾ Theorem.4 F k (G) is hamiltonian if and only if G is complete graph. Proof Let G be complete graph and let {v, v, v 3,, v n } be the vertices in a graph G. In complete graph all vertices are mutually adjacent and G is hamiltonian. By the definition of token graph, F k (G) contains ( n k) number of vertices and by the Lemma.3, F (G) and F n (G) are hamiltonian. Now, we have to prove for k =, 3, 4,, n tokens. We prove this by induction method, here (k + ) th term is n- token. If k= token then, V (F (G)) = {(v v ), (v v 3 ), (v v 4 ),, (v v n ) (v v3), (v v 4 ), (v v 5 ),, (v v n ),, (v n v n )}. Here we consider two vertices A = {v v n } and B = {v v n }. By the symmetric difference we get v v. That is, A B = (A B) (A B) =v v v n v n = v v. Therefore v v are adjacent in G then A and B are also adjacent in F (G). Similarly v v n is adjacent with v 3 v n and the same follows for all vertices.

140 Traversability and Covering Invariants of Token Graphs 35 Now if k=3 token, then V (F 3 (G)) = {(v v v 3 ), (v v v 4 ),, (v v v 3 ), (v v 3 v 4 ), (v v 3 v 5 ),, (v v 3 v n ),, (v n v n v n )}. Here also (v v v n ) is adjacent with (v v 3 v n ),(v v 3 v n ) with (v v 4 v n ), and (v n 3 v n v n ) with (v n v n v n ). Hence, we get spanning cycle in F 3 (G) as {v v n v n, v v n v n, v 3 v n v n,, v n v n v n, v v n v n }. Therefore, F 3 (G) is hamiltonian graph. Thus the result is true for all k=n. Similarly, If k=n- token, then V (F n (G)) = V (F (G)). By the Lemma.3 and Remark., F n (G) = F (G). That is, F k (G) = F n k (G), () F (G) = F n k (G), () and if k =, then F (G) = F n (G). (3) Then G = F (G) = F n (G). Thus F n (G) is hamiltonian. For the converse, assume F k (G) is hamiltonian, we have to prove G is complete. Suppose G is not complete graph then by the symmetric difference the vertices in F k (G); k=,3,...,n-, form a sub graph homiomorphic to K,3 a contradiction. Theorem.5 If G is wheel, then F k (G) is hamiltonian graph. Proof Let G be wheel, hence it contains spanning cycle and let {v, v, v 3,, v n } be the vertex of graph G. Here v n is a vertices of maximum degree in G. Let V, V, V 3,, V ( n k) be the vertices in graph F k (G). By the lemma.3, we know that G = F (G) = F n (G). Then G is hamiltonian then F (G) and F n (G) are also hamiltonian. Now we prove for k =, 3, 4,, n tokens. We know that F k (G) = F n (G). If k= then, V (F (G)) = {V, V, V 3,, V ( n k) } = {(v v ), (v v 3 ),, (v v n ), (v v 3 ),, (v v n ), (v 3 v 4 ), (v 3 v 5 ),..., (v 3 v n ),..., (v n v n )}. In graph G, the n th vertex is adjacent with remaining all the vertices. Therefore by the symmetric difference we get spanning cycle as {v v n, v v n, v v n,, v v, v v n, v v n,, v v 3, v 3 v n,, v 3 v 4, v 4 v n,, v n v n, v n v n, v v n }. Thus F (G) contains spanning cycle then F n (G) also contains spanning cycle. Clearly, by Remark., F (G) and F n (G) are hamiltonian. Similarly for all tokens we get spanning cycle. Hence F k (G) is hamiltonian. ¾

141 36 Keerthi G. Mirajkar and Priyanka Y. B 3. Covering Invariants Of Token Graphs In the following section, we determine the point covering number α (G),line covering number α (G), point independence number β (G) and line independence number β (G) of token graph of complete graph. Theorem 3. For any complete graph K n ; n >, ( n k α (F k (K n )) = ) ( n k, β (F k (K n )) = Proof Let K n be the complete graph with n-vertices and F k (K n ) be the token graph of complete graph with ( ) n ( n k number of vertices. k) lines are required cover all the points in F k (K n ). By Remark., F (K n ) = K n and F n (K n ) = K n, i.e., F k (K n ) = K n when k = or n. For k =, 3, 4,, n, the vertices F k (K n ) are adjacent but not mutually and by Remark., it contains ( ) n ( n k number of vertices. Hence k) number of lines are require to cover all ( n the points. α (F k (K n )) = k). From the Gallai result, we know that α (G) + β (G) = G ). In F k (K n ), α (F k (K n )) + β (F k (K n )) = β (F k (K n )) = = = ( ) n ) ( ) n k n ( k + β (F k (K n )) =, k ( ) ( n ) n k k ( ) ( n ( n ( n n k) (Note that k) k = k).) ( n k). But F k (G) contains odd number of vertices then, β (K n ) = ( n k) = ( n k) Theorem 3. For any complete graph K n ; n >, ( n k) (K n) β (F k (K n )) = ( n k) (K n), β (K n ) = ( n k) if ( n k) is even, + if ( n k) is odd. ¾

142 Traversability and Covering Invariants of Token Graphs 37 and α (F k (K n )) = ( n k) β (F k (K n )). Proof Let K n be the complete graph with n-vertices and F k (K n ) be the token graph of complete graph with ( n k) number of vertices. By the definition of complete graph, (Kn ) = n and α (K n ) = n, β (K n ) =. Therefore by the Remark., α (F k (K n )) = n and β (F k (K n )) = when k = or n. Now we have to prove for k =, 3, 4,, n tokens. F k (K n ) contains ( n k) number of ( n vertices, and k) ( n (K n) are required to cover the vertices in F k (K n ). k) (K n) number of vertices are non-adjacent to each other and adjacent with remaining ( ) n ( n k k) (K n) number of vertices in F k (K n ), when ( ( n k) is even. if n ( k) is odd, then n ) ( n k is covered by k) (K n) + vertices, which are non-adjacent to each other and are adjacent with remaining ( ) n ( n k) + vertices in F k (K n ). Thus independence number in F k (K n ) is ( n k) (K n) or k ( n k) (K n) (K n) +. So, β (F k (K n )) = ( n k) (K n) if ( n k) (K n) + if ( n ) k is even, ( n ) k is odd. From the Gallai result, we know that α (G) + β (G) = G. In F k (K n ), when ( n k) is even then, α (F k (K n )) + β (F k (K n )) = ) ( n k α (F k (K n )) + (G) = α (F k (K n )) = α (F k (K n )) = ( ) n k ( ) n k ( ) ( n ) n k k (G) ( ) n β (F k (K n )). k If ( n k) is odd then, α (F k (K n )) + β (F k (K n )) = ( n ) k α (F k (K n )) + + = (G) ( ) n k ( ) n k

143 38 Keerthi G. Mirajkar and Priyanka Y. B α (F k (K n )) = α (F k (K n )) = ( ) ( n ) n k + k (G) ( ) n β (F k (K n )). k Acknowledgement The authors are thankful to University Grants Commission (UGC), Govt.of India for financial support through research grant under UGC-RGNF-4-5-SC-KAR- 7598/(SA-III/ References [] Aulett, V., Monti A., Parente M. and Persiano P., A linear-time algorithm for the feasibility of pebble motion on trees. Algorithmica 3(3) (999), 3-45, doi:.7/pl959. [] Fabila-Monroy R., Flores-Penaloza D., Huemer C., Hurtado F., Urrutia J. and Wood D.R., Token graphs, Graphs and Combin., 8 (), [3] Harary F., Graph Theory, Addison-Wesley, Reading, 969. [4] Hurlbert G. A survey of graph pebbling, Congressus Numerantium, 39(999), 4-64, arxiv:math/464. [5] Hurlbert G., Recent progress in graph pebbling, Graph Theory Notes N Y, 49(5), [6] Jacobs D., Hendrickson B., An algorithm for two dimensional rigidity percolation: the pebble game, J. Comput. Phys., 37()(997), , doi:.6/jcph [7] Jacobs D. and Thorpe M.F., Generic rigidity percolation: the pebble game, Phys. Rev. Lett., 75(995), , doi:.3/physrevlett [8] Merino C., The chip-firing game, Discret. Math., 3(-3)(5), 88-, doi:.6/j. disc [9] Papadimitriou C.H., Raghavan P., Sudan M. and Tamaki H., Motion planning on a graph, In: Proc. 35th Annual Symp. on Foundations of Comput. Sci., (FOCS 94)(994), 5-5, IEEE, Washington, doi:.9/sfcs [] Savage J.E.,Models of Computation: Exploring the Power of Computing. Addison-Wesley, Boston. jes/book/home.html (997b).

144 International J.Math. Combin. Vol.(6), Different Labelings on Parallel Transformations of a Class of Trees Teena Liza John and Mathew Varkey T.K. (Department of Mathematics, T.K.M.College of Engineering, Kollam-695, Kerala, India) teenavinu@gmail.com, mathewvarkeytk@gmail.com Abstract: A graph G = (V, E) with p vertices and q edges is said to be a mean graph if there exists an injective function f : V {,,, q} that induces an edge labeling f : E {,,, q} defined by f (uv) = = f(u) + f(v) if f(u) + f(v) is even f(u) + f(v) + if f(u) + f(v) is odd for every edge uv of G.Further f is called a super-mean labeling if f(v (G)) {f (e) : e E(G)} = {,,, p + q}. If the vertex labels are all even numbers in {, 4,, q} so that f (e) = f(u)+f(v) then f is an even mean labeling of G and if the vertex labels are in {, 3,,q } so that f (e) = f(u)+f(v)+, then G is an odd-mean graph. In this paper, we investigate a typical class of trees based on this definition. Key Words: Mean labeling, super-mean labeling, even-mean labeling, odd-mean labeling, parallel transformation of trees. AMS(): 5C78.. Introduction Throughout this paper,by a graph we mean a simple finite undirected graph without isolated vertices.for basic notations and terminology in graph theory we follow []. The concept of mean labeling was introduced in [5], super-mean labeling in [4] and odd-mean labeling in [].. T n Class of Trees In [3], T n class of trees are defined as follows. Definition. Let T be a tree and x and y be two adjacent vertices in T.Let there be two end vertices(non-adjacent vertices of degree )x and y in T such that the length of x x is equal to the length of the path y y. If the edge xy is deleted and x, y are joined by an edge x y, Received April 4, 5, Accepted May 3, 6.

145 4 Teena Liza John and Mathew Varkey T.K. then such a transformation of edges from xy to x y is called a parallel transformation of an edge in T. Definition. A tree is said to be a T n tree if and only if a resultant parallel transformation of edges reduce T into a Hamiltonian path. Such Hamiltonian path is denoted as P T. T 7 is given below in figure.(a). Here e, e, e 3, e 4 and e 5 are the edges to be deleted and e i ; i =,, 5 ( shown in broken lines ) the corresponding edges to be added to generate P T from T 7 (Figure.(b)). e e 4 e 3 e e 4 e e 5 e e 3 e 5 Fig. (a) Fig. (b) Theorem. T n is a mean graph. Proof Let T n be a tree on n vertices and by definition there exist a path P T corresponding to T n. Let E = {e, e,, e n } be the edges of T n. Let E S = {e, e,, e s } be the set of edges to be deleted from and E s = {e, e,, e s} be the edges to be added to T n so as to obtain a Hamiltonian path P T with V (P T ) = V (T n ) and E(P T ) = {E(T n ) E S } E S. Label the vertices of P T as x, x,, x n starting from the initial pendant vertex.

146 Different Labelings on Parallel Transformations of a Class of Trees 4 Define an injective mapping f : V (P T ) {,,, n }, as f(x i ) = i for all i. Now f induces edge labeling f on E(P T ) as f (xy) = = f(x) + f(y) if f(x) and f(y) are of same parity f(x) + f(y) + otherwise where xy E(P T ). Since P T is a path, every edge of P T is of the form x i x i+. f (x i x i+ ) = f(x i) + f(x i+ ) +, since f(x i ) and f(x i+ ) are of different parity = i + i + = i for i =,,, n Obviously f is injective and f (G) = {,,, n }. So it is proved that f is a mean labeling on P T. We have to prove that f is a mean labeling on T n. For this, it is enough to prove that f (e s ) = f (e s) where e s = x i x j E(T n ) and e s = x i+r x j r E(T n ). Now, e s must be of the form x i+rx i+r+, since it is an edge of a path P T. So (x i+r, x j r ) = (x i+r, x i+r+ ) f(x i+r ) + f(x j r ) + = f(x i+r) + f(x i+r+ ) + Therefore So and j = i + r + f (e s ) = f(x i) + f(x i+r+ ) + = i + r f (e s) = f (x i+r, x i+r+ ) = f(x i+r) + f(x i+r+ ) + = i + r Therefore, f (e s ) = f (e s ). Thus, f admits mean labeling on T n. Hence we get the theorem. ¾ Definition.3 A graph with p vertices and q edges is said to be odd mean if there exists a function f : V (G) {,,, q } which is one-one and the induced map f : E(G) {, 3,, q } defined by f (uv) = f(u)+f(v), if f(u)+f(v) is even or f(u)+f(v)+ if f(u) + f(v) is odd, is a bijection. If a graph has an odd mean labeling, then we say that G is an odd mean graph.

147 4 Teena Liza John and Mathew Varkey T.K. Definition.4 A function f is called an even-mean labeling of a graph G with p vertices and q edges if f is an injection from the vertices of G to {, 4,, q} such that when each edge uv is assigned the label f(u)+f(v), then the resulting edge labels are distinct.a graph which admits an even mean labeling is said to be an even-mean graph. Theorem. T n satisfies both even and odd mean labeling. Proof To prove T n is an even-mean graph,we consider f e : V (G) {, 4,, q} such that f e (x i ) = i for i =,,, n. Now, to show T n is odd-mean, we take another injective mapping f o : V (G) {, 3,, q+ } as f o (x i ) = i for i =,,, n. ¾ Theorem.3 Parallel transformation of trees generate a class of super-mean graphs. Proof Consider a T n tree on n vertices. By definition there exist a P T corresponding to T n. Let E = {e,, e n } be the edges of T n. Let E r = {e, e,, e r } be the edges to be deleted from T n, E r E and E r = {e,, e r} be the set of edges to be added to T n to make a path P T, such that if e n is the deleted edge, e n is the corresponding edge added at a distance d n by parallel transformation. Now we have V (P T ) = V (T n ) and E(P T ) = {E(T n ) E r } E r. Now we label the vertices of P T by x, x,, x n successively starting at one end vertex of the path P T. Define a mapping f : V (P T ) {,,, n } such that f(x i ) = i for all i =,...,n. Now, by the definition itself, f is one-one. Let f be the induced mapping defined on the edge set of P T such that f f = {,,, n } as f (xy) = = f(x) + f(y) if f(x) + f(y) is even f(x) + f(y) + if f(x) + f(y) is odd where xy E(P T ). Since P T is a path, every edge of P T is of the form x i x i+ for i =,,, n f (x i x i+ ) = f(x i) + f(x i+ ) = i ; i =,,, n Hence it is clear that f is one one and f(g) f (G) = {,,, n }. Hence f is a super mean labeling on P T. Now it is to show f is super mean on T n. It is enough to show that f (e k ) = f (e k ). Let e k = x i x j where x i x j E(T n ). To get P T, we have to delete e k and adjoin e k at a distance d from x i such that e k = x i+rx j r. Since e k is an edge of P T, it must be of the form e k = x i+rx i+r+.

148 Different Labelings on Parallel Transformations of a Class of Trees 43 Hence (x i+r, x j r ) = (x i+r, x i+r+ ) f(x i+r ) + f(x j r ) = f(x i+r) + f(x i+r+ ) = j = i + r + f (e k ) = f (x i x j ) = f(x i) + f(x j ) = f(x i) + f(x i+r+ ) = (i + r) f (e k) = f (x i+r, x i+r+ ) = f(x i+r) + f(x i+r+ ) = (i + r) Therefore, f (e k ) = f (e k). Thus, f is super mean on T n also. Hence, T n is a super mean graph. Example. In Figure., we show a super-mean labeling on tree T Fig. References [] F.Harary, Graph Theory, Addison-Wesley Publishing Company, 97. [] K.Manickam and M.Marudai, Odd-mean labelings of graphs, Bulletin of Pure and Applied Sciences, 5E()(6), [3] Mathew Varkey T.K. and Shajahan A., On labeling of parallel transformation of a class of trees, Bulletin of Kerala Mathematics Association, Vol.5, No.(9), [4] R.Ponraj and D.Ramya, Super mean labeling of graphs, Preprint. [5] S.Somasundaram and R.Ponraj, Mean labelings of graphs, National Academy Science Letter, Vol.6(3), -3.

149 International J.Math. Combin. Vol.(6), 44-5 Cycle and Armed Cap Cordial Graphs A.Nellai Murugan (Department of Mathematics, V.O.Chidambaram College, Tamil Nadu, India) P.Iyadurai Selvaraj (Department of Computer Science, V.O.Chidambaram College, Tamil Nadu, India) Abstract: Let G = (V,E) be a graph with p vertices and q edges. A Cap ( ) cordial labeling of a Graph G with vertex set V is a bijection from V to, such that if each edge uv is assigned the label, if f(u)=f(v)=, f(uv) =, otherwise. with the condition that the number of vertices labeled with and the number of vertices labeled with differ by at most and the number of edges labeled with and the number of edges labeled with differ by at most. Otherwise, it is called a Smarandache cordial labeling of G. A graph that admits a cordial labeling is called a cordial graph (CCG). In this paper, we proved that cycle C n (n is even), bistar B m,n, P m P n and Helm are cordial graphs. Key Words: Cap cordial labeling, Smarandache cordial labeling, Cap cordial graph. AMS(): 5C78.. Introduction A graph G is a finite non-empty set of objects called vertices together with a set of unordered pairs of distinct vertices of G which is called edges. Each pair e = {uv} of vertices in E is called an edge or a line of G. In this paper, we proved that Cycle C n (n : even), Bi-star B m, n, P m P n and Helm are cordial graphs.. Preliminaries Let G = (V,E) be a graph with p vertices and q edges. A (cap) cordial labeling of a Graph G Received February, 5, Accepted December 5, 5.

150 Cycle and Armed Cap Cordial Graphs 45 with vertex set V is a bijection from V to (, ) such that if each edge uv is assigned the label, if f(u) = f(v) = f(uv) =, otherwise. with the condition that the number of vertices labeled with and the number of vertices labeled with differ by at most and the number of edges labeled with and the number of edges labeled with differ by at most. Otherwise, it is called a Smarandache cordial labeling of G. The graph that admits a cordial labeling is called a cordial graph (CCG). we proved that cycle C n (n is even), bistar B m,n, P m P n and Helm are cordial graphs Definition. A graph with sequence of vertices u, u,, u n such that successive vertices are joined with an edge, P n is a path of length n. The closed path of length n is Cycle C n. Definition. A P m P n graph is a graph obtained from a path P m by joining a path of length P n at each vertex of P m. Definition.3 A bistar is a graph obtained from a path P by joining the root of stars S m and S n at the terminal vertices of P. It is denoted by B m, n. Definition.4 A Helm graph is a graph obtained from a Cycle C n by joining a pendent vertex at each vertex of on C n. It is denoted by C n K. 3. Main Results Theorem 3. A cycle C n (n : odd) is a cordial graph Proof Let V (C n ) = {u i : i n}, E(C n ) = {[(u i u i+ ) : i n ] (u u n )}. A vertex labeling f : V (C n ) {, } is defined by with an induced edge labeling f (u u n ) =,, i n f(u i ) =, n+, i n f, i n (u i u i+ ) =, n+, i n, Here V (f) + = V (f) and E (f) = E (f) +. It satisfies the condition V (f) V (f), E (f) E (f).

151 46 A.Nellai Murugan and P.Iyadurai Selvaraj Hence, C n is cordial graph. ¾ For example, C 7 is cordial graph as shown in the Figure. u u 7 u 6 u u 5 u 3 u 4 Figure Graph C 7 Theorem 3. A star S n is a cordial graph. Proof Let V (S n ) = {u, u i : i n} and E(S n ) = {(uu i ) : i n}. Define f : V (S n ), with vertex labeling as follows: Case. If n is even, then f(u) =,, i n f(u i ) =, n, + i n and an induced edge labeling f, i n (uu i ) =, n, + i n. Here V (f) + = V (f) and E (f) = E (f). It satisfies the condition V (f) V (f) and E (f) E (f). Case. If n is odd, then f(u) =, and with an induced edge labeling, i n+ f(u i ) =, n+3, i n f, i n+ (uu i ) =, n+3, i n.

152 Cycle and Armed Cap Cordial Graphs 47 Here V (f) = V (f) and E (f) = E (f) +. It satisfies the condition V (f) V (f) and E (f) E (f). Hence, S n is cordial graph. ¾ For example, S 5 and S 6 are cordial graphs as shown in the Figures and 3. u u u 6 u u 5 u 3 u 4 Figure Graph S 6 u u u u 3 u 5 u 4 Figure 3 Graph S 5 Theorem 3.3 A bistar B m,n is a cordial graph. Proof Let V (B m,n ) = {(u, v), (u i : i m), (v j : j n)} and E(B m,n ) = {[(uu i ) : i m] [(vv i ) : i m] [(uv)]}. Define f : V (B m,n ) {, } by two cases. Case. If m = n, the vertex labeling is defined by f(u) = {}, f(v) = {}, f(u i ) = {, i m}, f(v i ) = {, i m} with an induced edge labeling f (uu i ) = {, i m}, f (vv i ) = {, i m} and f (uv) =. Here V (f) = V (f) and E (f) = E (f) +. It satisfies the condition V (f) V (f) and E (f) E (f). Case. If m < n, the vertex labeling is defined by f(u) = {}, f(v) = {}, f(u i ) = {, i m}, f(v i ) = {, i m},, i mod, f(v m+i ) =, i mod, i n m,

153 48 A.Nellai Murugan and P.Iyadurai Selvaraj with an induced edge labeling f (uu i ) = {, i m}, f (vv j ) = {, j m}, f (uv) =, f, i mod, (vvm + i) =, i mod, i n m. Here, if n m is odd, then V (f) + = V (f) and E (f) = E (f); if n m is even, then V (f) = V (f) and E (f) = E (f) +. It satisfies the condition V (f) V (f) and E (f) E (f). Case 3. If n < m, by substituting m by n and n by m in Case the result follows. Hence, B m,n is a cordial graph. ¾ For example B 3,3, B,6 and B 6, are cordial graphs as shown in the Figures 4, 5 and 6. u u u v v v u 3 v 3 Figure 4 Graph B 3,3 v u v u u v v 3 v 4 v 5 v 6 Figure 5 Graph B,6

154 Cycle and Armed Cap Cordial Graphs 49 u u u 3 u v v u 4 v u 5 u 6 Figure 6 Graph B 6, Theorem 3.4 A graph P m P n is cordial. Proof Let G be the graph P m P n with V (G) = {[u i : i m], [v ij : i m, j n ]} and E(G) = {[(u i u i+ ) : i m ] [(u i v i ) : i m] [(v ij v ij+ ) : i m, j n ]}. Define f : V (G) {, } by cases following. Case. If m is even, then the vertex labeling is defined by, i m f(u i ) =, with an induced edge labeling, m + i m, f(v ij ) = f, i m (u i u i+ ) =,, i m, j n, + i m, j n, m, m + i m, f (u i v i ) = f, i m, j n, (v ij v ij+ ) = + i m, j n., m Here V (f) = V (f) and E (f) = E (f) +. It satisfies the condition V (f) V (f) and E (f) E (f)., i m,, m + i m,

155 5 A.Nellai Murugan and P.Iyadurai Selvaraj Case. If m is odd and n is odd, the vertex labeling is defined by, i m f(u i ) =, m+, i m, f(v m+ j ) =, j n, with an induced edge labeling n, + j n f, i m (u i u i+ ) =, m+, + i m, f, i m, j n, (v ij v ij+ ) = m+3, i m, j n, f (v m+ j v m+ j+ ) =, j n 3,, n j n., t i m, j n, f(v ij ) = m+, i m, j n, f, i m (u i v i ) =, m+, + i m, Here V (f) + = V (f) and E (f) = E (f). It satisfies the condition V (f) V (f) and E (f) E (f). Case 3. If m is odd and n is even, the vertex labeling is defined by, i m f(u i ) =, m+, i m, with an induced edge labeling f, i m (u i u i+ ) =, m+, + i m, f, i m, j n, (v ij v ij+ ) = m+, i m, j n, f (v m+ j v m+ j+ ) =, j n 4,, n j n., i m, j n, f(v ij ) = m+, i m, j n f, i m (u i v i ) =, m+, + i m, Here V (f) = V (f) and E (f) = E (f) +. It satisfies the condition V (f) V (f) and E (f) E (f).

156 Cycle and Armed Cap Cordial Graphs 5 Hence, the graph P m P n is cordial. ¾ For example, P 4 P 5, P 5 P 5 and P 5 P 6 are cordial as shown in Figures 7, 8 and 9. u u u 3 u 4 v v v 3 v 4 v v v 3 v 4 v 3 v 3 v 33 v 43 v 4 v 4 v 34 v 44 Figure 7 Graph P 4 P 5 u u u 3 u 4 u 5 v v v 3 v 4 v 5 v v v 3 v 4 v 5 v 3 v 3 v 33 v 43 v 53 v 4 v 4 v 34 v 44 v 54 Figure 8 Graph P 5 P 5 u u u 3 u 4 u 5 v v v 3 v 4 v 5 v v v 3 v 4 v 5 v 3 v 3 v 33 v 43 v 53 v 4 v 4 v 34 v 44 v 54 v 5 v 5 v 35 v 45 v 55 Figure 9 Graph P 5 P 6

157 5 A.Nellai Murugan and P.Iyadurai Selvaraj Theorem 3.5 A Helm (C n K ) is cordial. Proof Let G be the graph (C n K ) with V (G) = {u i, v i : i m} and E(G) = {(u i v i ) : i m}. A vertex labeling on G is defined by f(u i ) = {, i m}, f(v i ) = {, i m} with an induced edge labeling f (u i u i+ ) = {, i m }, f (u m u ) =, f (u i v i ) = {, i m}. Here V (f) = V (f) and E (f) = E (f). It satisfies the condition V (f) V (f) and E (f) E (f). Hence, A Helm is cordial. For example, a Helm (C 6 K ) is cordial as shown in the Figure. ¾ v 6 v u u 6 u 5 v 5 u u 4 v u 3 v 4 v 3 Figure Graph (C 6 K ) References [] J.A.Gallian, A Dynamic Survey of graph labeling, The Electronic journal of Combinatorics, 6(), #DS6. [] F.Harary, Graph Theory, Addition - Wesley publishing company Inc, USA, 969. [3] A.Nellai Murugan and V.Baby Suganya, Cordial labeling of path related splitted graphs, Indian Journal of Applied Research, Vol.4, Issue 3, Mar. 4, -8. [4] A.Nellai Murugan and M.Taj Nisha, A study on divisor cordial labelling of star attached paths and cycles, Indian Journal of Research, Vol.3, Issue 3, Mar. 4, -7. [5] A.Nellai Murugan and V.Brinda Devi, A study on path related divisor cordial graphs, International Journal of Scientific Research, Vol.3, Issue 4, April. 4, [6] A.Nellai Murugan and A.Meenakshi Sundari, On Cordial Graphs, International Journal of Scientific Research, Vol.3, Issue 7, July. 4, [7] A.Nellai Murugan and P.Iyadurai Selvaraj, Path related cup cordial graphs, Indian Journal of Applied Research, Vol. 4, Issue 8, August 4.

158 International J.Math. Combin. Vol.(6), 53-6 Traffic Congestion A Challenging Problem in the World Mushtaq Ahmad Shah, Mridula Purohit and M.H.Gulzar 3. Department of Mathematics, Vivekananda Global University, Jaipur, India. Department of Mathematics, VIT East Jaipur, India. Department of Mathematics, University of Kashmir, India shahmushtaq8@gmail.com Abstract: The purpose of this paper is to describe the problems which involves in the reduction of traffic congestion. is quite appropriate. In particular we use graph theoretical approach which We use crossing number technique to reduce traffic congestion. The minimum number of crossing points in a complete graph is given by Cr(Kn) [ n ] [ n ] [ n ] [ n 3 ] 4 where[ ] represents greatest integer function. And we illustrate the result with counter examples. Key Words: crossing number, complete graph, traffic control, edge connectivity, vertex connectivity. AMS(): 5C9.. Introduction The crossing number (sometimes denoted as C(G)) of a graph G is the smallest number of pair wise crossings of edges among all drawings of G in the plane. In the last decade, there has been significant progress on a true theory of crossing numbers. There are now many theorems on the crossing number of a general graph and the structure of crossing critical graphs, whereas in the past, most results were about the crossing numbers of either individual graphs or the members of special families of graphs. The study of crossing numbers began during the Second World War with Paul Turan. In [], he tells the story of working in a brickyard and wondering about how to design an efficient rail system from the kilns to the storage yards. For each kiln and each storage yard, there was a track directly connecting them. The problem he Consider was how to lay the rails to reduce the number of crossings, where the cars tended to fall off the tracks, requiring the workers to reload the bricks onto the cars. This is the problem of finding the crossing number of the complete bipartite graph. It is also natural to try to compute the crossing number of the complete graph. To date, there are only conjectures for the crossing numbers of these graphs Called Guys conjecture which suggest that crossing number of complete Received October, 5, Accepted June 6, 6.

159 54 Mushtaq Ahmad Shah, Mridula Purohit and M.H.Gulzar graph Kn is given by V (K n ) = Z(n)[][3] where [ ] represents greatest integer function. which can also be written as z(n) = 4 [ n ] [ n ] [ ] [ ] n n 3 64 z(n) = n(n ) (n 4) n even 64 (n ) (n 3) n odd Guy prove it for n in 97 in 7 Richter prove it for n For any graph G, we say that the crossing number c(g) is the minimum number of crossings with which it is possible to draw G in the plane. We note that the edges of G need not be straight line segments, and also that the result is the same whether G is drawn in the plane or on the surface of a sphere. Another invariant of G is the rectilinear crossing number, c(g), which is the minimum number of crossings when G is drawn in the plane in such a way that every edge is a straight line segment. We will find by an example that this is not the same number obtained by drawing G on a sphere with the edges as arcs of great circles. In drawing G in the plane, we may locate its vertices wherever it is most convenient. A plane graph is one which is already drawn in the plane in such a way that no two of its edges intersect. A planar graph is one which can be drawn as a plane graph [6]. In terms of the notation introduced above, a graph G is planar if and only if c(g) =. The earliest result concerning the drawing of graphs in the plane is due to Fary [4] [7], who showed that any planar graph (without loops or multiple edges) can be drawn in the plane in such a way that every Edge is straight. Thus Farys result may be rephrased: if c(g) =, then c(g) =. In a drawing, the nodes of the graph are mapped into points of a plane, and the arcs into continuous curves of the plane, no three having a point in common. A minimal drawing does not contain an arc which crosses itself, nor two arcs with more than one point in common, [5],[8]. In general for a set of n line segments, there can be up to (n ) intersection points, since if every segment intersects every other segment, there would be n(n ) = (n ) Crossing points to compute them all we require (n ) algorithm. The traffic theory is a physical phenomenon that aims at understanding and improving automobile traffic, and the problem associated with it such as traffic congestion [9]. The traffic control problem is to minimize the waiting time of the public transportation while maintaining the individual traffic flow optimally []. Significant development of traffic control systems using traffic lights have been achieved since the first traffic controller was installed in London in 868. The first green wave was realized in Salt Lake City (U.K.) in 98, and the first area traffic controller was introduced in Toronto in 96. At the beginning, electromechanical devices were used to perform traffic control. Then Intelligent Transportation System (ITS) is used extensively in urban areas to control traffic at an intersection []. The traffic data in a particular region can be used to direct the traffic flow to improve traffic output without adding new roads. In order to collect accurate traffic data semi conductor-based controllers known as sensors were

160 Traffic Congestion A Challenging Problem in the World 55 placed in different places to collect traffic information are used in traffic control system [], [], [3]. Nowadays, microprocessor based controller are used in Traffic Control Systems. The combinatorial approach to the optimal traffic control problem was founded by Stoffers [4] in 968 by introducing the Compatibility Graph of traffic streams. One of the main uses of traffic theory is the development of traffic models which can be used for estimation, prediction, and control related tasks for the automobile traffic process. The term Intelligent Transportation System (ITS) refers to information and communication technology applied to transport infrastructure and vehicles, that improves transport outcomes such as transport safety, transport productivity, transport reliability, informed traveler choice, environmental performance etc. [5], [6]. ITS mainly comes from the problems caused by traffic congestion and synergy of new information technology for simulation, real time control and communication networks. Traffic congestion has been increased world wide as a result of increased motorization, urbanization, population growth and changes in population density. Congestion reduces efficiency of transportation infrastructure and increases travel time, air pollution and fuel consumption. At the beginning of 9, in United States large increase in both motorization and urbanization led to the migration of the population from sparsely populated rural areas and densely packed urban areas into suburbs (sub urban areas). Intelligent Transport Systems vary in technologies applied, from basic management system such as car navigation; traffic signal control systems; container management system; variable message sign; automatic number plate recognition or speed cameras to monitor applications; such as security CCTV systems; and to more advanced applications that integrate live data and feedback from a number of other sources, such as parking guidance and information systems; weather information etc. Additional predictive techniques are being developed to allow advanced modeling and comparison with historical data. The traffic flow predictions will be delivered to the drivers via different channels such as roadside billboards, radio stations, internet, and on vehicle GPS (Global Positioning Systems) systems. One of the components of an ITS is the live traffic data collection. To collect accurate traffic data sensors have to be placed on the roads and streets to measure the flow of traffic. Some of the constituent technologies implemented in ITS are namely, Wireless Communication, Computational technologies, Sensing technologies, Video Vehicle Detection etc. Urban traffic congestion is a significant and growing problem in many parts of the world. Moreover, as congestion continues to increase, the conventional approach of building more roads doesn t always work for a variety of political, financial, and environmental reasons. In fact, building new roads can actually compound congestion, in some cases, by inducing greater demands for vehicle travel that quickly eat away the additional capacity? Against this backdrop of serious existing and growing congestion traffic Control techniques and information systems are needed that can substantially increase capacity and Improve traffic flow efficiency. Application of ITS technologies in areas such as road user information and navigation systems, improved traffic control systems and vehicle guidance and control systems has significant potential for relieving traffic congestions. Theorem. The edge connectivity of a graph G cannot exceed the degree of the vertex with the smallest degree in G.

161 56 Mushtaq Ahmad Shah, Mridula Purohit and M.H.Gulzar Theorem. The vertex connectivity of any graph G can never exceed the edge connectivity of G. Theorem.3 The maximum vertex connectivity one can achieve in a graph of n vertices and e edges is e n Thus we conclude that vertexconnectivity edgeconnectivity e n. Definition.4 A graph G(v, e) where v is the set of vertices and e the set of edges is said to be complete if degree of each vertex is v. Definition.5 The number of edges incident on a vertex is said to be degree of the vertex. Proposed Solution.6 As congestion continues to increase, the conventional approach of building more roads doesn t always work for a variety of political, financial, and environmental reasons. In fact, building new roads can actually compound congestion. There is no particular technique which reduces traffic congestion. Numbers of techniques are simultaneously required to curb this problem. Traffic congestion is one of the challenging problem in the world the aim of this research paper is that how to curb this problem. Before giving the solution to the problem we would like to introduce you the graph theoretical approach of the problem, using underlying graphs. We represent various cities by vertices and roads connected them by edges. Since every city must be connected with all other cities in particular geographical area so first of all we are dealing with complete graphs then we shall remove all the edges in the graph in such a way that maximum crossing pointes will be removed and there is no effect in the connectivity. Following are the techniques require curbing traffic congestion. Definition.7 Let G(v, e) be a complete graph v the set of vertices and e the set of edges. the crossing number Cr(G) of a complete graph G(v, e) is the least number of crossings, common points of two arcs other than a vertex, in any drawing of graph in a plane (or on a sphere) in a drawing the vertex of the graph are mapped into points of a plane and the arcs into continue curves of the plane no three having a point in common, unless it be an end point (vertex) of the arc. A drawing which exhibits a crossing number is called minimal a minimal drawing does not contain an arc which crosses itself nor two arcs with more than one point in common. For any complete graph K n it has been shown that the minimum number of crossing points is given by [ Cr(Kn) n ][ n ][ n ][ n 3 ] 4, where[ ] represents greatest integer function. Since minimum number of crossing points means minimum number of interruptions on roads which minimize the waiting time of the traffic participants so traffic congestion is reduces. After drawing the graph with minimum crossings we remove all those edges in a graph which does not effect the connectivity of the graph but reduce more crossings so the graph becomes more efficient as shown in Figure here red dots represent vertices and yellow dots represent crossing points.

162 Traffic Congestion A Challenging Problem in the World 57 Figure We shall keep this point in mind that no three edges has a point in common if it is necessary then we have to keep other crossing point at least one kilometer away from one another as shown in Figure Figure Because more than one crossing at point will increase interruption on traffic flow, first of all we have to try our best to reduce the intensity of crossing points, if it is not possible then flyovers should be constructed at every crossing so that there is no interruption on traffic flow, in this case only slow moments are possible not traffic jam. The above crossing point technique will reduce traffic congestion in a large extent. if there are maximum crossing points on the roads then maximum traffic congestion is possible so minimum crossing means minimum traffic congestion so we have to reduce the crossing points then traffic congestion is reduced. The aim to reduce traffic congestion is to reduce the crossing points. There are certain crossing points where more than two roads cross each other and traffic lights are imposed to allow the traffic flow alternatively one by one. so traffic congestion is increased because more and more vehicles

163 58 Mushtaq Ahmad Shah, Mridula Purohit and M.H.Gulzar have to be stopped on the roads First of all we shall try over best to reduce those crossing points where more than two crossing points exist which is quite possible, if it is necessary then, on these intersection pointes where more than two crossings points exist we have to make flyovers (as shown in fig.3 below these flyovers have already designed in certain parts of the world for this purpose) in such a way that there should not be any crossing point and traffic flow should be in continuous manner may be some times there are slow moments but still it will reduce traffic congestion. Figure 3 3. Methods Method 3. There should be exclusive lanes for public transport so that private transport system does not affect the moment of the public transport. Method 3. A turning restriction is imposed on the traffic, and vehicles are allow to turn on certain places turning points are at least one kilometer away from the crossing points if the turning points are at crossing points it will definitely increase traffic congestion, and proper flyovers system is imposed as shown in Figure 4, which will automatically reduce traffic congestion in a large extent.

164 Traffic Congestion A Challenging Problem in the World 59 Figure 4 Method 3.3 A double win method has already imposed in certain cities of the world and has been proved to be very happy one. A congestion charge is essentially an economic method of regulating traffic by imposing fees on vehicle users that travels a city more crowed roads, but charge vary by city to city depending up on crowed on the city. Method 3.4 The Parking restrictions on road side should be banned to reduce congestion. Method 3.5 Remove some link roads at high efficiency points. Then we have to connect them other side so that minimum crossings are possible. References [] P.Turan, A note of welcome, J. Graph Theory, (977), 5. [] Erdos and Guy, Crossing number problem, American Math. Month, 8(973), [3] R.Guy, A combinatorial problem, Bull. Malayan Math. Soc., 7 (96),68-7. [4] I.Fary, On straight line representation of planar graphs, Acta Univ. Szeged, (948), [5] F.Harry and A. Hill, on the number of crossings in a complete graphs, Proc. Edinburgh Math. Soc., 3 (96), [6] J.W. Moon, On the distribution of crossing in random complete graphs, SAIM. J. Appl.Math., 3(965), [7] The minimum number of intersection in complete graph, Thomas L. Saaty, June 4, 964, Vol.5, Mathematics, T.L.SAATY Communicated by Einar Hille. [8] Crossing numbe,r A complete chapter R. B.Richter and G. Salazak Unpublished manuscript. 8.pdf [9] Gazis D. C., Traffic Science, John Wiley and Sons, New York, 97. [] Gazis, D. C., Traffic Theory, Kluwer Academic Publishers, London,. [] Gu W., Jia X., On a traffic control problem, 8th International Symposium on Parallel Architectures, Algorithms and Networks, ISPAN, 5.