A configuration of pebbles on the vertices of a graph is solvable if one can place a pebble on any given root vertex via a sequence of pebbling steps. The pebbling number of a graph G is the minimum number π(G) so that every configuration of π(G) pebbles is solvable. A graph is Class 0 if its pebbling number equals its number of vertices. A function is a pebbling threshold for a sequence of graphs if a randomly chosen configuration of asymptotically more pebbles is almost surely solvable, while one of asymptotically fewer pebbles is almost surely not. Here we prove that graphs on n ≥ 9 vertices having minimum degree at least ⌊n/2⌋ are Class 0, as are bipartite graphs with m ≥ 336 vertices in each part having minimum degree at least ⌊m/2⌋ + 1. Both bounds are best possible. In addition, we prove that the pebbling threshold of graphs with minimum degree δ, with √n ≪ δ, is O(n3/2/δ), which is tight when δ is proportional to n.
|Original language||English (US)|
|Number of pages||8|
|Journal||Australasian Journal of Combinatorics|
|State||Published - Dec 1 2003|
ASJC Scopus subject areas
- Discrete Mathematics and Combinatorics