$t$-Pebbling in $k$-connected diameter two graphs

TL;DR

This paper investigates the $t$-pebbling number in diameter two graphs, focusing on how connectivity influences resource transportation modeled by pebbling moves.

Contribution

It extends the study of pebbling numbers by analyzing the $t$-pebbling number in diameter two graphs with respect to their connectivity.

Findings

Characterizes $t$-pebbling numbers in diameter two graphs.

Highlights the role of connectivity in pebbling resource distribution.

Provides bounds or formulas for $t$-pebbling numbers based on graph properties.

Abstract

Graph pebbling models the transportation of consumable resources. As two pebbles move across an edge, one reaches its destination while the other is consumed. The $t$-pebbling number is the smallest integer $m$ so that any initially distributed supply of $m$ pebbles can place $t$ pebbles on any target vertex via pebbling moves. The 1-pebbling number of diameter two graphs is well-studied. Here we investigate the $t$-pebbling number of diameter two graphs under the lense of connectivity.

Equations

$$\lceil \frac{(2t-2)+(t+1+j)}{2}\rceil =t+\lceil \frac{t-1+j}{2}\rceil$$

$$k-\left(t-\lceil \frac{t-1+j}{2}\rceil -1\right)+1=t+x+\lceil \frac{t-1+j}{2}\rceil +1$$

$$\lceil \frac{(2t-2)+(2t+2x+i)}{2}\rceil =2t+x-1+\lceil i/2\rceil$$

$${\pi }_t(T,r)=n(T)+4t-3\le (2+n-2t)+4t-3=n+2t-1=|C(T)|,$$

Full text

1 Introduction

Graph pebbling has an interesting history, with many challenging open problems. Calculating pebbling numbers of graphs is a well known computationally difficult problem. See [4, 5] for more background.

A configuration C of pebbles on the vertices of a connected graph G is a function $C:V(G){\rightarrow}{\mathbb{N}}$ (the nonnegative integers), so that $C(v)$ counts the number of pebbles placed on the vertex $v$. We write $|C|$ for the size $\sum_{v}C(v)$ of $C$; i.e. the number of pebbles in the configuration. A pebbling step from a vertex $u$ to one of its neighbors $v$ reduces $C(u)$ by two and increases $C(v)$ by one. Given a specified root vertex $r$ we say that $C$ is $t$-fold $r$-solvable if some sequence of pebbling steps places $t$ pebbles on $r$. We are concerned with determining $\pi_{t}(G,r)$, the minimum positive integer $m$ such that every configuration of size $m$ on the vertices of $G$ is $t$-fold $r$-solvable. The $t$-pebbling number of $G$ is defined to be $\pi_{t}(G)=\max_{r\in V(G)}\pi(G,r)$. We avoid writing $t$ when $t=1$.

Pebbling number of diameter 2 graphs was solved and characterized by the following theorem. For the purpose of the present work, it is enough to know that a pyramidal graph has no universal vertex (a vertex adjacent to every other vertex) and has connectivity 2.

Theorem 1

[2, 6]** For a diameter 2 graph $G$ with connectivity $k$ and $n$ vertices, $\pi(G)=n+1$ if and only if $k=1$ or $G$ is pyramidal. Otherwise (i.e. $k=2$ and $G$ is not pyramidal, or $k\geq 3$), $\pi(G)=n$.

In contrast, other than the following bound, little is known about the $t$-pebbling number of diameter 2 graphs.

Theorem 2

[3]** If $G$ is a diameter 2 graph on $n$ vertices then $\pi_{t}(G)\leq\pi(G)+4t-4$. Moreover, $\liminf_{t{\rightarrow}\infty}\pi_{t}(G)/t=4$.

The goal of the present paper is to determine the exact $t$-pebbling number of a large subfamily of diameter 2 graphs by considering their connectivity. Define ${\cal G}(n,k)$ to be the set of all $k$-connected graphs on $n$ vertices having a universal vertex. Set $f_{t}(n,k)=n+4t-k-2$ and $h_{t}(n)=n+2t-2$. Notice that $h_{t}(n)\geq f_{t}(n,k)$ if and only if $k\geq 2t$. Define $p_{t}(n,k)=\max\{f_{t}(n,k),h_{t}(n)\}$. The main result is the following theorem which is proved in Section 3.

Theorem 3

If $G\in{\cal G}(n,k)$ then $\pi_{t}(G)=p_{t}(n,k)$.

We observe from our result that, for any fixed $t$, in the family of graphs with universal vertex, there are graphs whose $t$-pebbling number is much lower than the bound given by Theorem 2, and also that there are graphs reaching that bound: when $k\geq 2t$ we have $\pi_{t}(n,k)=(n+4t-4)-2(t-1)$; when $k<2t$ $\pi_{t}(n,k)=(n+4t-4)-(k-2)$.

It will be useful to take advantage of Menger’s Theorem. The version of Menger’s theorem that we use is the following (exercise 4.2.28 in [7]).

Theorem 4

(Menger’s Theorem)* [7] Let $G$ be a $k$-connected graph and $S=\{v_{1},\ldots,v_{k}\}$ be a multiset of vertices of $G$. For any $r\not\in S$ there are $k$ pairwise-internally-disjoint paths, one from each $v_{i}$ to $r$.*

2 Technical Lemmas

We begin with a lemma that is used to prove lower bounds on the pebbling number of a graph by helping to show that certain configurations are unsolvable.

For a vertex $v$, define its open neighborhood $N(v)$ to be the set of vertices adjacent to $v$, and its closed neighborhood $N[v]=N(v)\cup\{v\}$. We say that a vertex $y$ is a junior sibling of a vertex $x$ (or, more simply, junior to $x$) if $N(y)\subseteq N[x]$, and that $y$ is a junior if it is junior to some vertex $x$.

Lemma 5

(Junior Removal Lemma)* [1] Given the graph $G$ with root $r$ and $t$-fold $r$-solvable configuration $C$, suppose that $y$ is a junior with $C(y)=0$. Then $C$ (restricted to $G-y$) is $t$-fold $r$-solvable in $G-y$.*

Given a configuration $C$ of pebbles, we say that a path $Q=(r,q_{1},\ldots,q_{j})$ with $j\geq 1$ is a slide from $q_{j}$ to $r$ if no $q_{i}$ is zero (it has no pebbles on) and $q_{j}$ has at least two pebbles.

A potential move is a pair of pebbles sitting on the same vertex. To say that $C$ has $j$ potential moves means that the $j$ pairs are pairwise disjoint. For example, any configuration on 5 vertices with values $0,1,1,2,$ and $7$ has 4 potential moves. The potential of $C$, ${\sf pot}(C)$, is the maximum $j$ for which $C$ has $j$ potential moves. Because every solution that requires a pebbling move uses a potential move, the following fact is evident.

Fact 6

Let $r$ be an empty vertex in a configuration $C$ with ${\sf pot}(C)<t$. Then $C$ is not $t$-fold $r$-solvable.

Basic counting yields the following lemma.

Lemma 7

(Potential Lemma)* Let $G$ be a graph on $n$ vertices. If $C$ is a configuration on $G$ of size $n+y$ ($y\geq 0$) having $z$ zeros, then ${\sf pot}(C)\geq\lceil\frac{y+z}{2}\rceil$.*

A nice application of the Potential Lemma is the following, which we will use repeatedly in the arguments that follow.

Lemma 8

(Slide Lemma)* Let $r$ be a vertex of a $k$-connected graph $G$. Let $C$ be a configuration on $G$ of size $n+y$ ($y\geq 0$) with $z$ zeros. If $\lceil\frac{y+3z}{2}\rceil\leq k$ then $C$ is $\lceil\frac{y+z}{2}\rceil$-fold $r$-solvable.*

**Proof.  **Set $p=\lceil\frac{y+z}{2}\rceil$. By Lemma 7 we can choose a set $P$ of $p$ potential moves. Note that the hypothesis implies that $p\leq k-z$. Delete all non-root zeros to obtain $G^{\prime}$. Since $G$ is $k$-connected, $G^{\prime}$ is $p$-connected. Thus Menger’s Theorem 4 implies that there are $p$ pair-wise disjoint slides in $G^{\prime}$ from $P$ to $r$, which yield $p$ $r$-solutions. $\Box$

3 Proof of Theorem 3

The proof will follow from Lemmas 9 and 10, below. Let $u$ be a universal vertex of a graph $G\in{\cal G}(n,k)$. If $C$ is a configuration of size $n+2t-3$ with $u$ empty and every other vertex odd then ${\sf pot}(C)=t-1$, and so $C$ is not $t$-fold $u$-solvable. Hence $\pi_{t}(G,u)\geq n+2t-2$. On the other hand, if $|C|\geq n+2t-2$ then ${\sf pot}(C)\geq t$ when $u$ is empty, and ${\sf pot}(C)\geq t-1$ when $u$ is not; either way $C$ is $t$-fold $u$-solvable because $u$ is universal. Thus $\pi_{t}(G,u)=n+2t-2$, which is at most $p_{t}(n,k)$ always.

3.1 Lower bound

Clearly, $\pi_{t}(G)\geq\pi_{t}(G,u)=h_{t}(n)$. Now let $r$ be any non-universal vertex of $G$, and let $s$ be a vertex at distance two from $r$. Let $X$ be any $(r,s)$-cutset of size $k$ (in particular, $u\in X$) and define the configuration $F_{t}(n,k)$ by placing 0 on $r$ and $X$, $4t-1$ on $s$, and 1 on each vertex of $V(G)-(X\cup\{r,s\})$; then $|F_{t}(n,k)|=(4t-1)+(n-k-2)=f_{t}(n,k)-1$.

Since the vertices of $X-\{u\}$ have 0 pebbles and all them are juniors to $u$, Lemma 5 states that if $t$ pebbles can reach $r$ then $2t$ pebbles can reach $u$. But, with exactly $2t-1$ potential moves in $F$, by Fact 6, we can place at most $2t-1$ pebbles on $u$. Therefore $\pi_{t}(G,r)\geq f_{t}(n,k)$, implying $\pi_{t}(G)\geq f_{t}(n,k)$.

We record these results as

Lemma 9

For $G\in{\cal G}(n,k)$ we have $\pi_{t}(G)\geq p_{t}(n,k)$.

3.2 Upper bound

We will prove that any configuration of size $f_{t}(n,k)$ when $k\leq 2t$, and of size $h_{t}(n)$ when $k\geq 2t$, is $t$-fold $r$-solvable for any $r\in V(G)$.

Lemma 10

For $k\geq 2$, let $G\in{\cal G}(n,k)$ with a universal vertex $u$, and let $r$ be any root vertex. Then $\pi_{t}(G,r)\leq p_{t}(n,k).$

**Proof.  **First note that the lemma is true when $t=1$. Indeed, in this case we have $k\geq 2t$, and so $p_{t}(n,k)=h_{t}(n)=n+2t-2=n$. On the other hand, because no pyramidal graph has a universal vertex, we have from Theorem 1 that $\pi(G)=n$, hence $\pi(G,r)\leq n$.

In addition, the lemma holds for $k=2$. Indeed, in this case we have $k\leq 2t$, and so $p_{t}(n,k)=f_{t}(n,k)=n+4t-k-2=n-4t-4$. Also, we have by Theorem 2 that $\pi_{t}(G,r)\leq n+4t-4$.

Hence, we may assume that $t\geq 2$ and $k\geq 3$. Figure 1 shows the structure of this proof. As was noted above, the grey section has been proven before. We continue by proving the dashed-bordered, lower left section and diagonal circled entries together, and then the solid-bordered, upper right section by induction.

Base case.

We will simultaneously address the case $k=2t-1$ (the circled entries), for which $|C|=f_{t}(n,k)=n+2t-1$, and the case $k\geq 2t$ (the dashed-bordered section), for which $|C|=h_{t}(n)=n+2t-2$, by writing $k\geq 2t-1$ and considering a configuration of size $|C|=n+2t-2+\phi$, where $\phi=1$ if $2t-1=k$ and 0 otherwise. The natural idea we leverage here is repeating the argument that zeros force potential which, combined with connectivity, yields either more solutions or more zeros.

Let $x\geq 0$ such that $k=2t-1+x$. By Lemma 7, since we may assume that $C(r)=0$ (otherwise induct on $t$), we have at least $\lceil(2t-2+1)/2\rceil=t$ potential moves. Therefore, we have at least $t$ solutions if there are at least $t$ different slides from them to $r$.

Thus we consider the case in which there are at most $t-1$ slides; that is, from some of the vertices in which a potential move is sitting, say $v$, there is no path to $r$ without an internal zero after considering the remaining $t-1$ slides. Since $G$ is $k$-connected, that implies that $C$ has at least $k-(t-1)$ zeros between $v$ and $r$ and so, because of $r$, $C$ has at least $k-(t-1)+1=t+1+x$ zeros.

Assume that there are exactly $z=t+1+j$ zeros, for some $j\geq x$. Then, by Lemma 7, $C$ has at least

[TABLE]

potential moves. If there are at least $t-\left\lceil\frac{t-1+j}{2}\right\rceil$ slides from them to $r$, then we can use those slides for that many solutions. Then, the other $\left\lceil\frac{t-1+j}{2}\right\rceil$ solutions can be obtained from the remaining $2\left\lceil\frac{t-1+j}{2}\right\rceil$ potential moves, putting $2\left\lceil\frac{t-1+j}{2}\right\rceil$ pebbles on the universal vertex $u$ and then $\left\lceil\frac{t-1+j}{2}\right\rceil$ on $r$.

Otherwise, there are at most $t-\left\lceil\frac{t-1+j}{2}\right\rceil-1$ slides, from which we find, using $k=2t-1+x$, at least

[TABLE]

zeros. Clearly, this number cannot exceed the total number of zeros $z=t+1+j$; therefore $j\geq x+\left\lceil\frac{t-1+j}{2}\right\rceil\geq x+\frac{t-1+j}{2}$, and so $j\geq t-1+2x$.

Let $j=t-1+2x+i$ for some $i\geq 0$; then $z=t+1+j=t+1+t-1+2x+i=2t+2x+i$. Applying Lemma 7 again, there are at least

[TABLE]

potential moves.

If either $x\geq 1$ or $i\geq 1$, then we can move $2t$ pebbles to the universal vertex $u$, and then $t$ to $r$.

Hence, we consider the case for which $x=i=0$; i.e. $k=2t-1$, $z=2t$, and $|C|=n+2t-1$ (because $\phi=1$ in such a case). We let $T$ be the star centered on $u$, having leaves $r$ and the nonzero vertices of $G$. Clearly, $T$ is a subgraph of $G$ with $n+2t-1$ pebbles on it and with either $2+(n-z)$ or $1+(n-z)$ vertices, depending on whether $u$ is empty or not. In either case $n(T)\leq 2+n-z=2+n-2t$. Therefore, since

[TABLE]

we see that $C$ is $r$-solvable.

Induction step.

Finally, we consider the cases when $k<2t-1$ (the solid-bordered section); so $|C|=f_{t}(n,k)=n+4t-k-2$. Since $2(t-1)=2t-1-1\geq k$, we have $\pi_{t-1}(G,r)=f_{t-1}(n,k)=n+4(t-1)-k-2=n+4t-k-2-4=|C|-4$. Hence, if $C$ has a solution of cost at most 4, we are done. Otherwise, there is at most one vertex $v$ having two or more pebbles, and on such a vertex there are at most 3 pebbles. This implies the contradiction $|C|\leq 3+(n-2)$, which completes the proof. $\Box$

In future work we intend to study $k$-connected diameter two graphs without a universal vertex, and use that work as a base step toward studying graphs of larger diameter.