mathematical combinatorics

ISSN

1937 - 1055

Special Issue 1, 2018

INTERNATIONAL

MATHEMATICAL

JOURNAL

OF

COMBINATORICS

Proceedings of the International Conference on Discrete Mathematics and its Applications

EDITED BY THE MADIS OF CHINESE ACADEMY OF SCIENCES AND ACADEMY OF MATHEMATICAL COMBINATORICS & APPLICATIONS, USA

November,

2018

Special Issue 1, 2018

ISSN 1937-1055

International Journal of

Mathematical Combinatorics (www.mathcombin.com)


Edited By The Madis of Chinese Academy of Sciences and Academy of Mathematical Combinatorics & Applications, USA

November,

2018

Aims and Scope: The International J.Mathematical Combinatorics (ISSN 1937-1055) is a fully refereed international journal, sponsored by the MADIS of Chinese Academy of Sciences and published in USA quarterly comprising 110-160 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 10 Estes St. Ipswich, MA 01938-2106, USA Tel.: (978) 356-6500, Ext. 2262 Fax: (978) 356-9371 http://www.ebsco.com/home/printsubs/priceproj.asp and Gale Directory of Publications and Broadcast Media, Gale, a part of Cengage Learning 27500 Drake Rd. Farmington Hills, MI 48331-3535, USA Tel.: (248) 699-4253, ext. 1326; 1-800-347-GALE Fax: (248) 699-8075 http://www.gale.com 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 1.730), Institute for Scientific Information (PA, USA), Library of Congress Subject Headings (USA). Subscription A subscription can be ordered by an email directly to Prof.Linfan Mao PhD The Editor-in-Chief of International Journal of Mathematical Combinatorics Chinese Academy of Mathematics and System Science Beijing, 100190, P.R.China Email: [email protected]

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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 Email: [email protected]

Shaofei Du Capital Normal University, P.R.China Email: [email protected]

Deputy Editor-in-Chief

Yuanqiu Huang Hunan Normal University, P.R.China Email: [email protected]

Xiaodong Hu Chinese Academy of Mathematics and System Science, P.R.China Email: [email protected]

Guohua Song Beijing University of Civil Engineering and H.Iseri Architecture, P.R.China Mansfield University, USA Email: [email protected] Email: [email protected]

Editors Arindam Bhattacharyya Jadavpur University, India Email: [email protected] Said Broumi Hassan II University Mohammedia Hay El Baraka Ben M’sik Casablanca B.P.7951 Morocco

Xueliang Li Nankai University, P.R.China Email: [email protected] Guodong Liu Huizhou University Email: [email protected] W.B.Vasantha Kandasamy Indian Institute of Technology, India Email: [email protected]

Junliang Cai Beijing Normal University, P.R.China Email: [email protected]

Ion Patrascu Fratii Buzesti National College Craiova Romania

Yanxun Chang Beijing Jiaotong University, P.R.China Email: [email protected]

Han Ren East China Normal University, P.R.China Email: [email protected]

Jingan Cui Ovidiu-Ilie Sandru Beijing University of Civil Engineering and Politechnica University of Bucharest Architecture, P.R.China Romania Email: [email protected]

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International Journal of Mathematical Combinatorics

Mingyao Xu Peking University, P.R.China Email: [email protected] Guiying Yan Chinese Academy of Mathematics and System Science, P.R.China Email: [email protected] Y. Zhang Department of Computer Science Georgia State University, Atlanta, USA

Guest Editors of Special Issue 1 (2018) R. Kala Manonmaniam Sundaranar University, Tamilnadu, India Email: [email protected] K. Selvakumar Manonmaniam Sundaranar University, Tamilnadu, India Email: [email protected] S. Monikandan Manonmaniam Sundaranar University, Tamilnadu, India Email: [email protected]

Famous Words: Mathematics is less related to accounting than it is to philosophy. Leonard Adleman.

iii

Message from the Vice-Chancellor’s Desk

Message from the Vice-Chancellor’s Desk

Established in 1990, the Manonmaniam Sundaranar University provides high quality education, equipping students with the acumen to address both local and global challenges. Within a short span of its existence, the university has achieved remarkable success and continues to forge its future with new program initiatives. The department of mathematics is one of the oldest departments in the university. Since its inception in 1991, the department is known for its excellence in teaching and research. It is the first department in the university to get UGC-SAP(DRS-I) assistance and is now a UGC- SAP(DRS-II) department. It is also a DST-FIST department and is continuously supported by NBHM through library grant since 1998. I am proud to say that the department of mathematics is one among the 10 departments of various universities of Tamil Nadu that is upscaled by the state government as per PG center with high tech laboratory in 2012. I am extremely happy that the department of mathematics is organizing this International Conference on Discrete Mathematics and its Applications during January 18-21, 2018. Conferences of such nature provide great opportunity, not only to update knowledge and keep obsessed with latest development scenario in the respective field, but also an occasion for Resource persons/ delagates/students to exchange ideas and interact with each other so that new ideas encompassing the intricacies of Discrete Mathematics will emerge. As a university, we are aware that knowledge gained through research is a crucial element of any University’s mission all over the world. I have no doubt that this conference can promote such development and innovations in the field of Discrete Mathematics in this Country will go a long way enriching the field of Discrete Mathematics and thereby contribute to the developmental activities. I congratulate the head of the department and the faculty members for their commitment and contributions, without which this event would not have been possible. The publication of the proceedings of the conference in International journal of Combinatorial Mathematics is an additional feather in cap of the Department of Mathematics.

Prof. K. Baskar Vice-Chancellor Manonmaniam Sundaranar University

iv


Report of the Convener The discrete mathematics is a dynamic field in theory, application and in practice. Researchers in discrete mathematics have established important connections with mainstream areas of pure and applied mathematics, and as a consequence, research techniques and problems are drawn from a wide range of different fields, including algebra, topology, geometry, probability, analysis, and logic. The discrete mathematics has now burgeoned into one of the fastest growing branches of mathematics. The main aim of the conference is to bring together leading academic scientists, researchers and research scholars to exchange and share their experiences and research results about all aspects of discrete mathematics. It also provides a premier interdisciplinary forum for researchers, practitioners and educators to present and discuss the most recent innovations, trends, and concerns, practical challenges encountered and the solutions adopted in the field of discrete mathematics. The speakers and the topics are chosen in such a way that the overall input will cover areas in which current research in discrete mathematics is going on at national and international levels. 10 resource persons including Prof. Albert Faessler, from Bern university, Switzerland and Prof. S. Arumugam, leading graph theorist at international level gave plenary talks for the conference, 102 participants from various parts of the country attended the conference, 39 papers were presented and 30 were considered for publication. After careful refereeing process by the guest editors, only 21 were accepted for publication in the International Journal of Mathematical Combinatorics. I am happy to acknowledge the financial support of the university grants commission, New Delhi through SAP (DRS-II) and the Manonmaniam Sundaranar University for the successful conduct of the conference. It is heartening to thank all the members of the organizing committee and advisory committee for their valuable contribution to achieve the goal of the conference. It is my pleasure to thank the guest editors for their valuable support in the referee process. I take this opportunity to thank all my postgraduate students and research scholars of the department of mathematics in making necessary arrangements. My special thanks are due to Prof.Linfan Mao, the editor-in-chief for his immediate response to publish the proceedings of the conference in IJMC. Last but not the least, I thank our research scholar Mr.V.P. Asan Nagoor Meeran for his excellent service in changing the final version of the papers into the proceedings format and preparing the camera ready copy for the proceedings.

Prof. J. Paulraj Joseph Head, Department of Mathematics Manonmaniam Sundaranar University

Contents

Part I. Talks in Inaugural Sections 1. Message from the Vice-Chancellor’s Desk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii 2. Report of the Convener . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Part II. Papers in the Conference 1. Radial Radio Sequence of a Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2. Non-Isolated Resolving Number for Some Splitting Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3. Prime Labelings of Some Helm and Gear Related Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4. Maximum Number of Holes in Square of Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5. Clique-to-Clique Detour Distance in Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 6. Connected Double Monophonic Number of a Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 7. Analytic Odd Mean Labeling of Square and H Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 8. Learning Some Special Graphs Using Hyperedge Replacement . . . . . . . . . . . . . . . . . . . . . . 68 9. Parikh Matrices of Binary and Ternary Words Under Prouhet Morphism . . . . . . . . . . . 78 10. Nardhaus Gaddum Type Results for Total Resolving Number of Graphs and Their Derived Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 11. Bounds for the Non-Neighbor Harmonic Index of Subdivision Graphs . . . . . . . . . . . . 104 12. 4-Remainder Cordial Labeling Graphs Obtained From Ladder . . . . . . . . . . . . . . . . . . . 114 13. Non-Split Perfect Triple Connected Domination Number of Different Product of Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 14. The Connectivity Number of an Arithmetic Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 15. Q-Fuzzy Bi-ideals in Near Subtraction Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 16. Intuitionistic Fuzzy Operator A(m,n) and its Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 144 17. Intuitionistic Fuzzy Soft Multi Set and its Application. . . . . . . . . . . . . . . . . . . . . . . . . . . .155 18. Signed Product Cordial Labeling of Zero-Divisor Graphs . . . . . . . . . . . . . . . . . . . . . . . . . 164 19. On the Domination Number of a Graph and its Line Graph . . . . . . . . . . . . . . . . . . . . . . 170 20. Radio Mean D-Distance Number of Family of Tree and Cycle-Related Graphs . . . . 182 21. A Note on Support Neighbourly Irregular Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Part III. An Introduction to Manonamaniam Sundaranar University 1. Introduction to Manonamaniam Sundaranar University . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

International J.Math. Combin. Special Issue 1 (2018), 1-8

Radial Radio Sequence of a Graph Selvam Avadayappan, M. Bhuvaneshwari and S.Vimalajenifer (Research Department of Mathematics, VHNSN College, Virudhunagar - 626 001, India) E-mail: [email protected], [email protected], [email protected]

Abstract: Let G(V (G), E(G)) be a graph. A radial radio labeling, f , of a connected graph G is an assignment of positive integers to the vertices satisfying the following condition: d(u, v) + |f (u) − f (v)| ≥ 1 + r(G), for any two distinct vertices u, v ∈ V (G), where d(u, v) and r(G) denote the distance between the vertices u and v and the radius of the graph G, respectively. The span of a radial radio labeling f is the largest integer in the range of f and is denoted by span(f ). The radial radio number of G, rr(G), is the minimum span taken over all radial radio labelings of G. The sequence (µ1 (v))v∈V (G) arranged in decreasing order is called the (µ1 (v)) − rr sequence of G, where (µ1 (v)) is the radial radio number of the induced subgraph induced by the closed neighborhood of v in V (G). In this paper, we present some results on the (µ1 (v)) − rr sequence of a graph.

Key Words: Frequency assignment problem, radius, diameter, radio labeling, radial radio labeling, radio number, radial radio number, rr sequence.

AMS(2010): 05C78. §1. Introduction In this paper, we consider only simple, connected, undirected and finite graphs. For basic notations and terminology, we follow [3]. Let G(V (G), E(G)) be a simple connected graph. The number of vertices incident at a vertex v ∈ V (G) is called the degree of the vertex in G and it is denoted by deg(v). The maximum degree in G is denoted by ∆(G) or ∆ and the minimum degree in G is denoted by δ(G) or δ. A vertex v ∈ V (G) is called a full vertex, if it is adjacent to all other vertices. The distance d(u, v) between any two vertices u and v, is the length of a shortest (u, v)− path in G. The eccentricity, e(u), of a vertex u in V (G) is the distance of a vertex farthest from u. The radius of a graph G is the minimum eccentricity among all the vertices and is denoted by r(G) or r. The diameter of G is the maximum eccentricity among all the vertices and is denoted by diam(G) or d. The relation between r(G) and diam(G) is given by the inequality r(G) ≤ diam(G) ≤ 2r(G) [6]. For further details on distance in graphs, one can refer [4]. For a subset S of V (G), let < S > denote the induced subgraph of G induced by S. A 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 30, 2018, Accepted May 22, 2018, Edited by S. Monikandan.

2

Selvam Avadayappan, M. Bhuvaneshwari and S.Vimalajenifer

clique C is a subset of V (G) with maximum number of vertices such that < C > is complete. The clique number of a graph G, denoted by ω(G) or ω, is the number of vertices in a clique of G. For any vertex v ∈ V (G), the open neighborhood N (v) is the set of all vertices adjacent to v. That is, N (v) = {u ∈ V (G) : uv ∈ E(G)}. The closed neighborhood N [v] of v is defined by λ S N [v] = N (v) ∪ {v}. N0 (v) = {v}, Nλ (v) = {w : d(v, w) = λ}, for λ ≥ 1 and Nλ [v] = Ni (v). i=0

Note that, N1 [v] = N [v]. The centre of G is the subgraph of G induced by the set of vertices of minimum eccentricity. A graph G which is isomorphic to its centre is called a self centered graph. Note that, in a self centered graph r(G) = diam(G). A simple graph G is called vertex transitive if, for any two vertices u and v of G, there exits an automorphism g of G such that g(u) = g(v). Let f : A → B be a function. Let S ⊆ A. The restriction of f to S, denoted by f |S , is a function from S to B defined by (f |S )(x) = f (x), for all x ∈ S. In 1960’s Rosa [11] introduced the concept of graph labeling. A graph labeling is an assignment of numbers to the vertices or edges or both, satisfying some constraints. Rosa named the labeling introduced by him as β − valuation and later on it becomes a very famous interesting graph labeling called graceful labeling, which is the origin for any graph labeling problems. Motivated by real life problems, many mathematicians introduced various graph labeling concepts [8]. Here, we see one of the familiar graph labelings in graph theory. The problem of assigning frequencies to the channels for the FM radio stations is known as Frequency Assignment Problem (FAP). This problem was studied by W. K. Hale [9]. In a telecommunication system, the assignment of channels to FM radio stations play a vital role. Motivated by the FAP, Chartrand et al.[5] introduced the concept of radio labeling. For a given k, 1 ≤ k ≤ diam(G), a radio k- coloring, f , is an assignment of positive integers to the vertices satisfying the following condition: d(u, v) + |f (u) − f (v)| ≥ 1 + k (1) for any two distinct vertices u, v ∈ V (G). Whenever, diam(G) = k, the radio k- coloring is called a radio labeling [5] of G. The span of a radio labeling f is the largest integer in the range of f and is denoted by span(f ). The radio number of G is the minimum span taken over all radio labelings of G and is denoted by rn(G). Motivated by the work of Chartrand et al., on radio labeling, KM. Kathiresan and S. Vimalajenifer [10] introduced the concept of radial radio labeling. A radial radio labeling f of G is a function f : V → {1, 2, · · · , } satisfying the condition, d(u, v) + |f (u) − f (v)| ≥ 1 + r(G)

(2)

for any two distinct vertices u, v ∈ V (G). This condition is obtained by taking k = r(G) in (1). Condition (2) is known as radial radio condition. The span of a radial radio labeling f is the largest integer in the range of f and is denoted by span(f ). The radial radio number is the minimum span taken over all radial radio labelings of G and is denoted by rr(G).

3

Radial Radio Sequence of a Graph

That is, rr(G) = min max f (v), where the minimum runs over all radial radio labelings of f

v∈V

G. In [2], it has been proved that Theorem A([2]) Let G be a simple connected graph with a full vertex. Then rr(G) ≥ ω(G). The above result has greater importance in proving many of our theorems in this paper. The concept (µλ (v))−rr sequence of a graph, has been introduced by Selvam Avadayappan et al [1]. For a vertex v ∈ V (G), let µλ (v) denote the radial radio number of the induced subgraph induced by Nλ [v]. That is, µλ (v) = rr(< Nλ [v] >). Then the sequence (µλ (v))v∈V (G) arranged in decreasing order is called the (µλ (v)) − rr sequence of the graph G. In this paper, we discuss about (µ1 (v)) − rr sequence of some simple connected graph. A graph G is said to be µ − rr regular if the (µ1 (v)) − rr sequence of G is (µ, µ, · · · , µ), where µ = rr(< N [u] >), for all u ∈ V (G) and µ is called rr− constant. A µ − rr regular graph G is said to be µ − rr highly regular if rr(G) = µ. A graph G on n vertices is said to be rr irregular, if it is not µ − rr regular. For further details on the regularity of the (µ1 (v)) − rr sequence of graph, one can refer [1]. The minimum and maximum value of a (µ1 (v)) − rr 1 sequence are defined by δrr (G) = min µ1 (v) and ∆1rr (G) = max µ1 (v). v∈V (G)

v∈V (G)

In this paper, we discuss some results on (µ1 (v)) − rr sequence of a graph and also we construct a graph with given (µ1 (v)) − rr sequence. §2. Some Basic Results Throughout this section, we consider only non trivial simple connected graph. In this section, we prove some basic results which are used in the development of this paper. Theorem 2.1 For any G, rr(G) = 2 if and only if G ∼ = K1,n , n ≥ 1. Proof Assume that rr(G) = 2. Suppose r ≥ 2. Then we need at least two labels a and a + r, , in any radial radio labeling. This gives that, rr(G) ≥ a + r ≥ 3 , as a ≥ 1, which is a contradiction. Thus r = 1. This implies that, G contains a full vertex. If ω(G) ≥ 3, then by Theorem A, we have rr(G) ≥ 3, which contradicts our assumption. This implies that, ω(G) ≤ 2. Since G is non trivial, ω(G) 6= 1. Hence we end up with the conclusion that r = 1 and ω(G) = 2. This forces that, G ∼ 2 = K1,n , n ≥ 1. The converse is obvious. Now, we establish an upper bound for the radial radio number of a self centered graph. Theorem 2.2 Let G be any self-centered graph. Then for any radial radio labeling f , of G, we have |{f (v) : v ∈ V (G)}| = |V (G)|. In other words, if G is any self-centered graph, then under any radial radio labeling f , of G, no two distinct vertices receive same labels. Proof Assume that G is self-centered. Then r = d. Let f be any radial radio labeling of G. Let u, v ∈ V (G). Then f (u) = f (v) implies u and v are at distance at least r + 1. Hence no repetition of labels is possible, since e(v) = r, for all v ∈ V (G). Thus |{f (v) : v ∈ V (G)}| =

4


|V (G)|.

2

Corollary 2.3 For any self-centered graph G, rr(G) ≥ |V (G)|. Proof From Theorem 2.2, we have for any radial radio labeling, f of G, |{f (v) : v ∈ V (G)}| = |V (G)|. That is, |range(f )| = |V (G)|. But we have span(f ) = max{f (v) : v ∈ V (G)}. So span(f ) ≥ |V (G)|. This is true for any radial radio labeling of G. This implies that, rr(G) ≥ |V (G)|. 2 Next we prove some results on the minimum and the maximum values of any radial radio sequence. 1 Theorem 2.4 For any connected graph G, δrr (G) ≥ 2. 1 Proof On contrary, assume that there is a connected graph G with δrr (G) = 1. Then there exists a vertex u ∈ V (G) such that rr(< N [u] >) = 1. This implies that, < N [u] >∼ = K1 and so u is an isolated vertex in G. This forces that, G is disconnected, which is a contradiction. 1 Hence δrr (G) ≥ 2. 2 1 Theorem 2.5 For any connected graph G, δrr (G) = 2 if and only if there exists vertex, which is not in any triangle. 1 Proof Assume that δrr (G) = 2. Then there exists a vertex v ∈ V (G) such that rr(< N [v] >) = 2 and therefore by Theorem A, N [v] ∼ = K1,n , where n ≥ 1. That is, v lies in no triangle. Conversely, assume that there exists a vertex u ∈ V (G) such that u does not belong to a triangle. This forces that, < N [u] >∼ = K1,n and so rr(< N [u] >) = 2. Since G is non 1 2 trivial, δrr (G) = 2. This completes the proof.

From the above theorem, it is clear that 1 Corollary 2.6 For any triangle free graph G, δrr (G) = 2.

Since any tree is triangle free, we have 1 Corollary 2.7 For any tree G, δrr (G) = 2.

Theorem 2.8 Let G be any connected graph on n vertices. Then ∆1rr (G) = n if and only if G∼ = Kn . Proof Assume that ∆1rr (G) = n. Then there exists a vertex v ∈ V (G) such that rr(< N [v] >) = n and so < N [v] >∼ = Kn . This implies that, each vertex in the closed neighborhood 2 of v has degree n − 1. Thus G ∼ = Kn . The converse part is obvious. Corollary 2.9 For any connected graph G of order n which is not complete, ∆1rr (G) ≤ n − 1. Theorem 2.10 No graph has a (µ1 (v)) − rr sequence in which all the elements are distinct. Proof On contrary, assume that there exists a graph on n vertices for which the elements of

5


(µ1 (v)) − rr sequence are distinct. Then G is not isomorphic to Kn . From the above results, we 1 have δrr (G) ≥ 2 and ∆1rr (G) ≤ n − 1. But the number of elements in the (µ1 (v)) − rr sequence is n. Thus in the (µ1 (v)) − rr sequence of G, at least one positive integer appears twice, which is a contradiction to our assumption. Hence there does not exist (µ1 (v)) − rr sequence with distinct elements. 2 §3. Graphs with Given (µ1 (v)) − rr Sequence The following theorem constructs a family of graphs such that for each graph G, rr(G) repeats rr(G) times in (µ1 (v)) − rr sequence. Theorem 3.1 For any given m ≥ 3, there exists a connected graph G with (µ1 (v))−rr sequence is (m, m, · · · , m, m − 1, m − 2, · · · , 3, 2) with rr(G) = m. | {z } m times

Proof Assume that m ≥ 3. Consider the graph G on n = 2(m − 1) vertices. Let V (G) = {v1 , v2 , ..., vn } and let E(G) = {vj vk : 1 ≤ j ≤ m − 1, j + 1 ≤ k ≤ n − j + 1}. Here, ∆(G) = n and r = 1. Also, deg(v1 ) = ∆(G) = n. Define f : V (G) → {1, 2, 3, ...} such that f (vi ) = i, 1 ≤ i ≤ m and f (vj ) = m, m + 1 ≤ j ≤ n. It is obvious that, f is a radial radio labeling of G. Then span(f ) = m and so rr(G) ≤ m. Note that, we have ω(G) = m and so by Theorem A, rr(G) ≥ m. Combining the above two inequalities, we get rr(G) = m. Now, we shall find the radial radio number of the induced subgraphs induced by the closed neighborhoods of each vertex of G. Case 1. Consider the vertex v1 . We have, < N [v1 ] > ∼ = G and hence rr(< N [v1 ] >) = rr(G) = m. Case 2. Consider the vertex vi , 2 ≤ i ≤ m − 1. From the construction of G, we have N [vi ] = {v1 , v2 , · · · , vi , vj : i + 1 ≤ j ≤ n − i + 1}, 2 ≤ i ≤ m − 1. Also, the restriction function f |N [vi ] is the required radial radio labeling for the graph < N [vi ] > and hence rr(< N [vi ] >) ≤ m, 2 ≤ i ≤ m − 1. Also, ω(< N [vi ] >) = m and by Theorem A, we have rr(< N [vi ] >) ≥ m. Hence we conclude that rr(< N [vi ] >) = m, 2 ≤ i ≤ m − 1. Case 3. Consider the vertex vm . Here, < N [vm ] > ∼ = Km . This implies that, rr(< N [vm ] > ) = m. Case 4. Consider the vertex vm+j−1 , 2 ≤ j ≤ m − 1. We can see that, N [vm+j−1 ] = {v1 , v2 , ..., vm−j , vm+j−1 }, 2 ≤ j ≤ m − 1. This gives that, < N [vm+j−1 ] > ∼ = Km−j+1 and hence rr(< N [vm+j−1 ] >) = m − j + 1, 2 ≤ j ≤ m − 1. Thus we conclude that, the (µ1 (v)) − rr sequence of G is (m, m, · · · , m, m − 1, m − | {z } m times

2, · · · , 3, 2). This completes the proof.

2

Remark 3.2 The graph G, constructed in the above theorem is an example of rr− irregular graph.

6


If we consider the graph G on n = 2m − 1 vertices with V (G) = {v1 , v2 , · · · , vn } and E(G) = {vj vk : 1 ≤ j ≤ m − 1, j + 1 ≤ k ≤ n − j + 1} in the proof of the above theorem, then we have Theorem 3.3 For any m ≥ 3, there exists a connected graph G with (µ1 (v)) − rr sequence is (m, m, ..., m, m − 1, m − 2, ..., 3, 2). | {z } (m+1) times

Next, we construct a family of graphs, for a given constant m with µ1 (v) − rr sequence in which any number k appears k times in a sequence for 1 ≤ k ≤ m. Theorem 3.4 For any m ≥ 3, there exists a connected graph G with (µ1 (v)) − rr sequence is (m, m, · · · , m, m − 1, m − 1, · · · , m − 1, · · · i, i, · · · , i, · · · , 3, 3, 3, 2, 2). | {z } | {z } | {z } (m) times

(m−1) times

(i) times

Proof Consider the path Pm , whose vertex set is {v1 , v2 , · · · , vm }. Now, for each i, 3 ≤ i ≤ m, add (i − 1) new vertices vi1 , vi2 , · · · , vii−1 in such a way that the induced subgraph induced by the set of vertices {vi , vi1 , vi2 , · · · , vii−1 } is isomorphic to Ki . Denote the resulting graph as G. Then V (G) = {v1 , v2 , · · · , vm } ∪ {vi1 , vi2 , · · · , vii−1 : 3 ≤ i ≤ m} and E(G) = {vi vi+1 : 1 ≤ i ≤ m − 1} ∪ {vi vij : 3 ≤ i ≤ m, 1 ≤ j ≤ i − 1} ∪ {vik vil : 3 ≤ i ≤ m, 1 ≤ k 6= l ≤ i − 1}.

From the definition of G, we have < N [v1 ] > ∼ = K2 and < N [v2 ] > ∼ = K1,2 and so rr(< N [v1 ] >) = 2 and rr(< N [v2 ] >) = 2. First we consider the vertex vi , where 3 ≤ i ≤ m. Now N [vi ] = {vi−1 , vi , vi+1 , vi1 , vi2 , ..., vii−1 }. We will show that, rr(< N [vi ] >) = i, 3 ≤ i ≤ m. Define f : N [vi ] → {1, 2, 3, · · · } such that f (vi ) = 1; f (vij ) = j + 1, 1 ≤ j ≤ i − 1; f (vi−1 ) = f (vi+1 ) = 2.

Here, r(< N [vi ] >) = 1 and so the radial radio condition for f is d(u, v)+ |f (u)− f (v)| ≥ 2, for any two distinct vertices u and v in N [vi ]. It is obvious that, span(f ) = i. Hence rr(< N [vi ] >) ≤ i, 3 ≤ i ≤ m.

(3)

Also, Ki is the clique in < N [vi ] > and by Theorem A, we have rr(< N [vi ] >) ≥ i, 3 ≤ i ≤ m.

(4)

Combining the inequalities (3) and (4), we get rr(< N [vi ] >) = i, for any i 3 ≤ i ≤ m. Next we consider the vertex vjk , where 3 ≤ j ≤ m and 1 ≤ k ≤ j − 1. We have N [vjk ] = {vj , vjk : 1 ≤ k ≤ j − 1} and < N [vjk ] >∼ = Kj . Here also, r(< N [vjk ] >) = 1, for 1 ≤ k ≤ j − 1 and

7


3 ≤ i ≤ m. Since Kj is the maximal clique in N [vjk ], we have rr(< N [vjk ] >) ≥ j, 3 ≤ j ≤ m.

(5)

Now, for 3 ≤ j ≤ m define g : N [vjk ] → {1, 2, · · · } such that g(vj ) = 1; g(vjk ) = k + 1, 1 ≤ k ≤ j − 1.

Obviously, g is a radial radio labeling for < N [vjk ] >, 3 ≤ j ≤ m, 1 ≤ k ≤ j − 1. Then span(g) = j and hence rr(< N [vjk ] >) ≤ j (6) From the inequalities (1) and (2), we conclude that rr(< N [vjk ] >) = j, 3 ≤ j ≤ m and 1 ≤ k ≤ j − 1. Thus we have found that there are j vertices with rr(< N [vjk ] >) = j, for all 3 ≤ j ≤ m and 1 ≤ k ≤ j − 1. Hence the (µ1 (v)) − rr sequence of G is (m, m, · · · , m, m − 1, m − 1, · · · , m − 1, · · · i, i, · · · , i, ·, 3, 3, 3, 2, 2). {z } | {z } | {z } | (m) times

(m−1) times

(i) times

This completes the proof.

2

Theorem 3.5 For any given a1 , a2 , · · · , ak with a1 > a2 > · · · > ak ≥ 2 and k ≥ 1, there exists a simple connected graph whose (µ1 (v)) − rr sequence is (a1 , a1 , · · · , a1 , a2 , a2 , · · · , a2 , · · · , ai , ai , · · · , ai , · · · , ak , ak , · · · , ak ). | {z } | {z } | {z } | {z } a1 times

a2 times

ai times

ak times

Proof Consider the path Pk , whose vertex set is {v1 , v2 , · · · vk }. Now, for each ai , 1 ≤ i ≤ m add ai −1 new vertices vi1 , vi2 , · · · , vi ai −1 and join them in such a way that the induced subgraph induced by the set of vertices {vi , vi1 , vi2 , · · · , vi ai −1 } is isomorphic to Kai . Denote the resulting 2 graph as G. Now as in the proof of the above theorem, we obtain the required result. The following theorem gives the graph with radial radio number as twice the maximum value of its radial radio sequence. Theorem 3.6 For any given n ≥ 3, there exists an n − rr regular graph with rr(G) = 2n. Proof Consider the graph G = Kn × K2 , n ≥ 3. Now, we prove that G is the required graph. Let V (G) = {v1 , v2 , · · · , vn , v1′ , v2′ , · · · , vn′ } and let E(G) = {vi vj , vi′ vj′ : 1 ≤ i 6= j ≤ n} ∪ {vi vi′ : 1 ≤ i ≤ n}. We have r = 2. Define f : V (G) → {1, 2, · · · } such that f (vi ) = 2i − 1, 1 ≤ i ≤ n; f (vi′ ) = 2(⌊i/2⌋ − 1), ⌊n/2⌋ + 1 ≤ i ≤ n; f (v1′ ) = f (vn′ ) + 2; ′ f (vj′ ) = f (vj−1 ) + 2, 1 ≤ j ≤ ⌊n/2⌋. The radial radio condition for G is d(u, v) + |f (u) − f (v)| ≥ 3, for any two distinct vertices

8


u and v of G. It is obvious that, from the definition of f , f is a radial radio labeling of G. Thus span(f ) = 2n and so rr(G) ≤ 2n. Since G is self centered, from Corollary 2.3, we have rr(G) ≥ 2n. Thus rr(G) = 2n. Now, we shall find the radial radio numbers for the induced subgraphs induced by the closed neighborhood of each vertex of G. For any 1 ≤ i ≤ n, N [vi ] = {vi : 1 ≤ i ≤ n} ∪ {vi′ } and E(< N [vi ] >) = {vi vj : 1 ≤ i 6= j ≤ n} ∪ {vi vi′ }. Define g : N [vi ] → {1, 2, · · · } such that g(vi ) = i, 1 ≤ i ≤ n; g(vi′ ) = g(vi−1 ), 1 ≤ i ≤ n; g(vn′ ) = g(v1 ). Obviously, g is a radial radio labeling for < N [vi ] > and so rr(< N [vi ] >) ≤ n. Also, ∼ Kn , by Theorem A, rr(< N [vi ] >) ≥ n. Hence rr(< N [vi ] >) = n. Since G ω(< N [vi ] >) = is vertex transitive, rr(< N [u] >) = n, for all u ∈ V (G). Thus the (µ1 (v)) − rr sequence of G 2 is (n, n, · · · , n) and hence G is n − rr regular. This completes the proof. Acknowledgement The third author thanks U. G. C., India for supporting her by Maulana Azad National Fellowship for Minority Students. (Grant no. F1-17.1/2015-16/MANF-2015-17-TAM-66646). References [1] Selvam Avadayappan, M. Bhuvaneshwari, S. Vimalajenifer, A Note on Radial Radio Number of a Graph, International Journal of Applied and Advanced Scientific Research, 2017, 62-68. [2] Selvam Avadayappan, M. Bhuvaneshwari, S. Vimalajenifer, Radial radio number of a graph, Global Journal of Pure and Applied Mathematics, Vol.13, 5(2017), 347-357. [3] R. Balakrishnan and K. Ranganathan, A Text Book of Graph Theory, Springer- Verlag, New York, Inc(1999). [4] F. Buckley and F. Harary, Distance in Graphs, Addison Wesley Reading, 1990 [5] G. Chartrand, D. Erwin, F. Harary and P. Zhang, Radio labeling of graphs, Bulletin of the Institute of Combinatorics and its Applications, Vol.33(2001), 77-85. [6] G. Chartrand and P. Zhang, Introduction to Graph Theory, McGraw Hill Publications(India) Private Limited, New Delhi, India, 2001. [7] C. Fernandance, A. Flores, M. Tomova and C. Wyles, The Radio Number of Gear Graphs, reprint available at front.math.ucdvis.edu/0809.2623. [8] J. A. Gallian, A Dynamic Survey of Graph Labeling, The Electronic Journal of Combinatorics, 2015. [9] W. K. Hale, Frequency Assignment: Theory and Applications, Proceeding of the IEEE, Vol. 68, No. 12(1980). [10] KM. Kathiresan and S. Vimlajenifer, Radial radio number of graphs (Submitted). [11] A. Rosa, On certain valuations of the vertices of a graph, Theory of Graphs (Internat. Symposium, Rome, July 1966), Gordon and Breach, N. Y. and Dunod Paris (1967) 349355.


Non-Isolated Resolving Number for Some Splitting Graphs

Selvam Avadayappan, M. Bhuvaneshwari and P. Jeya Bala Chitra (Research Department of Mathematics, VHNSN College, Virudhunagar - 626 001, India) E-mail: [email protected], [email protected], [email protected]

Abstract:

Let G be a connected graph.

of V with an order imposed on it.

Let W = {w1 , w2 , ..., wk } be a subset

For any vertex v ∈ V , the vector r(v|W ) =

(d(v, w1 ), d(v, w2 ), ..., d(v, wk )) is called the metric representation of v with respect to W . If distinct vertices in V have distinct metric representation, then W is called a resolving set of G. The minimum cardinality of a resolving set of G is called the metric dimension of G and it is denoted by dim(G). A resolving set W is called a non-isolated resolving set if the induced sub graph hW i has no isolated vertices. The minimum cardinality of a non-isolated resolving set of G is called the non-isolated resolving number of G and is denoted by nr(G). In this paper, we determine the non-isolated resolving number for the splitting graph of some standard graphs.

Key Words: Resolving set, metric dimension, non-isolated resolving set, non-isolated resolving number, splitting graph.

AMS(2010): 05C12.

§1. Introduction

Throughout this paper, we consider only finite, simple and undirected graphs. The vertex set and edge set of a graph G are denoted by V (G) and E(G) respectively. The cardinality of the vertex set of a graph G is commonly denoted by n(G). For basic notations and terminology we refer [4]. For any vertex v ∈ V in a graph G, the neighborhood N (v) of v is the set of all vertices adjacent to v. That is, N (v) = {u ∈ V : uv ∈ E}. The concept of splitting graph was introduced by Sampath Kumar and Waliker [10]. The graph S(G), obtained from G, by adding a new vertex w for every vertex v ∈ V and joining w to all vertices of G adjacent to v, is called the splitting graph of G. For example, a graph G and its splitting graph S(G) are shown in Figure 1.1.

1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 28, 2018, Accepted March 22, 2018, Edited by S. Monikandan.

10

Selvam Avadayappan, M. Bhuvaneshwari and P. Jeya Bala Chitra

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Figure 1.1 Motivated by the problem of uniquely determining the locations of an intruder in a network, the concept of metric dimension of a graph was introduced by Slater in ([13] and [14]) and studied independently by Harary and Melter in [7]. The distance d(u, v) between two vertices u and v is the length of a shortest path between them. Let W = {w1 , w2 , · · · , wk } be an ordered set of vertices of G and let v be a vertex of G. The representation r(v|W ) of v with respect to W is the k-tuple (d(v, w1 ), d(v, w2 ), · · · , d(v, wk )). Moreover, the vertex v in G is said to be resolved by a vertex u in W , if the vertex u distinguishes the representation of v, r(v|W ) from the representation of all the other vertices of G. If distinct vertices of G have distinct representations with respect to W , then W is called a resolving set for G. A resolving set of minimum cardinality is called a basis for G and this cardinality is the metric dimension of G and it is denoted by dim(G). For example, in the graph G shown in Figure 1.2, W = {v1 , v5 } is a basis for G. Therefore, dim(G) = 2. v1 b

v2 b

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v4 v5 Figure 1.2

Applications of resolving set arise in various areas including coin weighing problem [12], drug discovery [6], robot navigation [9], network discovery and verification [3], connected joins in graphs [12] and strategies for master mind game [5]. For survey of results in metric dimension we refer to Chartrand and Zhang [5]. Several models of resolving set have been investigated by imposing conditions on the subgraph induced by a resolving set. Some of the well studied parameters of this type include connected resolving set [11] and independent resolving set. A resolving set W of G is said to be an independent resolving set if no two vertices in W are adjacent. A resolving set W of G is said to be a connected resolving set, if the induced subgraph induced by W is a non-trivial connected


11

subgraph of G. The minimum cardinality of a connected resolving set is the connected resolving number of G. It is denoted by cr(G). In a similar line, a non-isolated resolving set has been introduced in [8]. A resolving set W of G with at least two vertices is said to be a non-isolated resolving set, if the induced subgraph hW i induced by W has no isolated vertices. The minimum cardinality of a nonisolated resolving set in a graph G is the non-isolated resolving number of G and it is denoted by nr(G). A non-isolated resolving set of cardinality nr(G) is called an nr-set of G. v1

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For example, consider the graph G given in Figure 1.3. Here W ′ = {v1 , v2 , v3 } is an nr-set and W = {v1 , v2 } is a basis for G. Hence, dim(G) = 2 and nr(G) = 3. Since, every non-trivial connected graph has no isolated vertices, nr(G) ≤ cr(G). Also, it has been proved in [8] that, for any graph G, nr(G) ≤ 2 dim(G).

In [8], nr-values of some families of graphs, cartesian product of some graphs and the corona product of a graph G with K2 have been obtained. Further more, for any two positive integers k and n with 2 ≤ k ≤ n − 1, a graph G of order n with nr(G) = k has been constructed. In [1] the exact values of nr(G) for some more product graphs and the subdivision of some standard graphs have been obtained. Also, the nr-value for the generalized Petersen graph P (n, 2) have been obtained. Also in [2], the non-isolated resolving number for some standard graphs such as double broom, bistar and for the join of complete graphs and paths, etc have been determined. Further more, the relationship between the nr-value and the parameters χ(G) and ∆(G) have been discussed. In this paper, we determine the exact value of the non-isolated resolving number for the splitting graph of the graphs Pn , Cn , Kn and K1,n .

§2. nr-value for the splitting graph Theorem 2.1 For any positive integer n ≥ 2, nr(S(K1,n )) = 2n. Proof Let G = S(K1,n ). Let V (K1,n ) = {v, v1 , v2 , · · · , vn } where v is the center of K1,n and let v ′ , v1′ , v2′ , · · · , vn′ be the corresponding newly added vertices in G. Let W = ′ {v, v1 , v2 , · · · , vn ; v1′ , v2′ , · · · , vn−1 }. Then r(v ′ |W ) = (2, 1, 1, · · · , 1, 3, 3, · · · , 3 ) and r(vn′ |W ) = | {z } | {z } n−times

(n−1)−times

(1, 2, 2, · · · , 2). Also hW i is isomorphic to K1,2n−1 . Therefore W is a non-isolated resolving set for G. Hence nr(G) ≤ 2n. Next, we claim that nr(G) ≥ 2n. Let W be a non-isolated resolving set for G. If W contains all vi ’s and vj′ ’s for 1 ≤ i, j ≤ n, then we have done.

12


Suppose W contains all but one of {v1 , v2 , · · · , vn ; v1′ , v2′ , · · · , vn′ } since n ≥ 2, W contains at least one vi . If v ∈ / W , then vi′ is an isolated vertex in hW i, a contradiction. Therefore, v must be in W and so |W | ≥ 2n. Now, if two vertices of {v1 , v2 , · · · , vn }, say vi and vj , are not in W , then r(vi |W ) = r(vj |W ). Thus there exists at most one vi , say v1 ∈ / W . Similarly, there exists at most one vi′ , say v1′ ∈ / W . If v ′ ∈ / W , then r(v1 |W ) = r(v1′ |W ), a contradiction. Also, if v ∈ / W , then v2′ is an ′ isolated vertex in hW i, a contradiction. This forces that both v and v are in W . Therefore, |W | ≥ 2n. Hence nr(G) ≥ 2n. Thus nr(G) = 2n. 2 Illustration 2.2 The illustration of the above theorem when n = 4 is given in Figure 2.1

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Figure 2.1 nr−set for S(K1,4 ) Next we obtain the nr−value for the splitting graph of the complete graph Kn . Since S(K2 ) ∼ = P4 , nr(S(K2 )) = 2. When n = 3, nr(S(K3 )) = 3 which can be obtained from Figure 2.2. b

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Figure 2.2 nr−set for S(K3 ) For n theorem.

≥

4,

we

determine

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nr-value

for

S(Kn )

in

the

following

Theorem 2.3 For any positive integer n ≥ 4, nr(S(Kn )) = n − 1. Proof Let G = S(Kn ) and let V (G) = V1 ∪ V2 where V1 = {v1 , v2 , · · · , vn } and V2 = ′ , vn′ } are the corresponding newly added vertices. Let W = {v1 , v2′ , v3′ , · · · , vn−1 }.

{v1′ , v2′ , · · ·

13


Then for 2 ≤ i ≤ n − 1, r(vi |W ) = (1, 1, · · · , 1, 2, 1, 1, · · · , 1) where 2 appears in the ith place, r(vn |W ) = (1, 1, · · · , 1), r(v1′ |W ) = (2, 2, ..., 2) and r(vn′ |W ) = (1, 2, 2, · · · , 2). Also hW i is isomorphic to K1,n−2 . Therefore, W is a non-isolated resolving set for G. Hence nr(G) ≤ n − 1. Now, it is enough to prove that nr(G) ≥ n − 1. Let W be any non-isolated resolving set for G. On contrary assume that |W | = m < n − 1. Suppose that W ⊆ V1 . If vi ∈ / W , for ′ some i, 1 ≤ i ≤ n, then r(vi |W ) = r(vi |W ) = (1, 1, ..., 1). Therefore W = V1 and |W | = n, a c contradiction. If W ⊆ V2 , then hW i ∼ , a contradiction. = Km Let |W ∩ V1 | = 1. Without loss of generality, let v1 ∈ W . Then if v1′ ∈ W , then v1′ is an isolated vertex in hW i. Therefore, v1′ does not belong to W . Now if vi′ , vj′ ∈ / W for ′ ′ 2 ≤ i 6= j ≤ n, then r(vi |W ) = r(vj |W ). Hence there exists at most one i such that vi′ ∈ / W, 2 ≤ i ≤ n. Thus |W ∩ V2 | ≥ n − 2. In general, if |W ∩ V1 | = i, then |W ∩ V2 | ≥ n − (i + 1), 1 ≤ i ≤ n − 2. Therefore, |W | ≥ n − 1. Hence nr(G) ≥ n − 1. Thus nr(G) = n − 1. 2 Illustration 2.4 The illustration of the above theorem for the case when n = 5 is given in Figure 2.3. b

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Figure 2.3 nr-set for S(K5 ) ∼ P4 . Hence nr(S(P2 )) = 2. Next we consider the splitting graph of Pn . For n = 2, S(P2 ) = Also nr(S(P3 )) = 4 and nr(S(P4 )) = 3. the nr−set for the corresponding graphs are given in Figure 2.4.

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For n ≥ 5, we establish a formula for the nr−value of the splitting graph of the path in the following theorem.

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Theorem 2.5 For any positive integer n ≥ 5,   3n 5 nr(S(Pn )) =  3n + 1 5

if n ≡ 1, 2, 3(mod 5) if n ≡ 0, 4(mod 5)

.

Proof Let G = S(Pn ) and V (G) = {v1 , v2 , · · · , vn ; v1′ , v2′ , · · · , vn′ } where V (Pn ) = {v1 , v2 , · · · , vn } and v1′ , v2′ , · · · , vn′ be the corresponding splitted vertices. Case 1. Let n ≡ 0 (mod 5). ′ ′ ′ Take W = {v5i+1 , v5i+2 , v5(i+1) , vn−1 : 0 ≤ i ≤ ⌊ n6 ⌋}. Then |W | = ⌈ 3n 5 ⌉ + 1. Also v5i+1 is ′ ′ ′ ′ resolved by v5i+2 ; v5i+3 and v5i+4 are resolved by v5(i+1) and v5i+2 , v5i+3 , v5i+4 and v5i+4 are ′ resolved by v5(i+1) , 0 ≤ i ≤ ⌊ n6 ⌋. and hW i ∼ = (⌊ n6 ⌋P3 ) ∪ 2P2 . Therefore, W is a non-isolated resolving set for G.

Case 2. Let n ≡ 1 (mod 5). ′ ′ ′ Take W = {v5i+1 , v5i+2 , v5(i+1) , vn : 0 ≤ i ≤ ⌊ n6 ⌋}. Then |W | = ⌈ 3n 5 ⌉, vn is resolved by vn n and all the other vertices are resolved as in the Case 1. Also, hW i ∼ = (⌊ 6 ⌋P3 ) ∪ 2P2 . Therefore, W is a non-isolated resolving set for G.

Case 3. Let n ≡ 2, 3 (mod 5). ′ ′ Take W = {v5i+1 , v5i+2 , v5(i+1) : 0 ≤ i ≤ ⌊ n6 ⌋}. Then |W | = ⌈ 3n 5 ⌉. For n ≡ 2 (mod 5) all ′ the vertices are resolved as in Case 1 and for n ≡ 3 (mod 5), vn and vn′ are resolved by v5⌊ n 6 ⌋+2 n and all the other vertices are resolved as in Case 1. Also, hW i ∼ (⌊ ⌋P ) ∪ P . Therefore, W = 5 3 2 is a non-isolated resolving set for G.

Case 4. Let n ≡ 4 (mod 5). ′ ′ Take W = {v5i+1 , v5i+2 , v5(i+1) , vn−1 , vn′ : 0 ≤ i ≤ ⌊ n6 ⌋}. Then |W | = ⌈ 3n 5 ⌉ + 1 and vn ′ is resolved by vn and also all the remaining vertices are resolved as in Case 1. And hW i ∼ = (⌊ n6 ⌋P3 ) ∪ 2P2 . Therefore, W is a non-isolated resolving set for G. 3n Hence, from all the above cases nr(G) ≤ ⌈ 3n 5 ⌉, if n ≡ 1, 2, 3 (mod 5) and nr(G) ≤ ⌈ 5 ⌉ + 1, if n ≡ 0, 4 (mod 5). Let W be any non-isolated resolving set for G. Let n = 5s + t, 0 ≤ t ≤ 4. If the vertices v1 , v1′ , v2 and v2′ are not in W , then r(v1 |W ) = r(v1′ |W ). If v1 ∈ W , then r(v2 |W ) = r(v2′ |W ), a contradiction. If v2 ∈ W , then r(v1 |W ) = r(v1′ |W ), a contradiction. Hence v1′ or v2′ is in W . But to minimize the cardinality of W , we choose v2′ to be in W . Already, r(v3 |W ) 6= r(v3′ |W ) which does not forces the vertex v4′ to be in W . Thus v4′ ∈ / W. ′ ′ Now, r(v4 |W ) = r(v4 |W ) and v5 is the only vertex which differentiate the representation of the vertices v4 and v4′ . Hence v5′ ∈ W . Also, if v6 ∈ / W , then r(v2 |W ) = r(v4′ |W ) and r(v3′ |W ) = r(v5 |W ). Thus v6 ∈ W . Now to differentiate the representation of the vertices v7 and v7′ , either v7′ or v8′ must be in W . If v8′ ∈ W , then for non-isolation, we have to choose one more vertex. Hence in the case of t = 3, |W | is increased by one, a contradiction. Thus we choose v7′ to be in W . And

15


′ ′ then r(v9 |W ) = r(v9′ |W ) unless v10 ∈ W . Hence v10 ∈ W . And the pattern follows. Also, the n ′ vertices v1 and v5i+1 , 2 ≤ i ≤ ⌊ 6 ⌋ must be in W , otherwise v2′ and v5(i+1) are isolated vertices 3n n ′ in hW i. Thus, v1 , v5(i+1) ∈ W , where 2 ≤ i ≤ ⌊ 6 ⌋. Therefore, |W | ≥ ⌈ 5 ⌉ + 1 for n ≡ 0, 4 3n (mod 5) and |W | ≥ ⌈ 3n 5 ⌉ for n ≡ 1, 2, 3 (mod 5). Hence nr(G) ≥ ⌈ 5 ⌉ + 1, if n ≡ 0, 4 (mod 5) 3n and nr(G) ≥ ⌈ 5 ⌉, if n ≡ 1, 2, 3 (mod 5). 2

Illustration 2.6 The cases when n = 6, 7, 8, 9 and 10 are illustrated in Figure 2.5.

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Next we obtain the nr−value for the splitting graph of the cycle Cn . For n = 3, C3 ∼ = K3 . nr(S(C4 )) = 4, nr(S(C5 )) = 4 and nr(S(C6 )) = 5. The nr−set for the corresponding graphs are given in Figure 2.6.

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S(C6 ) Figure 2.6 For n ≥ 7, we determine a formula for the nr−value of the splitting graph of cycle Cn in the following theorem. Theorem 2.7 For any positive integer n ≥ 7, nr(S(Cn )) = ⌈ 3n 5 ⌉. Proof Let G = S(Cn ) and V (G) = {v1 , v2 , · · · , vn ; v1′ , v2′ , · · · , vn′ } where V (Cn ) = {v1 , v2 , · · · , vn } and v1′ , v2′ , · · · , vn′ be the corresponding newly added vertices. Case 1. Let n ≡ 0, 4 (mod 5). ′ ′ ∼ n Take W = {v5i+1 , v5i+2 , v5i+3 : 0 ≤ i ≤ ⌊ n6 ⌋}. Then |W | = ⌈ 3n 5 ⌉ and hW i = ⌈ 5 ⌉P3 . ′ ′ ′ ′ Also, r(v5i+4 |W ) is different from r(v5i+4 |W ), since d(v5i+4 , v5i+3 ) 6= d(v5i+4 , v5i+3 ). And also ′ ′ ′ ′ ′ v5i+1 , v5i+3 , v5i and v5i are resolved by v5i+1 and v5i+2 is resolved by v5i+1 or v5i+3 . Therefore, W is a non-isolated resolving set for G.

Case 2. Let n ≡ 1 (mod 5). 3n ′ ′ ′ ′ ′ Take W = {v5i+1 , v5i+2 , v5i+3 , vn−5 , vn−4 , vn−2 , vn−1 : 0 ≤ i ≤ ⌊ n−5 6 ⌋}. Then |W | = ⌈ 5 ⌉ and hW i ∼ = ((⌊ n5 ⌋ − 1)P3 ) ∪ 2P2 . Also vn′ and vn are resolved by v1′ and as in Case 1, all the other vertices are resolved. Therefore W is a non-isolated resolving set for G.

Case 3. Let n ≡ 2 (mod 5). 3n ′ ′ ′ ∼ , vn : 0 ≤ i ≤ ⌊ n−5 Take W = {v5i+1 , v5i+2 , v5i+3 , vn−1 6 ⌋}. Then, |W | = ⌈ 5 ⌉ and hW i = n ′ ((⌊ 5 ⌋ − 1)P3 ) ∪ P5 . Also, vn−1 and vn are resolved by vn and as in Case 1, all the other vertices are resolved. Therefore, W is a non-isolated resolving set for G.

Case 4. Let n ≡ 3 (mod 5).

17


3n ′ ′ ′ ′ Take W = {v5i+1 , v5i+2 , v5i+3 , vn−2 , vn−1 : 0 ≤ i ≤ ⌊ n−5 6 ⌋}. Then, |W | = ⌈ 5 ⌉ and n ′ ′ ′ hW i ∼ = ((⌊ 5 ⌋)P3 ) ∪ P2 . Also, vn−1 is resolved by vn−1 and vn−2 , vn are resolved by v1 and all the other vertices are resolved as in Case 1. Therefore, W is a non-isolated resolving set for G. 3n Hence, from all the above cases, we get |W | ≤ ⌈ 3n 5 ⌉. Thus, nr(G) ≤ ⌈ 5 ⌉. 3n Now it is enough to prove that, nr(G) ≥ ⌈ 5 ⌉. Let W be any non-isolated resolving set for G and let n = 5s + t, 0 ≤ t ≤ 4. Without loss of generality, we choose v1′ ∈ W . Now, r(v2′ |W ) = r(v4 |W ) and v3′ is the only vertex which differentiates the representation of v2′ and v4 . Hence v3′ ∈ W . Already, r(v4′ |W ) 6= r(v4 |W ) which does not forces the vertex v5′ to be in W . Hence v5′ ∈ / W . And then r(v5′ |W ) = r(v5 |W ) and r(v1 |W ) = r(v3 |W ) unless v6 , ∈ W . ′ Thus v6 ∈ W . Now to differentiate the representation of the vertices v9 and v9′ , the vertex v8′ must be in W . Hence v8′ ∈ W . Also the vertices v2 and v7 must be in W , otherwise hW i will contain v1′ , v3′ , v6′ and v8′ as isolated vertices. ′ ′ ′ ′ Thus for every 0 ≤ r ≤ s − 1, we choose the vertices v5r+1 , v5r+2 , v5r+3 , v5r+6 , v5r+7 , v5r+8 . ′ ′ Hence v5s+3 is the last chosen vertex. If t = 1, then {v5s+t−3 , v5s+t−2 , v5s+t−1 } ⊆ W . If ′ ′ t = 2, then {v5s+t−1 , v5s+t } ⊆ W . If t = 3, then {v5s+t−2 , v5s+t−1 } ⊆ W . If t = 4, then ′ ′ {v5s+t−3 , v5s+t−2 , v5s+t−1 } ⊆ W . Therefore, any non-isolated resolving set must contain ⌈ 3n 5 ⌉ 3n 3n 3n vertices. Hence |W | ≥ ⌈ 5 ⌉. That is, nr(G) ≥ ⌈ 5 ⌉. Thus, nr(G) = ⌈ 5 ⌉. 2

Illustration 2.8 The illustration of the above theorem for the cases n = 7, 8, 9, 10 and 11 are given in Figure 2.7.

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References [1] Selvam avadayappan, M. Bhuvaneshwari, P. Jeya Bala Chitra, More results on Non-isolated resolving number of a graph, International Journal of Applied and Advanced Scientific Research, Special issue, (2017) 49-53. [2] Selvam avadayappan, M. Bhuvaneshwari, P. Jeya Bala Chitra, Non-isolated resolving number of a graph, Global Journal of Pure and Applied Mathematics, Vol. 13, No. 5 (2017) 336-346. [3] Z. Beuliova, F. Eberhard, T. Erlebach, A. Hall, M. Hoffman, M. Mihalak, L. Ram, Network Discovery and Verification, IEEE J. Sel. Areas commun., 24 (2006) 2168-2181. [4] G. Chartrand and L. Lesniak, Graphs and Digraphs, 3rd ed., Chapman and Hall, London, (1996). [5] G. Chartrand, P. Zhang, The theory and application of resolvability in graphs: a survey, Congr. Numer., 160 (2003) 47-68. [6] V. Chavatal, Mastermind, combinatorica, 3 (1983) 325-329. [7] F. Harary, R.A. Melter, On the metric dimension of a graph, Ars. Combin., 2 (1976) 191-195. [8] P. Jeya Bala chitra, S. Arumugam, Resolving Sets without Isolated Vertices, Procedia Computer Science, 74 (2015) 38-42. [9] S. Khuller, B. Raghavachari, A. Rosenfield, Landmarks in graphs, Discrete Appl. Math., 70 (1996) 217-229. [10] E. Sampath Kumar, H.B. Waliker, on the splitting graph of a graph, Karnatak Uni. Sci., 25:13 (1980) [11] V. Saenpholphat and Ping Zhang, Connected Resolvability of graphs, Czechoslovak Mathematical Journal, 53 (2003), 827-840. [12] A. Sebo, E. Tannier, On metric generators of graphs, Math. Oper. Res, 29 (2004) 383-393. [13] P.J. Slater, Leaves of Trees, Congr. Numer., 14 (1975) 549-559. [14] P.J. Slater, Dominating and reference sets in graphs, J. Math. Phys. Sci., 22 (1988) 445-455.


Prime Labelings of Some Helm and Gear Related Graphs J. Naveen and S. Meena (Research Department of Mathematics, Government Arts College, C.Mutlur, Chidambaram, Tamil Nadu, India.) E-mail: e-mail: [email protected], [email protected]

Abstract: A Graph G with n vertices is said to admit prime labeling if its vertices can be labeled with distinct positive integers not exceeding n such that the labels of each pair of adjacent vertices are relatively prime. A graph G which admits prime labeling is called a prime graph. In this paper we investigate the existence of prime labeling of some graphs related to Helm Hn and Gear graph Gn . We discuss prime labeling in the context of the graph operation namely corona product.

Key Words: Graph Labeling, Prime Labeling, corona product, Prime Graph. AMS(2010): 05C78. §1. Introduction In this paper, we consider only finite simple undirected graph. The graph G is a set of vertices V (G), together with a set of edges E(G) and incidence relation. If u, v ∈ V (G) are connected by an edge, we say u and v are adjacent and the corresponding edge is denoted by uv or vu. The degree of a vertex u is the number of edges adjacent with u. A graph is connected if it does not consist of two or more disjoint “pieces”. The path Pn is the connected graph consisting of two vertices of degree 1 and n − 2 vertices of degree 2. An n-cycle Cn , is the connected graph consisting of n vertices each of degree 2. An n-star Sn , is the graph consisting of one vertex of degree n and n vertices of degree 1. But Sn consist of n + 1 vertices and n edges. A tree is a graph contains no cycle. Path and stars are example of trees. For notations and terminology we refer to Bondy and Murthy [1]. The notion of prime labeling was introduced by Roger Entringer and was discussed in a paper by Tout [8]. Two integers a and b are said to be relatively prime if their greatest common divisor is 1 denoted by (a, b) = 1. Every path Pn , cycle Cn and star Sn are prime [2]. Every wheel Wn iff n is even [2], all helm Hn , crown Cn∗ and gear graph Gn are prime [2]. We refer Gallian’s dynamic survey [2] for a comprehensive listing of the families of graphs that are known to have or known not to have prime vertex labeling. S.K.Vaidya et al [5,6,7] investigated about the existence of prime labeling for some path, cycle and wheel related graphs in the context of some graph operations namely duplication, fusion and vertex switching etc. Let G and H 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 25, 2018, Accepted May 22, 2018, Edited by S. Monikandan.

20

J. Naveen and S. Meena

be two graphs. The corona product G ⊙ H is obtained by taking one copy of G and |V (G)| copies of H and by joining each vertex of the i-th copy of H to the i-th vertex of G where 1 ≤ i ≤ |V (G)| [3]. For all m, n ∈ N with n ≥ 3, an m-hairy n-cycle , denoted by Cn ∗ Sm , is the cycle Cn with m pendants attached to each cycle vertex [4]. In other words this graph is corona product Cn ⊙ K m . The Helm Hn is obtained from a wheel Wn by attaching a pendent edge at each vertex of the n-cycle. The Gear graph Gn is, the graph obtained by wheel Wn = Cn + K1 by subdividing every spoke edge. We consider the graph obtained by, attaching the centre vertex of a copy of the star Sm to each vertex of Helm Hn and Gear graph Gn . The resulting graph will have m pendants at each vertex and it is denoted by Hn ∗ Sm and Gn ∗ Sm . The star notation indicates that we attach a copy of the star Sm at its vertex of the degree m to all the vertices of Hn and Gn . In other words Hn ∗ Sm is the corona product Hn ⊙ K m and Gn ∗ Sm is the corona product Gn ⊙ K m . In this paper, we proved that the graphs obtained by attaching m-star to each vertex of Helm Hn and Gear graph Gn are all prime graphs.

§2. Main Results Theorem 2.1 The graph Hn ∗ S2 admits prime labeling if n 6≡ 1(mod5). Proof Let c0 be the centre of the Helm Hn and let c1 , c2 , · · · , cn be the rim vertices of the Hn . Let p1 , p2 , · · · , pn be the pendant vertices attached at c1 , c2 , · · · , cn respectively. Let p10 , p20 be the pendent vertices attached at c0 and let pji and lij be the pendant vertices attached at ci and pi respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 2. The graph Hn ∗ S2 has 6n + 3 vertices and 7n + 2 edges. Define a labeling f : V → {1, 2, 3, · · · , 6n + 3} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 2 for 1 ≤ i ≤ n f (ci ) = 6i−1, f (pi) = 6i+1, f (p1i ) = 6i−2, f (p2i ) = 6i, f (li1) = 6i+2, f (li2) = 6i+3. It remains to show that the labels of each pair of adjacent vertices are relatively prime. gcd (f (c0 ), f (pj0 ) = gcd(1, j + 1) = 1, for 1 ≤ j ≤ 2. For 1 ≤ i ≤ n, gcd f (c0 ), f (ci ) = gcd(1, 6i − 1) = 1, gcd f (ci ), f (ci+1 ) = gcd(6i − 1, 6(i + 1) − 1) = gcd(6i − 1, (6i − 1) + 6) = 1 as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd f (c1 ), f (cn ) = gcd(5, 6n − 1) = 1, since n 6≡ 1(mod5), gcd f (ci ), f (pi ) = gcd(6i−1, 6i+1) = 1 as these two numbers are consecutive odd integers. gcd f (ci ), f (p1i ) = gcd(6i − 1, 6i − 2) = 1. gcd f (ci ), f (p2i ) = gcd(6i − 1, 6i) = 1. gcd f (pi ), f (li1 ) = gcd(6i + 1, 6i + 2) = 1 as these two numbers are consecutive integers. gcd f (pi ), f (li2 ) = gcd(6i+1, 6i+3) = 1 as these two numbers are consecutive odd integers. Thus f is a prime labeling. Hence Hn ∗ S2 is a prime graph.

2

Prime Labelings of Some Helm and Gear Related Graphs

21

Theorem 2.2 The graph Hn ∗ S3 admits prime labeling if n 6≡ 1(mod7). Proof Let c0 be the centre of the Helm Hn and let c1 , c2 , · · · , cn be the rim vertices of the Hn . Let p1 , p2 , · · · , pn be the pendant vertices attached at c1 , c2 , · · · , cn respectively. Let p10 , p20 , p30 be the pendent vertices attached at central vertex c0 and let pji and lij be the pendant vertices attached at ci and pi respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 3. The graph Hn ∗ S3 has 8n + 4 vertices and 9n + 3 edges. Define a labeling f : V → {1, 2, 3, · · · , 8n + 4} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 3.

For 1 ≤ i ≤ n, f (ci ) = 8i − 1, f (pi ) = 8i + 3, f (p1i ) = 8i − 3, f (p2i ) = 8i − 2, f (p3i ) = 8i, f (li1 ) = 8i + 1, f (li2 ) = 8i + 2, f (li3 ) = 8i + 4. It remains to show that the labels of each pair of adjacent vertices are relatively prime. gcd f (c0 ), f (pj0 ) = gcd(1, j + 1) = 1, for 1 ≤ j ≤ 3. For 1 ≤ i ≤ n, gcd f (c0 ), f (ci ) = gcd(1, 8i − 1) = 1, gcd f (ci ), f (c( i + 1)) = gcd(8i − 1, 8(i + 1) − 1) = gcd 8i − 1, (8i − 1) + 8 = 1 as these two numbers are odd and their difference is 8 and both are not multiples of 3, 5 or 7. gcd f (c1 ), f (cn ) = gcd(7, 8n − 1) = 1, since n 6≡ 1(mod7). gcd f (ci ), f (pi ) = gcd(8i − 1, 8i + 3) = 1 as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd f (ci ), f (p1i ) = gcd(8i − 1, 8i − 3) = 1 as these two numbers are consecutive odd integers. gcd f (ci ), f (p2i ) = gcd(8i − 1, 8i − 2) = 1. gcd f (ci ), f (p3i ) = gcd(8i − 1, 8i) = 1 as these two numbers are consecutive integers. gcd f (pi ), f (li1 ) = gcd(8i+3, 8i+1) = 1 as these two numbers are consecutive odd integers. gcd f (pi ), f (li2 ) = gcd(8i + 3, 8i + 2) = 1. gcd f (pi ), f (li3 ) = gcd(8i + 3, 8i + 4) = 1 as these two numbers are consecutive integers. Thus f is a prime labeling. Hence Hn ∗ S3 is a prime graph. 2 Illustration 2.3 10

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Figure 1. Prime labeling of H4 ∗ S3

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Theorem 2.4 The graph Hn ∗ S5 admits prime labeling if n 6≡ 1(mod11). Proof Let c0 be the centre of the Helm Hn and let c1 , c2 , · · · , cn be the rim vertices of the Hn . Let p1 , p2 , · · · , pn be the pendant vertices attached at c1 , c2 , · · · , cn respectively. Let pj0 , 1 ≤ j ≤ 5 be the pendent vertices attached at central vertex c0 and let pji and lij be the pendant vertices attached at ci and pi respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 5. The graph Hn ∗ S5 has 12n + 6 vertices and 13n + 5 edges. Define a labeling f : V → {1, 2, 3, · · · , 12n + 6} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 5.

For 1 ≤ i ≤ n,

f (ci ) = 12i − 1, f (pi ) = 12i + 5, f (p2i ) = 12i − 4, f (p3i ) = 12i − 3, f (p5i ) = 12i, f (li1 ) = 12i + 1, f (li3 ) = 12i + 3, f (li4 ) = 12i + 4,

f (p1i ) = 12i − 5, f (p4i ) = 12i − 2, f (li2 ) = 12i + 2, f (li5 ) = 12i + 6.

It remains to show that the labels of each pair of adjacent vertices are relatively prime. gcd f (c0 ), f (pj0 ) = gcd(1, j + 1) = 1 for 1 ≤ j ≤ 5. For 1 ≤ i ≤ n,

gcd f (c0 ), f (ci ) = gcd(1, 12i − 1) = 1, gcd f (ci ), f (ci+1 ) = gcd(12i − 1, 12(i + 1) − 1) = gcd(12i − 1, (12i − 1) + 12) = 1 as these two numbers are odd and their difference is 12 and both are not multiples of 3, 5, 7 or 11. gcd f (c1 ), f (cn ) = gcd(11, 12n − 1) = 1 since n 6≡ 1(mod11). gcd f (ci ), f (pi ) = gcd(12i − 1, 12i + 5) = 1 as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd f (ci ), f (p1i ) = gcd(12i − 1, 12i − 5) = 1 as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd f (ci ), f (p2i ) = gcd(12i − 1, 12i − 4) = 1 among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd f (ci ), f (p3i ) = gcd(12i − 1, 12i − 3) = 1 as these two numbers are consecutive odd integers. gcd f (ci ), f (p4i ) = gcd(12i − 1, 12i − 2) = 1. gcd f (ci ), f (p5i ) = gcd(12i − 1, 12i) = 1. gcd f (pi ), f (li1 ) = gcd(12i + 5, 12i + 1) = 1 as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd f (pi ), f (li2 ) = gcd(12i + 5, 12i + 2) = 1 among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd f (pi ), f (li3 ) = gcd(12i + 5, 12i + 3) = 1 as these two numbers are consecutive odd integers. gcd f (pi ), f (li4 ) = gcd(12i + 5, 12i + 4) = 1. gcd f (pi ), f (li5 ) = gcd(12i + 5, 12i + 6) = 1 as these two numbers are consecutive integers. Thus f is a prime labeling. Hence Hn ∗ S5 is a prime graph. 2


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Figure 2. Prime labeling of H4 ∗ S5 Theorem 2.6 The graph Gn ∗ S2 admits prime labeling if n 6≡ 1(mod7). Proof Let c0 be the centre of the Gear Gn and let c1 , c2 , · · · , cn be the rim vertices of Gn . Let p1 , p2 , · · · , pn be the subdividing vertices of Gn . Let p10 , p20 be the pendent vertices attached at c0 and let pji and lij be the pendant vertices attached at pi and ci respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 2 and the graph Gn ∗ S2 has 6n + 3 vertices and 7n + 2 edges. Define a labeling f : V → {1, 2, 3, · · · , 6n + 3} as follows: f (c0 ) = 1,

f (pj0 ) = j + 1,

for 1 ≤ j ≤ 2.

For 1 ≤ i ≤ n f (ci ) = 6i + 1, f (p2i ) = 6i,

f (pi ) = 6i − 1,

f (li1 ) = 6i + 2,

f (p1i ) = 6i − 2, f (li2 ) = 6i + 3.

It remains to show that the labels of each pair of adjacent vertices are relatively prime. For 1 ≤ i ≤ n,

gcd f (ci ), f (ci+1 ) = gcd(6i + 1, 6(i + 1) + 1)

= gcd(6i + 1, (6i + 1) + 6) = 1 as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd f (c1 ), f (cn ) = gcd(7, 6n + 1) = 1 since n 6≡ 1(mod7).

Similar to Theorem 2.1 all other gcd’s are one. Thus f is a prime labeling. Hence Gn ∗ S2 2 is a prime graph. Theorem 2.7 The graph Gn ∗ S3 admits prime labeling if n 6≡ 1(mod11). Proof Let c0 be the centre of the Gear Gn and let c1 , c2 , · · · , cn be the rim vertices of the Gn . Let p1 , p2 , · · · , pn be the subdividing vertices of Gn . Let p10 , p20 , p30 be the pendent vertices attached at central vertex c0 and let pji and lij be the pendant vertices attached at pi and ci respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 3.

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The graph Gn ∗ S3 has 8n + 4 vertices and 9n + 3 edges. Define a labeling f : V → {1, 2, 3, · · · , 8n + 4} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 3.

For 1 ≤ i ≤ n,

f (pi ) = 8i − 1,

f (p3i )

= 8i,

f (ci ) = 8i + 3, f (li1 )

= 8i + 1,

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f (p2i ) = 8i − 2,

f (li3 ) = 8i + 4.

= 8i + 2,


gcd f (ci ), f (ci+1 ) = gcd(8i + 3, 8(i + 1) + 3)

= gcd(8i + 3, (8i + 3) + 8) = 1 as these two numbers are odd and their difference is 8 and both are not multiples of 3, 5 or 7. gcd f (c1 ), f (cn ) = gcd(11, 8n + 3) = 1 since n 6≡ 1(mod11), 8n 6≡ 8(mod11), 8n + 3 6≡ 0(mod11), 8n + 3 is not multiple of 11. Similar to Theorem 2.2 all other gcd’s are one. Thus f is a prime labeling. Hence Gn ∗ S3 is a prime graph. 2 Illustration 2.8 10 9

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Figure 3. Prime labeling of G4 ∗ S3 Theorem 2.9 The graph Gn ∗ S5 admits prime labeling if n 6≡ 1(mod17). Proof Let c0 be the centre of the Gear Gn and let c1 , c2 , · · · , cn be the rim vertices of the Gn . Let p1 , p2 , · · · , pn be the subdividing vertices of Gn . Let pj0 , 1 ≤ j ≤ 5 be the pendent vertices attached at central vertex c0 and let pji and lij be the pendant vertices attached at pi and ci respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 5.

The graph Gn ∗ S5 has 12n + 6 vertices and 13n + 5 edges. Define a labeling f : V → {1, 2, 3, · · · , 12n + 6} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 5. For 1 ≤ i ≤ n,


f (pi ) = 12i − 1, f (p2i ) = 12i − 4, f (p5i ) = 12i, f (li3 ) = 12i + 3,

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f (ci ) = 12i + 5, f (p1i ) = 12i − 5, f (p3i ) = 12i − 3, f (p4i ) = 12i − 2, f (li1 ) = 12i + 1, f (li2 ) = 12i + 2, f (li4 ) = 12i + 4, f (li5 ) = 12i + 6.


gcd f (ci ), f (ci+1 ) = gcd(12i + 5, 12(i + 1) + 5) = gcd(12i + 5, (12i + 5) + 12) = 1 as these two numbers are odd and their difference is 12 and both are not multiples of 3, 5, 7 or 11. gcd f (c1 ), f (cn ) = gcd(17, 12n + 5) = 1, since n 6≡ 1(mod17). Similar to Theorem 2.3 all other gcd’s are one. Thus f is a prime labeling. Hence Gn ∗ S5 is a prime graph. 2

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Figure 4. Prime labeling of G4 ∗ S5 Theorem 2.11 The graph Hn ∗ S7 admits prime labeling if n 6≡ 1, 5, 8, 9(mod11). Proof Let c0 be the centre of the Helm Hn and let c1 , c2 , · · · , cn be the rim vertices of the Hn . Let p1 , p2 , · · · , pn be the pendant vertices attached at c1 , c2 , · · · , cn respectively. Let pj0 , 1 ≤ j ≤ 7 be the pendent vertices attached at central vertex c0 and let pji and lij be the pendant vertices attached at ci and pi respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 7. The graph Hn ∗ S7 has 16n + 8 vertices and 17n + 7 edges. Define a labeling f : V → {1, 2, 3, · · · , 16n + 8} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 7. For 1 ≤ i ≤ n, f (ci ) = 16i − 5, f (ci ) = 16i − 1, f (ci ) = 16i − 3, f (ci ) = 16i − 7,

for for for for

i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15), i ≡ 0, 2, 5, 8, 14(mod15), i ≡ 10(mod15) and i 6≡ 56(mod105), i ≡ 11(mod15) and i 6≡ 56(mod105),

26


f (pi ) = 16i + 3, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15), f (pi ) = 16i + 7, for i ≡ 0, 6, 7, 9, 12(mod15), f (pi ) = 16i + 5, for i ≡ 2, 3(mod15), f (p1i ) = 16i − 7, for i 6≡ 11(mod15), i 6≡ 56(mod105), f (p1i ) = 16i − 5, for i ≡ 11(mod15), i 6≡ 56(mod105), f (p2i ) = 16i − 6, f (p3i ) = 16i − 4, f (p4i ) = 16i − 3, for i 6≡ 10(mod15), i 6≡ 56(mod105), f (p4i ) = 16i − 5, for i ≡ 10(mod15), i ≡ 56(mod105), f (p5i ) = 16i − 2, f (p6i ) = 16i − 1, for i 6≡ 0, 2, 5, 8, 14(mod15), f (p6i ) = 16i − 5, for i ≡ 0, 2, 5, 8, 14(mod15), f (p7i ) = 16i, f (li1 ) = 16 + 1, f (li2 ) = 16i + 2, f (li3 ) = 16i + 4, f (li4 ) = 16i + 5, for i 6≡ 2, 3(mod15), f (li4 ) = 16i + 3, for i ≡ 2, 3(mod15), f (li5 ) = 16i + 6, f (li6 ) = 16i + 7, for i 6≡ 0, 6, 7, 9, 12(mod15), f (li6 ) = 16i + 3, for i ≡ 0, 6, 7, 9, 12(mod15), f (li7 ) = 16i + 8. It remains to show that the labels of each pair of adjacent vertices are relatively prime. For 1 ≤ i ≤ n, gcd(f (c0 ), f (ci )) = gcd(1, 16i − 5) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15). gcd(f (c0 ), f (ci )) = gcd(1, 16i − 1) = 1, for i ≡ 0, 2, 5, 8, 14(mod15). gcd(f (c0 ), f (ci )) = gcd(1, 16i − 3) = 1 since i ≡ 10(mod15) and i ≡ 56(mod105). gcd(f (c0 ), f (ci )) = gcd(1, 16i − 7) = 1 since i ≡ 11(mod15) and i 6≡ 56(mod105). gcd(f (c0 ), f (pj0 )) = gcd(1, j + 1) = 1, for 1 ≤ j ≤ 7. gcd(f (ci ), f (ci+1 )) = gcd(16i − 5, 16(i + 1) − 5) = gcd(16i − 5, (16i − 5) + 16) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15) as these two numbers are odd and their difference is 16 and both are not multiples of 3,5,7,11 or 13. gcd(f (ci ), f (ci+1 )) = gcd(16i − 1, 16(i + 1) − 5) = gcd(16i − 1, (16i − 1) + 12) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) as these two numbers are odd and their difference is 12 and both are not multiples of 3,5,7, or 11. gcd(f (ci ), f (ci+1 )) = gcd(16i − 1, 16(i + 1) − 1) = gcd(16i − 1, (16i − 1) + 16) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) as these two numbers are odd and their difference is 16 and both are not multiples of 3,5,7,11, or 13. gcd(f (ci ), f (ci+1 )) = gcd(16i − 5, 16(i + 1) − 1) = gcd(16i − 5, (16i − 5) + 20) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15) as these two numbers are odd and their difference is 20 and both are not multiples of 3,5,7,11,13 or 17. gcd(f (ci ), f (ci+1 )) = gcd(16i − 3, 16(i + 1) − 7)


27

= gcd(16i − 3, (16i − 3) + 12) = 1, for i ≡ 10(mod15) as these two numbers are odd and their difference is 12 and both are not multiples of 3,5,7, or 11. gcd(f (ci ), f (ci+1 )) = gcd(16i − 3, 16(i + 1) − 5) = gcd(16i − 3, (16i − 3) + 14) = 1, for i ≡ 56(mod105) as these two numbers are odd and their difference is 14 and both are not multiples of 3,5,7,11, or 13. gcd(f (ci ), f (ci+1 )) = gcd(16i − 7, 16(i + 1) − 5) = gcd(16i−7, (16i−7)+18) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105) as these two numbers are odd and their difference is 18 and both are not multiples of 3,5,7,11,13 or 17. gcd(f (c1 ), f (cn )) = gcd(11, 16n − 5) = 1, gcd(f (c1 ), f (cn )) = gcd(11, 16n − 1) = 1, gcd(f (c1 ), f (cn )) = gcd(11, 16n − 3) = 1, gcd(f (c1 ), f (cn )) = gcd(11, 16n − 7) = 1 since n 6≡ 1, 5, 8, 9(mod11). gcd(f (ci ), f (p2i )) = gcd(16i − 5, 16i − 6) = 1, since i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15). gcd(f (ci ), f (p2i )) = gcd(16i − 7, 16i − 6) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105). gcd(f (ci ), f (p3i )) = gcd(16i − 5, 16i − 4) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15). gcd(f (ci ), f (p3i )) = gcd(16i − 3, 16i − 4) = 1, for i ≡ 10(mod15) and i ≡ 56(mod105) as these two numbers are consecutive integers. gcd(f (ci ), f (p2i )) = gcd(16i − 1, 16i − 6) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) among these two numbers one is odd and other is even and their difference is 5 and they are not multiples of 5. gcd(f (ci ), f (p2i )) = gcd(16i − 3, 16i − 6) = 1, for i ≡ 10(mod15) and i ≡ 56(mod105). gcd(f (ci ), f (p3i )) = gcd(16i − 1, 16i − 4) = 1, for i ≡ 0, 2, 5, 8, 14(mod15). gcd(f (ci ), f (p3i )) = gcd(16i − 7, 16i − 4) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd(f (ci ), f (p7i )) = gcd(16i − 5, 16i) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15) among these two numbers one is odd and other is even and their difference is 5 and they are not multiples of 5. gcd(f (ci ), f (p7i )) = gcd(16i − 1, 16i) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) as these two numbers are consecutive integers. gcd(f (ci ), f (p7i )) = gcd(16i − 3, 16i) = 1, for i ≡ 10(mod15) and i ≡ 56(mod105) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd(f (ci ), f (p7i )) = gcd(16i − 7, 16i) = 1, for i ≡ 11(mod15) and i ≡ 56(mod105) among these two numbers one is odd and other is even and their difference is 7 and they are not multiples of 7. gcd(f (ci ), f (p1i )) = gcd(16i − 5, 16i − 7) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15), i 6≡ 11(mod15). gcd(f (ci ), f (p1i )) = gcd(16i − 7, 16i − 5) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105) as these two numbers are consecutive odd integers. gcd(f (ci ), f (p1i )) = gcd(16i − 1, 16i − 7) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) and i 6≡ 11(mod15) as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5.

28


gcd(f (ci ), f (p1i )) = gcd(16i − 3, 16i − 7) = 1 since i ≡ 10(mod15) and i ≡ 56(mod105) and i 6≡ 11(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. For i 6≡ 10(mod15) gcd(f (ci ), f (p4i )) = gcd(16i − 5, 16i − 3) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15). gcd(f (ci ), f (p4i )) = gcd(16i − 1, 16i − 3) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) as these two numbers are consecutive odd integers. gcd(f (ci ), f (p4i )) = gcd(16i − 7, 16i − 3) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105) as these two numbers are odd and their difference is 4 and both are not multiples of 3. For i ≡ 10(mod15) and i ≡ 56(mod105), gcd(f (ci ), f (p4i )) = gcd(16i − 3, 16i − 5) = 1 as these two numbers are consecutive odd integers. gcd(f (ci ), f (p5i )) = gcd(16i − 5, 16i − 2) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd(f (ci ), f (p5i )) = gcd(16i−1, 16i−2) = 1, for i ≡ 0, 2, 5, 8, 14(mod15). gcd(f (ci ), f (p5i )) = gcd(16i − 3, 16i − 2) = 1, for i ≡ 10(mod15) and i ≡ 56(mod105) as these two numbers are consecutive integers. gcd(f (ci ), f (p5i )) = gcd(16i − 7, 16i − 2) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105) among these two numbers one is odd and other is even and their difference is 5 and they are not multiples of 5. gcd(f (ci ), f (p6i )) = gcd(16i − 5, 16i − 1) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd(f (ci ), f (p6i )) = gcd?(16i − 3, 16i − 1) = 1, for i ≡ 10(mod15) and i ≡ 56(mod105) as these two numbers are consecutive odd integers. gcd(f (ci ), f (p6i )) = gcd(16i − 7, 16i − 1) = 1, for i ≡ 11(mod15) and i 6≡ 56(mod105) as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd(f (ci ), f (p6i )) = gcd(16i − 1, 16i − 5) = 1, for i ≡ 0, 2, 5, 8, 14(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd(f (pi ), f (li1 )) = gcd(16i + 3, 16i + 1) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15) as these two numbers are consecutive odd integers. gcd(f (pi ), f (li1 )) = gcd(16i + 7, 16i + 1) = 1, for i ≡ 0, 6, 7, 9, 12(mod15) as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd(f (pi ), f (li1 )) = gcd(16i + 5, 16i + 1) = 1, for i ≡ 2, 3(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd(f (pi ), f (li2 )) = gcd(16i + 3, 16i + 2) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15) as these two numbers are consecutive integers. gcd(f (pi ), f (li2 )) = gcd(16i + 7, 16i + 2) = 1, for i ≡ 0, 6, 7, 9, 12(mod15) among these two numbers one is odd and other is even and their difference is 5 and they are not multiples of 5. gcd(f (pi ), f (li2 )) = gcd(16i + 5, 16i + 2) = 1, for i ≡ 2, 3(mod15) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3.


29

gcd(f (pi ), f (li3 )) = gcd(16i + 3, 16i + 4) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15),

gcd(f (pi ), f (li3 )) = gcd(16i + 5, 16i + 4) = 1, for i ≡ 2, 3(mod15),

gcd(f (pi ), f (li3 )) = gcd(16i + 7, 16i + 4) = 1, for i ≡ 0, 6, 7, 9, 12(mod15) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. For i 6≡ 2, 3(mod15), gcd(f (pi ), f (li4 )) = gcd(16i + 3, 16i + 5) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15).

gcd(f (pi ), f (li4 )) = gcd(16i + 7, 16i + 5) = 1, for i ≡ 0, 6, 7, 9, 12(mod15).

gcd(f (pi ), f (li4 )) = gcd(16i + 5, 16i + 3) = 1, for i ≡ 2, 3(mod15) as these two numbers are consecutive odd integers. gcd(f (pi ), f (li5 )) = gcd(16i + 3, 16i + 6) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd(f (pi ), f (li5 )) = gcd(16i + 7, 16i + 6) = 1, for i ≡ 0, 6, 7, 9, 12(mod15).

gcd(f (pi ), f (li5 )) = gcd(16i + 5, 16i + 6) = 1, for i ≡ 2, 3(mod15) as these two numbers are consecutive integers. gcd(f (pi ), f (li6 )) = gcd(16i + 3, 16i + 7) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd(f (pi ), f (li6 )) = gcd(16i + 5, 16i + 7) = 1, for i ≡ 2, 3(mod15) as these two numbers are consecutive odd integers. gcd(f (pi ), f (li6 )) = gcd(16i + 7, 16i + 3) = 1, for i ≡ 0, 6, 7, 9, 12(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd(f (pi ), f (li7 )) = gcd(16i + 3, 16i + 8) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15) among these two numbers one is odd and other is even and their difference is 5 and they are not multiples of 5. gcd(f (pi ), f (li7 )) = gcd(16i + 7, 16i + 8) = 1, for i ≡ 0, 6, 7, 9, 12(mod15) as these two numbers are consecutive integers. gcd(f (pi ), f (li7 )) = gcd(16i + 5, 16i + 8) = 1, for i ≡ 2, 3(mod15) among these two numbers one is odd and other is even and their difference is 3 and they are not multiples of 3. gcd(f (ci ), f (pi )) = gcd(16i − 5, 16i + 3)

= gcd(16i − 5, (16i − 5) + 8) = 1, for i ≡ 1, 4, 13(mod15) as these two numbers are odd and their difference is 8 and both are not multiples of 3, 5 or 7. gcd(f (ci ), f (pi )) = gcd(16i − 1, 16i + 5)

= gcd(16i − 1, (16i − 1) + 6) = 1, for i ≡ 2(mod15) as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd(f (ci ), f (pi )) = gcd(16i − 5, 16i + 5)

= gcd(16i − 5, (16i − 5) + 10) = 1, for i ≡ 3(mod15) as these two numbers are odd and their difference is 10 and both are not multiples of 3, 5 or 7. gcd(f (ci ), f (pi )) = gcd(16i − 1, 16i + 3)

= gcd(16i − 1, (16i − 1) + 4) = 1, for i ≡ 5, 8, 14(mod15) as these two numbers are odd and their difference is 4 and both are not multiples of 3. gcd(f (ci ), f (pi )) = gcd(16i − 5, 16i + 7)

30


= gcd(16i − 5, (16i − 5) + 12) = 1, for i ≡ 6, 7, 9, 12(mod15) as these two numbers are odd and their difference is 12 and both are not multiples of 3,5,7 or 11. gcd(f (ci ), f (pi )) = gcd(16i − 3, 16i + 3) = gcd(16i − 3, (16i − 3) + 6) = 1, for i ≡ 10(mod15) as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. gcd(f (ci ), f (pi )) = gcd(16i − 7, 16i + 3) = gcd(16i − 7, (16i − 7) + 10) = 1, for i ≡ 11(mod15) as these two numbers are odd and their difference is 10 and both are not multiples of 3, 5 or 7. gcd(f (ci ), f (pi )) = gcd(16i − 1, 16i + 7) = gcd(16i − 1, (16i − 1) + 8) = 1, for i ≡ 0(mod15) as these two numbers are odd and their difference is 8 and both are not multiples of 3, 5 or 7. Thus f is a prime labeling. Hence Hn ∗ S7 is a prime graph.

2

Illustration 2.12 20 18 17

21

22 23 24

l1¹ 13 1914 12 15 10 16 9

p1¹ 11 72 71 70 69 68 66 65

64 63 62 67

c

1

25 2

59

8

61 60 58

co

3 1

7 6

27

26 28 29 35

4 5

57

30 31 32

33 34 36 37 38 39 40

43

48 47 46

41 42 44 45 51

56 55 54

49 50 52 53

Figure 5. Prime labeling of H4 ∗ S7 Theorem 2.13 The graph Gn ∗ S7 admits prime labeling if n 6≡ 1, 8(mod19) and n + 4 6≡ 0(mod19). Proof Let c0 be the centre of the Gear Gn and let c1 , c2 , · · · , cn be the rim vertices of the Hn . Let p1 , p2 , · · · , pn be the pendant vertices attached at c1 , c2 , · · · , cn respectively. Let pj0 , 1 ≤ j ≤ 7 be the pendent vertices attached at central vertex c0 and let pji and lij be the pendant vertices attached at ci and pi respectively for 1 ≤ i ≤ n and 1 ≤ j ≤ 7. The graph Gn ∗ S7 has 16n + 8 vertices and 17n + 7 edges. Define a labeling f : V → {1, 2, 3, · · · , 16n + 8} as follows: f (c0 ) = 1, f (pj0 ) = j + 1, for 1 ≤ j ≤ 7. For 1 ≤ i ≤ n, f (pi ) = 16i − 5, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15), f (pi ) = 16i − 1, for i ≡ 0, 2, 5, 8, 14(mod15) f (pi ) = 16i − 3, for i ≡ 10(mod15) and i ≡ 56(mod105),


31

f (pi ) = 16i − 7, for i ≡ 11(mod15) and i 6≡ 56(mod105), f (ci ) = 16i + 3, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15), f (ci ) = 16i + 7, for i ≡ 0, 6, 7, 9, 12(mod15), f (ci ) = 16i + 5, for i ≡ 2, 3(mod15), i 6≡ 78(mod105), f (p1i ) = 16i + 1, for i ≡ 78(mod105), f (p1i ) = 16i − 7, for i ≡ 11(mod15), i ≡ 56(mod105), f (p2i ) = 16i − 6, f (p3i ) = 16i − 4, f (p4i ) = 16i − 3, for i ≡ 10(mod15), i 6≡ 56(mod105), f (p5i ) = 16i − 2, f (p6i ) = 16i − 1, for i 6≡ 0, 2, 5, 8, 14(mod15), f (p6i ) = 16i − 5, for i ≡ 0, 2, 5, 8, 14(mod15), f (p7i ) = 16i, f (li1 ) = 16 + 1, f (li2 ) = 16i + 2, f (li3 ) = 16i + 4, f (li4 ) = 16i + 5, for i 6≡ 2, 3(mod15), f (li4 ) = 16i + 3, for i ≡ 2, 3(mod15), f (li5 ) = 16i + 6, f (li6 ) = 16i + 7, for i 6≡ 0, 6, 7, 9, 12(mod15), f (li6 ) = 16i + 3, for i ≡ 0, 6, 7, 9, 12(mod15), f (li7 ) = 16i + 8. It remains to show that the labels of each pair of adjacent vertices are relatively prime. For 1 ≤ i ≤ n, gcd(f (c0 ), f (pi )) = gcd(1, 16i − 5) = 1, for i ≡ 1, 3, 4, 6, 7, 9, 12, 13(mod15). gcd(f (c0 ), f (pi )) = gcd(1, 16i − 1) = 1, for i ≡ 0, 2, 5, 8, 14(mod15). gcd(f (c0 ), f (pi )) = gcd(1, 16i − 3) = 1, for i ≡ 10(mod15) and i ≡ 56(mod105). gcd(f (c0 ), f (pi )) = gcd(1, 16i − 7) = 1, for i ≡ 11(mod15) and i ≡ 56(mod105). gcd(f (c0 ), f (pj1 )) = gcd(1, j + 1) = 1, for 1 ≤ j ≤ 7. gcd(f (ci ), f (c( i + 1))) = gcd(16i + 3, 16(i + 1) + 3) = gcd(16i+3, (16i+3)+16) = 1, for i ≡ 1, 4, 5, 8, 10, 11, 13, 14(mod15) as these two numbers are odd and their difference is 16 and both are not multiples of 3,5,7,11 or 13. gcd(f (ci ), f (c( i + 1))) = gcd(16i + 3, 16(i + 1) + 5) = gcd(16i + 3, (16i + 3) + 18) = 1, for i ≡ 1, 2(mod15) as these two numbers are odd and their difference is 18 and both are not multiples of 3,5,7, 11,13 or 17. gcd(f (ci ), f (c( i + 1))) = gcd(16i + 5, 16(i + 1) + 5) = gcd(16i+5, (16i+5)+16) = 1, for i ≡ 2, 3(mod15), i 6≡ 78(mod105) as these two numbers are odd and their difference is 16 and both are not multiples of 3,5,7,11, or 13. gcd(f (ci ), f (ci+1 )) = gcd(16i + 5, 16(i + 1) + 3) = gcd(16i + 5, (16i + 5) + 14) = 1, for i ≡ 3, 4(mod15) as these two numbers are odd and their difference is 14 and both are not multiples of 3,5,7,11 or 13. gcd(f (ci ), f (ci+1 )) = gcd(16i + 3, 16(i + 1) + 7)

32


= gcd(16i + 3, (16i + 3) + 20) = 1, for i ≡ 0, 5, 6, 8, 9, 11, 12, 14(mod15) as these two numbers are odd and their difference is 20 and both are not multiples of 3,5,7, 11, 13 17 or 19. gcd(f (ci ), f (ci+1 )) = gcd(16i + 7, 16(i + 1) + 7) = gcd(16i + 7, (16i + 7) + 16) = 1, for i ≡ 6, 7(mod15) as these two numbers are odd and their difference is 16 and both are not multiples of 3,5,7,11 or 13. gcd(f (ci ), f (ci+1 )) = gcd(16i + 7, 16(i + 1) + 3) = gcd(16i + 7, (16i + 7) + 12) = 1, for i ≡ 7, 8, 9, 10, 12, 13(mod15) as these two numbers are odd and their difference is 12 and both are not multiples of 3,5,7 or 11. gcd(f (ci ), f (c( i + 1))) = gcd(16i + 1, 16(i + 1) + 3) = gcd(16i + 1, (16i + 1) + 18) = 1, for i ≡ 78(mod105) as these two numbers are odd and their difference is 18 and both are not multiples of 3,5,7,11,13 or 17. gcd(f (c( i − 1)), f (ci )) = gcd(16(i − 1) + 5, 16i + 1)

= gcd(16i + 1) − 12, 16i + 1) = 1, for i ≡ 78(mod105) as these two numbers are odd and their difference is 12 and both are not multiples of 3,5,7 or 11. gcd(f (c1 ), f (cn )) = gcd(19, 16n + 3) = 1, gcd(f (c1 ), f (cn )) = gcd(19, 16n + 7) = 1, gcd(f (c1 ), f (cn )) = gcd(19, 16n + 5) = 1. Since n 6≡ 1, 8(mod19) and n + 4 6≡ 0(mod19), gcd(f (pi ), f (ci )) = gcd(16i − 5, 16i + 1)

= gcd(16i − 5, (16i − 5) + 6) = 1, for i ≡ 78(mod105) as these two numbers are odd and their difference is 6 and both are not multiples of 3 or 5. Similar to Theorem 2.7 all other gcd’s are one. Thus f is a prime labeling. Hence Gn ∗ S7 is a prime graph. 2 Illustration 2.14 20 18 17

21 22 23 24

l1¹ 19

c1

13 12 1415 10 16 9 72 71 70 69 68 66 65

p1 ¹ 11 25 26 64 2 63 28 3 62 29 o 1 59 67 4 27 30 35 61 31 5 60 8 32 58 7 6 57 43 41 48 42 47 44 46 45 51

c

56 55 54

53

33 34 36 37 38 39 40

49 50 52

Figure 6. Prime labeling of G4 ∗ S7


33

§3. Conclusion Labeled graph is the topic of current due to its diversified application. We investigate eight new results on prime labeling of graphs. It is an effort to relate the prime labeling and some graph operations. This approach is novel as it provides prime labeling for the larger graph resulted due to certain graph operations on a given graph. Analogous work can be carried out for other families and in the context of different types of graph labeling techniques.

References [1] Bondy.J.A and Murthy. U.S.R, Graph Theory and its Application, (North - Holland). Newyork (1976). [2] J.Gallian, A dynamic survey of graph labeling, Elactron.J. Comb. 17(2014). [3] M. Tavakoli, F. Rahbarnia and A.R. Ashrafi, Studying the corona product of graphs under some graph invariants, —it Transactions on Combinatorics, Vol. 3, No. 3 (2014), 43-49. [4] Rose - Hulman, Prime vertex labelings of unicyclic graphs, Undergraduate Mathematical Journal, Volume 16, No.1,2015. [5] S K Vaidya, K K Kanani, Prime labeling for some cycle related graphs, Journal of Mathematics Research, Vol. 2, No. 2, May 2010. [6] S.K. Vaidya, U.M. Prajapati, Some results on prime and k-prime labeling, Journal of Mathematics Research, Vol. 3, No. 1; February 2011. [7] C. S K Vaidya, U M Prajapati, Prime labeling in the context of duplication of graph elements, International Journal of Mathematics and Soft Computing, Vol.3, No.1(2013),1320. [8] Tout. A, Dabboucy. A.N and Howalla. K, Prime labeling of graphs, Nat. Acad. Sci. Letters 11 (1982)365-368, Combinatorics and Application, Vol.1, Alari. J (Wiley. N.Y 1991) 299-359.


Maximum Number of Holes in Square of Cycles Srinivasa Rao Kola, Balakrishna Gudla and Niranjan P K Department of Mathematical and Computational Sciences National Institute of Technology Karnataka, Surathkal, India E-mail: [email protected], [email protected], [email protected]

Abstract: An L(2, 1)-coloring of a simple connected graph G is an assignment f of nonnegative integers to the vertices of G such that |f (u) − f (v)| > 2 for all edges uv in G and |f (u) − f (v)| > 1 if u and v are at distance 2 in G. The span of an L(2, 1)-coloring is the maximum color assigned by it. The minimum of spans over all L(2, 1)-colorings of G is called λ-number of G, denoted by λ(G). An L(2, 1)-coloring with span λ(G) is called a span coloring of G. A color h ∈ {0, 1, 2, . . . , k} is said to be a hole in an L(2, 1)-coloring f of G with span k if no vertex of G receives h. A no-hole coloring is an L(2, 1)-coloring without holes. An L(2, 1)-coloring of G is irreducible if none of the colors can be decreased to get another L(2, 1)-coloring of G. An irreducible no-hole coloring (or inh-coloring) of G is an L(2, 1)-coloring of G which is both irreducible and no-hole. The inh-span of a graph G, denoted by λinh (G) is the smallest number k such that there is an inh-coloring of G with span k. The maximum number of holes in G, denoted by Hλ (G) is the maximum number of holes over all irreducible span colorings of G. The λ-number of Cn2 , the square of cycle Cn , is known. In this paper, we determine the maximum number of holes in Cn2 , n > 7. Also, we show that Cn2 , n > 7 is inh-colorable, and λinh (Cn2 ) = λ(Cn2 ), n > 7 except n = 10, 12, 18.

Key Words: L (2, 1)-coloring; span of a graph; irreducible coloring; no-hole coloring; inhcoloring.

AMS(2010): 05C78, 05C15 §1. Introduction The frequency assignment problem is the problem of assigning frequencies to transmitters in such a way that the communications does not interfere. Griggs and Yeh introduced L (2, 1)coloring of graphs as a variation of frequency assignment problem. Here transmitters are represented as vertices and two vertices are joined by an edge if the corresponding transmitters are very close. Let Z∗ be the set of all non-negative integers. An L (2, 1)-coloring of a simple connected graph G is a mapping f : V (G) → Z∗ such that |f (u) − f (v)| > 2 if d(u, v) = 1 and |f (u) − f (v)| > 1 if d(u, v) = 2, where d(u, v) represent the distance between the vertices 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 17, 2018, Accepted June 24, 2018, Edited by S. Monikandan.

Maximum Number of Holes in Square of Cycles

35

u and v. The span of f , denoted by spanf is max{f (v) : v ∈ V (G)}. The span of G, denoted by λ(G) is min{span f : f is an L (2, 1) -coloring of G}. Griggs and Yeh [3] have found the span of paths and cycles. Also, they have proved that λ(T ) = ∆ + 1 or ∆ + 2 for any tree T with maximum degree ∆. Further, they have shown that λ(G) 6 ∆2 + 2∆ for any graph G with maximum degree ∆ and conjectured that λ(G) 6 ∆2 . Chang and Kuo [1] have improved the upper bound given by Griggs and Yeh [3] as λ(G) 6 ∆2 + ∆. Gonclaves [2] further improved the upper bound as ∆2 + ∆ − 2. Havet et al. [4] proved that the Griggs and Yeh’s conjecture is true for large values of ∆ (∆ > 1069 ). Zhang [10] have determined λ(Cn2 ) as below. Theorem 1.1([10]) For n > 4,    6 if n = 4 or n ≡ 0 (mod 7), 2 λ(Cn ) = 8 if n ∈ {5, 9, 10, 11, 17},    7 otherwise. Because frequencies are typically purchased in a block, one may want to use all available frequencies in that block. Fishburn and Roberts [6] have introduced the concept of no-hole coloring. For a graph G and an L (2, 1)-coloring f of it with span k, an integer h is called a hole in f if h ∈ {0, 1, 2, . . . , k} and there is no vertex in G such that f (v) = h. An L (2, 1)-coloring f : V (G) → {0, 1, 2, . . . , spanf } is said to be a no-hole coloring of G if it is onto. A graph G is no-hole colorable if there exists a no-hole coloring of G. Fishburn and Roberts [5] have shown that a graph G of order n is no-hole colorable if and only if n > λ(G)+ 1. Fishburn and Roberts [6] have defined a full coloring of a graph as a no-hole coloring with span λ(G). A graph is full colorable if it has a full coloring. An L (2, 1)-coloring f of a graph G is said to be irreducible if there is no L (2, 1)-coloring g of G such that g(v) 6 f (v) for all vertices v in G and g(u) < f (u) for at least one vertex u in G. Fishburn et al. [7] have introduced the concept of irreducibility of an L (2, 1)-coloring. Irreducibility will assure that we are not wasting frequencies and that each transmitter is using the lowest possible frequency allowable. A graph G is inh-colorable if there exists an irreducible no-hole coloring of G. The inh-span of a graph G, denoted by λinh (G) is the minimum number k such that there is an inh-coloring of G with span k. Fishburn et al. [7] have proved that paths Pn (n > 5), cycles Cn (n > 5) and non-star trees are inh-colorable. For any non-star tree T , Laskar et al. [9] have proved that is λinh (T ) = λ(T ). The problem of determining minimum number of different frequencies required for interference free communications in a given network is same as determining maximum number of holes in a span coloring. The maximum number of holes over all irreducible span colorings of G is denoted by Hλ (G). Laskar and Eyabi [8] have determined the exact values of maximum number of holes for paths, cycles, stars and complete multipartite graphs. In this paper, we find the maximum number of holes in Cn2 , n > 7. Also, we show that Cn2 , n > 7 is inh-colorable, and λinh (Cn2 ) = λ(Cn2 ), n > 7 except n = 10, 12, 18.

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Srinivasa Rao Kola, Balakrishna Gudla and Niranjan P K

§2. Results Let Cn2 be the square of a cycle Cn : v0 v1 v2 . . . vn−1 v0 . Throughout this paper, the subscripts are taken over modulo n. We use the following lemmas given by Zhang [10]. Lemma 2.1([10]) The following two statements will not occur for any L (2, 1)-coloring of Cn2 . (i) There are three consecutive colors or two pairs of consecutive colors on four consecutive vertices in Cn . (ii) There are five consecutive colors on five consecutive vertices in Cn . Lemma 2.2([10]) If f is any L (2, 1)-coloring of Cn2 with span 6 such that f (vi ) = f (vj ), then |i − j| > 7. The two lemmas below are used in the sequel. Lemma 2.3 In any L (2, 1)-coloring of Cn2 with span 7, any seven consecutive vertices in Cn receive at least six different colors. Proof Let f be an L (2, 1)-coloring of Cn2 with span 7. Suppose vi , vi+1 , vi+2 , vi+3 , vi+4 , vi+5 and vi+6 are seven consecutive vertices in Cn and they receive only five different colors. So, f (vi ) = f (vi+5 ) and f (vi+1 ) = f (vi+6 ). Since d(vi+7 , vi+2 ) = 3 and vi+2 is adjacent to vi , vi+1 , vi+3 and vi+4 , the color of vi+2 can be used to color vi+7 . Similarly, the colors of vi+3 and vi+4 can be used to color vi+8 and vi+9 respectively. Hence the colors of vi , vi+1 , vi+2 , vi+3 and vi+4 forms a coloring of C52 which is not possible as λ(C52 ) is 8. 2 Lemma 2.4 If f is a two hole coloring of Cn2 with span 7 and f (vi ) = f (vj ), then |i − j| > 6. Proof From the definition of L (2, 1)-coloring, it is clear that|i − j| > 5. Suppose that f (vi ) = f (vi+5 ) = c1 . Let c2 , c3 , c4 and c5 be the colors of vertices vi+1 , vi+2 , vi+3 and vi+4 respectively. By Lemma 2.1, it is easy to see that c1 6= 1 and c1 6= 6. Note that the colors c1 , c1 −1, c1 +1 cannot be used to vi+1 , vi+2 , vi+3 and vi+4 . The remaining colors contains either at least four consecutive colors or at least three consecutive colors and a pair of consecutive colors. Among these colors four different colors are used for vi+1 , vi+2 , vi+3 and vi+4 in which at least two of them are consecutive. By the adjacency of vi+1 , vi+2 , vi+3 and vi+4 in Cn2 , the consecutive colors must be the colors of vi+1 and vi+4 . By Lemma 2.3, a new color is required to color the vertex vi+6 , say c. Since vi+6 and vi+5 are adjacent, c is different from c1 , c1 + 1 and c1 + 1. Since f is a two hole coloring and c1 , c2 , c3 , c4 , c5 and c are six different colors, f assigns c1 , c2 , c3 , c4 , c5 and c only. Also, by Lemma 2.3, f (vi−1 ) is different from c1 , c2 , c3 , c4 , c5 . Therefore f (vi−1 ) = c. If c = 0 or 7, then we get contradiction to Lemma 2.1 as c2 , c3 , c4 and c5 contains at least three consecutive colors or two pairs of consecutive colors. If c is consecutive to c1 − 1 or c1 + 1, then we get a contradiction to Lemma 2.1 as c2 , c3 , c4 and c5 contains at least three consecutive colors or two pairs of consecutive colors. Therefore c must be consecutive to both c3 and c4 . Now the vertex vi+7 must receive c2 and hence it is not possible to color vi+8 2 with any of the colors c1 , c2 , c3 , c4 , c5 and c.


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The following theorem is the main result of the paper. Theorem 2.5 For n > 7,   0     2 Hλ (Cn2 ) =   4     1

if n = 7m or n ∈ {8, 9, 15, 16, 22, 23, 29}, if n = 6m, n 6= 7p or n ∈ {11, 17}, if n = 10, otherwise.

Proof The proof is divided into 4 cases following. Case 1. n = 7m or n ∈ {8, 9, 15, 16, 22, 23, 29}. Subcase 1.1 n = 7m. Let Cn : v0 v1 v2 . . . vn−1 v0 . From Theorem 1.1, λ(Cn2 ) = 6. Now by Lemma 2.2, any seven consecutive vertices of Cn receives seven different colors. Therefore any span coloring of Cn2 cannot have a hole. Subcase 1.2 n ∈ {8, 9, 15, 16, 22, 23, 29}. Since λ(C82 ) = 7, λ(C92 ) = 8 and in any L (2, 1)-coloring of C82 and C92 are diameter two graphs, we have Hλ (C82 ) = Hλ (C92 ) = 0. Since λ(Cn2 ) = 7, from Lemma 2.3, it is clear that 2 2 Hλ (Cn2 ) 6 2. Clearly, any color in C15 and C16 can be used at the most two times. Since 2 2 2 2 n > 2(7), we need at least eight colors to color C15 and C16 and hence Hλ (C15 ) = Hλ (C16 ) = 0. 2 Similarly we can prove that Hλ (Cn ) = 0 for other values of n. Case 2. n = 6m and n 6= 7p or n ∈ {11, 17}. Subcase 2.1 n = 6m and n 6= 7p. Since λ(Cn2 ) = 7, by Lemma 2.3, Hλ (Cn2 ) 6 2. Note that the sequence h0 4 7 1 3 6i can be used in a cyclic order to obtain a two hole irreducible span coloring of Cn2 . Subcase 2.2 n = 11 or 17. 2 2 2 For n = 11, λ(C11 ) = 8. Since no color can be appear three times in C11 , Hλ (C11 ) 6 3. 2 Suppose f is a three hole span coloring of C11 . Then out of six colors used, only one color is not repeated, say c. Without loss of generality, we assume that f (v0 ) = c. Then the colors of v1 , v2 , v3 , v4 and v5 can be used only to the vertices v6 , v7 , v8 , v9 and v10 respectively. So, hf (v1 ) f (v2 ) f (v3 ) f (v4 ) f (v5 )i is a C52 coloring. Also, any L (2, 1)-coloring of C52 with span 8 includes only the colors {0, 2, 4, 6, 8}. Since the color c is adjacent to four distinct colors f (v1 ), f (v2 ), f (v9 ) and f (v10 ), c must differ by two from four colors among {0, 2, 4, 6, 8}, which 2 is not possible. Therefore Hλ (C11 ) 6 2. The sequence h2 0 4 8 6 2 0 5 3 8 6i is an irreducible 2 two hole span coloring of C11 . 2 2 2 For n = 17, λ(C17 ) = 8. Since no color can be used four times in C17 , Hλ (C17 ) 6 3. 2 Suppose f is a three hole span coloring of C17 . Then f uses one color twice, say c. Let vi and 2 vj be the vertices of C17 that receive c. Then at least one of the paths vi+1 vi+2 . . . vj−2 vj−1 and

38


vj+1 vj+2 . . . vi−2 vi−1 contains at least seven vertices. Without loss of generality, we assume that the path vi+1 vi+2 . . . vj−2 vj−1 contains at least seven vertices. Since vi+1 , vi+2 , vi+3 , vi+4 , vi+5 , vi+6 , vi+7 receive only five different colors, f (vi+1 ) = f (vi+6 ) and f (vi+2 ) = f (vi+7 ). As in proof of Lemma 2.3, we can show that f (vi+1 ), f (vi+2 ), f (vi+3 ), f (vi+4 ),f (vi+5 ) is a coloring of C52 . 2 2 So, we get a contradiction to the color c similar to the case of C11 . Therefore Hλ (C17 ) 6 2. A 2 two hole irreducible span coloring of C17 is given in Figure 1. 0

8

2 4

6

6

1

8

4

0 2

7 4

2 0

8

6

2 Figure 1 An irreducible span coloring of C17 with 3 and 5 as holes.

Case 3. n = 10. 2 It is clear that Hλ (Cn2 ) 6 4. A four hole irreducible span coloring of C10 is a cyclic repetition of sequence h0 2 4 6 8i.

Case 4. n 6= 6m, n 6= 7p and n ∈ / {8, 9, 10, 11, 15, 16, 17, 22, 23, 29}. From Lemma 2.3, it is easy to see that three hole span coloring is not possible. Suppose f is a two hole span coloring of Cn2 . By Lemma 2.4, clearly f must be a cyclic repetition of six colors. Since n 6= 6m, it is not possible to color Cn2 . Therefore Hλ (Cn2 ) 6 1. By division algorithm, for n > 30, n = 6m + r, 0 < r < 6 and m > 5. Then n = 6(m − r) + 6r + r = 6(m − r) + 7r. Also, for n 6 30 we have n ∈ {13, 19, 20, 25, 26, 27} and hence it is easy to see that n can be written as 6(m − r) + 7r. By using the sequence h1 3 7 0 4 6i (m − r) times followed by the 2 sequence h1 3 5 7 0 4 6i r times we get an irreducible span coloring of Cn2 with 2 as hole. The theorems below show that Cn2 , n > 7 is inh-colorable. Theorem 2.6 If n > 7 and n 6= 10, 12, 18, then λinh (Cn2 ) = λ(Cn2 ). Proof From Theorem 2.5, it is clear that λinh (Cn2 ) = λ(Cn2 ) for n = 7m and n ∈ {8, 9, 15, 16, 22, 23, 29}. Let n 6= 7m and n ∈ / {8, 9, 15, 16, 22, 23, 29}. By division algorithm n = 7l + r, 1 6 r 6 6. Case 1. If l > r, then the coloring of using the sequence h4 2 0 5 7 3 1 6i r times followed by the sequence h4 2 0 5 3 1 6i (l − r) times which is given by Zhang [10] is an irreducible no-hole coloring of Cn2 . Hence λinh (Cn2 ) = λ(Cn2 ).


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Case 2. l < r. Subcase 2.1 If l + r > 6, then l + r = 6 + j and 0 < j 6 l. Now we use the sequence h0 2 4 6 1 3 5i (j − 1) times followed by h0 2 4 7 1 3 6i once followed by h0 4 7 1 3 6i (l − j) times and h0 4 7 1 3 5i one time to get an inh-coloring of Cn2 . Therefore λinh (Cn2 ) = λ(Cn2 ). Subcase 2.2 If l + r 6 6, then we have n = 11 or 17. Since h0 8 4 2 6 0 8 3 1 7 5i and 2 2 2 2 h0 2 4 6 1 3 5 7 0 2 4 6 8 1 3 5 7i are inh-coloring of C11 and C17 respectively, λinh (C11 ) = λ(C11 ) 2 2 and λinh (C17 ) = λ(C17 ). 2 Theorem 2.7 For n = 10, 12, 18, λinh (Cn2 ) = λ(Cn2 ) + 1. 2 Proof n = 10: Suppose that f is a no-hole coloring of C10 with span 8 = λ(G). Let c be the color which is repeated. Without loss of generality, we assume that f (v0 ) = c. The only possible repetition of c is for v5 , that is f (v5 ) = c. Since v1 , v2 , v8 , v9 are adjacent to v0 and v3 , v4 , v6 and v7 are adjacent to v5 , none of the colors c + 1 and c − 1 can be used to color any of 2 the vertices of C10 giving a contradiction to the fact that f is a no-hole coloring. The sequence 2 h8 6 0 2 4 7 9 5 3 1i is an inh-coloring of C10 with span 9 = λ(G) + 1.

2 n=12: Suppose that f is a no-hole coloring of C12 with span 7. Since in any span coloring 2 2 of C12 no color can be used three times, in a no-hole coloring of C12 exactly four colors will be used two times. First we show that no two consecutive colors can be repeated. Suppose c and c + 1 are used twice. Then at least one of c and c + 1 is different from 0 and 7. We assume c is different from 0 and 7. Without loss of generality, we assume f (v0 ) = c. Then c must be repeated to one of v5 , v6 and v7 . If c is repeated to v5 (or v7 ) then c + 1 can be used to either v8 or v9 (v4 or v3 ) only which implies c + 1 cannot be repeated. If c is repeated to v6 , then 2 c + 1 can be used to v3 and v9 . Then c − 1 cannot be assigned to any of the vertices in C12 . Therefore no two consecutive colors can be repeated. Since four colors are used twice, at least two colors differ by exactly two, say c and c + 2. Now, we prove that f is not a no-hole coloring. Without loss of generality, we assume that f (v0 ) = c. Suppose that f (v5 ) = c. Then by Lemma 2.3, f (v11 ) 6= f (v4 ) and f (v1 ) 6= f (v6 ). Since c + 2 is used twice, either f (v6 ) = f (v11 ) = c + 2 or one of the vertices v7 , v8 , v9 and v10 receives one copy of c + 2. In any case, c + 1 cannot be 2 used to any of the vertices in C12 which contradicts the fact that f is a no-hole coloring. An 2 inh-coloring of C12 is given in Figure 2.

6

0 v0

v11

4

v1 7

v10

v2

3

1

v3

v9 8

v8

3

v4

5

v7

2

v6

0

v5 6

2 2 Figure 2 An inh-coloring of C12 with λinh (C12 ) = 8.

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2 n=18: Suppose that f is a no-hole coloring of C18 with span 7. Then there are four ways of repetition of colors in f.

(i) Five colors appears three times and three colors once. (ii) Four colors three times, two colors two times and two colors one time. (iii) Three colors three times, four colors two times and one color one time. (iv) Two colors three times, six colors two times. First we show that no two consecutive colors can appear three times. Suppose c and c + 1 appear three times. Without loss of generality, we assume that f (v0 ) = c, then only possible way to use c and c + 1 is for vertices v6 , v12 and v3 , v9 , v15 respectively. Therefore, f cannot assign c − 1 and c + 2 (c + 2 if c = 0 and c − 1 if c = 6) to any vertex. Case 1. possible.

Since no two consecutive colors can be used three times, clearly this case is not

Case 2. The colors which are used twice are consecutive or one of them is consecutive to two of the colors appeared thrice. Suppose a two times used color c is consecutive to two of the three times used colors. Without loss of generality, f (v0 ) = c − 1. Let f (vi ) = f (vj ) = c − 1 and i < j. Clearly, 5 6 i, j 6 13 and 5 6 j − i 6 8. It is easy to see that either one of i, j − i, 18 − j is five or all of them are six. Suppose all of them are six. Then f (v6 ) = f (v12 ) = c − 1 and two of v3 , v9 , v15 receive c. In this case c + 1 cannot be appear three times. Suppose one of i, i − j, 18 − j is five, say i. Then one of j − i and 18 − j is 5 and the other is 8 or one of them is 6 and the other is 7. If j − i = 5 or 18 − j = 5, then c cannot be used two times. If j − i = 6, then f (v11 ) = c − 1 and f (v8 ) = c, and f (v14 ) = c or f (v15 ) = c. In any of these cases c + 1 cannot be used three times. Suppose the colors which are used two times are consecutive, say c and c + 1. Since no two consecutive colors appear three times and four colors appears three times, c 6= 0, c + 1 6= 7, and c − 1 and c + 2 appears three times. As in previous cases we get contradiction to repetition of colors. Case 3. In this case, apart from the possibilities given in Case (i) and Case (ii), we also get the following possibilities. A color c is used two times and its consecutive colors c − 1 and c + 1 are used twice; 0, 2, 7 (or 0, 5, 7) are used thrice and 1 (or 6) is used once. The case a color c is used two times and its consecutive colors c − 1 and c + 1 are used twice can be proved similar to the case where two consecutive colors cannot be used thrice. Suppose 0, 2, 7 are used thrice and 1 is used once. Without loss of generality, we assume that f (v0 ) = 2. If f (v6 ) = f (v12 ) = 2, then v3 , v9 and v15 must receive 1 and 3 (used twice). In this case 0 cannot be used three times. If f (v5 ) = f (v11 ) = 2, then f v8 = 3 and one of v14 and v15 receive 3 and the other receive 1. If f (v15 ) = 3, then f (v11 ) = f (v4 ) = 0 and v1 , v2 , v3 must receive three colors from {4, 5, 6, 7} which is not possible. If f (v14 ) = 3, then f (v11 ) = f (v4 ) = 4 and v1 , v2 , v3 must receive colors from {0, 6, 7} which is not possible. Case 4. If one of the colors repeated three times is different from both 0 and 7, then two of its consecutive colors will be used at least twice, as in the previous cases, we get contradictions.


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Suppose the colors repeated three times are 0 and 7. Without loss of generality, we assume that f (v0 ) = 0. From the previous cases, it is easy to see that the difference between the indices of two vertices receiving zero is six for at least one pair. We assume f (v6 ) = 0 and f (v3 ) = 1. Further, it is easy to see that one of v1 , v2 , v4 , v5 must receive 7. Suppose f (v1 ) = 7. Then at least one of v4 and v5 must receive 6, otherwise we get a contradiction to Lemma 2.1. If f (v4 ) = 6, then one of v8 and v9 receive 3 and one of v11 , v12 , v13 , v14 , v15 , v16 , v17 receive 4 which is not possible as the color 5 cannot be used two times. If f (v5 ) = 6, then f (v2 ) = 5 and f (v4 ) = 3 then at least one of 5 and 3 cannot be used two times. Similarly, we can prove when 2 f (v2 ) = 7. Using the sequence h0 2 4 6 8 1 3 5 7i two times we get an inh-coloring of C18 . 2 References [1] Gerard. J. Chang and David. Kuo, The L(2, 1)-Labeling Problem on Graphs, SIAM J. Discret. Math., Vol.9(1996) 309-316. [2] Ganclaves D, On the L(p,1)-labelling of graphs, Discrete Mathematics, Vol.308(2008) 14051414. [3] Gerrold R. Griggs and Roger K. Yeh, Labeling graphs with a condition at distance 2, SIAM J. Discrete Math., Vol.5(1992), 586-595. [4] Havet F, Reed B and Sereni J. S, L(2, 1)-labeling of graphs, Proceedings of the ACM-SIAM Symposium on Discrete Algorithm (SODA’08). [5] Peter C. Fishburn and Fred S. Roberts, No-hole L (2, 1)-colorings, SIAM J. Discrete Math., Vol.130(2003), 513-519. [6] Peter C. Fishburn and Fred S. Roberts, Full color Theorems for L (2, 1)-colorings, SIAM J. Discrete Math., Vol.20(2006), 428-443. [7] Peter C. Fishburn, Renu C. Laskar, Fred S. Roberts and Jhon Villalpando, Perameters of L (2, 1)-colorings, Manuscript. [8] Renu C. Laskar and Gilbert Eyabi, Holes in L(2, 1)-coloring on certain classes of graphs, AKCE International J. Graphs Comb., Vol.6(2009), 329-339. [9] Renu C. Laskar, Gretchen L. Matthews, Beth Novick and John Villalpando, On irreducible no-hole L(2, 1)-coloring of trees, Networks, Vol.53(2009), 206-211. [10] Xiaoling Zhang, Distance two labeling on the square of a cycle, Korean J. Math., Vol.23(2015), 607-618.


Clique-to-Clique Detour Distance in Graphs S. Athisayanathan (Department of Mathematics, Loyola College (Autonomous), Chennai - 600034, India)

I. Keerthi Asir (Department of Mathematics, St. Xavier’s College (Autonomous), Palayamkottai - 627002, India) E-mail: [email protected], [email protected]

Abstract: Let ζ be the set of all cliques in a connected graph G and C, C ′ ∈ ζ. A cliqueto-clique C − C ′ path P is a u − v path, where u ∈ C and v ∈ C ′ , in which P contains no vertices of C and C ′ other than u and v. The clique-to-clique detour distance, D(C, C ′ ) is the length of a longest C − C ′ path. A C − C ′ path of length D(C, C ′ ) is called a C − C ′ detour. The clique-to-clique detour eccentricity eD3 (C) of a clique C in G is the maximum clique-to-clique detour distance from C to a clique C ′ ∈ ζ in G. The clique-to-clique detour radius R3 of G is the minimum clique-to-clique detour eccentricity among the cliques of G, while the clique-to-clique detour diameter D3 of G is the maximum clique-to-clique detour eccentricity among the cliques of G. It is shown that R3 ≤ D3 for every connected graph G and that every two positive integers a and b with 2 ≤ a ≤ b are realizable as the clique-toclique detour radius and the clique-to-clique detour diameter of some connected graph. Also it is shown that for any two positive integers a and b with 2 ≤ a ≤ b are realizable as the clique-to-clique radius (diameter) and the clique-to-clique detour radius (diameter) of some connected graph.

Key Words: Distance, detour distance, clique-to-clique distance. AMS(2010): 05C12. §1. Introduction By a graph G = (V, E), we mean a finite undirected connected simple graph. For basic graph theoretic terminologies, we refer to Chartrand and Zhang [2]. If X ⊆ V , then hXi is the subgraph induced by X. A clique C of a graph G is a maximal complete subgraph and we denote it by its vertices. For example if one is locating an emergency facility like police station, fire station, hospital, school, college, library, ambulance depot, emergency care center, etc., then the primary aim is to minimize the distance between the facility and the location of a possible emergency. In 1964, Hakimi [3] considered the facility location problems as vertex-to-vertex distance in graphs. For 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received December 31, 2017, Accepted May 22, 2018, Edited by S. Monikandan.

Clique-to-Clique Detour Distance in Graphs

43

any two vertices u and v in a connected graph G, the distance d(u, v) is the length of a shortest u − v path in G. A u − v path of length d(u, v) is called a u − v geodesic in G. For a vertex v in G, the eccentricity e(v) of v is defined by e(v) = max{d(v, u) : u ∈ V }. A vertex u of G such that d(u, v) = e(v) is called an eccentric vertex of v. The radius r and diameter d of G are defined by r = rad(G) = min{e(v) : v ∈ V } and d = diam(G) = max{e(v) : v ∈ V } respectively. A vertex v in G is called a central vertex if e(v) = r and the center of G is defined by C(G) = Cen(G) = h{v ∈ V : e(v) = r}i. A vertex v in G is called a peripheral vertex if e(v) = d and the periphery of G is defined by P (G) = P er(G) = h{v ∈ V : e(v) = d}i. If every vertex of a graph G is central vertex then G is called self-centered. For example if one is making an election canvass or circular bus service the distance from the location is to be maximized. In 2005, Chartrand et. al. [1] studied the concepts of detour distance in graphs. For any two vertices u and v in a connected graph G, the detour distance D(u, v) is the length of a longest u − v path in G. A u − v path of length D(u, v) is called a u − v detour in G. For a vertex v in G, the detour eccentricity eD (v) of v is defined by eD (v) = max{D(v, u) : u ∈ V }. A vertex u of G such that D(u, v) = eD (v) is called a detour eccentric vertex of v. The detour radius R and detour diameter D of G are defined by R = radD G = min{eD (v) : v ∈ V } and D = diamD (G) = max{eD (v) : v ∈ V } respectively. A vertex v in G is called a detour central vertex if eD (v) = R and the detour center of G is defined by CD (G) = CenD (G) = h{v ∈ V : eD (v) = R}i. A vertex v in G is called a detour peripheral vertex if eD (v) = D and the detour periphery of G is defined by PD (G) = P erD (G) = h{v ∈ V : eD (v) = D}i. If every vertex of a graph G is detour central vertex then G is called detour self-centered. For example when a railway line, pipe line or highway is constructed, the distance between the respective structure and each of the communities to be served is to be minimized. In a social network a clique represents a group of individuals having a common interest. Thus the centrality with respect to cliques have interesting applications in social networks. In 2002, Santhakumaran and Arumugam [9] introduced and studied the facility locational problems as clique-to-clique distance in graphs. For the cliques C and C ′ in a connected graph G, the cliqueto-clique distance is defined by d(C, C ′ ) = min{d(u, v) : u ∈ C, v ∈ C ′ }. For our convenience a C − C ′ path of length d(C, C ′ ) is called a clique-to-clique C − C ′ geodesic or simply C − C ′ geodesic. The clique-to-clique eccentricity of C is defined by e3 (C) = max{d(C, C ′ ) : C ′ ∈ ζ}. A clique C of G such that e3 (C) = d(C, C ′ ) is called a clique-to-clique eccentric clique of C. The clique-to-clique radius r3 and clique-to-clique diameter d3 of G are defined by r3 = min{e3 (C) : C ∈ ζ} and d3 = max{e3 (C) : C ∈ ζ} respectively. A clique C in a graph G is called a clique-to-clique central clique if e3 (C) = r3 and the clique-to-clique center of G is defined by Z3 (G) = h{C ∈ ζ : e3 (C) = r3 }i. For our convenience Z3 (G) is denoted by C3 (G). A clique C in a graph G is called a clique-to-clique peripheral clique if e3 (C) = d3 and the clique-to-clique periphery of G is defined by P3 = h{C ∈ ζ : e3 (C) = d3 }i. If every clique of a graph G is clique-to-clique central clique then G is clique-to-clique self-centered. For example when a dam, lake, pond, river or channel is constructed, the maximum number of places should be covered between the respective structure and each of the communities to be served. These motivated us to introduce a distance called the clique-to-clique detour

44

S. Athisayanathan and I. Keerthi Asir

distance in graphs and investigate certain results related to clique-to-clique detour distance and other distances in graphs. Further these ideas have interesting applications in channel assignment problem in radio technologies. Also there are useful applications to security based communication network design. Throughout this paper, G denotes a connected graph with at least two vertices.

§2. Clique-to-Clique Detour Distance Definition 2.1 Let ζ be the set of all cliques in a connected graph G and C, C ′ ∈ ζ. A cliqueto-clique C −C ′ path P is a u−v path, where u ∈ C and v ∈ C ′ , in which P contains no vertices of C and C ′ other than u and v. The clique-to-clique detour distance D(C, C ′ ) is the length of a longest C − C ′ path in G. A C − C ′ path of length D(C, C ′ ) is called a clique-to-clique C − C ′ detour or simply C − C ′ detour. For our convenience a C − C ′ path of length d(C, C ′ ) is called a clique-to-clique C − C ′ geodesic or simply C − C ′ geodesic. y

x

z

w u

v

Fig. 2.1 The graph G

Example 2.2 Consider the graph G given in Fig 2.1. For the cliques C = {u, v}, C ′ = {x, y, z} and C ′′ = {v, z}, the C − C ′ paths are P1 : v, z and P2 : u, w, x while the paths P3 : u, v, z and P4 : u, w, x, y, z are not C − C ′ paths. Similarly the C − C ′′ paths are Q1 : v, Q2 : u, w, x, z and Q3 : u, w, x, y, z while the paths Q4 : u, v, z and Q5 : u, w, x, y, z, v are not C − C ′′ paths. Now the clique-to-clique distance d(C, C ′ ) = 1, d(C, C ′′ ) = 0 and the clique-to-clique detour distance D(C, C ′ ) = 2, D(C, C ′′ ) = 4. Also P1 is a C − C ′ geodesic, Q1 is a C − C ′′ geodesic and P2 is a C − C ′ detour, Q3 is a C − C ′′ detour.

Since the length of a C − C ′ path between any two cliques C and C ′ in a graph G of order n is at most n − 2, we have the following observation. Observation 2.3 For any two cliques C and C ′ in a non-trivial connected graph G of order n, 0 ≤ d(C, C ′ ) ≤ D(C, C ′ ) ≤ n − 2. The bounds in Observation 2.3 are sharp. If G = K2 , then 0 = d(C, C) = D(C, C) = n − 2. If G is a cycle of order n ≥ 4, then D(C, C ′ ) = n − 2 for any two distinct adjacent cliques C and C ′ in G and if G is a tree, then d(C, C ′ ) = D(C, C ′ ) for any two cliques C and C ′ in G and if G is an odd cycle, then d(C, C ′ ) < D(C, C ′ ) for any two cliques C and C ′ in G.


45

Theorem 2.4 Let C and C ′ be any two adjacent cliques (C 6= C ′ ) in a connected graph G. Then D(C, C ′ ) = n − 2 if and only if G is cycle. Proof Assume that G is a cycle Cn : u1 , u2 , . . . , un−1 , un , u1 . Since any edge in a cycle is a clique, without loss of generality we assume that C = {u1 , u2 }, C ′ = {un , u1 } be any two adjacent cliques. Then there exist two distinct C − C ′ paths, say P1 and P2 such that P1 : u1 is a trivial C − C ′ path of length 0 and P2 : u2 , u3 , . . . , un−1 , un is a C − C ′ path of length n − 2. It is clear that D(C, C ′ ) = n − 2. Conversely assume that for any two distinct adjacent cliques C and C ′ in a connected graph G, D(C, C ′ ) = n − 2. We prove that G is a cycle. Suppose that G is not a cycle. Then G must be either a tree or a cyclic graph. Case 1. If G is a tree, then C − C ′ path is trivial. So that D(C, C ′ ) = 0 < n − 2, which is a contradiction. Case 2. If G is a cyclic graph, then G must contain a cycle Cd : x1 , x2 , . . . , xd , x1 of length d < n. If C = {x1 , x2 } and C ′ = {xn , x1 } then D(C, C ′ ) < n − 2, which is a contradiction. 2 Since the maximum length of a C − C ′ path between any two cliques C and C ′ in Kn,m is 2n − 2, we have the following observation. Observation 2.5 Let Kn,m (n ≤ m) be a complete bipartite graph with the partition V1 , V2 of V (Kn,m ) such that |V1 | = n and |V2 | = m. Let C and C ′ be any two cliques in Kn,m , then D(C, C ′ ) = 2n − 2. Since every tree has unique C − C ′ path between any two cliques C and C ′ , we have the following observation. Observation 2.6 If G is a tree, then d(C, C ′ ) = D(C, C ′ ) for every cliques C and C ′ in G. The converse of the Observation 2.6 is not true. For the graph G obtained from a complete bipartite graph K2,n (n ≥ 2) by joining the vertices of degree n by an edge. In such a graph every clique C is adjacent and isomorphic to K3 and so d(C, C ′ ) = D(C, C ′ ) = 0, but G is not tree. §3. Clique-to-Clique Central Concepts Definition 3.1 Let G be a connected graph and ζ be the set of all cliques in G. The cliqueto-clique detour eccentricity eD3 (C) of a clique C in ζ is defined as eD3 (C) = max {D(C, C ′ ) : C ′ ∈ ζ}. A clique C ′ for which eD3 (C) = D(C, C ′ ) is called a clique-to-clique detour eccentric clique of C. The clique-to-clique detour radius of G is defined by R3 = radD3 (G) = min {eD3 (C) : C ∈ ζ} and the clique-to-clique detour diameter of G is defined as, D3 = diamD3 (G) = max {eD3 (C) : C ∈ ζ}. A clique C in G is called a clique-to-clique detour central clique if eD3 (C) = R3 and the clique-to-clique detour center of G is defined as, CD3 (G) = CenD3 (G) = h{C ∈ ζ : eD3 (C) = R3 }i. A clique C in G is called a clique-to-clique detour peripheral clique if eD3 (C) = D3 and the clique-to-clique detour periphery of G is defined as, PD3 (G) = P erD3 (G) = h{C ∈ ζ : eD3 (C) = D3 }i. If every clique of G is a clique-to-clique detour central clique, then G is called a clique-to-clique detour self centered graph. If G is a

46


clique-to-clique detour self centered graph then G is its own clique-to-clique detour periphery. Example 3.2 For the connected graph G given in Fig. 3.1, the set of all cliques in G are given by, ζ = {C1 , C2 , C3 , C4 , C5 , C6 , C7 , C8 , C9 , C10 } where C1 = {v1 , v2 , v3 }, C2 = {v3 , v4 }, C3 = {v4 , v5 }, C4 = {v5 , v6 }, C5 = {v6 , v7 }, C6 = {v7 , v8 }, C7 = {v8 , v10 }, C8 = {v9 , v10 }, C9 = {v4 , v9 }, C10 = {v10 , v11 , v12 , v13 , v14 }. The clique-to-clique eccentricity e3 (C), the cliqueto-clique detour eccentricity eD3 (C) of all the cliques of G are given in Table 1 and the cliqueto-clique detour eccentric clique of all the cliques of G are given in Table 2.

v5

v6

v7

v2

v4

v3

v8 v11

v1 v9

v12

v10


v14

v13

Cliques C

C1

C2

C3

C4

C5

C6

C7

C8

C9

C10

e3 (C)

4

3

2

3

3

4

3

2

2

3

eD3 (C)

6

5

5

6

5

5

5

6

5

6

Table 1

The clique-to-clique radius r3 = 2, the clique-to-clique diameter d3 = 4, the clique-to-clique detour radius R3 = 5 and the clique-to-clique detour diameter D3 = 6. Also the clique-to-clique center C3 (G) = h{C3 , C8 , C9 }i, the clique-to-clique periphery P3 (G) = h{C1 , C6 }i, the cliqueto-clique detour center CD3 (G) = h{C2 , C3 , C5 , C6 , C7 , C9 }i and the clique-to-clique detour periphery PD3 (G) = h{C1 , C4 , C8 , C10 }i.

47


Clique C

Clique-to-Clique Detour Eccentric Cliques

C1

C4 , C8 ,C10

C2

C4 , C8 ,C10

C3

C4 , C9

C4

C1

C5

C1 ,C4 , C6

C6

C5 , C7 ,C10

C7

C1 , C8

C8

C1

C9

C3 , C8 ,C10

C10

C1 Table 2

Remark 3.3 In a connected graph G, the clique-to-clique center C3 (G) and the clique-toclique detour center CD3 (G) are distinct. Also the clique-to-clique periphery P3 (G) and the clique-to-clique detour periphery PD3 (G) need not be same. For the the graph G given in Fig 3.1, it is shown that C3 (G) and CD3 (G) are distinct. Also P3 (G) and PD3 (G) are distinct. The clique-to-clique detour radius and the clique-to-clique detour diameter of some standard graphs are given in Table 3.

Graph G

Kn

R3

0

D3

0

Pn n−3 2

n−3

Cn (n ≥ 4)

Wn (n ≥ 4)

Kn,m (m ≥ n)

n−2

n−3

2n − 2

n−2

n−3

2n − 2

Table 3

Example 3.4 The complete graph Kn , the cycle Cn , the wheel Wn , star K1,n and the complete bipartitate graph Kn,m are clique-to-clique detour self centered graphs. Remark 3.5 A clique-to-clique detour self-centered graph need not be a vertex-to-clique detour self centered graph. For example, the star graph K1,n is the desired graph. Remark 3.6 A clique-to-clique detour self-centered graph need not be a clique-to-vertex detour self centered graph. For example, the graph obtained by joining a vertex of K2 and a vertex of K3 is the desired graph.

48


v1

v8

v2

v7 v6

v3

v5

v4

Fig. 3.2 The graph G Remark 3.7 (i) A vertex-to-clique self centered graph need not be a clique-to-clique detour self centered graph. (ii) A clique-to-vertex self centered graph need not be a clique-to-clique detour self centered graph. (iii) A clique-to-clique self-centered graph need not be a cliqueto-clique detour self centered graph. For the graph G given in Fig 3.2, C1 (G) = hV (G)i and C2 (G) = C3 (G) = hζi, but CD3 (G) = h{v3 , v4 , v5 , v6 }i. Since the clique-to-clique eccentricity is the maximum clique-to-clique distance and the clique-to-clique detour eccentricity is the maximum clique-to-clique detour distance, the following observation is a consequence of Observation 2.3. Observation 3.8 For every clique C in G of order n, 0 ≤ e3 (C) ≤ eD3 (C) ≤ n − 2.

The bounds in Observation 3.8 are sharp. If G = K2 , then 0 = e3 (C) = eD3 (C) = n − 2. If G = Cn (n ≥ 4), then e3 (C) < eD3 (C) = n − 2 for any clique C in G and if G is a tree, then e3 (C) = eD3 (C) for any clique C in G. Since the clique-to-clique radius (diameter) is the minimum (maximum) clique-to-clique eccentricity and the clique-to-clique detour radius (diameter) is the minimum (maximum) cliqueto-clique detour eccentricity, the following observation is a consequence of Observation 3.8. Observation 3.9 Let G be a connected graph. Then (i) 0 ≤ r3 ≤ R3 ≤ n − 2;

(ii) 0 ≤ d3 ≤ D3 ≤ n − 2. In [2] it is shown that, the radius and the diameter are related by r ≤ d ≤ 2r and the detour radius and the detour diameter are related by R ≤ D ≤ 2R. Also Santhakumaran et.al. [9] showed that the vertex-to-clique radius and the vertex-to-clique diameter are related by r1 ≤ d1 ≤ 2r1 + 1, the clique-to-vertex radius and the clique-to-vertex diameter are related by r2 ≤ d2 ≤ 2r2 + 1 and the clique-to-clique radius and the clique-to-clique diameter are related by r3 ≤ d3 ≤ 2r3 + 1. Keerthi Asir et.al. [5,6] showed that the similar inequality does not hold for the vertex-to-clique detour distance as well as the clique-to-vertex detour distance. The following example shows that the similar inequality does not hold for the clique-to-clique

49


detour distance too. Example 3.10 For the graph G of order n ≥ 8 obtained by identifying the central vertex of the wheel Wn−1 = K1 + Cn−2 and an end vertex of the path P2 . It is easy to verify that D3 > 2R3 + 1. Ostrand [7] showed that every two positive integers a and b with a ≤ b ≤ 2a are realizable as the radius and the diameter respectively of some connected graph and Chartrand et.al. [1] showed that every two positive integers a and b with a ≤ b ≤ 2a are realizable as the detour radius and the detour diameter respectively of some connected graph. Also Santhakumaran et.al. [9] showed that every two positive integers a and b with a ≤ b ≤ 2a + 1 are realizable as the vertex-to-clique radius and the vertex-to-clique diameter respectively of some connected graph, every two positive integers a and b with a ≤ b ≤ 2a + 1 are realizable as the clique-tovertex radius and the clique-to-vertex diameter respectively of some connected graph and every two positive integers a and b with a ≤ b ≤ 2a + 1 are realizable as the clique-to-clique radius and the clique-to-clique diameter respectively of some connected graph. Keerthi Asir et.al. [5] showed that every two positive integers a and b with 2 ≤ a ≤ b are realizable as the vertex-toclique detour radius and the vertex-to-clique detour diameter respectively of some connected graph and also Keerthi Asir et.al. [6] showed that every two positive integers a and b with 2 ≤ a ≤ b are realizable as the clique-to-vertex detour radius and the clique-to-vertex detour diameter respectively of some connected graph. Now, we have a similar realization theorem for the clique-to-clique detour radius and the clique-to-clique detour diameter. Theorem 3.11 For each pair a, b of positive integers with 2 ≤ a ≤ b, there exists a connected graph G with R3 = a and D3 = b. Proof The proof is divided into 3 cases. Case 1. a = b. Let G = Ca+2 : u1 , u2 , · · · , ua+2 , u1 be a cycle of order a + 2. If C = {ui , ui+1 } then eD3 (C) = a for 1 ≤ i ≤ a + 2. Thus R3 = a and D3 = b as a = b. ua

ua+1 ua+2

v1

v2

v3

u1

u2 u3

Fig. 3.3 The graph G Case 2. 2 ≤ a < b ≤ 2a.

vb−a+2

50


Let Ca+2 : u1 , u2 , · · · , ua+2 ,u1 be a cycle of order a + 2 and Pb−a+2 : v1 , v2 , · · · , vb−a+2 be a path of order b − a + 2. We construct the graph G of order b + 3 by identifying the vertex u1 of Ca+2 and v1 of Pb−a+2 as shown in Fig. 3.3. If C = {u1 , u2 } = {u1 , ua+2 }, then eD3 (C) = a. Also if C = {ui , ui+1 }, then eD3 (C) = b − i + 2 for 2 ≤ i ≤ a+2 and eD3 (C) = b − a + i − 1 2 a+2 for 2 < i ≤ a + 1. Also if C = {vi , vi+1 } then eD3 (C) = a + i − 1 for 1 ≤ i ≤ b − a + 1. In particular, if C = {u2 , u3 } = {ua+1 , ua+2 } = {vb−a+1 , vb−a+2 } then eD3 (C) = b. It is easy to verify that there is no clique S in G with eD3 (S) < a and there is no clique S ′ in G with eD3 (S ′ ) > b. Thus R3 = a and D3 = b as a < b. u1

v1

v2

v3

va

va+1 v2a+1 v2a+2

ub+2

Fig. 3.4 The graph G Case 3. b > 2a. Let G be the graph of order b + 2a + 4 obtained by identifying the central vertex of the wheel Wb+3 = K1 + Cb+2 and an end vertex of the path P2a+2 , where K1 = v1 , Cb+2 : u1 , u2 , . . . , ub+2 , u1 and P2a+2 : v1 , v2 , . . . , v2a+2 . The resulting graph G is shown in Fig. 3.4. If C = {v1 , ui , ui+1 }, then eD3 (C) = b for 1 ≤ i ≤ b+1. Also if C = {vi , vi+1 } then eD3 (C) = 2a−i for 1 ≤ i ≤ a and eD3 (C) = i − 1 for a < i ≤ 2a + 1. It is easy to verify that there is no clique S in G with eD3 (S) < a and there is no clique S ′ in G with eD3 (S ′ ) > b. Thus R3 = a and D3 = b as b > 2a. 2 Keerthi Asir et.al. [5] showed that for every two positive integers a and b with 2 ≤ a ≤ b, there exists a connected graph G with vertex-to-clique radius r1 = a and vertex-to-clique detour radius R1 = b and also Keerthi Asir et.al. [6] showed that for every two positive integers a and b with 2 ≤ a ≤ b, there exists a connected graph G with clique-to-vertex radius r2 = a and clique-to-vertex detour radius R2 = b. Now, we have a similar realization theorem for the clique-to-clique radius r3 and the clique-to-clique detour radius R3 . Theorem 3.12 For any two positive integers a, b with 2 ≤ a ≤ b, there exists a connected graph G such that r3 = a and R3 = b. Proof The proof is divided into 2 cases following. Case 1. a = b. Let P1 : u1 , u2 , · · · , ua+2 and P2 : v1 , v2 , · · · , va+2 be two paths of order a+2. We construct

51


the graph G of order 2a + 4 by joining u1 in P1 and v1 in P2 by an edge. If C = {u1 , v1 } then e3 (C) = eD3 (C) = a and if C = {ui , ui+1 } = {vi , vi+1 } then e3 (C) = a + i for 1 ≤ i ≤ a + 1. It is easy to verify that there is no clique S in G with e3 (S) < a and eD3 (S) < b. Thus r3 = a and R3 = b as a = b. Case 2. 2 ≤ a < b. Let P1 : u1 , u2 , · · · , , ua+1 , ua+2 and Q1 : v1 , v2 , · · · , va+1 , va+2 be paths of order a + 2. Let P2 : w1 , w2 , · · · , wb−a+2 and Q2 : z1 , z2 , · · · , zb−a+2 be two paths of order b − a + 2. We construct the graph G of order 2b + 4 as follows: (i) identify the vertices u1 in P1 with w1 in P2 and also identify the vertices v1 in Q1 with z1 in Q2 (ii) identify the vertices u3 in P1 with wb−a+2 in P2 and also identify the vertices zb−a+2 in Q2 with v3 in Q1 (iii) join each vertex wi (2 ≤ i ≤ b−a+1) in P2 with u2 in P1 and join each vertex zi (2 ≤ i ≤ b−a+1) in Q2 with v2 in Q1 (iv) join u1 in P1 with v1 in Q1 . The resulting graph G is shown in Fig. 3.5. If C = {u1 , v1 } then e3 (C) = a and eD3 (C) = b. Also if C = {ui , ui+1 } = {vi , vi+1 } then e3 (C) = a + i and eD3 (C) = 2b − a + i for 3 ≤ i ≤ a + 1. Also if C = {u2 , wi , wi+1 } = {v2 , zi , zi+1 } then e3 (C) = a + 1 for i = 1 and e3 (C) = a + 2 for 2 ≤ i ≤ b − a + 1. However eD3 (C) = b + i for 1 ≤ i ≤ b − a + 1. It is easy to verify that there is no clique S in G with e3 (S) < a and eD3 (S) < b. Thus r3 = a and R3 = b as a < b. 2 w3 w2

wb−a+1

u1

v1

u2

u3

v2

v3

u4

ua+1

ua+2

v4

va+1

va+2

zb−a+1

z2 z3


Keerthi Asir et.al. [5] showed that for every two positive integers a and b with 2 ≤ a ≤ b, there exists a connected graph G with vertex-to-clique diameter d1 = a and vertex-to-clique detour diameter D1 = b and Keerthi Asir et.al. [6] showed that for every two positive integers

52


a and b with 2 ≤ a ≤ b, there exists a connected graph G with clique-to-vertex diameter d2 = a and clique-to-vertex detour diameter D2 = b Now, we have a similar realization theorem for the clique-to-clique diameter d3 and the clique-to-clique detour diameter D3 . Theorem 3.13 For any two positive integers a, b with 2 ≤ a ≤ b, there exists a connected graph G such that d3 = a and D3 = b. Proof The proof is divided into 2 cases following. Case 1. a = b. Let Pa+2 : u1 , u2 , · · · , ua+3 be a path of order a + 3. If C = {ui , ui+1 } then e3 (C) = eD3 (C) = a − i + 1 for 1 ≤ i ≤ a+2 and e3 (C) = eD3 (C) = i − 2 for a+2 < i ≤ a + 1. In 2 2 particular if C = {u1 , u2 } = {ua+1 , ua+2 } then e3 (C) = eD2 (C) = a. It is easy to verify that there is no clique S in G with e3 (S) = eD3 (S) > a. Thus d3 = a and D3 = b as a = b. Case 2. 2 ≤ a < b. Let P1 : u1 , u2 , · · · , ua+2 be a path of order a + 2. Let P2 : w1 , w2 , · · · , wb−a+2 be a path of order b − a + 2. Let P3 : x1 , x2 be a path of order 2. We construct the graph G of order b + 3 as follows: (i) identify the vertices u1 in P1 , w1 in P2 with x1 in P3 and identify the vertices u3 in P1 with wb−a+2 in P2 (ii) join each vertex wi (2 ≤ i ≤ b − a + 1) in P2 with u2 in P1 . The resulting graph G is shown in Fig. 3.6. w3 wb−a+1

w2

ua+1

x1 u2

u3

ua+2

u4

x2 Fig. 3.6: The graph G

If C = {x1 , x2 } then e3 (C) = a and eD3 (C) = b. Also if C = {ui , ui+1 } then e3 (C) = i − 1 and eD3 (C) = b − a − 1 + i for 3 ≤ i ≤ a + 1. If C = {u2 , wi , wi+1 } then eD3 (C) = b − i − 1 for 1 ≤ i ≤ b−a+2 and eD3 (C) = i − 1 if i − 1 ≥ b − i − 1 and eD3 (C) = b − i − 1 if i − 1 ≤ b − i − 1 2 b−a+2 for < i ≤ b − a + 1. Also we have to find e3 (C) by the following subcases. 2 Subcase 2.1 a = 2.

In this case, e3 (C) = 1 for 1 ≤ i ≤ b − a + 1. Subcase 2.2 a ≥ 3. In this case, e3 (C) = a − 1 for 1 ≤ i ≤ b − a and e3 (C) = a − 2 for i = b − a + 1.


53

It is easy to verify that there is no clique S in G with e3 (S) > a and eD3 (S) > b. Thus d3 = a and D3 = b as a < b. 2 Observation 3.14 For every connected graph G of order n ≥ 4, R1 = R2 = R3 = D1 = D2 = D3 = n − 2 if and only if G is cycle. Observation 3.15 For every connected graph G of order n ≥ 2, R1 = R2 = R3 = D1 = D2 = D3 = 0 if and only if G is complete. Harary and Norman [4] showed that the center of every connected graph G lies in a single block of G, Chartrand et.al. [1] showed that the detour center of every connected graph G lies in a single block of G. Also Keerthi Asir et.al. [5] showed that the vertex-to-clique detour center of every connected graph G lies in a single block of G and Keerthi Asir et.al. [6] showed that the clique-to-vertex detour center of every connected graph G does not lie in a single block of G. The following example shows that the clique-to-clique detour center of every connected graph G does not lie in a single block of G. Example 3.16 For the Path P2n+1 , the clique-to-clique detour center is always P3 , which does not lie in a single block.

References [1] G. Chartrand, H. Escuadro and P. Zhang, Detour Distance in Graphs, J.Combin. Math. Combin. Comput., 53, (2005), 75-94. [2] G. Chartrand and P. Zhang, Introduction to Graph Theory, Tata McGraw-Hill,New Delhi, 2006. [3] S. L. Hakimi, Optimum location of switching centers and absolute centers and medians of a graph, Operations Research, 12 (1964). [4] F. Harary and R. Z. Norman, The dissimilarity characteristics of Husimi trees, Ann. of Math., 58, (1953), 134-141. [5] I. Keerthi Asir and S. Athisayanathan, Vertex-to-Clique detour distance in graphs, J. Prime Research in Mathematics, 12 (2006), 45-59. [6] I. Keerthi Asir and S. Athisayanathan, Clique-to-Vertex Detour Distance in Graphs, J. Prime Research in Mathematics, 11 (2015), 42-54. [7] Philip A. Ostrand, Graphs with specified radius and diameter, Discrete Mathematics, 4, (1973), 71-75. [8] P. J. Slater, Centrality of paths and vertices in a graph: Cores and Pits, Proceedings of fourth Int.Conf. on Theory and Appln. of Graphs, Western Michigan University, Wiley, NY(1980), 529-542. [9] A. P. Santhakumaran and S. Arumugam, Centrality with respect to cliques, Int. J. Management and Systems, 18, No.3(2002), 275-280.


Connected Double Monophonic Number of a Graph A.P.Santhakumaran Department of Mathematics Hindustan Institute of Technology and Science, Chennai - 603 103, India

T.Venkata Raghu (Department of Applied Sciences and Humanities Sasi Institute of Technology and Engineering, Tadepalligudem-534 101, India E-mail: [email protected], [email protected]

Abstract: A set S of a connected graph G = (V, E) of order n is called a double monophonic set of G if for every pair of vertices x, y in G there exist vertices u, v in S such that x, y lie on a u − v monophonic path. The double monophonic number dm(G) of G is the minimum cardinality of a double monophonic set. A double monophonic set S in a connected graph G is said to be connected if the subgraph G[S] induced by S is connected. The minimum cardinality of a connected double monophonic set of G is the connected double monophonic number of G, and is denoted by dmc (G). Some general properties satisfied by connected double monophonic sets are discussed and the connected double monophonic number of some standard graphs are obtained. It is proved that for a connected graph G of order n ≥ 2, dmc (G) = 2 if and only if G = K2 . For positive integers a, b and n such that 2 ≤ a < b ≤ n, it is shown that there exists a connected graph G of order n with dm(G) = a and dmc (G) = b.

Key Words: Double monophonic set, double monophonic number, connected double monophonic set, connected double monophonic number.

AMS(2010): 05C12. §1. Introduction By a graph G=(V,E ) we mean a finite, undirected connected graph without loops or multiple edges. The order and size of G are denoted by n and m, respectively. For basic graph theoretic terminology we refer to [3]. The distance d (x,y) is the length of the shortest x -y path in G. Any x -y path of length d (x,y) is called an x -y geodesic. A subset S of V is called a geodetic set of the graph G if every vertex x of G lies on a u-v geodesic for some vertices u,v in S. A geodetic set of minimum cardinality is a minimum geodetic set. The cardinality of a minimum geodetic set is the geodetic number of G and is denoted by g(G). The geodetic number of a graph was 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received February 04, 2018, Accepted June 20, 2018, Edited by S. Monikandan.

Connected Double Monophonic Number of a Graph

55

introduced and further studied in [1,2,4,8]. The neighbourhood of a vertex v is the set N (v ) consisting of all vertices u which are adjacent with v. A vertex v is an extreme vertex if the subgraph induced by N (v ) is complete. A connected geodetic set of G is a geodetic set S such that the subgraph G[S ] induced by S is connected. The minimum cardinality of a connected geodetic set of G is the connected geodetic number of G and is denoted by gc (G). A chord of a path P is an edge joining two non-adjacent vertices of P. A path P is called monophonic if it is a chordless path. A subset S of V is called a monophonic set of G if every vertex v of G lies on a x −y monophonic path for some vertices x and y in S. The minimum cardinality of a monophonic set of G is called the monophonic number of G and is denoted by m(G). A connected monophonic set of G is a monophonic set S such that the subgraph G[S ] induced by S is connected. The minimum cardinality of a connected monophonic set of G is the connected monophonic number of G and is denoted by mc (G). The connected monophonic number of a graph was introduced and studied in [9]. Results regarding monophonic sets and monophonic number of a graph are studied in [7]. A set S of vertices of G is called a double geodetic set of G if for each pair of vertices x, y in G, there exist vertices u,v in S such that x,y lie on a u-v geodesic. The double geodetic number dg(G) of G is the minimum cardinality of a double geodetic set. The double geodetic number of a graph was introduced and studied in [5]. A set S of vertices of G is called a double monophonic set of G if for each pair of vertices x,y in G there exist vertices u, v in S such that x,y lie on a u-v monophonic path. The double monophonic number dm(G) of G is the minimum cardinality of a double monophonic set. The double monophonic number of a graph was introduced and studied in [6]. The following theorems will be used in the sequel. Theorem 1.1 ([6]) Each extreme vertex of a connected graph G belongs to every double monophonic set of G. In particular, if the set of all extreme vertices of G is a double monophonic set, then it is the unique minimum double monophonic set of G. Theorem 1.2 ([6]) Let G be a connected graph with a cut-vertex v. Then each double monophonic set of G contains at least one vertex from each component of G-v.

§2. Connected Double Monophonic Number of a Graph Definition 2.1 A connected double monophonic set S of G is a double monophonic set such that the subgraph G[S] induced by S is connected. The minimum cardinality of a connected double monophonic set of G is the connected double monophonic number of G, and is denoted by dmc (G). A connected double monophonic set of cardinality dmc (G) is called a dmc -set of G. Example 2.2 For the graph G in Figure 2.1, S = {v 3 ,v 5 } is the only double monophonic set of cardinality 2 and so dm(G) = 2. Note that S not a connected double monophonic set. It is easily seen that the sets S1 = {v 1 ,v 3 ,v5 } , S2 = {v 2 ,v 3 ,v5 }, and S3 = {v 3 ,v 4 ,v5 } are the only connected double monophonic sets of cardinality 3 so that dmc (G)=3.

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A.P.Santhakumaran and T.Venkata Raghu

v4 r

v5

v3

r

r

r

r

v2

v1 Figure 2.1

Since every extreme vertex of a connected graph is either an initial vertex or a terminal vertex of a monophonic path containing it, the following result is clear. Theorem 2.3 Each extreme vertex of a connected graph G belongs to every connected double monophonic set of G. Result 2.4 It follows from Theorem 2.3 that for the complete graph Kn (n≥ 2), dmc (Kn ) = n. Remark 2.5 There are non-complete graphs G of order n with dm c (G)= n. For the graph G given in Figure 2.2, S = {v 2 , v 7 , v 8 , v 9 } is the set of all extreme vertices and S is not a double monophonic set. It is not hard to check that {v 2 , v 3 , v 4 ,v 5 ,v 6 ,v 7 ,v 8 ,v 9 } is the unique double monophonic set so that dm(G) = 8. Also, it is clear that the set of all vertices is the unique connected double monophonic set so that dm c (G) = 9. v6 v8

v7

r

r

r

v1 v9

r

v5

r

r

v4

r

r

v2

r

v3

Figure 2.2 Since every double monophonic set contains at least two vertices, it follows that dm(G)≥ 2. Also every connected double monophonic set is a double monophonic set so that dm(G) ≤ dm c (G). Hence the following result is got. Theorem 2.6 For any connected graph G of order n, 2 ≤ dm(G) ≤ dmc (G) ≤ n. The bounds in the Theorem 2.6 are sharp. For any non-trivial path P, dm(P )=2 and for the complete graph K n , dm c (K n )=n (n ≥ 2). Further, all the inequalities in the theorem are strict. For the graph given in Figure 2.3, the set S ={u,z,w } is a minimum double monophonic set, and S ′ ={u,z,y, x, w } is a minimum connected double monophonic set so that dm(G) = 3

57


and dmc (G) = 5. u

r

v

r

r

y

r

z

r

r

x

w

Figure 2.3 Theorem 2.7 Let G be a connected graph with a cut-vertex v. Then every connected double monophonic set of G contains at least one vertex from each component of G – v. Proof This follows from Theorem 1.2.

2

Theorem 2.8 Every cut-vertex of a connected graph G belongs to each connected double monophonic set of G. Proof Let v be any cut-vertex of G and let G1 ,G2 ,...,Gr (r ≥2) be the components of G−{v }. Let S be any connected double monophonic set of G. Then by Theorem 2.7, S contains at least one vertex from each Gi (1 ≤ i ≤ r ). Since G[S ] is connected, it follows that v ∈ S. 2 Corollary 2.9 For any non-trivial tree T of order n, dmc (T) = n. Proof This follows from Theorems 1.1 and 2.8.

2

Theorem 2.10 Let G be a connected graph of order n ≥ 2. Then dmc (G) = 2 if and only if G = K2 . Proof If G = K 2 , then by Result 2.4 dm c (G) = 2. Conversely, let dm c (G) = 2. Let S = {u, v } be a minimum connected double monophonic set of G. Then uv is an edge. If G 6= K 2 , then there exists a vertex w different from u and v, and hence no pair of vertices of G can lie on a u−v monophonic path. Hence S is not a dmc -set, which is a contradiction. Thus G = 2 K 2. Theorem 2.11 Let G be a connected graph. If every vertex of G is either a cut-vertex or an extreme vertex, then dmc (G)=n. Proof Let G be a connected graph with every vertex of G is either a cut-vertex or an extreme vertex. Then the result follows from Theorem 1.1 and Theorem 2.8. The converse of Theorem 2.11 need not be true. This is clear from the graph G given in Figure 2.2. In view of Theorem 2.6, we have the following realization result. 2 Theorem 2.12 If a, b and n are positive integers such that 2 ≤ a < b ≤ n, then there exists a connected graph G of order n with dm(G) = a and dmc (G) = b.

58


Proof This theorem is proved by considering 2 cases. Case 1. 2 ≤ a < b = n. Let G be any tree of order n with number of pendant vertices a. Then obviously dm(G) = a and by Corollary 2.9, dmc (G)= b. Case 2. 2 ≤ a < b < n. Construct a graph G order n by considering a path P b-a+2 : u 1 , u 2 , ..., u b-a+2 of length b-a+1, adding n-b+a-2 new vertices w 1 , w 2 , ..., w n−b , v 1 , v 2 , ..., v a−2 to P b-a+2 and joining each w i (1 ≤ i ≤ n - b) to both u 1 and u 3 , and also joining each v j (1 ≤ j ≤ a-2) to u b-a+1 . The graph G is shown in Figure 2.4. Now the set of all extreme vertices S = {u b-a+2 , v 1 , v 2 , ..., v a−2 } is not a double monophonic set of G. However, S ∪{u 1 } is a double monophonic set of G and it follows from Theorem 2.3 that dm(G)= a. By Theorems 2.3 and 2.8, every connected double monophonic set of G contains all extreme vertices and cut-vertices. Let S ′ = {u 3 , u 4 , ..., u b-a+1 } be the set of all cut-vertices of G. Now, S ∪S ′ is not a double monophonic set of G. It is easily verified that S ∪S ′ ∪{u 1 , u 2 } is a minimum connected double monophonic set of G so that dm c (G) = b. Thus the proof is complete. u1 r

u2 r

r

w1

u3 u4 r

r

r

r

r r

ub−a+1

r

ub−a+2

r

r

r

v1 v2

r

r

va−2

r

w2 r

wn−b Figure 2.4

§3. Connected Double Monophonic Numbers of Some Standard Graphs Theorem 3.1 (i) For the star G= K1,n−1 , dmc (G) = n. (ii) For the complete bipartite graph G = Kr,s (2≤r≤s), dmc (G)=r+1. (iii) For the cycle G = Cn (n≥4), dmc (G) = 3. (iv) For the wheel G = Wn (n≥5), dmc (G) = 3. (v) For the graph G = Kn - e, where e is an edge of G, dmc (G) = 3. Proof (i) This follows from Corollary 2.9. (ii) Let X and Y be the partite sets of K r,s . Let S be a double monophonic set of G. First we prove that X ⊆S or Y ⊆S. Otherwise, X *S and Y *S. This means that there exist vertices x ∈X, y∈Y such that x, y ∈S. / Since S is a double monophonic set, the vertices x, y lie on a monophonic path joining a pair of vertices of S. Since the monophonic paths have length either 1 or 2, it follows that the pair of vertices x, y lie only on monophonic paths of the type x -y, x -t,

59


and s-y for some t ∈X and s∈Y. Hence it is clear that x ∈S or y∈S. This is a contradiction to x, y ∈S. / Thus the claim that X ⊆S or Y ⊆S is proved. Also, it is easily seen that both X and Y are double monophonic sets and so dm(G)=min{r, s}=r. It is clear that S =X ∪{y} for any y∈Y is a minimum connected double monophonic set of G so that dm c (G)=r +1. (iii) Any set S of G consisting of three adjacent vertices is clearly a minimum connected double monophonic set so that dm c (G) = 3. (iv) Let Wn =C n−1 +K 1 , where C n−1 : v 1 , v 2 ,· · · ,v n−1 ,v 1 and K 1 ={v }. Let S be any set consisting of two non-adjacent vertices on the cycle C n−1 and the central vertex v. Then S is a minimum connected double monophonic set of G so that dm c (G) = 3. (v) Let e be the edge e = uv. Then u and v are the only extreme vertices of G and it is clear that S = {u, v } is a double monophonic set so that dm(G)= 2. Then S ′ = S ∪{w } for any w ∈S / is a minimum connected double monophonic set of G so that dm c (G)=3. 2 Theorem 3.2 Let G be the graph obtained from the cycle Cn (n≥ 4) by adding a pendant edge to Cn . Then dmc (G) = 4. Proof Let G be the graph in Figure 3.1 obtained from the cycle C n :v 1 ,v 2 ,...,vn ,v 1 by adding a pendant edge e=v 1 v at the vertex v 1 of Cn . Then it is clear that the set S ={v, v 3 } is a double monophonic set so that dm(G)=2. In fact, any set S ′ ={v, vi } with v i 6= v 1 , v 2 , v n is a double monophonic set. Also, it is clear that the set T ={v, v 1 , v 2 , v 3 } is a minimum connected double monophonic set so that dm c (G) = 4. v1 v2 v3

r

r

r

v

vn

r

q

s

q r

r

vi+1

r r

vi Figure 3.1 Theorem 3.3 Let G be the graph obtained from the wheel Wn (n≥5) by adding a pendant edge e at a vertex on Cn−1 . Then dm(G)=2 and dmc (G)=4. Proof Let Wn =C n−1 +K 1 , where C n−1 : v 1 , v 2 ,...,v n−1 ,v 1 and K 1 ={v }. Let G=W n +e(n≥4) with e=v 1 x. Then the set S ={x,v 3} is a minimum double monophonic set of G so that dm(G)=2. It can be easily verified that the set T ={x,v 1 ,v 2 ,v 3 } is a minimum connected double monophonic set of G and so dm c (G)=4. 2 Theorem 3.4 Let G be the graph obtained from the wheel Wn (n≥4) by adding a pendant edge at the central vertex v of Wn . Then dm(G)=n and dmc (G)=n+1.

60


Proof Let Wn =C n−1 +K 1 (n≥4) where C n−1 : v 1 , v 2 ,...,v n−1 ,v 1 and K 1 ={v }. Let G=W n +e(n≥4) with e=vx. Then it can be easily verified that the set S ={x,v 1 ,v 2 ,...,v n−1 } is the unique minimum double monophonic set of G so that dm(G)=n. Also, it is clear that the set T ={x,v 1 ,v 2 ,...,v n−1 ,v } is the unique minimum connected double monophonic set of G so that dm c (G)=n+1. 2 Theorem 3.5 For the graph G=Kn +e (n≥2), dm(G)=n and dmc (G)=n+1. Proof For the graph G=K n + e(n≥2) where e=v 1 x, the set S ={x,v 2,v 3 ,...,v n } is the unique minimum double monophonic set of G so that dm(G)=n. Also, the set of all vertices of G is the unique minimum connected double monophonic set of G so that dm c (G)= n+1. 2 Theorem 3.6 Let G be the graph obtained from the complete bipartite graph Kr,s (2≤r≤s) by adding a pendant edge to Kr,s . Then dmc (G)=r+2. Proof Let U ={u 1,u 2 ,...,u r } and W ={w 1 ,w 2 ,...,w s } be the partite sets of K r,s . Let e=u 1 x be a pendant edge. Then the set S ={x,u 2,...,u r } is a minimum double monophonic set of G so that dm(G)=r. Let S ′ =S ∪{u 1 ,w i }(1≤i≤s). It is clear that S ′ is a minimum connected double monophonic set of G so that dm c (G)=r +2. Similarly, if e = w1 x, it can be proved that dm c (G)=r +2. 2 References [1] F. Buckley and F. Harary, Distance in Graphs, Addison Wesley, Redwood City, CA, (1990). [2] G. Chartrand, F. Harary and P. Zhang, On the geodetic number of a graph, Networks 39 (2002), 1-6. [3] F. Harary, Graph Theory, Addison Wesley, USA(1969). [4] F. Harary, E. Loukakis and C. Tsouros, The geodetic number of a graph, Math. Comput. Modeling 17 (1993), 89 - 95. [5] A.P. Santhakumaran and T. Jebaraj, The double geodetic number of a graph, Discuss. Math. Graph Theory, 32(2012), 109-119. [6] A.P. Santhakumaran and T. Venkata Raghu, The double monophonic number of a graph, International Journal of Computational and Applied Mathematics,11(1)(2016), 21-26. [7] A.P. Santhakumaran, P. Titus and K. Ganesamoorthy, On the monophonic number of a graph, J. Appl. Math. Informatics, vol.32(2014), 255-266. [8] A.P. Santhakumaran and P. Titus, On the connected geodetic number of a graph, J. Combin. Math. Combin. Comput., 69 (2009), 219-229. [9] P. Titus and K. Ganesamoorthy, The connected monophonic number of a graph, Graphs and Combinatorics, 30(2014), 237-245.


Analytic Odd Mean Labeling of Square and H Graphs P. Jeyanthi and R.Gomathi Research Centre, Department of Mathematics Govindammal Aditanar College for Women, Tiruchendur 628215, Tamilnadu, India

Gee-Choon Lau Faculty of Computer and Mathematical Sciences Universiti Teknologi MARA (Segamat Campus) 85009, Johor, Malaysia) E-mail: [email protected], [email protected], [email protected]

Abstract: Let G = (V, E) be a graph with p vertices and q edges. A graph G is analytic odd mean if there exist an injective function f : V → {0, 1, 3, 5 . . . , 2q − 1} with an induce edge labeling f ∗ : E → Z such that for each edge uv with f (u) < f (v), l m 2 2  f (v) −(f (u)+1) , if f (u) 6= 0; 2 m f ∗ (uv) = l  f (v)2 , if f (u) = 0 2

is injective. We say that f is an analytic odd mean labeling of G. In this paper we prove that the square graph of Pn , Cn , Bn,n , H-graph and H ⊙ mK1 are analytic odd mean graphs.

Key Words: Mean labeling, analytic mean labeling, analytic odd mean labeling, analytic odd mean graph.

AMS(2010): 05C78. §1. Introduction Throughout this paper we consider only finite, simple and undirected graph G = (V, E) with p vertices and q edges and notations not defined here are used in the sense of Harary [1]. A graph labeling is an assignment of integers to the vertices or edges or both, subject to certain conditions. There are several types of labeling. An excellent survey of graph labeling is available in [2]. Somasundaram et al. introduced concept of mean labeling in [4]. A graph G is called a mean graph if there is an injective function f : Vm → {0, 1, 2, 3, · · · , q} with an induce edge l (v) labeling f ∗ : E → Z given by f ∗ (uv) = f (u)+f is injective. The notion of analytic mean 2 labeling was due to Tharmaraj et al. in [5]. A graph G is analytic mean graph if it admits a bijection f : V → 1, 2, . .m. , p − 1} such that the induced edge labeling f ∗ : E → Z l {0, f (u)2 −f (v)2 ∗ given by f (uv) = with f (u) > f (v) is injective. Motivated by the concept of 2 analytic mean labeling, we extend our study and introduced a new labeling called analytic odd mean labeling. A graph G is an analytic odd mean if there exist an injective function 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 25, 2018, Accepted April 04, 2018, Edited by S. Monikandan.

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P. Jeyanthi, R.Gomathi and Gee-Choon Lau

f : V → {0, 1, 3, 5, · · · , 2q − 1} with an induce edge labeling f ∗ : E → Z such that for each edge uv with f (u) < f (v), l m 2 2  f (v) −(f (u)+1) , if f (u) 6= 0; 2 f ∗ (uv) = l f (v)2 m  , if f (u) = 0 2

is injective. We say that f is an analytic odd mean labeling of G. We proved that path Pn , cycle Cn , complete graph Kn , complete bipartite graph Km,n , wheel graph Wn , flower graph F ln , ladder graph Ln , comb Pn ⊙ K1 , graph Ln ⊙ K1 and union of two cycles are analytic odd mean graphs in [2]. We use the following definitions in the subsequent section to prove the main results. Definition 1.1 Let u and v be (not necessarily distinct) vertices of a graph G. A u − v walk of G is a finite, alternating sequence u = u0 , e1 , u1 , e2 , · · · , en , un = v of vertices and edges beginning with vertex u and ending with vertex v such that ei = ui−1 ui , i = 1, 2, · · · , n. The number n is called the length of the walk. The walk is said to be open if u and v are distinct vertices; otherwise it is closed. A walk u0 , e1 , u1 , e2 , u2 , · · · , en , un is determined by the sequence u0 , u1 , u2 , · · · , un of its vertices and hence we specify this walk by (u0 , u1 , u2 , · · · , un ). A walk in which all the vertices are distinct is called a path. A closed walk (u0 , u1 , u2 , · · · , un = u0 ) in which u0 , u1 , u2 , · · · , un−1 are distinct is called a cycle. Pn denotes a path on n vertices and Cn denotes a cycle on n vertices. Definition 1.2 The n-bistar graph Bn,n is obtained from two copies of K1,n by joining the vertices of maximum degree by an edge. Definition 1.3 For a simple connected graph G the square of graph G is denoted by G2 and defined as the graph with the same vertex set as of G and two vertices are adjacent in G2 if they are at a distance 1 or 2 apart in G. Defintion 1.4 The H-graph of a path Pn , n ≥ 3 is obtained from two copies of Pn with vertices v1 , v2 , v3 , · · · vn and u1 , u2 , u3 , · · · un by joining the vertices v(n+1)/2 and u(n+1)/2 by an edge if n is odd and the vertices v(n/2)+1 and un/2 if n is even. Definition 1.5 The graph H ⊙ mK1 is obtained from the H-graph by attaching m pendent vertices at each ith vertex on the two paths on n vertices for 1 ≤ i ≤ n. §2. Main Results In this section we prove that square graph of Pn , Cn , Bn,n , H-graph and H ⊙ mK1 are analytic odd mean graphs. Theorem 2.1 The square graph of path Pn2 is an analytic odd mean graph when n ≥ 3. Proof Let the vertex set and edge set of square graph of path be V (Pn2 ) = {vi : 1 ≤ i ≤ n} and E(Pn2 ) = {vi vi+1 : 1 ≤ i ≤ n − 1} ∪ {vi vi+2 : 1 ≤ i ≤ n − 2} respectively. Hence |V (Pn )| = n and|E(G)| = 2n − 3.

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We define an injective map f : V ((Pn2 ) → {0, 1, 3, 5, · · · , 4n − 7} by f (v0 ) = 0 and f (vi ) = 4i − 3 for 1 ≤ i ≤ n − 1. The induced edge labeling f ∗ is defined as follows: f ∗ (v0 v1 ) = 1, f ∗ (v0 v2 ) = 13 f ∗ (vi vi+1 ) = 12i − 1 for 1 ≤ i ≤ n − 1. ∗ and f (vi vi+2 ) = 28i + 11 for 1 ≤ i ≤ n − 2. We observe that the edge labels of vi vi+1 are increased by 10,12,12,12,· · · and that of vi vi+2 are increased by 26,28,28,28,· · · as i increases. Clearly the edge labels are distinct and odd. Hence Pn2 admits an analytic odd mean labeling. 2 An analytic odd mean labeling of P82 is shown in Figure 1. 13

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Theorem 2.2 The square graph of cycle Cn2 is an analytic odd mean graph when n ≥ 4. Proof Let the vertex set and edge set of square graph of cycle be V (Cn2 ) = {vi : 1 ≤ i ≤ n} and E(Cn2 ) = {vi vi+1 : 1 ≤ i ≤ n − 1} ∪ {v1 vn } ∪ {vi vi+2 : 1 ≤ i ≤ n − 2} ∪ {vn v2 , vn−1 v1 } respectively. Hence |V (Cn )| = n and |E(Cn )| = 2n. We define an injective map f : V (Cn ) → {0, 1, 3, 5, . . . , 4n − 1} by f (vi ) = 4i − 3 for 1 ≤ i ≤ n. The induced edge labeling f ∗ is defined as follows: f ∗ (vn v1 ) = 8n2 − 12n + 3 f ∗ (vn v2 ) = 8n2 − 12n − 13 f ∗ (vn−1 v1 ) = 8n2 − 28n + 23 f ∗ (vi vi+1 ) = 12i − 1 for 1 ≤ i ≤ n − 1. and f ∗ (vi vi+2 ) = 28i + 11 for 1 ≤ i ≤ n − 2. We observe that the edge labels are distinct and odd. Hence Cn2 admits an analytic odd 2 mean labeling. An analytic odd mean labeling of C62 is shown in Figure 2. 203 143

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2 Theorem 2.3 The square graph of n-bistar Bn,n is an analytic odd mean graph. 2 Proof Let the vertex set and edge set of square graph of n-bistar be V (Bn,n ) = {u, v, ui , vi : 2 2 1 ≤ i ≤ n} and E(Bn,n ) = {uv, uui , vvi , uvi , vui : 1 ≤ i ≤ n} respectively. Hence |V (Bn,n )| = 2 2n + 2 and |E(Bn,n )| = 4n + 1. 2 We define an injective map f : V (Bn,n ) → {0, 1, 3, 5, · · · , 8n + 1} by f (u) = 0, f (v) = 1 f (ui ) = 2i + 1 for 1 ≤ i ≤ n and f (vi ) = 2n + 2i + 1 for 1 ≤ i ≤ n. The induced edge labeling f ∗ is defined as follows: f ∗ (uv) = 1 f ∗ (uui ) = 2i2 + 2i + 1 for 1 ≤ i ≤ n f ∗ (vvi ) = 2(n + i)2 + 2(n + i) − 1 for 1 ≤ i ≤ n f ∗ (uvi ) = 2(n + i)2 + 2(n + i) + 1 for 1 ≤ i ≤ n and f ∗ (vui ) = 2i2 + 2i − 1 for 1 ≤ i ≤ n. We observe that the edge labels of uui are increased by 4i and that of vvi are increased by 4(n + i) as i increases from 2 to n. The edge labels of uui and uvi are differ by 2 and that of vvi and vui are differ by 2 for each i. Clearly the edge labels are odd and distinct. Hence the 2 square graph of n-bistar Bn,n admits an analytic odd mean labeling. 2 2 An analytic odd mean labeling of square graph of n-bistar B6,6 is shown in Figure 3.

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Figure 3 Theorem 2.4 The H-graph of a path Pn (n ≥ 3) is an analytic odd mean graph. Proof Let G be the H-graph. Let the vertex set and edge set of H-graph be V (G) = {vi , ui : 1 ≤ i ≤ n} and E(G) = {vi vi+1 : 1 ≤ i ≤ n − 1} ∪ {ui ui+1 : 1 ≤ i ≤ n − 1} ∪ {v(n+1)/2 u(n+1)/2 if n is odd} ∪ {v(n/2)+1 un/2 if n is even} respectively. Hence |V (G)| = 2n and

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|E(G)| = 2n − 1. We define an injective map f : V (G) → {0, 1, 3, 5, · · · , 4n − 3} by f (v1 ) = 0, f (vi ) = 2i − 3 for 2 ≤ i ≤ n and f (ui ) = 2n + 2i − 3 for 1 ≤ i ≤ n. The induced edge labeling f ∗ is defined as follows: f ∗ (vi vi+1 ) = 2i − 1 for 1 ≤ i ≤ n f ∗ (ui ui+1 ) = 2n + 2i − 1 for 1 ≤ i ≤ n − 1 f ∗ (v(n+1)/2 u(n+1)/2 ) = 4n2 − 5n + 2 if n is odd. and f ∗ (v(n/2)+1 un/2 ) = 4n2 − 9n + 5 if n is even.

We observe that the edge labels are odd and distinct. Hence the H-graph admits an analytic odd mean labeling. 2 An analytic odd mean labeling of H-graph with n = 5 and n = 6 are shown in Figure 4.

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Figure 4 Theorem 2.5 The graph H ⊙ mK1 is an analytic odd mean graph. Proof Let G = H⊙mK1 . Let the vertex set and edge set of G be V (G) = {vi , ui , vij , uji : 1 ≤ i ≤ n and 1 ≤ j ≤ m} and E(G) = {vi vi+1 : 1 ≤ i ≤ n − 1} ∪ {ui ui+1 : 1 ≤ i ≤ n − 1} ∪ {v(n+1)/2 u(n+1)/2 if n is odd} ∪ {v(n/2)+1 un/2 if n is even} ∪ {vi vij , ui uji : 1 ≤ i ≤ n and 1 ≤ j ≤ m} respectively. Hence |V (G)| = 2n(m + 1) and |E(G)| = 2n(m + 1) − 1. We define an injective map f : V (G) → {0, 1, 3, 5, · · · , 4n(m + 1) − 3} by f (v1 ) = 0, f (vi ) = 2i − 3 for 2 ≤ i ≤ n f (ui ) = 2n + 2i − 3 for 1 ≤ i ≤ n f (vij ) = 4n − 3 + 2m(i − 1) + 2j for 1 ≤ i ≤ n and 1 ≤ j ≤ m and f (uji ) = 4n − 3 + 2m(n + i − 1) + 2j for 1 ≤ i ≤ n and 1 ≤ j ≤ m. The induced edge labeling f ∗ is defined as follows: f ∗ (vi vi+1 ) = 2i − 1 for 1 ≤ i ≤ n − 1, f ∗ (ui ui+1 ) = 2n + 2i − 1 for 1 ≤ i ≤ n − 1,

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f ∗ (v(n+1)/2 u(n+1)/2 ) = 4n2 − 5n + 2 if n is odd, f ∗ (vn/2+1 un/2 ) = 4n2 − 9n + 5 if n is even, f (vi vij ) = [(2(2n + j − i) + 2m(i − 1) − 1) (2(2n + j + i) + 2m(i − 1) − 5) + 1] /2, for 1 ≤ i ≤ n and 1 ≤ j ≤ m and f (ui uji ) = [(2(n + j − i) + 2m(n + i − 1) − 1) (2(3n + j + i) + 2m(n + i − 1) − 5) + 1] /2 for 1 ≤ i ≤ n and 1 ≤ j ≤ m. Clearly, all the edge labels are distinct and odd. Hence the graph H ⊙ mK1 admits an analytic odd mean labeling. 2 An analytic odd mean labeling of H ⊙ mK1 with n, m = 3 and n, m = 4 are shown in Figure 5 and Figure 6 respectively. 11 13

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References [1] F.Harary, Graph Theory,Addison-Wesley,Reading,Massachusetts,1972. [2] P.Jeyanthi,R.Gomathi and GeeChoon Lau, Analytic odd mean labeling of some standard graphs, Proceedings ofthe ICADM 2018 (ISSN-2348 - 6600), 62 - 68. [3] Joseph A. Gallian, A Dynamic Survey of Graph Labeling, The Electronic Journal of Combinatorics, (2016), # DS6. [4] M.Kannan, R.Vikrama Prasad and R.Gopi, Even Vertex Odd Mean Labeling of H-graph, International Journal of Mathematical Archive, 8(8)(2017), 162 -167. [5] S.Somasundaram and R. Ponraj,Mean labelings of graphs,National Academy Science Letter,26 (2003),210-213. [6] T.Tharmaraj and P.B.Sarasija, Analytic mean labelled graphs, International Journal of Mathematical Archive, 5(6)(2014), 136 -146.


Learning Some Special Graphs Using Hyperedge Replacement Emerald Princess Sheela J.D and Thanga Murugeshwari.V (Research Department of Mathematics, Queen Mary’s College,Chennai,India.) E-mail: [email protected], [email protected]

Abstract: In this paper, some special graphs namely Triangular Snakes, Quadrilateral Snakes, Double Triangular Snakes and Double Quadrilateral Snakes are generated using hyperedge replacement graph grammars and a learning algorithm is developed to learn these special graphs. This paper is the impact of Jeltsch and Kreowski work on hyperedge replacement [4]. The learning algorithm to learn these special graphs gives the exact decomposition of these graphs from which the production rules are identified.

Key Words: Graph Grammars, Hyperedge replacement,Triangular Snakes, Quadrilateral Snakes, Double Triangular Snakes and Double Quadrilateral Snakes.

AMS(2010): 68Q42, 68R10, 97R40 §1. Introduction Graph grammar is used as a synonym for graph rewriting system, especially in the context of languages. The main component of a graph grammar is a finite set of productions; a production is, in general, a triple (M, D, E) where M and D are graphs (the “mother” and “daughter” graph, respectively) and E is some embedding mechanism. Such a production can be applied to a (“host”) graph H whenever there is an occurrence of M in H. It is applied by removing this occurrence of M from H, replacing it by (an isomorphic copy of) D, and finally using the embedding mechanism E to attach D to the remainder of H [1]. The question of learning automata and grammar from strings was dealt with by researchers. D.Angulin,[3] used language learning as a central problem for computational learning theory which led to defining and studying different learning paradigms as learning with the help of oracles. Thus, the field of grammatical inference (also known as grammar induction) is transversal to a number of research areas including machine learning, formal language theory, syntactic and structural pattern recognition, computational linguistics, computational biology and speech recognition [2]. The Snake graphs have applications in Combinatorics, Networking, Block designs, CAD and Bioinformatics. In this paper, we generate Triangular Snakes, Quadrilateral Snakes, Double Triangular Snakes and Double Quadrilateral Snakes using hyperedge replacement graph grammars and give a learning algorithm to learn the above graphs from a given set of sample graphs. 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 25, 2018, Accepted May 19, 2018, Edited by R. Kala.

Learning Some Special Graphs Using Hyperedge Replacement

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§2. Preliminaries Let N denote the set of all natural numbers. For a finite set of alphabets A, A∗ denote the set of all strings over A including the empty string, For w ∈ A∗ , |w| denotes the length of w. Definition 2.1([1]) Let C be an arbitrary but finite set of labels and let type:C → N be a typing function. A hypergraph H over C is a tuple (V, E, att, lab, ext) Where V is a finite set of nodes, E is a finite set of hyper edges, att : E → V ∗ is a mapping assigning a sequence of pair wise distinct attachment node att(e) to each e ∈ E, lab : E → C is a mapping that labels each hyperedge such that type (lab(e)) = |att(e)|, ext ∈ V ∗ is a sequence of pair wise distinct external nodes. The Class of all hyper graphs over C is denoted by Hc . The components of a hypergraph H may be denoted by VH , EH , attH , labH , extH respectively. Definition 2.2([1]) A hyperedge replacement grammar is a system HRG = (N, T, P, S), Where N ⊆ C is a set of non-terminals. T ⊆ C with T ∩ N = ∅ is a set of terminals, P is a finite set of productions. A production over N is an ordered pair P = (A, R) with A ∈ N, R ∈ Hc and type(A) = type(R). A is called the left-hand side of P and is denoted by lhs(P ). R is called the right-hand side and is denoted rhs(P ). S ∈ N is a start symbol. We denote the class of all hyperedge replacement grammars by HRG. Definition 2.3 A m-hypergraph with m nodes that is all nodes are external and a single hyperedge e is said to be handle if aatH (e) = extH . If labH (e) = A, then H is said to be handle induced by A and is denoted by A0 . Definition 2.4([1]) The hypergraph language L(HRG) generated by HRG is Ls (HRG) where for all A ∈ N. LA (HRG) consists of all hypergraphs in HT derivable from A◦ by applying ∗ productions of P : LA (HRG) = {H ∈ HT /A◦ →P H} We denote the class of all hyperedge replacement grammars by HRL. Definition 2.5([5]) A Triangular snake Tn is obtained from a path u1 u2 · · · un by joining ui and ui+1 to a new vertex vi for 1 ≤ i ≤ n − 1. That is, every edge of a path is replaced by a triangle C3 . Definition 2.6([5]) A Double Triangular snake D(Tn ) consists of two Triangular snakes that have a common path. sDefinition 2.7([5]) A quadrilateral snake Qn is obtained from a path u1 u2 · · · un by joining ui and ui+1 to new vertices vi , wi , 1 ≤ i ≤ n−1. That is, every edge of a path is replaced by cycle C4 . Definition 2.8([5]) A double Quadrilateral snake D(Qn ) consists of two Quadrilateral snakes that have a common path.

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Emerald Princess Sheela J.D and Thanga Murugeshwari.V

§3. Generation of Some Special Graphs Using HRG 3.1 Generation of Triangular Snakes Using HRG Consider the hyperedge replacement grammar HRG = {{S, D}, {P1, P2 }, S} where the axiom and the productions are given in Figure 1.

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The Production rules contains a start symbol S and hyperedge D. If P1 is applied, the D−edge is replaced by another path of length two which again contains the hyperedge D. If P2 is applied the D edge is replaced by C3 . The following figure illustrates the generation of Triangular snakes from the above production rules. Then, L(HRG) = {Tn /n ≥ 2}

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3.2 Generation of Quadrilateral Snakes Using HRG Consider the hyperedge replacement grammar HRG = {{S, D}, {P1 , P2 }, S} where the axiom and the productions are given in Figure 3.

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Figure 3 The Production rules contains a start symbol S and hyperedge D. If P1 is applied, the D−edge is replaced by another path of length two which again contains the hyperedge D. If P2 is applied the D edge is replaced by C4 . The following figure illustrates the generation of Quadrilateral snakes from the above production rules. Then, L(HRG) = {Qn /n ≥ 2} S=

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Figure 4 3.3 Generation of Double Triangular Snakes Using HRG Consider the hyperedge replacement grammar HRG = {{S, D}, {P1 , P2 }, S} where the axiom and the productions are given in figure 5. S=

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Figure 5 The Production rules contains a start symbol S and hyperedge D.If P1 is applied, the D−edge is replaced by another path of length two which again contains the hyperedge D. If P2 is applied the D−edge is replaced by a Diamond Graph.The following figure illustrates the generation of Double Triangular snakes from the above production rules. Then, L(HRG) = {D(Tn )/n ≥ 2}.

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3.4 Generation of Double Quadrilateral Snakes Using HRG Consider the hyperedge replacement grammar HRG = {{S, D}, {P1 , P2 }, S} where the axiom and the productions are given in the following figures

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Figure 7 The Production rules contains a start symbol S and hyperedge D. If P1 is applied, the D−edge is replaced by another path of length two which again contains the hyperedge D. If P2 is applied the D−edge is replaced by a hexagon with vertices v1 and v4 are adjacent. The following figure illustrates the generation of Double Quadrilateral snakes from the above production rules. Then, L(HRG) = {D(Qn )/n ≥ 2} S=

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§4 Learning Algorithm for Special Graphs Using HRG. 4.1 A Learning Algorithm to Learn Triangular Snake Graphs. INPUT: A positive presentation G1 , G2 , ..., Gk where Gi ∈ Tn , n ≥ 2. OUTPUT: A Sequence of Production rules that generate input samples and the class Tn , n ≥ 2. PROCEDURE: Let G1 , G2 , · · · , Gk be the triangular snakes which are taken as Input Samples. Let Grammar, Gr = {{S}, C3, (S, Gi )/i = 1 to k, where k is the natural number (S, 0)◦ } NewProd represents the set of new productions that are identified. Initialize N ewprod = ∅ Initialize REN AM E N T = {S} Let i = 1 Begin For each Gi do Begin DECOMPOSE (Gi ) Introduce a new non-terminal Dj as a hyperedge of type 2, not occurring before, where j is a natural number such that (S, G1i ) ∪ (Dj , G2i ) = (S, Gi ) where j = 1, 2 DECOMPOSE (Gi ) = (S, G1i ) ∪ (Dj , G2i ) − (S, Gi ). N ewprod = N ewprod ∪ DECOM P OSE(Gi ) Again, Decompose G2i . Introduce a new non-terminal Dj+1 as a hyperedge of type 2 between adjacent vertices in path of length n, not occurring before such that (Dj , G3i ) ∪ (Dj+1 , G4i ) and G4i must be C3 . DECOMPOSE (G2i ) = (Dj , G3i ) ∪ (Dj+1 , G4i ) − (Dj , G2i ). N ewprod = N ewprod ∪ DECOM P OSE(G2i ) End DECOMPOSE (Gi ) Gr = Gr ∪ N ewprod End REN AM E N T (Dj ) = {S = S, D1 = D2 = D3 = D}. This function is similar to that in RENAME operation of Jelstch and Kreowski [4]. Gr = Gr ∪ N ewprod ∪ REN AM E N T REDUCE is where Repeated and redundant productions are identified and removed [4]. Gr = Gr − REDU CE Gr = {S, D, C3 , N ewP rod, S} is the grammar obtained from the above algorithm 4.2 Learning Tn , n ≥ 2 Using the Above Algorithm Consider the input graphs {G1 , G2 , G3 } Gr = {{S}, C3, (S, Gi )/i = 1 to 3, (S, 0)◦ } Initialize N ewprod = ∅ Initialize REN AM E N T = {S}

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and (D5 , G33 ) ∪ (D6 , G43 ). DECOM P OSE(G23 ) = (D5 , G23 ) ∪ (D6 , G43 ) − (D5 , G23 ). N ewprod = N ewprod ∪ DECOM P OSE(G23 ) Where G13 , G23 , G33 , G43 are shown in Figure 12. b

G13

:

b

G23

b

D5

:

b

b

b b

b

b

b

G33 :

b

b

D6

b

D6

D6

G43 :

b

b

b

Figure 12 Then REN AM E N T (Dj ) = {S = S, D1 = D2 = D3 = D4 = D5 = D6 = D} is given in Figure 13, 14 and 15. b

G11

:

b

G21

b

D

:

b

b

b b

b

b b

b

b

G31 :

b

b

D

b

D

b

D

D

b

G41 :

b

b

Figure 13 b

G12

:

b

G22

b

D

:

b

b b

b

b

G32 :

b

b

D

G42

b

D

:

b

b

Figure 14 b

G13

:

b

G23 :

b

D

b

b

b b

b

b

b

G33 :

b

D

b

D

b

D

b

G43 :

b

b

Figure 15 Gr = Gr ∪ N ewprod ∪ REN AM E N T. Repeated and Redundant productions are identified and removed using REDUCE. The rules will be repeated when RENAME NT is applied to the non-terminals. Hence removing the REDU CE productions derives the sample inputs and the class of all Triangular Snakes. Gr = Gr − REDU CE. Applying the above algorithm, the required production rules are obtained in Figure 16.

76


S=

P1 =

b

D

b

D

b

b

D

b

b

P2 = b

b

Figure 16 In general, Gri where r = 1, 2, 3, 4 is a factor got from the decomposition of Gi . In the above algorithm if the input graphs of Tn , n ≥ 2 is replaced by graphs of Qn , n ≥ 2 or graphs of D(Tn ), n ≥ 2 or graphs of D(Qn ), n ≥ 2 and if C3 is replaced by C4 or Diamond Graph or a Hexagon graph with vertices v1 and v4 are adjacent respectively, Quadrilateral snakes, Double Triangular Snakes and Double Quadrilateral Snakes can be learned respectively. 4.3 Correctness of the Algorithm [4] (i) Each sample can be derived from the axiom of an inferred grammar. (ii) Each production either being an initial one or one obtained by decompositions and renaming can be used for deriving one of the samples. (iii) The axiom of an inferred grammar is (S, 0)◦ or some renaming of it because the axiom has this form initially and the RENAME operation is the only one affecting the axiom. A grammar with the above properties is called samples composing. Hence our grammar is samples composing as it satisfies the above conditions. Since the HRG of Triangular Snakes Quadrilateral Snakes, Double Triangular Snakes and Double Quadrilateral Snakes are subclasses of the class of hyperedge replacement grammars in [4] it is decidable, that Triangular Snakes, Quadrilateral Snakes, Double Triangular Snakes and Double Quadrilateral Snakes can be learned. §5. Conclusion We have given an algorithm for learning Triangular Snakes, Quadrilateral Snakes, Double Triangular Snakes and Double Quadrilateral Snakes using Hyperedge Replacement Graph grammars. It is possible to extend it further for varied classes of graphs and graph grammars. References [1] G.Rozenberg, Handbook of Graph Grammars and Computing by Graph Transformation, World Scientific (1997).


77

[2] Colin de la Higuera, Current Trends in Grammatical Inference, Lecture Notes in Computer Science, 1876,28-31(2000). [3] D.Angulin, On the Complexity of Minimum Inference of Regular Sets, Information and Control, 39,337-350 (1978). [4] E.Jeltsch, H. Kreowski, Grammatical Inference based on Hyperedge Replacement, Lecture Notes in Computer Science, 32, 461-474(1990). [5] J.A. Gallian, A dynamic survey of graph labelling, The Electronic Journal of Combinatorics, 16 (2013) DS6.


Parikh Matrices of Binary and Ternary Words Under Prouhet Morphism K.G. Subramanian (Honorary Visiting Professor)

Atulya Nagar Department of Mathematics and Computer Science Faculty of Science, Liverpool Hope University, Liverpool, L16 9JD, UK

Sastha Sriram Department of Science and Humanities (Mathematics) Saveetha Institute of Medical and Technical Sciences, Chennai 602 105, India E-mail: [email protected], [email protected], [email protected]

Abstract: Properties of words that are finite sequences of symbols from a finite set, constitute an interesting topic of investigation in the field of “combinatorics on words”. Parikh matrix of a word w over an ordered alphabet is a relatively new notion introduced by Mateescu et al. (2000). The entries above the main diagonal in this upper triangular Parikh matrix provide counts of certain subwords in the word w that are subsequences of w. On the other hand several morphisms on words have also been explored for their properties. Here we consider a morphism, called Prouhet morphism and establish several properties of images of words under this morphism in the context of subwords and Parikh matrices.

Key Words: Combinatorics on words, morphism, subword, Parikh matrix. AMS(2010): 68R § 1. Introduction The pioneering work of Axel Thue [5] in the early years of twentieth century on combinatorial problems on words, can be considered as the basis for the beginning of the field of combinatorics on words. There has been an enormous growth in this field in the recent past with researchers establishing a number of interesting results on combinatorial properties of words [7] with motivation arising from different fields. A finite word or simply called a word, is a finite sequence of symbols belonging to a finite set, called an alphabet and a subword of a word α is a subsequence of α. The concept of Parikh matrix of a word, introduced by Mateescu et al. [9], is a recent research topic in this field. The 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 25, 2018, Accepted June 05, 2018, Edited by R. Kala.

Parikh Matrices of Binary and Ternary Words Under Prouhet Morphism

79

entries above the main diagonal in the Parikh matrix of a word α, which is an upper triangular matrix with ones in the diagonal, are numbers that provide the counts of certain subwords of the word α. In fact the entries in the diagonal above the main diagonal of the Parikh matrix of a word α over an alphabet Σ, give the number of each of the symbols of Σ occurring in α and is thus an extension of the classical Parikh vector, an important notion extensively used, especially in formal language theory[10]. Two words may have different Parikh matrices although they have the same Parikh vector. Yet the Parikh matrix is also not injective. With the introduction of the notion of Parikh matrix, a number of studies on various properties related to the Parikh matrix, such as injectivity, have been done. In particular, properties of subwords and Parikh matrices of image words under certain mappings, called morphisms on words [6] have been recently established. Here we consider Prouhet morphism [11] which is a natural generalization of the well-known Thue-Morse morphism and derive several properties of subwords and Parikh matrices of words over an ordered binary or ternary alphabet.

§2. Preliminaries Certain basic notions and known results are recalled based on [7,9] in this section. An ordered alphabet is an alphabet, which is a finite set of symbols, with an ordering on its elements. For example, {a < b < c} is an ordered alphabet, with an ordering a < b < c on its symbols. A scattered subword or simply called a subword, is a subsequence of a given word α, which itself is a finite sequence of symbols belonging to an alphabet. The number of subwords β in a given word α is denoted by |α|β . As an example, for the word α = acbbcacaba over {a < b < c}, the number of subwords β = aca in α is |α|β = 10. The Parikh vector [10] of a word α gives the number of occurrences (including repetition) in the word α, of each of the symbols of the alphabet. The Parikh vector of the word acbbcacaba over the ordered alphabet {a < b < c}, is the vector (4, 3, 3). An extension of the notion of Parikh vector is the Parikh matrix [9] of a word. For a word w over an ordered alphabet Σ, the Parikh matrix of w is a triangular matrix, with 1′ s on the main diagonal and 0′ s below it but the entries above the main diagonal provide information on the number of certain subwords in w. We now recall the definition of a Parikh matrix mapping [9]. A triangular matrix is a square matrix M = (mi,j )1≤i,j≤n , such that mi,j are non-negative integers for all 1 ≤ i, j ≤ n, mi,j = 0, for all 1 ≤ j < i ≤ n, and mi,i = 1, for all 1 ≤ i ≤ n. The set Mn of all triangular matrices of dimension n ≥ 1 is a monoid with respect to multiplication of matrices. Definition 2.1 ([9]) Let Σk = {a1 < a2 < · · · < ak } be an ordered alphabet. The Parikh matrix mapping is the monoid morphism Ψk : Σ∗k → Mk+1 , defined by the condition: Ψk (λ) = Ik+1 , the (k + 1) × (k + 1) unit matrix, and if Ψk (aq ) = (mi,j )1≤i,j≤(k+1) , then for each 1 ≤ i ≤ (k + 1), mi,i = 1, mq,q+1 = 1, all other elements of the matrix Ψk (aq ) are 0, 1 ≤ q ≤ k. For a word w = ai1 ai2 · · · aim , aij ∈ Σk for 1 ≤ j ≤ m, we have Ψk (w) = Ψk (ai1 )Ψk (ai2 ) · · · Ψk (aim ).

80

K.G. Subramanian, Atulya Nagar and Sastha Sriram

In other words Ψk (w) is computed by multiplication of matrices and the triangular matrix Ψk (w) is called the Parikh matrix of w. Remark 2.2 Note that it has been proved in [9] that in the matrix Ψk (w) = (mij )1≤i,j≤(k+1) , mi,j+1 = |w|ai ai+1 ···aj for all 1 ≤ i, j ≤ (k + 1) which is the number of subwords ai ai+1 · · · aj in the word w. We are concerned here mainly with binary or ternary alphabet. We illustrate the notion of Parikh matrix of a word over these alphabets. Let Σ2 = {a < b} and Σ3 = {a < b < c}. The Parikh matrix Ψ2 (α) of a word α = bababba over Σ2 is computed as follows: Since 

1

 Ψ2 (a) =   0 0

1 0





1

   1 0   , Ψ2 (b) =  0 0 1 0

0 0



 1 1  , 0 1

the Parikh matrix Ψ2 (α) is given by 

1 3

 Ψ2 (α) = Ψ2 (b)Ψ2 (a)Ψ2 (b)Ψ2 (a)Ψ2 (b)Ψ2 (b)Ψ2 (a) =   0 1 0 0

5



 4  . 1

by matrix multiplication. Note that |α|a = 3, |α|b = 4, |α|ab = 5. Two words u, v over an ordered alphabet are said to be M −equivalent [9], if their Parikh matrices are the same. Such a word u or v is called M −ambiguous. We note that the words u = abbaaba, v = baababa over {a < b}   1 4 5    have the same Parikh matrix Ψ2 (u) = Ψ2 (v) =   0 1 3  and hence are M −equivalent. 0 0 1 Also, both u and v are M −ambiguous.

We next recall the weak-ratio property considered in [12]. Two words u, v over {a < b < c} are said to satisfy the weak-ratio property, written u ∼wr v, if there is a rational constant k > 0, such that |u|a = k|v|a , |u|b = k|v|b and |u|c = k|v|c . If the alphabet is {a, b}, u ∼wr v, if there is a rational constant k > 0, such that |u|a = k|v|a and |u|b = k|v|b . We finally recall the notion of a morphism [7] on words as a mapping f : Σ∗ → Γ∗ , where Σ and Γ are two alphabets, such that f (uv) = f (u)f (v), for words u, v over Σ. It is shown in [2] that the weak ratio property is preserved by morphisms. Lemma 2.3 Let Σ, Γ be two finite ordered alphabets. Let α, β be two words over Σ in weak ratio property and f : Σ∗ −→ Γ∗ be a morphism. Then f (α) ∼wr f (β). § 3. Properties of Image Words Under Prouhet Morphism Prouhet morphism (see, for example, [11]) is a natural generalization of the well-known Thue morphism. We recall the definition of Prouhet morphism restricting to a 3-symbol alphabet.

81


Definition 3.1 ([11]) Let Σ3 = {a, b, c}. The Prouhet morphism p : Σ∗3 7→ Σ∗3 is defined by p(a) = abc, p(b) = bca, p(c) = cab.

Certain basic properties relating to subwords of images of words under p are known [13]. We recall these results [13] and for completeness recall also the proofs [13] of two of these results. We also need the following results (Lemmas 3.2 and 3.3) from [2]. Lemma 3.2 Let Γ1 , Γ2 be two finite ordered alphabets and f : Γ∗1 −→ Γ∗2 be a morphism. Then, for all w ∈ Γ∗1 , and for all a ∈ Γ2 , we have |f (w)|a =

X

r∈Γ1

|w|r · |f (r)|a .

Lemma 3.3 ([2]) Let Σ2 = {a < b}, Σ3 = {a < b < c} be two alphabets and f : Σ∗3 −→ Σ∗3 be a morphism. For x, y ∈ Σ3 , w ∈ Σ+ 2 , we have |f (w)|xy =

X

r∈Σ2

|w|r |f (r)|xy +

X

r,t∈Σ2

|w|rt |f (r)|x |f (t)|y .

Theorem 3.4 ([13]) Let Σ2 = {a < b} and Σ3 = {a < b < c}. For a word w over Σ2 , we have (i) |p(w)|a = |p(w)|b = |p(w)|c = |w| (ii) |p(w)|ab = 12 |w|(|w| + 1) − |w|b ; |p(w)|bc = 21 |w|(|w| + 1) (iii) |p(w)|abc = |w||w|a + 12 |w|b (|w|b − 1) + |w|ab + |w|aaa + |w|aab + |w|abb + |w|aba + |w|baa + |w|bab + |w|bba + |w|bbb where p : Σ∗3 7→ Σ∗3 is the Prouhet morphism. 

1 p

r



   Proof Let w ∈ Σ∗2 . Let the Parikh matrix of the word w be   0 1 q . 0 0 1 We make use of Lemma 3.3 and the known identity |w|ab + |w|ba = |w|a × |w|b , in deriving the formulae in the statement of the theorem. We have |p(w)|ab

=

=

|w|a |p(a)|ab + |w|b |p(b)|ab + |w|aa |p(a)|a |p(a)|b + |w|ab |p(a)|a |p(b)|b +|w|ba |p(b)|a |p(a)|b + |w|bb |p(b)|a |p(b)|b p(p − 1) q(q − 1) (p + q)2 + p − q |w|(|w| + 1) p+ + pq + = = − |w|b 2 2 2 2

since |w|a = p, |w|b = q, |p(a)|ab = 1, |p(b)|ab = 0, |p(a)|a = |p(b)|a = |p(a)|b = |p(b)|b = 1, and |w|aa = p(p−1) , |w|bb = q(q−1) . Note that |w| = |w|a + |w|b = p + q. This proves statement (ii). 2 2 The statement (iii) can be proved in a similar manner.

82


We now prove statement (iv). When we apply p on w, every a ∈ w yields abc in p(w). The subword aa in w also yields the subword abc in p(w) (with either a arising from the first a in aa and bc arising from the second a or ab arising from the first a and c arising from the second a). Likewise, the subword abc arises from ab, ba, bb. Also, the subword aaa in w yields the subword abc in p(w) (with a from the first a in aaa, b from the second a and c from the third a). Likewise,the subwords aab, aba, abb, baa, bab, bba, bbb in w yield the subword abc in p(w). Hence |p(w)|abc

=

=

|w|a + 2|w|aa + 2|w|ab + |w|ba + |w|bb + |w|aaa + |w|aab +|w|aba + |w|abb + |w|baa + |w|bab + |w|bba + |w|bbb 1 1 |w|a + 2 × (|w|a (|w|a − 1)) + |w|a |w|b + |w|ab + (|w|b (|w|b − 1)) + η 2 2

where η = |w|aaa + |w|aab + |w|aba + |w|abb + |w|baa + |w|bab + |w|bba + |w|bbb . Thus 1 1 |p(w)|abc = p + p(p − 1) + pq + |w|ab + q(q − 1) + η = p(p + q) + |w|ab + q(q − 1) + η. 2 2

2

This proves the statement (iv) as |w| = p + q. We provide an example following.

Example 3.5 Let w = abaab over the alphabet Σ2 = {a < b} so that p(w) = abcbcaabcabcbca. Here |w| = 5, |w|a = 3, |w|b = 2, |w|ab = 4. Also |w|aaa = 1, |w|aab = 3, |w|aba = 2, |w|abb = 1, w|baa = 1, |w|bab = 2, |w|bba = 0, |w|bbb = 0. Using Theorem 3.4, we have, |p(w)|a = |p(w)|b = |p(w)|c = 5; |p(w)|ab = 12 (|w|(|w| + 1)) − |w|b = 13, |p(w)|bc = 12 (|w|(|w| + 1)) = 15. Likewise, |p(w)|abc = 30. The Parikh matrix of p(w) = abcbcaabcabcbca is 

1

  0    0  0

5 13 30 1

5

0

1

0

0



 15   . 5   1

We now consider two sets of certain special words so that corresponding words in the two sets are M −equivalent and show (Theorem 3.7) that the Prouhet morphism retains M −equivalence. We first recall a needed known result in the following Lemma 3.6. Lemma 3.6 ([1]) Let Σ2 = {a < b} and Σ3 = {a < b < c}. Let φ : Σ∗2 7→ Σ∗3 be a morphism. Let α, β be any two M −equivalent words over Σ2 . Then 

1 0

  0 1  Ψ3 (φ(α)) − Ψ3 (φ(β)) =   0 0  0 0

0 n 0 1 0



 0    0   1


83

where n = |φ(α)|abc − |φ(β)|abc is an integer. Theorem 3.7 Let Σ2 = {a < b} and Σ3 = {a < b < c}. Let S1 = {uvδvu | δ ∈ Σ∗2 } and S2 = {vuδuv | δ ∈ Σ∗2 } where the words u, v over Σ2 = {a < b} satisfy the weak-ratio property, i.e u ∼wr v. Then the word p(uvδvu) is M −equivalent to the corresponding word p(uvδvu) where p : Σ∗3 7→ Σ∗3 is the Prouhet morphism. Proof Since u ∼wr v, let |v|a = k · |u|a , |v|b = k · |u|b where k, (k > 0) is a constant rational number. The words x = uvδvu and y = vuδuv are M −equivalent with the Parikh matrix 

1

  0  0

α 1 0

γ



 β  . 1

where α = 2(|u|a + |v|a ) + |δ|a , β = 2(|u|b + |v|b ) + |δ|b γ = |uv|ab + |vu|ab + (k + 1)2 |u|a |u|b + (k + 1)|u|a |δ|b + (k + 1)|u|b |δ|a + |δ|ab Then p(x) and p(y) are M −equivalent if Ψ3 (p(x)) = Ψ3 (p(y)), which holds, by Lemma 3.6, if |p(x)|abc − |p(y)|abc = 0. Now p(x) = p(uv)p(δ)p(vu), p(y) = p(vu)p(δ)p(uv) so that |p(x)|abc

= |p(uv)|abc + |p(vu)|abc + |p(δ)|abc + |p(uv)|ab |p(δ)|c + |p(uv)|ab |p(vu)|c +|p(uv)|a |p(δ)|bc + |p(uv)|a |p(vu)|bc + |p(δ)|ab |p(vu)|c +|p(δ)|a |p(vu)|bc + |p(uv)|a |p(δ)|b |p(vu)|c

and, |p(y)|abc

=

|p(vu)|abc + |p(uv)|abc + |p(δ)|abc + |p(vu)|ab |p(δ)|c + |p(vu)|ab ||p(uv)|c +|p(vu)|a |p(δ)|bc + |p(vu)|a ||p(uv)|bc + |p(δ)|ab |p(uv)|c +|p(δ)|a |p(uv)|bc + |p(vu)|a ||p(δ)|b |p(uv)|c .

But from Theorem 3.4, |p(uv)|a = |uv| = |vu| = |p(vu)|a , |p(uv)|ab = 12 |uv|(|uv| + 1) − |uv|b = 12 |vu|(|vu|+1)−|vu|b = |p(uv)|ab . Likewise |p(uv)|c = |p(vu)|c and |p(uv)|bc = |p(uv)|bc . This proves that |p(x)|abc = |p(y)|abc . Hence the theorem. 2 Corollary 3.8 Let Σ2 = {a < b} and Σ3 = {a < b < c}. Let S1 = {abδba | δ ∈ Σ∗2 } and S2 = {baδab | δ ∈ Σ∗2 }. Then the word p(abδba) is M −equivalent to the corresponding word p(baδab) where p : Σ∗3 7→ Σ∗3 is the Prouhet morphism. Theorem 3.9 Let Σ2 = {a < b} and Σ3 = {a < b < c}. Let α, β be two M −equivalent words

84


P P over Σ2 such that x∈X |α|x = x∈X |β|x for X = {aab, aba, abb, baa, bab, bba}. Then p(α) and p(β) are M −equivalent, where p : Σ∗3 7→ Σ∗3 is the Prouhet morphism. Proof Due to Lemma 3.6, it is enough to show that |p(α)|abc = |p(β)|abc . Now, from Theorem 3.4, we have |p(α)|abc

=

=

1 |α||α|a + |α|b (|α|b − 1) + |α|ab + |α|aaa + |α|aab 2 +|α|abb + |α|aba + |α|baa + |α|bab + |α|bba + |α|bbb 1 1 |α||α|a + |α|b (|α|b − 1) + |α|a (|α|a − 1)(|α|a − 2) 2 6 1 + |α|b (|α|b − 1)(|α|b − 2) + |α|ab + ζ 6

where ζ = |α|aab + |α|abb + |α|aba + |α|baa + |α|bab + |α|bba . Since α and β are M −equivalent, P |α| = |β|, |α|a = |β|a , |α|b = |β|b and |α|ab = |β|ab . Hence, using the hypothesis, x∈X |α|x = P 2 x∈X |β|x for X = {aab, aba, abb, baa, bab, bba}, the assertion is proved.

Remark 3.10 The hypothesis in Theorem 3.9 is not vacuous. The words α = aababa and β = abaaab are M −equivalent words over Σ2 and satisfy the hypothesis. In fact p(α) = abcabcbcaabcbcaabc, p(β) = abcbcaabcabcabcbca. Also, ζ = 16. The words p(α), p(β) have the same Parikh matrix, namely, 

1

  0    0  0

6 19 50 1

6

0

1

0

0



 21   . 6   1

Theorem 3.11 Let Σ3 = {a < b < c}. For a word γ over Σ3 , we have (i) |p(γ)|a = |p(γ)|b = |p(γ)|c = |γ| (ii) |p(γ)|ab = |γ|c |γ|a

1 2

(|γ|a (|γ|a + 1)) +

1 2

(|γ|b (|γ|b − 1)) +

1 2

(|γ|c (|γ|c + 1)) + |γ|a |γ|b + |γ|b |γ|c +

(iii) |p(γ)|bc = |γ|c |γ|a

1 2

(|γ|a (|γ|a + 1)) +

1 2

(|γ|b (|γ|b + 1)) +

1 2

(|γ|c (|γ|c − 1)) + |γ|a |γ|b + |γ|b |γ|c +

where p : Σ∗3 7→ Σ∗3 is the Prouhet morphism. Proof The proof of statement (i) is similar to the proof of Theorem 3.4. For proving


85

statement (ii), we have |p(γ)|ab

= =

|γ|a + |γ|c + |γ|aa + |γ|ab + |γ|ac + |γ|ba + |γ|bb + |γ|bc + |γ|ca + |γ|cb + |γ|cc 1 1 1 |γ|a + |γ|c + (|γ|a (|γ|a − 1)) + (|γ|b (|γ|b − 1)) + (|γ|c (|γ|c − 1)) 2 2 2 +|γ|ab + |γ|ba + |γ|bc + |γ|cb + |γ|ac + |γ|ca .

Using |γ|ab + |γ|ba = |γ|a |γ|b and two similar identities and simplifying, we obtain (ii). The proof of statement (iii) is similar. Given two words u, v over Σ∗3 , conditions are obtained in [8], for the product words uv and vu to have the same Parikh matrix. We make use of these conditions in order to derive corresponding conditions for image words under Prouhet morphism to have the same Parikh 2 matrix. Lemma 3.12 ([8]) Let u, v be two words over Σ3 = {a < b < c} satisfying weak-ratio property and the conditions (i) |u|a |u|bc = |u|ab |u|c and (ii) |v|a |v|bc = |v|ab |v|c . Then Ψ3 (uv) = Ψ3 (vu). Theorem 3.13 Let γ, δ be two words over Σ3 = {a < b < c} satisfying weak-ratio property. Let |γ|b = |γ|c and |δ|b = |δ|c . Then Ψ3 [p(γδ)] = Ψ3 [p(δγ)] where p : Σ∗3 7→ Σ∗3 is the Prouhet morphism. Proof By Lemma 2.3, p(γ) and p(δ) satisfy weak-ratio property. From Theorem 3.11 (ii) and (iii), |p(γ)|ab = |p(γ)|bc , since |γ|b = |γ|c . Also, from Theorem 3.11 (i), |p(γ)|a = |p(γ)|c . Hence |p(γ)|a |p(γ)|bc =|p(γ)|ab |p(γ)|c . Likewise, |p(δ)|a |p(δ)|bc =|p(δ)|ab |p(δ)|c . Using Lemma 2 3.12, it follows that Ψ3 [p(γδ)] = Ψ3 [p(δγ)]. § 4. Shuffle on Words and Prouhet Morphism Atanasiu and Teh [3] considered a restricted shuffle operator SShuf on two binary words over Σ2 as follows: If u, v ∈ Σ+ 2 such that u = a1 a2 · · · an , v = b1 b2 · · · bn for n ≥ 1, then SShuf (u, v) = a1 b1 a2 b2 · · · an bn where ai , bi ∈ Σ2 , (1 ≤ i ≤ n). Properties on M −equivalence in the context of such binary words SShuf (u, v) are obtained in [3]. We recall two of these properties in the following lemmas. Lemma 4.1 ([3]) If v1 , v2 , w1 , w2 are binary words over Σ2 , such that |v1 | = |v2 | and v1 ≡M w1 , v2 ≡M w2 , then SShuf (v1 , v2 ) ≡M SShuf (w1 , w2 ). Lemma 4.2 ([3]) Let v, w be two binary words over Σ2 , such that |v| = |w|. Then SShuf (v, w) ≡M SShuf (w, v) if and only if ψ(v) = ψ(w) where ψ(x) is the Parikh vector of the word x. Remark 4.3 Results corresponding to Lemma 4.1 and Lemma 4.2 are not known for ternary words. In fact it is pointed out in [3] in Page 7, that these results cannot be immediately extended to the ternary alphabet. An extension of the operator SShuf f [3], denoted by Sm,n , is introduced in [4]. Informally expressed, this shuffle operator Sm,n , m, n ≥ 1, in operating on a pair of binary words (u, v)

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forms a binary word Sm,n (u, v) obtained by concatenating alternately consecutive factors of u and v, with each factor in u of length m and in v of length n respectively. Here we recall this extension operator but with m = n. Definition 4.4 Let u, v be two binary words over Σ2 such that u = u1 u2 · · · ul and v = v1 v2 · · · vl , where ui , vi ∈ Σ∗2 , |ui | = |vi | = k, for 1 ≤ i ≤ l. The shuffle operator, denoted as Sk,k is defined on the pair (u, v) as follows: Sk,k (u, v) = u1 v1 u2 v2 · · · ul vl . In particular, when m = n = 1, we have S1,1 (u, v) = SShuf (u, v). Theorem 4.5 If v1 , v2 , w1 , w2 are binary words over Σ2 , such that |v1 | = |v2 | and v1 ≡M w1 , v2 ≡M w2 , then S3,3 (p(v1 ), p(v2 )) ≡M S3,3 (p(w1 ), p(w2 )), if Σx∈X |SShuf (v1 , v2 )|x = Σx∈X |SShuf (w1 , w2 )|x , where X = {aab, aba, abb, baa, bab, bba}. Proof It can be seen that S3,3 (p(v1 ), p(v2 )) = p(SShuf (v1 , v2 )), S3,3 (p(w1 ), p(w2 )) = p(SShuf (w1 , w2 )). By Lemma 4.1, we have SShuf (v1 , v2 ) ≡M SShuf (w1 , w2 ). Since by hypothesis |SShuf (v1 , v2 )|aab + |SShuf (v1 , v2 )|aba + |SShuf (v1 , v2 )|abb +|SShuf (v1, v2 )|baa + |SShuf (v1 , v2 )|bab + |SShuf (v1 , v2 )|bba = |SShuf (w1 , w2 )|aab + |SShuf (w1 , w2 )|aba + |SShuf (w1 , w2 )|abb +|SShuf (w1 , w2 )|baa + |SShuf (w1 , w2 )|bab + |SShuf (w1 , w2 )|bba , it follows from Theorem 3.9 that p(SShuf (v1 , v2 )) ≡M p(SShuf (w1 , w2 )) which proves the Theorem.

2

We illustrate Theorem 4.5 by considering v1 = abaabb, w1 = aabbab, v2 = bbabba, and w2 = bbbaab so that p(v1 ) = abcbcaabcabcbcabca, p(v2) = bcabcaabcbcabcaabc and so S3,3 (p(v1 ), p(v2 )) = abcbcabcabcaabcabcabcbcabcabcabcaabc. Likewise, S3,3 (p(w1 ), p(w2 )) = abcbcaabcbcabcabcabcaabcabcabcbcabca.


87

The words S3,3 (p(v1 ), p(v2 )) and S3,3 (p(w1 ), p(w2 )) have the same Parikh matrix 

1 12 71 320

  0    0  0

1

12

0

1

0

0



 78   . 12   1

2

Theorem 4.6 Let v, w be two binary words over Σ2 , such that |v| = |w|. Then S3,3 (p(v), p(w)) ≡M S3,3 (p(w), p(v)) if ψ(v) = ψ(w) where ψ(x) is the Parikh vector of the word x and if Σx∈X |SShuf (v, w)|x = Σx∈X |SShuf (w, v)|x , where X = {aab, aba, abb, baa, bab, bba}. Proof As in the proof of Theorem 4.5, we have S3,3 (p(v), p(w)) = p(SShuf (v, w)), S3,3 (p(w), p(v)) = p(SShuf (w, v)). By Lemma 4.2, SShuf (v, w) ≡M SShuf (w, v). Also, by hypothesis |SShuf (v, w)|aab + |SShuf (v, w)|aba + |SShuf (v, w)|abb +|SShuf (v, w)|baa + |SShuf (v, w)|bab + |SShuf (v, w)|bba = |SShuf (w, v)|aab + |SShuf (w, v)|abb . It follows from Theorem 3.9 that p(SShuf (v, w)) ≡M p(SShuf (w, v)) which proves the result. 2 § 5. Conclusion We have considered here Prouhet morphism and derived certain properties related to subwords and Parikh matrices of the image words under this morphism. Investigating properties of special words under an individual morphism, such as the Prouhet morphism considered here, is of significance in knowing the ability of a specific morphism in distinguishing or not distinguishing M −equivalent words. It will be of interest to construct M − equivalent words whose images under Prouhet morphism are not M −equivalent. References [1] A. Atanasiu, Morphisms on Amiable Words, Annals of Bucharest University, Vol.LIX(2010), 99–111. [2] A. Atanasiu, K. Mahalingam and K. G. Subramanian, Morphisms and Weak-Ratio Prop-

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[3] [4]

[5] [6] [7] [8] [9] [10] [11] [12] [13]


erty, In: (Eds. Gh. Paun et al.) Discrete Mathematics and Computer Science, The Publishing House of the Romanian Academy, Romania, 2014, pp 13–21. A. Atanasiu, W.C. Teh, A New Operator over Parikh Languages, Int. J. Found. Comput. Sci., Vol.27(6)(2016), 757–770. S. Bera, K. Mahalingam and K.G. Subramanian, Properties of Parikh Matrices of Binary Words Obtained by an Extension of a Restricted Shuffle Operator, Int. J. Found. Comp. Sci. Vol.29 (2018), 403–414. J. Berstel, Axel Thue’s papers on repetitions in words: a translation, Publications du LaCIM 20, Université du Québec à Montréal, 1995. J. Berstel, and P. Séébold, A characterization of overlap-free morphisms, Discrete App. Math. Vol.46(3)(1993), 275–281. M. Lothaire, Combinatorics on Words, Cambridge Mathematical Library, Cambridge university Press, 1997. K. Mahalingam and K. G. Subramanian, Product of Parikh matrices and commutativity, Int. J. Found. Comput. Sci., Vol.23(2012), 207–223. A. Mateescu, A. Salomaa, K. Salomaa and S. Yu, A Sharpening of the Parikh Mapping, RAIRO-Theor. Inform. and Appl., Vol.35(6)(2001), 551–564. G. Rozenberg and A. Salomaa, Handbook of Formal Languages, Springer-Verlag New York, Inc., 1997, Vols. 1–3. P. Séébold, Lyndon factorization of the Prouhet words. Theo. Comput. Sci., Vol.307(2003), 179–197. K. G. Subramanian, A. M. Huey and A. K. Nagar, On Parikh Matrices, Int. J. Found. Comput. Sci., Vol.20(2)(2009), 211–219. K.G. Subramanian, I. Venkat, P. Isawasan, A.T. Khader and A.K. Nagar, On Prouhet morphisms and Parikh matrices, Paper presented at the 4th International Conference on Mathematical sciences, Nov. 2016.

International J.Math. Combin. Special Issue 1, (2018) 89-103

Nardhaus Gaddum Type Results for Total Resolving Number of Graphs and Their Derived Graphs N. Shunmugapriya and J. Paulraj Joseph (Department of Mathematics, Manonmaniam Sundaranar University, Tirunelveli - 627 012, India.) E-mail: [email protected], [email protected]

Abstract: Let G = (V, E) be a simple connected graph. An ordered subset W of V is said to be a resolving set of G if every vertex is uniquely determined by its vector of distances to the vertices in W. The minimum cardinality of a resolving set is called the resolving number of G and is denoted by r(G). Total resolving number as the minimum cardinality taken over all resolving sets in which hW i has no isolates and it is denoted by tr(G). In this paper, we obtain the bounds on the sum of the total resolving number of a graph and its derived graph such as complement of a graph, block graph, line graph, power graph, subdivision graph and total graph. Also, we characterize the extremal graphs.

Key Words: Resolving number, total resolving number, complement of a graph, block graph, line graph, power graph, subdivision graph and total graph.

AMS(2010): 05C12, 05C35. §1. Introduction Let G = (V, E) be a finite, simple, connected and undirected graph. The join G + H of two graphs G and H consists of G ∪ H and all edges joining a vertex of G and a vertex of H. For a cut vertex v of a connected graph G, suppose that the disconnected graph G \ {v} has k components G1 , G2 , · · · , Gk (k ≥ 2). The induced subgraphs Bi = G[V (Gi )∪{v}] are connected and referred to as the brances of G at v. A cut vertex v is a path support if there is a nontrivial path as a branch at v; a simple path support if there is exactly one path support at v; a multi path support if there are more than one path support at v. A (k, l)-kite is a graph obtained by identifying any vertex of a cycle Ck with an end vertex of a path Pl . A block of a graph G is a maximal connected subgraph of G that has no cut vertex. G is called a triangle free graph if it has no triangle. To identify non adjacent vertices x and y of a graph G is to replace these vertices by a single vertex incident to all the edges which were incident in G to either x or y. If W = {w1 , w2 , · · · , wk } ⊆ V (G) is an ordered set, then the ordered k-tuple (d(v, w1 ), · · · , d(v, wk )) is called the representation of v with respect to W and it is denoted by r(v|W ). Since the representation for each wi ∈ W contains exactly one 0 in the ith position, all the vertices of 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 16, 2018, Accepted June 21, 2018, Edited by R. Kala.

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W have distinct representations. W is called a resolving set for G if all the vertices of V (G) \ W also have distinct representations. The minimum cardinality of a resolving set is called the resolving number of G and it is denoted by r(G). In 1975, Slater introduced these ideas and used locating set for what we have called resolving set. He referred to the cardinality of a minimum resolving set in G as its location number. In 1976, Harary and Melter discovered these concepts independently as well but used the term metric dimension rather than location number. In 2003, Ping Zhang and Varaporn Saenpholphat [12] and [13] studied connected resolving number and in 2015, we introduced and studied total resolving number in [7]. In this paper, we use the term resolving number to maintain uniformity in the current literature. If W is a resolving set and the induced subgraph hW i has no isolates, then W is called a total resolving set of G. The minimum cardinality taken over all total resolving sets of G is called the total resolving number of G and is denoted by tr(G). In 2013, Aouchiche and Hansen [1] gave a survey of Nordhaus - Gaddum type relations for a variety of graph invariants with reference to a graph and its complement. A new direction to extend Nordhaus - Gaddum type results to other derived graphs was discussed in [6]. Hence we studied total resolving number of derived graphs in [8],[10] and [11]. In this paper, we investigate the sum of the total resolving number of graphs and some derived graphs such as block graph, line graph, power graph, subdivision graph, total graph and characterize the extremal graphs.

§2. Literature Review The following results are used in subsequent sections. Observation 2.1 ([7]) Let {w1 , w2 } ⊂ V (G) be a total resolving set in G. Then the degrees of w1 and w2 are at most 3. Theorem 2.2 ([7])

For n ≥ 3, tr(Pn ) = 2 and tr(Cn ) = 2.

Observation 2.3 ([7]) For any graph G of order n ≥ 3, 2 ≤ tr(G) ≤ n − 1. Theorem 2.4 ([7]) For n ≥ 3, tr(G) = n − 1 if and only if G ∼ = Kn or K1,n−1 . Observation 2.5 ([7]) For any tree T, tr(T ) = 2 if and only if T is a path. Observation 2.6 ([7]) Let G be a unicyclic graph with even cycle Ck . Then tr(G) = 2 if and only if at most two adjacent vertices of Ck are simple path supports and no vertex of G is a multi path support. Observation 2.7 ([7]) Let G be a unicyclic graph with odd cycle Ck . Then tr(G) = 2 if and only if no vertex of G is a multi path support and either at most three consecutive vertices of Ck are simple path supports or exactly two non adjacent vertices of Ck at a distance two are simple path supports. Remark 2.8 ([7]) Let G be a graph of order at least 4. If W = {v1 , v2 } is a total resolving

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set, then the following are true. (a) W, W ≤ 4;

(b) if W, W = 1, then W must be a path; (c) if W, W = 2, then d(v1 ) = d(v2 ) = 2; (d) if 3 ≤ W, W ≤ 4, then a unique vertex v ∈ W is adjacent to both v1 and v2 .

Remark 2.9 ([7]) If G is not a path and {w1 , w2 } is a total resolving set of G, then w1 and w2 lie on some cycle of G. Notation 2.10 ([9]) If H and K are two graphs, then the graph obtained by identifying one center of H with one center of K is denoted by H ∗ K and the graph obtained by joining one center of H to one center of K is denoted by HeK. Theorem 2.11 ([9]) Let G be a 1-connected graph of order at least 4. Then tr(G) = n − 2 if and only if G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. §3. Complement of Graphs In this section, we obtain the bounds on the sum of total resolving number of a graph and its complement and characterize the extremal graphs. Definition 3.1 The complement G of a graph G is that graph whose vertex set is V (G) and such that for each pair u, v of vertices of G, uv is an edge of G if and only if uv is not an edge of G. Theorem 3.2 Let G be a graph of order n ≥ 4. Then 4 ≤ tr(G) + tr(G) ≤ 2n − 4. Proof The lower bound follows from Observation 2.3. Since G and G are connected graphs, ∆(G) ≤ n − 2. Using Theorem 2.4, tr(G) + tr(G) ≤ 2n − 4. 2 Observation 3.3 For n ≥ 3, tr(Cn ) + tr(Cn ) = 4 if and only if n = 5 or 6. Theorem 3.4 Let G be a unicyclic graph. Then tr(G)+tr(G) = 4 if and only if G is isomorphic to one of the graphs given in Figure 3.1.

H1

H6

H2

H7

H3

H8

H4

H9

Fig. 3.1 The graphs G satisfying tr(G) + tr(G) = 4

H5

H10

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Proof Let G be a unicyclic graph with cycle Ck , k ≥ 3. Let V (Ck ) = {v1 , v2 , . . . , vk } and E(Ck ) = {ei = vi vi+1 / 1 ≤ i ≤ k − 1} ∪ {ek = vk v1 }. Assume that tr(G) + tr(G) = 4. Then tr(G) = 2 and tr(G) = 2. Using Observation 2.1, k ≤ 6 and n ≤ 7. We consider the following four cases. Case 1. k = 6. We claim that n = 6. Suppose n = 7. Then G has a vertex of degree 1, a vertex of degree 3 and all other vertices are of degree 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. Thus n = 6 and hence in this case G∼ = H10 . Case 2. k = 5. We claim that n = 5 or 6. Suppose n = 7. We consider the following two subcases. Subcase 2.1 G has only one pendant edge. Then G has a vertex of degree 1, a vertex of degree 3 and all other vertices are of degree 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. Subcase 2.2 G has two pendant edges. By Observation 2.7, they are not adjacent edges. If they are incident with two adjacent vertices of C5 , then G has two vertices are of degree 1, two adjacent vertices are of degree 3 and all other vertices are of 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. If they are incident with two non adjacent vertices of C5 , say v1 and v3 . Since d(v1 ) = d(v3 ) = 3, d(v2 ) = d(v4 ) = d(v5 ) = 4 and d(v6 ) = d(v7 ) = 5, {v1 , v3 } is the only 2-element set in which each vertex is of degree 3. But there does not exist a vertex adjacent to both v1 and v3 , which is a contradiction to Remark 2.8. Hence n = 5 or 6. If n = 5, then G ∼ = H4 and n = 6, then H9 . Case 3. k = 4. We claim that n = 4, 5 or 6. Suppose n = 7. By Observation 2.6, G has one or two pendant edges. If G has only one pendant edge, then G has a vertex of degree 1, a vertex of degree 3 and all other vertices are of degree 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. If G has two pendant edges, then by Observation 2.6, two adjacent vertices of C4 are simple path supports. But in G, there does not exist two adjacent vertices of degree at most 3 and hence by Observation 2.1, tr(G) 6= 2. If n = 4, then G ∼ = C4 , but G is disconnected. If n = 5, then G ∼ = H3 . If n = 6, then by ∼ Observation 2.6,G = H7 or H8 . Case 4. k = 3. We claim that n = 3, 4, 5 or 6. Suppose n = 7. By Observation 2.7, G has one, two or three pendant edges. If G has only one pendant edge, then G has a vertex of degree 1, a vertex of degree 3 and all other vertices are of degree 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. If G has two pendant edges,


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then by Observation 2.7, two vertices of C3 are simple path supports and G has two vertices are of degree 1, two adjacent vertices are of degree 3 and all other vertices are of degree 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. If G has three pendant edges, then G has three vertices are of degree 1, three vertices are of degree 3 and all other vertices are of degree 2. Hence there does not exist two adjacent vertices of degree at most 3 in G and hence by Observation 2.1, tr(G) 6= 2. Hence 3 ≤ n ≤ 6. If n = 3 or 4, then G is disconnected. If n = 5, then by Observation 2.7, G ∼ = H1 or ∼ H2 . If n = 6, then by Observation 2.7, G = H5 or H6 . The converse can be easily verified.

2

Observation 3.5 ([13]) If a connected graph G contains a set S of vertices of G of cardinality p ≥ 2 such that d(u, x) = d(v, x) ∀ u, v ∈ S and x ∈ V (G) \ {u, v}, then every resolving set must contain at least p − 1 vertices of S. Notation 3.6 The graph obtained from Ks + Kt by introducing two non adjacent vertices u and v such that u is adjacent to all the vertices of Ks and v is adjacent to all the vertices of Kt is denoted by (Ks + Kt ) ◦ (uv). Theorem 3.7 For s ≥ 1 and t ≥ 2, tr[(Ks + Kt ) ◦ (uv)] = s + t − 1. Proof Let V (Ks ) = {u1 , u2 , . . . , us } and V (Kt ) = {v1 , v2 , . . . , vt }. Let W = {u1 , u2 , . . . , us } ∪ {v1 , v2 , . . . , vt−1 }. Then each coordinate of the representation of vt is 1, first s coordinates of the representation of u are 1 and others are 2 and first t coordinates of the representation of v are 2 and others are 1. It follows that W is a resolving set. Since hW i has no isolates, tr[(Ks + Kt ) ◦ (uv)] ≤ s + t − 1. Let W be any total resolving set of (Ks + Kt ) ◦ (uv). Using Observation 3.5, at least s − 1 vertices of Ks must belong to W and at least t − 1 vertices of Kt must belong to W. But hKs ∪ Kt i is complete, s + t − 1 vertices of Ks ∪ Kt must belong to W and hence tr[(Ks + Kt ) ◦ (uv)] ≥ s + t − 1. Thus 2 tr[(Ks + Kt ) ◦ (uv)] = s + t − 1. Theorem 3.8 Let G be a 1-connected graph order n ≥ 4. Then tr(G) + tr(G) = 2n − 4 if and only if G ∼ = P4 . Proof Assume that tr(G) + tr(G) = 2n − 4. Then tr(G) = n − 2 and tr(G) = n − 2. By Theorem 2.11, G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. If G ∼ = K1,s ∗ Kt or K3 ∗ Kt , then G is disconnected, which is a ∼ contradiction. If G = Bs,t , s ≥ 1, t ≥ 2, then G ∼ = (Ks + Kt ) ◦ (uv) and hence by Theorem 3.7, tr(G) ≤ n − 3, which is a contradiction. If G ∼ = 2K3 + e, then let V (G) = {v1 , v2 , v3 , v4 , v5 , v6 } and E(G) = {v1 v2 , v2 v3 , v3 v1 , v3 v4 , v4 v5 , v4 v6 , v5 v6 }. Let W = {v1 , v4 , v5 }. Then r(v2 |W ) = (2, 1, 1), r(v3 |W ) = (2, 3, 1), r(v6 |W ) = (1, 2, 2). Since hW i has no isolates, tr(G) ≤ 3 = n − 3, which is a contradiction. If G ∼ = K3 e K1,s , then let V (G) = {v1 , v2 , . . . , vn } and E(G) = {v1 v2 , v2 v3 , v1 v3 , v3 v4 } ∪ {v4 vi / 5 ≤ i ≤ n}. Let W = {v1 , v2 } ∪ {v5 , . . . , vn−1 }. Then each coordinate of the representation of vn is 1, 1st and 2nd coordinates of the representation of v3 are 2 and others are 1 and 1st and 2nd coordinates of the representation of v4 are 1 and others

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are 2. It follows that W is a resolving set of G. Since hW i has no isolates, tr(G) ≤ n − 3, which is a contradiction. Hence G ∼ 2 = P4 . The converse can be easily verified. Theorem 3.9 Let G be a 1-connected graph of order n ≥ 5. Then tr(G) + tr(G) = 2n − 5 if and only if G ∼ = Bs,t (s ≥ 1, t ≥ 2) or 2K3 + e or K3 e K1,s , s ≥ 3. Proof Assume that tr(G) + tr(G) = 2n − 5. Then tr(G) = n − 2 and tr(G) = n − 3. By Theorem 2.11, G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. But we claim that G ∼ = Bs,t (s ≥ 1, t ≥ 2) or 2K3 + e or K3 e K1,s , s ≥ 3. If G ∼ = K1,s ∗ Kt or K3 ∗ Kt , then G is disconnected, which is a contradiction. By Theorem 3.8, G ∼ 6 B1,1 . Hence G ∼ = = Bs,t (s ≥ 1, t ≥ 2) or 2K3 + e or K3 e K1,s , s ≥ 3. The converse can be easily verified. 2 Problem 3.10 If G is not a unicyclic graph of order n ≥ 4, then characterize G for which tr(G) + tr(G) = 4. §4. Block Graphs In this section, we obtain the bounds on the sum of total resolving number of a graph and its block graph and characterize the extremal graphs. Definition 4.1 The block graph B(G) of a graph G is a graph whose vertices are the blocks of G and two of these vertices are adjacent whenever the corresponding blocks contain a common cut vertex of G. Theorem 4.2 ([8])

Let G be a graph with n′ ≥ 3 blocks. Then 2 ≤ tr(B(G)) ≤ n′ − 1.

Theorem 4.3 ([5]) G are complete.

A graph G is a block graph of some graph H if and only if the blocks of

Definition 4.4 A graph G is said to be a block path if B(G) is a path. Theorem 4.5 ([8]) Let G be a graph with n′ ≥ 3 blocks. Then tr(B(G)) = 2 if and only if G satisfies one of the following conditions: (i) G is a block path (ii) G has either a unique cut vertex lying in exactly three blocks or a cut vertex v lying in exactly three blocks, all other cut vertices other than v lying in exatly two blocks and every block contains at most two cut vertices. Theorem 4.6 ([8]) Let G be a graph with n′ ≥ 3 blocks. Then tr(B(G)) = n′ − 1 if and only if either G has exactly one cut vertex or G has at least two cut vertices and all of them lie in one block such that each cut vertex lying in exactly two blocks. Theorem 4.7 Let G be a graph of order n ≥ 4 and n′ ≥ 3 blocks. Then 4 ≤ tr(G)+tr(B(G)) ≤ n + n′ − 2. Proof The proof follows from Observation 2.3 and Theorem 4.2.

2


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Notation 4.8 ([9]) Let G be the collection of graphs G such that G is the union of two distinct paths P1 : x1 x2 . . . xr , P2 : y1 y2 . . . ys , r ≤ s and x1 y1 ∈ E(G), xi yi ∈ E(G) for at least one i, 2 ≤ i ≤ r. Theorem 4.9 ([9]) If G is a bipartite graph that is not a path, then tr(G) = 2 if and only if G ∈ G. Theorem 4.10 Let G be a bipartite graph of order n ≥ 4 and n′ ≥ 3 blocks. Then tr(G) + tr(B(G)) = 4 if and only if G ∼ = Pn , n ≥ 4 or G ∈ G . Proof Assume that tr(G)+tr(B(G)) = 4. Then tr(G) = 2 and tr(B(G)) = 2. If G is a tree, then by Observation 2.5, G ∼ = Pn . Now we may assume that G contains a cycle. By Theorem 4.9, G ∈ G . Thus G is the block path and hence by Theorem 4.5, tr(B(G)) = 2. Hence G ∼ = Pn , n ≥ 4 or G ∈ G . The converse can be easily verified. 2 Theorem 4.11 Let G be a unicyclic graph with cycle Ck of order n and n′ ≥ 3 blocks. Then tr(G) + tr(B(G)) = 4 if and only if G ∼ = (k, l)-kite or ∆(G) = 3 and G satisfies one of the following conditions: (i) G has exactly two adjacent vertices of degree 3 which are in Ck (ii) G has exactly two non adjacent vertices at a distance two of degree 3 which are in Ck , k is odd. Proof Assume that tr(G) + tr(B(G)) = 4. Then tr(G) = 2 and tr(B(G)) = 2. By Observations 2.6 and 2.7, ∆(G) ≤ 3. Since n′ ≥ 3, ∆(G) = 3. By Observations 2.6 and 2.7, G has at most three vertices of degree 3 which are in Ck . Now, we claim that G has at most two vertices of degree 3. Suppose that G has three vertices of degree 3. Then B(G) is a spider with ∆(G) = 3. By Observation 2.5, tr(B(G)) ≥ 3, which is a contradiction. Hence G has at most two vertices of degree 3. If G has one vertex degree 3, then G ∼ = (k, l)-kite. If G has two vertices of degree 3, then by Observations 2.6 and 2.7, G satisfies either (i) or (ii). The converse can be 2 easily verified. Theorem 4.12 Let G be a graph of order n ≥ 4 and n′ ≥ 3 blocks. Then tr(G) + tr(B(G)) = n + n′ − 2 if and only if G is a star. Proof Assume that tr(G)+tr(B(G)) = n+n′ −2. Then tr(G) = n−1 and tr(B(G)) = n′ −1. By Theorem 2.4, G ∼ = Kn or K1,n−1 . Since n′ ≥ 3, G ∼ = K1,n−1 . The converse can be easily verified. 2 Theorem 4.13 Let G be a graph of order n ≥ 4 and n′ ≥ 3 blocks. Then tr(G) + tr(B(G)) = n + n′ − 3 if and only if G ∼ = K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or 2K3 + e. Proof Assume that tr(G)+tr(B(G)) = n+n′ −3. Then tr(G) = n−2 and tr(B(G)) = n′ −1 or tr(G) = n − 1 and tr(B(G)) = n′ − 2. Suppose tr(G) = n − 1 and tr(B(G)) = n′ − 2. Then by Theorem 2.4, tr(G) = n − 1 if and only if G ∼ = Kn or K1,n−1 . Since n′ ≥ 3, G ∼ = K1,n−1 and ′ hence using Theorem 4.12, tr(B(G)) = n − 1, which is a contradiction. Therefore tr(G) = n− 2 and tr(B(G)) = n′ − 1. By Theorem 2.11, G ∼ = Bs,t , s, t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1

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and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. Since tr(B(G)) = n′ − 1 using Theorem 4.6, G∼ 2 = K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or 2K3 + e. The converse can be easily verified. Problem 4.14 If G is a connected graph of order n ≥ 4 and n′ ≥ 3 blocks, then characterize G for which tr(G) + tr(B(G)) = n + n′ − 4. §5. Line Graphs In this section, we obtain the bounds on the sum of total resolving number of a graph and its line graph and characterize the extremal graphs. Definition 5.1 The line graph L(G) of a graph G is a graph whose vertex set is E(G) and two vertices e and f of L(G) are adjacent if and only if they are adjacent in G. Theorem 5.2 ([8]) Let G be a graph of order n ≥ 3 and size m ≥ 3. Then 2 ≤ tr(L(G)) ≤ m−1. Theorem 5.3 Let G be a graph of order n ≥ 3 and size m ≥ 3. Then tr(L(G)) = m − 1 if and only if G ∼ = K3 or P4 or K1,n−1 . Theorem 5.4 Let G be a graph of order n ≥ 3 and size m ≥ 3. Then 4 ≤ tr(G) + tr(L(G)) ≤ n + m − 2. Proof The proof follows from Observations 2.3 and 5.2.

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Theorem 5.5 Let G be a unicyclic graph with cycle Ck of order n ≥ 3 and size m ≥ 3. Then tr(G) + tr(L(G)) = 4 if and only if G ∼ = Cn or (k, l)-kite or ∆(G) = 3 and G satisfies one of the following conditions: (i) G has exactly two adjacent vertices of degree 3 which are in Ck , k is odd (ii) G has exactly two non adjacent vertices of degree 3 which are in C5 . Proof Let V (Ck ) = {v1 , v2 , . . . , vk } and E(Ck ) = {ei = vi vi+1 / 1 ≤ i ≤ k − 1} ∪ {ek = vk v1 }. Assume that tr(G) + tr(L(G)) = 4. Then tr(G) = 2 and tr(L(G)) = 2. Let W = {w1 , w2 } be a total resolving set of L(G). By Observations 2.6 and 2.7, ∆(G) ≤ 3. If ∆(G) = 2, then G∼ = Cn . If ∆(G) = 3, then by Observations 2.6 and 2.7, G has at most three vertices of degree 3 which are in Ck . But we claim that G has at most two vertices of degree 3. Suppose that G has three vertices of degree 3. By Observations 2.6 and 2.7, such vertices are consecutive and k is odd. Let vk+1 , vk+2 and vk+3 be the neighbors of v1 , v2 and v3 respectively in G. Let ek+1 = v1 vk+1 , ek+2 = v2 vk+2 and ek+3 = v3 vk+3 in G. By Remark 2.9, w1 and w2 are in some cycle of L(G) and by Observation 2.1, d(w1 ) ≤ 3 and d(w2 ) ≤ 3. Therefore k ≥ 5. If k = 5, then we can easily verify that tr(L(G)) ≥ 3. So we may assume that k ≥ 7. Since w1 and w2 are in some cycle of L(G), W ⊂ {e1 , e2 , . . . , ek+3 }, d(w1 ) = 2 and d(w2 ) = 2 or 3. If d(w1 ) = 2 and d(w2 ) = 3, then by Remark 2.9, either w1 = ek+1 and w2 = ek or w1 = ek+3 and w2 = e3 . Without loss of generality, let w1 = ek+1 and w2 = ek . Then r(ek+2 |W ) = r(e2 |W ) = (2, 2), which is a contradiction. So we may assume that d(w1 ) = 2


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and d(w2 ) = 2. Then either W ⊂ {e4 , e5 , . . . , e⌈ k ⌉ } or W ⊂ {e⌈ k ⌉+2 , . . . , ek−1 }. Without 2 2 loss of generality, let W = {e4 , e5 , . . . , e⌈ k ⌉ }. Then r(ek+3 |W ) = r(e2 |W ) in L(G), which is a 2 contradiction. Thus G has at most two vertices of degree 3. ∼ (k, l)-kite. If G has two vertices of degree 3 If G has one vertex degree 3 in Ck , then G = in Ck , then we consider the following two cases. Case 1. Degree 3 vertices are adjacent. Now, we claim k is odd. Suppose k is even. Let d(v1 ) = d(v2 ) = 3 and vk+1 , vk+2 be the neighbors of v1 and v2 respectively. Let ek+1 = v1 vk+1 and ek+2 = v2 vk+2 . By Observation 2.1, d(w1 ) ≤ 3 and d(w2 ) ≤ 3. By Remark 2.9, w1 and w2 are in some cycle of L(G). If k = 4, then we can easily verify that tr(L(G)) ≥ 3. So we may assume that k ≥ 6. Since w1 and w2 are in some cycle of L(G), W ⊂ {e1 , e2 , . . . , ek+3 }, d(w1 ) = 2 and d(w2 ) = 2 or 3. If d(w1 ) = 2 and d(w2 ) = 3, then by Remark 2.9, either w1 = ek+1 and w2 = ek or w1 = ek+2 and w2 = e2 . Without loss of generality, let w1 = ek+1 and w2 = ek . Then r(ek+2 |W ) = r(e2 |W ) = (2, 2), which is a contradiction. So we may assume that d(w1 ) = 2 and d(w2 ) = 2. Then either W ⊂ {e3 , e4 , . . . , e k , e k +1 } or W ⊂ {e k +2 , . . . , ek−1 }. Without loss 2 2 2 of generality, let W = {e3 , e4 , . . . , e k , e k +1 }. Then r(ek+2 |W ) = r(e2 |W ) in L(G), which is a 2 2 contradiction. Thus k is odd and hence G satisfies (i). Case 2. Degree 3 vertices are non adjacent. By Observations 2.6 and 2.7, k is odd. Then the proof is similar to Case 1 and hence we can easily verify that k = 5. Thus G satisfies (ii). The converse can be easily verified.

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Theorem 5.6 Let G be a graph of order n ≥ 3 and size m ≥ 3. Then tr(G) + tr(L(G)) = n + m − 2 if and only if G is either K3 or K1,n−1 . Proof Assume that tr(G)+tr(L(G)) = n+m−2. Then tr(G) = n−1 and tr(L(G)) = m−1. By Theorem 5.3, G ∼ = K3 or P4 or K1,n−1 . But tr(P4 ) = n − 2. Hence G is either K3 or K1,n−1 . The converse can be easily verified. 2 Theorem 5.7 Let G be a graph of order n ≥ 4 and size m ≥ 3. Then tr(G) + tr(L(G)) = n + m − 3 if and only if G ∼ = P4 . Proof Assume that tr(G)+tr(L(G)) = n+m−3. Then tr(G) = n−2 and tr(L(G)) = m−1 or tr(G) = n − 1 and tr(L(G)) = m − 2. Suppose tr(G) = n − 1 and tr(L(G)) = m − 2. By Theorem 2.4, tr(G) ∼ = K1,n−1 and hence by Theorem 5.6, tr(L(G)) = m − 1, which is a contradiction. Therefore tr(G) = n − 2 and tr(L(G)) = m − 1. Then by Theorem 5.3, G ∼ = K3 ∼ or P4 or K1,n−1 . If G = K3 or K1,n−1 , then by Theorem 2.4, tr(G) = n − 1, which is a contradiction. Thus G ∼ 2 = P4 . The converse can be easily verified. Problem 5.8 If G is a connected graph of order n ≥ 3 and size m ≥ 3, then characterize G for which tr(G) + tr(L(G)) = n + m − 4.

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§6. Power Graphs In this section, we obtain the bounds on sum of total resolving number of a graph and its power graph and characterize the extremal graphs. Definition 6.1 Let G be a graph of order n ≥ 3 and diameter d. If k is a positive integer such that 2 ≤ k ≤ d, then the k th power of G, denoted by Gk is a graph whose vertex set is V (G) and two distinct vertices of Gk are adjacent if their distance in G is at most k. Theorem 6.2 ([10]) Let G be a graph of order n ≥ 3 and diameter d. Then (a) 2 ≤ tr(Gk ) ≤ n − 1, 2 ≤ k ≤ d. (b) tr(Gk ) = 2 if and only if G ∼ = Pn . (c) tr(Gk ) = n − 1 if and only if diam(G) = k. Theorem 6.3 ([10]) Let T be a spider. Then tr(T 2 ) = ∆(T ). 2 Theorem 6.4 ([10]) For s, t ≥ 2, tr(Bs,t ) = s + t − 1.

Theorem 6.5 If d is the diameter of G and 2 ≤ k ≤ d, then 4 ≤ tr(G) + tr(Gk ) ≤ 2n − 2. In particular, lower bound is attained if and only if G is a path and upper bound is attained if and only if G is a star. Proof By Observation 2.3 and Theorem 6.2, 4 ≤ tr(G) + tr(Gk ) ≤ 2n − 2. Assume that tr(G) + tr(Gk ) = 4. Then tr(G) = 2 and tr(Gk ) = 2. By Theorem 6.2, tr(G2 ) = 2 if and only if G ∼ = Pn . By Theorem 2.2, tr(G) = 2 and hence G ∼ = Pn . The converse of the lower bound can be easily verified. Assume that tr(G)+ tr(Gk ) = 2n− 2. Then tr(G) = n− 1 and tr(Gk ) = n− 1. By Theorem 2.4, G ∼ = Kn or K1,n−1 . Using Theorem 6.2, G ∼ = K1,n−1 and k = 2. The converse of the upper bound can be easily verified. 2 Theorem 6.6 Let G be a 1-connected graph of order n ≥ 4. Then tr(G) + tr(G2 ) = 2n − 3 if and only if G ∼ = K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3. Proof Assume that tr(G) + tr(G2 ) = 2n − 3. Then tr(G) = n − 2 and tr(G2 ) = n − 1. By Theorem 2.11, G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. By Theorem 6.2, tr(Gk ) = n − 1 if and only if diam(G) = k. Since k = 2, diam(G) = 2 and hence G ∼ = K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3. The converse can be easily verified. 2 Theorem 6.7 Let G be a 1-connected graph of order n ≥ 4. Then tr(G) + tr(G3 ) = 2n − 3 if and only if G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K3 e K1,s , s ≥ 3. Proof Assume that tr(G) + tr(G3 ) = 2n − 3. Then tr(G) = n − 2 and tr(G3 ) = n − 1. By Theorem 2.11, G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. By Theorem 6.2, tr(Gk ) = n − 1 if and only if diam(G) = k. Since k = 3, diam(G) = 3 and hence G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K3 e K1,s , s ≥ 3. The


converse can be easily verified.

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Theorem 6.8 Let T be a tree of order n ≥ 4. Then tr(T ) + tr(T 2 ) = 2n − 4 if and only if T ∼ = P4 or B1,2 . Proof Assume that tr(T ) + tr(T 2 ) = 2n − 4. Then tr(T ) = n − 2 and tr(T 2 ) = n − 2 or tr(T ) = n − 3 and tr(T 2 ) = n − 1. If tr(T ) = n − 3 and tr(T 2 ) = n − 1, then by Theorem 6.2, tr(Gk ) = n − 1 if and only if diam(G) = k. Since k = 2, diam(T ) = 2 and hence T ∼ = K1,n−1 . But tr(K1,n−1 ) = n − 1, which is a contradiction to tr(T ) = n − 3. Therefore tr(T ) = n − 2 2 and tr(T 2 ) = n − 2. By Theorem 2.11, T ∼ ) = s + t − 1, = Bs,t , s, t ≥ 1. By Theorem 6.4, tr(Bs,t 2 ∼ s, t ≥ 2. It follows that tr(T ) = n − 3, which is a contradiction and hence T = P4 or B1,2 . The converse can be easily verified. 2 §7. Subdivision Graphs In this section, we obtain the bounds on the sum of total resolving number of a graph and its subdivision graph and characterize the extremal graphs. Definition 7.1 The subdivision graph of a graph G is the graph obtained from G by deleting every edge uv of G and replacing it by a vertex w of degree 2 that is joined to u and v. Theorem 7.2 ([11]) Let G be a graph of order n ≥ 3. Then 2 ≤ tr(S(G)) ≤ n − 1. Theorem 7.3 ([11]) Let G be a graph of order n ≥ 3. Then tr(S(G)) = 2 if and only if G is isomorphic to Pn or Cn or (k, l)-kite. Theorem 7.4 ([11]) For n ≥ 3, tr(S(Kn )) = n − 1. Remark 7.5 tr(S(K1,n−1 )) = n − 1 and tr(S(Bs,t )) = s + t. Theorem 7.6 Let G be a graph of order n ≥ 3. Then 4 ≤ tr(G) + tr(S(G)) ≤ 2n − 2. Proof The proof follows from Observation 2.3 and Theorem 7.2.

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Theorem 7.7 Let G be a graph of order n ≥ 3. Then tr(G) + tr(S(G)) = 4 if and only if G is isomorphic to Pn or Cn or (k, l)-kite. Proof Assume that tr(G) + tr(S(G)) = 4. Then tr(G) = 2 and tr(S(G)) = 2. By Theorem 7.3, Pn or Cn or (k, l)-kite. By Theorem 2.2 and Observation 2.6, tr(G) = 2. Hence G ∼ = Pn or Cn or (k, l)-kite. The converse can be easily verified. 2 Theorem 7.8 Let G be a unicyclic graph with the cycle Ck . Then tr(G) + tr(S(G)) = 5 if and only if ∆(G) = 3 and either G has exactly two adjacent vertices of degree 3 which are in Ck or G has three vertices of degree 3 which are in C3 . Proof Let V (Ck ) = {v1 , v2 , . . . , vk } and E(Ck ) = {ei = vi vi+1 / 1 ≤ i ≤ k − 1} ∪ {ek = vk v1 }.

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Let vij be the new vertex of an edge vi vj in S(G). Assume that tr(G) + tr(S(G)) = 5. Then tr(G) = 2 and tr(S(G)) = 3 or tr(G) = 3 and tr(S(G)) = 2. Suppose tr(G) = 3 and tr(S(G)) = 2. By Theorem 7.3, Pn or Cn or (k, l)-kite. By Theorem 2.2 and Observations 2.6, 2.7, tr(S(G)) = 2, which is a contradiction. Therefore tr(G) = 2 and tr(S(G)) = 3. Let W = {w1 , w2 , w3 } be a total resolving set of S(G). By Observations 2.6 and 2.7, ∆(G) ≤ 3. If ∆(G) = 2, then G ∼ = Cn and hence by Theorem 7.3, tr(S(G)) = 2, which is a contradiction. If ∆(G) = 3, then by Observations 2.6 and 2.7, G has at most three vertices of degree 3 which are in Ck . Now, we claim that G has two or three vertices of degree 3 in Ck . If G has one vertex of degree 3 in Ck , then G ∼ = (k, l)-kite and hence by Theorem 7.3, tr(S(G)) = 2, which is a contradiction. Thus G has two or three vertices of degree 3. We consider the following two cases. Case 1. G has two vertices of degree 3. Then by Observations 2.6 and 2.7, either they are adjacent or non adjacent with distance two and k is odd. Now, we claim that they are adjacent. Suppose they are non adjacent with distance two and k is odd. Let d(v1 ) = d(v3 ) = 3. Let vk+1 , vk+2 be the neighbors of v1 and v3 respectively in G. Then we can easily verify that W ⊂ V (S(Ck )). If either W = {v12 , v2 , v23 } or W = {v⌈ k ⌉+1 , v(⌈ k ⌉+1)(⌈ k ⌉+2) , v⌈ k ⌉+2 }, then r(v1(k+1) |W ) = r(v1k |W ) and r(v3(k+2) |W ) = 2 2 2 2 r(v34 |W ), which is a contradiction. If W = {v(⌈ k ⌉)(⌈ k ⌉+1) , v⌈ k ⌉+1 , v(⌈ k ⌉+1)(⌈ k ⌉+2) } or 2 2 2 2 2 W = {v(⌈ k ⌉+1)(⌈ k ⌉+2) , v⌈ k ⌉+2 , v(⌈ k ⌉+2)(⌈ k ⌉+3) }, then without loss of generality, let W = 2 2 2 2 2 {v(⌈ k ⌉)(⌈ k ⌉+1) , v⌈ k ⌉+1 , v(⌈ k ⌉+1)(⌈ k ⌉+2) }. Then r(v23 |W ) = r(v3(k+1) |W ), which is a contra2 2 2 2 2 diction. Let X = {v3 , v4 , . . . , v⌈ k ⌉+1 } and Y = {v⌈ k ⌉+2 , v⌈ k ⌉+3 , . . . , vk }. If W ⊂ V (S(hXi)) 2 2 2 or W ⊂ V (S(hY i)), then without loss of generality, let W ⊂ V (S(hXi)). Then r(v23 |W ) = r(v3(k+1) |W ), which is a contradiction. Thus G has exactly two adjacent vertices of degree 3 which are in Ck . Case 2. G has three vertices of degree 3. Then by Observations 2.6 and 2.7, such vertices are consecutive vertices. Now, we claim that k = 3. If k ≥ 5, then let d(v1 ) = d(v2 ) = d(v3 ) = 3. Let vk+1 , vk+2 and vk+3 be the neighbors of v1 , v2 and v3 respectively in G. Then the proof is similar to Case 1 and hence we can easily verify that tr(S(G)) ≥ 3. Thus k = 3. The converse can be easily verified.

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Theorem 7.9 Let G be a graph of order n ≥ 3. Then tr(G) + tr(S(G)) = 2n − 2 if and only if G ∼ = Kn or G ∼ = K1,n−1 . Proof Assume that tr(G) + tr(S(G)) = 2n − 2. Then tr(G) = n − 1 and tr(S(G)) = n − 1. By Theorem 2.4, G ∼ 2 = Kn or G ∼ = K1,n−1 . The converse is obvious. Theorem 7.10 Let G be a graph of order n ≥ 3. Then tr(G) + tr(S(G)) = 2n − 3 if and only if G ∼ = K3 ∗ K3 . Proof Assume that tr(G) + tr(S(G)) = 2n − 3. Then tr(G) = n − 1 and tr(S(G)) = n − 2 or tr(G) = n − 2 and tr(S(G)) = n − 1. If tr(G) = n − 1 and tr(S(G)) = n − 2, then by


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Theorem 2.4, ∼ = Kn or K1,n−1 . But by Theorem 7.4 and Remark 7.5, tr(S(G)) = n − 1, which is a contradiction. Therefore tr(G) = n − 2 and tr(S(G)) = n − 1. By Theorem 2.11, G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 3 or K3 e K1,s , s ≥ 3. If G ∼ = Bs,t , s ≥ 1 and t ≥ 1 or 2K3 + e or K1,s ∗ Kt , s ≥ 1 and t ≥ 3 or K3 ∗ Kt , t ≥ 4 or K3 e K1,s , s ≥ 3, then we can easily verify that tr(S(G)) ≤ n − 2, which is a contradiction. Hence G ∼ 2 = K3 ∗ K3 . The converse can be easily verified. Problem 7.11 If G is a connected graph of order n ≥ 3, then characterize G for which tr(G) + tr(S(G)) = 2n − 4. §8. Total Graphs In this section, we obtain bounds on the sum of total resolving number of a graph and its total graph and characterize the extremal graphs. Definition 8.1 The total graph T (G) of a graph G is a graph whose vertex set is V (T (G)) = V (G) ∪ E(G) and two distinct vertices x and y of T (G) are adjacent if x and y are adjacent vertices of G or adjacent edges of G or x is a vertex incident with edge y. Theorem 8.2 ([2]) The total graph T (G) is isomorphic to the square of the subdivision graph S(G). It is trivial that total resolving is at least two for any graph and hence tr(G)+tr(T (G)) ≥ 4. Since it has been proved [11], Pn is the only graph with tr(T (G)) = 2. Therefore tr(G) + tr(T (G)) = 4 if and only if G ∼ = Pn . Hence if G is not a path, then tr(G) + tr(T (G)) ≥ 5. If T is a tree, then by Observation 2.5, tr(T ) = 2 if and only if T is a path. Hence tr(T )+tr(T (T )) 6= 5. In the next theorem, we investigate unicyclic graph. Theorem 8.3 Let G be a unicyclic graph with cycle Ck of order n ≥ 3. Then tr(G)+tr(T (G)) = 5 if and only if G ∼ = Cn or (k, l)-kite or ∆(G) = 3 and G has exactly two adjacent vertices of degree 3 which are in Ck . Proof Let V (Ck ) = {v1 , v2 , . . . , vk } and E(Ck ) = {ei = vi vi+1 / 1 ≤ i ≤ k − 1} ∪ {ek = vk v1 }. Assume that tr(G) + tr(T (G)) = 5. Then tr(G) = 2 and tr(T (G)) = 3 or tr(G) = 3 and tr(T (G)) = 2. If tr(G) = 3 and tr(T (G)) = 2, then the above discussion, G ∼ = Pn , which is a contradiction. Therefore tr(G) = 2 and tr(T (G)) = 3. By Observations 2.6 and 2.7, ∆(G) ≤ 3. If ∆(G) = 2, then G ∼ = Cn . Therefore n′ = 1 and hence total resolving number does not exist. If ∆(G) = 3, then by Observations 2.6 and 2.7, G has at most three vertices of degree 3 which are in Ck . We claim that G has at most two vertices of degree 3. Suppose that G has three vertices of degree 3. By Observations 2.6 and 2.7, such vertices are consecutive and k is odd. Suppose k = 3. By our assumption, d(vi ) = 3, i = 1, 2, 3. Let v4 , v5 , v6 be the neighbors of v1 , v2 , v3 respectively. Let vij be the new vertex of the edge vi vj in T (G). If either W ⊂ V (G) or W ⊂ E(G), then we can easily verify that two vertices of V (T (G)) \ W have the same

102


representations. So we assume that neither W ⊂ V (G) nor W ⊂ E(G). If W = {v13 , v12 , v2 }, then r(v1 |W ) = r(v23 |W ) = (1, 1, 1), W = {v1 , v2 , v23 }, then r(v36 |W ) = r(v12 |W ) = (1, 1, 1), W = {v4 , v1 , v12 }, then r(v2 |W ) = r(v13 |W ) = (2, 1, 1), W = {v14 , v1 , v13 }, then r(v23 |W ) = r(v36 |W ) = (2, 2, 1), which is a contradiction. Now, we may assume that k ≥ 5. Let d(vk−2 ) = d(vk ) = 3 and vk+1 , vk+2 be neighbors of vk−2 and vk respectively in G. Let w2 be the neighbor of w1 andw3 in T (G). If W ⊂ V (G), then there exists either d(vk−2 , w1 ) < d(vk , w1 ) or d(vk−2 , w3 ) < d(vk , w3 ). Without loss of generality let d(vk−2 , w1 ) < d(vk , w1 ). Then r(v(k+1)(k−2) |W ) = r(vk+1 |W ), which is a contradiction. If W ⊂ E(G), then the neighbors of w1 which is not adjacent to w2 have the same representations, which is a contradiction. So we may assume that one vertex W in E(G) and others are in V (G) or one vertex of W in V (G) and others are in E(G). Then there exists d(w1 , vk−2 ) ≤ d(w1 , vk ) or d(w1 , vk−2 ) ≤ d(w, vk ). Withous loss of generality, let d(w1 , vk−2 ) ≤ d(w1 , vk ). Then r(v(k+1)(k−2) |W ) = r(v(k−1)(k−2) |W ), which is a contradiction. Thus G has at most two vertices of degree 3. If G has one vertex of degree 3, then G ∼ = (k, l)-kite. If G has two vertices of degree 3, then by Observations 2.6 and 2.7, they are either adjacent or non adjacent with distance two and k is odd. If they are non adjacent with distance two and k is odd, then the proof is similar to G has three consecutive vertices of degree 3, and hence we can easily verify that tr(T (G)) ≥ 4, which is a contradiction. Thus G has two adjacent vertices of degree 3 which are in Ck . The converse can be easily verified. 2 Theorem 8.4 Let G be a tree of order n ≥ 5. Then tr(G) + tr(T (G)) = 6 if and only if G is a spider with ∆(G) = 3. Proof Assume that tr(G) + tr(T (G)) = 6. Then tr(G) = 3 and tr(T (G)) = 3. Since tr(T (G)) = 3, by Theorems 6.3 and 8.2, G is a spider with ∆(G) = 3. The converse is obvious.

2

Theorem 8.5 ([8]) Let G be a graph of order n ≥ 3 and δ(G) ≥ 2. Then tr(T (G)) ≤ n − 1. Theorem 8.6 ([11]) Let G be a graph of order n ≥ 3 and δ(G) ≥ 2. Then tr(G) + tr(T (G)) ≤ 2n − 2. Proof The proof follows from Theorems 2.3 and 8.5.

2

Theorem 8.7 Let G be a graph of order n ≥ 3 and δ(G) ≥ 2. Then tr(G) + tr(T (G)) = 2n − 2 if and only if G ∼ = Kn . Proof Assume that tr(G) + tr(T (G)) = 2n − 2. Then tr(G) = n − 1 and tr(T (G)) = n − 1. By Theorem 2.4, if tr(G) = n−1, then G ∼ = Kn or K1,n−1 . Since δ(G) ≥ 2, G ∼ = Kn . Conversely, if G ∼ = Kn , then by Theorem 2.4, tr(G) = n − 1. By Theorem 8.5, tr(T (G)) ≤ n − 1. Next, we claim that tr(T (G)) ≥ n − 1. Suppose that tr(T (G)) ≤ n − 2. If W ⊂ V (G), then without loss of generality, let W = {v1 , v2 , · · · , vn−2 }, then r(vn−1 |W ) = r(vn |W ) = (2, 2, · · · , 2) in T (G), which is a contradiction. If W ⊂ E(G), then we can easily verify that one vertex of V (G) and one vertex of E(G) have the same representations in T (G), which is a contradiction. If neither


103

W ⊂ V (G) nor W ⊂ E(G), then neighbors of two vertices of V (G) are not in W ∩ E(G). Let vn−1 and vn be such vertices. Then r(vn |W ) = r(vn−1 |W ), which is a contradiction. Thus tr(T (G)) ≥ n − 1 and hence tr(T (G)) = n − 1. Hence tr(G) + tr(T (G)) = 2n − 2. 2 Problem 8.8 If G is a connected graph of order n ≥ 3 and δ(G) = 1, then find the upper bound for tr(G) + tr(T (G)) and characterize the extremal graphs.

Acknowledgement The research work of the second author is supported by the University Grants Commission, New Delhi through Basic Science Research Fellowship (vide Sanction No.F.7-201/2007(BSR).

References [1] M. Aouchiche,P. Hansen, A survey of Nordhaus-Gaddum type relations, Discrete Applied Mathematics, 161, 466-546, 2013. [2] M. Behzad, A criterion for the planarity of a total graph, Proc. Cambridge Philos. Soc., 63, 679-681, 1967. [3] G. Chartrand, L.Eroh, M. A. Johnson and O. R. Oellermann, Resolvability in graphs and metric dimension of a graph, Discrete Appl. Math. 105(2000), 99-113. [4] Cong X. Kang, Eunjeong Yi and Linda Eroh, On metric dimension of graphs and their complements, JCMCC 83(2012), 193-203. [5] Frank Harary, A characterization of block graphs, Canad. Math. Bull., 6(1-6)1963. [6] J. Paulraj Joseph, New directions in graph theory, National Conference on Discrete Mathematics and its Applications, Department of Mathematics, Manonmaniam Sundaranar University, Tirunelveli, 2015. [7] J. Paulraj Joseph and N. Shunmugapriya, Total resolving number of a graph, Indian Journal of Mathematics, Vol 57, No. 3(2015), 323-343. [8] J. Paulraj Joseph and N. Shunmugapriya, Total resolving number of block graphs and line graphs, International Journal of Pure and Applied Mathematical Sciences, Vol. 10, No. 2(2017), 189-195. [9] J. Paulraj Joseph and N. Shunmugapriya, Total resolving number of graphs - some characterizations, International Journal of Applied Graph Theory, Vol.2, No 1(2018). [10] J. Paulraj Joseph and N. Shunmugapriya, Total resolving number of power graphs, International Journal of Mathematics and its Applications, Vol. 4, Issue 4- F, (2017), 909-913. [11] N. Shunmugapriya and J. Paulraj Joseph, Total resolving number of subdivision graphs and total graphs, International Journal of Advances in Mathematics, Vol. 2018, No.3(2018), 34-40. [12] Ping Zhang and Varaporn Saenpholphat, Connected resolvability of graphs, Czechoslovak Mathematical Journal, Vol 53(2003), No. 4, 827-840. [13] Ping Zhang and Varaporn Saenpholphat, On connected resolvability of graphs, Australasian Journal of Combinatorics, Vol 28(2003), 25-37.


Bounds for the Non-Neighbor Harmonic Index of Subdivision Graphs A.Rizwana (Department of Mathematics, Sadakathullah Appa College, Tirunelveli - 627 011, India)

G.Jeyakumar (Department of Mathematics, St.John’s College, Palayamkottai - 627 002, India) E-mail: [email protected], [email protected]

Abstract: P

uv∈E(G)

The non-neighbor harmonic index of a graph G is defined as H(G) =

2 d(u)+d(v)

where d(u) denotes the number of non-neighbors of a vertex u in G.

In this paper, we determine the upper and lower bounds for the non- neighbor harmonic index of subdivision and t-subdivision graphs.

Key Words: Non-neighbor harmonic index, subdivision graphs, t-subdivision graphs. AMS(2010): 05C07, 92E10. §1. Introduction Molecular graph is connected undirected graph one-to-one corresponded to structural formula of chemical compound so that vertices of the graph correspond to atoms of the molecule and edges of the graph correspond to chemical bonds between these atoms. In the fields of chemical graph theory, molecular topology, and mathematical chemistry, a topological index also known as a connectivity index is a type of a molecular descriptor that is calculated based on the molecular graph of a chemical compound. Topological indices are numerical parameters of a graph which characterize its topology and are usually graph invariant [8]. Topological indices are used for example in the development of quantitative structure-activity relationships (QSARs) in which the biological activity or other properties of molecules are correlated with their chemical structure [10]. The Hosoya index is the first topological index recognized in chemical graph theory, and it is often referred to as the topological index [6]. Other examples include the Wiener index [16] , Randic’s molecular connectivity index [7], Sum connectivity index [18], Zagreb Indices and Zagreb Coindices [9]. In this paper we are concerned with simple graphs G, having no directed or weighted edges, and no self loops. The degree of the vertex v ∈ V (G), written d(v), is the number of first neighbors of v in the underlying graph G. In the 1980’s, Siemion Fajtlowicz created a vertex-degree-based quantity which was re-introduced by Zhong in 2012 called Harmonic Index 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 16, 2018, Accepted May 14, 2018, Edited by R. Kala.

105

Bounds for the Non-Neighbor Harmonic Index of Subdivision Graphs

[14],[15]. The harmonic index is defined as H(G) =

P

uv∈E(G)

2 d(u)+d(v)

where d(u) is the degree

of the vertex u ∈ V (G) [1],[11],[12]. The vertices that are not adjacent to a vertex v ∈ V (G) are called non-neighbors of the vertex v. In this paper we define d(v) as the number of the nonneighbors of the vertex v ∈ V (G); where d(v) = n−1−d(v). Depending on the non-neighbors of P 2 the vertices of G the non-neighbor harmonic index [17] is defined as H(G) = d(u)+d(v) uv∈E(G)

where d(u) = n − 1 − d(u). For any vertex v ∈ V (G), we use N (v) to denote the set of neighbors of v in G. The minimum degree of G is denoted by δ and the maximum degree of G is denoted by ∆. We use Sn , Pn and Kn to denote the star, path and complete graph on n vertices, respectively. For any graph G and for two distinct vertices u and v of G, the distance between u and v in G is the number of edges in a shortest path joining u and v. The eccentricity e(v) of a vertex v in a connected graph G is the maximum graph distance between v and any other vertex u of G. The maximum eccentricity is the graph diameter denoted by diam(G). The minimum graph eccentricity is called the graph radius denoted by rad(G). For terms and definition not given in this paper, we refer to [3],[4],[5],[13]. For a graph G, the subdivision graph S(G) is a graph obtained from G by replacing each edge of G by a path of length 2. The t-subdivision graph St (G) of G is a graph obtained from G by replacing each edge of G by a path of length t + 1. Obviously, S1 (G) = S(G). In this paper, we present some sharp bounds for the non-neighbor harmonic index of the subdivision and t-subdivision graphs. We also characterize graphs with these sharp bounds.

§2. Preliminaries Lemma 2.1(Schweitzer’s Inequality, [2]) Let x1 , x2 , ...xn be positive real numbers such that for 1 ≤ i ≤ n holds m ≤ xi ≤ M . Then ! n ! n X X 1 n2 (m + M )2 xi ≤ . x 4mM i=1 i=1 i The equality holds if and only if x1 = ... = xn = m = M or n is even , x1 = ... = x n2 = m and x n2 +1 = ... = xn = M , where m < M and x1 ≤ ... ≤ xn . Definition 2.2([2]) A graph G is called a (∆, δ)-bidegreed if whose vertices have degree either ∆ or δ(∆ 6= δ). P d(u) Definition 2.3 The degree ratio of a graph G is defined as D(G) = , where u∈V (G)

d(u)

d(u) = n − 1 − d(u) > 0.

Definition 2.4([17]) For any graph G, the non-neighbor first Zagreb index is defined as P M1 (G) = d(u) + d(v) . uv∈E(G)

Definition 2.5([2]) Let G be a graph of order n and size m. If m = n − 1 then G is called a tree, if m = n then G is called uincyclic graph and if m = n + 1 then G is called bicyclic graph.

106

A.Rizwana and G.Jeyakumar

§3. The Non-Neighbor Harmonic Index of Subdivision and t-Subdivision Graphs In this section, we present the formula for computing the non-neighbor harmonic index of subdivision and t-subdivision graphs. The non-neighbor harmonic index of subdivision graphs depends on the two quantities p and q where p = m + n − 2 and q = 2m + n − 3 while the non-neighbor harmonic index of t-subdivision graphs depends on the two quantities pt and qt where pt = tm + n − 2 and qt = 2tm + n − 3. Let G be a graph of order n and size m ≥ 1. The non-neighbor harmonic index of the subdivision graph G is given by H(S(G)) =

X

uv∈E(G)

"

2 d(u) + q

+

2 q + d(v)

#

where, q = 2m + n − 3.

,

For the t-subdivision graph G, We have H(St (G)) =

X

uv∈E(G)

"

2 d(u) + 2tm + n − 3 +

X

uv∈E(G)

H(St (G)) =

X

uv∈E(G)

H(St (G)) =

X

uv∈E(G)

" "

2

+  

d(u) + 2tm + n − 3

2 2tm + n − 3 + d(v)

2 2(tm + n − 3)



(t−1) times

z }| { +...+ #

2

+

#

+

2 , 2(tm + n − 3)

2(t − 1)m , 2(tm + n − 3)

2tm + n − 3 + d(v) # 2 2 (t − 1)m where qt = 2tm + n − 3. + + (tm + n − 3) d(u) + qt qt + d(v)

Example 3.1 We already know the following results. (1) For any n ≥ 2, H(St (Kn )) =

4m 2tm+n−3

(2) For any n ≥ 3, H(St (Cn )) =

m(t+1) tm+n−3 .

(3) For any n ≥ 2, H(St (Pn )) =

4 2tm+2n−5

+

(t−1)m tm+n−3 .

+

tm+2n−m−4 tm+n−3 .

Example 3.2 Let G be a k- regular graph with order n and size m, then H(S(G)) =

4m 4m (t − 1)m and H(St (G)) = + . 2m + 2n − k − 4 2tm + 2n − k − 4 tm + n − 3

Theorem 3.3 Let G be a graph with order n and size m, then H(S(G)) =

X

u∈V (G)

4p − 2n, where, p = m + n − 2. 2p − d(u)

Proof Let u be a vertex of G. For each neighbor of u,

2 d(u)+2m+n−3

appears exactly once

107


in

P

uv∈E(G)

h

2 d(u)+2m+n−3

H(S(G))

+

2 2m+n−3+d(v)

i . Thus



2

X

=

uv∈E(G)

X

=

u∈V (G)

X

=

u∈V (G)

X

=

u∈V (G)

X

=

u∈V (G)



d(u) + 2m + n − 3 2d(u)

d(u) times

z }| { +...+

2 2m + n − 3 + d(u)

 

d(u) + 2m + n − 3 2d(u) 2(m + n − 2) − d(u) 4(m + n − 2) − 2n 2(m + n − 2) − d(u) 4p − 2n, where, p = m + n − 2. 2p − d(u)

2


Corollary 3.4 Let G be a graph with order n and size m, then H(St (G)) =

X

u∈V (G)

4pt (t − 1)m − 2n + where pt = tm + n − 2. 2pt − d(u) tm + n − 3

§4. Upper Bounds for the Non-Neighbor Harmonic Index of Subdivision and t-Subdivision Graphs

In this section, the upper bounds for the non-neighbor harmonic index of subdivision and tsubdivision graphs based on ∆, δ, D(G) and rad(G) is provided using Schweitzer’s inequality. Also their corresponding extremal graphs are given.

Theorem 4.1 Let G be a graph with order n and size m, then H(S(G)) ≤

pn2 (4p − ∆ − δ)2 − 2n where p = m + n − 2, 2(np − m)(2p − δ)(2p − ∆)

and with the equality if and only if G is a regular graph or a (∆, δ)-bidegreed graph.

Proof For every vertex u in G, we have 2p − ∆ ≤ 2p − d(u) ≤ 2p − δ, where p = m + n − 2. P Using Schweitzer’s inequality and 2p − d(u) = 2pn − 2m we get u∈V (G)

108


X

u∈V (G)

2p − d(u)

X

u∈V (G)

X

u∈V (G)

X

u∈V (G)

1 n2 (2p − ∆ + 2p − δ)2 ≤ 2p − d(u) 4(2p − δ)(2p − ∆) 1 n2 (4p − ∆ − δ)2 ≤ 2p − d(u) 8(np − m)(2p − δ)(2p − ∆) 4p pn2 (4p − ∆ − δ)2 ≤ . 2p − d(u) 2(np − m)(2p − δ)(2p − ∆)

By Theorem 3.3, We have

pn2 (4p − ∆ − δ)2 H(S(G)) ≤ − 2n. 2(np − m)(2p − δ)(2p − ∆) Moreover, By Schweitzer’s inequality, equality holds in Theorem 4.1 if and only if δ = ∆ or vertices of G have degree δ and the remaining n2 vertices have degree ∆, (i.e) G is regular or (∆, δ)-bidegreed. 2 n 2

Corollary 4.2 Let G be a graph with order n and size m, then H(St (G)) ≤

pt n2 (4pt − ∆ − δ)2 (t − 1)m − 2n + where pt = tm + n − 2. 2(npt − m)(2pt − δ)(2pt − ∆) tm + n − 3

with equality if and only if G is a regular graph or a (∆, δ)-bidegreed graph. Theorem 4.3 Let G be a graph with order n and size m, then H(S(G)) ≤

1 m D(G) + where q = 2m + n − 3. 2 q

Proof Using Schweitzer’s inequality, for every vertex u ∈ V (G) we have 1 d(u)

+

4 1 ≥ q d(u) + q

with equality if and only if d(u) = q. Thus X

  X d(u) 1  X d(u)  ≤ + 2 q d(u) + q d(u) u∈V (G) u∈V (G) 2d(u)

u∈V (G)

By Theorem 4.3 we have H(S(G)) ≤ 21 D(G) +

m q ,

where q = 2m + n − 3.

Corollary 4.4 Let G be a graph with order n and size m, then H(St (G)) ≤

1 m (t − 1)m D(G) + + , where, qt = 2tm + n − 3. 2 qt tm + n − 3

2


109

Corollary 4.5 Let G be a graph with order n and size m, then 2n(n − 1) where q = 2m + n − 3 and q 2n(n − 1) (t − 1)m H(St (G)) ≤ + where qt = 2tm + n − 3 qt tm + n − 3 H(S(G)) ≤

In both cases, equality holds if and only if G ∼ = Kn . Theorem 4.6 Let G be a graph with order n and size m, then H(S(G)) ≤

2n(n − rad(G)) , where, p = m + n − 2 2p + rad(G) − n

and with equality if and only if G ∼ = Kn or G ∼ = CPn . Proof For every vertex u ∈ V (G), we have d(u) ≤ n − e(u) 1 1 1 ⇒ ≤ ≤ 2p − d(u) 2p + e(u) − n 2p + radG − n X 4p 4pn ⇒ ≤ . 2p − d(u) 2p + radG − n u∈V (G)

By Theorem 3.3 we have H(S(G)) ≤

2n(n − rad(G)) . 2p + radG − n

Moreover equality holds if and only if G is self-centered and for every vertex u ∈ V (G), equality holds in d(u) ≤ n − e(u). If e(u) = 1 for some vertex u ∈ V (G) then d(u) = n − 1 and e(v) ≤ 2 for all vertices v 6= u. Since G is self-centered, e(u) = 1 for all vertices u ∈ V (G). Thus G ∼ = Kn . Suppose e(u) ≥ 2 for all vertices u ∈ V (G). If e(u) ≥ 3 for some vertex v then diam(G) = 3 and G ∼ = P4 . This contradicts that G is self-centered. So e(u) = 2 for all vertices u ∈ V (G) and then d(u) = n − 2 for all vertices u ∈ V (G). Hence G ∼ = CPn . Hence equality ∼ ∼ holds if and only if G = Kn or G = CPn . 2 Corollary 4.7 Let G be a graph with order n and size m, then H(St (G)) ≤

2n(n − rad(G)) (t − 1)m + , where, pt = m + n − 2 2pt + rad(G) − n tm + n − 3

and with equality if and only if G ∼ = Kn or G ∼ = CPn . Theorem 4.8 Let G be a graph with order n, size m and r pendant vertices, then H(S(G)) ≤

4pr 4p(n − r) + − 2n where p = m + n − 2, 2p − 1 2p − ∆

110


and with equality if and only if G is regular graph (or) (∆, 1)-bidegreed graph. Proof X

u∈V (G)

  r times z }| { 1 1 1 1  = + +...+ 2p − d(u) 2p − 1 2p − 1 2p − 1 X

+

u∈V (G) and d(u)

Proof Let f (x) =

1 x.

4m 2 − 2 M1 (G) where q = 2m + n − 3. q q

Since f is a convex function on (0, ∞), by Jensen’s inequality for

111


every edge uv ∈ E(G), we have 2 d(u) + q

2

+

d(v) + q

≥

8 d(u) + d(v) + 2q

with equality if and only if d(u) = d(v). Hence H(S(G)) =

X

uv∈E(G)

≥ =

2

+

d(u) + q

2 q + d(v)

#

8

X

uv∈E(G)

X

uv∈E(G)

=

"

4m 2 − 2 q q

d(u) + d(v) + 2q !−1 !−1 d(u) + d(v) 4 4 X d(u) + d(v) > 1− 1+ 2q q q 2q uv∈E(G) X 4m 2 d(u) + d(v) = − 2 M1 (G). q q

uv∈E(G)

2

Corollary 5.2 Let G be a graph with order n and size m, then H(St (G)) >

4m (t − 1)m 2 − 2 M1 (G) + where qt = 2tm + n − 3. qt qt tm + n − 3

Theorem 5.3 Let G be a graph with order n and size m, then H(S(G)) ≥ p = m + n − 2 with equality if and only if G is regular graph.

2mn pn−m ,

where

Proof By Cauchy-Schwartz inequality, we have X

u∈V (G)

(2p − d(u))

X

u∈V (G)

2 2p − d(u)



≥

with equality if and only if all the d(u)′ s are equal. P Since (2p − d(u)) = 2pn − 2m, we get

X

u∈V (G)

2 p 1  2p − d(u) p 2p − d(u)

u∈V (G)

X

u∈V (G)

By theorem 3.3, We have H(S(G)) ≥

4pn2 4p ≥ 2p − d(u) 2pn − 2m

2

2mn np−m .

Corollary 5.4 Let G be a graph with order n and size m, then H(S(G)) ≥ where pt = tm + n − 2 with equality if and only if G is regular graph.

2mn pt n−m

+

(t−1)m tm+n−3 ,

112


Corollary 5.5 Let G be a tree of order n, then 2n(n − 1) , where, p = m + n − 2, n(p − 1) + 1 2n(n − 1) (t − 1)(n − 1) H(St (G)) ≥ + , where, pt = tm + n − 2. n(pt − 1) + 1 t(n − 1) + n − 3 H(S(G)) ≥

In both cases, equality holds if and only if G ∼ = P2 . Corollary 5.6 Let G be an unicyclic graph of order n, then 2n , where, p = m + n − 2, p−1 2n (t − 1)n + , where, pt = tm + n − 2. H(St (G)) ≥ pt − 1 n(t + 1) − 3 H(S(G)) ≥

In both cases, equality holds if and only if G ∼ = Cn . Corollary 5.7 Let G be a bicyclic graph of order n, then 2n(n + 1) , where, p = m + n − 2, n(p − 1) − 1 2n(n + 1) (t − 1)(n − 1) H(St (G)) ≥ + , where, pt = tm + n − 2. n(pt − 1) − 1 t(n − 1) + n − 3 H(S(G)) ≥

References [1] Aleksandar Ilic, Note on the harmonic index of a Graph, arXiv:1204.3313v1 [math.CO] 15 Apr (2012). [2] Bibi Naime Onagh, The Harmonic Index of Subdivision Graphs, Transactions on Combinatorics Vol. 6 No. 4 (2017), pp. 15-27. [3] J.A Bondy and U.S.R. Murty, Graph Theory, Springer Berlin, (2008). [4] G.Chartrand and P.Zang, Introduction to Graph Theory, Tata McGraw-Hill, New-Delhi, (2006). [5] Douglas B.West, Introduction to Graph Theory, Second Edition, PHI Learning Private Limited, New Delhi. [6] Hosoya, Haruo, Topological index. A newly proposed quantity characterizing the topological nature of structural isomers of saturated hydrocarbons, Bulletin of the Chemical Society of Japan, 44 (9): 2332–2339, (1971). [7] Huiqing Liu, Mei Lu, Feng Tian, On the Randic index, Journal of Mathematical Chemistry Vol. 38, No. 3, October (2005). [8] Ivan Gutman, Degree-based topological indices, Croat. Chem. Acta, 86(4)(2013), 251-361. [9] Ivan Gutman, Boris Furtula, Zana Kovijanic Vukicevic and Goran Popivoda, On Zagreb Indices and Coindices, MATCH Commun. Math. Comput. Chem., 74(2015), 5-16.


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[10] James Devillers and Alexandru T.Balaban, Topological Indices and Related Descriptors in QSAR and QSPAR, Gordon and Breach Science Publishers. [11] Jianxi Li, Jian-Bo Lv, Yang Liu, The Harmonic Index of Some Graphs, Bull. Malays. Math. Sci. Soc. (2016) 39:S331S340. [12] Juan Carlos Hernandez-Gomez, Jesus Romero-Valencia, R. Reyes Carreto, Mathematical Aspects on the Harmonic Index International Journal of Mathematical Analysis, Vol. 11, (2017), no. 2, 85 - 95. [13] Krishnaiyan Thulasiraman, Subramanian Arumugam, Andreas Brandstadt and Takao Nishizeki, Handbook of Graph Theory, Combinatorial Optimization, and Algorithms, CRC Press, A Chapman & Hall Book. [14] Lingping Zhong, The harmonic index for Graphs, Applied Mathematics Letters, 25(2012)561566. [15] Lingping Zhong, Kexiang Xu, The harmonic index for Bicyclic Graphs, Article in Utilitas Mathematica, March (2013). [16] Martin Knor, Riste Skrekovski, Aleksandra Tepeh, Mathematical aspects of Wiener index, arXiv:1510.00800v1 [math.CO] 3 Oct (2015). [17] A.Rizwana, G.Jeyakumar, S.Somasundaram, On the Non-Neighbor Zagreb Indices and Non-Neighbor Harmonic Index, International Journal of Mathematics and its Applications, Volume 4, Issue 2-D (2016), 89-101. [18] Zhibin Du, Bo Zhou, On Sum-Connectivity Index of Bicyclic Graphs, Bull. Malays. Math. Sci. Soc.(2)35(1) (2012), 101-117.


4−Remainder Cordial Labeling Graphs Obtained From Ladder Ponraj. R Department of Mathematics Sri Paramakalyani College, Alwarkurichi–627 412, Tamil Nadu, India

Annathurai. K Department of Mathematics, Thiruvalluvar College Affiliated to Manonmaniam Sundaranar University, Tirunelveli, Tamil Nadu, India

Kala. R Department of Mathematics Manonmaniam Sundaranar University, Tirunelveli, Tamil Nadu, India E-mail: [email protected], [email protected], [email protected]

Abstract: Let G be a (p, q) graph. Let f be a map from V (G) to the set {1, 2, · · · , k} where k is an integer 2 < k ≤ |V (G)|. For each edge uv assign the label r where r is the remainder when f (u) is divided by f (v) (or) f (v) is divided by f (u) according as f (u) ≥ f (v) or f (v) ≥ f (u). The function f is called a k-remainder cordial labeling of G if |vf (i) − vf (j)| ≤ 1, i, j ∈ {1, · · · , k} where vf (x) denote the number of vertices labelled with x and |ef (0) − ef (1)| ≤ 1 where ef (0) and ef (1) respectively denote the number of edges labeled with even integers and number of edges labelled with odd integers. A graph with a k-remainder cordial labeling is called a k-remainder cordial graph. In this paper we investigate the 4−remainder cordial labeling behavior of Ln ⊙ K1 , Ln ⊙ 2K1 , Ln ⊙ K2 , and some ladder related graphs, where G1 ⊙ G2 denotes the corona of G1 with G2 .

§1. Introduction Graphs considered here are finite and simple. The corona of G with H, G ⊙ H is the graph obtained by taking one copy of G and p copies of H and joining the ith vertex of G with an edge to every vertex in the ith copy of H. Ponraj et al. [4], introduced remainder cordial labeling of graphs and investigate the remainder cordial labeling behavior of path, cycle, star, bistar, complete graph, etc, and also the concept of k-remainder cordial labeling introduced in [6] and investigate the 4-remainder cordial labeling behaviour of certain graphs. In this paper we present the 4−remainder cordial labeling behavior of Ln ⊙ K1 , Ln ⊙ 2K1 , Ln ⊙ K2 , and some ladder related graphs. Terms are not defined here follows from Harary [3] and Gallian [2]. 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 08, 2018, Accepted May 17, 2018, Edited by R. Kala.

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§2. k−Remainder Cordial Labeling Definition 2.1 Let G be a (p, q) graph. Let f be a map from V (G) to the set {1, 2, · · · , k} where k is an integer 2 < k ≤ |V (G)|. For each edge uv assign the label r where r is the remainder when f (u) is divided by f (v) (or) f (v) is divided by f (u) according as f (u) ≥ f (v) or f (v) ≥ f (u). The function f is called a k-remainder cordial labeling of G if |vf (i) − vf (j)| ≤ 1, i, j ∈ {1, · · · , k} where vf (x) denote the number of vertices labelled with x and |ef (0) − ef (1)| ≤ 1 where ef (0) and ef (1) respectively denote the number of edges labeled with even integers and number of edges labelled with odd integers. A graph with a k-remainder cordial labeling is called a k-remainder cordial graph. Notation 1 The ladder is the graph Ln = Pn XK2 with vertex set V (Ln ) = {ui , vi : 1 ≤ i ≤ n} and E(Ln ) = {ui vi } ∪ {ui ui+1 , vi vi+1 : 1 ≤ i ≤ n − 1}. Clearly the order and size of the graph Ln are 2n and 3n − 2 respectively. Now we investigate the 4−remainder cordial labeling behavior of the graph Ln ⊙ K1 . Theorem 2.2 The graph Ln ⊙ K1 is 4−remainder cordial for all n. Proof Let V (Ln ⊙K1 ) = V (Ln )∪{xi , yi : 1 ≤ i ≤ n} and E(Ln ⊙K1 ) = E(Ln )∪{ui xi , vi yi : 1 ≤ i ≤ n}. Note that Ln ⊙ K1 has 4n vertices and 5n − 2 edges. We now give a 4−remainder cordial labeling to the graph Ln ⊙ K1 . The proof of this theorem can be proved in the following two cases.

Case 1. n is even. Assign the label 2 to the vertices u1 , u3 , · · · , un−1 and 3 to the vertices u2 , u4 , · · · , un . Next assign the label 1 to the vertices v1 , v3 , · · · , vn−1 and 4 to the vertices v2 , v4 , · · · , vn . We now consider the pendant vertices xi and yi . Assign the label 1 to the vertices x1 , x3 , · · · , xn−1 and 4 to the vertices x2 , x4 , · · · , xn . Finally assign the label 2 to the vertices y1 , y3 , · · · , yn−1 and 3 to the vertices y2 , y4 , · · · , yn . Case 2. n is odd. As in Case 1, assign the labels to the vertices ui , vi , xi , yi , (1 ≤ i ≤ n − 1). Next finally assign the labels 1, 2, 4 and 3 respectively to the vertices un , vn , xn , and yn . Thus the table given below establish that this vertex labeling f is 4-remainder cordial labeling of Ln ⊙ K1 graph. 2 Nature of n

vf (1)

vf (2)

vf (3)

vf (4)

ef (0)

ef (1)

5n−2 2 5n−1 2

5n−2 2 5n−3 2

n is even

n

n

n

n

n is odd

n

n

n

n

Next we investigate the 4−remainder cordial labeling behavior of the graph Ln ⊙ 2K1 . Theorem 2.3 The graph Ln ⊙ 2K1 is 4−remainder cordial for all n. ′

′

Proof Let V (Ln ⊙ 2K1 ) = V (Ln ) ∪ {xi , xi , yi , yi : 1 ≤ i ≤ n} and E(Ln ⊙ 2K1 ) =

116

Ponraj. R, Annathurai. K and Kala. R

′

′

E(Ln ) ∪ {ui xi , ui xi , vi yi , vi yi : 1 ≤ i ≤ n}. Then it is easy to verify that the order and size of the Ln ⊙ 2K1 are 6n and 7n − 2 respectively. Case 1. n is even. Let f be the 4-remainder cordial labeling of Ln ⊙ K1 . Define a map g : V (Ln ⊙ K1 ) → ′ ′ {1, 2, 3, 4} given by g(ui ) = f (ui ), g(vi ) = f (vi ), g(xi ) = g(xi ) = f (xi ), g(yi = g(yi )) = f (yi ), (1 ≤ i ≤ n). Case 2. n is odd. ′

′

Assign the labels to the vertices ui , vi , xi , xi , yi , yi , (1 ≤ i ≤ n − 1) as in case(i). Finally ′ ′ assign the labels 2, 4, 3, 3, 1 and 1 to the vertices un , vn , xn , xn , yn , and yn respectively. The table given below establish that the induced vertex labeling g is a 4-remainder cordial labeling of Ln ⊙ 2K1 . 2 Nature of n

vf (1)

vf (2)

vf (3)

vf (4)

ef (0)

ef (1)

n is even

3n 2 3n+1 2

3n 2 3n−1 2

3n 2 3n+1 2

3n 2 3n−1 2

7n−2 2 7n−1 2

7n−2 2 7n−3 2

n is odd

Finally we investigate the 4−remainder cordial labeling behavior of the graph Ln ⊙ K2 . Theorem 2.4 The graph Ln ⊙ K2 is 4−remainder cordial for all n. ′

′

Proof Let V (Ln ⊙ K2 ) = V (Ln ) ∪ {xi , xi , yi , yi : 1 ≤ i ≤ n} and E(Ln ⊙ K2 ) = E(Ln ) ∪ ′ ′ ′ ′ {ui xi , ui xi , xi xi , vi yi , vi yi , yi yi : 1 ≤ i ≤ n}. Then clearly the order and size of the Ln ⊙ K2 are 6n and 9n − 2 respectively. Case 1. n ≡ 0 (mod 4). Assign the labels 1, 2, 3, 4 to the vertices u1 , u2 , u3 , u4 . Next assign the labels 1, 2, 3, 4 respectively to the vertices u5 , u6 , u7 , u8 . Continuing in this pattern until we reach the vertex un . Note that un received the label 4. Now assign the labels 4, 3, 2, 1 to the vertices v1 , v2 , v3 , v4 . Next assign the labels 4, 3, 2, 1 respectively to the vertices v5 , v6 , v7 , v8 . Proceeding like this until ′ ′ we reach the vertex vn which is received the label 1. We now consider the vertices xi , xi , yi , yi of degree two. Assign the label 1 to the vertices x1 , x3 , x5 , · · · and 3 to the vertices x2 , x4 , x6 and ′ ′ ′ ′ ′ ′ so on. Next assign the label 2 to the vertices x1 , x3 , x5 , · · · and 4 to the vertices x2 , x4 , x6 , and ′ ′ ′ so on. Next assign the label 4 to the vertices y1 , y3 , y5 , · · · and 3 to the vertices y1 , y3 , y5 , · · · . ′ ′ ′ Assign the label 2 to the vertices y2 , y6 , y10 , · · · and y2 , y6 , y10 , · · · . Finally assign the label 1 ′ ′ ′ to the vertices y4 , y8 , y12 , · · · and y4 , y8 , y12 , and so on. Case 2. n ≡ 1 (mod 4). ′

′

In this case assign the labels to the vertices ui , vi , xi , xi , yi , yi , (1 ≤ i ≤ n − 1) as in case(i). ′ ′ Finally assign the labels 3, 2, 4, 2, 1 and 3 to the vertices un , vn , xn , xn , yn , and yn respectively. Case 3. n ≡ 2 (mod 4). ′

′

′

Assign the label 1 to the vertices x1 , x5 , x9 , · · · and x1 , x5 , x9 , · · · . Next assign the label ′ ′ ′ 2 to the vertices x3 , x7 , x11 , · · · and x3 , x7 , x11 , and so on. Assign the label 3 to the vertices

117

4−Remainder Cordial Labeling Graphs Obtained From Ladder

′

′

′

x2 , x4 , x6 , · · · and 4 to the vertices x2 , x4 , x6 , and so on. We now consider the vertices of degree 4. Assign the labels 1, 2, 3, 4 to the vertices u1 , u2 , u3 , u4 . Next assign the labels 1, 2, 3, 4 respectively to the vertices u5 , u6 , u7 , u8 . Continuing like this until we reach the vertex un−2 . Next assign the labels 1, 2 to the vertices un−1 and un respectively. Assign the labels 4, 3, 2, 1 respectively to the vertices v1 , v2 , v3 , v4 . Next assign the labels 4, 3, 2, 1 to the vertices v5 , v6 , v7 , v8 respectively. Proceeding like this until we reach the vertex vn−2 which is received the label 1. Finally assign the labels 4, 3 to the vertices vn−1 and vn respectively. We now consider ′ the vertices yi , and yi . Assign the label 4 to the vertices y1 , y3 , y5 , · · · and 3 to the vertices ′ ′ ′ ′ ′ ′ y1 , y3 , y5 , · · · . Next assign the label 2 to the vertices y2 , y6 , y10 , · · · and y2 , y6 , y10 , · · · . Finally ′ ′ ′ assign the label 1 to the vertices y4 , y8 , y12 , · · · and y4 , y8 , y12 , and so on. Case 4. n ≡ 3 (mod 4). ′

′

Assign the labels to the vertices ui , vi , xi , xi , yi , yi , (1 ≤ i ≤ n − 1) as in Case 3. Finally ′ ′ assign the labels 3, 2, 1, 4, 1 and 3 to the vertices un , vn , xn , xn , yn , and yn respectively. Thus the table following shows that this vertex labeling f is 4-remainder cordial labeling of Ln ⊙ K2 . Nature of n

vf (1)

vf (2)

vf (3)

vf (4)

ef (0)

ef (1)

n ≡ 0, 2 (mod 4)

6n 4 6n−2 4

6n 4 6n+2 4

6n 4 6n+2 4

6n 4 6n−2 4

9n−2 2 9n−3 2

9n−2 2 9n−1 2

n ≡ 1, 3 (mod 4)

For illustration, 4-remainder cordial labeling of L7 ⊙ K2 is shown in Figure 1. 1

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Figure 1 References [1] Cahit I., Cordial graphs : A weaker version of graceful and harmonious graphs, Ars combin., 23 (1987), 201–207. [2] Gallian J.A., A dynamic survey of graph labeling, The Electronic Journal of Combinatorics., 19, (2016). [3] Harary F., Graph Theory, Addision wesley, New Delhi, 1969. [4] Ponraj R., Annathurai K., and Kala R., Remainder cordial labeling of graphs, Journal of Algorithms and Computation, Vol.49, (2017), 17–30. [5] Ponraj R., Annathurai K. and Kala R., Remaider cordiality of some graphs, Accepted for publication in Palestin Journal of Mathematics . [6] Ponraj R., Annathurai K. and Kala R., k-Remaider cordial graphs, Journal of Algorithms and Computation,Vol.49(2), (2017), 41-52.


Non-Split Perfect Triple Connected Domination Number of Different Product of Paths G. Mahadevan (Department of Mathematics, Gandhigram Rural Institute - Deemed to be University, Gandhigram, Dindigul, India.)

T. Ponnuchamy (Research Department of Mathematics, Vivekananda College,, Tiruvedakam west, Madurai,Tamilnadu, India)

Selvam Avadayappan (Research Department of Mathematics, VHNSN College, Virudhunagar, India) E-mail: [email protected], [email protected], selvam [email protected]

Abstract: A graph G is said to be triple connected if any three vertices lie on a path in G. The concept of triple connected domination number was introduced by J. Paulraj Joseph et.al. recently. A dominating set S is said to be triple connected dominating set, if the sub graph hSi is triple connected. The minimum cardinality taken over all triple connected dominating sets is called triple connected domination number of a graph G and it is denoted by γtc (G). Motivated by the above recently the concept of non-split perfect triple connected domination number was introduced in [1]. A subset S of V of a non-trivial graph G is said to be non-split perfect triple connected dominating set, if S is a triple connected dominating set and hV − Si is connected and has at least one perfect matching. The minimum cardinality taken over all non-split perfect triple connected sets in G is called the non-split perfect triple connected domination number of G and is denoted by γnsptc (G). In this paper we investigate this number for different product of the paths.

Key Words: Non-split perfect triple connected domination number, Cartesian product, corona product, lexicographic product and strong product.

AMS(2010): 05C78. §1. Introduction By a graph we mean a finite, simple, connected and undirected graph G(V, E), where V denotes vertex set and E edge set. Unless otherwise stated, the graph G has p vertices and q edges. We denote a path on p vertices by Pp . The concept of connectedness plays an important role in any network. A variety of connectedness has been studied in the literature by considering the existence of a path between any two vertices. The concept of triple connected graphs was introduced by J. Paulraj Joseph et.al., A graph G is said to be triple connected if any three 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 30, 2018, Accepted August 02, 2018, Edited by R. Kala.

Non-Split Perfect Triple Connected Domination Number of Different Product of Paths

119

vertices lie on a path in G. A dominating set S is said to be triple connected dominating set, if the sub graph hSi is triple connected. The minimum cardinality taken over all triple connected dominating sets is called triple connected domination number of a graph G and it is denoted by γtc (G). The concept of complementary perfect domination number was introduced by J. Paulraj Joseph et.al., A set S ⊆ V is called complementary perfect dominating set, if S is a dominating set of G and the induced sub graph hV − Si has a perfect matching. The minimum cardinality taken over all complementary perfect dominating sets is called the complementary perfect domination number of G and is denoted by γcp (G). The concept of complementary perfect triple connected domination number was introduced by G. Mahadevan. et. al., A subset S of V of a non-trivial graph G is said to be complementary perfect triple connected dominating set, if S is a triple connected dominating set and the induced sub graph hV − Si has a perfect matching. The minimum cardinality taken over all complementary perfect triple connected dominating sets is called the complementary perfect triple connected domination number of G and is denoted by γcptc (G). The concept of non-split perfect triple connected domination number was introduced by G. Mahadevan. et. al., A subset S of V of a non-trivial graph G is said to be non-split perfect triple connected dominating set, if S is a triple connected dominating set and the induced sub graph hV − Si is connected and has a perfect matching. The minimum cardinality taken over all non-split perfect triple connected dominating sets is called the non-split perfect triple connected domination number of G and is denoted by γnsptc (G). The Cartesian product of the graphs G and H is denoted by G2H, ′ ′ ′ whose V (G2H) = {(g, h)/ g ∈ V (G) and h ∈ V (H)} E(G2H) = {(g, h)(g , h )/ g = g and ′ ′ ′ hh ∈ E(H) (or) h = h and gg ∈ E(G)}. The number of vertices in the Cartesian product of G2H is |V (G)||V (H)|. The Lexicographic Product of the graphs G and H is denoted by G ◦ H, ′ ′ ′ whose V (G ◦ H) = {(g, h)/ g ∈ V (G) and h ∈ V (H)} E(G ◦ H) = {(g, h)(g , h )/ gg ∈ E(G) ′ ′ (or) g = g and hh ∈ E(H)}. The number of vertices in the Lexicographic product of G ◦ H is |V (G)||V (H)|. The corona of two graphs G and H is the graph formed from one copy of G and |V (G)| copies of H where the ith vertex of G is adjacent to every vertex in the ith copy of H. It is denoted by G[H]. The number of vertices in the corona product of G and H is |V (G)|(|V (H) + 1|). The Strong product of the graphs G and H is denoted by G ⊠ H, whose V (G ⊠ H) = {(g, h)/g ∈ V (G) and h ∈ V (H)}. E(G ⊠ H) = E(G2H) ∪ E(G × H). The number of vertices in the Strong product of G ⊠ H is |V (G)||V (H)|. Motivated by the above, in this paper we investigate the non-split perfect triple connected domination number of different product of paths. §2. Lexicographic product In this section we obtain the non-split perfect triple connected domination number of the Lexicographic product of paths. Theorem 2.1 For any 2 ≤ m ≤ n and m < 5,  3 γnsptc (Pm ◦ Pn ) = 4

if m, n are odd, otherwise.

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G. Mahadevan, T. Ponnuchamy and Selvam Avadayappan

Proof Let Pm and Pn be path on m vertices and n vertices respectively. Then the Lexicographic product of Pm and Pn is denoted by Pm ◦ Pn . Here V (Pm ◦ Pn ) = {ui , vij /1 ≤ i ≤ m, 1 ≤ j ≤ n}. By assumption 2 ≤ m ≤ n and m < 5. Case 1. m is even. For any value of n the resultant graph contains even number of vertices. Also by the definition of non-split perfect triple connected dominating set S, |S| ≥ 3. Suppose the non-split perfect triple connected dominating set S contains 3 vertices, V − S has an odd number of vertices and hence hV − Si has no perfect matching. Therefore for any non-split perfect triple connected dominating set has at least 4 vertices. Hence γnsptc (Pm ◦ Pn ) = 4. Case 2. m is odd. Subcase 2.1 n is even. The resultant graph contains even number of vertices. Also by the definition of non-split perfect triple connected dominating set S, |S| ≥ 3. Suppose the non-split perfect triple connected dominating set S contains 3 vertices. In this case V − S has an odd number of vertices and hence hV − Si has no perfect matching. Therefore for any non-split perfect triple connected dominating set has at least 4 vertices. Hence γnsptc (Pm ◦ Pn ) = 4. Subcase 2.2 n is odd. The resultant graph contains odd number of vertices. Also by the definition of non-split perfect triple connected dominating set S, |S| ≥ 3. Suppose the non-split perfect triple connected dominating set S contains 3 vertices. In this case V − S has an even number of vertices and hence hV − Si has a perfect matching, which is minimum.Hence,  3 if m, n are odd γnsptc (Pm ◦ Pn ) = . 2 4 otherwise Theorem 2.2 For any 5 ≤ m ≤ n,

γnsptc (Pm ◦ Pn ) =

 m − 1 m − 2

if m is odd, n is even, otherwise.

Proof Let Pm and Pn be path on m vertices and n vertices respectively. Then the Lexicographic product of Pm and Pn is denoted by Pm ◦ Pn . Here V (Pm ◦ Pn ) = {ui , vij / 1 ≤ i ≤ m, 1 ≤ j ≤ n}. By assumption 5 ≤ m ≤ n Case 1. m is even. Let S = {(uj , v1 ) : 2 ≤ j ≤ m − 1}. Then S contains m − 2 vertices, hSi = Pm−2 which is triple connected. Also for any vertex in V − S is adjacent to some vertex in S and hence S is a triple connected dominating set. hV − Si = Pm ◦ P2 ∪ Pm ◦ P3 ∪ . . . ∪ Pm ◦ Pn ∪ 2P1 is connected, it has an even number of vertices, it has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ◦ Pn . Hence γnsptc (Pm ◦ Pn ) ≤ m − 2.


121

Claim 1.1 γnsptc (Pm ◦ Pn ) = m − 2. Suppose γnsptc (Pm ◦ Pn ) < m − 2. In this case there ′ ′ exists a set S such that S is a non-split perfect triple connected dominating set. Assume that ′ |S | = m − 3. ′

′

Subcase 1.1 Assume S ⊂ S. In this case (uk , v1 ) ∈ S − S . a) Suppose k = 2 or m − 1, ′ ′ there is a vertex (u1 , v1 ) or (um , v1 ) is not adjacent to any vertex in S . Therefore S is not ′ a dominating set and hence S is not a non-split perfect triple connected dominating set in ′ ′ Pm ◦ Pn . b) Suppose k 6= 2 and m − 1. In this case hS i is disconnected and hence S is not a non-split perfect triple connected dominating set in Pm ◦ Pn . ′

′

′

Subcase 1.2 Assume S ∩ S 6= Φ or S ∩ S = Φ. In this case S has an odd number of ′ ′ vertices, V − S has an odd number of vertices and hence hV − S i has no perfect matching. ′ Therefore S is not a non-split perfect triple connected dominating set in Pm ◦ Pn . Hence γnsptc (Pm ◦ Pn ) = m − 2. Case 2. m is odd. Subcase 2.1 n is odd. Let S = {(uj , v1 ) : 2 ≤ j ≤ m−1}. Then S contains m−2 vertices, hSi = Pm−2 which is triple connected. Also for any vertex in V − S is adjacent to some vertex in S, S is a triple connected dominating set. hV − Si = Pm ◦ P2 ∪ Pm ◦ P3 ∪ . . . ∪ Pm ◦ Pn ∪ 2P1 is connected, it has an even number of vertices, it has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ◦ Pn . Hence γnsptc (Pm ◦ Pn ) ≤ m − 2. ′

Claim 2.1 γnsptc (Pm ◦Pn ) = m−2. Suppose γnsptc (Pm ◦Pn ) < m−2, there exists a set S ′ ′ such that S is a non-split perfect triple connected dominating set. Assume that |S | = m − 3. ′

′

Subcase i Assume S ⊂ S. In this case (uk , v1 ) ∈ S − S . (i) Suppose k = 2 or m − 1, ′ ′ there is a vertex (u1 , v1 ) or (um , v1 ) is not adjacent to any vertex in S . Therefore S is not a ′ ′ dominating set. (ii) Suppose k 6= 2 and m − 1, hS i is disconnected. Hence S is not a non-split perfect triple connected dominating set in Pm ◦ Pn . ′

′

′

Subcase ii Assume S ∩ S 6= Φ or S ∩ S = Φ. In this case S has an even number of ′ ′ vertices, V − S has an odd number of vertices. Hence < V − S > has no perfect matching. ′ Therefore S is not a non-split perfect triple connected dominating set in Pm ◦ Pn . Hence γnsptc (Pm ◦ Pn ) = m − 2. Subcase 2.2 n is even. Let S = {(uj , v1 ) : 2 ≤ j ≤ m}. Then S contains m − 1 vertices, hSi = Pm−1 which is triple connected. Also for any vertex in V − S is adjacent to some vertex in S, S is a triple connected dominating set. hV − Si = Pm ◦ P2 ∪ Pm ◦ P3 ∪ · · · ∪ Pm ◦ Pn ∪ P1 is connected, it has an even number of vertices and hence it has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ◦ Pn . Hence γnsptc (Pm ◦ Pn ) ≤ m − 1. ′

Claim 2.2 γnsptc (Pm ◦Pn ) = m−1. Suppose γnsptc (Pm ◦Pn ) < m−1, there exists a set S ′ ′ such that S is a non-split perfect triple connected dominating set. Assume that |S | = m − 2. ′ ′ Then V − S has an odd number of vertices. In this case hV − S i has no perfect matching. ′ Therefore S is not a non-split perfect triple connected dominating set in Pm ◦ Pn . Hence

122


 m − 1 if m is odd, n is even γnsptc (Pm ◦ Pn ) = m − 1. Hence γnsptc (Pm ◦ Pn ) = . m − 2 otherwise This completes the proof.

2

Example 2.3 Consider the graphs. P2 :

u1

u2

b

b

P3 :

v1

v2

b

b

v3 b

Figure 2.1 Then, the Lexicographic Product P2 ◦ P3 is given in Figure 2.2. u1 , v1

u1 , v2

b

b

b

b

u2 , v1

u2 , v2

u1 , v3 b

b

u2 , v3

Figure 2.2 Here V (P2 ◦ P3 ) = {(ui , vj )/ 1 ≤ i ≤ 2, 1 ≤ j ≤ 3}. Let S = {(u1 , v1 ), (u1 , v2 ), (u2 , v1 ), (u2 , v2 )} be the subset of V (P2 ◦ P3 ). Every vertex in V − S is adjacent to some vertex in S. Therefore S is a dominating set. Also hSi = P2 ◦ P2 , it is triple connected. hV − Si = P2 , it is connected and has a perfect matching in P2 ◦ P3 . Therefore S is the non-split perfect triple connected dominating set in P2 ◦ P3 , which is minimum. Hence γnsptc (P2 ◦ P3 ) = 4. Similarly, γnsptc (P6 ◦ P6 ) = 4 = 6 − 2, γnsptc (P7 ◦ P8 ) = 6 = 7 − 1. Remark 2.4 In the corona product of paths Pm [Pn ], m, n ≥ 3, the non-split perfect triple connected domination number does not exist. Theorem 2.5 For any n ≥ 2,

 n + 3 if n is odd, γnsptc (P2 [Pn ]) = n + 2 if n is even.

Proof Let P2 and Pn be path on 2 vertices and n vertices respectively. Then the Corona product of P2 and Pn is denoted by P2 [Pn ]. Here V (P2 [Pn ]) = {ui , vij / 1 ≤ i ≤ 2, 1 ≤ j ≤ n}. Case 1. n is even. Consider the set S = V (P1 [Pn ] ∪ P1 ) and |S| = n + 2. Claim 1.1 S is a non-split perfect triple connected dominating set in P2 [Pn ]. Here

123


every vertex in V − S is adjacent to one vertex in S, S is a dominating set in P2 [Pn ]. Also hSi = P1 [Pn ]∪K1 , S is a triple connected set. hV − Si = Pn which is connected. Here n is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 [Pn ]. Hence γnsptc (P2 [Pn ]) ≤ |S|. ′

Claim 1.2 γnsptc (P2 [Pn ]) = |S|. Suppose γnsptc (P2 [Pn ]) < |S|, there exists a set S ′ such that S is a non-split perfect triple connected dominating set in P2 [Pn ]. Assume that ′ |S | = n + 1.vskip 2mm ′

′

Subcase A Assume S ⊂ S. Then at least one vertex vk ∈ S − S .

Subcase A.1 Suppose vk = u2 . In this case every vertex in the 2nd copy of Pn is not ′ ′ adjacent to any vertex in S , and hence S is not a dominating set in P2 [Pn ]. Subcase A.2 Suppose vk ∈ {v1j / 1 ≤ j ≤ n}. In this case hV − Si = Pn ∪ P1 , which is disconnected. ′

′

Subcase B Assume S ∩ S 6= Φ. Then the set S contains the vertices of 1 copy of Pn ′ except the vertices of P2 . But hS i is disconnected. Therefore S is the minimal non-split perfect triple connected dominating set in P2 [Pn ]. Hence γnsptc (P2 [Pn ]) = |S| = n + 2. Case 2. n is odd. Consider the set S = V (P1 [Pn ] ∪ P2 ) and |S| = n + 3. Claim 2.1 S is a non-split perfect triple connected dominating set in P2 [Pn ]. Here every vertex in V − S is adjacent to some vertex in S, S is a dominating set in P2 [Pn ]. Also hSi = P1 [Pn ] ∪ P2 , S is a triple connected set. hV − Si = Pn−1 which is connected. Here n − 1 is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 [Pn ]. Hence γnsptc (P2 [Pn ]) ≤ |S|. ′

Claim 2.2 γnsptc (P2 [Pn ]) = |S|.Suppose γnsptc (P2 [Pn ]) < |S|, there exists a set S such ′ ′ that S is a non-split perfect triple connected dominating set in P2 [Pn ]. Assume that |S | = n+2. ′

′

Subcase A Assume S ⊂ S. Then at least one vertex vk ∈ S − S . Subcase A.1 Suppose vk = u2 . In this case every vertex in the 2nd copy of Pn is not ′ ′ adjacent to any vertex in S , S is not a dominating set in P2 [Pn ]. Subcase A.2 Suppose vk ∈ {v1j / 1 ≤ j ≤ n}. In this case hV − Si = Pn ∪ P1 , which is disconnected. ′

′

Subcase B Assume S ∩ S 6= Φ. Then the set S contains the vertices of 1 copy of Pn ′ except the vertices of P2 . But hS i is disconnected. Therefore S is the minimal non-split perfect triple connected dominating set in P2 [Pn ]. Hence γnsptc (P2 [Pn ]) = |S| = n + 3. Therefore,

 n + 3 if n is odd γnsptc (P2 [Pn ]) = . n + 2 if n is even


2

124


Example 2.6 Consider the graphs.

P2 :

u1 b

u2 b

v1

v2

b

P3 :

b

v3 b

Figure 2.3 Then the Corona Product P2 [P3 ] is given in Figure 2.4. v11 b v12 b

bv21

u1

u2

b

b

v13 b

bv22

bv23

Figure 2.4 Here V (P2 [P3 ]) = {u1 , v11 , v12 , v13 , u2 , v21 , v22 , v23 }. Let S = {u1 , v11 , v12 , v13 , u2 , v21 } be the subset of V (P2 [P3 ]). Then every vertex in V − S is adjacent to some vertex in S. Therefore S is a dominating set in P2 [P3 ]. Also hSi = P1 [P3 ] ∪ P2 , it is triple connected. hV − Si = P2 , it is connected and has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 [P3 ] which is minimum. Hence γnsptc (P2 [P3 ]) = 6 = 3 + 3. Similarly, γnsptc (P2 [P4 ]) = 6 = 4 + 2. §3. Strong Product In this section we obtain the non-split perfect triple connected domination number of the Strong product of paths. Definition 3.1 The direct product of the graphs G and H is denoted by G × H, whose V (G × ′ ′ ′ H) = {(g, h)/ g ∈ V (G) and h ∈ V (H)}, E(G × H) = {(g, h)(g , h )/ gg ∈ E(G) and ′ hh ∈ E(H)}. The number of vertices in the direct product of G × H is |V (G)||V (H)|. Observation 3.2 Direct product of graphs the non-split perfect triple connected domination number does not exists, since for any triple connected dominating set S, hV −Si is disconnected. Hence γnsptc (G × H) doesn’t exists. Theorem 3.3 For any n ≥ 2, γnsptc (P2 ⊠ Pn ) =

 n

n − 1

if n is even if n is odd

.

Proof Let Pm and Pn be path on m vertices and n vertices respectively. Then the Strong product of Pm and Pn is denoted by Pm ⊠Pn has mn vertices. Here V (Pm ⊠Pn ) = {(ui , vj )/ 1 ≤ i ≤ m, 1 ≤ j ≤ n}. Let m = 2. Case 1. n is even.

125


Consider the set S = {(u1 , vj1 )/ 1 ≤ j ≤ n} and |S| = n. Claim 1.1 S is a non-split perfect triple connected dominating set in P2 ⊠ Pn . Here every vertex in V −S is adjacent to some vertex in S, S is a dominating set in P2 ⊠Pn . Also hSi = P1 ⊠ Pn , S is a triple connected set. hV − Si = Pn which is connected. Here n is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pn . Hence γnsptc (P2 ⊠ Pn ) ≤ |S|. Claim 1.2 γnsptc (P2 ⊠ Pn ) = |S| = n. ′

′

Suppose γnsptc (P2 ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect ′ triple connected dominating set in Pm ⊠ Pn . Assume that |S | = n − 1. ′

′

Subcase A Assume S ⊂ S. Then there exists a vertex (uk , vk ) ∈ S − S . Subcase A.1 Suppose (uk , vk ) = (u1 , vn ). In this case V − S has an odd number of vertices, hV − Si has no perfect matching in P2 ⊠ Pn . ′

Subcase A.2 Suppose (uk , vk ) ∈ {(u1 , vj )/ 1 ≤ j ≤ n−1}. In this case hV −S i = Pn ∪P1 , which is disconnected. ′

′

Subcase B Assume S ∩ S 6= Φ. (i) Then at least one vertex (ul , vl ) ∈ S ∩ S. Here ′ ′ |S | = n − 1, |V − S | = n + 1, has an odd number of vertices. Therefore, hV − S i has no ′ ′ perfect matching. Also there is a vertex (um , vj ) is not adjacent to any vertex in S , S is not ′ a dominating set in P2 ⊠ Pn . (ii) (ul , vl ) ∈ S − S. Suppose (ul , vl ) ∈ {(u2 , vj )/ 1 ≤ j ≤ n − 1}, ′ ′ there exists a vertex (u1 , vi ) is not adjacent to any vertex in S , S is not a dominating set in P2 ⊠ Pn . Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pn , which is minimum and hence γnsptc (P2 × Pn ) = |S| = n. ′

Case 2 n is odd. Consider the set S = {(u1 , v1j )/ 1 ≤ j ≤ n − 1} and |S| = n − 1. Claim 2.1 S is a non-split perfect triple connected dominating set in P2 ⊠ Pn . Here every vertex in V − S is adjacent to some vertex in S, S is a dominating set in P2 ⊠ Pn . Also hSi = P1 ⊠ Pn−1 , S is a triple connected set. hV − Si = Pn+1 which is connected. Here n + 1 is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pn and hence γnsptc (P2 ⊠ Pn ) ≤ |S|. Claim 2.2 γnsptc (P2 ⊠ Pn ) = |S| = n − 1. ′

′

Suppose γnsptc (P2 ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect ′ triple connected dominating set in P2 ⊠ Pn . Assume that |S | = n − 2. ′

′

Subcase A Assume S ⊂ S. In this case there exists a vertex (uk , vk ) ∈ S − S . Subcase A.1 Suppose (uk , vk ) = (u1 , vn ), V − S has an odd number of vertices. hV − Si has no perfect matching in P2 ⊠ Pn . ′

Subcase A.2 Suppose (uk , vk ) ∈ {(u1 , vj )/ 1 ≤ j ≤ n − 2}, hV − S i = Pn−1 ∪ P1 , which is disconnected.

126


′

′

Subcase B Assume S ∩ S 6= Φ. (i) Then at least one vertex (ul , vl ) ∈ S ∩ S. Here ′ ′ ′ |S | = n − 2, |V − S | = n + 2, has an odd number of vertices. Therefore hV − S i has no perfect ′ ′ matching. Also there is a vertex (um , vj ) is not adjacent to any vertex in S and S is not a ′ dominating set in P2 ⊠ Pn . (ii) (ul , vl ) ∈ S − S. Suppose (ul , vl ) ∈ {(u2 , vj )/ 1 ≤ j ≤ n − 2}, ′ ′ there exists a vertex (um , vj ) is not adjacent to any vertex in S . In this case S is not a dominating set in P2 ⊠ Pn . Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pn , which is minimumand hence γnsptc (P2 ⊠ Pn ) = |S| = n − 1. n if n is even Hence γnsptc (P2 ⊠ Pn ) = . n − 1 if n is odd This completes the proof. 2 Theorem 3.4 For any n ≥ m ≥ 2, γnsptc (Pm ⊠ Pn ) =

 n + 2

if m = 3,

(m − 2)n if m ≥ 4.

Proof Let Pm and Pn be path on m vertices and n vertices respectively. Then the Strong product of Pm and Pn is denoted by Pm ⊠Pn has mn vertices. Here V (Pm ⊠Pn ) = {(ui , vj )/ 1 ≤ i ≤ m, 1 ≤ j ≤ n}. Case 1. m = 3. Consider the set S = {(u2 , vj1 ) : 1 ≤ j ≤ n − 1} ∪ {(ui , v1 ) : 1 ≤ i ≤ 3} ∪ (u3 , v2 ) and |S| = n + 2. Claim 1.1 S is a non-split perfect triple connected dominating set in P3 ⊠ Pn . Here every vertex in V − S is adjacent to some vertex in S, S is a dominating set in P3 ⊠ Pn . Also hSi = P4 ∪ Pn−2 , S is a triple connected set. hV − Si = Pn−1 ∪ Pn−2 ∪ P1 which is connected. Here n is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P3 ⊠ Pn . Hence γnsptc (P3 ⊠ Pn ) ≤ |S|. Claim 1.2 γnsptc (P3 ⊠ Pn ) = |S| = n + 2. ′

′

Suppose γnsptc (P3 ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect triple ′ ′ connected dominating set in P3 ⊠ Pn . Assume that |S | = n + 1. For any set S contains n + 1 ′ ′ vertices, V − S contains 2n − 1 vertices which is odd. Hence hV − S i has no perfect matching in P3 ⊠ Pn . Therefore S is the minimum cardinality of non-split perfect triple connected dominating set in P3 ⊠ Pn . Hence γnsptc (P3 ⊠ Pn ) = n + 2. Case 2. m ≥ 4. Subcase A m is even. Consider the set S = {(uk , v1 ) : 1 ≤ k ≤ m} ∪ {(ui , vj ) : 2 ≤ i ≤ m − 1, 1 ≤ j ≤ n − 1} and |S| = (m − 2)(n). Claim 2.1 S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Here


127

every vertex in V − S is adjacent to some vertex in S, S is a dominating set in Pm ⊠ Pn . Also hSi = Pm−2 ⊠ Pn−1 ∪ 2P1 , S is a triple connected set. hV − Si = 2Pn−1 ∪ Pm−2 which is connected, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) ≤ |S|. Claim 2.2 γnsptc (Pm ⊠ Pn ) = |S| = (m − 2)n. ′

′

Suppose γnsptc (Pm ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect ′ ′ triple connected dominating set in Pm ⊠ Pn . Assume that |S | = (m − 2)(n) − 1. For any set S ′ ′ contains (m−2)(n)−1 vertices, V −S contains 2n+1 vertices which is odd. Hence hV −S i has no perfect matching in Pm ⊠ Pn . Therefore S is the minimum cardinality of non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm × Pn ) = (m − 2)n. Subcase B m is odd. Consider the set S = {(uk , v1 ) : 1 ≤ k ≤ m} ∪ {(ui , vj ) : 2 ≤ i ≤ m − 1, 1 ≤ j ≤ n − 1} ∪ (um , v2 ) and |S| = (m − 2)n. Claim 2.3 S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Here every vertex in V −S is adjacent to some vertex in S, S is a dominating set in Pm ⊠Pn . Also hSi = Pm−2 ⊠ Pn−1 ∪ P1 ∪ P2 , S is a triple connected set. hV − Si = 2Pn−1 ∪ Pm−2 which is connected. hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) ≤ |S|. Claim 2.4 γnsptc (Pm ⊠ Pn ) = |S| = (m − 2)n. ′

′

Suppose γnsptc (Pm ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect ′ ′ triple connected dominating set in Pm ⊠ Pn . Assume that |S | = (m − 2)(n) − 1. For any set S ′ ′ contains (m−2)(n)−1 vertices, V −S contains 2n+1 vertices which is odd. Hence hV −S i has no perfect matching in Pm ⊠ Pn . Therefore S is the minimum cardinality of non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) = (m − 2)n. Therefore,

γnsptc (Pm ⊠ Pn ) = This completes the proof.

 n + 2

if m = 3,

(m − 2)n

ifm ≥ 4.

2

Theorem 3.5 For any m > 2,

γnsptc (Pm ⊠ P2 ) =

 m

m − 1

if m is even, if m is odd.

Proof Let Pm and Pn be path on m vertices and n vertices respectively. Then the Strong product of Pm and Pn is denoted by Pm ⊠Pn has mn vertices. Here V (Pm ⊠Pn ) = {(ui , vj )/ 1 ≤ i ≤ m, 1 ≤ j ≤ n}. Let n = 2. Case 1. m is even.

128


Consider the set S = {(u1 , vj1 )/ 1 ≤ j ≤ m} and |S| = m. Claim 1.1 S is a non-split perfect triple connected dominating set in P2 ⊠ Pm . Here every vertex in V − S is adjacent to some vertex in S, S is a dominating set in P2 ⊠ Pm . Also hSi = P1 ⊠ Pm , S is a triple connected set. hV − Si = Pm which is connected. Here m is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pm . Hence γnsptc (P2 ⊠ Pm ) ≤ |S|. Claim 1.2 γnsptc (P2 ⊠ Pm ) = |S| = m. ′

′

Suppose γnsptc (P2 ⊠ Pm ) < |S|, there exists a set S such that S is a non-split perfect ′ triple connected dominating set in P2 ⊠ Pm . Assume that |S | = m − 1. ′

′

Subcase A Assume S ⊂ S. Then there exists a vertex (uk , vk ) ∈ S − S . ′

Subcase A.1 Suppose (uk , vk ) = (u1 , vm ). In this case V − S has an odd number of ′ vertices, hV − S i has no perfect matching P2 ⊠ Pm . ′

Subcase A.2 Suppose (uk , vk ) ∈ {(u1 , vj )/ 1 ≤ j ≤ n−1}. In this case hV −S i = Pm ∪P1 , which is disconnected. ′

′

Subcase B Assume S ∩ S 6= Φ. a) Then at least one vertex (ul , vl ) ∈ S ∩ S. Here ′ ′ ′ |S | = m − 1, |V − S | = mn − m + 1, has an odd number of vertices. Therefore hV − S i has no ′ ′ perfect matching. Also there is a vertex (un , vj ) is not adjacent to any vertex in S . Hence S is ′ not a dominating set in P2 ⊠Pm .b) (ul , vl ) ∈ S −S. Suppose (ul , vl ) ∈ {(u2 , vj )/ 1 ≤ j ≤ n−1}, ′ ′ there exists a vertex (u1 , vi ) is not adjacent to any vertex in S . In this case S is not a dominating set in P2 ⊠ Pm . Hence S is a non-split perfect triple connected dominating set in P2 ⊠ Pm , which is minimum. Hence γnsptc (P2 ⊠ Pm ) = |S| = m. Case 2. m is odd. Consider the set S = {(u1 , vj1 )/ 1 ≤ j ≤ m} and |S| = m − 1. Claim 2.1 S is a non-split perfect triple connected dominating set in P2 ⊠ Pm . Here every vertex in V −S is adjacent to some vertex in S, S is a dominating set in P2 ⊠Pm . Also hSi = P1 ⊠ Pm−1 , S is a triple connected set. hV − Si = Pm−1 which is connected. Here m−1 is even, hV −Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pm . Hence γnsptc (P2 ⊠ Pm ) ≤ |S|. Claim 2.2 γnsptc (P2 ⊠ Pm ) = |S| = m − 1. ′

′

Suppose γnsptc (P2 ⊠ Pm ) < |S|, there exists a set S such that S is a non-split perfect ′ triple connected dominating set in P2 ⊠ Pm . Assume that |S | = m − 2. ′

′

Subcase A Assume S ⊂ S. Then there exists a vertex (uk , vk ) ∈ S − S . ′

Subcase A.1 Suppose (uk , vk ) = (u1 , vm ). In this case V − S has an odd number of ′ vertices. Hence hV − S i has no perfect matching in P2 ⊠ Pm . ′

Subcase A.2 Suppose (uk , vk ) ∈ {(u1 , vj )/ 1 ≤ j ≤ m − 2}. In this case hV − S i = Pm−1 ∪ P1 , which is disconnected.

129


′

′

Subcase B Assume S ∩ S 6= Φ. (i) Then at least one vertex (ul , vl ) ∈ S ∩ S. Here ′ ′ ′ |S | = m − 2, |V − S | = mn − m + 2 has an odd number of vertices. Therefore hV − S i has no ′ ′ perfect matching. Also there is a vertex (un , vj ) is not adjacent to any vertex in S , S is not a ′ dominating set in P2 ⊠Pm . (ii) (ul , vl ) ∈ S −S. Suppose (ul , vl ) ∈ {(u2 , vj )/ 1 ≤ j ≤ m−2}. In ′ ′ this case there exists a vertex (un , vj ) is not adjacent to any vertex in S , S is not a dominating set in P2 ⊠ Pm . Therefore S is a non-split perfect triple connected dominating set in P2 ⊠ Pm , which is minimum. Hence γnsptc (P2 ⊠ Pm ) = |S| = m − 1. Therefore,  m if m iseven, γnsptc (Pm ⊠ P2 ) = m − 1 if m isodd.

2


Theorem 3.6 For any m > n and n ≥ 3, γnsptc (Pm ⊠ Pn ) =

 m + 2

if n = 3

(n − 2)m if n ≥ 4

.

Proof Let Pm and Pn be path on m vertices and n vertices respectively. Then the Strong product of Pm and Pn is denoted by Pm ⊠Pn has mn vertices. Here V (Pm ⊠Pn ) = {(ui , vj )/ 1 ≤ i ≤ m, 1 ≤ j ≤ n}. Case 1. n = 3. Consider the set S = {(u2 , vj1 ) : 1 ≤ j ≤ m − 1} ∪ {(ui , v1 ) : 1 ≤ i ≤ 3} ∪ (u3 , v2 ) and |S| = m + 2. Claim 1.1 S is a non-split perfect triple connected dominating set in P3 ⊠ Pm . Here every vertex in V −S is adjacent to some vertex in S, S is a dominating set in P3 ⊠Pm . Also hSi = P4 ∪ Pm−2 , S is a triple connected set. hV − Si = Pm−1 ∪ Pm−2 ∪ P1 which is connected. Here n is even, hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in P3 ⊠ Pm . Hence γnsptc (P3 ⊠ Pm ) ≤ |S|. Claim 1.2 γnsptc (P3 ⊠ Pm ) = |S| = m + 2. ′

′

Suppose γnsptc (P3 ⊠ Pm ) < |S|, there exists a set S such that S is a non-split perfect ′ ′ triple connected dominating set in P3 ⊠ Pm . Assume that |S | = m + 1. For any set S ′ ′ contains m + 1 vertices, V − S contains 2m − 1 vertices which is odd. Hence hV − S i has no perfect matching in P3 ⊠ Pm . Therefore S is the minimum cardinality of non-split perfect triple connected dominating set in P3 ⊠ Pm . Hence γnsptc (P3 ⊠ Pm ) = m + 2. Case 2. n ≥ 4. Subcase A n is even. Consider the set S = {(uk , v1 ) : 1 ≤ k ≤ n} ∪ {(ui , vj ) : 2 ≤ i ≤ n − 1, 1 ≤ j ≤ m − 1} and |S| = (m)(n − 2). Claim 2.1 S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Here

130


every vertex in V − S is adjacent to some vertex in S, S is a dominating set in Pm ⊠ Pn . Also hSi = Pn−2 ⊠ Pm−1 ∪ 2P1 , S is a triple connected set. hV − Si = 2Pm−1 ∪ Pn−2 which is connected. hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) ≤ |S|. Claim 2.2 γnsptc (Pm ⊠ Pn ) = |S| = (n − 2)m. ′

′

Suppose γnsptc (Pm ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect ′ ′ triple connected dominating set in Pm ⊠ Pn . Assume that |S | = (n − 2)(m) − 1. For any set S ′ ′ contains (n − 2)(m) − 1 vertices, V − S contains 2m + 1 vertices which is odd. Hence hV − S i has no perfect matching in Pm ⊠Pn . Therefore S is the minimum cardinality of non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) = (n − 2)m. Subcase B n is odd. Consider the set S = {(uk , v1 ) : 1 ≤ k ≤ n} ∪ {(ui , vj ) : 2 ≤ i ≤ n − 1, 1 ≤ j ≤ m − 1} ∪ (un , v2 ) and |S| = (n − 2)m. Claim 2.3 S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Here every vertex in V −S is adjacent to some vertex in S, S is a dominating set in Pm ⊠Pn . Also hSi = Pn−2 ⊠ Pm−1 ∪ P1 ∪ P2 , S is a triple connected set. hV − Si = 2Pm−1 ∪ Pn−2 which is connected. hV − Si has a perfect matching. Therefore S is a non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) ≤ |S|. Claim 2.4 γnsptc (Pm ⊠ Pn ) = |S| = (n − 2)m. ′

′

Suppose γnsptc (Pm ⊠ Pn ) < |S|, there exists a set S such that S is a non-split perfect ′ ′ triple connected dominating set in Pm ⊠ Pn . Assume that |S | = (n − 2)(m) − 1. For any set S ′ ′ contains (n − 2)(m) − 1 vertices, V − S contains 2m + 1 vertices which is odd. Hence hV − S i has no perfect matching in Pm ⊠Pn . Therefore S is the minimum cardinality of non-split perfect triple connected dominating set in Pm ⊠ Pn . Hence γnsptc (Pm ⊠ Pn ) = (n − 2)m. Therefore, γnsptc (Pm ⊠ Pn ) = This completes the proof.

 m + 2

ifn = 3,

(n − 2)m ifn ≥ 4.

2

Example 3.7 Consider the graphs.

P2 :

u1 b

u2 b

P3 :

v1 b

Figure 3.1 Then, the strong product P2 ⊠ P3 is given in Figure 3.2.

v2 b

v3 b


u1 , v1 b

b

u2 , v1

u1 , v2 b

b

u2 , v2

131

u1 , v3 b

b

u2 , v3

Figure 3.2 Here V (P2 ⊠ P3 ) = {(ui , vj )/ 1 ≤ i ≤ 2, 1 ≤ j ≤ 3}. Let S = {(u1 , v1 ), (u1 , v2 ), (u1 , v3 ), (u2 , v1 )} be the subset of V (P2 ⊠ P3 ). Every vertex in V − S is adjacent to some vertex in S. Therefore S is a dominating set. Also hSi = K3 ∪ K1 , it is triple connected. hV − Si = P2 , it is connected and has a perfect matching in P2 ⊠ P3 . Therefore S is the non-split perfect triple connected dominating set in P2 ⊠P3 , which is minimum. Hence γnsptc (P2 ⊠P3 ) = 4 = (2−1)3+1. Similarly, γnsptc (P3 ⊠ P3 ) = 5 = 3 + 2, γnsptc (P4 ⊠ P4 ) = 8 = (4 − 2)4.

References [1] Mahadevan G, IravithulBasira A, and Sivagnanam C, Non-split perfect triple connected domination number of a graph, Asian Journal of Research in Social Sciences and Humanities, Vol 6 (2016), pp 1954-1966. [2] Paulraj Joseph J, Angel Jebitha M.K, Chithra Devi P And Sudhana G, Triple connected graphs, Indian Journal Mathematics and Mathematical Sciences, Vol 8(1) (2012), pp 61-75. [3] Paulraj Joseph J, Mahadevan G, and Selvam A, Triple Connected Domination number of a Graph, International Journal of Mathematical Combinatorics, Vol 3 (2012), pp 93-104. [4] Richard H. Hammack, Wilfried Imrich and Sandi Klavzar Handbook of Product Graphs, Second edition, CRC Press, Boca Raton, FL(2011).


The Connectivity Number of an Arithmetic Graph Mary Jenitha L and Sujitha S (Department of Mathematics, Holy Cross College (Autonomous), Tirunelveli, Tamil Nadu, India) E-mail: [email protected], [email protected]

Abstract: The arithmetic graph Vn is defined as a graph with its vertex set is the set consists of the divisors of n (excluding 1) where n is a positive integer and n = pa1 1 pa2 2 pa3 3 . . . par r where p′i s are distinct primes and ai ’s > 1 and two distinct vertices a, b which are not of the same parity are adjacent in this graph if (a, b) = pi , for some i, 1 ≤ i ≤ r . In this paper, we study some results related to the connectivity κ of an arithmetic graph. It is also shown ′

that, the edge connectivity κ and the connectivity κ are equal in arithmetic graph Vn .

Key Words: Arithmetic graph, connectivity, edge connectivity. AMS(2010): 05C12. §1. Introduction By a graph G = (V, E ), we mean a finite undirected connected graph without loops or multiple edges. The order and size of G are denoted by ν and ǫ respectively. We consider connected graphs with at least three vertices. For basic definitions and terminologies we refer to [2]. The arithmetic graph Vn is defined as a graph with its vertex set is the set consists of the divisors of n (excluding1) where n is a positive integer and n = pa1 1 pa2 2 pa3 3 . . . par r where p′i s are distinct primes and a′i s > 1 and two distinct vertices a,b which are not of the same parity are adjacent in this graph if (a, b) = pi for some i, 1 ≤ i ≤ r. The vertices a and b are said to be of the same parity if both a and b are the powers of the same prime, for instance a = p 2 , b = p 5 . The construction of an arithmetic graph with a given integer was introduced and studied by Vasumathi and Vangipuram in [4]. The domination parameters of an arithmetic graph were further studied by various authors in [3]. Connectivity is one of the basic concepts of graph theory. It is closely related to the theory of network flow problems. In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v. Otherwise, they are called disconnected. A graph is said to be connected if every pair of vertices in the graph is connected. The degree of a vertex v in a graph G is the number of edges of G incident with v and is denoted by degG (v) or d (v ). A vertex of degree zero in G is called an isolated vertex and a vertex of degree one is called a pendent vertex or an end-vertex of G. The maximum and minimum degree of a graph G 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received February 24, 2018, Accepted March 16, 2018, Edited by R. Kala.

The Connectivity number of an Arithmetic Graph

133

is denoted by ∆(G) and δ(G) respectively. A cut-vertex (cut-edge) of a graph G is a vertex (edge) whose removal increases the number of components. A vertex cut, or separating set of a connected graphG is a set of vertices whose removal renders G disconnected. The connectivity or vertex connectivity κ(G) is the number of vertices of a minimal vertex cut. A graph is called k-connected or k-vertex-connected if its vertex connectivity is k or greater. Any graph G is said to be k-connected if it contains at least k +1 vertices, but does not contain a set of k − 1 vertices whose removal disconnects the graph and κ(G) is defined as the largest k such thatG is k -connected. An edge cut of G is a set of edges whose removal renders the graph G ′ disconnected. The edge-connectivity κ (G) is the number of edges of a minimal edge cut. A graph is said to be maximally connected if its connectivity equals its minimum degree. A graph is said to be maximally edge-connected if its edge-connectivity equals its minimum degree. For vertices u and v in a connected graph G, the distance d (u, v ) is the length of a shortest u v path in G. Two vertices u and v of G are antipodal if d (u,v ) = diam G or d (G). A vertex v is an extreme vertex of a graph G if the subgraph induced by its neighbors is complete. The following theorems are used in sequel. ′

Theorem 1.1 ([2]) For a connected graph G, κ(G) 6 κ (G) 6 δ(G). Theorem 1.2 ([2]) A connected graph is a tree if and only if every edge is a cut edge. Theorem 1.3 ([2]) A vertex v of a tree G is a cut vertex of G if and only if d(v) >1. Theorem 1.4 ([1]) it The number of vertices of an arithmetic graph G = Vn , n = pa1 1 pa2 2 pa3 3 · · · par r where p′i s are distinct primes, are [(a1 + 1)(a2 + 1) · · · (ar + 1)] − 1. §2. Main Results Theorem 2.1 For an arithmetic graph G = Vn , n = pa1 1 pa2 2 where p1 and p2 are distinct primes and ai = 1 for all i = 1, 2; then the connectivity and the edge connectivity numbers are equal to 1. Proof Consider the arithmetic graph G = Vn , where n is the product of two distinct primes. The vertex set of Vn contains three vertices namely p1 , p2 , p1 × p2 .Clearly the arithmetic graph Vn is a tree containing two end vertices and an internal vertex. By theorem 1.3, the end vertices p1 and p2 are not cut vertices. It is clear that the internal vertex p1 × p2 is the only cut vertex of Vn . Hence connectivity number κ(Vn ) = 1. Also by theorem 1.2, every edge of Vn is a cut ′ edge and hence the edge connectivity number κ (Vn ) = 1. 2 Theorem 2.2 For an arithmetic graph G = Vn , n = pa1 1 pa2 2 where p1 and p2 are distinct primes, then  1 f or a = 1 & a > 1; i, j = 1, 2, ′ i j κ(Vn ) = κ (Vn ) = 2 f or ai > 1; i = 1, 2.

Proof Consider the arithmetic graph G = Vn , where n is the product of two distinct

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Mary Jenitha L and Sujitha S

primes. Case 1. ai = 1 and aj > 1; i, j = 1, 2. a a The vertex set of Vn is V (Vn ) = p1 , p21 , p31 , · · · , p1 j , p2 , p1 × p2 , p21 × p2 , · · · , p1 j × p2 . a Clearly p1 and p2 are adjacent to the vertices p1 × p2 , p21 × p2 , ..., p1 j × p2 , so that d(p1 ) > 1 a and d(p2 ) > 1. The vertices p21 , p31 , ..., p1 j are non adjacent to each other and are adjacent to exactly one vertex p1 × p2 since otherwise it contradicts the definition of an arithmetic graph. a Therefore d(p21 ) = d(p31 ) = · · · = d(p1 j ) = 1. Since the graph has no isolated vertices, the ′ minimum degree of the graph is one. Hence by theorem 1.1, κ(G) ≤ κ (G) ≤ δ(G) = 1. Hence ′ it is clear that, κ(Vn ) = 1 = κ (Vn ) . Case 2. ai > 1; i = 1, 2. In this case the vertex set of Vn is V (Vn ) = {{p1 , p2 , p21 , p22 , p31 , · · · , pa1 1 , pa2 2 , p1 × p2 , p21 × p2 , · · · , pa1 1 × p2 , p1 × p22 , p1 × p32 , · · · , pa1 1 × pa2 2 } . By the definition of an arithmetic graph, p1 and p2 are adjacent to the vertices p1 × p2 , p21 × p2 , · · · , pa1 1 × p2 , p1 × p22 , p1 × p32 , · · · , pa1 1 × pa2 2 . Therefore we have d(p1 ) > 1 and d(p2 ) > 1. Since the arithmetic graph is free from isolated vertices, d(v) > 0 for all v ∈ V (Vn ). The vertices p21 , p31 , · · · , pa1 1 which are in the product of themselves with many times (till the maximum power) are adjacent to at least the vertices p1 × p2 , p1 × p22 ,· · · ,p1 × pa2 2 and the vertices p22 , p32 , · · · , pa2 2 are adjacent to at least the vertices p1 × p2 ,p21 × p2 , · · · , pa1 1 × p2 hence its degrees are greater than one. Also, the vertices which are in the combination of two distinct primes have at least the vertices p1 and p2 are adjacent. Therefore δ(Vn ) ≥ 2. But the vertex pa1 1 × pa2 2 is adjacent to exactly the two vertices p1 and p2 , since otherwise it contradicts definition. Hence we find that δ(Vn ) = 2. By theorem 1.1, ′ κ(G) ≤ κ (G) ≤ δ(G) = 2. Let S = {p1 , p2 } be the set of vertices which are adjacent to pa1 1 × pa2 2 . Clearly the deletion of S from Vn , isolates the vertex pa1 1 × pa2 2 . Hence κ(Vn ) = 2. ′ Also, by theorem 1.1, it is clear that κ (Vn ) 6 δ(Vn ) = 2. Since d(pa1 1 × pa2 2 ) = 2, the removal ′ of two edges incident at this vertex disconnects the graph. Hence κ (Vn ) = 2. 2 Theorem 2.3 For an arithmetic graph G = Vn , n = pa1 1 pa2 2 · · · par r where pi , i = 1, 2, · · · , r ′ (r > 2) are distinct primes and ai = 1 for all i = 1, 2, · · · , r then κ(Vn ) = κ (Vn ) = r. Proof Consider the arithmetic graph G = Vn , where n is the product of more than two ′ distinct primes and ai s are equal to 1. By result 1.4, the arithmetic graph Vn contains 2r − 1 vertices. Among the 2r − 1 vertices, the vertex p1 × p2 × · · · × pr is adjacent to exactly r vertices namely p1 , p2 , · · · , pr . Therefore d(p1 × p2 × · · · × pr ) = r. Suppose it is adjacent to more than r vertices. Then there exists a vertex vi 6= pi , which is adjacent to p1 × p2 × · · · × pr and hence (p1 × p2 × · · · × pr , vi ) 6= pi which contradicts the definition of an arithmetic graph. So d(p1 × p2 × · · · × pr ) = r. Also we can easily seen that the minimum degree δ(Vn ) = r. By theorem 1.1, it is observe that κ(Vn ) 6 r. To prove κ(Vn ) = r. Suppose κ(Vn ) < r. Let S = {p1 , p2 , · · · , pr−1 } be the vertex cut of Vn such that |S| 6 r − 1. If S is deleted from Vn then it is easily seen that the vertex pr is adjacent to at least the vertex p1 × p2 × · · · × pr and the vertex pi × pj is adjacent to either p1 × p2 × · · · × pi−1 × pj × pj+1 × · · · × pr or p1 × p2 × · · · × pi × pj−1 × pj+1 × · · · × pr and the vertex pi × pj × pk is adjacent to either p1 × p2 × · · · × pi−1 × pj−1 × pk × · · · × pr or p1 × p2 × ... × pi × pj−1 × pk−1 × · · · × pr or

135

The Connectivity number of an Arithmetic Graph

p1 × p2 × · · · × pi−1 × pj × pk−1 × · · · × pr and so on. This implies that the induced graph hVn − Si is connected. Therefore we need at least r vertices to disconnect the graph. But the deletion of S ∪ {pr }, the graph is disconnected. Hence κ(Vn ) = r. ′

′

Also, by Theorem 1.1 it is clear that κ (Vn ) 6 δ(Vn ). Since δ(Vn ) = r, we have κ (Vn ) 6 r. Since d(p1 × p2 × · · · × pr ) = r, the removal of r edges incident at the vertex p1 × p2 × · · · pr , the graph Vn is disconnected and it is clear that the edge cut of Vn contains exactly r edges ′ namely p1 × p2 × · · · × pr p1 , p1 × p2 × · · · × pr p2 , · · · p1 × p2 × · · · × pr pr . Therefore κ (Vn ) = r ′ and hence κ(Vn ) = κ (Vn ) = r. 2 Theorem 2.4 For an arithmetic graph G = Vn , n = pa1 1 pa2 2 · · · par r where p1 , p2 , · · · , pr are ′ ′ distinct primes and ai s ≥ 1 for all i = 1, 2, 3, · · · , r and pi > 2 then κ(Vn ) = κ (Vn ) = r. Proof We prove the theorem by considering the following four cases. ′

Case 1. All the ai s, i = 1, 2, 3, · · · r are equal to one. In this case we follow Theorem 2.3 and arrived the result. ′

Case 2. Some of the ai s are equal to one and the others are greater than 1. Consider the vertex set of Vn as V (Vn ) = {p1 , p2 , · · · , pr , p1 × p2 , · · · , pa1 1 × pa2 2 × · · · × par r } . ′ Let the last vertex be pa1 1 ×pa2 2 ×· · ·×par r say v1 , where ai s are the maximum powers of the given distinct primes. By the definition of an arithmetic graph, we see that the only vertices which are adjacent to v 1 are p1 , p2 , · · · , pr . Hence d(v1 ) = r. Also the minimum degree of Vn occurs ′ at the vertex v1 . That is, δ(Vn ) = r = d(v1 ). By theorem 1.1, κ(Vn ) = κ (Vn ) 6 δ(Vn ) = r. But the removal of r vertices adjacent to v1 makes the graph disconnected. Hence we obtained ′ the result κ(Vn ) = r. The edge connectivity κ (Vn ) = r is same as Theorem 2.3. ′

Case 3. All the ai s are equal and greater than 1. Here also consider the last vertex of V (Vn ), say pa1 1 × pa2 2 × · · · × par r where the ai s are the maximum power of given distinct primes. By the definition of an arithmetic graph, it is clear that p1 , p2 , · · · , pr are the only vertices which are adjacent to the vertex pa1 1 × pa2 2 × . . . . × par r . The remaining proof is similar to Case 2 ′

′

Case 4. All the ai s are distinct and greater than one. Consider the last vertex in the vertex set of Vn , say pa1 1 × pa2 2 × · · · . × par r where the ai s are the maximum power of the given distinct primes. By the definition of an arithmetic graph, this vertex is adjacent to exactly r vertices namely p1 , p2 , p3 , · · · , pr . Suppose it is adjacent to any other vertex except pi then it contradicts the definition of an arithmetic graph. The remaining proof is similar to Case 2. 2 ′

Corollary 2.5 For an arithmetic graph G = Vn , n = pa1 1 pa2 2 · · · par r where p1 , p2 , · · · , pr are ′ distinct primes and ai s ≥ 1 for all i = 1, 2, 3, · · · , r the connectivity number and the edge connectivity number are equal. Proof It is obvious from Theorems 2.1, 2.2, 2.3 and 2.4. 2 Remark 2.6 The arithmetic graph Vn is a maximally connected graph.

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Mary Jenitha L and Sujitha S

Example 2.7 Consider the arithmetic graph G = V60 . since 60 = 22 × 3 × 5. The vertex set of G is V (V60 ) = {2, 3, 5, 22, 2 × 3, 2 × 5, 3 × 5, 22 × 3, 22 × 5, 2 × 3 × 5, 22 × 3 × 5}. Clearly the vertex cut and edge cut of G is S = {2, 3, 5} and S 1 = {22 × 3 × 52, 22 × 3 × 53, 22 × 3 × 55} ′ respectively. Hence κ(G) = κ (G) = δ(G) = 3.

22 x 3 x 5

2x3x5

22 x 5

22 x 3

b

b2 b

b

3

b

b b

b

3x5

b

b

5

22

2x3

b2 x 5

Figure 1 Arithmetic graph G = V60

Conclusion From the above theorems, it is clear that the connectivity number, the edge connectivity number and the minimum degree of the given arithmetic graph are equal. Also, if the given integer n ′ is the product of more than two distinct primes then κ(Vn ) and κ (Vn ) depend on the number of distinct primes and they do not depend upon the powers of primes.

References [1] Apostol, T.M., Introduction to Analytic Number Theory, Springer - verlag, Berlin, Heidelberg (1980). [2] Bondy and Murty, Graph Theory with Applications, Macmillan (1976). [3] Suryanarayana Rao K.V., Sreenivansan .V, The Split Domination in Arithmetic Graphs, International Journal of Computer Applications, Volume 29, No.3, (2011). [4] Vasumathi. N., and Vangipuram. S., Existence of a graph with a given domination Parameter, Proceedings of the Fourth Ramanujan Symposium on Algebra and its Applications, University of Madras,Madras, 187-195 (1995).


Q-Fuzzy Bi-ideals in Near Subtraction Semigroups P. Annamalai Selvi, R. Sumitha and S. Jayalakshmi (Department of Mathematics, Sri Parasakthi College for Women, Courtallam,Tamilnadu, India.) E-mail: [email protected], [email protected], [email protected]

Abstract: In this paper we introduce the notation of Q- fuzzy bi-ideals in near subtraction semigroup and give some characterizations of Q- fuzzy bi-ideals in near subtraction semigroups .

Key Words: Near subtraction semigroups, bi-ideal, Q- fuzzy bi-ideals, Q-anti fuzzy biideals.

AMS(2010): 20M12. §1. Introduction B.M.Schein [8] considered systems of the form (X; ◦; /), where X is a set of functions closed under the composition “◦”of functions (and hence (X; ◦) is a function semigroup) and the set theoretic subtraction “/”(and hence (X; /) is a subtraction algebra in the sense of [8]).Y.B.Jun et al [5] introduced the notation of ideals in subtraction algebras and discussed the characterization of ideals.Dheena et al [3],discussed the fundamental properties related to ideals and subtraction algebras. The concept of fuzzy set was first initiated by Zadeh [10].Mahalakshmi et al. [6] studied the notation of bi-ideals in near subtraction semigroups. Manikandan [7] studied fuzzy bi-ideals in near-rings.Many researches who are involved in studying,applying, refining and teaching fuzzy sets have successfully applied this theory in many different fields. The purpose of this paper to introduce the notation of Q-fuzzy bi-ideals in near subtraction semigroups.We investigate some basic results,examples and We show that every Q-fuzzy bi-ideal of a near subtraction semigroup is a Q-fuzzy sub near- subtraction semigroup.

§2. Preliminaries Definition 2.1 A non empty set X together with two binary operations “−”and “·”is said to be a near subtraction semigroup(right) if it satisfies the following: (i) (X, −) is a subtractionalgebra; (ii) (X, ·) is a semigroup; 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received February 10, 2018, Accepted September 18, 2018, Edited by K. Selvakumar.

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P. Annamalai Selvi, R. Sumitha and S. Jayalakshmi

(iii) (x − y)z = xz − yz for every x, y, z ∈ X. It is clear that 0x = 0, for all x ∈ X. Similarly we can define a left near-subtraction semigroup. Here after a near-subtraction semigroup means only a right near-subtraction semigroup. Definition 2.2 A near subtraction semigroup X is said to be zero - symmetric if x0 = 0 for every x ∈ X. Definition 2.3 A nonempty subset S of a subtraction semigroup X is said to be a subalgebra of X, if x − y ∈ S, for all x, y ∈ S. Definition 2.4 A Q-fuzzy set µ in X is a Q-fuzzy sub near- subtraction semigroup of X if for all x, y ∈ X and q ∈ Q, (i) µ(x − y, q) ≥ min{µ(x, q), µ(y, q)}; (ii) µ(xy, q) ≥ min{µ(x, q), µ(z, q)}. § 3. Q-fuzzy bi-ideals Definition 3.1 A Q-fuzzy subalgebra µ in X is called a Q- fuzzy bi-ideal of X if (i) µ(x − y, q) ≥ min{µ(x, q), µ(y, q)}; (ii) µ(xyz, q) ≥ min{µ(x, q), µ(z, q)} for all x, y, z ∈ X and q ∈ Q. Example 3.2 Let X = {0, a, b, c} be the Klein’s four group. Define subtraction and multiplication in X as follows:

−

0

a

b

c

·

0

a

b

c

0

0

0

0

0

0

0

0

0

0

a

a

0

c

b

a

0

a

0

a

b

b

0

0

b

b

0

b

0

b

c

c

0

c

0

c

0

c

0

c

Then (X, −, ·) is a near subtraction semigroup(see [ [2], pg.407] scheme 6). We define an Q-fuzzy set in X as follows: µ(0, q) = 0.9, µ(a, q) = 0.7, µ(b, q) = 0.4 = µ(c, q). Then µ is a Q-fuzzy bi-ideal of X. Definition 3.3 A Q-fuzzy subalgebra µ in X is called a Q- anti fuzzy bi-ideal of X if (i) µ(x − y, q) ≤ max{µ(x, q), µ(y, q)} and

Q-Fuzzy Bi-ideals in Near Subtraction Semigroups

139

(ii) µ(xyz, q) ≤ max{µ(x, q), µ(z, q)} for all x, y, z ∈ X and q ∈ Q. Definition 3.4v A Q-fuzzy subalgebra µ in X and t ∈ [0, 1], we define ∪(µ; t) = {x ∈ X/ µ(x, q) ≥ t, q ∈ Q} is called a upper t-level cut of µ. Definition 3.5 A family of Q-fuzzy set {µi /i ∈ ∧} where i ∈ ∧ ( index set ) is a nearsubtraction semigroup X, the intersection ∩i∈∧ µi of {µi /i ∈ ∧} is defined by (∩i∈∧ µi )(x, q) = inf {µi (x, q)/i ∈ ∧, q ∈ Q} for each x ∈ X. Definition 3.6 A Q-fuzzy bi-ideal of a near-subtraction subgroup X is said to be normal if µ(0, q) = 1. Definition 3.7 Let A and B be two fuzzy subsets of a semigroup X. We define the product of A and B as follows:  sup x=yz {min{A(y), B(z)}} if x = yz, for all y, z ∈ X (A · B)(x) = 0 otherwise

Lemma 3.8 A Q- fuzzy set µ is a Q- fuzzy bi-ideal of X iff µc is a Q- anti-fuzzy bi-ideal of X, for all x, y, z ∈ X and q ∈ Q. Proof Notice that µc (x − y, q) =

1 − µ(x − y, q)

≤

1 − min{µ(x, q), µ(y, q)}

=

max{1 − µ(x, q), 1 − µ(y, q)}

=

max{µc (x, q), µc (y, q)}.

Therefore, µc (x − y, q) ≤ max{µc (x, q), µc (y, q)}, and µc (xyz, q) =

1 − µ(xyz, q)

≤

1 − min{µ(x, q), µ(z, q)}

=

max{1 − µ(x, q), 1 − µ(z, q)}

=

max{µc (x, q), µc (z, q)}.

Whence, µc (xyz, q) ≤ max{µc (x, q), µc (z, q)}. Hence µc is a Q-anti-fuzzy bi-ideal in X.

Conversely assume that µc is a Q-anti-fuzzy bi-ideal in X. For all x, y, z ∈ X and q ∈ Q, notice that µ(x − y, q) = ≥

1 − µc (x − y, q)

1 − max{µc (x, q), µc (y, q)}

=

min{1 − µc (x, q), 1 − µc (y, q)}

=

min{µ(x, q), µ(y, q)}.

140


So, µ(x − y, q) ≥ min{µ(x, q), µ(y, q)}, and µ(xyz, q) = ≥

1 − µc (xyz, q)

1 − max{µc (x, q), µc (z, q)}

=

min{1 − µc (x, q), 1 − µc (z, q)}

=

min{µ(x, q), µ(z, q)}.

Whence, µ(xyz, q) ≥ min{µ(x, q), µ(z, q)}. Hence µ is a Q- fuzzy bi-ideal of X.

2

Theorem 3.9 Let µ be a Q-fuzzy set in a near subtraction semigroup X. Then µ is a Q-fuzzy bi-ideal of X iff the upper level cut ∪(µ; t) of X is a bi-ideal of X for each t ∈ [0, 1]. Proof Let µ is a Q-fuzzy bi-ideal of X. Let x, y ∈ ∪(µ; t).Then µ(x) ≥ t and µ(y) ≥ t. Now, µ(x − y, q) ≥ min{µ(x, q), µ(y, q)} = t ⇒ µ(x − y, q) ≥ t and so x − y ∈ ∪(µ; t). Hence ∪(µ; t) is a subalgebra of X. Let x, z ∈ ∪(µ; t) and y ∈ X. Then µ(x) ≥ t and µ(z) ≥ t. Now µ(xyz, q) ≥ min{µ(x, q), µ(z, q)} = t ⇒ µ(xyz, q) ≥ t, and so xyz ∈ ∪(µ; t). Hence ∪(µ; t) is a bi-ideal of X. Conversely assume that ∪(µ; t) is a bi-ideal of X. To prove that µ is a Q- fuzzy bi-ideal of X, for all t ∈ [0, 1]. Suppose µ is not a fuzzy bi-ideal of X. Suppose x, y ∈ X and q ∈ Q and µ(x − y, q) < min{µ(x, q), µ(y, q)}. Then x, y ∈ ∪(µ; t). Choose t such that µ(x − y, q) < t < min{µ(x, q), µ(y, q)}. We get x − y ∈ / ∪(µ; t). which is a contraction. Hence µ(x − y, q) ≥ min{µ(x, q), µ(y, q)}. Suppose x, y, z ∈ X and q ∈ Q and µ(xyz, q) < min{µ(x, q), µ(z, q)}. Then x, z ∈ ∪(µ; t).Choose t such that µ(xyz, q) < t < min{µ(x, q), µ(z, q)}. We get xyz ∈ / ∪(µ; t).which is a contraction. Hence µ(xyz, q) ≥ min{µ(x, q), µ(z, q)}. Hence µ is a Q-fuzzy bi-ideal of X.

2

Theorem 3.10 If {µi /i ∈ ∧} is a family of Q- fuzzy bi-ideals in a near subtraction semigroup X. Then ∩i∈∧ µi is a Q- fuzzy bi-ideal of X. Proof Let {µi /i ∈ ∧} is a family of Q-fuzzy bi-ideals of a near subtraction semigroup X. Let x, y, z ∈ X and q ∈ Q. (i) ∩i∈∧ µi (x − y, q) =

inf{µi (x − y, q)/i ∈ ∧, q ∈ Q}

≥

inf{min{µi (x, q), µi (y, q)/i ∈ ∧, q ∈ Q}}

=

min{inf{µi (x, q)/i ∈ ∧, q ∈ Q}, inf {µi(y, q)/i ∈ ∧, q ∈ Q}}

=

min{(∧i∈∧ µi )(x, q), (∧i∈∧ µi )(y, q)}.

141


Therefore, ∩i∈∧ µi (x − y, q) ≥ min{(∧i∈∧ µi )(x, q), (∧i∈∧ µi )(y, q)}. (ii) ∩i∈∧ µi (xyz, q) =

inf {µi (xyz, q)/i ∈ ∧, q ∈ Q}

≥

inf{min{µi (x, q), µi (z, q)/i ∈ ∧, q ∈ Q}}

=

min{inf{µi (x, q)/i ∈ ∧, q ∈ Q}, inf {µi(z, q)/i ∈ ∧, q ∈ Q}

=

min{(∧i∈∧ µi )(x, q), ((∧i∈∧ µi )(z, q).

Therefore, ∩i∈∧ µi (xyz, q) ≥ min{(∧i∈∧ µi )(x, q), (∧i∈∧ µi )(z, q)}.

2

Hence, ∩i∈∧ µi is a Q- fuzzy bi-ideal of X.

Theorem 3.11 Let µ be a Q-fuzzy bi-ideal of a zero-symmetric near subtraction semigroup X. Then the following are equivalent: (i) µ(xyz, q) ≥ min{µ(x, q), µ(z, q)}; (ii) µXµ ⊆ µ, for all x, y, z ∈ X and q ∈ Q. Proof (i) ⇒ (ii). Let µ be a Q-fuzzy bi-ideal of a zero-symmetric near subtraction semi′ group X. Let x ∈ X. Let µ(xyz, q) ≥ min{µ(x, q), µ(z, q)}. If there exists x, y, x1 , x2 ∈ X and ′ q ∈ Q such that x = xy and x = x1 x2 . Then by hypothesis µ(x1 x2 y, q) ≥ min{µ(x1, q), µ(y, q)} ′

we have (µ · X · µ)(x , q) = =

sup min{µX(x, q), µ(y, q)} x′ =xy

sup min{ sup {min{µ(x1 , q), X(x2 , q)}, µ(y, q)}} x=x1 x2

x′ =xy

=

sup min{ sup {min{µ(x1 , q), 1}, µ(y, q)}} x=x1 x2

x′ =xy

=

sup

min{µ(x, q), µ(y, q)}

x′ =xy=x1 x2 y

≤

′

sup

µ(x1 x2 y, q) = µ(x , q)

′

x =x1 x2 y

Hence µXµ ⊆ µ.

′

(ii) ⇒ (i). Assume that µXµ ⊆ µ. Let x, y, z ∈ X and q ∈ Q such that x = xyz. ′

′

Then µ(xyz, q) = µ(x , q) ≥ (µXµ)(x , q) =

sup min{(µX)(a, q), µ(b, q)} x′ =ab

=

sup min{ sup min{µ(a1 , q), X(a2 , q), µ(b, q)} x′ =ab

=

a=a1 a2

sup

min{µ(a1 , q), X(a2 , q), µ(b, q)}

x′ =a1 a2 b=xyz

≥ min{µ(x, q), X(y), µ(z, q)} = min{µ(x, q), 1, µ(z, q)} = min{µ(x, q), µ(z, q)}

142


Hence µ(xyz, q) ≥ min{µ(x, q), µ(z, q)}

2

Lemma 3.12 Let µ be a Q-fuzzy bi-ideal of a zero-symmetric near subtraction semigroup X. Then the following are equivalent: (i) µ is a Q-fuzzy bi-ideal of X; (ii) µ(xyz, q) ≥ min{µ(x, q), µ(z, q)}; (iii) µXµ ⊆ µ, for all x, y, z ∈ X and q ∈ Q.

2

Proof Straightforward

Theorem 3.13 Let µ be a Q-fuzzy bi-ideal of a near subtraction semigroup X, and let f : [0, 1] → [0, 1] be an increasing function. Then the fuzzy set µf (x, q) = f (µ(x, q)) is a Q-fuzzy bi-ideal of X. In particular, if f [µ(0, q)] = 1 then µf is normal. Proof For any x, y ∈ X and q ∈ Q, notice that µf (x − y, q) =

f (µ(x − y, q))

≥

f (min{µ(x, q), µ(y, q)})

=

min{f (µ(x, q)), f (µ(y, q))} = min{µf (x, q), µf (y, q)}.

Therefore, µf (x − y, q) ≥ min{µf (x, q), µf (y, q)}. For any x, y, z ∈ X and q ∈ Q, we have µf (xyz, q) =

f (µ(xyz, q))

≥

f (min{µ(x, q), µ(z, q)})

=

min{f (µ(x, q)), f (µ(z, q))} = min{µf (x, q), µf (z, q)}.

Therefore, µf (xyz, q) ≥ min{µf (x, q), µf (z, q)}. Hence µf is a Q-fuzzy bi-ideal of X. If f [µ(0, q)] = 1 then µf (0, q) = 1. Hence µf is normal.

2

References [1] J. C. Abbott, Sets, Lattices, and Boolean Algebras, Allyn and Bacon, Inc., Boston, Mass, 1969. [2] Chinnadurai.V, Fuzzy Ideals in Algebraic Structures Lap Lambert Academic pulishing,2013. [3] P.Dheena and G.Satheesh Kumar, On strongly regular near subtraction semigroups, Commun. Korean Math.Soc. 22(2007), no.3, pp.323-330. [4] Gunter Pilz, Near Rings, the Theory and its Applications, North Holland Publishing Company, Amsterda, 1983.


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[5] Y. B. Jun and H. S. Kim, On ideals in subtraction algebras, Sci. Math. Jpn. 65(2007), no.1, 129-134. [6] V. Mahalakshmi, S. Maharasi and S. Jayalakshmi, Bi-ideals of near subtraction semigroup, Indian Advances in Algebra 6(1)(2013)35-48. [7] T. Manikandan, fuzzy bi-ideals of near-rings, J. Fuzzy Math., 17(3)(2009) 659-671. [8] B. M. Schein, Difference semigroups, Comm. Algebra 20, (1992), no. 8, 2153-2169. [9] Seyadali Fathima. S and Balakrishnan. R, A Study on Special Cases of Near Subtraction Semigroups, Ph.D Dissertation, M.S.University, Tamilnadu, India (2013). [10] L.A. Zadeh , Fuzzy sets, information and control 8, (1965) 338-353 [11] B. Zelinka, Subtraction semigroups, Math. Bohem. 120 (1995), no. 4, 445-447.


Intuitionistic Fuzzy Operator A(m,n) and its Extensions R. Nagalingam and S. Rajaram (Research Department of Mathematics, Sri S.R.N.M College, Sattur - 626 203, India) E-mail: [email protected], rajaram [email protected]

Abstract: An Operator is a special symbol performing specific operations. Many Operators have been defined over intuitionistic fuzzy sets. In this paper we propose a new Operator A(m,n) and its extensions over the intuitionistic fuzzy sets. Based on A(m,n) some theorems related to set relations, set operations, modal and level operators have been proved. To demonstrate its application one numerical example in decision making problem is discussed.

Key Words: Fuzzy sets, Intuitionistic fuzzy sets, Intuitionistic fuzzy numbers, Intuitionistic fuzzy operators, Decision making.

AMS(2010): 03E72, 03E55, 03E75, 62C86. §1. Introduction L.A.Zadeh [1] introduced fuzzy set in 1965 was a generalisation of crisp set. Further the fuzzy sets were generalized by K.T.Atanassov ([3] ,[4])in which non-membership values were considered into the introduction of intuitionistic fuzzy sets(IFS). In set theory, the basic operations are the union and the intersection of sets. At the beginning of the Zadeh’s fuzzy set theory, Min operator for the intersection and the Max operator for the union were given. In [2] Atanassov (1986) and in [9] De et al. (2000) introduced several basic operations on IFS. Classical methods are inadequete to solve problems with uncertain data in the field of engineering, social science, economics, medical science, operations research and variety of fields including decision making. Fuzzy set operators can be considered as a suitable mathematical tool for solving problems with uncertainities. Researchers introduced many intuitionistic fuzzy operators. We motivated by the operators available in literature and an operator A(m,n) introduced by Nagalingam. R and Rajaram. S[8].In [11] Xu (2007) defined operational laws of intuitionistic fuzzy information, including the intuitionistic fuzzy averaging operators and intuitionistic fuzzy aggregation operators. In [5] Ejegwa Paul Augustine used IFS in career determination, medical diagnosis and pattern recognition. Co k han Cuvalcioglu and Esra Aykut [6] applied some intuitionistic fuzzy modal operators to agriculture. In [7] Eulalia Szmidt and Janusz Kacprzyk used IFS in some medical applications. In [8] Nagalingam R and Rajaram S used intuitionistic fuzzy operator A(m,n) for career determination. 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received March 11, 2018, Accepted June 22, 2018, Edited by K. Selvakumar.

145

Intuitionistic Fuzzy Operator A(m,n) and its Extensions

This paper is organized as follows: Some basic definitions related to IFS are presented in section 2. New operator A(m,n) and its extensions are introduced and some theorems related to the introduced operators are proved in section 3. An application problem on decision making (α,β) related to A(m,n) ∩ A(m,n) is discussed in section 4. §2. Preliminaries An IFS A defined over a non-empty set X is the form of elements hx, µA (x), νA (x)i where µA : X → [0, 1] and νA : X → [0, 1] are the degree of membership and degree of non-membership for every x ∈ X respectively be such that 0 ≤ µA (x) + νA (x) ≤ 1. The complement of µA (x) + νA (x) from 1 namely πA (x) = 1 − µA (x) − νA (x) is called hesitancy degree of x in A.

Let IF S(X) denote the family of all IFS in the universe X. For every A, B ∈ IF S(X)which are represented by A = {hx, µA (x), νA (x)i | x ∈ X}, B = {< x, µB (x), νB (x) >| x ∈ X}. Some relations and operations are defined as follows ([2]): (i) A = {< x, νA (x), µA (x) >| x ∈ X}; (ii) A ⊂ B iff µA (x) ≤ µB (x) and νA (x) ≥ νB (x)∀x ∈ X; (iii) A = B iff µA (x) = µB (x) and νA (x) = νB (x)∀x ∈ X; (iv) A ∩ B = {< x, min(µA (x), µB (x)), max(νA (x), νB (x)) >| x ∈ X}; (v) A ∪ B = {< x, max(µA (x), µB (x)), min(νA (x), νB (x)) >| x ∈ X}; (vi) A + B = {< x, µA (x) + µB (x) − µA (x)µB (x), νA (x)νB (x) >| x ∈ X}; (vii) A · B = {< x, µA (x)µB (x), νA (x) + νB (x) − νA (x)νB (x) >| x ∈ X};

B (x) νA (x)+νB (x) (viii) A@B = {< x, µA (x)+µ , >| x ∈ X}; 2 2 p p (ix) A$B = {< x, µA (x)µB (x), νA (x)νB (x) >| x ∈ X};

A (x)µB (x) , 2νA (x)νB (x) >| x ∈ X}. (x) A#B = {< x, µ2µ A (x)+µB (x) νA (x)+νB (x)

If µA (x) = µB (x) = 0 then

µA (x)µB (x) µA (x)+µB (x)

= 0 and if νA (x) = νB (x) = 0 then

νA (x)νB (x) νA (x)+νB (x)

= 0.

µA (x)+µB (x)) (xi) A ∗ B = {< x, 2(µ , νA (x)+νB (x) >| x ∈ X}. A (x)µB (x)+1) 2(νA (x)νB (x)+1)

The level operators are defined by Pα,β (A) = {< x, max(µA (x), α), min(νA (x), β) >| x ∈ X} and

Qα,β (A) = {< x, min(µA (x), α), max(νA (x), β) >| x ∈ X} where α + β ≤ 1. The operator Gα,β (A) is defined by Gα,β (A) = {< x, αµA (x), βνA (x) >| x ∈ X} The empty

IFS, the totally uncertain IFS and the unit IFS are defined by O∗ = {< x, 0, 1 > |x ∈ X} ,

U ∗ = {< x, 0, 0 > |x ∈ X} and E ∗ = {< x, 1, 0 > |x ∈ X} respectively.

Definition 2.1 An Intuitionistic Fuzzy Number A is defined as follows: (i) Intuitionistic fuzzy subset of the real line R; (ii) Normal i.e there is an x0 ∈ R be such that µA (x0 ) = 1 ( so νA (x0 ) = 0);

146

R. Nagalingam and S. Rajaram

(iii) A convex set for the membership function µA (x) i.e, µA (λx1 + (1 − λ)x2 ) ≥ min(µA (x1 ), µA (x2 )) for all x1 , x2 ∈ R, λ ∈ [0, 1]; (iv) A convex set for the non-membership function νA (x) i.e, νA (λx1 + (1 − λ)x2 ) ≥ max(νA (x1 ), νA (x2 )) for all x1 , x2 ∈ R, λ ∈ [0, 1].

§ 3. New Operators on IFS For every A ∈ IFS(X) and for every m, n ∈ N, we define A(m,n) = {< x, 1 −

(1 − µA (x))m (νA (x))m , >| x ∈ X}. n n

It is clearly an IFS. Based on A(m,n) the extensions are defined by (i) A(m,(n)) = {< x, 1 − ( 1−µnA (x) )m , ( νAn(x) )m >| x ∈ X}; (α,β)

(ii) A(m,n) = {< x, 1 −

m (1−αµA (x))m , β (νA (x)) n n

(α,(β))

(iii) A(m,n) = {< x, 1 −

>| x ∈ X};

(1−αµA (x))m (βνA (x))m , n n

>| x ∈ X};

(α,β)

(iv) A(m,(n)) = {< x, 1 − ( 1−αµnA (x) )m , β( νAn(x) )m >| x ∈ X}; (α,(β))

(v) A(m,(n)) ) = {< x, 1 −

1−αµA (x) m βνA (x) m ) ,( n ) n

>| x ∈ X}.

Clearly, all of the above are IFS. Theorem 3.1 Let A ∈ IF S(X) and m, n ∈ N. Then (i) (A(m,1) )(n,1) = A(mn,1) = (A(n,1) )(m,1) ; (ii) (A(1,m) )(1,n) = A(1,mn) = (A(1,n) )(1,m) ; (iii) (A(1,m) )(n,1) = A(n,mn ) ; (iv) (A(n,1) )(1,m) = A(n,m) ; (v) (A(m,n) )(m1 ,n1 ) = A(mm1 ,n1 nm1 ) ; (vi) (A(m,n) )(m1 ,n1 ) + (A(n,m) )(n1 ,m1 ) = A(mm1 +nn1 ,m1 n1 mn1 nm1 ) . Theorem 3.2 Let A, B ∈ IF S(X) and m, n ∈ N. Then (i) (A ∪ B)(m,n) = A(m,n) ∪ B(m,n) ; (ii) (A ∩ B)(m,n) = A(m,n) ∩ B(m,n) ; (iii) (A ∩ B)(m,n) = (A)(m,n) ∩ (B)(m,n) ; (iv) (A ∪ B)(m,n) = (A)(m,n) ∪ (B)(m,n) .


147

Proof We know that (A ∪ B)(m,n) = {< x, max(µA (x), µB (x)), min(νA (x), νB (x)) >| x ∈ X}(m,n)

(1 − max(µA (x), µB (x)))m (min(νA (x), νB (x)))m , >| x ∈ X} n n (1 − µA (x))m (1 − µB (x))m (νA (x))m (νB (x))m = {< x, 1 − min( , ), min( , ) >| x ∈ X} n n) n n (1 − µA (x))m (1 − µB (x))m (νA (x))m (νB (x))m = {< x, max(1 − ,1 − ), min( , ) >| x ∈ X} n n n n = A(m,n) ∪ B(m,n) = {< x, 1 −

This proves (i). Proof of (ii), (iii) and (iv) are similar to that of (i).

2

Theorem 3.3 Let A, B ∈ IF S(X) and m, n ∈ N. Then (i) (A + B)(m,n) ⊂ A(m,n) + B(m,n) ; (ii) (A + B)(m,n) ⊂ (A)(m,n) + (B)(m,n) ; (iii) A(m,n) @B(m,n) ⊂ (A@B)(m,n) ; (iv) (A)(m,n) @(B)(m,n) ⊂ (A@B)(m,n) . If n = 1 then equality holds in (i) and (ii) and if m = 1 then equality holds in (iii) and (iv). Proof Calculation shows that (A + B)(m,n) (1 − (µA (x) + µB (x) − µA (x)µB (x)))m (νA (x)νB (x))m , >| x ∈ X} n n ((1 − µA (x))(1 − µB (x)))m (νA (x))m (νB (x))m ) = {< x, 1 − , >| x ∈ X}. n n (1 − µA (x))m (1 − µB (x))m (νA (x))m (νB (x))m = {< x, 1 − , >| x ∈ X} n n

= {< x, 1 −

and A(m,n) + B(m,n) = {< x, 1 −

(1 − µA (x))m (1 − µB (x))m (νA (x))m (νB (x))m , >| x ∈ X}. n2 n2

Hence, we get the proof of (i). Proof of (ii),(iii) and (iv) are similar to that of (i). Theorem 3.4 Let A, B, C ∈ IF S(X) and m, n ∈ N. Then (i) [(A ∪ B)@C](m,n) = (A@C)(m,n) ∪ (B@C)(m,n) ; (ii) [(A ∩ B)@C](m,n) = (A@C)(m,n) ∩ (B@C)(m,n).

2

148


Proof Calculation shows that (A ∪ B)@C

=

µA (x) + µC (x) µB (x) + µC (x) , ), 2 2 νA (x) + νC (x) νB (x) + νC (x) min( , ) >| x ∈ X}. 2 2 {< x, max(

Therefore, [(A ∪ B)@C](m,n)

C (x) µB (x)+µC (x) , ))m (1 − max( µA (x)+µ 2 2 , n C (x) νB (x)+νC (x) (min( νA (x)+ν , ))m 2 2 >| x ∈ X} n C (x) C (x) (1 − ( µA (x)+µ (1 − ( µB (x)+µ ))m ))m 2 2 = {< x, max(1 − ,1− ), n n C (x) m C (x) m ( νA (x)+ν ) ( νB (x)+ν ) 2 2 min( , ) >| x ∈ X} n n C (x) C (x) m (1 − ( µA (x)+µ ))m ( νA (x)+ν ) 2 2 = {< x, 1 − , >| x ∈ X} n n C (x) C (x) m [ (1 − ( µB (x)+µ ))m ( νB (x)+ν ) 2 2 {< x, 1 − , >| x ∈ X} n n [ = (A@C)(m,n) (B@C)(m,n) .

= {< x, 1 −

This proves (i). Proof of (ii) is similar to that of (i).

Theorem 3.5 Let A ∈ IF S(X) and m, n, m1 , n1 ∈ N and α, β, γ, δ ∈ [0, 1]. Then h i (γ,δ) (γ,δ) (i) [P1−(1−α)n ,β n (A)](m1 ,n1 ) = P1− (1−γ+γ(1−α)n )m1 , δβm1 n (A)(m1 ,n1 ) ; n1

n1

h i (γ,δ) (γ,δ) (ii) [Q1−(1−α)n ,β n (A)](m1 ,n1 ) = Q1− (1−γ+γ(1−α)n )m1 , δβm1 n (A)(m1 ,n1 ) ; n1

(iii) [Q

(iv) [P

1 1−(1−α) n

1 1−(1−α) n

,β

1 n

1 ,β n

n1

(γ,δ)

(A)](m1 ,n1 ) = Q

f rac1n )m1 1− (1−γ+γ(1−α) n1

(γ,δ)

(A)](m1 ,n1 ) = P

f rac1n )m1 1− (1−γ+γ(1−α) n1

m1 n , δβn 1

m1 n , δβn 1

h i (γ,δ) (A)(m1 ,n1 ) ;

h i (γ,δ) (A)(m1 ,n1 ) .

Proof Calculation shows that (γ,δ)

[P1−(1−α)n ,β n (A)](m1 ,n1 ) = {< x, max(1 − (1 − α)n , µA (x)),

(γ,δ)

min(β n , νA (x)) >| x ∈ X}(m1 ,n1 )

= {< x, 1 −

(1 − γ max(1 − (1 − α)n , µA (x)))m1 (min(β n , νA (x)))m1 ,δ >| x ∈ X} n1 n1

2


(1 − max(γ(1 − (1 − α)n ), γµA (x)))m1 , n1 (β n )m1 (νA (x))m1 δ min( , ) >| x ∈ X} n1 n1 (min(1 − γ(1 − (1 − α)n ), 1 − γνA (x)))m1 = {< x, 1 − , n1 (β n )m1 (νA (x))m1 δ min( , ) >| x ∈ X} n1 n1 (1 − γ(1 − (1 − α)n ))m1 (1 − γµA (x))m1 = {< x, 1 − min( , ), n1 n1 δ(β n )m1 δ(νA (x))m1 min( , ) >| x ∈ X} n1 n1 (1 − γ(1 − (1 − α)n ))m1 (1 − γµA (x))m1 = {< x, max(1 − ,1 − ), n1 n1 δβ m1 n δ(νA (x))m1 min( , ) >| x ∈ X} n n1 h i1 (γ,δ) = P1− (1−γ+γ(1−α)n )m1 , δβm1 n (A)(m1 ,n1 ) .

149

= {< x, 1 −

n1

n1

This proves (i). Proof of (ii), (iii) and (iv) are similar to that of (i)

2

§ 4. Application on Decision Making After completion of 12th standard examination, five students from a school wanted to join a degree course in a metropolitan city college . For admission of degree course they applied seven degree courses in six popular colleges which are located in different places in the metropolitan city. The decision making problem is to select best and convenient college to join for completion of the degree course successively for each student. Let Su = {sub1 , sub2 , sub3 , sub4 , sub5 , sub6 } be a set of subjects in 12th standard examination, St = {st1 , st2 , st3 , st4 , st5 } be a set of students, C = {c1 , c2 , c3 , c4 , c5 , c6 } be a set of popular colleges in the metropolitan city, D = {d1 , d2 , d3 , d4 , d5 , d6 , d7 } denotes the common degree courses in all the six colleges. Here sub1 denotes Mathematics, sub2 denotes Physics, sub3 denotes Chemistry, sub4 denotes Computer science, sub5 denotes English, sub6 denotes Language part, d1 denotes B.Sc (Maths), d2 denotes B.Sc (Physics), d3 denotes B.Sc (Chemistry), d4 denotes B.Sc (Computer Science), d5 denotes B.A (English Literature), d6 denotes B.A (Political Science) and d7 denotes B.A (History). Based on the results of previous examination conducted by the school the following table shows the membership value and non-membership value of each student corresponding to each

150


subject score to get more than 70 percent marks. Table 1. The relation between students and subjects (P) P

sub1

sub2

sub3

sub4

sub5

sub6

st1

(0.8, 0.2)

(0.8, 0.1)

(0.7, 0.1)

(0.6, 0.2)

(0.6, 0.2)

(0.8, 0.2)

st2

(0.7, 0.1)

(0.7, 0.2)

(0.8, 0.1)

(0.6, 0.3)

(0.6, 0.3)

(0.5, 0.4)

st3

(0.7, 0.2)

(0.8, 0.1)

(0.6, 0.3)

(0.7, 0.2)

(0.5, 0.3)

(0.7, 0.2)

st4

(0.5, 0.1)

(0.5, 0.2)

(0.5, 0.2)

(0.8, 0.1)

(0.5, 0.4)

(0.7, 0.2)

st5

(0.8, 0.1)

(0.5, 0.3)

(0.7, 0.3)

(0.5, 0.4)

(0.4, 0.3)

(0.7, 0.2)

The following Table shows the membership value and non-membership value of the possible allotment of degree courses in colleges based on each subject score. Table 2. The relation between subjects and basic degrees Q Q

d1

d2

d3

d4

d5

d6

d7

sub1

(0.8, 0.1)

(0.5, 0.2)

(0.5, 0.3)

(0.5, 0.3)

(0.7, 0.2)

(0.7, 0.3)

(0.6, 0.1)

sub2

(0.6, 0.2)

(0.7, 0.2)

(0.6, 0.2)

(0.5, 0.1)

(0.6, 0.2)

(0.6, 0.3)

(0.5, 0.2)

sub3

(0.5, 0.3)

(0.5, 0.2)

(0.7, 0.2)

(0.5, 0.3)

(0.6, 0.2)

(0.5, 0.3)

(0.5, 0.2)

sub4

(0.5, 0.2)

(0.5, 0.3)

(0.6, 0.1)

(0.8, 0.2)

(0.5, 0.3)

(0.5, 0.2)

(0.4, 0.2)

sub5

(0.4, 0.2)

(0.5, 0.3)

(0.4, 0.2)

(0.8, 0.1)

(0.8, 0.2)

(0.6, 0.1)

(0.4, 0.3)

sub6

(0.4, 0.1)

(0.4, 0.2)

(0.6, 0.3)

(0.4, 0.3)

(0.8, 0.1)

(0.7, 0.2)

(0.7, 0.1)

Based on the previous university examination results the following table shows the membership value and non-membership value of the success rate of completion of each degree courses in six different colleges. Table 3 The relation between basic degrees and success rate of degrees in colleges (R) R

c1

c2

c3

c4

c5

c6

d1

(0.5, 0.2)

(0.6, 0.2)

(0.3, 0.1)

(0.4, 0.1)

(0.3, 0.5)

(0.4, 0.5)

d2

(0.4, 0.5)

(0.2, 0.1)

(0.6, 0.1)

(0.7, 0.1)

(0.2, 0.5)

(0.3, 0.4)

d3

(0.4, 0.2)

(0.4, 0.3)

(0.8, 0.1)

(0.6, 0.2)

(0.5, 0.3)

(0.4, 0.1)

d4

(0.2, 0.3)

(0.3, 0.5)

(0.4, 0.2)

(0.8, 0.1)

(0.6, 0.2)

(0.5, 0.3)

d5

(0.4, 0.2)

(0.4, 0.2)

(0.3, 0.1)

(0.3, 0.1)

(0.6, 0.2)

(0.5, 0.2)

d6

(0.3, 0.2)

(0.4, 0.1)

(0.6, 0.1)

(0.7, 0.1)

(0.4, 0.1)

(0.2, 0.3)

d7

(0.5, 0.1)

(0.6, 0.1)

(0.2, 0.1)

(0.6, 0.1)

(0.6, 0.2)

(0.4, 0.2)

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Consider the composition S = Q ◦ R. It gives the data in which the subjects subi (i = 1 to 6) in terms of colleges ck (k = 1 to 6). The composition S = Q ◦ R is defined by µS (subi , ck ) = max{min[µQ (subi , dj ), µR (dj , ck )]} and ν(S) (subi , ck ) = min{max[νQ (subi , dj ), νR (dj , ck )]} for ∀ j = 1 to 7. By using this composition we get the following table. Table 4. The relation between subjects and colleges (S) S =Q◦R

c1

c2

c3

c4

c5

c6

sub1

(0.5, 0.1)

(0.6, 0.1)

(0.6, 0.1)

(0.7, 0.1)

(0.6, 0.2)

(0.5, 0.2)

sub2

(0.5, 0.2)

(0.6, 0.2)

(0.6, 0.2)

(0.7, 0.1)

(0.6, 0.2)

(0.5, 0.2)

sub3

(0.5, 0.2)

(0.5, 0.2)

(0.5, 0.2)

(0.6, 0.2)

(0.6, 0.2)

(0.5, 0.2)

sub4

(0.5, 0.2)

(0.5, 0.2)

(0.5, 0.1)

(0.8, 0.2)

(0.6, 0.2)

(0.5, 0.1)

sub5

(0.4, 0.2)

(0.4, 0.1)

(0.6, 0.1)

(0.8, 0.1)

(0.6, 0.1)

(0.5, 0.2)

sub6

(0.4, 0.1)

(0.6, 0.1)

(0.6, 0.1)

(0.7, 0.1)

(0.6, 0.1)

(0.5, 0.2)

Consider the composition T = P ◦ (Q ◦ R) = P ◦ S. It gives the data in which the students sti in terms of admission of courses in colleges ck . The composition T is defined by µT (sti , ck ) = max{min[µP (sti , subj ), µS (subj , ck )]} and νT (sti , ck ) = min{max[νP (sti , subj ), νS (subj , ck )]} ∀ i = 1 to 5, k = 1 to 6 and j = 1 to 6. By applying this composition T we get the following table. Table 5. The relation between students and colleges (T) c1

c2

c3

c4

c5

c6

st1

(0.7, 0.2)

(0.7, 0.2)

(0.7, 0.2)

(0.7, 0.1)

(0.7, 0.2)

(0.7, 0.2)

st2

(0.7, 0.1)

(0.7, 0.1)

(0.7, 0.1)

(0.7, 0.1)

(0.7, 0.2)

(0.7, 0.2)

st3

(0.7, 0.2)

(0.7, 0.2)

(0.7, 0.2)

(0.7, 0.1)

(0.7, 0.2)

(0.6, 0.2)

st4

(0.7, 0.1)

(0.6, 0.1)

(0.7, 0.1)

(0.8, 0.1)

(0.7, 0.2)

(0.6, 0.1)

st5

(0.7, 0.1)

(0.6, 0.2)

(0.7, 0.1)

(0.7, 0.2)

(0.7, 0.2)

(0.7, 0.2)

The following factors are important for students to join a degree course in a college: (i) Distance from their home; (ii) Economic situation of their family; (iii) Students willingness to join a particular degree course; (iv) Fees structure of the course; (v) Faculty position of each department in the college;

152


(vi) Discipline, teaching, coaching, valuation method, internal marks, transport facility, library facility, hostel facility of the colleges, P.G degree courses offered in the same UG courses, and campus interview facility. Based on the above factors, the pair of values (αi , βj ) are given to each student corresponding to each colleges where αi and βj denote the possible membership value, non-membership value respectively to join in a particular college cj .

Table 6: (αi , βj ) values for i = 1 to 5, j=1 to 6 c1

c2

c3

c4

c5

c6

st1

(0.5, 0.1)

(0.4, 0.2)

(0.2, 0.2)

(0.3, 0.4)

(0.5, 0.3)

(0.4, 0.2)

st2

(0.6, 0.2)

(0.5, 0.1)

(0.5, 0.2)

(0.4, 0.5)

(0.5, 0.4)

(0.6, 0.2)

st3

(0.6, 0.3)

(0.5, 0.2)

(0.6, 0.2)

(0.5, 0.1)

(0.5, 0.3)

(0.7, 0.1)

st4

(0.7, 0.2)

(0.6, 0.2)

(0.6, 0.1)

(0.5, 0.1)

(0.5, 0.1)

(0.3, 0.2)

st5

(0.6, 0.1)

(0.4, 0.2)

(0.3, 0.1)

(0.5, 0.2)

(0.3, 0.2)

(0.6, 0.1)

Let m be the number of students and n be the number of basic degrees. We considered five subjects (i.e., m = 5) and six degree courses (i.e., n = 6). Apply the operator A(5,6) = {< x, 1 −

(1 − µA (x))5 (νA (x))5 , >| x ∈ X} 6 6

to each entry of Table 5 we get the following table. Table 7. The values of A(5,6) A(5,6)

c1

c2

c3

c4

c5

c6

st1

(0.994791, 5.33E-05)

(0.999595, 5.33E-05)

(0.999595, 5.33E-05)

(0.999595, 1.67E-06)

(0.998293, 5.33E-05)

(0.994791, 5.33E-05)

st2

(0.994791, 1.67E-06)

(0.999595, 1.67E-06)

(0.999595, 1.67E-06)

(0.999595, 1.67E-06)

(0.998293, 5.33E-05)

(0.994791, 5.33E-05)

st3

(0.994791, 5.33E-05)

(0.999595, 5.33E-05)

(0.998293, 5.33E-05)

(0.999595, 1.67E-05)

(0.998293, 5.33E-05)

(0.994791, 5.33E-05)

st4

(0.994791, 1.67E-06)

(0.998293, 1.67E-06)

(0.998293, 1.67E-06)

(0.999946, 1.67E-06)

(0.998293, 5.33E-05)

(0.994791, 1.67E-06)

st5

(0.994791, 1.67E-06)

(0.998293, 5.33E-05)

(0.999595, 1.67E-06)

(0.999595, 5.33E-05)

(0.999595, 5.33E-05)

(0.994791, 5.33E-05)

(α,β)

Apply the operator A(m,n) to each entry of Table 5 and the corresponding values (αi , βj ) of Table 6.

153


(α,β)

Table 8. The values of A(m,n) c1

c2

c3

c4

c5

c6

st1

(0.9604422, 5.33E-10)

(0.9677514, 1.71E-08)

(0.9215955, 1.71E-08)

(0.9487160, 1.72E-08)

(0.9719883, 1.30E-07)

(0.9453867, 1.71E–08)

st2

(0.971988, 5.33E-10)

(0.9806618, 1.67E-11)

(0.9806620, 5.33E-10)

(0.9677514, 5.21E-08)

(0.9719883, 5.46E-07)

(0.9719883, 1.71E-08)

st3

(0.9719883, 1.30E-07)

(0.9806618, 1.71E-08)

(0.9821043, 1.71E-08)

(0.9806618, 1.67E-05)

(0.9719883, 1.30E-07)

(0.9806618, 5.33E-10)

st4

(0.9806618, 5.33E-10)

(0.9821043, 5.33E-10)

(0.9821043, 1.67E-11)

(0.9870400, 1.67E-11)

(0.9719883, 5.33E-10)

(0.9260491, 5.33E-10)

st5

(0.9719883, 1.67E-11)

(0.9577412, 1.71E-18)

(0.9487157, 1.67E-11)

(0.9806618, 1.71E-08)

(0.9487157, 1.71E-08)

(0.9719883, 5.33E-10)

Apply the intersection operator on the values of Tables 7 and 8 we get the following table. (α,β)

Table 9. The values of Am,n ∩ A(m,n) c1

c2

c3

c4

c5

c6

st1

(0.9604422, 5.33E-05)

(0.9677514, 5.33E-05)

(0.9215955, 5.33E-05)

(0.9487160, 1.67E-06)

(0.9719883, 5.33E-05)

(0.9453867, 5.33E-05)

st2

(0.971988, 1.67E-06)

(0.9806618, 1.67E-06)

(0.9806620, 1.67E-06)

(0.9677514, 1.67E-06)

(0.9719883, 5.33E-05)

(0.9719883, 5.33E-05)

st3

(0.9719883, 5.33E-05)

(0.9806618, 5.33E-05)

(0.9821043, 5.33E-05)

(0.9806618, 1.67E-06)

(0.9719883, 5.33E-05)

(0.9806618, 5.33E-05)

st4

(0.9806618, 1.67E-06)

(0.9821043, 1.67E-06)

(0.9821043, 1.67E-06)

(0.9870400, 1.67E-06)

(0.9719883, 5.33E-05)

(0.9260491, 1.67E-06)

st5

(0.9719883, 1.67E-06)

(0.9577412, 5.33E-05)

(0.9487157, 1.67E-06)

(0.9806618, 5.33E-05)

(0.9487157, 5.33E-05)

(0.9719883, 5.33E-05)

From the maximum membership value and minimum non membership value of each row we conclude that st1 will join in college c5 , st2 will join in college c2 , or c3 , st3 will join in college c3 , st4 will join in college c2 , or c3 , st5 will join in college c4 .

§5. Conclusion In this paper a new IFS operator A(m,n) and its extensions are defined.The theorems related to the operator A(m,n) and its extensions are discussed. An application based on these operator to select a suitable college and a suitable course is discussed. Finally we conclude that A(m,n) ∩ (α,β) A(m,n) is one of the optimum tool for selecting a suitable college to join a degree course. The scope of this paper is to introduce such a new operation with new IFS operator in decision making problems.

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References [1] [2] [3] [4] [5]

[6]

[7] [8]

[9] [10] [11]

Zadeh.L.A, (1965), Fuzzy sets, Information and control, Vol. 8, 338-353. Atanassov.K, (1986), Intuitionistic fuzzy sets, Fuzzy Sets and Systems, 20, 87-96. Atanassov.K, (1999), Intuituionistic Fuzzy Sets, Springer, Heidelberg. Atanassov. K, (2012), On Intuituionistic Fuzzy Set Theory, Springer, Berlin. Ejegwa P.A., Akubo A.J., Joshua D.M., (2014), Intuitionistic fuzzy sets and its application in career determination via normalised Euclidean distance method, European Scientific Journal, 10(15) , 529-536. Cokhan Cuvalcioglu and Esra Aykut, (2015), An application of some intuitionistic fuzzy modal operators to agriculture, 19th conference on IFS, Burgas, Notes on Intuitionistic Fuzzy Sets , ISSN 1310 - 4926, Vol21, No.2, 140-149. Eulalia Szmidt and Janusz Kacprzyk, (2001), Intuitionistic fuzzy sets in some medical applications, Fifth Int. conf. on IFSs, Sofia, NIFS, 7, 4, 58-64. Nagalingam.R and Rajaram.S, (2017), New Intuitionistic fuzzy operator A(m,n) and an application on decision making, Advances in Fuzzy Mathematics, ISSN 0973-533X Volume 12, Number 4, pp881-897. De S.K., Biswas.R and Roy.A.R, (2000), Some operations on intuitionistic fuzzy sets, Fuzzy Sets and Systems, Vol 114, No-4, 477-484. Beloslav Riecan and Krassimir T. Atanassov, (2006), n-Extraction Operation over intuitionistic fuzzy sets, NIFS, 12, 4, 9 - 11. Xu, (2007), Intuitionistic fuzzy aggregation operators, IEEE Transaction on Fuzzy Systems, 15, 1179 - 1187.


Intuitionistic Fuzzy Soft Multiset and its Application K Reji Kumar (Department of Mathematics,N.S.S College, Cherthala,India)

S A Naisal (Research Scholar,Marthoma College, Thiruvalla, M G University, Kottayam, Kerala, India) E-mail: [email protected], [email protected]

Abstract: This paper explains extension of fuzzy soft Multi sets. The extension explain on intuitionistic form of fuzzy soft multi set. We here define intuitionistic fuzzy soft multi set relations, its various type including equivalence relation. Composition of relations and the equivalence of the inverse relation when relation satisfies the equivalence property are discussed. The notion of intuitionistic fuzzy soft multi sets are tried to apply in a decision making problem which study the spread of a plant known as Makaniya Micrantha. Finally we use normalized Euclidean distance formula for intuitionistic fuzzy soft multi set, which lead to a conclusion to which area has maximum and minimum spread of this plant there by giving a result, to which area the destruction of the plant is require and in which area its to be promoted. We establish several important results with adequate examples.

Key Words: Fuzzy sets, soft sets, multisets, fuzzy topology. AMS(2010): 03E72, 54A40. §1. Introduction Uncertainties rule the human mind and perceptions. In most of the engineering, physics, chemical, computer sciences, biological economics, social sciences and medical sciences problems human mind deal with these uncertainties. Classical or ordinary set theory, which is based on the crisp and wee can see exact case may not be fully suitable for handling such kind of problems of uncertainty. A number of mathematical tools like probability theory, fuzzy sets [1] , soft sets Molodtsov [3] rough sets [2] are well known and are efficient and effective models for dealing with vagueness and uncertainties. Then also each of them has distinguished advantages and certain limitations. Parametrization was not done before 1990s. This inadequacy of parametrization tools was perfectly cleared by Molodtsov [3] by introducing a new concept named soft set theory. He applied soft set theory in many fields, game theory, smoothness of a function, the Riemann integral, Perron integral and measurement theory. Maji [4] presented an application 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received March 30, 2018, Accepted September 24, 2018, Edited by K. Selvakumar.

156

K Reji Kumar and S A Naisal

of soft sets in decision making problems for the first time, that was based on the reduction of parameters to keep the optimal choice of objects. The concept of soft set relations are introduced as a sub soft set of the cartesian product of the soft sets in [10]. Combining soft sets with fuzzy sets , Maji et al [6] defined fuzzy soft sets which are used for solving decision making problems. As a generalization of Molodtsov soft set Alkhazaleh [7] presented the definition of a soft multi set and he defined basic operations such as complement, union, and intersection etc. Salleh and Alkhazaleh presented the application of soft multi set in decision making problem. AND, OR operator on soft multiset and DeMorgans laws was proved by [11] Babitha. In 2011 Salleh gave a brief literature survey from soft sets to intuitionistic fuzzy soft sets. In 2012 Alkhazaleh and Salleh [8] introduced the concept of fuzzy soft multi set theory and studied the application of the set. Mukherjee and Das [14] in 2014 introduced the concepts of intuitionistic fuzzy soft multi sets and studied intuitionistic fuzzy soft multi topological spaces in details.While going through the above references it seems to be important to go through the non membership of the relations of fuzzy soft multi sets. This paper can be considered as an extention of the paper [12] ’Interior exterior and boundary of fuzzy soft multi set topology on decision making.

§2. Preliminaries Definition 2.1 ([3]) Let U be an initial universe set and E be set of parameters. Let P (U ) denotes the power set of U and A ⊆ U . A pair (F, A) is called a soft set over U , where F is a mapping given by F : A → P (U ). Definition 2.2 ([9]) A multiset M drawn from the set X is represented by a function Count M or CM defined as CM : X → N , where N represents the set of non negative integers. Definition 2.3 ([4]) Let U be an initial universal set and let E be a set of parameters. Let I U denote the power set of all fuzzy subsets of U . Let A ⊆ E. A pair (F, E) is called a fuzzy soft set over U , where F is a mapping given by F :A → I U . Definition 2.4 ([8]) Let {Ui : i ∈ I} be a collection of universes such that ∩i∈I Ui = φ and let {EUi : i ∈ I} be a collection of sets of parameters. Let U = πi∈I P (Ui ) where P (Ui ) denotes the power set of Ui , E = πi∈I EUi and A ⊆ E. A pair (F, A) is called a fuzzy soft multiset over U , u where F is a mapping given by F : A → U , for all e in A F (e) = ({ µF (e) (u) }; i ∈ I). Definition 2.5 ([8]) A soft multiset (F, A) over U is called a seminull soft multiset, denoted by (F, A) ≈ φi , if at least one of the soft multiset parts of (F, A) equals φ.

A soft multiset (F, A) over U is called a null soft multiset, denoted by (F, A)phi , if all of the soft multiset parts of (F, A) equals φ.

Definition 2.6 ([8]) A soft multiset (F, A) over U is called an absolute soft multiset, denoted by (F, A)A , if (eUi,j , FeUi,j ) = Ui , ∀i. Definition 2.7 ([8]) The union of two soft multisets (F, A) and (G, B) over U , denoted by

Intuitionistic Fuzzy Soft Multiset and its Application

157

(F, A) ∪ (G, B) is the soft multiset (H, C) where C = A ∪ B, and ∀ǫ ∈ C, H(ǫ)

= F (ǫ), if ǫ ∈ A − B, = G(ǫ), if ǫ ∈ B − A, = F (ǫ) ∪ G(ǫ), if ǫ ∈ A ∩ B.

Definition 2.8 ([8]) The intersection of two soft multisets (F, A) and (G, B) over U , denoted by (F, A) ∪ (G, B) is the soft multiset (H, C) where C = A ∪ B, and ∀ǫ ∈ C, H(ǫ) = F (ǫ), if ǫ ∈ A − B, = {G(ǫ), if ǫ ∈ B − A}, = {F (ǫ) ∩ G(ǫ), if ǫ ∈ A ∩ B}. Definition 2.9 ([8]) The union of two fuzzy soft multisets (F, A) and (G, B) over U , denoted by (F, A) ∪ (G, B) is the soft multiset (H, C) where C = A ∪ B, and ∀ǫ ∈ C, H(ǫ)

= F (ǫ), if ǫ ∈ A − B, = G(ǫ), if ǫ ∈ B − A, = ∪(F (ǫ), G(ǫ))if ǫ ∈ A ∩ B,

where

[ (F (ǫ), G(ǫ)) = s(FǫUi ,j , GǫUi ,j ),

for all i ∈ {1, 2 · · · , m} with s as an s- norm.

Definition 2.10 ([8]) The intersection of two fuzzy soft multisets (F, A) and (G, B) over U , denoted by (F, A) ∩ (G, B) is the soft multiset (H, C) where C = A ∪ B, and f orallǫ ∈ C, H(ǫ)

= F (ǫ), if ǫ ∈ A − B, = G(ǫ), if ǫ ∈ B − A, = ∩(F (ǫ), G(ǫ))if ǫ ∈ A ∩ B,

where

\ (F (ǫ), G(ǫ)) = t(FǫUi ,j , GǫUi ,j ),

for all i ∈ {1, 2, · · · , m} with t as an t- norm.

Definition 2.11 ([13]) The normalized Euclidean distance dn−H between two intuitionistic fuzzy set A and B is defined as

dn=H (A, B) =

!1 i=1 i 2 1 Xh 2 2 2 (µA (xi ) − µB (xi )) + (νA (xi ) − νB (xi )) + (πA (xi ) − πB (xi )) , 2n n

158


where, X = {x1 , x2 , · · · , xn } for i = 1, 2, · · · , n. §3. Intuitionistic Fuzzy Soft Multiset Relations Definition 3.1 Let (F, A) and (G, B) be two intuitionistic fuzzy soft multi sets over U. Then the Cartesian product of (F, A) and (G, B) is defined as (F, A) × (G, B) = {(ai , bj ) : min((ηF (a) (ak ), ηG(b) (ak )), max(µF (a) (ak ), µG(b) (ak )))/u ∈ Ui }. Example 3.2 Consider two intuitionistic fuzzy soft multi sets (F, A) and (F2 , B)as follows p1 p2 h1 h2 h3 h4 c1 c2 c3 (F1 , A) = {(a1 , ({ 0.5,0.3 , 0.4,0.3 , 0.1,0 , 0,0 , })), { 0.8,0.1 , 0,5,0.3 , 0,0 , }, { 0.8,0.1 , 0,0 }), p1 p2 h1 h2 h3 h4 c1 c2 c3 (a2 , ({ 0.9,0.1 , 0.5,0.4 , 0,0 , 0,0 }, { 0.7,0.2 , 0,0 , 0,0 , }, { 0.5,0.4 , 0,0 }))},

p1 p2 h1 h2 h3 h4 c1 c2 c3 (F2 , B) = {(a1 , ({ 0.7,0.3 , 0.7,0.3 , 0,0 , 0,0 }, { 0.8.0.1 , 0,0 , 0,0 , }, { 0.5,0.3 , 0,0 })),

p1 p2 h1 h2 h3 h4 c1 c2 c3 , 0.5,0.4 , 0,0 , 0,0 }, { 0,0 , 0,0 , 0,0 , }, { 0.8,0.2 , 0,0 }))}, (a2 , ({ 0.9,0.1

h1 h2 h3 h4 c1 c2 c3 (F1 , A) × (F2 , B) = {((a1 , b1 ), ({ 0.2,0.3 , 0.2 , 0,0 , 0,0 }, { 0.8,0.2 , 0,0 , 0,0 },

p1 p2 h1 h2 h3 h4 c1 c2 c3 { 0.5,0.4 , 0.0 })), ((a1 , b2 ), ({ 0.2,0.3 , 0.2,0.3 , 0,0 , 0,0 }, { 0,0 , 0,0 , 0,0 },

p1 p2 h1 h2 h3 h4 c1 c2 c3 { 0.8,0.2 , 0,0 })), ((a2 , b1 ), ({ 0.7,0.2 , 0.5,0.3 , 0,0 , 0,0 }, { 0.7,0.2 , 0,0 , 0,0 }, p1 p2 h1 h2 h3 h4 c1 c2 c3 { 0.5,0.4 , 0,0 })), ((a2 , b2 ), ({ 0.9,0.1 , 0.5,0.3 , 0,0 , 0,0 }, { 0,0 , 0,0 , 0,0 },

p1 p2 { 0.5,0.4 , 0,0 }))}.

Definition 3.3 Consider two intuitionistic fuzzy soft multi sets (F, A) and (G, B) over U. Then the intutionistic fuzzy soft multi set relation R on (F, A) and (G, B) is a fuzzy soft multi subset of the set (F, A) × (G, B). Where R is given by R : A × B → U . Definition 3.4 An intuitionistic fuzzy soft multi relation R is said to be reflexive if and only if µR(a,a) (u) = 1 for all u ∈ U and a ∈ A. p1 h1 h2 h3 h4 c1 c2 c3 Example 3.5 Consider a relation R in (F, A) as {((a1 , a1 ), ({ 1,0 , 1,0 , 1,0 , 1,0 }, { 1,0 , 1,0 , 1,0 }, { 1,0 , p1 p2 p2 h1 h2 h3 h4 c1 c2 c3 h1 h2 1,0 })), ((a1 , a2 ), ({ 0.2,0.3 , 0.2,0.5 , 0,0 , 0,0 }, { 0,0 , 0,0 , 0,0 }, { 0.8,0.2 , 0,0 })), ((a2 , a1 ), ({ 0.7,0.2 , 0.5,0.4 , p1 p2 p1 p2 h3 h4 c1 c2 c3 h1 h2 h3 h4 c1 c2 c3 0,0 , 0,0 }, { 0.7,0.3 , 0,0 , 0,0 }, { 0.5,0.3 , 0,0 })), ((a2 , a2 ), ({ 1,0 , 1,0 , 1,0 , 1,0 }, { 1,0 , 1,0 , 1,0 }, { 1,0 , 1,0 }))}, then R is a reflexive relation.

Definition 3.6 Consider an intuitionistic fuzzy soft multi sets (F, A), then a relation R on (F, A) is said to be symmetric if and only if (a, b) ∈ R then (b, a) ∈ R. Example 3.7 Consider a relation R in (F, A) as p1 p2 h1 h2 h3 h4 c1 c2 c3 h1 h2 h3 {((a1 , b1 ), ({ 0.2,0.4 , 0.2,0.5 , 0,0 , 0,0 }, { 0.8,0.1 , 0,0 , 0,0 }, { 0.5,0.4 , 0,0 })), ((a1 , b2 ), ({ 0.2,0.6 , 0.6,0.3 , 0,0 , p1 p2 h4 c1 c2 c3 h1 h2 h3 h4 c1 c2 c3 }, { , , }, { , })), ((b , a ), ({ , , , }, { , , }, 1 1 0,0 0.3,0.5 0.8,0.1 0,0 0.8,0.1 0,0 0.2,0.7 0.2,0.7 0,0 0,0 0.3,0.5 0.8,0.1 0,0 p1 p2 p1 p2 h1 h2 h3 h4 c1 c2 c3 { 0.8,0.1 , 0,0 })), ((a2 , b2 ), ({ 0.9,0.1 , 0.5,0.3 , 0,0 , 0,0 }, { 0,0 , 0,0 , 0,0 }, { 0.5,,0.4 , 0,0 }))}, then R is a Symmetric relation. Definition 3.8 Consider two intuitionistic fuzzy soft multi set relations R1 , R2 in (F, A), the composition of R1 and R2 is defined as R1 o R2 : A × A → U . That is max

µR1 oR2 (a,b) (u) = k {min(ηR1 (a,k) (u), ηR2 (k,b) (u)), max(µR1 (a,k) (u), µR2 (k,b) (u))},


159

for all u ∈ Ui , a, b, k ∈ A. Definition 3.9 Consider an intuitionistic fuzzy soft multi sets (F, A), then a relation R on (F, A) is said to be transitive if and only if RoR ⊆ R. Example 3.10 Consider a relation R in (F, A) as p1 p2 h1 h2 h3 h4 c1 c2 c3 h1 h2 {((a1 , b1 ), ({ 0.3,0.4 , 0.7,0.2 , 0,0 , 0,0 }, { 0.8,0.2 , 0.8,0.1 , 0,0 }, { 0.5,0.4 , 0,0 })), ((a1 , b2 ), ({ 0.2,0.5 , 0.6,0.3 , p1 p2 c1 c2 c3 h1 h2 h3 h4 c1 c2 c3 h3 h4 0,0 , 0,0 }, { 0.3,0.5 , 0.8,0.2 , 0,0 }, { 0.5,0.3 , 0,0 })), ((a2 , b1 ), ({ 0.3,0.5 , 0.6,0.3 , 0,0 , 0,0 }, { 0.3,0.5 , 0.8,0.1 , 0,0 }, p1 p2 p1 p2 h1 h2 h3 h4 c1 c2 c3 { 0.5,0.1 , 0,0 })), ((a2 , b2 ), ({ 0.9,0.1 , 0.6,0.2 , 0,0 , 0,0 }, { 0.6,0.3 , 0.8,0.1 , 0,0 }, { 0.5,0.4 , 0,0 }))}, then R is a transitive relation. Definition 3.11 Consider a relation R in an intuitionistic fuzzy soft multi set (F, A). Then R is said to be an equivalence relation if R is reflexive, symmetric, and transitive. Definition 3.12 Consider a relation R in an intuitionistic fuzzy soft multi set (F, A). Then the inverse of R, denoted by R−1 is defined as R−1 (a, b) = R(b, a) for all a, b ∈ A. Example 3.13 Consider a relation R in (F, A) as p1 p2 h1 h2 h3 h4 c1 c2 c3 h1 h2 h3 {((a, a), ({ 0.3,0.5 , 0.7,0.2 , 0,0 , 0,0 }, { 0.8,0.2 , 0.8,0.1 , 0,0 }, { 0.5,0.4 , 0,0 })), ((a, b), ({ 0.2,0.7 , 0.6,0.3 , 0,0 , p1 p2 h4 c1 c2 c3 h1 h2 h3 h4 c1 c2 c3 0,0 }, { 0.3,0.5 , 0.8,0.1 , 0,0 } , { 0.5,0.4 , 0,0 })), ((b, a), ({ 0.2,0.7 , 0.6,0.3 , 0,0 , 0,0 }, { 0.3,0.5 , 0.8,0.1 , 0,0 }, p1 p2 p1 p2 h1 h2 h3 h4 c1 c2 c3 { 0.5,0.4 , 0,0 })), ((b, b), ({ 0.9 , 0.6 , 0 , 0 }, { 0.6 , 0.8 , 0 }, { 0.5 , 0 }))}, here (a, b) is the inverse of (b, a). Theorem 3.14 A relation R is a symmetric intuitionistic fuzzy soft multi relation on (F, A) if and only if the inverse of R, R−1 is symmetric. Proof Consider the relation R as a symmetric intuitionistic fuzzy soft multi relation on (F, A). Then µR−1 (a,b) = µR(b,a) = µR(a,b) = µR−1 (b,a) . Therefore µR−1 (a,b) = µR−1 (b,a) . 1 Therefore R is symmetric. 1 Conversely assume that R is a symmetric fuzzy soft multi set on (F, A). Then µR(a,b) = µR−1 (b,a) = µR−1 (a,b) = µR(b,a) . Since µR(a,b) = µR(b,a) , R is a symmetric intuitionistic fuzzy 2 soft multi relation on (F,A). Theorem 3.15 A relation R is a reflexive ituitionistic fuzzy soft multi relation on (F, A) if and only if the inverse of R, R−1 is reflexive. Proof Consider the relation R as a reflexive intuitionistic fuzzy soft multi relation on (F, A). Then µR−1 (a,a) = µR(a,a) . Thus R−1 is reflexive. Conversely let R−1 is a reflexive relation on (F, A). Then µR(a,a) = µR−1 (a,a) Therefore R is reflexive. 2 Theorem 3.16 A relation R is a transitive intuitionistic fuzzy soft multi relation on (F, A) if and only if the inverse of R, R−1 is transitive. Proof Consider the relation R as a transitive intuitionistic fuzzy soft multi relation on (F, A). For every a, b ∈ A µR−1 (a,b) = µR(b,a) ≥ µRoR(b,a) max

= k {min(ηR(b,k) , ηR( k,a) , max(µR(b,k) , µR( k,a) )}

160


max

= k {min(ηR(k,a) , ηR( b,k) ), max(µR(k,a) , µR( b,k) )} max

= k {min(ηR−1 (a,k) , ηR−1 (k,b) ), max(µR−1 (a,k) , µR−1 (k,b) )} = µR−1 oR−1 (a,b) . Hence R−1 is a transitive fuzzy soft multi relation on (F, A). Conversely assume that R−1 is a transitive fuzzy soft multi relation on(F, A). For any (a, a) ∈ A × A, µR(a,b) = µR−1 (b,a) ≥ µR−1 oR−1 max

= k {min(ηR−1 (b,k) , ηR−1 (k,a) ), max(µR−1 (b,k) , µR−1 (k,a) )} max

= k {min(ηR−1 (k,a) , ηR−1 (b,k) ), max(µR−1 (k,a) , µR−1 (b,k) )} max

= k {min(ηR(a,k) , ηR( k,b) ), max(µR(a,k) , µR( k,b) )} = µRoR(a,b) .

2

Theorem 3.17 Consider a relation R in an intuitionistic fuzzy soft multi set (F, A). Then R is transitive if and only if Rn ⊆ R, for n ≥ 1. Proof Let the relation R be transitive. We have to prove that Rn ⊆ R, for n ≥ 1. For this we use the method of induction. Suppose n = 1. Then R1 = R, therefore it is obviously true. Assume that Rn ⊆ R for n ≥ 1. Consider (x, y) ∈ Rn+1 . We know from the definition that Rn+1 = Rn oR, then there must be an k ∈ A such that (x, k) ∈ R also (k, y) ∈ Rn . Our assumption was Rn ⊆ R, therefore (k, y) ∈ R. As R is transitive (x, k) ∈ R, (k, y) ∈ R which implies (x, y) ∈ R. Also we know (x, y) is an arbitrary element of Rn+1 , which shows that Rn+1 ⊆ R. Now we have to show the converse part, that is if Rn ⊆ R then R is transitive. Assume that (x, y), (y, z) ∈ R, by the definition of composition (x, z) ∈ R2 . But R2 ⊆ R implies (x, z) ∈ R. 2 Therefore R is transitive.

Theorem 3.18 Consider a relation R from an intuitionistic fuzzy soft multi set (F, A) to a fuzzy soft multi set (F, B), then we have the following R((F, A)∪(F, B)) 6= R((F, A))∪R((F, B)). Proof

We discuss it with 3 cases as following.

Case 1. Consider the intuitionistic fuzzy soft multi sets (F, A) and (F, B) with same elements. Then it is obvious that R((F, A) ∪ (F, B)) = R((F, A)) ∪ R((F, B)). Case 2. Let every element of intuitionistic fuzzy soft multi set (F, A) is in the intuitionisk tic fuzzy soft multi set (F, B). If (F, A) = {ai , { kxpa , lyqal · · · }, i, k, l ∈ N , and kpa , lqa · · · are k corresponding fuzzy values} and (F, B) = {bi , { kxpb , lyqbl · · · }, i, k, l ∈ N , and kpb , lqb · · · are corresponding fuzzy values}, then (F, A) ∪ (F, B)

=

xk yl , · · · }, i, k, l ∈ N, kpab lqab where kpab = max{kpa , kpb }, lqab = max{lqa , lqb } · · · ,

{ci , {

are the corresponding fuzzy values. In kpb , k represents the position of multi content,

pb


161

represents the fuzzy value and it correspond to the intuitionistic fuzzy soft multi set (F, B). Then R(F, A) ⊆ R(F, B) and therefore R((F, A) ∪ (F, B)) = R(F, B). Case 3. Let some elements in (F, A) are in (F, B). k If (F, A) = {ai , { kxpa , lyqal , · · · }, i, k, l ∈ N , and kpa , lqa , · · · are corresponding fuzzy values} xk zm and (F, B) = {bi , { kpb , mrb · · · }, i, k, m ∈ N , and kpb , mrb , · · · are corresponding fuzzy values} .

zm k Now (F, A) ∪ (F, B) = {ci , { kxpab , lyqal , m , i, k, l, m ∈ N, and lqa , mrb , · · · are corresponding rb fuzzy values}, where kpab is the max {kpa , kpb }.

xk zm R((F, A)∪(F, B)) = {(ci , cj ), { kpab , lyqal , m , i, k, l, m ∈ N, and lqa , mrb · · · are correspond∗ rb ing fuzzy values}, where kpab∗ is the minimum of kpab ’s, if i 6= j. Now R(F, A) and R(F, B) are found by taking the minimum of kpa ’s or kpb ’s respectively. Then R((F, A)) ∪ R((F, B)) will be the maximum {kpa ’s and kpb ’s. So it is clear that R((F, A) ∪ (F, B)) 6= R((F, A)) ∪ R((F, B)).2

Theorem 3.19 Let (F,A), (G,B), (H,C) be intuitionistic fuzzy soft multi sets in (X, E, τ ) and let (F, A) ⊆ (G, B) ⊆ (H, C). Then (G, B) − (F, A) ⊆ (H, C) − (F, A). Proof Here (F, A) ⊆ (G, B) and (G, B) ⊆ (H, C) ⇒ (F, A) ⊆ (H, C). That is η(F,A) (x) ≤ η(G,B) (x) ≤ η(H,C) (x) and µ(F,A) (x) ≥ µ(G,B) (x) ≥ µ(H,C) (x) for all x ∈ X As (F,A) is the smallest of other (G,B) and (H,C) substract (F,A) from both sides of (G, B) ⊆ (H, C). Hence (G, B) − (F, A) ⊆ (H, C) − (F, A).

2

§ 4. An Application of Intuitionistic Fuzzy Soft Multiset Theory Invasion of alien plant species is considered as a serious problem that which different countries have to face. Alien plant species invade and badly affect both natural and semi natural ecosystems. Many weed plants contain a variety of metabolites which are harmful as well as beneficial to biotic stresses, whereas same compounds in plants are beneficial to insects also for their better growth and development to continue their life style. Mikania micrantha is one of the 100 worst alien species. It is among the ten worst exotic species in South-east and South Asia, and one of the 16 exotic species in China. Mikania micrantha, an invasive alien species, is native to Central and South America is one of the serious bio diversity. Let S = {s1 , s2 , s3 , s4 } be the different universes or countries. Consider C={Biological, Medical, Shade, Spread} as the set of decision parameters. andSu = {Biological property, Medicinal advantage for animals, Medicinal disadvantage for animals, Shade for air fields, quick spread} are the set of subjects relating to the above parameters. That is Micania Micrantha have many advantages as well as disadvantages. Such as we have many many biological aspects we have to learn from this plant. These plants are good to some extend to some animals, but shows serious kidney decease to many animals. Another thing is it spreads as fast as other plants.

162


Table 1. Parameters vs Subject .

Biological

Medical(A)

Medical(D)

Shade

Spread

Biological

(0.81,0.1,0.1)

(0.7,0.21,0.1)

(0.9,0.0,0.11)

(0.6,0.31,0.1)

(0.8,0.11,0.1)

Medical

(0.91,0.1,0.0)

(0.8,0.11,0.1)

(0.8,0.11,0.1)

(0.51, 0.3,0.2)

(0.7,0.21,0.1)

Shade

(0.5,0.31,0.2)

(0.51,0.2,0.3)

(0.9,0.0,0.1)

(0.5,0.41,0.1)

(0.71,0.1,0.2)

Spread

(0.71,0.2,0.1)

(0.5,0.41,0.1)

(0.9,0.11,0.0)

(0.61,0.3, 0.1)

(0.8,0.0,0.21)

Each of these parameter is described by three numbers such as soft multi membership, soft multi non membership, and soft multi hesitation margin. Here the parameter vs subject describes while biological, biological is considered the possibility that the plant is used for biological aspects is 0.8 and its non usage is 0.1 and that of degree of hesitancy at that time is 0.1. Similarly all the other triplets are defined. After going through a serious calculations we have the following table. Table 2. Universe vs subject .

Biological

Medical(A)

Medical(D)

Shade

Spread

S1

(0.61,0.3,0.1)

(0.5,0.41,0.1)

(0.6,0.21,0.2)

(0.5,0.3,0.2)

(0.5,0.51,0.0)

S2

(0.9,0.11,0.0)

(0.81,0.1,0.1)

(0.8,0.1,0.1)

(0.5, 0.3,0.2)

(0.7,0.2,0.1)

S3

(0.5,0.3,0.2)

(0.5,0.2,0.3)

(0.9,0.0,0.1)

(0.5,0.4,0.1)

(0.7,0.1,0.2)

S4

(0.7,0.21,0.1)

(0.51,0.4,0.1)

(0.9,0.1,0.0)

(0.6,0.31, 0.1)

(0.81,0.0,0.2)

Each of four countries different usage and non usage in different subjects are given. Using the definition cited in the preliminary section we have the following table Table 3. Universe vs Parameter .

Biological

Medical

Shade

Spread

S1

0.1007

0.0898

0.0905

0.0806

S2

0.0739

0.0857

0.0592

0.0955

S3

0.0783

0.0806

0.0812

0.0858

S4

0.0832

0.0750

0.0929

0.1015

From the above table, it is clear that the shortest distance gives the proper area determination for each universe. The result is as follows S1 neads to take the measure to stop the spread of Macania Micrantha. S2 needs to take step to use Micania Micrantha as a shade for air field etc. S3 needs to take step to use Micania Micrantha as a research area in biological studies S4 needs to take step to take Micania Micrantha as a medicinal plant. So we have seen that each country have different aspects about the same plant, that is each country will have economic growth if the plant is treated in different way.


163

§4. Conclusion Here we establish many properties of intuitionistic fuzzy soft multi set theory. Here we have gone through the properties of fuzzy soft multiset relations. The variation between two intuitionistic fuzzy soft multi set relations are also proved. Szmidt’s distance between intuitionistic fuzzy sets was discussed and extended to intuitionistic fuzzy soft multi sets through an application on a decision making problem. The distance formula can be used for various uncertainty problems, and can be used as a perfect tool for finding the apt decision.

References [1] L. A. Zadeh, Fuzzy sets, Inf. Control, 8 (1965) 378-352. [2] Z. Pawlak, Rough Sets, Int. J. Inf. Comput. Sci. 11 (1982) 341-356. [3] D. Molodtsov, Soft set theory-First results, Computers and Mathematics with Applications, 37(4/5) (1999), 19-31. [4] P. K. Maji, R. Biswas and A. R. Roy; Fuzzy Soft Sets, Journal of Fuzzy Mathematics, Vol 9 , no.3,pp.589-602,2001. [5] D. Chen, E. C. C. Tsang, D. S. Yeung and X. Wang; The parameterized reduction of soft sets and its applications, Comput. Math. Appl., 49 (2005), 757-763. [6] P. K. Maji et al, Fuzzy soft sets, J. Fuzzy Math., 9(3)(2001), 589-602. [7] S. Alkhazaleh, A. R. Saleh, N. Hassan, Soft Multiset Theory. Applied Mathematical Sciences vol. 5, 2011, no. 72, 3561- 3573. [8] S. Alkhazaleh and A. R. Salleh, Fuzzy soft multisets theory Abstract and Applied Analysis (2012) Article ID 350603. [9] K. P. Girish, and Sunil Jacob John. Multiset topologies induced by multiset relations. Information Sciences 188(0) (2012), 298–313. [10] K. V. Babitha, J. J.Sunil, Soft set Relations and Functions, Computers and Mathematics with Applications, 60 (2010) 1840-1849. [11] K. V. Babitha, Sunil Jacob John, On soft Multisets, Annals of Fuzzy Mathematics and Informatics,Volume 5, No. 1, (2013) 35-4. [12] K. Rejikumar and S. A. Naisal, Interior exterior and boundary of fuzzy soft multi set topology on decision making. IEEE Xplore. Control, Instrumentation, Communication and Computational Technologies.(2017): 147- 152. [13] E. Szmidt, Distances and Similarities in Intuitionistic Fuzzy Sets, Springer (2014). [14] A. Mukherjee and A. K. Das, Parameterized Topological Space Induced by an Intuitionistic Fuzzy Soft Multi Topological Spaces, Ann. Pure and Applied Math., 7, 7-12 (2014).


Signed Product Cordial Labeling of Zero-Divisor Graphs C. Subramanian and T. Tamizh Chelvam (Department of Mathematics, Manonmaniam Sundaranar University, Tirunelveli, Tamil Nadu, India.) E-mail: [email protected], [email protected]

Abstract: Let G = (V, E) be a simple graph with n vertices. A function f : V (G) → {−1, 1} is called as a signed product cordial labeling if, for each edge e = uv, the induced map f ∗ (uv) = f (u)f (v) satisfies the condition |vf (−1)−vf (1)| ≤ 1 and |ef (−1)−ef (1)| ≤ 1 where vf (i) and ef (i) are the number of vertices and edges with label i, i ∈ {−1, 1} respectively. A graph G is said to be signed product cordial graph if it has a signed product cordial labeling. In this paper, we prove that certain classes of zero-divisor graph of commutative rings are signed product cordial.

Key Words: Commutative rings, zero-divisor graph, cordial labeling, signed product cordial labeling.

AMS(2010): 05C78, 05C25. §1. Introduction Let R be a commutative ring with non-zero identity, Z(R) its set of all zero-divisors in R and Z ∗ (R) = Z(R) \ {0}. The zero-divisor graph of R is the simple undirected graph Γ(R) with vertex set Z ∗ (R) and two distinct vertices x and y are adjacent if xy = 0. All graphs considered in this paper are finite, simple and undirected. Labeling of graphs has been studied by several authors. In fact Gallian[2] has given a dynamic survey of labeling. Let G = (V, E) be a simple graph on n vertices. The complement G of G is the graph with vertex set V (G) and two vertices are adjacent in G if and only if they are not adjacent in G. For graph theoretic standard notation we follow F.Harray . A function f : V (G) → {−1, 1} is said to be a signed product cordial labeling if for each edge e = uv, the induced map f ∗ (uv) = f (u)f (v) satisfies the condition |vf (−1) − vf (1)| ≤ 1 and |ef (−1) − ef (1)| ≤ 1 where vf (i) and ef (i)) are the number of vertices and edges with label i, i ∈ {−1, 1} respectively. A graph G is said to be signed product cordial graph if it has a signed product cordial labeling. The join G1 + G2 of G1 and G2 is a graph with vertex set V (G1 ) ∪ V (G2 ) and edge set E[G1 + G2 ] = E(G1 ) ∪ E(G2 ) ∪ {uv : u ∈ V (G1 ) and v ∈ V (G2 )}. In this paper, we prove that the zero-divisor graph Γ(Zn ), where n = 4p and n = pq for two distinct primes p and q and join of some of them are signed product cordial graphs. 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received February 02, 2018, Accepted June 22, 2018, Edited by K. Selvakumar.

165

Signed Product Cordial Labeling of Zero-divisor Graphs

§2. Signed Product Cordial Labeling In this section, we prove that certain classes of zero-divisor graphs are signed product cordial. Theorem 2.1 For any prime number p > 2, the zero-divisor graph Γ(Z2p ) is signed product cordial. Proof Let p > 2 be a prime number. Then the vertex set of Γ(Z2p ) is V (Γ(Z2p )) = {v1 , · · · , vp−1 , vp } = {2, 4, . . . , 2(p − 1), p} and the edge set E(Γ(Z2p )) = {vi vp /1 ≤ i ≤ p − 1}. Define f : V → {−1, 1} by f (vi ) = (−1)i for 1 ≤ i ≤ p − 1 and f (vp ) = −1. Clearly vf (−1) =

p−1 p−1 + 1; vf (1) = . 2 2

Then the induced edge labeling function f ∗ : E(G) → {−1, 1} is given by f ∗ (vi vp ) =

Hence, we have ef (−1) =

p−1 2

 1

if i is odd;

−1, if i is even.

= ef (1) and so Γ(Z2p ) is a signed product cordial graph.

2

Theorem 2.2 For any prime number p > 3, the zero-divisor graph Γ(Z3p ) is signed product cordial. Proof Let p > 3 be a prime number. Then Z ∗ (Z3p ) = {p, 2p, 3, 6, . . . , 3(p − 1)} and E(Γ(Z3p )) = {u1 vi , u2 vi , 1 ≤ i ≤ p − 1}. Note that |V (Γ(Z3p ))| = p + 1. Define f : V (G) → {−1, 1} by f (p) = −1, f (2p) = 1 and f (3i) = (−1)i for 1 ≤ i ≤ p − 1. It is clear that ∗ vf (−1) = p+1 2 = vf (1). Now the induced edge labeling function f : E(G) → {−1, 1} is given by   1 −1 if j is odd; if j is odd; f ∗ (p 3j) = and f ∗ (2p 3j) = −1 if j is even. 1, if j is even. From this we have that ef (−1) = p−1 2 + Γ(Z3p ) is a signed product cordial graph.

p−1 2

= p − 1, ef (1) =

p−1 2

+

p−1 2

= p − 1 and so

2

Theorem 2.3 For any prime number p ≥ 3, the zero-divisor graph Γ(Z4p ) is signed product cordial. Proof One can partition the vertex set of Γ(Z4p ) into V1 = {p, 2p, 3p} = {u1 , u2 , u3 } and V2 = {2, 4, · · · , 2(p − 1), 2(p + 1), · · · , 2(2p − 1)} = {v1 , v2 , · · · , vp−1 , vp+1 , · · · , v2p−1 }. Note that the edge set of Γ(Z4p ) is nothing but E(Γ(Z4p )) = {u1 v2 , u1 v4 , · · · , u1 vp−1 , u1 vp+1 , · · · , u1 v2p−2 , u2 v1 , u2 v2 , u2 v3 , · · · , u2 vp−1 ,

u2 vp+1 , · · · , u2 v2p−1 , u3 v2 , u3 v4 , · · · , u3 vp−1 , u3 vp+1 , · · · , u3 v2p−2 }.

Clearly |V | = 2p + 1, |E| = 2p − 2 + 2(2p−2) = 4p − 4. Define f : V (Γ(Z4p )) → {−1, 1} 2 by f (u1 ) = −1, f (u2 ) = 1, f (u3 ) = −1; f (vj ) = (−1)j ; 1 ≤ j ≤ p − 1 and f (vj ) = −(−1)j ;

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p + 1 ≤ j ≤ 2p − 1.

From this, we have that vf (−1) = p − 1 + 2 = p + 1; vf (1) = p − 1 + 1 = p and the induced edge labeling function f ∗ : E→ {−1, 1} is given by −1 if i is odd and j is even, 1 < j < 2p − 1; f ∗ (ui vj ) = 1 if i is even and all j, 1 ≤ j < 2p − 1, j 6= p. ⇒ ef (−1) =

2p−2 2

+

2p−2 2

= 2p − 2 and ef (1) = 2p − 2.

Hence |ef (−1) − ef (1)| ≤ 1 and so Γ(Z4p ) is a signed product cordial.

2

Theorem 2.4 For two distinct primes p and q with p < q, the zero-divisor graph Γ(Zpq ) is signed product cordial. Proof The vertex set of Γ(Zpq ) can be partitioned into V1 and V2 where, V1 = {p, 2p, 3p, · · · , (q− 1)p} = {u1 , u2 , · · · , uq−1 } and V2 = {q, 2q, 3q, · · · , (p − 1)q} = {v1 , v2 , · · · , vp−1 }.

The edge set of Γ(Zpq ) is given by E(Γ(Zpq )) = {ui vj : ui ∈ V1 and vj ∈ V2 , 1 ≤ i ≤ q − 1, 1 ≤ j ≤ p − 1}.

Consider the vertex labeling f : V (Γ(Zpq )) → {−1, 1} defined by f (ui ) = (−1)i for 1 ≤ i ≤ q − 1 and f (vj ) = (−1)j for 1 ≤ j ≤ p − 1. Further note that |E| = (p − 1)(q − 1) and the induced edge labeling f ∗ : E → {−1, 1} is given by   both i & j are odd 1 ≤ i < q − 1, 1 ≤ j ≤ p − 1; 1  ∗ f (ui vj ) = and both i & j are even;    −1 otherwise.

Clearly ef (−1) = (p−1)(q−1) and ef (1) = 2 Γ(Zpq )) is a signed product cordial.

(p−1)(q−1) . 2

Hence |ef (−1) − ef (1)| ≤ 1 and so

2

Theorem 2.5 For any prime number p > 2, the join graph Γ(Z2p ) + Γ(Z4 ) is signed product cordial. Proof Let G = Γ(Z2p ) + Γ(Z4 ). The vertex and edge sets of the graph G, are respectively V (G) = {u1 , u2 , . . . up−1 , up , x} = {2, 4, . . . , 2(p − 1), p, x}, where x = 2 ∈ Z4 and E = {ui up , ui x, up x/1 ≤ i ≤ p − 1}. Note that |V | = p + 1 and |E| = p − 1 + p − 1 + 1 = 2p − 1. Consider the vertex labeling f : V (G) → {−1, 1} by f (uk ) = (−1)k for 1 ≤ k ≤ p and f (x) = 1. Then the induced edge labeling f ∗ : E(G) → {−1, 1} is given by   1 1 i is odd; i is even; f ∗ (ui up ) = , f ∗ (ui x) = −1 i is even −1 i is odd and f ∗ (up x) = 1.

From the above

ef (−1) =

(p − 1) (p − 1) (p − 1) (p − 1) + = p − 1, ef (1) = + +1=p 2 2 2 2

and |ef (−1) − ef (1)| ≤ 1. Therefore G is a signed product cordial.

2

167


Theorem 2.6 For any prime number p > 2, the join graph Γ(Z2p ) + Γ(Z9 ) is signed product cordial. Proof Let G = Γ(Z2p ) + Γ(Z9 ). The vertex and edge sets of the graph G are V (G) = {u1 , . . . up−1 , up , x, y} = {2, 4, . . . , 2(p − 1), p, x, y}, where x = 3 and y = 6 ∈ Z9 and E(G) = {ui up , ui x, ui y, up x, up y, xy/1 ≤ i ≤ p − 1}. Note that |V | = p + 2 and |E| = p − 1 + p − 1 + p − 1 + 3 = 3p. Define the vertex labeling f : V (G) → {−1, 1} by f (uk ) = (−1)k for 1 ≤ k ≤ p, p−1 f (x) = −1 and f (y) = 1. Clearly vf (−1) = p−1 2 + 1 and vf (1) = 2 + 1. Then the induced edge labeling f ∗ : E(G) → {−1, 1} is given by   1 −1 i is even; i is odd; f ∗ (ui up ) = f ∗ (ui x) = −1 i is even, 1 i is odd,  1 i is even; f ∗ (ui y) = f ∗ (up x) = 1, , f ∗ (up y) = −1 −1 i is odd, + 2 = p, ef (1) = and f ∗ (xy) = −1. From the above ef (−1) = 3(p−1) 2 ef (1)| ≤ 1. Hence G is a signed product cordial.

3(p−1) 2

+ 1 and |ef (−1) −

2

Theorem 2.7 For any prime number p > 2, the join graph Γ(Z2p ) + Γ(Z6 ) is signed product cordial. Proof Let G = Γ(Z2p ) + Γ(Z6 ). Then V (G) = {u1 , . . . up−1 , up , x, y, z} = {2, 4, . . . , 2(p − 1), p, x, y, z}, where x = 2, y = 3 and z = 3 ∈ Z6 and E(G) = {ui up , ui x, ui y, ui z, up x, up y, up z, xy, yz/1 ≤ i ≤ p − 1}. Note that |V | = p + 3 and |E| = p − 1 + p − 1 + p − 1 + p − 1 + 5 = 4p + 1. Define the vertex labeling f : V (G) → {−1, 1} by f (uk ) = (−1)k for 1 ≤ k ≤ p, f (x) = p−1 −1, f (y) = −1 and f (z) = 1. Note that vf (−1) = p−1 2 + 2; vf (1) = 2 + 1. Then the induced edge labeling f ∗ : E(G)  → {−1, 1} is given by 1 i is odd; f ∗ (ui up ) = −1 i is even,  −1 i is even; f ∗ (ui x) = f ∗ (ui y) = 1 i is odd,  1 i is even; f ∗ (ui z) = −1 i is odd,

and f ∗ (up x) = 1, f ∗ (up y) = 1, f ∗ (up z) = −1, f ∗ (xy) = 1, f ∗ (yz) = −1 +2, ef (1) = 4(p−1) +3 and |ef (−1)−ef (1)| ≤ 1. Therefore From the above ef (−1) = 4(p−1) 2 2 G is signed product cordial. 2 Corollary 2.8 For any prime number p > 2, the join graph Γ(Zp2 ) + Γ(Z4 ) is signed product cordial. Proof Since the graph Γ(Zp2 ) + Γ(Z4 ) ∼ = Γ(Z2p ), by Theorem ??, Γ(Zp2 ) + Γ(Z4 ) is signed

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2

product cordial.

Theorem 2.9 For any prime number p > 2, the join graph Γ(Zp2 ) + Γ(Z9 ) is signed product cordial. Proof Let G = Γ(Zp2 )+Γ(Z9 ). Then V (G) = {u1 , . . . up−1 , x, y} = {p, 2p, . . . , (p−1)p, x, y}, where x = 3 and y = 6 ∈ Z9 and E(G) = {ui x, ui y, xy/1 ≤ i ≤ p − 1}. Note that |V | = p + 1 and |E| = p − 1 + p − 1 + 1 = 2p − 1.

Define the vertex labeling f : V (G) → {−1, 1} by f (uk ) = (−1)k for 1 ≤ k ≤ p − 1, p−1 f (x) = −1 and f (y) = 1. Clearly vf (−1) = p−1 2 + 1 and vf (1) = 2 + 1. Then the induced edge labeling f ∗ : E(G) → {−1, 1} is given by   −1 i is even; 1 i is even; f ∗ (ui x) = f ∗ (ui y) = and f ∗ (xy) = −1. 1 −1 i is odd i is odd, From the above ef (−1) = (p−1) + (p−1) + 1 = p, ef (1) = 2 2 |ef (−1) − ef (1)| ≤ 1. Hence G is signed product cordial.

(p−1) 2

+ (p−1) = p − 1 and satisfies 2

2

Theorem 2.10 For any prime number p > 2, the join graph Γ(Zp2 ) + Γ(Z6 ) is signed product cordial. Proof Let G = Γ(Zp2 ) + Γ(Z6 ). Then V (G) = {u1 , . . . up−1 , x, y, z} = {p, 2p, . . . , (p − 1)p, x, y, z}, where x = 2, y = 3 and z = 3 ∈ Z6 and E(G) = {ui x, ui y, ui z, xy, yz/1 ≤ i ≤ p − 1}. Note that |V | = p + 2 and |E| = p − 1 + p − 1 + p − 1 + 2 = 3p − 1.

Define the vertex labeling f : V (G) → {−1, 1} by f (uk ) = (−1)k for 1 ≤ k ≤ p − 1, p−1 f (x) = −1, f (y) = −1 and f (z) = 1. Clearly vf (−1) = p−1 2 + 2; vf (1) = 2 + 1. Then the ∗ induced edge labeling f : E(G) → {−1, 1} is given  by −1 i is even; f ∗ (ui x) = f ∗ (ui y) = 1 i is odd,  1 i is even; f ∗ (ui z) = −1 i is odd,

f ∗ (xy) = 1 and f ∗ (yz) = −1 From the above,

ef (−1) = ef (1) =

(p − 1) (p − 1) (p − 1) 3(p − 1) + + +1= + 1, 2 2 2 2 (p − 1) (p − 1) (p − 1) 3(p − 1) + + +1= + 1. 2 2 2 2

and |ef (−1) − ef (1)| ≤ 1. Hence G is signed product cordial.

2

Corollary 2.11 For any prime number p > 2, the join graph Γ(Zp2 ) + Γ(Z4 ) is signed product cordial.


169

Proof Since the graph Γ(Z4 ) ∼ = Γ(Z4 ), by Corollary 2.8. Γ(Zp2 ) + Γ(Z4 ) is a signed product cordial. 2 References [1] I. Cahit, Cordial graphs: A weaker version of graceful and harmonious graphs, ARS Combinatoria, 23 (1987), 201–207. [2] J.A. Gallian, A dynamic survey of graph labeling, Electronic Journal of Combinatorics, 17(2010), # DS6, 1–246. [3] F. Harary, Graph Theory, Addison-Wesley Publishing Company Inc, USA, 1969. [4] Jayapal Baskar Babujee, Shobana Loganathan, On signed product cordial labeling, Applied Mathematics, 2(2011), 1525–1530. [5] M. Sundaram, R. Ponraj and S. Somasundram, Total product cordial labeling of graphs, Bulletin of Pure and Applied Sciences: Section E. Mathematics and Statistics, 25(2006), 199–203.


On the Domination Number of a Graph and its Line Graph E. Murugan and J. Paulraj Joseph Department of Mathematics Manonmaniam Sundaranar University, Abishekapatti, Tirunelveli - 627 012, Tamil Nadu, India E-mail: [email protected], [email protected]

Abstract: Let G be a connected graph of order n and size m and L(G) be its line graph. Let γ(G) denote the domination number of G. In this paper, we obtain lower and upper bounds for the sum γ(G) + γ(L(G)) and characterize the extremal graphs.

Key Words: Domination number, domination number of line graph. AMS(2010): 05C69, 05C70. §1. Introduction Let G = (V, E) be a simple graph of order n and G be its complement. If α(G) is a graph parameter, then the lower and upper bounds on the sum α(G) + α(G) in terms of n are of prime importance in graph theory. The first of its kind with reference to chromatic number χ(G) of G was studied by Nordhaus and Gaddum on complementary graphs and published in American Mathematical Monthly in 1956. They proved lower and upper bounds on the sum and on the product of χ(G) and χ(G) in terms of the order n of G. Since then, any bound on the sum and / or the product of an invariant in a graph G and the same invariant in the complement G of G is called a Nordhaus-Gaddum type inequality or relation. The original relations presented by Nordhaus and Gaddum [12] in 1956 are as follows. Theorem 1.1 ([12]) If G is a graph of order n, then √ 2 n ≤ χ(G) + χ(G) ≤ n + 1 and

(n + 1)2 . 4 Furthermore, these bounds are best possible for infinitely many values of n. n ≤ χ(G).χ(G) ≤

The theory of dominating sets was formally introduced by C. Berge [3] at the late fifties and by Ore [13] in the early sixties. A subset S of V is called a dominating set of G if every vertex not in S is adjacent to some vertex in S. The domination number γ(G) (or γ for short) 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received February 13, 2018, Accepted June 28, 2018, Edited by K. Selvakumar.

On the Domination Number of a Graph and its Line Graph

171

of G is the minimum cardinality taken over all dominating sets of G. The relation of NordhausGaddum type for domination in graphs were proved by Jaeger and Payan [10] in 1972 are as follows. Theorem 1.2 ([10]) For any graph G with at least two vertices, 3 ≤ γ(G) + γ(G) ≤ n + 1 and 2 ≤ γ(G).γ(G) ≤ n. This has been extended to other graph theoretic parameters. A survey of these results is found in [1]. There are several derived graphs in the literature such as line graphs, total graphs, subdivision graphs, power graphs. It is noted that complement of the graph is one of the derived graphs of G. Rosalin et.al [16] and Shunmugapriya et.al [17] extended Nordhaus-Gaddum type results to some of these derived graphs for their concepts induced cycle path number of graphs and total resolving number of graphs respectively. The line graph L(G) of a graph G is a graph whose vertex set is E(G) and two vertices of L(G) are adjacent if and only if the corresponding edges are adjacent in G. The concept of ′ edge domination was introduced by Mitchell and Hedetniemi [11]. A subset S of E is called ′ ′ an edge dominating set of G if every edge not in S is adjacent to some edge in S . The edge ′ ′ domination number γ (G) (or γ for short) of G is the minimum cardinality taken over all edge dominating sets of G. The domination number of a line graph L(G) of a graph G is the same ′ as an edge domination number of a graph G, that is γ (G) = γ(L(G)). In this paper, we extend the Nordhaus-Gaddum type result to line graph for the parameter domination number. We need the following definitions for our results. The graph S(G) obtained from G by subdividing each edge of G exactly once is called the subdivision of G. The degree of an edge e = uv of G is defined by deg e = deg u + deg v − 2. Corona of two graphs G1 and G2 , denoted by G1 ◦ G2 is the graph obtained by taking one copy of G1 and |V (G1 )| copies of G2 in which ith vertex of G1 is joined to every vertex in the ith copy of G2 . The diameter of G is the maximum distance between two vertices of G and is denoted by diam(G). A tree of diameter three is called a bistar. We denote by C3,k the graph obtained from a C3 and k(≥ 0) copies of K2 by joining one end of each K2 with a fixed vertex of C3 . We denote by C4,k the graph obtained from C4 by joining a vertex of C4 with the center of S(K1,k ). The edge independence number β1 (G) of a graph G is the maximum cardinality of an independent set of edges. The following theorems will be used in the sequel. Theorem 1.3 ([9]) If G is a graph of order n, then γ(G) = 1 iff ∆(G) = n − 1. Theorem 1.4 ([7,14]) For a graph G with even order n and no isolated vertices, γ(G) = n/2 if and only if the components of G are the cycle C4 or the corona H ◦ K1 for any connected graph H.

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E. Murugan and J. Paulraj Joseph

Graphs in family A

B1

B4

B3

B2

B5

Graphs in family B

Figure 1 In [6,15] E. J. Cockayne, T. W. Haynes and S. T. Hedetniemi characterized connected graphs for which γ(G) = ⌊n/2⌋. For this characterization, they defined six classes of graphs by using the following families of graphs. Let G1 = {C4 } ∪ {G : G = H ◦ K1 , where H is connected} and G2 = A ∪ B − {C4 } For any graph H, let S(H) denote the set of connected graphs, each of which can be formed from H ◦ K1 by adding a new vertex x and edges joining x to one or more vertices of H. Then define S G3 = S(H) , H

where the union is taken over all graphs H. Let y be a vertex of a copy of C4 and, for G ∈ G3 , let θ(G) be the graph obtained by joining G to C4 with the single edge xy, where x is the new vertex added in forming G. Then define G4 = {θ(G) : G ∈ G3 } . Next, let u, v, w be a vertex sequence of a path P3 . For any graph H, let P(H) be the set of connected graphs which may be formed from H ◦ K1 by joining each of u and w to one or more vertices of H. Then define S G5 = P(H) . H

Let H be a graph and X ∈ B. Let R(H, X) be the set of connected graphs which may be formed from H ◦ K1 by joining each vertex of U ⊆ V (X) to one or more vertices of H such that no set with fewer than γ(X) vertices of X dominates V (X) − U . Then define G6 =

[

H,X

R(H, X).

Theorem 1.5 ([6,15]) A connected graph G satisfies γ(G) =

n 2

if and only if G ∈ G =

6 S

i=1

Gi .

.

173


Theorem 1.6 ([13]) If a graph G has no isolated vertices, then γ(G) ≤ n/2. ′

Theorem 1.7 ([5]) For any (n, m) graph G, γ ≤ ⌊n/2⌋. ′

Theorem 1.8 ([2]) For any connected graph G of even order n, γ = n/2 if and only if G is isomorphic to Kn or Kn/2,n/2 . ′

Theorem 1.9 ([2]) For any tree T of order n 6= 2, γ ≤ (n − 1)/2; equality holds if and only if T is isomorphic to the subdivision of a star. ′

Theorem 1.10 ([2]) Let G be a connected unicyclic graph. Then γ = ⌊n/2⌋ if and only if G is isomorphic to either C4 , C5 , C7 , C3,k , C4,k for some k ≥ 0.

§2. Main Results ′

′

Theorem 2.1 For any connected graph G of size m, γ (G) = 1 if and only if ∆ (G) = m − 1 ′ where ∆ (G) is a maximum edge degree of G. ′

Proof It is easy to see that γ(G) = 1 if and only if ∆ = n − 1. Since γ (G) = γ(L(G)). ′ ′ Hence by Theorem 1.3, γ (G) = 1 if and only if ∆ (G) = m − 1. 2 ′

Corollary 2.2 For any tree T of order n, γ (T ) = 1 if and only if T is isomorphic to either star or bistar.

2

Proof It follows from Theorem 2.1. ′

Theorem 2.3 Let G be a connected graph of order n ≥ 5 in which ∆ (G) 6= m − 1. Then ′ γ (G) = 2 if and only if there exists two edges e1 = uv and e2 = wx such that G \ {u, v, w, x} is empty. ′

Proof Assume γ (G) = 2. Suppose there does not exist two edges e1 = uv and e2 = wx such that G \ {u, v, w, x} is empty. Then there is an edge e of G, which is not dominated by any two edges of G which is a contradiction to our assumption. Conversely, assume that there exists two edges e1 = uv and e2 = wx such that G \ {u, v, w, x} is empty. By hypothesis, every edge of G is adjacent to either e1 or e2 and hence {e1 , e2 } is an ′ ′ ′ ′ edge dominating set. Hence γ (G) ≤ 2. Since ∆ (G) 6= m − 1, γ (G) 6= 1. Thus γ (G) = 2. 2 ′

′

′

Lemma 2.4 If H is a subgraph of G, then γ (G) ≤ γ (H) + γ (G \ E(H)). Proof Let H be a subgraph of G. Then G \ E(H) is also a subgraph of G which is edge ′ ′ ′ disjoint from H and let denote it by G . Let S1 , S2 be a γ -sets of H and G respectively. Then S1 ∪ S2 is an edge dominating set of G.

174


Therefore, ′

γ (G)

≤ |S1 ∪ S2 | = |S1 | + |S2 | − |S1 ∩ S2 | ′

′

′

≤ |S1 | + |S2 | = γ (H) + γ (G ) ′

′

= γ (H) + γ (G \ E(H)).

2

This completes the proof. ′

Observation 2.5 γ (Ck ◦ K1 ) =

 k,

if k is even, 2  k+1 , if k is odd. 2

′ Lemma 2.6 Let G ∼ = H ◦ K1 for any connected graph H. Then γ (G) = H is either C3 or a star.

n 2

− 1 if and only if

′

Proof Assume that γ (G) = n2 − 1. We claim that H contains no P4 . Suppose H contains ′ ′ ′ a P4 : v1 v2 v3 v4 (See Figure 2). Then H ◦ K1 contains the pendent edges v1 v1 , v2 v2 , v3 v3 and ′ ′ ′ v4 v4 and P4 ◦ K1 is a subgraph of G. Then γ (P4 ◦ K1 ) = 2 and G = G \ V (P4 ◦ K1 ) is a graph on n − 8 vertices. v1′

v2′

v3′

v4′

r

r

r

r

r

r

r

r

v1

v2

v3

v4

Figure 2 Therefore, by Lemma 2.4 and Theorem 1.7, ′

γ (G) ≤ =

′

′

′

γ (P4 ◦ K1 ) + γ (G ) ≤ 2 + ⌊ n n − 2 < − 1, 2 2

n−8 ⌋ 2

which is a contradiction. Hence the claim and H is either C3 or a star. Conversely, if H ∼ = C3 , then G ∼ = C3 ◦K1 . Then by Observation 2.5, γ(L(G)) = 2 = 62 −1 = n ∼ n 2 − 1. If G = K1, 2 −1 , then the set of all edges incident to the centre of H is a minimum edge dominating set of cardinality n2 − 1 and hence γ(L(G)) = n2 − 1. 2 Theorem 2.7 If G is a connected graph of order n, then 2 ≤ γ(G) + γ(L(G)) ≤ n. Moreover, ′

(i) the lower bound is attained if and only if ∆(G) = n − 1 and ∆ (G) = m − 1; (ii) the upper bound is attained if and only if G ∼ = C4 .

175


Proof Since 1 ≤ γ(G) ≤ n/2 and 1 ≤ γ(L(G)) ≤ n/2, 2 ≤ γ(G) + γ(L(G)) ≤ n. (i) Assume that γ(G) + γ(L(G)) = 2. Then γ(G) = 1 and γ(L(G)) = 1. Then by Theorem ′ 1.3 and Theorem 2.1, ∆(G) = n − 1 and ∆ (G) = m − 1. Converse is obvious by Theorem 1.3 and Theorem 2.1. (ii) Assume that γ(G) + γ(L(G)) = n. By Theorem 1.6 and Theorem 1.7, γ(G) = n/2 and γ(L(G)) = n/2 and it follows from Theorem 1.4 and Theorem 1.8, G ∼ = C4 . Converse is 2 obvious by verification. Theorem 2.8 Let G be a connected graph of even order n ≥ 4, then γ(G) + γ(L(G)) = n − 1 if and only if G is isomorphic to either K4 or K3,3 or C3 ◦ K1 or K1, n2 −1 ◦ K1 . Proof Assume γ(G) + γ(L(G)) = n − 1. We consider two cases. Case 1. γ(G) =

n 2

− 1 and γ(L(G)) = n/2.

By Theorem 1.8, γ(L(G)) = n/2 if and only if G is isomorphic to Kn or Kn/2,n/2 . Since γ(Kn ) = 1 = n2 − 1, we get n = 4 and hence G ∼ = K4 . Since γ(L(Kn/2,n/2 )) = 2 = n2 − 1, we get n = 6 and hence G ∼ = K3,3 . Case 2. γ(G) = n/2 and γ(L(G)) =

n 2

− 1.

By Theorem 1.4, γ(G) = n/2 iff G is isomorphic to C4 or H ◦ K1 where H is a connected graph. If G ∼ = C4 , then γ(L(G)) = 2 6= n2 − 1. If G ∼ = H ◦ K1 , then by Lemma 2.6, H ∼ = C3 or n K1, 2 −1 . Conversely, if G ∼ = K4 , then by Theorem 1.8, γ(L(G)) = 2 and Theorem 1.3, γ(G) = 1 which gives γ(G) + γ(L(G)) = 1 + 2 = 3 = 4 − 1 = n − 1. If G ∼ = K3,3 , then by Theorem 1.8, γ(L(G)) = 3. Also, γ(G) = 2 which gives γ(G) + γ(L(G)) = 2 + 3 = 5 = 6 − 1 = n − 1. If G ∼ = C3 ◦ K1 , then by Observation 2.5, γ(L(G)) = 2 and by Theorem 1.4, γ(G) = 3 which gives γ(G) + γ(L(G)) = 3 + 2 = 5 = 6 − 1 = n − 1. If G ∼ = K1, n2 −1 ◦ K1 , then by Theorem 1.4, n n γ(G) = 2 . Also, γ(L(G)) = 2 − 1 which gives γ(G) + γ(L(G)) = n − 1. 2 Theorem 2.9 Let G be a connected graph of an odd order n ≥ 5, then γ(G) + γ(L(G)) = n − 1 if and only if G ∈ G2 or isomorphic to S(K1, n−1 ), C3,k , C4,k or one of the graphs G1 , G2 , G3 2 given in Figure 3. u1

...

...

u1

...

u1

G2

G1

G3

Figure 3 Proof Given γ(G) + γ(L(G)) = n − 1. Since n is odd, γ(G) = γ(L(G)) must be n−1 2 . Since 6 S ′ γ(G) = n−1 Gi . For all G ∈ G2 , it is easy to verify that γ (G) = n−1 2 , by Theorem 1.5, G ∈ 2 . We consider the following cases.

i=2

176


Case 1. G ∈ G3 . Since G is connected, x is adjacent to at least one vertex of H. We claim that H is empty. Suppose not. Let e = uv be an edge in H. Then e dominates two pendent edges which are incident with u and v(See Figure 4). x H

u e

v

...

Figure 4 ′

′

′

′

If E is the set of edges adjacent to e, then |V (E )| = 5 and e dominates E . Let G = ′ ′ G \ V (E ). Then there is a one-one correspondence between the pendant edges of G and the ′ edges in any minimum edge dominating set of G . Therefore by Lemma 2.4 and Theorem 1.7, ′

′

′

′

′

γ (G) ≤ γ (E ) + γ (G ) ≤ 1 + ⌊

n−5 n−1 ⌋< , 2 2

which is a contradiction. Thus H is empty and hence G ∼ = S(K1, n−1 ). 2

Case 2. G ∈ G4 . Since G is connected, x is adjacent to at least one vertex of H. We claim that H is independent. Suppose not. Let e = uv be an edge in H. Then e dominates two pendent edges ′ which are incident with u and v. If E = E(C4 ) ∪ {xy} ∪ {edges adjacent to e = uv}, then ′ ′ ′ ′ ′ |V (E )| = 9. Let G = G \ V (E ). Clearly γ (E ) = 3. Then there is a one-one correspondence ′ ′ between the pendant edges of G and the edges in any minimum edge dominating set of G , we get (using Lemma 2.4 and Theorem 1.7) ′

′

′

′

′

γ (G) ≤ γ (E ) + γ (G ) ≤ 3 + ⌊

n−1 n−9 ⌋< , 2 2

which is a contradiction. Thus H is empty and hence G ∼ = C4,k . Case 3. G ∈ G5 . ′

′ ′

′

′

′

We claim that H is empty. Let E = huvwsts t i. Then |V (E )| = 7 and γ (E ) = 2. Let ′ ′ G = G \ V (E ). Then there is a one-one correspondence between the pendant edges of G and ′ the edges in any minimum edge dominating set of G , we get (using Lemma 2.4 and Theorem 1.7) ′ ′ ′ ′ ′ n−7 n−1 γ (G) ≤ γ (E ) + γ (G ) ≤ 2 + ⌊ ⌋< , 2 2 which is a contradiction. Thus H is empty. ′

Claim A G ∼ = S(K1, n−1 ). That is to prove that either u or w but not both is adjacent to 2

′

all the vertices of H. Suppose not. Then there exists two internally disjoint P4 ’s say P = s suv ′ ′ ′′ ′′ ′′ and Q = vwrr . Then {su, wr} is a γ -set of P ∪ Q. Consider G = G \ V (E ) where E = ′ ′ ′′ ′ ′′ hs suvwrr i. Clearly |V (E )| = 7 and γ (E ) = 2. Then there is a one-one correspondence

177


′′

′′

between the pendant edges of G and the edges in any minimum edge dominating set of G , we get (using Lemma 2.4 and Theorem 1.7) ′

′

′′

′

′′

γ (G) ≤ γ (E ) + γ (G ) ≤ 2 + ⌊

n−7 n−1 ⌋< , 2 2

which is a contradiction. Hence w is a pendent vertex. Since G is connected, u must be adjacent to all the vertices of H. Hence G ∼ = S(K1, n−1 ). 2

Case 4. G ∈ G6 . We claim that H is empty. Suppose not. Let e = uv be an edge in H. Then e dominates two pendent edges which are incident with u and v. Since G is connected, u is adjacent to ′ atleast one vertex of X say u1 . Let E denote the subgraph induced by the edges adjacent to ′ ′ ′ ′ e. Then |V (E )| ≥ 5 and e dominates the edges of E . Let G = G \ V (E ). Then there is a ′ one-one correspondence between the pendant edges of G and the edges in any minimum edge ′ dominating set of G , we get (using Lemma 2.4 and Theorem 1.7) ′

′

′

′

′

γ (G) ≤ γ (E ) + γ (G ) ≤ 1 + ⌊

n−5 n−1 ⌋< , 2 2

which is a contradiction. Hence H is empty. Next we claim that |V (U )| = 1. Suppose |V (U )| ≥ 2, then there exist two vertices u1 , u2 ∈ V (X) such that either u1 , u2 are adjacent or u1 , u2 are not adjacent. If |V (X)| = 3, then there exist two vertices u1 , u2 ∈ V (X) such that ′ u1 , u2 are adjacent. This gives γ (G) < n−1 2 which is a contradiction. If |V (X)| = 5, then there exist two vertices u1 , u2 ∈ V (X) such that either u1 ,u2 are adjacent or u1 , u2 are not adjacent. Subcase 4.1 u1 , u2 are adjacent. Then X \ {u1 , u2 } is P3 by definition of G6 , this is not possible. On the other hand, ′ X \ {u1 , u2 } is two components K1 and K2 , this gives γ (G) < n−1 2 which is a contradiction. Subcase 4.2 u1 , u2 are not adjacent. Then X \ {u1 , u2 } is P3 by definition of G6 , this is not possible. On the other hand, ′ X \ {u1 , u2 } is two components K1 and K2 , this gives γ (G) < n−1 which is a contradiction. 2 Hence |V (U )| must be 1. Let denote it by a vertex u1 . ′ Let B = {B1 , B2 , B3 , B4 , B5 }. If X = B1 , then γ (G) = γ(G) = n−1 . Hence G ∼ = C3,k . If 2

′

X = B2 , then γ (G) < n−1 2 which is a contradiction. Now we assume that X = B3 , B4 , B5 . We ∼ claim that X \ {u1 } = C4 . Suppose U 6= {u1 }. Then there is a single edge e which is adjacent ′ ′ ′ ′ to all the edges of X \ {u1 } in X and let it be E . Clearly |V (E )| = 4. Let G = G \ V (E ). ′ Then there is a one-one correspondence between the pendant edges of G and the edges in any ′ minimum edge dominating set of G , we get (using Lemma 2.4 and Theorem 1.7) ′

′

′

′

′

γ (G) ≤ γ (E ) + γ (G ) ≤ 1 + ⌊

n−4 n−1 ⌋< , 2 2

which is a contradiction and since G is connected, u1 is adjacent to all the vertices of H. Hence G∼ = G1 , G2 , G3 .

178


′ ∼ Conversely, if G ∈ G2 , then it can be easily verify that γ(G) = γ (G) = n−1 2 . If G = ′ n−1 n−1 S(K1, n−1 ), then by Theorem 1.9, γ (G) = 2 and also we can check γ(G) = 2 . If G ∼ = C3,k 2 ′ n−1 n−1 ∼ G1 or or C4,k , then by Theorem 1.10, γ (G) = and also we can check γ(G) = . If G =

2

′

G2 or G3 , then it can be easily verify that γ(G) = γ (G) =

2

n−1 2 .

′ Lemma 2.10 Let G ∼ = H ◦K1 for any connected graph H. If γ (G) =

n 2 −2,

2

then diam(H) ≤ 4.

Proof Suppose diam(H) ≥ 5. Let P6 be a path of length 5 in H. Then P6 ◦ K1 is a ′ ′ subgraph of G and γ (P6 ◦ K1 ) = 3 and G = G \ V (P6 ◦ K1 ) is a graph on n − 12 vertices. Therefore, by Lemma 2.4 and Theorem 1.7, ′

′

′

′

≤ γ (P6 ◦ K1 ) + γ (G ) ≤ 3 + ⌊

γ (G)

=

n n − 3 < − 2, 2 2

n − 12 ⌋ 2

2

which is a contradiction. Hence diam(H) ≤ 4. ′ Lemma 2.11 Let G ∼ = H ◦ K1 for any connected graph H. Then γ (G) = β1 (H) = 2.

n−4 2

if and only if

′

Proof It is clear that n = 2k for some k. Assume that γ (G) = n−4 2 . We claim that ′ β1 (H) = 2. If β1 (H) = 1, then H is either C3 or star. Then by Lemma 2.6, γ (G) = n2 − 1 which is a contradiction. If β(H) ≥ 3, then H contains a P6 . By Proof of Lemma 2.10, ′ γ (G) ≤ n−6 2 which is a contradiction. Hence β1 (H) = 2. ′

Conversely, assume that β1 (H) = 2. Let D be a γ -set of G. Clearly G has n2 pendant edges which are independent edges and we may take that as e1 , e2 , . . . , ek . Since β1 (H) = 2, then there are two independent edges of H say ei , ej each of them are adjacent to two distinct pendant edges in G. Without loss of generality we may take ei is adjacent to e1 and e2 and ej is adjacent to e3 and e4 . Then D = {e1 , · · · , ek } \ {e1, e2 , e3 , e4 } ∪ {ei , ej } = {ei , ej , e5 , e6 , . . . , ek } ′ is an edge dominating set of G and hence γ (G) ≤ k − 4 + 2 = k − 2. Now let S be any edge dominating set of G. Let S1 ⊆ {e1 , e2 , · · · , ek } and S2 ⊆ E(H) such that S1 ∪ S2 = S. Assume that S2 contains l edges. Since β1 (H) = 2, S2 covers at most l + 2 vertices of H. Since these l + 2 vertices are incident with l + 2 edges of {e1 , e2 , . . . , ek }, S1 must contain at least k − (l + 2) edges. Therefore, |S| = |S1 ′

′

[

S2 | = |S1 | + |S2 | ≥ k − (l + 2) + l = k − 2.

So γ (G) ≥ k − 2. Hence γ (G) = k − 2 =

2

n−4 2 . ′

Theorem 2.12 Let G be a connected graph of even order n ≥ 6 with γ(G) 6= γ (G). Then γ(G) + γ(L(G)) = n − 2 if and only if G is isomorphic to K6 or K4,4 or G = H ◦ K1 in which β1 (H) = 2. Proof We consider the following cases. Case 1. γ(G) =

n−4 2

′

and γ (G) =

n 2.

179


By Theorem 1.8, G is isomorphic to Kn or Kn/2,n/2 . Since γ(Kn ) = 1 = n−4 2 , we get n = 6 ∼ and hence G ∼ , we get n = 8 and hence G K4,4 . = K6 . Since γ(Kn/2,n/2 ) = 2 = n−4 = 2 ′

Case 2. γ(G) = n/2 and γ (G) =

n−4 2 . ′

By Theorem 1.4 and hypothesis, G is isomorphic to H ◦ K1 . Since γ (G) = Lemma 2.11, β1 (H) = 2.

n−4 2 ,

then by

′

Conversely, if G = K6 , then by Theorem 1.3, γ(K6 ) = 1 and by Theorem 1.8, γ (K6 ) = 3. Hence γ(G) + γ(L(G)) = 4 = 6 − 2 = n − 2. If G = K4,4 , then γ(K4,4 ) = 2 and by Theorem 1.8, ′ γ (K4,4 ) = 2. Hence γ(G) + γ(L(G)) = 4 = 6 − 2 = n − 2. If G = H ◦ K1 in which β1 (H) = 2, then by Lemma 2.11, γ(L(G)) = n−4 2 2 and by Theorem 1.4, γ(G) + γ(L(G)) = n − 2. Theorem 2.13 Let G be a tree of odd order n ≥ 5. Then γ(G) + γ(L(G)) = n − 2 if and only if G is isomorphic to one of the following graphs F1 , F2 , . . . , F9 given in Figure 5.

...

F2

F1

...

...

...

...

...

F3

F4

...

F5

...

...

F6

F7

...

...

...

...

...

F8

F9

Figure 5 Proof We consider two cases. Case 1. γ(G) =

n−3 2

′

and γ (G) =

n−1 2 .

By Theorem 1.9, G ∼ = S(K1, n−1 ). But γ(S(K1, n−1 )) = 2

Case 2. γ(G) =

n−1 2

′

and γ (G) =

By Theorem 1.5, G ∈

6 S

i=2

Subcase 2.1 G ∈ G3 .

2

n−1 2

6=

n−3 2 .

n−3 2 .

Gi . Since G2 , G4 , G6 contains cycles, G must be in G3 or G5 .

Assume that H is connected. We claim that diam(H) ≤ 2. Suppose diam(H) ≥ 3. Let ′ ′ ′ ′ P4 : v1 v2 v3 v4 be a shortest path of length 3 in H and v1 v1 , v2 v2 , v3 v3 , v4 v4 are pendant edges

180


in G such that x is adjacent to exactly one vertex of H and P4 ◦ K1 is a subgraph of G and ′ ′ γ (P4 ◦ K1 ) = 2. Then G = G \ V (P4 ◦ K1 ) is a graph on n − 8 vertices. Therefore by Lemma 2.4 and Theorem 1.7, ′

γ (G) ≤ =

′

′

′

γ (P4 ◦ K1 ) + γ (G ) ≤ 2 + ⌊ n−4 n−3 < , 2 2

n−8 ⌋ 2

which is a contradiction. Hence diam(H) ≤ 2. Hence G ∼ = F1 or F2 .

Assume that H is disconnected. Let H1 , H2 , · · · , Hs be the components of H. Clearly x is adjacent to exactly one vertex of each component Hi of H and diam(Hi ) ≤ 2. Claim A At most one component of H is non-trivial.

Suppose there are two non trivial components of H.D Let e1 = x1 y1 and e2 E= x2 y2 be ′ ′ ′ ′ ′ ′ two edges from the two non trivial components and H = x1 y1 , x2 y2 , x1 , y1 , x2 , y2 and G = ′

′

G \ V (H ). Then there is is a one-one correspondence between the pendant edges of G and the ′ edges in any minimum edge dominating set of G , we get (using Lemma 2.4 and Theorem 1.7) ′

′

′

′

′

γ (G) ≤ γ (H ) + γ (G ) ≤ 2 + ⌊

n−8 n−3 ⌋< , 2 2

which is a contradiction. Hence we get Claim A. If no component has diameter 1 or 2, then G ∼ = S(K1, n−1 ) whose edge domination number is n−1 . Therefore there is exactly one component 2 2 Hi has diam(Hi ) ≤ 2 and Hj = K1 (j 6= i). Hence G ∼ = F3 or F4 . Subcase 2.2 G ∈ G5 . Assume that H is connected. As in Subcase 2.1, diam(H) ≤ 2. It follows from the definition of G5 , G ∼ = F5 or F6 . Assume that H is disconnected. Let H1 , H2 , ..., Hs be the components of H. Subcase 2.2.1 Either u or w(say w) is a pendant vertex. Then u is adjacent to exactly one vertex of each component of H. As in Subcase 2.1, exactly one component Hi has diam(Hi ) ≤ 2 and Hj = K1 (j 6= i). Hence G ∼ = F3 or F4 . Subcase 2.2.2 Both u and w are not pendant vertices. Clearly u and w are adjacent to different components of H. As in Subcase 2.1, at most one component Hi has diam(Hi ) ≤ 2 and Hj = K1 , (j 6= i). If no component has diameter 1 or 2, then G ∼ = F7 whose edge domination number is n−3 2 . If exactly one component Hi has diameter 1 or 2 and Hj = K1 (j 6= i). Hence G ∼ F or F = 8 9. Conversely assume that G = F1 , F2 , · · · , F9 . Then it can be easily verify that γ(G) + γ(L(G)) = n − 2. 2

Problem 2.14 Let G be a connected graph of odd order n ≥ 5 which is not a tree. Characterize graphs for which γ(G) + γ(L(G)) = n − 2.


181

References [1] M. Aouchiche and P. Hansen, A survey of Nordhaus-Gaddum type relations, Discrete Applied Mathematics, 161(4-5) (2013), 466-546. [2] S. Arumugam and S. Velammal, Edge Domination in Graphs, Taiwanese Journal of Mathematics, 2(2)(1998), 173-179. [3] C. Berge, Theory of Graphs and Its Applications, London, Methuen and Co, 1966. [4] J. A. Bondy and U. S. R. Murty, Graph Theory, India, Springer, 2008. [5] G. Chartrand and S. Schuster, On the independence number of complementary graphs, Trans. New York. Acad. Sci., Series II, 36(3) (1974), 247 - 251. [6] E. J. Cockayne, T. W. Haynes and S. T. Hedetniemi, Extremal graphs for inequalities involving domination parameters, Discrete. Math., 216(2000) 1-10. [7] J. F. Frank, M. S. Jacobson, L. F. Kinch and J. Roberts, On graphs having domination number half their order, Period. Math. Hungar., 16:287-293. [8] Gary Chartrand and Ping Zhang, Introduction To Graph Theory, India, Tata McGraw-Hill, 2006. [9] T. W. Haynes, S. T. Hedetnimi and P. J. Slater, Fundamentals of domination in graphs, New York, Marcel Dekkar Inc., 1998. [10] F. Jaeger and C. Payan, Relations due Type Nordhaus-Gaddum pour le Nombre d’Absorption d’un Graphe Simple, C. R. Acad. Sci. Paris Ser. A, 274(1972), 728-730. [11] S. Mitchell and S. T. Hedetniemi, Edge domination in trees, Congr. Numer, 19(1977), 489-509. [12] E. A. Nordhaus and J. Gaddum, On complementary graphs, Amer. Math. Monthly, 63(1956) 177-182. [13] O. Ore, Theory of Graphs, Am. Math. SOC. Colloq. Publ. 38, Providence, RI, 1962. [14] C. Payan and N. H. Xuong, Domination-balanced graphs, J. Graph Theory, 6:23-32, 1982. [15] B. Randerath and L. Volkmann, Characterization of graphs with equal domination and covering number, Discrete. Math., 191(1998) 173-179. [16] S. Rosalin, A study on induced cycle path number of graphs, Ph. D Thesis, Manonmaniam Sundaranar University, Tirunelveli (2015). [17] N. Shunmugapriya, A study on total resolving number of graphs, Ph. D Thesis, Manonmaniam Sundaranar University, Tirunelveli (2016).


Radio Mean D-Distance Number of Family of Tree and Cycle-Related Graphs T. Nicolas and K. JohnBosco (Department of Mathematics, St. Jude’s College, Thoothoor, Kanyakumari, Tamilnadu, India.) E-mail: [email protected], [email protected]

Abstract: A radio mean D-distance labeling of a connected graph G is an injective map f from the vertex set V (G) to N such that for two distinct vertices u and v of G, dD (u, v) + (v) ⌉ ≥ 1 + diamD (G), where dD (u, v) denotes the D-distance between u and v and ⌈ f (u)+f 2

diamD (G) denotes the D-diameter of G. The radio mean D-distance number of f , rmnD (f ) is the maximum label assigned to any vertex of G. The radio mean D-distance number of G, rmnD (G) is the minimum value of rmnD (f ) taken over all radio mean D-distance labeling f of G. In this paper we find the radio mean D-distance number of family of tree and cycle-related graphs.

Key Words: D-distance, radio D-distance coloring, radio D-distance number, radio mean D-distance, radio mean D-distance number.

AMS(2010): 05C78. §1. Introduction By a graph G = (V, E) we mean a finite undirected graph without loops or multiple edges. The order and size of G are denoted by p and q respectively. Let G be a connected graph of diameter d and let k an integer such that 1 ≤ k ≤ d. A radio k-coloring of G is an assignment f of colors (positive integers) to the vertices of G such that d(u, v) + |f (u) − f (v)| ≥ 1 + k for every two distinct vertices u, v of G. The radio k-coloring number rck (f ) of a radio k-coloring f of G is the maximum color assigned to a vertex of G. The radio k-chromatic number rck (G) is min{rck (f )} over all radio k-colorings f of G. A radio k-coloring f of G is a minimum radio k-coloring if rck (f ) = rck (G). A set S of positive integers is a radio k-coloring set if the elements of S are used in a radio k-coloring of some graph G and S is a minimum radio k-coloring set if S is a radio k-coloring set of a minimum radio k-coloring of some graph G. The radio 1-chromatic number rcl (G) is then the chromatic number χ(G). When k = Diam(G), the resulting radio k-coloring is called radio coloring of G. The radio number of G is defined as the minimum span of a radio coloring of G and is denoted as rn(G). Radio labeling (multi-level distance labeling) can be regarded as an extension of distance1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 30, 2018, Accepted May 19, 2018, Edited by R. Kala.

Radio Mean D-Distance Number of Family of Tree and Cycle-Related Graphs

183

two labeling which is motivated by the channel assignment problem introduced by Hale [6]. Chartrand et al. [2] introduced the concept of radio labeling of graph. Chartrand et al. [3] gave the upper bound for the radio number of Path. The exact value for the radio number of Path and Cycle was given by Liu and Zhu [10]. However Chartrand et al. [2] obtained different values than Liu and Zhu [10]. They found the lower and upper bound for the radio number of Cycle. Liu [9] gave the lower bound for the radio number of Tree. The exact value for the radio number of Hypercube was given by R. Khennoufa and O. Togni [8]. M. M. Rivera et al. [20] gave the radio number of Cn × Cn , the Cartesian product of Cn . In [4] C. Fernandez et al. found the radio number for complete graph, star graph, complete bipartite graph, wheel graph and gear graph. M. T. Rahim and I. Tomescu [16] investigated the radio number of Helm Graph. The radio number for the generalized prism graphs were presented by Paul Martinez et.al. in [11]. A wheel graph Wn is defined to be the join K1 + Cn . The vertex corresponding to K1 is known as the apex vertex and the vertices corresponding to cycle are known as rim vertices while the edges corresponding to cycle are known as rim edges. A gear graph Gn is obtained from wheel Wn by subdividing each of its rim edge, and a helm graph Hn is the graph obtained from wheel Wn by attaching a pendant edge to each rim vertex. A Shadow graph D2 (G) of a connected graph G is constructed by taking two copies of G ′ ′′ ′ ′ ′′ say G and G . Join each vertex u in G to the neighbours of the corresponding vertex u in ′′ G . The concept of D-distance was introduced by D. Reddy Babu et al. [17, 18, 19]. If u, v are vertices of a connected graph G, the D-length of a connected u − v path s is defined as P lD (s) = l(s) + deg(v) + deg(u) + deg(w) where the sum runs over all intermediate vertices w of s and l(s) is the length of the path. The D-distance, dD (u, v) between two vertices u, v of a connected graph G is defined a dD (u, v) = min{lD (s)} where the minimum is taken over all P u − v paths s in G. In other words, dD (u, v) = min{l(s) + deg(v) + deg(u) + deg(w)} where the sum runs over all intermediate vertices w in s and minimum is taken over all u − v paths s in G. In [12], we introduced the concept of Radio D-distance. The radio D-distance coloring is a function f : V (G) → N ∪ {0} such that dD (u, v) + |f (u) − f (v)| ≥ diamD (G) + 1. It is denoted by rnD (G). A radio D-distance coloring f of G is a minimum radio D-distance coloring if rnD (f ) = rnD (G), where rnD (G) is called radio D-distance number. Radio mean labeling was introduced by R. Ponraj et al [13,14,15]. A radio mean labeling is a one to one mapping f from V (G) to N satisfying the condition d(u, v) + ⌈

f (u) + f (v) ⌉ ≥ 1 + diam(G) 2

(1.1)

for every u, v ∈ V (G). The span of a labeling f is the maximum integer that f maps to a vertex of 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 (1.1) is called radio mean condition. In this paper, we introduce the concept of radio mean D-distance number. A radio mean

184

T. Nicolas and K. JohnBosco

D-distance labeling is a one to one mapping f from V (G) to N satisfying the condition dD (u, v) + ⌈

f (u) + f (v) ≥ 1 + diamD (G) 2

(1.2)

for every u, v ∈ V (G). The span of a labeling f is the maximum integer that f maps to a vertex of G. The radio mean D-distance number of G, rmnD (G) is the lowest span taken over all radio mean D-distance labelings of the graph G. The condition (1.2) is called radio mean D-distance condition. In this paper we determine the radio mean D-distance number of family of tree and cycle-related graphs. The function f : V (G) → N always represents injective map unless otherwise stated. We observe that for any two vertices u, v of G we have d(u, v) ≤ dD (u, v). The equality holds if and only if u, v are identical. If G is any connected graph then the D-distance is a metric on the set of vertices of G. We can check easily rD (G) ≤ dD (G) ≤ 2rD (G). The lower bound is clear from the definition and the upper bound follows from the triangular inequality.

§2. Radio Mean D-Distance Number of Family of Tree and Cycle-Related Graphs Observation 2.1 Let P = v1 , v2 , · · · , vn be a diametrical path in G with respect to D-distance. Let f be a radio mean D-distance on the graph G. Let f (v1 ) = a and f (vn ) = b. Then b = a+1. Theorem 2.2 The radio mean D-distance number of a bistar, rmnD (B(n, n)) ≤ 3(n + 1) if n ≥ 2. Proof Let V (B(n, n)) = {xj / j = 1, 2} ∪ {vi , ui / i = 1, 2, · · · , n} and E(B(n, n)) = {x1 x2 , x1 vi , x2 ui / i = 1, 2, 3, · · · , n − 1}. It is obvious that diamD (B(n, n)) = 2n + 7. The radio (v) mean D-distance condition dD (u, v) + ⌈ f (u)+f ⌉ ≥ 2n + 8, for every pair of vertices (u, v) 2 where u 6= v. If f (ui ) = a and f (vj ) = b for some i, j then a + b ≥ 1 (by Observation 2.1).

(1)

Similarly, If f (u1 ) = a and f (ui ) = b for any i > 1 then a + b ≥ 2n + 5.

(2)

If f (ui ) = a and f (x1 ) = b for any i > 1 then a + b ≥ 5.

(3)

If f (ui ) = a and f (x2 ) = b for any i > 1 then a + b ≥ 2n + 9.

(4)


185

If f (x1 ) = a and f (x2 ) = b then a + b ≥ 9.

(5)

For f to be minimal if we set b = a + 1, then the equations from (1) to (5) respectively give the values for a as a ≥ 1 or a ≥ n + 2 or a ≥ 2 or a ≥ n + 4 or a ≥ 4. Case 1. f (u1 ) = a = 1 and f (v1 ) = 2. If f (u2 ) = k, we know that n+5+⌈ 1+k 2 ⌉ ≥ 2n+8 for the edge u1 u2 . Then f (ui ) = 2n+2+i, 3 ≤ i ≤ n. If f (v2 ) = s, we have n + 5 + ⌈ 2+s 2 ⌉ ≥ 2n + 8 for the edge v1 v2 . Then f (vi ) = 3n + i, 4 ≤ i ≤ n. Therefore f (vn ) = 4n. Define f (x1 ) = 4n + 1 and f (x2 ) = 4n + 2. In this case f (x2 ) = 4n + 2.

(6)

Case 2. f (u1 ) = a = n + 2 and f (u2 ) = n + 3 Then f (ui ) = n + 1 + i, 3 ≤ i ≤ n. Therefore f (un ) = 2n + 1.

If f (x2 ) = k, we get that n + 3 + ⌈ n+2+k ⌉ ≥ 2n + 8 for the edge ui x2 for some i. Then 2 f (vi ) = 2n + 3 + i, 1 ≤ i ≤ n. Therefore f (vn ) = 3n + 3. In this case f (vn ) = 3n + 3.

(7)

Case 3. f (u1 ) = a = 2 and f (x1 ) = 3. Then f (ui ) = 2 + i, 2 ≤ i ≤ n. Therefore f (un ) = n + 2.

If f (x2 ) = k. We know that n + 3 + ⌈ 2+k 2 ⌉ ≥ 2n + 8 for the edge ui x2 for some i. That is k ≥ 2n + 7 implies that f (x2 ) = 2n + 7.

If f (v1 ) = s, we have n + 3 + ⌈ 3+s 2 ⌉ ≥ 2n + 8 for the edge vi x1 for some i. Then f (vi ) = 2n + 6 + i, 2 ≤ i ≤ n. Therefore f (vn ) = 3n + 6. In this case f (vn ) = 3n + 6.

(8)

Case 4. f (u1 ) = a = n + 4 and f (x2 ) = n + 5. Define f (ui ) = n+4+i, 2 ≤ i ≤ n. Therefore f (un ) = 2n+4. And also define f (x1 ) = 2n+5 and f (vi ) = 2n + 5 + i, 1 ≤ i ≤ n. Therefore f (vn ) = 3n + 5. In this case f (vn ) = 3n + 5. (9) Case 5. f (x1 ) = a = 4 and f (x2 ) = 5. Let f (v1 ) = k. We get that n + 3 + ⌈ 4+k 2 ⌉ ≥ 2n + 8 for the edge x1 vi for some i. Then f (vi ) = 2n + 4 + i, 2 ≤ i ≤ n. Therefore f (vn ) = 3n + 4. Define f (u1 ) = s. We have that n + 3 + ⌈ 5+s 2 ⌉2n + 8 for the edge x1 ui for some i.

186


Define f (ui ) = 3n + 3 + i, 2 ≤ i ≤ n. Therefore f (un ) = 4n + 3. In this case f (un ) = 4n + 3.

(10)

By comparing the equations (6), (7), (8), (9) and (10) we get the minimal range of the function f as {1 ≤ f (x) ≤ 3n + 3 for all x ∈ V (B(n, n))} which corresponds to Case 2. Thus, rmnD B(n, n) ≤ 3n + 3. Hence, rmnD B(n, n) ≤ 3n + 3 if n ≥ 2.

2

Theorem 2.3 The radio mean D-distance number of a shadow graph of star, rmnD (D2 (K1,n )) ≤ 4n − 2 if n ≥ 2. Proof Let V (D2 (K1,n )) = {xj / j = 1, 2} ∪ {vi , ui / i = 1, 2, · · · , n} be the vertex set and E(D2 (K1,n )) = {xj xj+1 , xj vi , xj+1 ui / i = 1, 2, 3, · · · , n and j = 1} be the edge set. It is obvious that diamD (D2 (K1,n )) = 4(n + 1). The radio mean D-distance condition implies that (v) dD (u, v) + ⌈ f (u)+f ≥ 4n + 5, for every pair of vertices (u, v) where u 6= v. 2 If f (x1 ) = a and f (x2 ) = b then a + b ≥ 1 (by Observation 2.1).

(1)

Similarly, if f (x1 ) = a and f (vi ) = b for any i ≥ 1 then a + b ≥ 4n + 3.

(2)

If f (v1 ) = a and f (vi ) = b for any i > 1 then a + b ≥ 4n − 3.

(3)

If f (vi ) = a and f (uj ) = b for some i, j then a + b ≥ 4n − 3.

(4)

For f to be minimal if we set b = a + 1, then the equations from (1) to (4) respectively give the values for a as a ≥ 1 or a ≥ 2n + 1 or a ≥ 2n − 2 or a ≥ 2n − 2. Case 1. f (x1 ) = a = 1 and f (x2 ) = 2. Let f (v1 ) = k. We know that 2n + 3 + ⌈ 1+k 2 ≥ 4n + 5 for the edge x1 v1 . Then f (vi ) = 4n + 1 + i, 2 ≤ i ≤ n. Therefore f (vn ) = 5n + 1. And also define f (ui ) = 5n + 1 + i, 1 ≤ i ≤ n. Therefore f (un ) = 6n + 1. In this case f (un ) = 6n + 1. Case 2. f (x1 ) = a = 2n + 1 and f (v1 ) = 2n + 2. Then f (vi ) = 2n + 1 + i, 2 ≤ i ≤ n. Therefore f (vn ) = 3n + 1.

(5)


187

Let f (u1 ) = k. We know that 2n + 3 + ⌈ 2n+1+k ⌉ ≥ 4n + 5 for the edge x1 ui for some i. 2 Then f (ui ) = 3n + 1 + i, 2 ≤ i ≤ n. Therefore f (un ) = 4n + 1 and f (x2 ) = 4n + 2. In this case f (x2 ) = 4n + 2.

(6)

Case 3. f (v1 ) = a = 2n − 2 and f (v2 ) = 2n − 1. We define f (vi ) = 2n − 3 + i, 3 ≤ i ≤ n. Therefore f (vn ) = 3n − 3. Let f (x1 ) = k We have that 2n + 3 + ⌈ 2n−2+k ⌉ ≥ 4n + 5 for the edge vi x1 for some i. 2

We define f (ui ) = 3n + i − 3, 1 ≤ i ≤ n. Therefore f (un ) = 4n − 3 and f (x2 ) = 4n − 1. In this case f (x1 ) = 4n − 2. (7) Case 4. f (v1 ) = a = 2n − 2 and f (u1 ) = 2n − 1. We define f (vi ) = 2n − 4 + 2i, 2 ≤ i ≤ n. Therefore f (vn ) = 4n − 4. And we also define f (ui ) = 2n − 3 + 2i, 2 ≤ i ≤ n. Then f (vn ) = 4n − 3.

⌉ ≥ 4n + 5 for the edge vi x1 for some i. Let f (x1 ) = k. We know that 2n + 3 + ⌈ 2n−2+k 2 Then f (x2 ) = 4n + 2. In this case f (x2 ) = 4n + 2. (8) By comparing the equations (5), (6), (7) and (8) we get the minimal range of the function f as {1 ≤ f (x) ≤ 4n − 2 for all x ∈ V (D2 (K1,n ))} which corresponds to Case 3.

2

Hence, rmnD (D2 (K1,n )) ≤ 4n − 2 if n ≥ 2.

Remark 2.4 rmnD (C3 ) = 3. Theorem 2.5 The radio mean D-distance number of a cycle,    6 D rmn (Cn ) ≤ 5    5

if n = 4 n

2 − 5, n ≥ n−1 − 4, n 2

6

if n is even

≥5

if n is odd.

Proof Let V (Cn ) = {v1 , v2 , v3 , · · · , vn } be the vertex set and E(Cn ) = {vi vi+1 , v1 vn / i = 1, 2, 3, · · · , n − 1} be the edge set with even n. It is obvious that diamD (Cn ) = 3 n2 + 2. The (v) radio mean D-distance condition implies that dD (u, v) + ⌈ f (u)+f ≥ 3 n2 + 3, for every pair 2 of vertices (u, v) where u 6= v. If f (v1 ) = a and f (v( n2 ) ) = b then

a + b ≥ 1 (by Observation 2.1).

(1)

Similarly, if f (v1 ) = a and f (v2 ) = b then a+b≥6

n 2

− 5.

(2)

188


If f (v1 ) = a and f (v3 ) = b then n

a+b≥6

2

− 11.

(3)

For f to be minimal if we set b = a + 1, then the equations from (1) to (3) respectively give the values for a as a ≥ 1 or a ≥ 3 n2 − 3 or a ≥ 3 n2 − 4. Case 1. f (v1 ) = a = 1 and f (v( n2 ) ) = 2. Let f (v2 ) = k. That is k ≥ 6 n2 − 6 which implies f (v2 ) = 6 n2 − 6. Then f (vi ) = 6 n2 − 8 + i, 3 ≤ i ≤ n2 − 1 and f (v( n )+i ) = 7 n2 − 9 + i, 1 ≤ i ≤ n2 . Therefore, the 2 largest label f (vn ) = 4n − 9. In this case f (vn ) = 4n − 9.

(4)

Case 2. f (v1 ) = a = 3 n2 − 3 and f (v2 ) = 3 n2 − 2. Then f (vi ) = 3 n2 − 4 + i, 3 ≤ i ≤ n. Therefore the largest label f (vn ) = 5 n2 − 4. In this case n f (vn ) = 5 − 4. (5) 2 Case 3. f (v1 ) = a = 3 n2 − 4 and f (v3 ) = 3 n2 − 3. Then f (v2i−1 ) = 3 n2 − 5 + i, 1 ≤ i ≤ n2 . Therefore f (vn−1 ) = 4 n2 − 5. Let f (vn−2 ) = k. That is k ≥ 2 n2 , since k ≥ 3 n2 − 4, we set f (vn−2 ) = 4 n2 − 4. Therefore f (vn−2i ) = 4 n2 − 5 + i, 1 ≤ i ≤ n2 − 1 and f (vn ) = 5 n2 − 5. In this case f (vn ) = 5

n

− 5..

2

(6)

By comparing the equations (4), (5) and (6) we get the minimal range of the function f as {1 ≤ f (x) ≤ 5 n2 − 5 for all x ∈ V (Cn )} which corresponds to Case 3. Therefore, rmnD (Cn ) ≤ 5 n2 − 5 if n is odd. It is obvious that diamD (Cn ) = 3 n−1 + 2. The radio mean D-distance condition implies 2 f (u)+f (v) n−1 D ⌉ ≥ 3 2 + 3 for every pair of vertices (u, v) where u 6= v. that d (u, v) + ⌈ 2 If f (v1 ) = a and f (v( n−1 ) ) = b then 2

a + b ≥ 1. (by Observation 2.1)

(7)

Similarly, if f (v1 ) = a and f (v2 ) = b then a+b≥6

n−1 2

− 5.

(8)

− 11.

(9)

If f (v1 ) = a and f (v3 ) = b then a+b≥6

n−1 2


189

For f to be minimal if we set b = a + 1, then the equations from (7) to (9) respectively give the values for a as a ≥ 1 or a ≥ 3 n−1 − 3 or a ≥ 3 n−1 − 4. 2 2

Case 1. f (v1 ) = a = 1 and f (v( n−1 ) ) = 2. 2 Let f (v2 ) = k. That is k ≥ 6 n−1 − 6 which implies f (v2 ) = 6 n−1 − 6. Therefore, 2 2 n−1 n−1 f (vi ) = 6 n−1 − 8 + i, 3 ≤ i ≤ − 1 and f (v ) = 7 − 9 + i, 1 ≤ i ≤ n−1 + 1. n−1 2 2 2 2 ( 2 )+i In this case f (vn ) = 4n − 12. (10) Case 2. f (v1 ) = a = 3 Then f (vi ) = 3

n−1 2

n−1 2

− 3 and f (v2 ) = 3

− 2.

− 4 + i, 3 ≤ i ≤ n. Therefore f (vn ) = 5 f (vn ) = 5

Case 3. f (v1 ) = a = 3

n−1 2

n−1 2

n−1 2

− 4 and f (v3 ) = 3

n−1 2

− 3. In this case (11)

− 3.

− 5 + i, 1 ≤ i ≤ n−1 . Therefore f (vn−2 ) = 4 n−1 − 5. 2 2 n−1 n−1 −4. Let f (vn−1 ) = k. That is k ≥ 2 2 . Since k ≥ 3 2 −4, we set f (vn−1 ) = 4 n−1 2 n−1 n−1 Therefore f (vn−2i+1 ) = 4 n−1 − 5 + i, 1 ≤ i ≤ and f (v ) = 5 − 4. In this case n 2 2 2 Then f (v2i−1 ) = 3

n−1 2

− 3.

n−1 2

f (vn ) = 5

n−1 2

− 4.

(12)

By comparing the equations (10), (11) and (12) we get the minimal range of the function f as {1 ≤ f (x) ≤ 5 n−1 − 4 for all x ∈ V (Cn )} which corresponds to Case 3. 2   if n = 4  6 Hence, rmnD (Cn ) ≤ 5 n2 − 5, n ≥ 6 2 if n is even    n−1 5 2 − 4, n ≥ 5 if n is odd.

Theorem 2.6 The radio mean D-distance number of a wheel graph, rmnD (Wn ) ≤ 2n + 1 if n ≥ 6.

Proof Let V (Wn ) = {v0 , v1 , v2 , v3 , · · · , vn } be the vertex set, where v0 is the central vertex and E(Wn ) = {v0 vi , vi vi+1 / i = 1, 2, 3, · · · , n − 1} be the edge set. It is obvious that (v) diamD (Wn ) = n+8. The radio mean D-distance condition implies that dD (u, v)+⌈ f (u)+f ⌉≥ 2 n + 9, for every pair of vertices (u, v) where u 6= v. If f (v1 ) = a and f (v⌈ n2 ⌉ ) = b then

a + b ≥ 1 (by Observation 2.1).

(2)

Similarly, if f (v1 ) = a and f (v2 ) = b then a + b ≥ 2n + 3.

(2)

190


If f (v1 ) = a and f (v3 ) = b then a + b ≥ 2n − 5.

(3)

a + b ≥ 9.

(4)

If f (v0 ) = a and f (v1 ) = b then

For f to be minimal if we set b = a + 1, then the equations from (1) to (4) respectively give the values for a as a ≥ 1 or a ≥ n + 1 or a ≥ n − 3 or a ≥ 4. Case 1. f (v1 ) = a = 1 and f (v⌈ n2 ⌉ ) = 2. Define f (v2 ) = k. That is k ≥ 2n + 2 which implies that f (v2 ) = 2n + 2. Therefore, f (v0 ) = 2n + 1, f (vi ) = 2n + i, 2 ≤ i ≤ n2 − 1 and f (v( n )+i ) = 5 n2 − 1 + i, 1 ≤ i ≤ n2 . In 2 this case f (vn ) = 3n − 1. (5) Case 2. f (v1 ) = a = n + 1 and f (v2 ) = n + 2. Then f (vi ) = n + i, 3 ≤ i ≤ n. Therefore f (vn ) = 2n and f (v0 ) = 2n + 1. In this case f (v0 ) = 2n + 1.

(6)

Case 3. Consider 2 subcases following. Subcase 3.1 n is even, f (v1 ) = a = n − 3 and f (v3 ) = n − 2. Then f (v2i−1 ) = n − 4 + i, 3 ≤ i ≤ n2 . Therefore f (vn−1 ) = 3 n2 − 4. Let f (vn−2 ) = k. That is k ≥ n2 + 7, since k > n − 3, only possible k ≥ 3 n2 + 2, we set f (vn−2 ) = 3 n2 + 2. Therefore f (vn−2i ) = 3 n2 + 1 + i, 1 ≤ i ≤ n2 − 1, f (vn ) = 4 n2 + 1 and f (v0 ) = 4 n2 + 2. In this case f (v0 ) = 2n + 2.

(7a)

Subcase 3.2 n is odd, f (v1 ) = a = n − 3 and f (v3 ) = n − 2. Then f (v2i−1 ) = n − 4 + i, 3 ≤ i ≤ n−1 . Therefore f (vn−2 ) = 3 n−1 − 3. 2 2 n−1 Let f (vn−1 ) = k. That is k ≥ 2 + 8, since k > n − 3, only possible k ≥ 3 n−1 + 2, we 2 n−1 n−1 n−1 set f (vn−1 ) = 3 2 + 2. Therefore f (vn−2i+1 ) = 3 2 + 1 + i, 1 ≤ i ≤ 2 , f (vn ) = 4 n−1 + 2 and f (v0 ) = 4 n−1 + 3. In this case 2 2 f (v0 ) = 2n + 1.

(7b)

Case 4. Consider 2 subcases following. Subcase 4.1 n is even, f (v0 ) = a = 4 and f (v1 ) = 5. Then f (v2i−1 ) = 4 + i, 1 ≤ i ≤ Therefore f (vn−1 ) = n2 + 4.

n 2

.

Let f (v2 ) = k. That is k ≥ 2n − 2 which implies that f (v2 ) = 2n − 2. Therefore,


f (v2i ) = 2n − 3 + i, 1 ≤ i ≤

n 2

− 1 and f (vn ) = 5 f (vn ) = 5

n 2

n 2

191

− 4. In this case

− 4.

Subcase 4.2 n is odd, f (v0 ) = a = 4 and f (v1 ) = 5. Then f (v2i−1 ) = 4+i, 1 ≤ i ≤ Therefore f (vn−2 ) = n−1 + 4. 2

(8a) n−1 2

.

Let f (v2 ) = k. That is k ≥ 2n − 2 which implies that f (v2 ) = 2n − 2. Therefore, f (v2i ) = 2n − 3 + i, 1 ≤ i ≤ n−1 and f (vn ) = 5 n−1 − 1. In this case 2 2 f (vn ) = 5

n−1 2

− 1.

(8b)

By comparing the equations (5), (6), (7a), (7b), (8a) and (8b) we get the minimal range of the function f as {1 ≤ f (x) ≤ 2n + 1 for all x ∈ V (Wn )} which corresponds to Case 2. Hence rmnD (Wn ) ≤ 2n + 1 if n ≥ 6. 2 Remark 2.7 rmnD (Wn ) = 7 if n = 4, 5. Theorem 2.8 The radio mean D-distance number of a gear graph, rmnD (Gn ) ≤ 3n + 4, n ≥ 6. Proof Let {v0 , v1 , v2 , v3 , · · · , vn } and {u1 , u2 , u3 , · · · , un } are the vertex set, where v0 is the central vertex and E(Gn ) = {v0 vi , vi ui / i = 1, 2, 3, · · · , n} be the edge set. It is clear that diamD (Gn ) = n + 14 (n ≥ 6). The radio mean D-distance condition implies that dD (u, v) + (v) ⌉ ≥ n + 15, for every pair of vertices (u, v) where u 6= v. ⌈ f (u)+f 2 If f (u1 ) = a and f (u⌈ n2 ⌉ ) = b then

a + b ≥ 1 (Observation 2.1).

(1)

Similarly, if f (u1 ) = a and f (u2 ) = b then a + b ≥ 2n + 11.

(2)

a + b ≥ 2n + 17.

(3)

a + b ≥ 2n + 3.

(4)

a + b ≥ 21.

(5)

a + b ≥ 15.

(6)

If f (u1 ) = a and f (v1 ) = b then

If f (u1 ) = a and f (v2 ) = b then


If f (v0 ) = a and f (u1 ) = b then

192


If f (v1 ) = a and f (v2 ) = b then a + b ≥ 2n + 9.

(7)

For f to be minimal if we set b = a + 1, then the equations from (1) to (7) respectively give the values for a as a ≥ 1 or a ≥ n + 5 or a ≥ n + 8 or a ≥ n + 1 or a ≥ 10 or a ≥ 7 or a ≥ n + 4. Case 1. f (u1 ) = a = 1 and f (u⌈ n2 ) = 2. Define f (v1 ) = k. That is k ≥ 2n + 16 which implies f (v1 ) = 2n + 16. Then f (vi ) = 2n + 15 + i, 1 ≤ i ≤ n. Therefore f (vn ) = 3n + 15. Let f (u2 ) = s. That is s ≥ 2n + 10. Since we have already used the label 2n + 10, we set f (u2 ) = 3n + 16 and f (v0 ) = 2n + 15. Then f (ui ) = 3n + 14 + i, 2 ≤ i ≤ n2 − 1 and f (u( n )+i ) = 7 n2 + 13 + i, 1 ≤ i ≤ n2 . Therefore the largest label f (un ) = 4n + 13. In this 2 case f (un ) = 4n + 13. (8) Case 2. f (u1 ) = a = n + 5 and f (u2 ) = n + 6. Then f (ui ) = n + 4 + i, 3 ≤ i ≤ n. Therefore f (un ) = 2n + 4. Let f (vn ) = k. That is k ≥ 13, since k > n + 5, we set f (vn ) = 2n + 5. Therefore f (vn−i+1 ) = 2n + 4 + i, 1 ≤ i ≤ n and f (v0 ) = 3n + 5. In this case f (v0 ) = 3n + 5.

(9)

Case 3. f (u1 ) = a = n + 8 and f (v1 ) = n + 9. We define f (ui ) = n + 2i + 6, 1 ≤ i ≤ n, f (vi ) = n + 2i + 7, 1 ≤ i ≤ n and f (v0 ) = 3n + 8. In this case f (v0 ) = 3n + 8. (10) Case 4. f (u1 ) = a = n + 1 and f (v2 ) = n + 2. Define f (v1 ) = k. We know that 6+⌈ n+1+k ⌉ ≥ n+15 for the edge u1 v1 . That is k ≥ n+16 2 which implies f (v1 ) = n + 16. Then f (vi ) = n + 14 + i, 3 ≤ i ≤ n. Therefore f (vn ) = 2n + 14. Let f (u2 ) = s. That is s ≥ n + 15 which implies that f (u2 ) = n + 15 and f (v0 ) = n + 14. Since we have already used the label n + 16, we set f (u3 ) = 2n + 15. Therefore f (ui ) = 2n + 12 + i, 3 ≤ i ≤ n. In this case f (un ) = 3n + 12.

(11)

Case 5. f (v0 ) = a = 10 and f (v1 ) = 11. Define f (v2 ) = k. That is k ≥ 2n − 2 which implies f (v2 ) = 2n − 2. Then f (vi ) = 2n − 4 + i, 2 ≤ i ≤ n. Therefore f (vn ) = 3n − 4.


193

Let f (u1 ) = s. That is s ≥ 2n + 6, since we have already used the label 2n + 6, we set f (u1 ) = 3n − 3. Therefore f (ui ) = 3n − 4 + i, 1 ≤ i ≤ n. In this case f (un ) = 4n − 4.

(12)

Case 6. f (v0 ) = a = 7 and f (u1 ) = 8. Define f (v1 ) = k. That is k ≥ 2n + 9 which implies that f (v1 ) = 2n + 9. Therefore, f (vi ) = 2n + i + 8, 1 ≤ i ≤ n. Let f (u2 ) = s. That is s ≥ 2n + 3 which implies that f (u2 ) = 2n + 3. Therefore f (ui ) = 2n + 5 − i, 2 ≤ i ≤ n. In this case f (vn ) = 3n + 8.

(13)

Case 7. f (v1 ) = a = n + 4 and f (v2 ) = n + 5. We define f (vi ) = n + 3 + i, 3 ≤ i ≤ n. Therefore f (vn ) = 2n + 3. Let f (un ) = k. That is k ≥ 14, since k > n + 4, we set f (un ) = 2n + 4. Therefore f (un−i+1 ) = 2n + i + 3, 1 ≤ i ≤ n and f (v0 ) = 3n + 4. In this case f (v0 ) = 3n + 4.

(14)

By comparing the equations (8) to (14) we get the minimal range of the function f as {1 ≤ f (x) ≤ 3n + 4 for all x ∈ V (Gn )} which corresponds to Case 7. Hence rmnD (Gn ) ≤ 3n + 4 if n ≥ 6.    11 if n = 3

Remark 2.9 rmnD (Gn ) =

16    17

2

if n = 4 if n = 5

Theorem 2.10 The radio mean D-distance number of a Helm graph, rmnD (Hn ) ≤ 3n + 2 if n ≥ 6. Proof Let {v0 , v1 , v2 , v3 , · · · , vn } and {u1 , u2 , u3 , · · · , un } are the vertex set, where v0 is the central vertex and also {u1 , u2 , u3 , · · · un } are the pendent vertices and E(Hn ) = {v0 vi , vi ui / i = 1, 2, 3, · · · , n} be the edge set. It is obvious that diamD (Hn ) = n + 14. The radio mean D(v) distance condition implies that dD (u, v) + ⌈ f (u)+f ⌉ ≥ n + 15, for every pair of vertices (u, v) 2 where u 6= v. If f (u1 ) = a and f (u⌈ n2 ⌉ ) = b then

a + b ≥ 1.

(1)

Similarly, if f (u1 ) = a and f (u2 ) = b then a + b ≥ 2n + 3.

(2)

194


If f (v0 ) = a and f (v1 ) = b then a + b ≥ 19.

(3)

a + b ≥ 15.

(4)

a + b ≥ 2n + 11.

(5)

a + b ≥ 2n + 17.

(6)

a + b ≥ 2n + 7.

(7)





For f to be minimal if we set b = a + 1, then the equations from (1) to (7) respectively give the values for a as a ≥ 1 or a ≥ n + 2(we choose) or a ≥ 9 or a ≥ 8 or a ≥ n + 5 or a ≥ n + 3 or a ≥ n + 8. Case 1. f (u1 ) = a = 1 and f (u⌈ n2 ⌉ ) = 2. Define f (u2 ) = k. That is k ≥ 2n+2, which implies that f (u2 ) = 2n+2 and f (v0 ) = 2n+1. Then f (ui ) = 2n + i, 2 ≤ i ≤ n2 − 1 and f (u( n )+1 ) = 5 n2 − 1 + i, 1 ≤ i ≤ n2 . Therefore 2 f (un ) = 3n − 1.

Let f (v1 ) = s. That is s ≥ 2n + 16, since we have already used the label 2n + 16, we set f (v1 ) = 3n. Therefore, f (vi ) = 3n + i − 1, 1 ≤ i ≤ n. In this Case f (vn ) = 4n − 1.

(8)

Case 2. f (u1 ) = a = n + 3, f (u2 ) = n + 4 and f (v0 ) = n + 2. Then f (ui ) = n + i + 2, 3 ≤ i ≤ n. Therefore f (un ) = 2n + 2. Let f (vn ) = k. That is k ≥ 15, since k > n + 2, we set f (vn ) = 2n + 3. Therefore f (vn−i+1 ) = 2n + i + 2, 1 ≤ i ≤ n. In this case f (v1 ) = 3n + 2.

(9)

Case 3. f (v0 ) = a = 9 and f (v1 ) = 10. Define f (v2 ) = k. That is k ≥ 2n + 1, which implies that f (v2 ) = 2n + 1. Then f (vi ) = 2n + i − 1, 2 ≤ i ≤ n. Therefore f (vn ) = 3n − 1. Let f (u1 ) = s. That is s ≥ 2n + 7, since we have already used the label 2n + 7, we set


195

f (u1 ) = 3n. Therefore f (ui ) = 3n − 1 + i, 1 ≤ i ≤ n. In this case f (un ) = 4n − 1.

(10)

Case 4. f (v0 ) = a = 8 and f (u1 ) = 9. Define f (v1 ) = k. That is k ≥ 2n + 8 which implies that f (v1 ) = 2n + 8. Then f (vi ) = 2n − i + 9, 1 ≤ i ≤ n. Therefore f (vn ) = n + 9.

Let f (u2 ) = s. That is s ≥ 2n − 6, since we have already used the label 2n − 6, we set f (u2 ) = 2n + 9. Therefore f (ui ) = 2n + i + 7, 2 ≤ i ≤ n. In this case f (un ) = 3n + 7.

(11)

Case 5. f (v1 ) = a = n + 5 and f (v2 ) = n + 6. Then f (vi ) = n + 4 + i, 3 ≤ i ≤ n. Therefore f (vn ) = 2n + 4.

Let f (u1 ) = k. That is k ≥ n + 12, since we have already used the label n + 12, we set f (u1 ) = 2n + 5. Therefore f (ui ) = 2n + i + 4, 1 ≤ i ≤ n and f (v0 ) = 3n + 5. In this case f (v0 ) = 3n + 5.

(12)

Case 6. f (v1 ) = a = n + 3 and f (u2 ) = n + 4. Define f (u1 ). That is k ≥ n+14 which implies f (u1 ) = n+14. Then f (ui ) = n+12+i, 3 ≤ i ≤ n. Therefore f (un ) = 2n + 12. Let f (v2 ) = s. We get that 6 + ⌈ n+4+s ⌉ ≥ n + 15 for the edge v2 u2 . That is s ≥ n + 13 2 which implies that f (v2 ) = n + 13 and f (v0 ) = n + 12. Since we have already used the label from n + 14, we set f (v3 ) = 2n + 13. Therefore f (vi ) = 2n + 10 + i, 3 ≤ i ≤ n. In this case f (un ) = 3n + 10.

(13)

Case 7. f (v1 ) = a = n + 8 and f (u1 ) = n + 9. We defined f (vi ) = n + 2i + 6, 1 ≤ i ≤ n, f (ui ) = n + 2i + 7, 1 ≤ i ≤ n and f (v0 ) = 3n + 8. In this case f (v0 ) = 3n + 8. (14) By comparing the equations (8) to (14) we get the minimal range of the function f as {1 ≥ f (x) ≥ 3n + 2 for all x ∈ V (Hn )} which corresponds to Case 2. Hence rmnD (Hn ) ≤ 3n + 2, n ≥ 6.

Remark 2.11 rmnD (Hn ) =

   10 if n = 3   

16 if n = 4 17 if n = 5

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References [1] F. Buckley and F. Harary, Distance in Graphs, Addition- Wesley, Redwood City, CA, 1990. [2] G. Chartrand, D. Erwinn, F. Harary, and P. Zhang, Radio labeling of graphs, Bulletin of the Institute of Combinatorics and Its Applications, Vol. 33, pp. 77-85, 2001. [3] G. Chartrand, D. Erwin, and P. Zhang, Graph labeling problem suggested by FM channel restrictions, Bull. Inst. Combin.Appl., 43, 43-57(2005). [4] C. Fernandaz, A. Flores, M. Tomova, and C. Wyels, The radio number of gear graphs, arXiv:0809. 2623, September 15, (2008). [5] J.A. Gallian, A dynamic survey of graph labeling, Electron. J. Combin. 19 (2012) #Ds6. [6] W.K. Hale, Frequency assignment: Theory and applications, Proc. IEEE 68 (1980), pp. 1497-1514. [7] F.Harary, Graph Theory, Addision Wesley, New Delhi (1969). [8] R. Khennoufa and O. Togni, The radio antipodal and radio numbers of the hypercube, Ars Combinatoria (accepted). [9] D. Liu, Radio number for trees, Discrete Math. 308 (7) (2008) 1153-1164. [10] D. Liu, X. Zhu, Multilevel distance labelings for paths and cycles, SIAM J. Discrete Math. 19 (3) (2005) 610-621. [11] P. Murtinez, J. OrtiZ, M. Tomova, and C. Wyles, Radio numbers for generalized prism graphs, Kodai Math. J., 22,131-139(1999). [12] T. Nicholas and K.JohnBosco , Radio D-distance number of some graphs International Journal of Engineering & Scientific Research, Vol.5 Issue 2, February 2017. [13] R. Ponraj, S. Sathish Narayanan and R. Kala, Radio mean labeling of graphs, AKCE International Journal of Graphs and Combinatorics, 12 (2015) 224-228. [14] R. Ponraj, S. Sathish Narayanan and R. Kala, On radio mean number of some graphs, International J.Math. Combin., Vol.3(2014), 41-48. [15] 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), 2014, pp.273 - 286. [16] M. T. Rahim, I. Tomescu, On multi-level distance labelings of helm graphs, accepted for publication in Ars Combinatoria. [17] Reddy Babu, D., Varma, P.L.N., D-distance in graphs, Golden Research Thoughts, 2 (2013), 53-58. [18] Reddy Babu, D., Varma, P.L.N., Average D-Distance Between Vertices Of A Graph , Italian Journal Of Pure And Applied Mathematics - N. 33(2014), 293-298. [19] Reddy Babu, D., Varma, P.L.N., Average D-Distance Between Edges Of A Graph, Indian Journal of Science and Technology, Vol 8(2), 152-156, January 2015. [20] M. M. Rivera, M. Tomova, C. Wyels, and A. Yeager, The radio number of Cn /Cn , Submitted to Ars Combinatoria, 2009.


A Note on Support Neighbourly Irregular Graphs Selvam Avadayappan, M. Bhuvaneshwari and R. Sinthu (Research Department of Mathematics, VHNSN College, Virudhunagar - 626 001, India) E-mail: [email protected], [email protected], [email protected]

Abstract: In any graphG, the support of a vertex is the sum of degrees of its neighbours. A connected graph G is said to be support neighbourly irregular (or simply SNI), if no two adjacent vertices in G have same support. In this paper, the necessary and sufficient conditions for some known families of graphs to be SNI have been discussed.

Key Words: Irregular graphs, support neighbourly irregular graphs, subdivision graphs, splitting graphs.

AMS(2010): 05C25. §1. Introduction Only finite, simple, connected, undirected graphs are considered in this paper. We refer [11] for further notations and terminology. The degree of a vertex v is denoted by d(v). A full vertex of G is a vertex which is adjacent to every other vertices of G. A graph G is said to be r regular, if every vertex of G has degree r. For r 6= k, a graph G is said to be (r,k) - biregular if d(v) is either r or k for any vertex v in G. In a graph G(V, E), for any vertex v ∈ V, the open neighbourhood of v is the set of all vertices adjacent to v. That is, N (v) = {u ∈ V (G)/ uv ∈ E(G)}. The closed neighbourhood of v is defined by N[v] = N(v) ∪ v. Clearly, if N[u]=[v], then u and v are adjacent and d(u) = d(v). Let G1 and G2 be any two graphs. The graph G1 ◦ G2 obtained from one copy of G1 and |V (G1 )| copies of G2 by joining each vertex in the ith copy of G2 to the ith vertex of G1 is called the corona of G1 and G2 . The concept of support of a vertex has been introduced and studied by Selvam Avadayappan and G. Mahadevan [6]. The support s(v) of a vertex v is the sum of degrees of its neighbours. P That is, s(v) = u∈N (v) d(u). Note that the support of any vertex in an r - regular graph is 2 r . A graph G is said to be a balanced graph, if any two vertices in G have the same support. It is easy to observe that the complete bipartite graphs Km,n and any regular graphs are balanced graphs. A graph G is said to be highly unbalanced, if distinct vertices of G have distinct 1 Proceedings of the International Conference on Discrete Mathematics and its Applications, Manonmaniam Sundaranar University, January 18-20, 2018. 2 Received January 30, 2018, Accepted May 19, 2018, Edited by R. Kala.

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supports. For example, a highly unbalanced graph is shown in Figure 1.

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The following results have been proved in [6]: P P 2 Result 1 v∈V s(v) = v∈V d(v) .

Result 2 For any balanced graph G, δ(G) = 1 if and only if G ∼ = K1,n , n ≥ 1. The study on this parameter has its own importance as any two vertices of same degree need not be of same importance in any graph unless they are isomorphic images of each other. The degrees of its neighbours contribute much in determining the weightage of a vertex in a graph. Hence it becomes essential to study about the support of the vertices also. In a graph G, deleting an edge uv and introducing a new vertex w and the new edges uw and vw is called the subdivision of the edge uv. The subdivision graph denoted by S 1 (G) is obtained from the graph G by subdividing every edge of G. For example, S 1 (K 1,5 ) is shown in Figure 2. b

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The concept of splitting graph of a graph was introduced by Sampath Kumar and Walikar [14]. For a graph G, the graph S(G), obtained from G, by adding a new vertex w for every vertex v ∈V and joining w to all vertices of G adjacent to v, is called the splitting graph of G. This resembles the method of taking clone of every vertex in a graph. For example, a graph G and its splitting graph S(G) are shown in Figure 3. A necessary and sufficient condition for a graph to be a splitting graph has been given in [14].

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A Note on Support Neighbourly Irregular Graphs

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Regarding the study about irregularity in graphs, there is no clear cut boundary defined so far to classify the non regular graphs into exact classes. Several attempts have been made to group some non regular graphs having similar properties. On that line, the concept of highly irregular graphs has been studied by Yousef Alavi and others in [2]. A connected graph G is said to be a highly irregular graph or simply a HI graph, if no two neighbours of every vertex have same degree. For more results on HI graph, we can refer [1], [2], [3]. A connected graph G is said to be a totally segregated graph, if no two adjacent vertices have the same degree. It has been introduced by Jackson and Roger in [13]. Later it has been studied independently by Gnaana Bhragsam and Ayyaswamy in [12]. They called this graph as neighbourly irregular graphs. An HI graph G and an NI graph H are shown in Figure 4. b

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In [15], Swaminathan and Subramanian have introduced a new type of irregular graphs called neighbourhood highly irregular graphs. A connected graph G is said to be neighbourhood highly irregular (or simply NHI), if any two distinct vertices in the open neighbourhood of v, have distinct closed neighbourhood sets. One can refer [4], [7],[8],[9],[10] for further reading on irregularity in graphs. Examples of NHI graphs are given in Figure 5.

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Figure 5 As an addition to this kind of classification, a new family of irregular graphs namely support neighbourly irregular graphs has been introduced in [5]. A connected graph is said to be support neighbourly irregular (or simply SNI ), if no two vertices having same support are adjacent. A graph proving the existence of SNI graphs is shown in Figure 6. b b

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Figure 6 In this paper, we establish some results on SNI graphs. We characterise many families of graphs which are SNI.

§2. Main Results Barbell graphs Bm,n are graphs obtained by joining central vertices of star graphs K1,m and K1,n by means of an edge. For clear reference B3,4 is shown in Figure 7. b

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Figure 7 Fact 1 Barbell graphs are not SNI for any m, n ≥ 1. Fact 2 Balanced graphs are not SNI. The following theorem characterises all balanced graphs. Theorem 2.1 A graph G is a balanced graph if and only if G is a regular graph or a biregular bipartite graph with each partition having vertices of same degree.

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Proof Let G be a balanced graph. Then every component of G has vertices of same support. Hence without loss of generality, we may assume that G is connected. Now, on contrary, assume that G is neither regular nor biregular bipartite of given type. The support of any vertex of degree δ is at most δ∆ and that of any vertex of degree ∆ is at least δ∆. Since G is balanced, we conclude that the support of any vertex in G is δ∆. This forces that if G is (δ, ∆) - biregular, then every vertex of degree δ must be adjacent only to vertices of degree ∆ and vice versa. That is, G is a biregular bipartite graph with each partition having vertices of same degree, which is a contradiction to our assumption. Therefore G has a vertex v of degree neither ∆ nor δ. IfV1 denotes the set of all vertices of degree ∆ or δ, then < V1 > = H is biregular bipartite with each partition having vertices of same degree. Now v ∈ / N(w), for any vertex w ∈ V(H). This is true for every vertex v of degree equal to neither ∆ nor δ. Hence G is disconnected. This is a contradiction. Hence the result. The converse is obvious. 2 Hence regular graphs and biregular bipartite graphs with each partition having vertices of same degree are not SNI. Theorem 2.2 The subdivision graph S1 (G) of any graph G is SNI if and only if G is an NI graph. Proof Assume that for a graph G, the subdivision graph S1 (G)is SNI. Then any two adjacent vertices in S1 (G) have distinct supports. Suppose G is not an NI graph. Then G contains at least two adjacent vertices, say u and v of same degree. Now let u’ be the newly added vertex in S1 (G) when edge uv is subdivided. Now one can easily verify that in S1 (G), the vertices u, u’ and v have the same support 2dG (u). Since S1 (G) contains two adjacent vertices of same support, it is not SNI, which is a contradiction. Hence G is an NI graph. Conversely, assume that G is an NI graph. Then G contains no two adjacent vertices of same degree. Now consider S1 (G). For a vertex u ∈ V(G), all neighbours of u in S1 (G) are new vertices added when subdividing all edges incident with u in G. Therefore no two vertices in G are adjacent in S1 (G). Hence it is enough to check the supports of vertices in G and the new vertices added in subdividing the edges incident with them. Now support of u in S1 (G) is 2dG (u) as all new vertices adjacent to it are of degree 2. Let uv∈ E(G). If u’ is the vertex newly added while subdividing the edge uv in S1 (G), then in S1 (G), s(u’) = dS1 (u) + dS1 (v). Suppose G is not SNI. Then for at least one edge uu’ ∈ E(S1 (G)), supports of u and u’ are equal. That is, 2d(u) = d(u) + d(v). Hence we get d(u) = d(v), which is a contradiction, since G is NI. Hence the result. 2 For the NI graph K2,3 , the respective subdivision graph which is SNI is shown in Figure 8. b b b b

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Recall that a caterpillar G is a tree which results in a path on removal of all pendant vertices of G. For example, one can refer a caterpillar given in Figure 9. u

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Figure 9 Note that the stars and barbell graphs are caterpillars. Here we note that the caterpillar shown in Figure 9 is not SNI as s(u) =s(v) =10. But there are SNI caterpillars also. The following theorem characterises the caterpillars which are SNI. Theorem 2.3 Any caterpillar T which is neither a star nor a barbell graph is SNI if and only if no two non pendant vertices in T at a distance three are of same degree. Proof Let T be any caterpillar which is neither a star nor a barbell graph. Suppose T is SNI. On contrary, assume that T contains two non pendant vertices say u and v of same degree with a distance three between them. Let uww’v be the path of length three joining u and v in T. Let the number of pendant vertices at w and w’ be x and y respectively. Now since T is a caterpillar, d(w) = 2 + x and d(w’) = 2 + y. The neighbour vertices of w in T are u, w’ together with x pendant vertices. Hence s(w) = 2 + y + x + d(u). Similarly, s(w’) = 2 + x + y + d(v). Since d(u) = d(v), we have supports of w and w’ are the same, which is a contradiction since T is SNI. Hence no such u and v exist in T. Conversely, assume that T has no two non pendant vertices in T at a distance three which are of same degree. Suppose T is not SNI. Then there exists an edge uv in T such that support of u and v are equal. Now since T is neither a star nor a barbell graph, at least one of the vertices of u and v contains non pendant vertex other than u or v as its neighbour. Case 1. Exactly one of the vertices u and v contains a non pendant neighbour w other than u and v as its neighbour. Without loss of generality, let w ∈ N(u). Since T is a caterpillar, other neighbours of u and v are pendant vertices. Now s(u)= d(w) + d(v) + d(u) - 2. And s(v) = d(v) - 1 + d(u). By our assumption, support of u and v are equal. Therefore d(w) = - 1, which is impossible. Hence this case does not exist. Case 2. Both u and v contain a non pendant neighbour other than u and v. Let u’ and v’ be non pendant neighbours of u and v, respectively. Since T is a caterpillar, apart from u, u’, v, v’ other neighbours of u and v are pendant vertices. Now s(u) = d(u’) + d(v) + d(u) - 2. Also s(v) = d(v’) + d(u) + d(v) - 2.

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But by our assumption, s(u) = s(v). This forces that d(u’)= d(v’). Also, it is obvious that u’uvv’ is the only path in T connecting u’ and v’. Hence distance between v’ and u’ is three, which is a contradiction to our assumption. Therefore T is SNI. 2 An SNI caterpillar is illustrated in Figure 10. b

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Figure 10 Lemma 2.4 In any connected graph G, the support of any vertex v in G ◦ Krc , sG◦Krc

 s (v) + r(d(v) + 1), if d(v) > 1 G = dG (w) + r, if d(v) = 1 and vw ∈ E(G ◦ K c ) r

Proof LetG be any connected graph with V(G) = {v1 , v2 , . . . ,vn }. In G◦Krc , the newly added vertices are the only pendant vertices and let uij , 1 ≤ j ≤ r,denote the pendant vertices attached to vi in G. Then we have,sG◦Krc (uij ) = dG◦Krc (vi ) = dG (vi ) + r. Now it is easy to note that for any vertex vi , there are r pendant vertices as new neighbours and degree of each neighbour of vi in G gets increased by r in G ◦ Krc . Hence sG◦Krc (v) = P 2 w∈N (v) dG◦Krc (w) = sG (v) + r(d(v) + 1). Hence we get the lemma.

Theorem 2.5 The corona graph G ◦ Krc , for any connected graph G is SNI, if and only if s(v) - s(u) + r(d(v) - d(u)) 6= 0, for any edge uv ∈ E(G).

Proof Let G be any graph such that G ◦ Krc is SNI. Suppose s(v) - s(u) + r(d(v) - d(u)) = 0, for any edge uv ∈ E(G). Then we have, sG (v) + r(d(v) + 1)= sG (u) + r(d(u) + 1). Hence sG◦Krc (v) = sG◦Krc (u) and uv ∈ E(G ◦ Krc ). This is a contradiction, since G ◦ Krc is SNI. Hence s(v) - s(u) + r(d(v) - d(u)) 6= 0, for any edge uv ∈ E(G). Conversely, suppose that s(v) - s(u) + r(d(v) - d(u)) 6= 0, for any edge uv ∈ E(G). We claim that in G ◦ Krc , all edges have end vertices of distinct supports. By our assumption, sG (v) + r(d(v) + 1) 6= sG (u) + r(d(u) + 1) for any edge uv in G ◦ Krc , where u and v ∈ V(G). Hence sG◦Krc (v) 6= sG◦Krc (u) for edges with both end vertices as vertices of G. Also no edge exists between newly added vertices in G◦Krc . It remains to check for supports of end vertices of edges between a vertex in G and a newly added vertex. But by the above lemma, support of a newly added vertex w is dG (v) + r where v is the vertex adjacent to w in G ◦ Krc . Now if s(w) = s(v), then dG (v) + r = sG (v) + rdG (v) + r, which is impossible, since r ≥ 1 and G is connected. 2 In all cases, we find that G ◦ Krc is SNI. Hence the converse part is true. Refer the graph G and its corona G ◦ Krc shown in Figure 11. Here G is SNI but G ◦ Krc is not SNI as sG (v) + 2(d(v) + 1)= sG (u) + 2(d(u) + 1).

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Figure 11 Next we characterize graphs for which the splitting graphs are SNI. Theorem 2.6 The splitting graph S(G) of any graph G is SNI if and only if G is SNI and contains no two adjacent vertices u and v such that 3s(u) = 2s(v). Proof Let G be a graph and S(G) be its splitting graph with the vertex set V(G) = {v1 , v2 , · · · , vn } and V(S(G)) = V(G) {w1 , w2 , · · · , wn } such that wi is the corresponding newly added vertex with respect to vi . Now let us first calculate the support of any vertex vi in S(G). sS(G) (vi ) =

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support. Since newly added vertices form an independent set in S(G), the edge e is between a newly added vertex and a vertex of G or two vertices in G itself. Case 1. e = vi wj . Then sS(G) (vi ) = sS(G) (wj ). That is, 3sG (vi ) = 2sG (vj ), which is a contradiction to our assumption. Case 2. e = vi vj . In this case, sS(G) (vi ) = sS(G) (vj ) which implies sG (vi ) = sG (vj ). This leads to a contradiction to the fact that G is SNI. Hence we conclude that S(G) is SNI. 2 For reference, a graph G and its splitting graph S(G) is shown in Figure 12. Here G is SNI but it contains vertices u and v such that 3s(v) = 2 s(u). Hence S(G) is not SNI. b b b b

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S(G) Figure 12 References [1] Yousef Alavi, F.Buckley, M.Shamula and S.Ruiz, Highly irregular m-chromatic graphs, Discrete Mathematics 72(1988), 3-13. [2] Yousef Alavi, Gary Chartand, F.R.K.Chung, Paul Erdös, R.L.Graham and O.R.Oellermann, Highly irregular graphs, Journal of Graph Theory 11(1987), 235-249. [3] Yousef Alavi, J.Liu, and J.Wang, Highly irregular digraphs, Discrete Mathematics 111(1993), 3-10. [4] Selvam Avadayappan and M. Bhuvaneshwari, Pairable graphs, International Journal of Innovative Science, Engineering & Technology, Vol.1 Issue 5, July 2014, pp.23-31. [5] Selvam Avadayappan, M.Bhuvaneshwari and R. Sinthu, Support neighbourly irregular graphs, International Journal of Scientific Research and Management, Vol. 4, Issue 3, pp. 4009 - 4014, 2016. [6] Selvam Avadayappan and G. Mahadevan, Highly unbalanced graphs, ANJAC Journal of Sciences, 2(1), 23-27, 2003. [7] Selvam Avadayappan, P. Santhi and R. Sri Devi, Some results on neighbourly irregular graphs, International journal of Acta Ciencia Indica, Vol. XXXII M, No.3, 10071012,(2006). [8] Selvam Avadayappan, P.Santhi, Some results on neighbourhood highly irregular graphs, Ars Combinatoria 98(2011), pp. 399-414. [9] Selvam Avadayappan, P.Santhi, Neighbourly irregular product graphs, Mathematics, computing and Modeling, Allied Publishers Pvt. Ltd., 2007, pp. 96-101. [10] Selvam Avadayappan, P.Santhi, Irregularity of product graphs, (Submitted).

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[11] R.Balakrishnan and K. Ranganathan, A Text Book of Graph Theory, Springer-Verlag, New York, Inc(1999). [12] S.Gnana Bhragsam and S.K.Ayyaswamy, Neighbourly irregular graphs, Indian Journal of Pure and Applied Mathematics, 35(3): 389-399, March 2004. [13] D.E.Jackson and Roger Entringer, Totally segregated graphs, Congressus Numerantium 55(1986),pp. 159 - 165. [14] SampathKumar.E, Walikar.H.B, On the Splitting graph of a graph, J.Karnatak Uni. Sci., 25: 13(1980). [15] V.Swaminathan and A.Subramanian, Neighbourhood highly irregular graphs, International Journal of Management and Systems, 8(2): 227-231, May-August 2002.


Introduction to Manonamaniam Sundaranar University

1. An Introduction to the Manonamaniam Sundaranar University The Manonamaniam Sundaranar University (MSU) is a dynamic institution of higher learning, set in Tirunelveli, the educational hub of Southern Tamil Nadu, with a campus area of 550 acres. The university was established in 1990, by the government of Tamil Nadu, as a teaching-cum-affiliating University. The university is named after the Tamil Poet litterateur Prof. Manonamaniam Sundaram Pillai (1855-1897), the author of the famous verse drama, Manonmaniam. The University caters to the needs of the three southern districts of Tamil Nadu, viz., Tirunelveli, Thoothukudi and Kanyakumari. The university is re-accredited by National Assessment and Accreditation Council(NAAC). The university is located at Abishekatti (on Tirunelveli - Tenkasi Road) at a distance of 8 kilometres from Tirunelveli. There are 22 departments and 5 centres in the university besides the Directorate of Distance and Continuing Education and Directorate of Vocational Education. Through the community colleges and ELP centres, the Directorate of Vocational Education is offering skill based diploma and certificate courses in various skill sectors. P.G Extension Centre is at Nagercoil. Sri Paramakalyani Centre of Excellence in Environmental Sciences is functioning at Alwarkurichi and Centre for Marine Science and Technology is at Rajakkamangalam.

Figure 1 Around 1,00,800 students are studying in the university departments / centres, seventy three affiliated colleges, four constituent colleges and six university colleges. The university is one of the pioneer institutions that offers Choice-Based Credit System (CBCS). The courses and

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research programmes offered by the university have been carefully chosen to the contemporary needs of the the region. The university departments offer Ph.D., M.Phil. and integrated P.G programmes in addition to regular P.G programmes. It also offers exemplary library services. Excellent opportunities in extracurricular activities are extended by the physical education and sports department, national service scheme and youth welfare department. The university has a women’s hostel and a men’s hostel with wi-fi facility for entire hostel campus. The university has sofisticated auditorium to accommodate around thousand students. An air conditioned internet centre is available for the students to access internet. The university provides affordable quality education to the underprivileged and unreached sections of the society in this region. The university has an indoor stadium and gymnasium. gnavani radio station is functioning in the university. The university has a health centre, guest house, post office, bank, ATM and students’ amenities centre. The university offers merit scholarships for the top rankers of the integrated PG programmes. Outdoor games, sports yoga and meditation classes are conducted regularly for the students. Annual sports day is also conducted for the staff and students.

2. Department of Mathematics of MSU 2.1 An Introduction to the Department of Mathematics of MSU The department of mathematics was established in the year 1992 with a mission to offer post graduate degree and research programs in mathematics. The department has been offering a full-time 2 year M.Sc., M.Phil., and Ph.D in mathematics since its inception. 5 years integrated M.Sc program has been introduced from the academic year 2016-17 with an intake of 28 students and Sundaranar Merit Scholarship of Rs.2000/- per month of top ten students. At present the department has 4 professors, 3 assistant professors and 2 temporary assistant professors.

Figure 2 The faculty members of the department are highly qualified researchers and the department has a good number of externally funded research projects granted by funding agencies including UGC, DST and DRDO etc. Advanced Academic Programs offered by the department

Introduction to Manonamaniam Sundaranar University

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have been streamlined to make the students competitive and accountable in accordance with the criteria stipulated for national accreditation. The department has academic links with premier national and international institutions. Adequate infrastructure facilities and the peaceful campus atmosphere are conductive to the students to pursue their academic endeavors in an excellent manner. 2.2 Endowments in the Department The department is blessed with the following two endowments in the year 2004. 1.Dr. S.S. Pillai Memorial Endowment 2.Professor S. Arumugam Endowment The former was created in memory of the renowned number theorist Dr.S.Sivasankara Narayana Pillai by the villagers of vallam, a small village near Shencottah in Tirunelveli district from where Dr.Pillai hailed. The latter was created by the students and well wishers of Professor S. Arumugam, the former head of the department who is known to the international community through his research publications and responsible for the growth of discrete mathematics in India. The Department is conducting annual endowment lectures by inviting eminent professors form leading institutions in India and abroad. 2.3 Research Activities in the Department The department of mathematics has been engaged in research in the areas of algebra, graph theory, harmonic analysis, optimization techniques and algorithms. The department has produced 69 Ph.Ds as on date. At present 21 full time research scholars are working for Ph.D and they supported by Basic Science Research Fellowship, Rajiv Gandhi National Fellowship, Maulana Azad National Fellowship, Inspire Fellowship and MAjor Research Project Fellowship. 2.4 Achievements of the Department The national board of higher mathematics under the ages of the department of atomic energy has been sanctioning annual library grant to the department from the year 1999. A sum of Rs. 4.5 lakhs has been sanctioned for the financial years 2014-15. In view of the research work done by the department during the past years, the university grants commission recognized and sanctioned UGC-SAP (DRS-I) support to the department for a period of five years (20072012). Having completed Phase-I, the department has been sanctioned UGC-SAP(DRS-II) for a period of five year from (2013-2018) and an amount of Rs. 57 lakhs has been allocated. The department has also been recognized by department of science and technology under FIST program for a period of five years (2014-2019) and an amount of Rs. 47 lakhs has been alloted. In 2012, the Tamil Nadu government recognized the department and ordered to upscale the department with high tech laboratories at a cost of Rs. 2.5 crores under various Heads. The department is functioning in a separate building from 12.6.2015. The department is taking special efforts for the success of students in NET/JRF/TNSET as on date 36 students of the department have passed these qualifying examinations for the post of appointment of assistant professor in colleges/ universities.

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2.5 Special Course The Department conducts endowment lectures by inviting national and international experts. But for the first time in the history of the University, Professor Albert F. Faessler from Bern University, Switzerland conducted a special course on algebraic introduction to cryptography for 10 days during 17-28 October 2016 for the II year M.Sc students to earn 2 extra credits.

Anyone can count the seeds in an apple, but no one can count the apples in a seed. Anonymous.

International Journal on Mathematical Combinatorics, Special Issue 1(2018)
