Abstract
A pebbling move on a graph consists of the removal of two pebbles from one vertex and the placement of one pebble on an adjacent vertex. Rubbling is a version of pebbling where an additional move is allowed, which is also called the strict rubbling move. In this new move, one pebble each is removed from and adjacent to a vertex , and one pebble is added on . The rubbling number of a graph is the smallest number , such that one pebble can be moved to each vertex from every distribution with pebbles. The optimal rubbling number of a graph is the smallest number , such that one pebble can be moved to each vertex from some distribution with pebbles. In this paper, we give short proofs to determine the rubbling number of cycles and the optimal rubbling number of paths, cycles, and the grid ; moreover, we give an upper bound of the optimal rubbling number of .
1. Introduction
Pebbling in graphs was first introduced by Chung [1]. It has its origin in number theory and also can be viewed as a model for the transportation of resources, starting from a pebble distribution on the vertices of a connected graph.
Let be a simple connected graph; we use and to denote the vertex set and edge set of , respectively. is the distance of and , and we write if they are adjacent. is the neighbour of , and is the degree of . Let be a subgraph of , and we use to denote the degree of in . A pebble distribution on is a function ( is the set of nonnegative integers), where is the number of pebbles on and is the size of .
A pebbling move consists of the removal of two pebbles from a vertex and the placement of one pebble on an adjacent vertex. For two pebble distributions and of , we say that contains if for all ; and is reachable from if we can obtain a distribution which contains from by pebbling moves. For a graph and a vertex , we call a root (or target vertex) if the goal is to place pebbles on . If pebbles can be moved to from by a sequence of pebbling moves, then we say that is -fold-solvable, and is -reachable from . If is -fold -solvable for every vertex , we say that is -solvable. Let be a subset of . If is -solvable for all , then we say is-solvable.
The-pebbling number of a graph, denoted by , is the smallest number , such that every distribution with size is -solvable. While , we use instead of , which is called the pebbling number of graph . The optimal pebbling number of a graph is the minimum number such that there exists a solvable distribution with size .
Rubbling is a version of pebbling where an additional move is allowed, which is also called the strict rubbling move. In this new move, one pebble each is removed from and adjacent to a vertex , and one pebble is added on . The-rubbling number of a graph, denoted by , is the smallest number , such that every distribution with size is -solvable by rubbling moves. Similarly, while , we use instead of , which is called the rubbling number of graph. The optimal rubbling number of a graph is the minimum number such that there exists a solvable distribution with size by rubbling moves.
There are many papers about pebbling on graphs; one can view the survey paper [2] written by Hurlbert. Rubbling is a new parameter with few results. The basic theory about rubbling and optimal rubbling was developed by Belford and Sieben [3]; they determined the rubbling numbers of trees and cycles, the optimal rubbling numbers of paths and cycles, and so on. The rubbling numbers of complete -ary trees are studied in [4], the rubbling numbers of caterpillars are given in [5], the optimal rubbling numbers of ladders, prisms, and Möbius ladders are determined in [6], and in [7], they give the bounds for the rubbling numbers of diameter 2 graphs.
In this paper, we give some new proofs to determine the rubbling numbers of cycles and the optimal rubbling numbers of paths, cycles, ladders, prisms, and Möbius ladders. Also, we give an upper bound of the optimal rubbling number of , from which we also get an upper bound of the optimal rubbling number of .
2. Main Result
Assume that . A pebbling move from to is the removal of two pebbles from and addition of one pebble on , denoted by ; a rubbling move from to is the removal of one pebble from and one pebble from and addition of one pebble on , denoted by .
The following lemma holds clearly.
Lemma 1. (see [3]). .
Definition 1. (see [3]). Given a multiset of rubbling moves on , the transition digraph is a directed multigraph whose vertex set is , and each move in isrepresented by two directed edges and . The transition digraph of a rubbling sequence is , where is the multiset of moves in .
Lemma 2. (see [3, 8], no-cycle lemma). Let be a sequence of rubbling moves on , reaching a distribution . Then, there exists a sequence of rubbling moves, reaching a distribution , such that(1)On each vertex , .(2) does not contain any directed cycles.A thread in a graph is a path containing vertices of degree 2. By Lemma 2, we can get the following.
Lemma 3. (see [3]). Let be a thread in ; if vertex is -reachable from the pebble distribution by a sequence of rubbling moves, then is -reachable from a rubbling sequence in which there is no strict rubbling move of the form for .
Definition 2. (see [9]). For a given distribution on , assume and . A smoothing move from changes by removing two pebbles from and adding one pebble on each neighbour of .
Lemma 4. (see [9]). Let be a distribution on a graph with distinct vertices and , where , . If is -reachable under , then is -reachable under the distribution obtained by making a smoothing move from .
A distribution is smooth, if it has at most two pebbles on every vertex with degree 2. A vertex is unoccupied under if .
Lemma 5. (smoothing lemma, [9]). If is a connected -vertex graph, with , then has a smooth optimal distribution with all leaves unoccupied.
Lemma 6. (see [9]). In a path with a smooth distribution having at most two pebbles on each endpoint, let be an unoccupied vertex. If is an endpoint, then is not 2-reachable under . If is an inner vertex, then no pebble can be moved out from without using an edge in both directions.
Definition 3. (see [9]). A graph is a quotient of a graph if the vertices of correspond to the sets in a partition of , and distinct vertices of are adjacent if at least one edge of has endpoints in the sets corresponding to both vertices of . In other words, each set in the partition of collapses to a single vertex of . If is a quotient of via the surjective map and is a distribution on , then the quotient distribution is the distribution on defined by .
Lemma 7. (collapsing lemma, [9]). Let be a quotient of and be the quotient distribution in . Then, is -fold -solvable by a sequence of pebbling moves is -fold -solvable by a sequence of pebbling moves. In particular, .
Similarly, we can get the collapsing lemma on rubbling.
Lemma 8. Let be a quotient of and be the quotient distribution on . Then, is -fold -solvable by a sequence of rubbling moves is -fold -solvable by a sequence of rubbling moves. In particular, .
Proof. The proof is similar to the proof of Lemma 7 in [9].
2.1. Rubbling and Optimal Rubbling in Paths and Cycles
Lemma 9. (see [10, 11]). The-pebbling numbers of the cycles and areIn this section, we give short proofs about the rubbling number of cycles and the optimal rubbling numbers of paths and cycles, which are determined in [3].
Theorem 1. (see [3]). The rubbling numbers of cycles are
Proof. For the even cycle , by Lemma 1, we have , over.
For the odd cycle , assume the target vertex is which is adjacent to and . Let be a pebble distribution on , and . We collapse the vertices into one vertex to get a new graph and the quotient distribution on , with and for .
By Lemma 3, we may assume that there is no strict rubbling move of the form where . So, one pebble can be moved to from on by a sequence of rubbling moves two pebbles can be moved to from on by a sequence of pebbling moves. Thus, (the last equality holds from Lemma 9). A simple calculation can show that .
Lemma 10. (see [9]). The optimal pebbling numbers of paths and cycles areMoreover, we can get the following lemma.
Lemma 11. Let be a pebble distribution with pebbles on . If is solvable by pebbling moves, then for , and otherwise.
Proof. We show it by induction on ; it holds for .
We can move pebbles from to , so by induction, , and thus . Similarly, .
We make smoothing moves on from to obtain a smooth pebble distribution .
Since , there exist many vertices unoccupied under . Assume is a vertex unoccupied under . Let and . By Lemma 6, we cannot move two pebbles to using only one direction.
If is solvable by , then we can get that is -solvable, and thus . Assume for . If , then . If , then , a contradiction to . If , then , a contradiction to , and thus .
By a similar argument, if is solvable by , then we can get .
Thus, we can always partition the path into two paths and , and is solvable in for , respectively.
By induction, we know that for , and otherwise.
Note that making a smooth move on the vertex leaves at least one pebble on , which means that if we can make a smooth move under , then at least two adjacent vertices have pebbles under , a contradiction to , and otherwise. So, , which completes the proof.
In [3], Belford and Sieben determined the optimal rubbling number of paths and cycles; here we give a short proof.
Theorem 2. (see [3]). The optimal rubbling number of path is , and the optimal rubbling number of cycle is for .
Proof. Upper Bound. For , let be a distribution so that for is odd or , and otherwise; clearly is solvable, and . For , let be a distribution so that for is odd, and otherwise; clearly is solvable, and . Lower Bound. First we show it for the path . We use induction on . By a simple calculation, one can show that it holds for .If , assume it holds for for all . From the upper bound and Lemma 10, we know that . Let be a solvable pebble distribution with pebbles on ; there must exist a vertex , which is reachable only by the strict rubbling move . Let , be two subpaths of ; then, by Lemma 3, we do not need the strict rubbling move to solve the vertices in for . That means is -solvable for all (), so for . So, .
Now we show it for the cycle ; similarly, by a simple calculation, one can show that it holds for .
If , from the upper bound and Lemma 10, we know that . Thus, let be an optimal pebble distribution with pebbles on , and there must exist a vertex , which is reachable only by a strict rubbling move . Then, by Lemma 3, we do not need the strict rubbling move to solve the vertices in , and thus is solvable on , so , which completes the proof.
3. Optimal Rubbling in Ladders and Prisms
Let and be simple connected graphs; we define the Cartesian product to be the graph with vertex set and edge set the union of and . We call a ladder and a prism. It is clear that a prism can be obtained from a ladder by joining the 4 endvertices by two edges to form two vertex-disjoint subgraphs. If the four endvertices are joined by two new edges in a switched way to get a subgraph, then a Möbius ladder is obtained.
The optimal rubbling numbers of ladders, prisms, and Möbius ladders are determined in [6]; here we give new proofs of these results.
Let and . The vertices of are denoted by and for . Let . Then, we have the following.
Theorem 3. Let , ; the optimal rubbling number of the ladder [6] isThe optimal rubbling number of is
Proof. Upper Bound. Let be a distribution on or . If , then for ; for and ; and otherwise. If , then for and ; for ; and otherwise. If , for , then for ; for ; otherwise. For , we add one more pebble on . One can check that is solvable for all and . Lower Bound. Note that the diameter of is , and let be the set of vertices with distance from (), ; if we collapse into one vertex for , then we can get a path (see Figure 1). By Lemma 10, the optimal pebbling number of is . By Lemma 8, . So, we only need to show that and .We use induction on for both (only need to show while ) and ; it holds for clearly.
Assume it holds for .
Case 1. First we consider , and let be a solvable distribution on with pebbles; then, we collapse into one vertex for to get a path with length and the quotient distribution on . By Lemma 8, is solvable in . We will show that .
Assume ; since is a solvable distribution on , there must exist some of which can be reachable only by the strict rubbling move . By Lemmas 3 and 8, we claim the following.
Claim 1. At most one pebble can be moved to () under and at most one pebble can be moved to () under .
Let , be two subpaths of and and be the subgraphs of which induce and , respectively.
Now we will show that for . Note that is isomorphic to and is isomorphic to .
Notation. For simplicity, a rubbling move in means a rubbling move for some and . And similarly, we can define a rubbling move and so on.
If is -solvable, then , and we are done. Assume is not -solvable for some ; without loss of generality, assume is not -solvable. If , then , and the strict rubbling move means that , which means , a contradiction to , is not -solvable. Thus, .
Since is not -solvable, to solve some vertex of , we must use the pebbles on . Note that we can move at most one pebble on from , and we must use the strict rubbling move under .
Consider the vertex , . Since is reachable from , we can move two pebbles to . From Claim 1, we can get that we can move one pebble to and one pebble to at the same time. Thus, is reachable under .
Subcase 1. . Then, we can move two pebbles on under (for is just adjacent to one vertex of ). So, is reachable under , too.
Note that the pebble moved on cost at least two pebbles on and we can move at most one pebble to by using this pebble. We can use instead of the rubbling sequence which contains the strict rubbling move and aim to move one pebble to , so is -solvable.
Subcase 2. . Then, at most one pebble can be moved to under ; otherwise, one more pebble can be moved from to , and then there are two pebbles on , and one pebble can be moved to under , a contradiction.
By a similar argument, we can show there exists some vertex in so that we must use the strict rubbling move to solve this vertex, and the next rubbling move must be . Since at most one pebble can be moved to under , exactly one pebble can be moved to under .(a)The vertex of used in is ; then, if it continues, the rubbling move must be , but then we can use instead, and thus we must use the strict rubbling move to solve at most one vertex . Let ; then, the vertices of are reachable without the strict rubbling moves and the pebble on . So, , and by induction, , over.(b)The vertex of used in is ; since at most one pebble can be moved to , we may assume (if not, then we can move two pebbles to , and then one pebble can be moved to ; this is just (a)). Thus, we use one pebble each on , , and to move one pebble to . Then, we rearrange the distribution on as follows: remove the pebble on , and add it on ; one can view that we can still solve and . Thus, in the new distribution, we do not need the strict rubbling move to solve , so .Thus, .
Assume that and for ; by induction, we can get Table 1, where the lower bound is given by .
From Table 1, we can find , which is a contradiction to the assumption that , and we are done.
Case 2. Now we consider . Note that the diameter of is . Let be the vertices set for , and . If we collapse into one vertex for , then we can get a path (see Figure 2). By Lemma 10, the optimal pebbling number of is . So, we only need to show that .
Let be a solvable distribution with pebbles on and be the quotient distribution on . So, is solvable on by Lemma 8.
Subcase 3. If , since is a solvable distribution on , there must exist some vertex of which can be reachable Assume that and be two disjoint subgraphs of which induce and , respectively.
We will show that is -solvable for . Otherwise, we must use the strict rubbling move under to solve one vertex of ; this move must be one of and . Without loss of generality, assume the rubbling move is ; then, by Lemma 2, the target vertex of the next rubbling move using the pebble on cannot be or ; if the target vertex is , we need one more pebble on or , a contradiction to the condition that can be reachable only by the strict rubbling move .
So, is -solvable, and thus for . Note that is isomorphic to and is isomorphic to . Thus, . Assume that for ; by induction, we can get Table 2, where the lower bound is given by .
From Table 2, we can find , a contradiction to the assumption , over.
Subcase 4. If , then is solvable on . If there exists a vertex which is reachable only by a strict rubbling move under , then we can get a contradiction by a similar argument of Subcase 3.
So, we only need to consider the case that all vertices of are reachable by only pebbling moves. By Lemma 11, we have for , and otherwise. Thus, at least one of is not solvable under , which completes the proof.
Theorem 4. (see [6]). The optimal rubbling number of prism isexcept .
Proof. It holds for ; we only need to show for . Upper Bound. can be viewed as a subgraph of ; it is clear that a solvable distribution with pebbles given in the proof of Theorem 3 is solvable on for , and we are done. Lower Bound. Let , . Assume is a solvable distribution with pebbles, and we collapse each set into one vertex (see Figure 3); then, we get a cycle and the quotient distribution . By Lemma 8, is solvable on . By Lemma 10, , and we only need to show .Assume ; then, there must exist a vertex which is reachable under only by the strict rubbling move . By a similar argument of Case 3 in the proof of Theorem 3, if we remove from (which is just isomorphic to ), then is still solvable. So, , which is a contradiction to the assumption , and we are done.
Theorem 5. (see [6]). The optimal rubbling number of the Möbius ladder is
Proof. The proof is similar to the proof of prisms.
4. Bounds on the Optimal Rubbling of
In [12], the optimal pebbling number of is determined.
Theorem 6. (see [12]). The optimal pebbling number of is .
In [13], the lower bound of the optimal pebbling number of is given.
Theorem 7. (see [13]). The optimal pebbling number of is no less than when .
Here we give an upper bound of the optimal rubbling number of by giving a solvable distribution on .
Theorem 8. .
Proof. We give a solvable distribution on with exactly pebbles. While , the solvable distribution is given in Figure 4.
While , assume for . Then, the solvable distribution on can be obtained from on while by placing five more pebbles on which is added on the left side of (see Figure 5(a)). So, for with .
Assume and . Then, the solvable distribution on can be obtained from on by placing four more pebbles on which is added on the left side of (see Figure 5(b)). So, for .
By a simple calculation, we have , and we are done.
One can check that the equality holds for , and here we give a conjecture.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(a)
(b)
Conjecture 1. .
Now we can get a bound of the optimal rubbling number of . We give a partition of , for example, ; then, we can get three disjoint subgraphs of called , and , and then we put some pebbles on and , so that the pebbles on can solve for . Clearly, this is a solvable distribution of . Thus, we have the following.
Theorem 9. , .
Theorem 10. If , , , , .
Proof. If , then which gives a partition of . We put the pebbles on except the inner subpaths , so that the pebbles on each can solve themselves, and thus we need pebbles, which is a solvable distribution. Similarly, we can get the other inequalities, which completes the proof.
Corollary 1. If , then .
Proof. From Theorems 2, 3, 8, and 10, we can get the result by a simple calculation.
Data Availability
No data were used to support the findings of the study.
Disclosure
This manuscript has been released as a pre-print at https://arxiv.org/pdf/1909.01703v1.pdf.
Conflicts of Interest
The authors declare that they have no conflicts of interest regarding the publication of the article.
Acknowledgments
This research was supported by Key Projects in Natural Science Research of Anhui Provincial Department of Education (nos. KJ2018A0438 (to Zheng-Jiang Xia) and KJ2020A0015 (to Zhen-Mu Hong)) and NSFC (no. 11601002 (to Zhen-Mu Hong)).