Bounds on the Rubbling and Optimal Rubbling Numbers of Graphs

# Bounds on the Rubbling and Optimal Rubbling Numbers of Graphs

Gyula Y. Katona  and  NÁndor Sieben Budapest University of Technology and Economics Faculty of Electrical Engineering and Informatics, Department of Computer Science and Information Theory, H-1521 Budapest Po. box. 91, Hungary Northern Arizona University, Department of Mathematics and Statistics, Flagstaff AZ 86011-5717, USA
###### Abstract.

A pebbling move on a graph removes two pebbles at a vertex and adds one pebble at an adjacent vertex. Rubbling is a version of pebbling where an additional move is allowed. In this new move, one pebble each is removed at vertices and adjacent to a vertex , and an extra pebble is added at vertex . A vertex is reachable from a pebble distribution if it is possible to move a pebble to that vertex using rubbling moves. The rubbling number is the smallest number needed to guarantee that any vertex is reachable from any pebble distribution of pebbles. The optimal rubbling number is the smallest number needed to guarantee a pebble distribution of pebbles from which any vertex is reachable. We give bounds for rubbling and optimal rubbling numbers. In particular, we find an upper bound for the rubbling number of -vertex, diameter graphs, and estimates for the maximum rubbling number of diameter graphs. We also give a sharp upper bound for the optimal rubbling number, and sharp upper and lower bounds in terms of the diameter.

###### Key words and phrases:
pebbling, rubbling
05C99

7/30/2019

## 1. Introduction

Graph pebbling has its origin in number theory. It is a model for the transportation of resources. Starting with a pebble distribution on the vertices of a simple connected graph, a pebbling move removes two pebbles from a vertex and adds one pebble at an adjacent vertex. We can think of the pebbles as fuel containers. Then the loss of the pebble during a move is the cost of transportation. A vertex is called reachable if a pebble can be moved to that vertex using pebbling moves. There are several questions we can ask about pebbling. How many pebbles will guarantee that every vertex is reachable (pebbling number), or that all vertices are reachable at the same time (cover pebbling number)? How can we place the smallest number of pebbles such that every vertex is reachable (optimal pebbling number)? For a comprehensive list of references for the extensive literature see the survey papers [8, 9].

Graph rubbling is an extension of graph pebbling. In this version, we also allow a move that removes a pebble each from the vertices and that are adjacent to a vertex , and adds a pebble at vertex . The basic theory of rubbling and optimal rubbling is developed in [1]. The rubbling number of complete -ary trees are studied in [6], while the rubbling number of caterpillars are determined in [13].

The current paper extends the theory of graph rubbling by providing bounds for the rubbling numbers of graphs. In Section 3, we give an upper bound for the rubbling number in terms of the number of vertices and the diameter of the graph. In Sections 4–5, we investigate how big the rubbling number of diameter 2 graphs can be. Let be the maximum rubbling number of diameter graphs with vertices. We construct a family of graphs whose rubbling numbers match all known values of . We also prove an upper bound for . Similar questions for pebbling are studied in [2, 5], more details are given in Section 5. In Section 6, we give a sharp upper bound for the optimal rubbling number of a graph in terms of the number of vertices. We also give sharp upper and lower bounds in terms of the diameter. Similar results for the optimal pebbling number are presented in [3, 11]. Our results are extensions of these.

## 2. Preliminaries

Throughout the paper, let be a simple connected graph. We use the notation for the vertex set and for the edge set. A pebble function on a graph is a function where is the number of pebbles placed at . A pebble distribution is a nonnegative pebble function. The size of a pebble distribution is the total number of pebbles . We are going to use the notation to indicate that for and for all .

Consider a pebble function on the graph . If then the pebbling move removes two pebbles at vertex , and adds one pebble at vertex to create a new pebble function

 p(v,v\shortrightarrowu)(v,u,∗)=(p(v)−2,p(u)+1,p(∗)).

If and , then the strict rubbling move removes one pebble each at vertices and , and adds one pebble at vertex to create a new pebble function

 p(v,w\shortrightarrowu)(v,w,u,∗)=(p(v)−1,p(w)−1,p(u)+1,p(∗)).

A rubbling move is either a pebbling move or a strict rubbling move. A rubbling sequence is a finite sequence of rubbling moves. The pebble function gotten from the pebble function after applying the moves in is denoted by . The pebble function gotten after applying the moves in a multiset of rubbling moves in any order is denoted by . The concatenation of the rubbling sequences and is denoted by .

A rubbling sequence is executable from the pebble distribution if is nonnegative for all . A vertex of is reachable from the pebble distribution if there is an executable rubbling sequence such that . The rubbling number of a graph is the minimum number with the property that every vertex of is reachable from any pebble distribution of size .

The optimal rubbling number of a graph is the size of a distribution with the least number of pebbles from which every vertex is reachable.

Given a multiset of rubbling moves on , the transition digraph is a directed multigraph whose vertex set is , and each move in is represented by two directed edges and . The transition digraph of a rubbling sequence is where is the multiset of moves in . Let represent the in-degree and the out-degree in . We simply write and if the transition digraph is clear from context.

A multiset of rubbling moves on is balanced with a pebble distribution at vertex if . We say is balanced with if is balanced with at all , that is, . A multiset of rubbling moves is called acyclic if the corresponding transition digraph has no directed cycles. An element is called an initial move of if in the transition digraph.

An important tool is the following result of [1].

###### Lemma 2.1.

(No Cycle) Let be a pebble distribution on and . The following are equivalent.

1. is reachable from .

2. There is a multiset of rubbling moves such that is balanced with and .

3. There is an acyclic multiset of rubbling moves such that is balanced with and .

4. Vertex is reachable from through an acyclic rubbling sequence.

## 3. Upper bound on the rubbling number

All the known upper bounds for the pebbling number are also upper bounds for the rubbling number since . The following result is the rubbling version of the upper bound [4, Theorem 1]. The difference between the pebbling upper bound and the rubbling upper bound is . The improvement is for (the path on vertices) since then .

###### Theorem 3.1.

If is a graph with vertices and diameter , then

 ρ(G)≤(n−d+1)(2d−1−1)+2.
###### Proof.

The statement clearly holds if since then , so we may assume that . Suppose is a distribution of pebbles from which vertex is not reachable. Let be a vertex whose distance is maximal from and let be the shortest path between and . Recursively define to be a vertex in whose distance is maximal from , and define to be the shortest path between and . The recursion must stop after steps. Let be the length of . Then we have and . If

 |p|≥m∑i=1(2li−1−1)+2l1−1+1

then either some has at least pebbles or there are some and with at least and pebbles respectively. In either case is reachable. So we must have

 |p| < m∑i=1(2li−1−1)+2l1−1+1≤m∑i=1(2l1−1−1)+2l1−1+1 = (m+1)(2l1−1−1)+2≤(n−l1+1)(2l1−1−1)+2 ≤ (n−d+1)(2d−1−1)+2.

The last inequality follows from the fact that is increasing for . ∎

The upper bound is sharp for since and for since for . It is also sharp for since . If the diameter of is 2, then the upper bound becomes . This is no surprise since and we know [5] that is either or . However, this upper bound is not sharp.

## 4. Lower bound for f(n,2)

There is no lower bound that forces to grow with the number of vertices of the graph. In fact for all . The only known lower bound for the rubbling number is coming from the diameter of the graph . So we could ask whether is a finite set for all . The family of star shaped graphs constructed in [2] can be used to show that this is not the case for . For we need a more elaborate construction.

Our aim is to construct a graph for any given with diameter 2 and a high rubbling number. Since the graph is not so easy to describe directly by giving the vertex and edge sets, we first define a simpler graph, then we make modifications on it to reach the final construction.

For a positive integer , let be the simple graph defined by

 V(Hs) ={(i,j)∣1≤i≤j≤s} E(Hs) ={{(i1,j1),(i2,j2)}∣i1=i2 or j1=j2}.

Clearly . Now we show that . Take two different vertices and where . If either or , then they are adjacent. Otherwise is a common neighbor of and , so their distance is 2.

Now we modify by deleting a few vertices and adding a few more edges in the following way. If is odd and , then delete the vertices

 (s−1,s),(s−3,s−2),(s−5,s−4),…,(2,3)

 {(s,s),(s−1,s−1)},{(s−2,s−2),(s−3,s−3)},…,{(3,3),(2,2)}.

If is even, then delete the vertices

 (s−1,s),(s−3,s−2),(s−5,s−4),…,(3,4)

 {(s,s),(s−1,s−1)},{(s−2,s−2),(s−3,s−3)},…,{(4,4),(3,3)}.

Let denote the graph that is obtained. Clearly . One can see that holds too, since any pair of vertices whose unique common neighbor was deleted, now is either connected by an edge or has a new common neighbor on the spine.

###### Definition 4.1.

If for a given we have for some , then let . Thus, we have the construction of for . For the values of where , the construction is given by adding some vertices and edges to . We add the vertices one by one until we reach the required vertices. A new vertex is adjacent to another vertex if either or .

A visualization of the graph family is shown in Figures 4.1 and 4.2. Roughly speaking, we add the new vertices on the left of the graph, starting at the top row and continuing towards the bottom. We create new edges to keep the general edge structure of the graph. We stop adding new vertices before we reach the number of vertices in . This means that we stop at vertex if is odd and stop at if is even. Graphs and shown in Figure 4.1 illustrate these differently placed last new vertices. Note that and .

Note that

 |V(H′s)|+s =(s+1)s/2−⌊(s−1)/2⌋+s =(s+2)(s+1)/2−⌊s/2⌋−1=|V(H′s+1)|−1

if is odd, and if is even.

The -th row of is while the -th column of is . The spine of is . A short calculation shows that has rows. It is easy to see that .

For we have .

###### Proof.

Let be the number of rows in . Let be a pebble distribution on containing pebbles and suppose that a goal vertex is not reachable. We are going to define some collections of rows and columns, but in one case it is just part of the columns. So let us first define the partial columns: Now let

 R={Rj∣j>j0},R=∪R, C={C′i∣i≠i0},C=∪C, X=Rj0∪Ci0,

as shown in Figure 4.3. Then , (note that may be empty) and .

If and there are two or three pebbles on , then we can apply a rubbling move on those pebbles and move a pebble to . Row cannot have or more pebbles, because then we can move two pebbles to so would be reachable. Similarly, if and has two or three pebbles, then we can move a pebble to , while if it has or more pebbles, then we can move two pebbles to . In the exceptional case, when has no vertex in the -th column (like the 5th column on Figure 4.3), we can move the pebbles along the spine to instead of .

If we can move two pebbles to , then is clearly reachable. That means that if or has one or two more pebbles than the size of or respectively, then we can move one pebble to . If it has three more pebbles, then we can move two pebbles to So if either or contains enough pebbles to move two pebbles to or both of them contains enough pebbles to move one pebble to , then is reachable.

We know that can have at most one pebble on . If has no pebbles, then the total number of pebbles cannot be more than

 |R|+|C|+2≤(k−j0)+(j0−1)+2=k+1

which is not possible. If has exactly one pebble, then the total number of pebbles cannot be more than

 |R|+|C|+1≤(k−j0)+(j0−1)+1=k

which is not possible either. ∎

In the proof of our next result, we are going to need to keep track of the movement of pebbles during rubbling moves. For this purpose, we need to replace our pebbles with dependency sets.

###### Definition 4.3.

A dependency distribution is a partition of an initial pebble set together with a location function . The elements of are called dependency sets or simply pebbles.

We think of a dependency set as a pebble with some additional information about the history of the pebble. Dependency distributions replace pebble distributions. Given a pebble distribution containing pebbles, we can create a corresponding dependency distribution such that for all .

###### Definition 4.4.

If and are both adjacent to , then the rubbling move removes the dependency sets and from at the vertices and respectively, and adds a new dependency set to with location . A vertex is reachable from a dependency distribution if a dependency set with can be created using rubbling moves.

Note that the rubbling move is essentially the rubbling move with some additional information about the history of the pebbles. It is clear that a vertex is reachable from a pebble distribution if and only if it is reachable from the corresponding dependency distribution. Also note that in a rubbling move we must have .

###### Definition 4.5.

Let be a sequence of rubbling moves in a dependency distribution. We say is dependent on if . We say that and are independent if and are disjoint.

Roughly speaking, and are independent if they rely on two disjoint sets of pebbles. It is clear that dependence of rubbling moves is a transitive relation. Also note that and are independent if and only if neither is dependent on nor is dependent on .

###### Example 4.6.

Consider the initial dependency distribution and the sequence of rubbling moves

 s1=({1},{2}\shortrightarrowu),s2=({3},{4}\shortrightarrowv),s3=({3,4},{5}\shortrightarrowu),s4=({1,2},{3,4,5}\shortrightarrowx).

Then and are independent but depends on .

For we have .

###### Proof.

Let be the largest column index in . We show that the goal vertex is not reachable from the pebble distribution that has a single pebble on vertex for and three pebbles on vertex as shown in Figures 4.1 and 4.2. To see this, we show that is not reachable from the dependency distribution with

 l({1},…,{k−2},{k−1},{k},{k+1})=((2,2),…,(k−1,k−1),(k,k),(k,k),(k,k)).

For a contradiction suppose that is reachable from , that is, there is a sequence of rubbling moves that creates a dependency set at .

Let us call a rubbling move horizontal if , and are all contained in the same row of . To reach the goal vertex , we must use a rubbling move involving vertices on . There are no pebbles on these vertices originally and the only way to move a new pebble there is to use a horizontal rubbling move. So there must be at least two independent horizontal moves and in our rubbling sequence. We show that this is not possible.

Since and are independent, at least one of the sets and contains at most one element of . Roughly speaking, this means that both and cannot rely on more than one pebble available at vertex in since there are only three pebbles there. So we can assume that .

Now we create a new dependency distribution by removing the two pebbles from that does not rely on for sure. We also remove the rubbling moves from that are dependent on or . More precisely, we remove the rubbling moves of the form for which . The resulting rubbling sequence is executable from and contains .

Let us call for a line of . Line contains the spine vertex . Note that all vertices on the spine are contained in exactly one line, and all other vertices are contained in at most two lines. We say that a pebble configuration is forbidden if there is a line with more than one pebble. Note that two pebbles on a row or on a column with is a forbidden configuration. It is clear that is not a forbidden pebble configurations.

We are going to show that a rubbling move cannot create a forbidden configuration if there was no forbidden configuration before the rubbling move. Suppose that a rubbling move creates a pebble on vertex of line . If is not on the spine, then it is contained in another line , so all of its neighbors are in . Both and cannot be on the same line since there was no forbidden configuration before this step. Thus one of and must be on while the other must be outside of as shown on Figure 4.4(a). If is on the spine, then it has a neighbor which is on the spine, too. Again, both of and cannot be on this vertex, because that is a forbidden configuration. Therefore one of and must be on again as shown on Figure 4.4(b) since in this case all other neighbors of are in .

Thus we can assume that is in . Since is in , we cannot have any other pebble on before the rubbling move. So is the only pebble on after the rubbling move and so the rubbling move did not create any forbidden configurations.

We saw that has the horizontal move . A horizontal move is only possible if there are two pebbles on a row which is a forbidden configuration. This is a contradiction since we do not have any forbidden configurations during the execution of .∎

For we have .

## 5. Upper bound for f(n,2)

Table 1 shows the maximum rubbling numbers

 f(n,2)=max{ρ(G)∣n=|V(G)| and 2=diam(G)}

of diameter 2 graphs with vertices. The values were calculated by a computer program [14]. The program checked all diameter 2 graphs with a given number of vertices. These graphs were generated by Nauty [10]. We have for . It is not clear whether this is true for all .

###### Problem 5.1.

Is it true that for all ?

There are more existing results for similar questions about pebbling. It is known [12] that since the pebbling number of a diameter 2 graph is either or . A classification of diameter 2 graphs with pebbling number is also known from [5]. Diameter 3 graphs are also studied. In [2], it is shown that .

The proof of the following result uses the method of [7].

###### Lemma 5.2.

Let be a 3-uniform hypergraph on vertices. If for all , then .

###### Proof.

Let denote the characteristic vectors of the sets in . We claim that the characteristic vectors are linearly independent over . This clearly implies the result.

Since every set contains elements, we have for all since . On the other hand, if then since the product is the cardinality of the intersection of the two corresponding sets. If then multiplying by we obtain . This proves our claim. ∎

The set of vertices adjacent to a given vertex of a graph is denoted by .

###### Proposition 5.3.

Let be a diameter two graph with goal vertex . Let be a pebble distribution on containing pebbles. If is not reachable using only five pebbles of , then has at least vertices.

###### Proof.

For let . Since the diameter of is two, we must have

 V(G)∖{x}=⋃v∈N(x)R(v).

Since is not reachable using only five pebbles, no can contain more than two pebbles otherwise we could move a pebble to . So there is a maximal subset of such that contains exactly one pebble on a vertex for all . Define and note that and are clearly disjoint. To simplify notation we write for . Figure 5.1 shows a visualization of these sets.

If two vertices and of are not adjacent, then the diameter condition implies that we can pick a common neighbor of and . Note that . If , , and are all different, then , otherwise would move two pebbles to from which is reachable using only four pebbles.

If then we write for . Define

 J={{i,j}∣{bi,bj}∈E(¯¯¯¯G)},C={c{i,j}∣{i,j}∈J}.

A vertex cannot be adjacent to two different vertices and of , otherwise would create two pebbles on , making reachable using only three pebbles. Hence the number of edges between the elements of is at most and so .

We also introduce a subset of by letting

 K={{i,j,k}∣c{i,j}=c{i,k}},C′={c{i,j,k}|{i,j,k}∈K}.

Though it may happen that , we have . Note that and implies . If then otherwise we could move two pebbles to using only four pebbles. Hence .

We can clearly move a pebble to any element of and so . If then otherwise would move a pebble to using only three pebbles. In particular, and so .

Suppose that . Then for since is in . We also have , otherwise would move a pebble to using only three pebbles. Similar arguments show that . Hence

 c{i,j,k}∈R(d{i,j,k})∖m⋃l=1Rlfor% somed{i,j,k}∈N(x)∖A.

Let . Then is disjoint from by definition. We also have . If then , otherwise would have at least four vertices with pebbles so we could move a pebble to using only these four pebbles. So and has the same number of elements.

The intersection of two elements and of cannot be a singleton set, otherwise

 (bj,bk\shortrightarrowc{j,k})(bj′,bk′\shortrightarrowc{j′,k′})(c{j,k},c{j′,k′}\shortrightarrowbi)

would move two pebbles to from where would be reachable using only five pebbles. Hence the 3-uniform hypergraph satisfies the conditions of Lemma 5.2 and so .

The result now follows from the calculation

 |VG| ≥|{x}|+|A|+|B|+|C|+|D|=1+m+m+|C∖C′|+|C′|+|D| ≥1+2m+|J|−3|C′|+|C′|+|C′|≥1+2m+(m2)−⌊m2⌋−|C′| =2+3m+m22−⌊m2⌋−|K|≥2+3m+m22−⌊m2⌋−m=⌊m2+32⌋.

###### Proposition 5.4.

Let be a diameter two graph with goal vertex . Let be a pebble distribution on containing pebbles. If is not reachable from then has at least vertices.

###### Proof.

If satisfies the conditions of Proposition 5.3, then must have at least vertices.

Otherwise there are five pebbles of from where is reachable. We can remove these pebbles to create a new pebble distribution containing pebbles. Then must satisfy the conditions of Proposition 5.3, otherwise we could move two pebbles to and so would be reachable. So must have at least vertices.∎

###### Proposition 5.5.

The rubbling number of a diameter 2 graph with vertices cannot be larger than .

###### Proof.

From Proposition 5.4, we know that if we have pebbles on a diameter 2 graph with vertices one of which is not reachable, then . The contrapositive gives that if a graph has vertices and , then any goal vertex is reachable from any placement of pebbles and so the rubbling number is larger than . The result now follows since we have

 n<⌊(m−5)2+32⌋ ⟺n+1≤(m−5)2+32 ⟺√2n−1+5≤m.

We have .

## 6. Bounds on the optimal rubbling number

We saw in [1] that . We show that the path requires the most pebbles for optimal rubbling amongst the graphs with a given number of vertices. The proof follows the ideas of [3].

###### Proposition 6.1.

If is a tree with vertices, then .

###### Proof.

We use induction on . The statement is clearly true for . In the inductive step let and let be the consecutive vertices of a longest path of . Note that . We are going to find a subtree of with vertices as shown in Figure 6.1. Then there is a pebble distribution on with size from which every