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Pebbling in Flower Snark Graph Global Journal of Pure and Applied Mathematics. ISSN 0973-1768 Volume 13, Number 6 (2017), pp. 1835–1843 © Research India Publications http://www.ripublication.com/gjpam.htm Pebbling in Flower Snark Graph Chris Monica M. Department of Mathematics, Loyola College, Chennai, India. Sagaya Suganya A. Department of Mathematics and Actuarial Science, B. S. Abdur Rahman Crescent University, Chennai, India. Abstract A pebbling move on a connected graph G = (V, E), is the removal of two pebbles from one vertex and placing one pebble on its adjacent vertex. The pebbling number f (G) is the least number of pebbles required in moving one pebble to an arbitrary vertex by a sequence of pebbling moves. In this paper, we have determined the pebbling number of an n-dimensional Flower Snark Jn for n ≥ 5. AMS subject classification: 05C99, 05C35. Keywords: Pebbling, Cubic Graph, Flower Snark. 1. Introduction Combinatorics is the study of arrangement of objects according to specified rules, and combinatorial techniques are needed to count, enumerate or represent possible solutions. Many complex problems in real world needs to find an optimal solution in a finite solution space. Although, certain practical problems such as finding shortest or cheapest round trips, internet data packet routing, planning, scheduling and time tabling which appears to be NP - complete, the literature has a vast number of problems which could be solved in polynomial time. To accomplish this, there are certain combinatorial games available such as pebbling, peg solitaire, chip firing and checker jumping [2]. Graph pebbling is a network optimization model for the transportation of resources that are consumed in the transit. For instant, if we consider electricity, heat, energy or communication of information there may be some loss as it moves from one location to 1836 Chris Monica M. and Sagaya Suganya A. another. The pebbling steps analyze the cost in loss of pebbles and it has been the subject of deep and extensive research in the context of proving lower bounds for computation on graphs [17]. In 1956, Erdos [19] initiated the study of zero sum sequences. On the subject of this study, Lemke and Klietman [20] proved the conjecture of Erdos. It was Lagarias and Saks who suggested graph pebbling as a tool for solving this number theoretical conjecture [9]. Chung successfully used this concept of pebbling and introduced graph pebbling into the literature where he obtained the pebbling number of a hypercube [3]. The concept of pebbling has applications in reduction of memory traffic in computers, register allocation problem and transportation of resources that are consumed in the transit [16, 17]. For a connected graph G, a pebbling configuration is the distribution of pebbles on the vertices of G. A pebbling move consists of removing two pebbles from a vertex and placing one pebble on the adjacent vertex. We say, one pebble is moved to any arbitrarily chosen target vertex say v, if one can repeatedly apply pebbling move so that in the resulting distribution v has at least one pebble. The pebbling number f (G) is the minimum number of pebbles that ensures that every vertex of the graph G can be pebbled, regardless of the initial configuration of pebbles. In case, one pebble is placed on all the vertices of the graph G except the target vertex then there is no pebbling move which means that f (G) ≥ n(G), where n(G) is the number of vertices of G [3] . For w, v ∈ V (G),ifw is at distance d from v and 2d − 1 pebbles are placed on w, then no pebble is moved to v which leads to the fact that f (G) ≥ 2d , where d is the diameter of G. Thus, we can say that f (G) ≥ max{n(G), 2d } [3]. A transmitting subgraph of a graph G is a path v0,v1,v2 ...vn in which one pebble is transmitted from v0 to vn with the distribution of at least two pebbles in v0 and at least one pebble on each of the other vertices in the path, except possibly vn. With this distribution of pebbles one can transmit a pebble from v0 to vn [8]. Graphs with pebbling number equal to the number of vertices are classified as class 0 graphs. Otherwise, graph G is class i if f (G) = n(G) + i [4]. There are some known graphs for which the pebbling number is computed such as path Pn on n vertices, complete graph Kn, hypercube Qn [3], product graph C5 ×C5 [8], fan graph Fn and wheel graph Wn [6], complete graph Ks,t [5], graphs with diameter 2 [4], cycle [15]. All these graphs are classified as class 0 graphs. Further, there are also results pertaining to graphs which does not fall under class 0 such as Jahangir graph J2,m (m ≥ 8) [13], square of paths [15], n-star graph [12], split graph [11]. Cubic graphs are natural models for real system and isa3-regular graph. Issacs [10] found an infinite family of snarks. A snark is a simple graph with girth at least 5, and cyclically 4-edge-connected and is a class of cubic graph. In the following section, we have obtained the pebbling number of a particular family of snark namely Flower Snark of dimension n which is denoted by Jn, for n ≥ 5. 2. Flower Snark Graph Let n ≥ 5 be an odd integer. A flower snark Jn is such a graph that its vertex set is {ai,bi,bij } for i = 0, 1, 2 ...n− 1, j = 1, 2 and edge set is {biai,bibi1,bibi2 : i = Pebbling in Flower Snark Graph 1837 0, 1, 2 ...n− 1}∪{aiai+1,bi1b(i+1)1,bi2b(i+1)2 : i = 0, 1, 2 ...n− 1}∪{b(n−1)2b01, b02b(n−1)1} [10]. The number of vertices and edges of an n - Flower snark are 4n and 6n respectively. See Figure 1. Figure 1: Flower Snark Graph J9 3. Pebbling in Flower Snark Theorem 3.1. For a Flower Snark Jn, f(Jn) = 4n + 1, n ≥ 5. Proof. The vertex set of Jn is partitioned into three disjoint subsets namely S1,S2 and S3 where the subsets are referred as S1 ={a0,a1 ...an−1}, S2 ={b0,b1 ...bn−1} and S3 ={b01,b02,b11,b12 ...b(n−1)1,b(n−1)2}. Each bi ∈ S2 is adjacent to ai ∈ S1 and bij ∈ S3 for all 0 ≤ i<nand j = 1, 2. (i) Let pi, p(xi) and p , i = 1, 2, 3 denote the number of pebbles distributed over each vertex of Si, number of pebbles initially placed on a particular vertex xi,0≤ i<nand the total number of pebbles on the set Si respectively. We consider three cases by fixing the target vertex in the sets S1, S2 and S3. Case 1: Let a0 ∈ S1 be the target vertex. If p(a1) = 2orp(an−1) = 2, then the proof is trivial. Therefore, we assume that p(a1)<2 and p(an−1)<2. Case 1.1: p1 > 1 In this case, assume that every vertex in S1 is distributed with at least two pebbles. Then there exists i, such that p(ai) ≥ 2, 2 ≤ i<n− 2 will initiate the pebbling move. The transmitting subgraph in this case would be {ai,ai−1 ...a1,a0}. On placing two pebbles on (n − 3) vertices of S1 and in particular at least one pebble on a1 and an−1, the number 1838 Chris Monica M. and Sagaya Suganya A. (1) of pebbles required to pebble the target vertex a0 is p ≥ 2n − 4. The target is also pebbled by placing 2d pebbles on a single vertex, where d is the diameter to the target vertex. Since S1 forms an n-cycle which can be either even or odd, = d the required number of pebbles in S1 to pebble the target vertex x0 is, f(C2n) 2 and 2d+1 f(C + ) = 2 + 1, where d is the diameter of the cycle [15]. Therefore 2n 1 3 ⎧ d ⎨ 2 if n is even p(1) ≥ 2d+1 ⎩ 2 + 1ifn is odd 3 Case 1.2: p1 ≤ 1 In this case, the set S1 has inadequate number of pebbles to start a pebbling move. In order to pebble the target a0, pebbles are extracted from either S2 or S3 which is discussed in the following subcases. Throughout the proof of the subcases, assume that p(b0) = 0 otherwise the solution is trivial. Case 1.2.1: p2 ≥ 2 Each ai is adjacent to bi for i ∈{0, 1, 2 ...n− 1}. Considering p(b0) = 0, for p2 ≥ 2, one pebble is moved to S1 so that p1 ≥ 2. After this move, we can pebble the target vertex as discussed in Case 1.1. Suppose p1 = 0, then there will be no vertex in S1 to trigger the pebbling move. Hence to initiate a pebbling move there should exists some i, for 1 ≤ i ≤ n − 1 such that p(bi) ≥ 4 which allows two pebbles moved to ai ∈ S1 and {bi,ai,ai−1,ai−2 ...a0} or {bi,ai,ai+1,ai+2 ...an−1,a0} ,1≤ i ≤ n − 1 forms a transmitting subgraph. There are atmost one pebble on (n − 1) vertices of S1 and minimum two pebbles on (n − 1) vertices of S2. Thus the number of pebbles required for pebbling in this case is p(1) + p(2) ≥ 3n − 3. Case 1.2.2: p2 ≤ 1 As p2 ≤ 1, S2 has inadequate number of pebbles to move one pebble to some vertex in S1.
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