Extensions of Graph Pebbling

1. Introduction Deep Graphs Results: Cover Pebbling Random Cover Pebbling Extensions of Graph Pebbling 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. We found the exact value of the cover pebbling number for complete multipartite graphs and wheel graphs. For n ≥3, γ(Wn) = 4n - 5 = 4v - 9. • 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 Kn, the complete graph on n vertices, is cover solvable? • Theorem: (Maxwell Boltzmann distribution)Set A0 = 1.5238…, where A0 is the solution of A - .5e2A= 3/2 . Then • t = A0n+φ(n)n½ → P(Knis cover solvable ) → 1 as n → ∞. • and • t = A0n - φ(n)n½ → P(Knis 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 γequal the golden ratio, (1+51/2) / 2. Then • t = γn + φ(n)n½ → P(Knis cover solvable ) → 1 as n → ∞. • and • t = γn - φ(n)n½ → P(Knis cover solvable ) → 0 as n → ∞. • where φ(n) → ∞ is arbitrary. A graph isdeep 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 K4, the complete graph on 4 vertices. This is W6, the wheel graph with six outer vertices. Here γ(W6) = 4(6) – 5 = 19. Theorem: Let G be a complete multipartite graph with partition sizes of s1≥ s2, ≥ … ≥ sr Then γ(G) = 4s1 + 2s2 + … + 2sr -3. Essentially, this bound occurs from stacking all the γ(G) pebbles on one vertex in the partition with s1 vertices. Theorem: Let G be a graph of order n and diameter d, and let C be a configuration on G containing at least 2d(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. 0 1 1 3 This is a sequence of subgraphs of K4 whose pebbling numbers are 4, 3, 2, and 1 respectively. Thus, K4 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. Here is an example of a pebbling move. What is pebbling? • 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 Kn, the complete graph on n vertices is n. First, notice that, π(Kn) > 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(Kn) = n. Cover Pebbling Complexity Carl R. Yerger*, Prof. Francis Edward Su (Advisor), Prof. Anant P. Godbole (Second Reader) 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 sielements using groups of four elements with no overlaps. References Domination Cover Pebbling Cover Pebbling [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. A set of vertices in a graph is a dominating setof 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: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. The following graph corresponds to the exact cover by four 4-sets problem, a1 = {s1, s2, s3, s4}, a2 = {s3, s4, s5, s6}, a3 = {s5, s6, s7, s8}. 0 0 0 0 0 0 0 0 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) ≤ 2d – 2 (n – 2) + 1. a3 a1 a2 9 9 9 1 1 1 0 9 7 2 1 5 • *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/thesis 1 1 1 0 2 0 0 0 0 0 0 1 1 3 3 1 3 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!). 0 1 0 0 1 1