1. Extensions of Graph Pebbling Carl R. Yerger*, Prof. Francis Edward Su (Advisor), Prof. Anant P. Godbole (Second Reader) Introduction Given a graph, G, assign non-negative integers to the vertices. These integers can be viewed as pebbles. A pebbling move takes two pebbles from some vertex and places one pebble on an adjacent vertex. In my thesis, I examined various problems that are derived from making pebbling moves. My research involved studying three main areas: deep graphs, cover pebbling and domination cover pebbling. Deep Graphs The domination cover pebbling number, ψ(G), of this graph is 3. The exact values for the domination cover pebbling number for path graphs, cycle graphs and complete binary trees were determined. Theorem: For all graphs G of order n with maximum diameter 2, ψ(G) ≤ n - 1. Theorem: Let G be a graph of diameter d ≥ 3 and order n. Then ψ(G) ≤ 2 d – 2 (n – 2) + 1. References Cover Pebbling: Instead of pebbling only one vertex, we must now simultaneously place a pebble on EVERY vertex. Example: What is the cover pebbling number of this graph? The answer is 9. It turns out that initially stacking all the pebbles on one vertex is always the worst case. Below is an example of a cover solution, a series of pebbling moves that forces each vertex to contain at least one pebble. What is pebbling? 1 A pebble can be moved to a root vertex v if given a sequence of pebbling moves it is possible to place one pebble on v. Define the pebbling number, π(G) to be the minimum number of pebbles required so that for any initial distribution of pebbles, it is possible to move to any root vertex v in G. As a simple example, the pebbling number of K n, the complete graph on n vertices is n. First, notice that, π(K n ) > n-1 since placing one pebble on all but one vertex means that the unpebbled last vertex cannot be reached. However, given n pebbles, we are forced to either put a pebble on every vertex, or place a pair of pebbles on a vertex, thus allowing every vertex to be pebbled in a pebbling move. So π(K n ) = n. 0 *Some of this research was completed at the 2004 East Tennessee State University REU. Collaborators on various topics include Nathaniel Watson, Anant Godbole, James Gardner, Alberto Teguia and Annalies Vuong. The REU was supported by NSF Grant DMS- 0139286. Thanks to Akemi Kashiwada and the HMC Math Department for this poster template. For more information contact cryerger@gmail.com A copy of my thesis can be found at www.math.hmc.edu/~cyerger/thesiswww.math.hmc.edu/~cyerger/thesis Cover Pebbling 3 Here is an example of a pebbling move. 1 Results: Cover Pebbling Random Cover Pebbling We found the exact value of the cover pebbling number for complete multipartite graphs and wheel graphs. For n ≥ 3, γ(W n ) = 4n - 5 = 4v - 9. 79 0 0 0 3 0 0 1 5 0 2 0 0 3 0 3 0 1 1 1 1 1 1 This is W 6, the wheel graph with six outer vertices. Here γ(W 6 ) = 4(6) – 5 = 19. Theorem: Let G be a complete multipartite graph with partition sizes of s 1 ≥ s 2, ≥ … ≥ s r Then γ(G) = 4s 1 + 2s 2 + … + 2s r -3. Essentially, this bound occurs from stacking all the γ(G) pebbles on one vertex in the partition with s 1 vertices. Theorem: Let G be a graph of order n and diameter d, and let C be a configuration on G containing at least 2 d (n - d + 1) – 1 pebbles. Then G is cover solvable. Recall that the diameter of a graph is the maximum shortest distance between two vertices. Domination Cover Pebbling A set of vertices in a graph is a dominating set of vertices if every vertex of G is either in the dominating set or adjacent to at least one vertex in the dominating set. Domination Cover Pebbling: Now a pebble must be placed on a dominating set of vertices. What is the domination cover pebbling number of this graph? Cover Pebbling Complexity Theorem: The cover pebbling decision problem is NP- complete! In other words, given a graph G and a configuration of vertices C, the question of whether there exists a sequence of pebbling moves that allows for a cover solution of G to exist is NP-complete. We can translate the perfect cover by 4-sets problem into the cover pebbling decision problem. This problem asks whether it is possible to pick all the s i elements using groups of four elements with no overlaps. The following graph corresponds to the exact cover by four 4-sets problem, a 1 = {s 1, s 2, s 3, s 4 }, a 2 = {s 3, s 4, s 5, s 6 }, a 3 = {s 5, s 6, s 7, s 8 }. A solution to the perfect cover by 4-sets problem described in the example above exists if and only if there is a cover solution to this graph (which there is!). 1 2 0 1 1 11 1 9 9 9 0 0 0 0 0 0 0 0 Given a connected graph G, distribute t pebbles on its vertices in some configuration. If the pebbles are indistinguishable, this is called Bose Einstein cover pebbling. However, if the pebbles are distinct, we shall refer to our process as Maxwell Boltzmann cover pebbling. Question: For each distribution, what is the probability that K n, the complete graph on n vertices, is cover solvable? Theorem: (Maxwell Boltzmann distribution) Set A 0 = 1.5238…, where A 0 is the solution of A -.5e 2A = 3/2. Then t = A 0 n+φ(n)n ½ → P(K n is cover solvable ) → 1 as n → ∞. and t = A 0 n - φ(n)n ½ → P(K n is cover solvable ) → 0 as n → ∞. where φ(n) → ∞ is arbitrary. More surprisingly, for Bose Einstein cover pebbling, we find the golden ratio is the special first-order ratio that is the “threshold” for cover solvability. Theorem: (Bose Einstein Distribution) Let γ equal the golden ratio, (1+5 1/2 ) / 2. Then t = γn + φ(n)n ½ → P(K n is cover solvable ) → 1 as n → ∞. and t = γn - φ(n)n ½ → P(K n is cover solvable ) → 0 as n → ∞. where φ(n) → ∞ is arbitrary. A graph is deep if for each positive integer n ≤ π(G), there exists an induced subgraph H of G such that π(H) = n. Example: The family of complete graphs are all deep. For instance, consider K 4, the complete graph on 4 vertices. This is a sequence of subgraphs of K 4 whose pebbling numbers are 4, 3, 2, and 1 respectively. Thus, K 4 is a deep graph. Notation: Let G(n, p) be a random graph on n vertices where each edge is placed independently with probability p. Theorem: The probability that G(n, p) is a deep graph, where p = 1/ (log log n), approaches 1 almost surely as n → ∞. In fact, G(n, p) is also almost surely a class 0 graph, that is, a graph whose pebbling number is equal to n. Other structural results about deep and class 0 graphs were also shown. [1] B. Crull, T. Cundiff, P. Feltman, G. H. Hurlbert, L. Pudwell, Z. Szaniszlo, Z. Tuza, The cover pebbling number of graphs, (2004), submitted. [2]A. Godbole, N. Watson, C. Yerger, Threshold and complexity bounds for the cover pebbling game, (2004), in preparation. [3] A. J. Hetzel, G. Isaksson, Deep graphs, (2004), submitted. [4] G. Hurlbert, A survey of graph pebbling, Congressus Numerantium 139 (1999), 41-64. [5] G. Hurlbert, B. Munyan, The cover pebbling number of hypercubes, (2004), in preparation. [6] N. Watson, C. Yerger, Cover pebbling numbers and bounds for certain families of graphs, (2004), submitted. a3a3 a2a2 a1a1