A Linear Optimization Technique for Graph Pebbling
Abstract
Graph pebbling is a network model for studying whether or not a given supply of discrete pebbles can satisfy a given demand via pebbling moves. A pebbling move across an edge of a graph takes two pebbles from one endpoint and places one pebble at the other endpoint; the other pebble is lost in transit as a toll. It has been shown that deciding whether a supply can meet a demand on a graph is NPcomplete. The pebbling number of a graph is the smallest such that every supply of pebbles can satisfy every demand of one pebble. Deciding if the pebbling number is at most is complete.
In this paper we develop a tool, called the Weight Function Lemma, for computing upper bounds and sometimes exact values for pebbling numbers with the assistance of linear optimization. With this tool we are able to calculate the pebbling numbers of much larger graphs than in previous algorithms, and much more quickly as well. We also obtain results for many families of graphs, in many cases by hand, with much simpler and remarkably shorter proofs than given in previously existing arguments (certificates typically of size at most the number of vertices times the maximum degree), especially for highly symmetric graphs.
Here we apply the Weight Function Lemma to several specific graphs, including the Petersen, Lemke, weak Bruhat, Lemke squared, and two random graphs, as well as to a number of infinite families of graphs, such as trees, cycles, graph powers of cycles, cubes, and some generalized Petersen and Coxeter graphs. This partly answers a question of Pachter, et al., by computing the pebbling exponent of cycles to within an asymptotically small range. It is conceivable that this method yields an approximation algorithm for graph pebbling.
1 Introduction
Graph pebbling is like a number of network models, including network flow, transportation, and supply chain, in that one must move some commodity from a set of sources to a set of sinks optimally according to certain constraints. Network flow constraints restrict flow along edges and conserve flow through vertices, and the goal is to maximize the amount of commodity reaching the sinks. The transportation model includes per unit costs along edges and aims to minimize the total cost of shipments that satisfy the source supplies and sink demands. At its simplest, the supply chain model ignores transportation costs while seeking to satisfy demands with minimum inventory. The graph pebbling model introduced by Chung [7] also tries to meet demands with minimum inventory, but constrains movement across an edge by the loss of the commodity itself, much like an oil tanker using up the fuel it transports, not unlike heat or other energy dissipating during transfer.
Specifically, a configuration of pebbles on the vertices of a connected graph is a function (the nonnegative integers), so that counts the number of pebbles placed on the vertex . We write for the size of ; i.e. the number of pebbles in the configuration. A pebbling step from a vertex to one of its neighbors reduces by two and increases by one (so that one can think of it as moving one pebble at the cost of another as toll). Given two configurations and we say that is solvable if some sequence of pebbling steps converts to . In this paper we study the traditional case in which the target distribution consists of a single pebble at some root vertex (one can peruse [14, 15, 17] for a wide array of variations on this theme). We are concerned with determining , the minimum number of pebbles so that every configuration of size is solvable. Then the pebbling number of equals . Alternatively, is one more than the maximum such that there is some root and some size configuration so that does not solve . The primary focus of this paper is to exploit this duality with newly discovered algebraic constraints.
1.1 Calculating Pebbling Numbers
Given a graph , configuration , and root , one can ask how difficult it is to determine if solves . In [18] it was determined that this problem is NPhard. Subsequently, [19, 22] proved that the problem is NPcomplete, with [19] showing further that answering the question “is ?” is complete (and hence both NPhard and coNPhard, and therefore in neither NP nor coNP unless NP coNP). Finding classes of graphs on which we can answer more quickly is therefore relevent, and there is some evidence that one can be successful in this direction. Besides what we share in this introduction, we show later that many graphs can have very short certificates that .
The unsolvable configuration with one pebble on every vertex other than the root shows that , where denotes the number of vertices of . In [20] it is proved that graphs of diameter two satisfy , with a characterization separating the two classes (Class 0 means and Class 1 means ) given in [5, 8]. One of the consequences of this is that 3connected diameter two graphs are Class 0. As an extension it is proved in [11] that connected diameter graphs are also Class 0, and they use this result to show that almost every graph with significantly more than (for any fixed ) edges is Class 0. Consequently, it is a very (asymptotically) small collection of graphs that cause all the problems.
Knowing the pebbling number of a graph and actually solving a particular configuration are two different things, as even a configuration that is known to be solvable (say, one of size equal to the pebbling number) can be difficult to solve. Evidence that most configurations are not so difficult, though, comes in the following form. The work of [2] shows that every infinite graph sequence has a pebbling threshold , which yields the property that almost every configuration on of size is solvable (and almost every configuration of size is not). In papers such as [3, 9, 10] we find that is significantly smaller than — for example, as opposed to for the complete graph , and roughly as opposed to for the path . Moreover, the proof techniques reveal that almost all of these solvable configurations can be solved greedily, meaning that every pebbling step reduces the distance of the pebble to the root. So the hardness of the problem stems from a rare collection of configurations.
With these results as backdrop, [1] presents a polynomial algorithm for determining the solvability of a configuration on diameter two graphs of connectivity some fixed . Furthermore, [21] contains an algorithm that calculates pebbling numbers, and is able to complete the task for every graph on at most 9 vertices. Also, the proof in [7] that the dimensional cube is of Class 0 is a polynomial algorithm (actually bounded by its number of edges ). Along these lines, our main objective is to develop algorithmic tools that will in a reasonable amount of time yield good upper bounds on for much larger graphs, and in particular decide in some cases whether or not a graph is of Class 0.
This latter determination is motivated most by the following conjecture of Graham in [7]. For graphs and , let denote the Cartesian product whose vertices are , with edges whenever in and whenever in .
Conjecture 1
(Graham) Every pair of graphs and satisfy .
The conjecture has been verified for many graphs; see [13] for the most recent work. However, as noted in [16], there is good reason to suspect that might be a counterexample to this conjecture, if one exists, where is the Lemke graph of Figure 1. Since is Class 0, Graham’s conjecture requires that is also, but it is a formiddable challenge to compute the pebbling number of a graph on 64 vertices. One hopes that graph structure and symmetry will be of use, but purely graphical methods have failed to date. The methods of this paper represent the first strides toward the computational resolution of the question^{1}^{1}1Yes, I’ll pay if you beat me to it!, “Is ?”. Certainly, these methods alone will not suffice^{2}^{2}2We obtain evidence that in Theorem 10 — in fact, for one root we show ., but if they produce a decent upper bound then the methods of [21] might be able to finish the job.
1.2 Results
The main tool we develop is the Weight Function Lemma 2. This lemma allows us to define a (very large) integer linear optimization problem that yields an upper bound on the pebbling number. This has several important consequences, including the following.

The pebbling numbers of reasonably small graphs often can be computed easily. Moreover, it is frequently the case that the fractional relaxation suffices for the task, allowing the computation for somewhat larger graphs.

It is also common that only a small portion of the constraints are required, expanding the pool of computable graphs even more.^{3}^{3}3We present some findings along these lines in Section 3.2, with graphs on and vertices. One can restrict the types of constraints to greedy, bounded depth, and so on, with great success, seemingly because of the comments above. Potentially, this allows one to begin to catalog special classes of graphs such as Class 0, (semi)greedy, and treesolvable.

The dual solutions often yield very short certificates of the results, in most cases quadratic in the number of vertices, and usually at most the number of vertices times the degree of the root. These certificates are remarkably simple compared to the usual solvability arguments that chase pebbles all over the graph in a barrage of cases. One can sometimes find such certificates for infinite families of graphs by hand, without resorting to machine for more than the smallest one or two of its members. This was our approach in Section 3.3, for example.

Our method gives trivial proofs of

and , which we write as , and

In this paper we apply the Weight Function Lemma to several specific graphs, including the Petersen, Lemke, weak Bruhat, Lemke squared, and two random graphs, as well as to a number of infinite families of graphs, such as trees, cycles, graph powers of cycles, cubes, and some generalized Petersen and Coxeter graphs.
2 The Weight Function Lemma
Let be the path on vertices. Then is easily proved by induction. In particular, any configuration of at least pebbles solves . But one can say more about smaller solvable configurations as well, with the use of a weight function . Define on by , and extend the weight function to configurations by . Then a pebbling step can only preserve or decrease the weight of a configuration. Since the weight of a configuration with a pebble on is at least , we see that is a lower bound on every solvable configuration. In fact, induction shows that every unsolvable configuration has weight at most , which equals . That is, this inequality characterizes unsolvable configurations on . The Weight Function Lemma 2 generalizes this result on trees, and we explore the applications of the lemma in the following sections.
2.1 Linear Optimization
Let be a graph and be a subtree of rooted at vertex , with at least two vertices. For a vertex let denote the parent of ; i.e. the neighbor of that is one step closer to (we also say that is a child of ). We call a strategy when we associate with it a nonnegative, nonzero weight function with the property that and for every other vertex that is not a neighbor of (and for vertices not in ). Let be the configuration with , for all , and everywhere else. With this notation note that the path result above can be restated: is unsolvable if and only if and , where is the strategy with associated weight function .
Lemma 2
[Weight Function Lemma] Let be a strategy of rooted at , with associated weight function . Suppose that is an unsolvable configuration of pebbles on . Then .
Proof. By contrapositive and induction.
The base case is when is a path, which is proved above.
Suppose , let be a leaf of , and define to be the
path from to in , with being the subpath from to its
closest vertex on of degree at least 3 in (or if none exists).
Denote by the tree , and among all such unsolvable
configurations, choose to be the one having largest weight on .
The restriction of to witnesses that is a strategy
for the root , so induction requires that .
Likewise, the restriction of to witnesses that
is a strategy for the root .
Because and ,
we must have which by induction means that the
restriction of to solves (and so ).
Let be the resulting configuration after moving a pebble from
to .
Since is unsolvable, so is .
Now , but , which contradicts the initial
choice of .
For a graph and root vertex , let be the set of all strategies in , and denote by the optimal value of the integer linear optimization problem :
(1) 
We also let be the optimum of the relaxation, which allows configurations to be rational. We will find the relation useful at times. The following corollary is straightforward.
Corollary 3
Every graph and root satisfies .
Proof. By definition, the pebbling number is one more than the size of the
largest unsolvable configuration.
Until now, one could only use trees in an individual manner: for every spanning tree rooted at . The Weight Function Lemma allows one to consider all subtrees rooted at (not only spanning trees) simultaneously, which we will see is significantly more powerful. One strength of the method is that the relaxation frequently has an integer optimum. This means that the dual solution will point out which tree constraints certify the result, and because the dual problem has only constraints there are at most that many such trees in the certificate. Experience has shown, however, that usually one can find a certificate with only trees (or sometimes a few extra). We will see this behavior starting in Section 3.
2.2 Basic Applications
We begin with the pebbling number of trees, whose formula was first discovered and proved in [7]. View a tree with root as a directed graph with every edge directed toward . Then a path partition of is a set of edgedisjoint directed paths whose union is . One path partition majorizes another if its nonincreasing sequence of path lengths majorizes that of the other. A path partition is maximum if it majorizes all others. We can use Corollary 3 to give a new proof of the following result of [7].
Theorem 4
For a tree and root we have , where denotes the length (number of edges) of .
Proof. We begin by showing that a maximum size unsolvable configuration has pebbles on leaves only, and in fact on all leaves. Indeed, if has a pebble on the nonleaf , then we define a pushback of at to be any configuration obtained by removing the pebbles from , adding pebbles to one of the children of , and adding pebble to all other children of . Certainly, if is unsolvable and has no pebbles past (on the subtree of rooted at , minus itself), then the pushback will also be unsolvable, and thus satisfy the constraints of . It will also be larger than . The configuration that places pebbles on the leaf of the path is one possible result of pushing back the empty configuration, and so satisfies the constraints of . Hence .
For the upper bound we prove that is optimal by using induction to show that the optimal configuration has pebbles on the leaf of for every . This is true if , so suppose and let denote one of the paths in whose leaf has the highest weight in , with a tie going to one of the shortest length. This is to guarantee that the graph , where is the root of , is a tree (i.e. is connected). In order to maximize the number of pebbles that satisfy we would transfer as many pebbles from to other leaves as possible because their weights are at most and we could add extra pebbles when the weight is smaller. But by induction on we know from the constraint that each , with equality for all if and only if is maximum. Therefore, since for all , we have
which implies that .
A slight weakness of these tree constraints is that they do not classify unsolvable configurations on trees the way that they do on paths. This is because they let in a few solvable configurations. For example, consider the star on four vertices with one of its leaves as root . Then the configuration with pebbles on each of the other two leaves is solvable and satisfies all tree contraints. Since it is the average of the two unsolvable configurations that place either and or and on those other leaves, it cannot be cut out by the tree constraints that don’t cut out at least one of these two other constraints. In this case it doesn’t hurt us, since the strategy bound yields and the actual pebbling number is 5, but it can cause trouble on graphs in general. For example, we know that the 3cube in Figure 2 has pebbling number 8, so that the shown configuration is solvable (pebbles from to top must be split in two directions in its solution). However, no strategy recognizes its solution, and Corollary 3 yields only (the three rotations of the strategy in the center in Figure 2 certify this). One can see where the aforementioned star appears in the Figure 2 configuration on and is exploited accordingly: moving pebbles from the along one edge yields a configuration, while splitting the moves along two edges yields a configuration.
3 General Applications
In this section we illustrate the method more fully by presenting short proofs of both known and new results. We begin by relaxing strategies in the following way. We now use the term basic to describe the strategies as currently defined. A nonbasic strategy will satisfy the inequality in place of the equality used in a basic strategy (see Figure 2). The following lemma shows that we can use nonbasic strategies in an upper bound certificate since they are conic combinations of a nested family of basic strategies. Thus the use of nonbasic strategies can simplify and shorten certificates significantly.
Lemma 5
If is a nonbasic strategy for the rooted graph , then there exists basic strategies for and nonnegative constants so that .
Proof. We use induction, as the result is true when has two vertices
since is basic then.
Given , let be a basic strategy on the edge set of ,
define to be the largest constant for which , and
denote .
Then some vertex of satisfies , so has
fewer vertices than .
Also, because is basic, any vertex whose unique path
contains also satisfies , which means that
is connected, and hence a strategy.
Moreover, is nonbasic since every nonneighbor of has
.
By induction, is a conic combination of basic strategies,
and so therefore is .
We use conic combinations of strategies to derive, for some , the inequality for unsolvable configurations . From this we surmise that . Instead of writing our strategies algebraically, it will be somewhat easier to show them graphically. We will display them so as to derive for some sequence with , and let the reader divide by . In fact, in many instances we will derive for all , which makes for the following observation.
Lemma 6
[Uniform Covering Lemma]
Let be a set of strategies for the root of the graph .
If there is some such that, for each vertex , we have
, then .
3.1 Specific Graphs
It has been said in jest that every graph theory paper should contain the Petersen graph, so we get it out of the way first.
Theorem 7
Let denote the Petersen graph. Then .
Of course, the vertex lower bound implies , but since the focus of this paper regards upper bounds, we prove them only.
Proof. The strategies shown in Figure 3 certify the result.
Without such nice symmetry, the Lemke graph requires a different certificate for each possible root.
Theorem 8
Let denote the Lemke graph and suppose . Then .
Proof. We show the strategies for each root vertex in turn, below.
For :
For :
Without the edge , the resulting graph would view and symmetrically. Using that symmetry, the solutions for become solutions for .
For :
For we use the solutions from the case given by the appropriate symmetry.
For :
Because of the cubelike configuration with on , the best that our tree strategies can muster is . Thus, to show that is Class 0 one must handle by more traditional methods.
To illustrate that larger graphs can be tackled, we move on to one of order 24 that is not Class 0. Define the (weak) Bruhat graph of order (see Figure 4) to have all permutations of as vertices, with an edge between pairs of permutations that differ by an adjacent transposition. One can recognize it as the Cayley graph of , generated by adjacent transpositions, and also note that is the cubic Ramanujan (expander) graph of [6]. Intuitively, expander graphs would seem to have low pebbling numbers, but because has diameter 6 we have the lower bound . We give here a fairly tight bound.
Theorem 9
Let be the Bruhat graph of order 4. Then .
Proof. Because the graph is vertex transitive, only one root must be checked.
The strategies shown in Figure 4 certify the result.
We combined them into one figure, separated by edge styles.
Next we consider , the square of the Lemke graph . Because has diameter 3, has diameter 6, and so . Strategies deliver the following upper bounds. Since the bounds are not that tight, we do not pursue bounds for all vertices, although it is likely (since is the most problematic root in ) that the upper bound of works for all roots .
Theorem 10
Let be the square of the Lemke graph . Then
Proof. For one can verify that a quarter of the sum of the four strategies in Figure 5 yields the weights
giving the bound of . One verifies that each is a strategy by making sure that each nonzero entry has a corresponding entry in its column or row with at least twice its weight that is joined to it by an edge in the appropriate copy of .
We use a similar argument when , using the strategies from Figure 6, along with their transposes, and divide by 8 to obtain the bound .
Likewise, for , the strategies and their transposes from Figure
7 yield an upper bound of .