Optimal pebbling number of graphs with given minimum degree
Andrzej Czygrinow, Glenn Hurlbert, Gyula Y. Katona, László F. Papp
aa r X i v : . [ m a t h . C O ] A p r Optimal pebbling number of graphs with given minimum degree
A. Czygrinow ∗ , G. Hurlbert † , G. Y. Katona ‡ , L. F. Papp § August 13, 2018
Abstract
Consider a distribution of pebbles on a connected graph G . A pebbling move removes twopebbles from a vertex and places one to an adjacent vertex. A vertex is reachable under a pebblingdistribution if it has a pebble after the application of a sequence of pebbling moves. The optimalpebbling number π ∗ ( G ) is the smallest number of pebbles which we can distribute in such a waythat each vertex is reachable. It was known that the optimal pebbling number of any connectedgraph is at most nδ +1 , where δ is the minimum degree of the graph. We strengthen this bound byshowing that equality cannot be attained and that the bound is sharp. If diam( G ) ≥ π ∗ ( G ) ≤ . nδ +1 . On the other hand, we show that for arbitrary largediameter and any ǫ > (cid:0) − ǫ (cid:1) n ( δ +1) . Graph pebbling is a game on graphs initialized by a question of Saks and Lagarias, which was answeredby Chung in 1989[1]. Its roots are originated in number theory. Each graph in this paper is simple. Wedenote the vertex set and the edge set of graph G with V ( G ) and E ( G ), respectively. We use n and δ for the order and the minimum degree of G , respectively.A pebbling distribution D on graph G is a function mapping the vertex set to nonnegative integers.We can imagine that each vertex v has D ( v ) pebbles. A pebbling move removes two pebbles from a vertexand places one to an adjacent one. We do not want to violate the definition of pebbling distribution,therefore a pebbling move is allowed if and only if the vertex loosing pebbles has at least two pebbles.A vertex v is k -reachable under a distribution D , if there is a sequence of pebbling moves, suchthat each move is allowed under the distribution obtained by the application of the previous moves andafter the last move v has at least k pebbles. We say that a subgraph H is k-solvable under distribution D if each vertex of H is k-reachable under D . When the whole graph is k-solvable under a pebblingdistribution, then we say that the distribution is k-solvable.When S is a subset of the vertex set then let D ( S ) denotes the total number of pebbles placed onthe elements of S . We say that D ( V ( S )) is the size of D . We use the standard | X | notation to denotethe size of X when X is a pebbling distribution or a set. A pebbling distribution D on a graph G will becalled k -optimal if it is k -solvable and its size is the smallest possible. This size is called as the k -optimalpebbling number and denoted by π ∗ k ( G ). When k equals one we omit the k part from all of the previousdefinitions and notations. ∗ School of Mathematical and Statistical Sciences, Arizona State University, [email protected]. † Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, [email protected]. ‡ Department of Computer Science and Information Theory, Budapest University of Technology and Economics, MTA-ELTE Numerical Analysis and Large Networks Research Group, Hungary, [email protected]. § Department of Computer Science and Information Theory, Budapest University of Technology and Economics, MTA-ELTE Numerical Analysis and Large Networks Research Group, Hungary, [email protected]. π ∗ ( G ) ≤ n ( δ +1) and they alsofound a version utilizing the girth of the graph. A construction for infinite number of graphs withoptimal pebbling number (2 . − δ +15 − o ( n )) nδ +1 is also given in that article.In the present paper we continue the study of graphs with fixed minimum degree. We prove thatthere are infinitely many diameter two graphs whose optimal pebbling number are close to the nδ +1 upperbound. More precisely, for any ǫ > (4 − ǫ ) δ +1 .One can ask that what happens if we consider bigger diameter? In the second part of Section 2 weuse the previous graphs as building blocks to construct a family of graphs with arbitrary large diameter,fixed minimum degree, and high optimal pebbling number. For any d and ǫ > d and its optimal pebbling number is more than (cid:0) − ǫ (cid:1) n ( δ +1) .In the case when the diameter is at least three we also prove a stronger upper bound in Section 3. Itis shown that π ∗ ( G ) ≤ n δ +1) holds in this case. Unfortunately, we do not know that if it sharp or not,but this is enough to conclude that the original upper bound of [2] does not hold with equality.Finally, the authors of [2] prove, for k ≥
6, that the family of connected graphs G with n verticesand minimum degree k has π ∗ ( G ) /n → girth ( G ) → ∞ . They asked if the same could be true for k ∈ { , , } . We prove in Section 4 that this is true for k ∈ { , } , leaving the case k = 3 unresolved. We say that a vertex v is dominated by a set of vertices S , if v is contained in S or there is a vertex in S which is adjacent to v . A vertex set S dominates G if each vertex of G is dominated by S . An edge { u, v } dominates G if the set of its endpoints dominates G . G (cid:3) H denotes the Cartesian product of graphs G and H , so V ( G (cid:3) H ) = V ( G ) × V ( H ) and { ( g, h ) , ( g ′ , h ′ ) } ∈ E ( G (cid:3) H ) if either g = g ′ , { h, h ′ } ∈ E ( H ) or { g, g ′ } ∈ E ( G ), h = h ′ .The distance between vertices u and w is the number of edges contained in a shortest path betweenthem. We denote this quantity with d ( u, v ). The distance- k open neighborhood of a vertex v , denotedby N k ( v ), contains all vertices whose distance from v is exactly k . On the other hand, the distance- k closed neighborhood of v contains all vertices whose distances from v is at most k . We denote this set by N k [ v ]. When k = 1 we omit the distance-1 part from the name and the upper index 1 from the notation.We are going to use small graphs as a building blocks in our construction. We call these graphs special . Definition 2.1
Graph H is special if the following properties are fulfilled:1. The diameter of H is two.2. H does not have a dominating edge.3. H has two vertices, called a special pair , u and v such that:(a) Distance between u and v is two.(b) If we remove u or v from H then the obtained subgraph does not have a dominating edge.(c) ( N ( u ) ∩ N ( v )) ∪ { u, v } is a dominating set of H . Claim 2.2
Let K m be the complete graph with V ( K m ) = { x , . . . , x m } . Then K m (cid:3) K m , the complementof K m (cid:3) K m , is special for any m ≥ . roof: Two vertices are adjacent in K m (cid:3) K m if and only if both of their coordinates are different. Itsdiameter is two since any two vertices share a common neighbor.If ( x i , x j ) and ( x k , x l ) are two adjacent vertices, then i = k , j = l , and so neither ( x i , x l ) nor( x k , x j ) is dominated by { ( x i , x j ) , ( x k , x l ) } . Hence two adjacent vertices do not dominate V ( K m (cid:3) K m ).Furthermore if we remove a vertex, at least one of the two not dominated vertices remains, which is stillnot dominated.Let u := ( x , x ) and v := ( x , x ) be the special pair. Then ( x m , x m ) is a common neighbor of u and v and it dominates everything except ( x i , x m ) and ( x m , x i ) where i < m . Vertices u and v dominateall of these except ( x , x m ), but this last one is dominated by ( x , x ) which is also a common neighborof u and v . Therefore ( N ( u ) ∩ N ( v )) ∪ { u, v } is a dominating set of K m (cid:3) K m . (cid:3) There are many other special graphs. In this paper we mention just one more example: the circulantgraph where the vertices are labeled from 1 to 2 m , m ≥ i and j are adjacent if andonly if i − j m, m ± m .Muntz et al. showed that the optimal pebbling number of a diameter two graph lacking a dominatingedge is 4. Therefore the optimal pebbling number of any special graph is 4. Using this we can prove thefollowing theorem: Theorem 2.3
For any ǫ > there is a graph G on n vertices with diameter two such that π ∗ ( G ) > (4 − ǫ ) nδ +1 . Proof:
For a fixed ǫ > K m (cid:3) K m where m > max( aa − ,
2) and a = q − ǫ . Eachdegree in K m (cid:3) K m is ( m − and its order is m . Since both a and m are greater than 1 we have that a > mm − , which implies the following: π ∗ ( K m (cid:3) K m ) = 4 = (4 − ǫ ) a > (4 − ǫ ) m ( m − > (4 − ǫ ) nδ + 1 (cid:3) Now we turn our attention to the case when the diameter is at least three. For any d and ǫ we showa graph whose diameter is 9 d − (cid:0) − ǫ (cid:1) nδ +1 . In factwe connect 3 d special graphs sequentially using the special pairs and prove that the optimal pebblingnumber of the obtained graph is 8 d . To obtain the desired result we choose the special graphs to be K m (cid:3) K m again.Let H , H , . . . , H l be (not necessarily the same) special graphs. Denote the vertex set of H i by B i andlet u i and v i be a special pair of H i . Connect these graphs sequentially by placing edges between v i and u i +1 to obtain a new graph G l . Consequently V ( G l ) = ∪ li =1 B i and E ( G l ) = ∪ li =1 E ( H i ) S ∪ l − i =1 { v i , u i +1 } .We say that a H i subgraph of G l is a block . Claim 2.4
Let k ∈ Z + , r ∈ { , , } , and let G k + r be the graph defined above. Then π ∗ ( G k + r ) ≤ k if r = 08 k + 4 if r = 18 k + 6 if r = 2 . Proof:
To construct a solvable distribution with 8 k pebbles in the r = 0 case we place 4 pebbles atvertices v j − and u j where j = 1 , , . . . , k . If a vertex of an H i has 4 pebbles, then each vertex of H i is reachable because H i is a diameter two graph. Otherwise u i and v i are both adjacent to verticeshaving 4 pebbles. Therefore each element of ( N ( u i ) ∩ N ( v i )) ∪ { u i , v i } is 2-reachable and since this isa domination set of H i , we can move a pebble to any vertex of H i . When r = 1 we use the same3onstruction and place 4 additional pebbles at a vertex of H k +1 , which solves H k +1 . In the last r = 2case we start again with the r = 0 construction and place 3 pebbles at both v k +1 and u k +2 . Thesetwo vertices are 4-reachable, therefore all vertices in the last two blocks are solvable. (cid:3) Claim 2.5
Let G l be a graph defined above, then π ∗ ( G ) = 4 , π ∗ ( G ) = 6 , π ∗ ( G ) = 8 . Proof:
The upper bounds have been shown. G is a special graph, therefore its optimal pebblingnumber is 4.Let D be a pebbling distribution of G having 5 pebbles. We can assume that H has less pebblesthan H , therefore D ( B ) ≤
2. Since π ∗ ( H ) = 4, we can not reach each vertex of H without theusage of the pebbles placed at B . H and H are connected by only one edge, therefore the only wayto use the pebbles of B is to move as many pebbles through this edge as possible. Furthermore, if wedo nothing with the pebbles of B and we move as many pebbles from B to v as possible, then theobtained distribution of H has to be solvable if D is solvable on G . But this obtained distribution canhave at most D ( B ) + D ( B ) / ≤ D is not solvable.To prove the last statement we argue that a distribution with 7 pebbles is not solvable on G . Let l := D ( B ). If l ≥
3, then we can accumulate on B ∪ B at most l/ − l ) < l = 0,then 4 pebbles must be obtained from 7 on B ∪ B which is not possible and if l = 1, then to beable to solve for B the remaining 6 pebbles must placed on u which is not solvable for B . Finally if D ( B ) = 2 = D ( B ), then only 3 pebbles are on B and it is not possible to move additional two to B . (cid:3) Claim 2.6
Let G l be a graph defined above, then π ∗ ( G l ) ≥ l . To prove this claim we need some preparation. We are going to use a cut argument which requiresthat in an optimal distribution several edges can not transfer pebbles. Therefore these edges can beremoved from the graph without changing the optimal pebbling number. We are going to remove edgesin such a way that the obtained graphs has two connected components and each of them is a smallerinstance of G l or it is almost a G l .Let G − l be the subgraph of G l which we obtain by deleting v l . Let G + l be the following graph: Wetake G l and add a leaf to u . Let G +0 be the one vertex graph. Lemma 2.7
Let D be a solvable distribution D on G l such that | D | < l − . Then there is a decom-position of G l into parts isomorphic to either G k and G l − k , with ≤ k < l , or G − k and G + l − k , with ≤ k ≤ l , such that no pebbles can be moved through the edges connecting the two parts. We are going to use the collapsing technique which was introduced in [2]. Let G and H be simplegraphs. We say that H is a quotient of G , if there is a surjective mapping φ : V ( G ) → V ( H ) suchthat { h , h } ∈ E ( H ) if and only if there are g , g ∈ V ( G ), where { g , g } ∈ E ( G ), h = φ ( g ) and h = φ ( g ). We say that φ collapses G to H , and if D is a pebbling distribution on G , then the collapseddistribution D φ on H is defined in the following way: D φ ( h ) = P g ∈ V ( G ) | φ ( g )= h D ( g ). Claim 2.8 If v is k -reachable under D , then φ ( v ) is k -reachable under D φ . This claim is a generalization of the Collapsing Lemma [2]. The proof given by Bunde et al. can beused to prove our claim. We are going to collapse G l to the path containing 3 l vertices, denoted by P l ,therefore we state some results about the path. Theorem 2.9 ([2])
A 2-optimal distribution of the n -vertex path contains n + 1 pebbles. Claim 2.10
If an inner vertex of the path is not 2-reachable, then one of its neighbors is not either. roof: If a vertex v is not 2-reachable, then v cuts the path to two, and no pebble can be movedthrough v . If both neighbors of v are 2-reachable, then we can move 2 pebbles to them simultaneously,since v can receive two pebbles in total, one from each neighbor, which is a contradiction. (cid:3) Proof of Lemma 2.7:
Let D ′ be the following distribution on G l : D ′ ( u ) = D ( u ) + 1, D ′ ( v l ) = D ( v l ) + 1 and D ′ ( x ) = D ( x ) if x / ∈ { u , v l } . Since D is a solvable distribution, u and v l are 2-reachableunder D ′ .Let P l be a path on 3 l vertices and denote its vertices with p , p , . . . p l . Let φ be a mapping whichmaps G l to P l in such a way that φ ( u i ) = p i − , φ ( v i ) = p i and φ ( x ) = p i − if x ∈ B i \ { u i , v i } . P l is a quotient of G l . D ′ φ has less than 3 l + 1 pebbles, therefore according to Theorem 2.9 it is not a 2-solvable distributionof P l . There is a vertex p i in P l which is not 2-reachable under D ′ φ . Claim 2.8 implies that both p and p l are 2-reachable, so 1 < i < l . Because p i is not 2-reachable, no pebble can move from p i toeither p i − or p i +1 . Similarly, pebbles cannot move from both p i − and p i +1 to p i . Thus, for one ofthe edges incident with p i , without loss of generality { p i , p i +1 } , no pebble can move across it. This alsomeans that p i +1 is not 2-reachable.Claim 2.8 yields that the vertices of φ − ( p i ) ∪ φ − ( p i +1 ) are also not 2-reachable under D . Thereforeno pebbles can be passed between φ − ( p i ) and φ − ( p i +1 ). Deleting the edges between them makes thegraph disconnected and leaves D solvable.The two connected components are isomorphic to either G k and G l − k or G − k and G + l − k , where i ∈ { k, k − , l − k ) + 1 } . This comes from the collapsing function. (cid:3) Lemma 2.7 guides us to make an induction argument. However we also need some information about π ∗ ( G + l ) and π ∗ ( G − l ). The 3(b) property of special graphs guarantees that these values are at least π ∗ ( G l ). Lemma 2.11 π ∗ ( G + l ) ≥ π ∗ ( G l ) and π ∗ ( G − l ) ≥ π ∗ ( G l ) . Proof:
Adding a leaf can not decrease the optimal pebbling number, hence the first inequality holds. G − is a diameter two graph without a dominating edge, therefore π ∗ ( G − ) = 4 = π ∗ ( G ). To prove therest of the assertion, we show that there is an optimal distribution D ′ of G − l which is also a solvabledistribution on G l (where we interpret D ′ to be a distribution on G l with no pebbles on v l ).Let D be an optimal distribution of G − l . Denote the last block of G − l , where the removed vertex waslocated, by H ′ l . Since H l was special, H ′ l does not have a dominating edge and its diameter is at leasttwo, therefore π ∗ ( H ′ l ) ≥ H ′ l can obtain pebbles from the rest of the graph only through the v l − , u l edge. Let k be themaximum number of pebbles which can arrive at u l using this edge. Since D was solvable, k + D ( V ( H ′ l )) ≥
4. If we relocate all pebbles of H ′ l at u l and put back v l , the obtained D ′ distribution of G l also satisfiesthat k + D ′ ( u l ) ≥
4, therefore each vertex of B l is reachable under D ′ . The other vertices remainedreachable since we moved the pebbles closer to them in one pile. (cid:3) Proof of Claim 2.6:
Assume the contrary. Let G l be a minimal counterexample. Claim 2.5 implies l ≥
4. Let D be an optimal distribution of G l .Since G l is a counterexample, | D | < l ≤ l −
1. Therefore we can apply Lemma 2.7. Accordingto this lemma, we can break G l to G k and G l − k or G − k and G + l − k such that no pebbles can be movedbetween the two parts. This means that D induces solvable distributions on both parts.In the second case when k = l , using Lemma 2.11 gives the following chain of inequalities: π ∗ ( G l ) = D ( G − k ) + D ( G + l − k ) ≥ π ∗ ( G − k ) + π ∗ ( G + l − k ) ≥ π ∗ ( G k ) + π ∗ ( G l − k ) G l was a minimal counterexample, therefore: 5 ∗ ( G l ) ≥ π ∗ ( G k ) + π ∗ ( G l − k ) ≥ k + 83 ( l − k ) = 83 l This contradicts with our assumption. When l = k we have: π ∗ ( G l ) = D ( G − l ) + D ( G +0 ) ≥ π ∗ ( G − l ) + π ∗ ( G +0 ) ≥ π ∗ ( G l ) + 1 , which is also a contradiction. Therefore there is no counterexample. (cid:3) Corollary 2.12 π ∗ ( G k ) = 8 k Theorem 2.13
For any ǫ > and any integer d , there is a graph G , such that its diameter is biggerthan d and π ∗ ( G ) ≥ ( − ǫ ) nδ +1 . Proof:
Consider G d with K m (cid:3) K m blocks, where m > max( aa − ,
3) and a = q / / − ǫ . Its diameter is3 d − δ ( G d ) = δ ( K m (cid:3) K m ) = ( m − and | V ( G d ) | = 3 dm . The optimal pebbling number of G d is 8 d . If we repeat the calculation of Theorem 2.3 we receive the desired result: π ∗ ( G d ) = 8 d = 3 d (cid:18) − ǫ (cid:19) a > d (cid:18) − ǫ (cid:19) m ( m − > (cid:18) − ǫ (cid:19) nδ + 1 (cid:3) In this section we give a construction of a pebbling distribution having at most n δ +1) pebbles for anygraph whose diameter is at least three.We are going to talk about several graphs on the same labeled vertex set. To make it clear whichgraph we are considering in a formula we write the name of the graph as a lower index, i.e. d G ( u, v ) isthe distance between vertices u and v in graph G .We define distances between subgraphs in the natural way: If H and K are subgraphs of G , then d G ( H, K ) = min u ∈ V ( H ) ,v ∈ V ( K ) ( d G ( u, v )).We can think about a vertex as a subgraph, therefore let distance- k open neighborhood of a subgraph H be the set of vertices whose distance from H is exactly k . We define the closed neighborhood similarly.Note that N d ( H ) = N d [ H ] \ N d − [ H ].The following property will be useful in our investigations: A vertex v ∈ V ( G ) is strongly reachable under D if each vertex from the closed neighborhood of v is reachable under D . This property togetherwith traditional reachability partition the vertex set to three sets T ( D ), H ( D ) and U ( D ), where T ( D )includes the strongly reachable vertices, the vertices of H ( D ) are reachable but not strongly reachableand U ( D ) contains the rest of the vertices. Theorem 3.1
Let G be a connected graph, such that its diameter is bigger than two and δ is its minimumdegree. We have π ∗ ( G ) ≤ n δ + 1) . Let D and D ′ be pebble distributions. D ′ is an expansion of D ′ ( D ≤ D ′ ) if ∀ v ∈ V ( G ) D ( v ) ≤ D ′ ( v ).If D = D ′ , then we write that D < D ′ . If D ′ is an expansion of D , then let ∆ D,D ′ be the pebblingdistribution defined as: ∆ D,D ′ ( v ) = D ′ ( v ) − D ( v ) ∀ v ∈ V ( G ).6f we would like to create a solvable distribution, then we can do it incrementally. We start withthe trivial distribution with no pebbles and add more and more pebbles to it. So we have a sequence ofdistributions 0 < D < D < · · · < D k − < D k where D k is solvable. The number of reachable vertices isgrowing during this process. We can ask which vertices are reachable, strongly reachable, or not reachableafter the i th step. Let T ( D i ), H ( D i ), U ( D i ) denote these sets respectively. Note that T ( D i ) ⊆ T ( D i +1 ),while U ( D i ) ⊇ U ( D i +1 ). Furthermore we know that T ( D k ) = V ( S ) and H ( D k ) = U ( D k ) = ∅ .If for each i the difference | ∆ D i ,D i +1 | is relatively small and |T ( D i +1 ) \ T ( D i ) | is relatively big, thenit yields that | D k | is not so big.To make this intuitive idea precise we define the strengthening ratio . Definition 3.2
Suppose that we have distributions D and D ′ on graph G , such that D < D ′ . Denotethe difference of the size of these distribution by ∆ p D,D ′ = | D ′ | − | D | = | ∆ D,D ′ | . We use ∆ T D,D ′ for set T ( D ′ ) \ T ( D ) and ∆ t D,D ′ denotes the cardinality of this set.We say that the strengthening ratio of the expansion D < D ′ is: E ( D, D ′ ) = ∆ t D,D ′ ∆ p D,D ′ The strengthening ratio of distribution D = 0 is E (0 , D ) , and the strengthening ratio of D = 0 is ∞ . Fact 3.3 If D is solvable, then | D | = n E (0 ,D ) . This fact shows that if we want to give a solvable distribution whose size is close to the optimum,then its strengthening ratio is also close to the optimum. Furthermore, a smaller solvable distributionhas bigger strengthening ratio. The next lemma shows that if we break D k to a sequence of expansions0 < D < D < · · · < D k − < D k , then the strengthening ratio of each expansion is a lower boundfor E (0 , D k ). Therefore we are looking for an expansion chain where the minimum strengthening ratioamong all expansion steps is relatively big. Lemma 3.4
Let D , D and D are distributions on G . If D < D and D < D , then E ( D , D ) ≥ min( E ( D , D ) , E ( D , D )) . Proof:
Let a, b, c, d be nonnegative real numbers, then the following inequality can be easily proven byelementary tools: a + bc + d ≥ min (cid:18) ac , bd (cid:19) . Using this and the definition of strengthening ratio, we obtain E ( D , D ) = ∆ t D ,D ∆ p D ,D = ∆ t D ,D + ∆ t D ,D ∆ p D ,D + ∆ p D ,D ≥≥ min (cid:18) ∆ t D ,D ∆ p D ,D , ∆ t D ,D ∆ p D ,D (cid:19) = min ( E ( D , D ) , E ( D , D )) . (cid:3) In the next lemma we state that we can construct a distribution D with some special properties.This lemma formalizes the following idea: If there are pairs of adjacent vertices, such that the closedneighborhood of each pair is large, then we can make all vertices of these pairs reachable with few pebbles,while lots of other vertices become reachable. The connection between few and lots of is established bystrengthening ratio. 7 emma 3.5 Let G be an arbitrary simple connected graph. There is a pebbling distribution D on G which satisfies the following conditions:1. The strengthening ratio of D is at least ( δ + 1) .2. If ( u, v ) is an edge of G and | N [ u ] ∪ N [ v ] | ≥ ( δ + 1) , then both of u and v are reachable under D . Proof:
Our proof is a construction for such a D :We say that and edge ( u, v ) has * property if and only if | N [ u ] ∪ N [ v ] | ≥ ( δ + 1).First of all, if there is no ( u, v ) edge in G with * property, then the trivial distribution 0 is good tobe D . Otherwise, we have to make reachable each vertex of any edge which has * property. To makethis we search for these edges, and if we find such an edge such that at least one of its vertices is notreachable, then we add some pebbles on D to make it reachable.We will define sets H, A, B ⊂ V ( G ), P, R ⊂ V ( G ) (cid:3) V ( G ) and let L p be a set containing vertices of G for each p ∈ P .These sets, except H , will contain the edges with * property or their vertices. They will have thefollowing semantics at the end of the construction: • Each element of H will be reachable under D , but not necessarily all of the reachable verticescontained in it. • Each vertex of B has a neighbor who has at least two 4-reachable distance-2 neighbors, or has an8-reachable distance-3 neighbor. • The elements of P are edges whose vertices will be 4-reachable. • The elements of R are edges whose vertices will be 8-reachable. • L p contains vertices from A whose distance from p is exactly 3.Then do the following:1. Choose an edge ( u, v ) which has * property and u, v / ∈ H . If we can not choose such an edge, thenmove to step 3.2. Add the elements of N [ u ] ∪ N [ v ] to H . Add ( u, v ) to P . Move to step 1.3. Search for an edge ( u, v ) which has * property and v / ∈ H . If we can not find one, then move tostep 6.4. Add the elements of N [ v ] to H .5. Count the number of pairs p in P whose distance from u is 2. If we get more than one, then add v to B and move to step 3. Otherwise, add v to A and add v to the set L p where p is the only pairwhose distance from u is 2.6. Do for each p ∈ P : If | L p | ≥
5, then move the elements of L p from A to B and also move p from P to R .7. Let D be the following: D ( v ) = v ∈ A, v ∈ B or v is an element of pair p ∈ P, v is the first element of pair p ∈ R, v is the second element of pair p ∈ R, . D . To seethis consider H . Each vertex of H is reachable under D by construction. We expanded H by distance-2closed neighborhoods of vertices which are 4-reachable in each step. Each vertex of an edge with *property is contained in H , A or B .Hence the second condition is satisfied. So we just need to verify the first one.The vertices of sets A , B , and vertices of edges contained in P and R are all 4-reachable. Hence eachvertex belongs to their neighborhood is strongly reachable. This implies that:∆ t ,D = |T ( D ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ p ∈ P ∪ R N [ p ] [ [ v ∈ A ∪ B N [ v ] !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = X p ∈ P ∪ R | N [ p ] | + X v ∈ A ∪ B | N [ v ] | For the second equality we need that these neighborhoods are disjoint, but this is true because of theconstruction: The distance between a vertex of A ∪ B and a pair p of P ∪ R is at least 3. The distancebetween p, p ′ ∈ P ∪ R is also at least 3. Both of these are guaranteed by step 2. d ( u, v ) ≥ u, v ∈ A ∪ B because of step 4.Using the * property of edges contained in P and R gives:∆ t ,D ≥ X p ∈ P ∪ R | N [ p ] | + X v ∈ A ∪ B | N [ v ] | ≥ ( | P | + | R | ) · δ + 1) + ( | A | + | B | )( δ + 1) , ∆ p ,D = | D | = 4 | A | + 3 | B | + 6 | P | + 11 | R | , E (0 , D ) = ∆ t ,D ∆ p ,D ≥ ( | P | + | R | ) · ( δ + 1) + ( | A | + | B | )( δ + 1)4 | A | + 3 | B | + 6 | P | + 11 | R | . Using ( a + b ) / ( c + d ) ≥ min( a/c, b/d ): E (0 , D ) ≥ min (cid:18) ( | P | + | A | )( δ + 1)6 | P | + 4 | A | , ( | R | + | B | )( δ + 1)11 | R | + 3 | B | (cid:19) . Step 6 of the construction implies that | A | ≤ | P | and | B | ≥ | R | .Let | A | = 4 x | P | . In this case 0 ≤ x ≤ x for the first part,which gains its minimum at x = 1:( | P | + | A | )( δ + 1)6 | P | + 4 | A | = ( + 4 x )( δ + 1)6 + 16 x ≥ δ + 1) . Let | B | = 5 y | R | . | B | ≥ | R | implies that 1 ≤ y . The function which we get from the second partgains its minimum at y = 1:( | R | + | B | )( δ + 1)11 | R | + 3 | B | = ( + 5 y )( δ + 1)11 + 15 y ≥
415 ( δ + 1) . This completes the proof of the Lemma. (cid:3)
During the proof we will show that a non solvable distribution whose strengthening ratio is abovethe desired bound always can be expanded to a bigger one whose strengthening ratio is still reasonable.To do this we want to decrease the number of vertices which are not strongly reachable. Usually weplace some pebbles at not reachable vertices. We know that if a vertex v is not reachable under D andwe make it 4-reachable, then all vertices of its closed neighborhood, which were not strongly reachable,become strongly reachable.We usually consider a connected component S of the graph induced by U ( D ).9here are several reasons why we do this. First of all, a chosen S is a small connected part of G where none of the vertices are reachable, hence it is much simpler to work with S instead of the wholegraph.A vertex from S has the property that none of its neighbors are strongly reachable. Thus, if we makea vertex from S S is that if we add some additional pebbles to S and makesure that all of its vertices become reachable, then these vertices become strongly reachable, too.If we make u and v both 4-reachable with at most 7 pebbles and their closed neighborhoods aredisjoint then this is good for us. The disjointness of the neighborhoods happens when d ( v, u ) ≥ S , which is a connected component of U ( D ). On the one hand,it is beneficial, but on the other hand it makes some trouble when we consider distances. Let u and v be vertices of S . Their distance can be different in G and S . For example if G is the wheel graph on n vertices and we place just one pebble at the center vertex, then S is the outer circle and the distancebetween two vertices of S can be (cid:4) n − (cid:5) , while their distance in G is not larger than 2.This difference is important because this shows that we can not decide the disjointness of closedneighborhoods by distance induced by S . The first idea to handle this is considering the original distancegiven by G , but then we have to consider the whole graph, which we would like to avoid. To overcomethis problem we make the following compromise:We count distances in graph N [ S ]. Clearly, this distance also can be smaller than the correspondingdistance in G , but it happens only for values higher than 3. Hence this N [ S ] distance determinedisjointness of the neighborhoods, and it will be enough for our investigation.The following lemmas will be used in the proof. Fact 3.6
Let S and B be induced subgraphs of G such that V ( B ) = N G [ V ( S )] . If max u,v ∈ V ( S ) ( d B ( u, v ))= 3 , then either there exist vertices a, b, c, d ∈ V ( S ) , such that they are neighbors in this order and d B ( a, d ) = 3 , or there exist vertices a, d ∈ V ( S ) such that d B ( a, d ) = 3 and there is a path P between a and d whose length is 3 and P contains a vertex from V ( B ) \ V ( S ) . Lemma 3.7
Let δ be the minimum degree of graph G . Let S and B are connected induced subgraphsof G , such that V ( B ) = N G [ V ( S )] . If max u,v ∈ V ( S ) ( d B ( u, v )) = 3 and exist a, d ∈ V ( S ) , such that d B ( a, d ) = d S ( a, d ) = 3 , then there is an u, v edge in S whose closed neighborhood has size at least ( δ + 1) . Proof:
Let a, b, c, d be the vertices of a shortest path between a and d which lies in S . If the statementholds for edge ( a, b ) or ( c, d ), then we have found the edge which we are looking for. Thus assume thecontrary. The Inclusion-exclusion principle gives us the following result for vertex pair a, b : | N [ a ] ∩ N [ b ] | = | N [ a ] | | {z } ≥ δ +1 + | N [ b ] | | {z } ≥ δ +1 − | N [ a ] ∪ N [ b ] | | {z } < δ + > δ + 23 . The same is true for pair c, d .The distance of a, d implies that N [ a ] ∩ N [ d ] = ∅ . Thus ( N [ a ] ∩ N [ b ]) ∩ ( N [ c ] ∩ N [ d ]) = ∅| N [ b ] ∪ N [ c ] | ≥| ( N [ b ] ∩ N [ a ]) ∪ ( N [ c ] ∩ N [ d ]) | == | N [ b ] ∩ N [ a ] | + | N [ c ] ∩ N [ d ] | − | ( N [ a ] ∩ N [ b ]) ∩ ( N [ c ] ∩ N [ d ]) | >> (cid:18) δ + 23 (cid:19) − δ + 43 . So edge b, c has the required property. (cid:3) emma 3.8 Let S and B be connected induced subgraphs of G , such that V ( B ) = N G [ V ( S )] . Assumethat there are vertices u and v in S whose distance in B is 4. Then at least one of the following conditionsholds:1. There exists a, b ∈ V ( S ) , such that d S ( a, b ) = d B ( a, b ) = 4
2. There exists c, d ∈ S , such that d B ( c, d ) = 3 and some of the shortest paths between c , d contain avertex from set V ( B ) \ V ( S ) . Proof:
Consider a pair of vertices u, v ∈ V ( S ) whose distance in B is four. It is clear that d S ( u, v ) ≥ S between u and v isgreater than four. Let P be a shortest path between u and v which lies in S . The length of P is at leastfive. Label the vertices of P as u = p , p , p , . . . , p k = v . Let i be the smallest value such p i does nothave a neighbor in N B [ u ]. The minimality of i implies that d B ( u, p i ) = 3. If i >
3, then the shortestpath between u and p i , which has length three, has to contain a vertex from V ( B ) \ V ( S ). This givesus the second condition.Otherwise i = 3. Let j be the smallest value, such that p j does not have a neighbor in N B [ u ]. Thecase j = 4 gives us d B ( u, p j ) = 4 = d S ( u, p j ) which fulfills the first condition. The other case is j > d B ( p , p ) = 3. It can happen if and only if the second condition holds. (cid:3) Lemma 3.9
Let δ be the minimum degree of G , S and B be induced subgraphs of G , such that V ( B ) = N [ V ( S )] . If max u,v ∈ V ( S ) ( d B ( u, v )) ≥ and exist vertices a, e ∈ V ( S ) such that d B ( a, e ) = d S ( a, e ) = 4 ,then one of the following two conditions holds:1. There exist u, v ∈ S such that d B ( u, v ) = 2 and | N B [ u ] ∪ N B [ v ] | ≥ ( δ + 1) .2. | N [ a ] ∪ N [ e ] ∪ ( N [ b ] ∩ N [ d ]) | ≥ ( δ + 1) , where a, b, c, d, e are the vertices of a path lying in S . Proof:
Assume that 1. does not hold. This gives us the following estimate on the size of the commonneighborhood of b and d : | N G [ b ] ∩ N G [ d ] | = | N G [ b ] | | {z } ≥ δ +1 + | N G [ d ] | | {z } ≥ δ +1 − | N G [ b ] ∪ N G [ d ] | | {z } < ( δ +1) >
215 ( δ + 1)since a and d do not have a common neighbor. The same is true for the pairs of b , e , and a , e whichimplies: | N [ a ] ∪ N [ e ] ∪ ( N [ b ] ∩ N [ d ]) | = | N [ a ] | + | N [ e ] | + | N [ b ] ∩ N [ d ] | ≥ δ + 1) . So if 1. does not hold, then 2. does. (cid:3)
The next lemma will be useful to give a lower bound on the number of vertices becoming stronglyreachable after the addition of some pebbles to S . Lemma 3.10
Let D be a pebbling distribution on G . Let S be a connected component of the subgraphwhich is induced by U ( D ) . Consider D ′ such that D ≤ D ′ . If there is an s ∈ V ( S ) such that s is2-reachable under ∆ D,D ′ and each vertex of S is reachable under D ′ , then N [ s ] ⊆ T ( D ′ ) , furthermore N [ s ] ⊆ ∆ T D,D ′ . Proof:
We show that each neighbor of s is strongly reachable under D ′ . Let v be a neighbor of s , and u be a neighbor of v . s is 2-reachable under D ′ hence v is reachable.11f u is reachable under D or it is a vertex of S , then it is reachable under D ′ . Else u is in U ( D ) \ V ( S ).So v separates two connected components in the induced subgraph by U ( D ), so v is reachable under D . s is 2-reachable under ∆ D,D ′ , thus v is also 2-reachable under D ′ and u is reachable. s was not reachable under D hence its neighbors were not strongly reachable under D . Therefore N [ s ] ⊆ ∆ T D,D ′ (cid:3) Proof of Theorem 3.1:
Indirectly assume that there is a graph G such that π ∗ ( G ) > n δ +1) anddiam( G ) >
2. This means that each solvable distribution has strengthening ratio below δ +1)15 .Let D be a pebbling distribution which satisfies the properties of Lemma 3.5. Let D be an expansionof D such that the strengthening ratio of D is at least δ +1)15 and subject to this requirement | D | ismaximal. According to our first assumption D is not solvable. We will show that either | D | is notmaximal or D is not an expansion of D . The first one is shown if we give a distribution D ′ such that D < D ′ and E D,D ′ ≥ δ +1)15 . We will give ∆ D,D ′ instead of D ′ . Clearly D , and ∆ D,D ′ together aredetermine D ′ .At each case we will assume that the conditions of the previous cases do not hold. Case A:
There exist u, v ∈ U ( D ) such that d ( u, v ) = 3 and there is a vertex w on a shortest pathbetween u and v which is contained in H ( D ).W.l.o.g. assume that w is a neighbor of v . Then let ∆ D,D ′ be the following:∆ D,D ′ ( x ) = x = u, x = v, .v is 4-reachable under D ′ , because w is reachable under D and it gets a pebble from u under δ D,D ′ , so w is 2-reachable without the three pebbles of v . This means that each vertex of the closed neighborhoodof u and v are strongly reachable. N [ u ] and N [ v ] are disjoint vertex sets and they are subsets of ∆ T ( D, D ′ ). Hence E ( D, D ′ ) ≥ | N [ u ] ∪ N [ v ] || ∆ D,D ′ | ≥ δ + 1)7 >
415 ( δ + 1) , so | D | was not maximal. Case B: max u,v ∈ V ( S ) d B ( u, v ) ≥ S whose length is four in both S and B by Lemma 3.8.Apply Lemma 3.9. If there are vertices u and v from V ( S ) such that d B ( u, v ) = 2 and | N B [ u ] ∪ N B [ v ] |≥ ( δ + 1), then let w be a common neighbor of u and v and choose ∆ D,D ′ as follows:∆ D,D ′ ( x ) = x ∈ { u, v } , x = w, . Each of u, v, w is 4-reachable, hence:∆ t D,D ′ ≥ | N [ u ] ∪ N [ v ] | ≥ δ + 1) , | ∆ D,D ′ | = 7, thus E ( D, D ′ ) ≥ ( δ + 1).If there is no such an u, v pair, then by Lemma 3.9 there is a path a, b, c, d, e in S such that d B ( a, e ) = d S ( a, e ) = 4, and | N ( b ) ∩ N ( c ) | ≥ ( δ + 1). Consider ∆ D,D ′ as follows:∆ D,D ′ ( x ) = ( x ∈ { a, e } , . N [ a ] ∪ N [ e ] ∪ ( N [ b ] ∩ N [ d ]) are 2-reachable, thus they are also strongly reachable.∆ t ≥ | N [ a ] ∪ N [ e ] ∪ ( N [ b ] ∩ N [ d ]) | = | N [ a ] | + | N [ e ] | + | N [ b ] ∩ N [ d ] | ≥ δ + 1) , E ( D, D ′ ) ≥ δ + 1)8 ·
15 = 4( δ + 1)15 . Case C: max u,v ∈ V ( S ) d B ( u, v ) = 3.If the conditions of Case A do not hold, then we can use Lemma 3.7 because of Fact 3.6. Let ( u, v )be the edge whose neighborhood size is at least | ( δ + 1) | . We will use this property only in the fourthsubcase.Consider the set K , which is a set of vertex sets. K is an element of K if and only if K is a subsetof V ( S ) such that for all k, j ∈ K , k = j implies that d B ( k, j ) ≥ | K | ≥ K is maximal (we cannot add an element to K ). max u,v ∈ V ( S ) d B ( u, v ) = 3 implies that K is not empty.The objective in this case is to use Lemma 3.10 for the vertices of K . Because this means that thevertices of ∪ k ∈ K N [ k ] are strongly reachable. Furthermore, N [ k ] and N [ k ] are disjoint if k , k ∈ K and k = k . These imply that ∆ t ≥ ∪ k ∈ K | N ( k ) | ≥ | K | ( δ + 1). To use this Lemma we need to give a proper∆ D,D ′ distribution and check that each vertex of S is reachable and each vertex of K is 2-reachableunder it.There are four subcases here: Subcase 1: ∀ s ∈ V ( S ) d B ( v, s ) ≤ K be an arbitrary element of K . Note that v / ∈ K .∆ D,D ′ ( x ) = x = v, x ∈ K, . Each vertex of S is reachable with the pebbles placed at v and the vertices of K are 2-reachable. E ( D, D ′ ) ≥ | K | ( δ + 1)4 + | K | ≥
13 ( δ + 1) . Subcase 2: ∀ s ∈ V ( S ) min( d B ( u, s ) , d B ( v, s )) ≤
2, but ∃ w ∈ V ( S ) d B ( v, w ) = 3.Choose K such that v ∈ K . Such a K is exists.∆ D,D ′ ( x ) = x ∈ { u, v } , x ∈ K \ { v } , .u and v are 4-reachable, hence all vertices of S are reachable. Furthermore, each vertex of K is 2-reachable. E ( D, D ′ ) ≥ | K | ( δ + 1)6 + | K | − ≥
27 ( δ + 1) . Subcase 3: ∃ s ∈ V ( S ) d B ( s, u ) = d B ( s, v ) = 3 and { s, v } / ∈ K . { s, v } is a subset of some elements of K . Choose K as one of these, | K | ≥ D,D ′ ( x ) = x = v, x ∈ K \ { v } , S is reachable with the pebbles placed at v and the vertices of K are 2-reachable. E ( D, D ′ ) ≥ | K | ( δ + 1)8 + | K | − ≥
310 ( δ + 1)13 ubcase 4: ∃ s ∈ V ( S ) d B ( s, u ) = d B ( s, v ) = 3 and { s, v } ∈ K . K = { s, v } ∆ D,D ′ ( x ) = ( x ∈ K, .K = { s, v } means that each vertex of S is in N [ s ] ∪ N [ v ], thus each vertex of S is reachable. N [ s ] ∩ ( N [ u ] ∪ N [ v ]) = ∅ hence:∆ t D,D ′ ≥ | N [ s ] ∪ N [ v ] ∪ N [ u ] | = | N [ s ] | + | N [ v ] ∪ N [ u ] | ≥
73 ( δ + 1) , E ( D, D ′ ) ≥
724 ( δ + 1) . Case D: max u,v ∈ V ( S ) d B ( u, v ) ≤ s of S , then all vertices of S and N [ s ] becomesstrongly reachable. Subcase 1: | V ( S ) | ≥ ( δ + 1).Let v be a vertex of S . ∆ D,D ′ ( x ) = ( x = v, . E ( D, D ′ ) ≥ δ + 1)15 · δ + 1) Subcase 2a: ∃ u, v ∈ V ( s ) such that | N [ u ] ∪ N [ v ] | ≥ ( δ + 1) and u and v are neighbors.∆ D,D ′ ( x ) = ( x = v, . Each vertex of S is reachable and u and v are 2-reachable under ∆ D,D ′ . Using Lemma 3.10 we get thatthe neighborhoods of u and v are both strongly reachable. E ( D, D ′ ) ≥ | N [ u ] ∪ N [ v ] | ≥ δ + 1)15 · δ + 1) . Subcase 2b: ∃ u, v ∈ V ( s ) such that | N [ u ] ∪ N [ v ] | ≥ and u and v share a common neighbor w ∈ V ( s ). ∆ D,D ′ ( x ) = ( x = w, . We can say the same like in the previous case.
Subcase 3a: | V ( S ) | ≤ ( δ + 1) and ∀ u, v ∈ V ( S ) ∃ h ∈ H ( D ) h ∈ N ( u ) ∩ N ( v ).Choose v as an arbitrary vertex of S .∆ D,D ′ ( x ) = ( x = v, . If s is a vertex of S other than v , then there is h ∈ H ( D ) which is a neighbor of both of them. h is reachable under D , under D ′ we have two additional pebbles so we can move a pebble to s from v through h . Thus each vertex of S is reachable under D ′ , so we can apply Lemma 3.10 for vertex v . E ( D, D ′ ) ≥ | N [ v ] | ≥ δ + 12 . ubcase 3b: | V ( S ) | ≤ ( δ + 1) and ∃ u, v ∈ V ( S ) ∄ h ∈ H ( D ) h ∈ N ( u ) ∩ N ( v ).The diameter of S (with respect to the distance defined in B ) guarantees that either u and v areneighbors or they share a common neighbor w ∈ V ( B ). Furthermore, in this subcase w ∈ V ( S ). u hasat least δ − ( ( δ + 1) −
1) = ( δ + 1) neighbors in H ( D ), but none of them is a neighbor of v . Hence | N [ u ] ∪ N [ v ] | ≥ ( δ + 1). This is subcase 2a or 2b. Subcase 4: ∃ v ∈ V ( S ), such that ∀ s ∈ V ( S ) d B ( v, s ) = 1.∆ D,D ′ ( x ) = ( x = v, . Each vertex of S is reachable under D ′ so we apply Lemma 3.10 again and get that E ( D, D ′ ) ≥ δ +12 .We have handled all cases when | V ( S ) | ≥ ( δ + 1) or | V ( S ) | ≤ ( δ + 1). So in the next sectionswe assume that ( δ + 1) < | V ( S ) | < ( δ + 1). Before we continue, we need one more definition. Let S be the set of connected components of the graph which is induced by U ( D ). Then we say that S ∈ S is isolated in S if for any other S ′ in S d G ( S, S ′ ) ≥ Subcase 5: ∃ S ∈ S such that S is not isolated.Exists S ′ and u ∈ S , v ∈ S such that d ( u, v ) = 2.∆ D,D ′ ( x ) = x = u, x = v, .u and v are both 4-reachable, hence all vertices of S and S ′ are reachable, furthermore they arestrongly reachable. E ( D, D ′ ) ≥ | V ( S ) ∪ V ( S ′ ) | ≥ · ( δ + 1)7 = 4( δ + 1)15 Subcase 6: S is isolated in S and | N [ S ] | ≥ ( δ + 1).Let s be an arbitrary vertex of S . Then:∆ D,D ′ ( x ) = ( x = s S becomes strongly reachable. We show that the same is true for any vertex of N ( S ).Consider h ∈ N ( S ). h is not strongly reachable under D , but all of its non reachable neighbors under D are contained in S , because S is isolated. Thus under D ′ h is strongly reachable which gives thefollowing result: E ( D, D ′ ) ≥ | N [ V ( S )] | ≥ ( δ + 1)4 = 4( δ + 1)15 . Subcase 7: S is isolated in S and ∃ h ∈ H ( D ) such that for each s ∈ V ( S ) d G ( h, s ) ≤ D,D ′ ( x ) = ( x = h, . Each vertex of S becomes strongly reachable thus: E ( D, D ′ ) ≥ | V ( S ) | > ( δ + 1)3 ≥ δ + 1)15 . Subcase 8:
None of the previous cases hold.In this case we will get a contradiction with D ≤ D . We summarize what we know about D :15 ∀ S ∈ S max u,v ∈ V ( S ) ( d B ( u, v )) = 2, • ∀ S ∈ S ( δ + 1) < | V ( S ) | < ( δ + 1), • ∀ S is isolated in S , • ∀ S ∈ S | N [ S ] | < ( δ + 1), • ∄ h ∈ H ( D ) such that the distance between h and any vertex of S is at most two, where S ∈ S .First of all, the diameter of G is at least 3, hence some pebbles have been placed, so H ( D ) is nonempty.Fix a component S ∈ S . H ( D ) ∩ N ( S ) is also nonempty, because G is connected. Consider an h ∈H ( D ) ∩ N ( S ). The last property guarantees that there is a vertex v in V ( S ) such that d ( v, h ) = 3. Aneighbor of h is in V ( S ). Denote this vertex with u . The first property and d ( v, h ) = 3 together implythat u ∈ N ( v ) ∩ S . h is in N ( v ), hence N [ h ] ∩ N [ v ] = ∅ . N [ v ] ⊆ N [ S ]: | N [ h ] ∩ N [ S ] | ≤ | N [ S ] \ N [ v ] | = | N [ S ] | − | N [ v ] | < δ + 1) − ( δ + 1) = 115 ( δ + 1) . So we have: | N [ h ] \ N [ S ] | ≥ ( δ + 1). u is contained in S , hence all of its neighbors are contained in N [ S ], thus: | N [ u, h ] | ≥ | N [ u ] | + | N [ h ] \ N [ S ] | ≥ δ + 1) .u is in S , so it is not reachable under D , but D is an expansion of D where u has to be reachablebecause | N [ u, h ] | ≥ ( δ + 1) and ( u, h ) ∈ E ( G ). This is a contradiction.We have seen that in each case we have a contradiction, so our assumption was false, hence thetheorem is true. (cid:3) Using this theorem we can prove that the upper bound of Bunde et al. can not hold with equality.
Claim 3.11
There is no connected graph G such that π ∗ ( G ) = nδ +1 . Proof:
Theorem 3.1 shows that the optimal pebbling number of graphs whose diameter is at least threeis smaller. So we have to check only diameter two and complete graphs whose optimal pebbling numberis at most 4. nδ +1 ≥ π ∗ ( K n ) = 2. (cid:3) Corollary 3.12
For any connected graph
G π ∗ ( G ) < nδ +1 and this bound is sharp. Muntz et al. [7] characterize diameter three graph graphs whose optimal pebbling number is eight.Their characterization can be reformulated in the following weird statement:
Claim 3.13
Let G be a diameter graph. π ∗ ( G ) = 8 if and only if there are no vertices x, u, v and w such that N [ x ] ∪ N [ u ] ∪ N [ v ] ∪ N [ w ] = V ( G ) . Theorem 3.1 can be used to establish a connection between this unusual domination property andthe minimum degree of the graph. Note that this is just a minor improvement of the trivial n − Corollary 3.14
Let G be a diameter graph on n vertices. If there are no vertices x, u, v and w suchthat N [ x ] ∪ N [ u ] ∪ N [ v ] ∪ N [ w ] = V ( G ) , then the minimum degree of G is at most n − . Graphs with high girth and low optimal pebbling number
The authors of [2] proved, for k ≥
6, that the family of connected graphs having n vertices, minimumdegree k , and girth at least 2 t + 1 has π ∗ ( G ) /n → t → ∞ . They ask (Question 6.3) whether thesame is true for k ∈ { , , } . We answer this affirmatively for k ∈ { , } . The case k = 3 remains open. Theorem 4.1
Suppose G is a connected graph of order n with δ ( G ) ≥ k and girth ( G ) ≥ t + 1 . If k ≥ then lim t →∞ π ∗ ( G ) /n = 0 . Proof:
For k, t ∈ Z + with k ≥
3, let L = 1 + k ( k − t − k − . Suppose G is a connected graph of order n with δ ( G ) ≥ k and girth ( G ) ≥ t + 1, and consider the following experiment consisting of two steps. Inthe first step, place 2 t pebbles on a vertex v with probability p , independently for each v . In the secondstep, one pebble on every vertex not reachable by pebbles placed in the first step. Clearly the pebblingdistribution is solvable.Let X be the expected number of pebbles used in the experiment. The probability that v is notreachable by the pebbles places in the first step is at most (1 − p ) L because, if at least one vertex in theclosed ball of radius t around v has 2 t pebbles on it, we can solve v . Thus E ( X ) ≤ t pn + (1 − p ) L n, and so π ∗ ( G ) ≤ (2 t p + (1 − p ) L ) n . In particular, for p = ln ( L/ t ) /L , using the bound 1 + x ≤ e x , we get π ∗ ( G ) ≤ (ln( L/ t ) + 1) 2 t nL . We have ( k − t < L < k − t , and so π ∗ ( G ) < (2 + t ln ( k − k − ) t · n . (cid:3) Acknowledgment
The research of Andrzej Czygrinow is supported in part by Simons Foundation Grant
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