Extensions of Graph Pebbling
Carl R. Yerger*, Prof. Francis Edward Su (Advisor), Prof. Anant P. Godbole (Second Reader)
Introduction
Results: Cover 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.
3
1
0
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.
1
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
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.
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.
Cover Pebbling Complexity
Theorem: The cover pebbling decision problem is NPcomplete! 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 si elements using
groups of four elements with no overlaps.
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
a1
9
9
1
9
0
7
1
5
0
0
0 0
0
a2
0
9
0
a3
1
2
0
3
0
3
0
3
0
1 1
1 1
1
1
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(Kn is cover solvable ) → 1 as n → ∞.
and
t = A0n - φ(n)n½ → P(Kn 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+51/2) / 2. Then
t = γn + φ(n)n½ → P(Kn is cover solvable ) → 1 as n → ∞.
and
t = γn - φ(n)n½ → P(Kn is cover solvable ) → 0 as n → ∞.
where φ(n) → ∞ is arbitrary.
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?
0
2
1
• 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?
The domination cover pebbling number, ψ(G), of this graph
is 3.
Deep Graphs
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 K4, the complete graph on 4 vertices.
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.
References
[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.
1
1
1
0
0
Random Cover Pebbling
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!).
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.
*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 DMS0139286.
Thanks to Akemi Kashiwada and the HMC Math Department for this
poster template.
For more information contact [email protected]
A copy of my thesis can be found at www.math.hmc.edu/~cyerger/thesis