Global Journal of Pure and Applied . ISSN 0973-1768 Volume 13, Number 6 (2017), pp. 1835–1843 © Research India Publications http://www.ripublication.com/gjpam.htm

Pebbling in Flower 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 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 Jn for n ≥ 5.

AMS subject classification: 05C99, 05C35. Keywords: Pebbling, , 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, 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- graph [12], split graph [11]. Cubic graphs are natural models for real system and isa3-. Issacs [10] found an infinite family of snarks. A snark is a simple graph with 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

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

(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. Therefore pebbles are extracted from S3 and transmitted to S2 inorder to pebble the target vertex in S1. Each vertex bi ∈ S2 is adjacent to ai ∈ S1 and bij ∈ S3 for 0 ≤ i

Case 1.2.3: p3 ≥ 2 In this case, the target is pebbled by extracting pebbles from S3. We eliminate the Pebbling in Flower Snark Graph 1839

pebbling move through the transmitting path {b0j ,b0,a0} for j = 1, 2 as it is trivial to pebble the target vertex and this case is identical to Case 1.2.2. The set S1 was initially distributed with p1 = 1 and in this distribution instead of placing one pebble on all the vertices of S1 it is sufficient to have any one arbitrary vertex say ai ∈ S1 with p(ai) = 1. Now this vertex ai ∈ S1 can initiate the pebbling move within S1 so that the target can be pebbled. Hence with only one pebble on any vertex of S1 and minimum two pebbles on all the 2(n − 1) vertices of S3 pebbling the target vertex from any arbitrary vertex is possible. This implies p(1) + p(3) ≥ 4n − 1 pebbles are required.

The pebbling number of Jn including the transmitting path {b0j ,b0,a0} for j = 1 and 2 will be p(1) + p(3) ≥ 4n + 1.

Case 2: b0 ∈ S2 is the target vertex. Without loss of generality, assume that p(a0) = 1 and p(b0j ) = 1 for j = 1, 2.

Case 2.1: p2 ≥ 4 Consider the case, when pebbles are to be transmitted from any arbitrary vertex bi ∈ S2 to the target vertex b0 ∈ S2. The vertex bi for 0 ≤ i ≤ n − 1 and b0 are non adjacent vertices. In order to pebble the target vertex, the transmitting path consists of vertices either from S1 or S3 which is discussed in the following subcases.

Case 2.1.1: Extraction of pebbles from S1 Two pebbles can be moved to the vertices of S1 as it is assumed that p2 ≥ 4. Thus, p1 ≥ 2 allows two pebbles moved to a0 ∈ S1 through the transmitting path {bi,ai,ai−1,ai−2 ...a0} or {bi,ai,ai+1,ai+2 ...an−1,a0},1≤ i ≤ n − 1 and hence the target b0 ∈ S2 is peb- bleable.

Case 2.1.2: Extraction of pebbles from S3 With p2 ≥ 4, one pebble is moved to every bij ∈ S3,1≤ i ≤ n − 1 for j = 1, 2. Considering p3 ≥ 1, the pebbling move within S3 is not possible. Instead, consider the movement of pebbles from p2 ≥ 4 to every vertex of bi1 ∈ S3 for 1 ≤ i ≤ n−1 such that p(bi1) = 2. The transmitting path will be {bi1,b(i−1)1,b(i−2)1 ...b01},1≤ i ≤ n − 1. Hence p(b01) = 1. As there is already one pebble on b0j by the initial assumption, we have p(b01) = 2. Thus the target b0 ∈ S2 is pebbleable. In the above mentioned cases of 2.1, we require minimum four pebbles on (n − 1) (2) (1) vertices of S2 and one pebbles on a0. Thus p + p ≥ 4n − 3 pebbles.

Case 2.2: p2 ≤ 3 As discussed, pebbling move within the set S2 is not possible and p2 with either two or three pebbles can transmit only one pebble to the vertices in S1 and S3. Due to insufficient number of pebbles to trigger the pebbling move we need to extract some pebbles from S1 or S3 which is discussed in the following subcases. 1840 Chris Monica M. and Sagaya Suganya A.

Case 2.2.1: p1 ≥ 4 With p1 ≥ 4, two pebbles are moved to every vertex in S2. Now consider the movement of pebbles to S3 in particular to the vertex bi1 ∈ S3. With p3 ≥ 1,1≤ i ≤ n − 1, p(bi1) = 2 will initiate the pebbling move in S3 so that two pebbles are moved to bi0. Hence one pebble is moved to the target b0 ∈ S2. In this case, we require four pebbles on n vertices of S1, three pebbles on (n − 1) vertices of S2 and minimum one pebble on (1) (2) (3) 2n vertices of S3. Thus p + p + p ≥ 4n + 3(n − 1) + 2n = 9n − 3.

Case 2.2.2: p1 ≥ 2 With p1 ≥ 2 , two pebbles are moved to a0 ∈ S1. To reach the target vertex b0 the trans- mitting path will be {ai,ai−1,ai−2 ...a0,b0},1≤ i ≤ n − 1 . The number of pebbles required to pebble the target vertex is p(1) + p(2) ≥ 4n − 3.

Case 2.2.3: p1 ≤ 1 As there are lack of pebbles in S1, consider pebbling the target vertex by extracting pebbles from S3. Now assume p3 ≥ 2, so that any vertex bij ∈ S3, j = 1or2 will trigger the pebbling move and the transmitting path to pebble the target vertex will be {bij ,b(i−1)j ,b(i−2)j ...b0j ,b0} for some j = 1or2,1≤ i ≤ n − 1. Thus summarizing there are atmost one pebble on S1, three pebbles on S2 and minimum two pebbles over the vertices of S3, the number of pebbles required in this case is p(1) + p(2) + p(3) ≥ (n − 1) + 3(n − 1) + 2(2(n − 1)) + 2 = 8n − 8.

Case 2.2.4: p3 ≥ 2 The transmitting path for pebbling the target vertex will be initiated from S3 with p3 ≥ 2, so that at least two pebbles can be moved to S2. Further, with the initial number of pebbles in S2 one pebble is transmitted to S1. Now assume that there exists at least one vertex ai ∈ S1 with p(ai) = 1 which can initiate the pebbling move in S1 so that one pebble is moved to a0 ∈ S1. Initially it was assumed that p(a0) = 1 . Now, after transmission of pebbles there are two pebbles on a0. Hence the target is pebbled through the transmitting path {bij ,bi,ai,ai−1,ai−2 ...a0,b0} or {bij ,bi,ai,ai+1,ai+2 ...an−1,a0,b0},1≤ i ≤ n−1. Thus there are atmost one pebble on S1, three pebbles on (n−1) vertices of S2 and (1) (2) (3) minimum two pebbles on 2(n − 1) vertices of S3, hence we require p + p + p ≥ 1 + 3(n − 1) + 2(2(n − 1)) = 7n − 6 pebbles to move a pebble to the target. The possibility of pebbling the target vertex in this case is same on excluding the distribution over S2. The pebbling number of Jn on including the the transmitting path (1) (3) {b0j ,b0,a0} for j = 1, 2 will be p + p ≥ 4n + 1.

Case 3: Let b0j ∈ S3 be the target vertex for 1 ≤ j ≤ 2. Let p(a0)<4 and p(b0)<2, otherwise the solution is trivial.

Case 3.1: p3 ≥ 1 S3 forms a 2n - cycle. With p3 ≥ 1, there should exists some bij ∈ S3 for j = 1or2, 1 ≤ i ≤ n − 1 such that p(bij ) ≥ 2. Since p(bij ) ≥ 2, one pebble can be moved to the Pebbling in Flower Snark Graph 1841

(3) target vertex through a sequence of pebbling move in S3. Thus we require p ≥ 2n pebbles.

Case 3.2: p3 < 1 Due to lack of pebbles in S3 the following subcases are discussed by retrieving pebbles from S1 and S2.

Case 3.2.1: p2 ≥ 8 With p2 ≥ 8, two pebbles are transmitted to every vertex of S3 and so the target is pebbled through the transmitting path {bi,bij ,b(i−1)j ,b(i−2)j ...b0j },1 ≤ i ≤ n − 1. The number of pebbles required is p(2) ≥ 8n − 7.

Case 3.2.2: p2 < 8 Pebbles are extracted from S1 and transmitted to S2, in order to move a pebble to b0j ∈ S3,1≤ j ≤ 2. Now assume, p1 ≥ 8. Then four pebbles can be moved to every vertex of S2. Thus from S2 with p2 < 8 minimum two pebbles can be transmitted to bij ∈ S3. Therefore the target is pebbled through the transmitting path {ai,bi,bij ,b(i−1)j ,b(i−2)j ...b0j },1≤ i ≤ n − 1. In this case, we require atmost seven pebbles on (n − 1) vertices of S2 and minimum eight pebbles on (n − 1) vertices of S1, thus the number of pebbles required is p(2) +p(1) ≥ 7(n−1)+8(n−1)+2 = 15n−13.

Case 3.2.3: p2 = 0 In this case we discuss the possibility of pebbling move from S1 to S3. Assume p1 ≥ 4, then after a sequence of pebbling move p(a0) ≥ 4. Hence the target is pebbled through the tranmitting path {ai,ai−1,ai−2 ...a0,b0,b0j }. Thus the number of pebbles required is p(1) ≥ 4n − 2.

In view of a result by Chung [3], f(Jn) cannot be less than 4n. Hence f(Jn) ≥ 4n. As all the possibilities of pebbling the target vertex in the sets S1, S2 and S3 are discussed we conclude the proof of the theorem by considering the least possiblity of pebbling from the three cases. The least possibilty of {2n − 4, 3n − 3, 6n − 6, 4n − 1, 4n + 1} and {4n − 3, 8n − 8, 4n + 1, 9n − 3} in Case 1 and Case 2 respectively is 4n + 1 and the least possibility of {2n, 8n − 7, 15n − 3, 4n − 2} in Case 3 is 8n − 7.

Thus we conclude that f(Jn) = 4n + 1 gives us the optimized value of pebbling any target vertex from any arbitrarily chosen vertex and so f(Jn) = 4n + 1. 

4. Conclusion and Open Problem

In this paper, we have obtained the pebbling number of an n-dimensional Flower Snark Jn for n ≥ 5. Determining the pebbling number of other variants of infinite family of snarks such as Double star snark, Goldberg Snark, Watkins Snark [1] is left open for the readers. 1842 Chris Monica M. and Sagaya Suganya A.

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