Modified Linear Programming and Class 0 Bounds for Graph Pebbling

Transcription

1 Modified Linea Pogamming and Class 0 Bounds fo Gaph Pebbling Daniel W. Canston Luke Postle Chenxiao Xue Cal Yege August 8, 05 Abstact Given a configuation of pebbles on the vetices of a connected gaph G, a pebbling move emoves two pebbles fom some vetex and places one pebble on an adjacent vetex. The pebbling numbe of a gaph G is the smallest intege k such that fo each vetex v and each configuation of k pebbles on G thee is a sequence of pebbling moves that places at least one pebble on v. Fist, we impove on esults of Hulbet, who intoduced a linea optimization technique fo gaph pebbling. In paticula, we use a diffeent set of weight functions, based on gaphs moe geneal than tees. We apply this new idea to some gaphs fom Hulbet s pape to give impoved bounds on thei pebbling numbes. Second, we investigate the stuctue of Class 0 gaphs with few edges. We show that evey n-vetex Class 0 gaph has at least 5 n edges. This dispoves a conjectue of Blasiak et al. Fo diamete gaphs, we stengthen this lowe bound to n 5, which is best possible. Futhe, we chaacteize the gaphs whee the bound holds with equality and extend the agument to obtain an identical bound fo diamete gaphs with no cut-vetex. Intoduction Gaph pebbling was intoduced by Chung in 989. Following a suggestion of Lagaias and Saks, she computed the pebbling numbe of Catesian poducts of paths to give a combinatoial poof of the following numbe-theoetic esult of Kleitman and Lemke. Depatment of Mathematics and Applied Mathematics, Viginia Commonwealth Univesity, Richmond, VA, 8, dcanston@vcu.edu; This autho is patially suppoted by NSA Gant Depatment of Combinatoics and Optimization, Univesity of Wateloo, Wateloo, Ontaio, Canada, NL G, lpostle@uwateloo.ca Depatment of Mathematics and Compute Science, Davidson College, Davidson, NC 805, chxue@davidson.edu Depatment of Mathematics, Davidson College, Davidson, NC 805, cayege@davidson.edu Keywods: pebbling, linea pogamming, Lemke

2 Theoem. [, ] Let Z n be the cyclic goup on n elements and let g denote the ode of a goup element g Z n. Fo evey sequence g, g,..., g n of (not necessaily distinct) elements of Z n, thee exists a zeo-sum subsequence (g k ) k K, such that k K g k. Hee K is the set of indices of the elements in the subsequence. Chung developed the pebbling game to give a moe natual poof of this theoem. Results of this type ae impotant in this aea of numbe theoy, as they genealize zeo-sum theoems such as the Edős-Ginzbug-Ziv theoem [6]. Ove the past two decades, pebbling has developed into its own subfield [, ], with ove 80 papes. We conside a connected gaph G with pebbles (indistinguishable makes) on some of its vetices. Moe pecisely, a configuation p on a gaph G is a function fom V (G) to N {0}. The size of p, denoted p, is v V (G) p(v). A pebbling move emoves two pebbles fom some vetex and places one pebble on an adjacent vetex. A ooted gaph is a pai (G, ) whee G is a gaph and V (G) is the oot vetex. A pebbling configuation p is solvable fo a ooted gaph (G, ) if some configuation p has at least one pebble on, and p can be obtained fom p by a sequence of pebbling moves. Othewise, p is unsolvable (o -unsolvable, when the oot is specified.) The pebbling numbe π(g) is the least intege k such that, fo any vetex v V (G) and any initial configuation p of k pebbles, p is solvable fo (G, v). Likewise π(g, ) is the pebbling numbe of G, when the oot vetex must be. A tivial lowe bound fo π(g) is V (G) : fo some oot, we place one pebble on each vetex othe than, fo a total of V (G) pebbles, but we cannot each. The path, P n, on n vetices has π(p n ) = n. Moe geneally, if gaph G has diamete d, then π(g) d. Let f(n, d) denote the maximum pebbling numbe of an n-vetex gaph with diamete d. Pachte, Snevily, and Voxman [5] poved that f(n, ) = n +, and Clake, Hochbeg, and Hulbet [5] classified all gaphs G of diamete with π(g) = n +. Bukh [] poved that f(n, ) = n/ + O(), and Postle, Steib, and Yege [7] stenthened Bukh s esult, poving the exact bound f(n, ) = n/ +. They also gave [7] an asymptotic bound fo f(n, ). Section gives some necessay peliminaies; in paticula it descibes a technique of Hulbet [0] based on linea pogamming. In Section, we impove Hulbet s method by using a diffeent set of weight functions, based on gaphs moe geneal than tees. We apply this new idea to some gaphs fom his pape to give impoved bounds on thei pebbling numbes. In Section, we investigate the stuctue of Class 0 gaphs (gaphs G with π(g) = V (G) ) with few edges. We show that evey n-vetex Class 0 gaph has at least 5 n edges. This dispoves a conjectue of Blasiak et al []. Fo diamete gaphs, we stengthen this bound to n 5 edges, which is best possible. We chaacteize the gaphs whee it holds with equality and extend the agument to obtain an identical bound fo diamete gaphs with no cut-vetex.

3 Linea Pogamming Peliminaies Computing a gaph s pebbling numbe is had. Watson [8] and Clak and Milans [] studied the complexity of gaph pebbling and some of its vaiants, including optimal pebbling and cove pebbling. Watson showed that it is NP-complete to detemine whethe a given configuation is solvable fo a given ooted gaph (G, ). Clak and Milans efined this esult, showing that deciding whethe π(g) k is Π P -complete; this means that pebbling is in the class of poblems computable in polynomial time by a co-np machine equipped with an oacle fo an NP-Complete language. Hulbet [0] intoduced a new linea pogamming technique, in hopes of moe efficiently computing bounds on pebbling numbes. Befoe we descibe ou impovements on it, we biefly explain his method. Let G be a gaph and let T be a subtee of G ooted at. Fo each v V (T ), let v + be the paent of v, the neighbo of v in T that is close to. A tee stategy is a tee T and an associated nonnegative weight function w T (o w if the context is clea) whee w() = 0 and w(v + ) = w(v) fo evey vetex not adjacent to. Futhe, w(v) = 0 if v V (T ). Let G be the vecto on V (G) in which evey enty is. Hulbet [0] poposed a geneal method fo defining such a weight function though tee stategies. He poved the following esult (hee denotes dot poduct). Lemma. Let T be a tee stategy of G ooted at, with associated weight function w. If p is an -unsolvable configuation of pebbles on V (G), then w p w G. The poof idea is easy. Suppose that p is a configuation with w p > w G. This implies that some vetex v in T has at least two pebbles. Now we make a pebbling move fom v towad the oot, i.e., fom v to v +, to get a new configuation p. Since w(v + ) = w(v), we have w p = w p > w G. By epeating this pocess, we can eventually move a pebble to the oot,. Since evey -unsolvable pebbling configuation p satisfies w p w G, it follows that π(g, ) is bounded above by one plus the numbe of pebbles in the lagest configuation p such that w p w G. Let T be the set of all tee stategies in G associated with oot vetex. By applying Lemma to all of T simultaneously, we aive at the following intege linea pogam: max v p(v) s.t. w p w G fo all T T. Let z G, be the optimal value of this intege linea pogam and let ẑ G, be the optimum of the linea elaxation, so that configuations can be ational. Since z G, ẑ G,, we get the bound π(g, ) z G, + ẑ G, +. Let w,..., w k be weight functions of tee stategies fo tees (possibly diffeent) ooted at, and let w be a convex combination of w,..., w k. If p is an -unsolvable configuation, then w p w G (othewise w i p > w i G, fo some i, a contadiction). Futhe, if w (v) fo all v, then p v w (v) p(v) w G. Fo ease of application, we state this obsevation in a slightly moe geneal fom. We call this the Coveing Lemma.

4 Lemma (Coveing Lemma). Fo a gaph G and a oot V (G), let w be a convex combination of tee stategies fo, and let C and M be positive constants. If w (v) C fo all v V (G)\{} and v V (G)\{} w (v) < M, then π(g, ) M C +. In paticula, if v V (G)\{} w (v) < C V (G), then π(g, ) = V (G). Fo any bound on π(g) aising fom such a w, a cetificate of the bound consists of the stategies w i and thei coefficents in the convex combination foming w. Hulbet applies this linea pogamming method moe boadly by consideing stategies on tees whee w(v + ) w(v), called nonbasic stategies. Since nonbasic stategies ae conic combinations of basic stategies [0, Lemma 5], this extension does not stengthen the method. Howeve, it often yields simple cetificates. Moe Geneal Weight Functions Hee we genealize the notion of weight function fom the pevious section to allow weight functions fo gaphs G that ae not tees. A weight function is a map w : V (G) R + {0}. A weight function fo a gaph G and oot is valid if w() = 0 and evey -unsolvable configuation p satisfies w p w G. Although it is hade to show that one of these moe geneal weight functions is valid, when we can, this often leads to impoved pebbling bounds fo a vaiety of gaph families. Given a gaph G and a oot, it is staightfowad to check that the theoy developed in the pevious section extends to any weight function w such that w p w G fo evey configuation p that is not -solvable. Ou next esult establishes a new family of such weight functions. A k-vetex is a vetex of degee k. Lemma. Fom G fom an even cycle C t by identifying one vetex with the endpoint of a path of length s t. Let x t be the esulting -vetex and x 0 be the -vetex fathest fom x t ; now x 0 and x t split the even cycle into two paths, P and P. Label the intenal vetices of P as x, x,..., x t, x t and the intenal vetices of P as x, x,..., x t, x t. Call the -vetex, and let P be the path fom x t to. Label the intenal vetices of P as x t+, x t+,..., x s, x s. Fo each i 0, give weight i to vetex x i o vetices x i and x i. Let α = s + t and give weight α to x s 0. Fix some ode on the vetices, and let w be the vecto of length V (G) whee enty i is the weight of vetex i. If p is an -unsolvable configuation, then w p w G. Poof. Figues and 6 both show examples of this lemma, which we apply late. Let p be an -unsolvable configuation. We will show that w p w G. Let M = α+ s+ + t. Note that w G = M. Let W 0 = αp(x 0 ), W L = v P \{x 0,x w(v)p(v), t} W R = v P \{x 0,x w(v)p(v), and W t} C = v P \{} w(v)p(v). (Hee L, R, and C stand fo left, ight, and cente.) We will show that W 0 + W L + W R + W C M. Claim. If W L = 0 o W R = 0, then the lemma is tue. By symmety, assume that W L = 0. Now Lemma implies that α W 0 + W R + W C S+. Multiplying by α gives W 0 +W L +W R +W C W 0 +W L +α(w R +W C )

5 α( s+ ) = α + α( s+ ) = α + ( s + t ) = α + s+ + t = M. This poves the claim. Claim. If W L + α W 0 > t, then the lemma is tue. Suppose that W L + α W 0 > t. We can assume that W L < t ; othewise we can move weight down to x t, without changing the sum W L +W C. Now we move some pebbles towad the oot and educe to the case in Claim. Specifically, we emove t W L pebbles fom x 0 and place half that many on x. Call the new configuation p and define W 0, W L, W R, and W C, analogously. Note that α W 0 + W L + W R + W C = α W 0 + W L + W R + W C. Since W L = t, we can move all weight fom intenal vetices of P to x t. This gives a new configuation p with W L = 0. Again α W 0 + W L + W R + W C = α W 0 + W L + W R + W C. So the claim holds by Claim. By Claim, we now assume that W L + α W 0 t. By symmety, we assume that W R + α W 0 t. By Lemma, we can also assume that W C s+ t. Adding these inequalities yields W C + W L + W R + α W 0 ( t ) + s+ t = s+ + s t. () We can assume that W 0 > 0, since othewise the lemma holds by Lemma. Thus, we have W 0 α, so (α )W 0 α. Subtacting this inequality fom () gives the desied esult. The following obsevation extends ou class of valid weight functions a bit futhe. Obsevation. Let G be a gaph and a oot; let w be a weight function on G such that w p w G fo evey -unsolvable configuation p. Fom G fom G by adding a new vetex u adjacent to some vetex u + of G (with u + ), and fom w fom w, whee w (u) = w(u+ ) and w (v) = w(v) fo evey v V (G). Fo evey -unsolvable configuation p in G, we have w p w G. Futhe, we can allow, moe geneally, that w (u) w(u+ ). We can also attach tees, athe than single vetices. Poof. If the new vetex u has moe than one pebble, we move as much weight as possible fom u to u +, which does not decease the total weight on G. This poves the fist statement. The second statement follows fom taking convex combinations of w and w. The final statement follows by induction on the size of the tee T that we attach (we just poved the induction step, and the base case, T = 0, is tivial).. The Cube and the Lemke Gaph To illustate the usefulness of Lemma and Obsevation, we give two easy applications of this method. We show that π(q ) = 8 and π(l) = 8, whee Q is the -dimensional cube and L is the Lemke gaph, shown in Figue. When using tee stategies alone, Hulbet s method cannot handle these gaphs. 5

6 Figue : A valid non-tee weight function. Poposition. If Q is the -cube, then π(q ) = 8. Poof. Evey gaph G satisfies π(g) V (G), so π(q ) 8. Thus, we focus on poving that π(q ) 8. To show that π(q ) 8, we fist note that the weight function in Figue is valid. Since a valid weight function emains valid when multiplied by a positive constant (in this case ), this statement follows fom Lemma, with t = and s = 0. Figue : A cetificate that π(q ) 8. The convex combination of the thee stategies shown in Figue (each taken with weight ) yields w such that w (v) = fo all v. Thus, the Coveing Lemma shows that π(q ) 8, so Q is Class 0. These thee stategies in Figue also seve as a cetificate that π(q ) 8, and they yield an efficient algoithm fo getting a pebble to, stating fom any configuation p with p 8. The most famous long-standing pebbling poblem is Gaham s conjectue: fo all gaphs G and G, π(g G ) π(g )π(g ); hee denotes the Catesian poduct. This conjectue has been veified only fo a few classes of gaphs. Specifically, it holds when G and G ae both cycles [9], both tees [], both complete bipatite gaphs [7], o a fan and a wheel [8]. When consideing Gaham s conjectue, we ae inteested in the Lemke Gaph, denoted L and shown on the left in Figue. This gaph is of inteest because it is the 6

7 smallest gaph without the -pebbling popety. The exact definition is unimpotant fo us hee; what mattes, is that if G has this popety, then π(g H) π(g)π(h) fo evey gaph H. This makes L L a natual candidate fo dispoving Gaham s conjectue. Figue : The Lemke gaph. Hulbet asseted that it is impossible, using tee stategies alone, to obtain the pebbling numbe of the Lemke gaph via the linea pogamming technique. Howeve, by using this method with moe geneal weight functions, we pove that π(l) = 8. Figue : A weight function useful fo the Lemke gaph. Theoem. If L is the Lemke gaph, then π(l) = 8. Poof. Note that π(l) V (L) = 8, so we focus on poving the uppe bound. Hulbet [0] showed that π(l, v) = 8 fo all vetices v L except fo, as shown on the left in Figue. So we only need to show that π(l, ) = 8. Now we need the weight function in Figue. Claim. The weight function in Figue is valid. The poof of this claim is vey simila to the poof of Obsevation, so we just sketch the ideas. If any vetex weighted 6 has no pebbles, then we invoke the weight 7

8 function in Figue, and multiply the esulting inequality by 5 to get one that implies what we want; so we assume that each vetex weighted 6 has a pebble. If the vetex weighted has a pebble, then the vetex weighted 5 has at most one pebble, so we ae done. Othewise, the vetex weighted 5 has at most pebbles; again, we ae done. This poves the claim. The poof that π(l, ) 8 uses the two stategies in Figue. The ightmost is a nonbasic tee stategy. The cente stategy is deived fom the weight function in Figue by adding a vetex with weight adjacent to some vetex with weight 6. This weight function is valid, by Obsevation. When we sum the weights of the two stategies, each vetex has weight at least 7 and the total weight is 55. Now the Coveing Lemma implies that π(l, ) 55 + = Lage Gaphs In this section we detemine the pebbling numbe of the Buhat gaph of ode. The (weak) Buhat gaph of ode m has as its vetices the pemutations of {,..., m}; two vetices ae adjacent if the coesponding pemutations diffe by an adjacent tansposition. Since this gaph is vetex-tansitive, we can choose the oot vetex abitaily. Using the linea pogamming method, Hulbet poved that π(b ) 7. By using moe geneal weight functions, we calculate the pebbling numbe of this gaph exactly. 8/, 0/, 6/,0,,,0 0,,,0,0,0,0 0,,0 0,0, 0,0, 8,0,0 0,8,0 0,0,8 0,0,(8,0) 8,0,0 0,0,(0,8) 0,0,6 6,0,0 0,6,0 0,6,0 0,0,6,0,0 0,,0 0,0, Figue 5: The Buhat gaph of ode and a set of stategies poving π(b ) = 6. 8

9 Theoem. If B is the Buhat gaph of ode, then π(b ) = 6. Poof. The diamete of B is 6, so π(b ) 6 = 6. We need to show that π(b ) 6. Note that the ightmost gaph in Figue 6 descibes two stategies, as we explain below. We combine these fou stategies, as shown in Figue 5, (weighted with multiplicities,,, ) to get a weight function 8 8 w, such that w (v) fo all v and w B = 6. This poves the desied uppe bound π(b ) 6 +. Thus, we only need to show that the fou stategies in Figue 6 ae valid. 8/ 0/ 6/ (0,8) (8,0) Figue 6: Moe geneal weight functions. The weight function on the ight denotes two diffeent weight functions on G; the fist includes the vetex labeled (8, 0) but not the one labeled (0, 8), and the second vice vesa. By Lemma, the leftmost and ightmost stategies ae valid (the fome with t = and s = ; the latte with t = 5 and s = 0). The poof that the middle stategy is valid is simila to the poof of Lemma, so we just sketch the ideas. Note that weights 0, 5, 0, and 5 (with the othe vetices unweighted) ae consistent with Lemma (mulitplied by 5 ), when t = and s = 0, and adding a vetex by Obsevation. Thus, we know that if p is -unsolvable, then these five vetices have weight at most 75. The key obsevation is that these five vetice can play the ole of P in the poof of Lemma. Let x 0, x, and x denote the vetex labeled 0 and its two neighbos, espectively. We fist conside the case whee 9

10 x o x, say x, has no pebbles. In this case we move as much weight as possible to the vetex labeled 5 fom x 0 and x. We also conside the case whee both x and x have pebbles. Eithe we can educe to the pevious case, o else we get two inequalities. We add these two to the inequality fo the bottom 5 vetices, which gives the desied inequality. Class 0 Gaphs. Peliminaies In this section, we study Class 0 gaphs. We focus on gaphs with diamete at least since those with diamete 0, a single vetex, and diamete, a complete gaph, ae well undestood. A gaph G is Class 0 if its pebbling numbe is equal to its numbe of vetices, i.e., π(g) = V (G). Recall that always π(g) V (G), so Class 0 gaphs ae those whee this tivial lowe bound holds with equality. Fo each vetex v, we wite N(v) fo the set of vetices adjacent to v, and we wite N[v] to denote N(v) {v}. Fo a gaph G, let e(g) denote the the numbe of edges in G. In this section, we pove lowe bounds on e(g) fo all Class 0 gaphs. Blasiak et al. [] showed that evey n-vetex Class 0 gaph G has e(g) n. They also conjectued (see [0, p. 9]) that fo some constant C and fo all sufficiently lage n thee exist n-vetex Class 0 gaphs with e(g) n + C. In paticula, they defined a family of genealized Petesen gaphs of abitay size and diamete with one vetex of some fixed degee m and all othe vetices of degee. They conjectued that these gaphs ae all Class 0. We dispove this conjectue in a vey stong sense. Shotly, we pove that fo fixed m, all sufficiently lage gaphs of this fom ae not Class 0. (Figue 7 shows P 8,, one of these genealized Petesen gaphs that is not Class 0.) Late in this section, we extend this idea to show that evey n-vetex Class 0 gaph G has e(g) 5n. To conclude the section, fo all diamete gaphs G we stengthen this lowe bound to e(g) n 5. Futhe, we chaacteize the gaphs whee this bound holds with equality (which include two infinite families). u V Figue 7: The genealized Petesen gaph, P 8,, is not Class 0. 0

11 Ou main tool fo poving bounds on e(g) is the following lemma. Lemma (Small Neighbohood Lemma). Let G be a Class 0 gaph. If u, v V (G), d(u) =, and u and v ae distance at least apat, then d(v). Similaly, if u, v V (G), d(u) =, u and v ae distance at least apat, and each neighbo of v is a -vetex, then d(v). Poof. The poofs fo both statements ae simila. In each case, we assume the statement is false and constuct a configuation with V (G) vetices that is u-unsolvable. Conside the fist statement fist. Suppose, to the contay, that u and v ae as equied, but d(v). Fom configuation p by putting 7 pebbles on v, 0 pebbles on each vetex of N[u] N(v), and pebble on each othe vetex. Since N[u] N[v] 7, this configuation has at least V (G) pebbles. Now no pebble can each u, since at most one pebble can leave N[]. This contadicts that G is Class 0, so d(v). Now conside the second statement. Suppose, to the contay, that u and v ae as equied, but d(v). Fom configuation p by putting 5 pebbles on v, 0 pebbles on each vetex of N[u] (N[N[v]] \ {v}), and pebble on each othe vetex. Since N[u] N[N[v]] 5, the configuation has at least V (G) pebbles, but no pebble can each v, since at most one pebble can leave N[N[v]]. This contadicts that G is Class 0. Thus, d(v). Coollay. Fo each intege C, thee exists an intege n 0, such that if G is any n-vetex gaph with δ(g) =, n n 0, and e(g) n + C, then G is not Class 0. Poof. We can choose n 0 sufficiently lage so that thee exists some pai of vetices u, v violating the second statement of the Small Neighbohood Lemma. Specifically, it suffices to find a -vetex v such that evey vetex within distance fou of v is a -vetex. To guaantee such a vetex v, we can take, fo example, n 0 = C 5.. Diamete at least Now we use the Small Neighbohood Lemma to pove, in Theoem, that evey n- vetex Class 0 gaph G with diamete at least has e(g) 5n. The case δ(g) = is complicated, so we handle it sepaately, in Lemma 6. Fo the case δ(g), we use the following easy lemma fom [5]. Lemma 5 ([5]). Evey Class 0 gaph G has no cut-vetices. Specifically, δ(g). Poof. Let G be a gaph with a cut-vetex u and neighbos v and v that ae in diffeent components of G u. Conside the distibution p with pebbles on v, 0 pebbles on each of u and v, and pebble on each othe vetex. Distibution p has V (G) pebbles, but no pebble can each v, which we now show. If a pebble eve moves to u, then at that point each vetex has at most one pebble, and v has no pebbles. Othewise, evey pebbling move is within the component of G u containing v, so no pebble eaches v. Thus no pebble can each v, so G is not Class 0.

12 Lemma 6. If an n-vetex Class 0 gaph G has diamete at least and δ(g) =, then e(g) 5 n. Poof. Let G be an n-vetex Class 0 gaph with diamete at least and δ(g) =. We assign each vetex v a chage ch(v), whee ch(v) = d(v). Now we edistibute these chages, without changing thei sum, so that all but a few vetices finish with chage at least 0; the chage of each vetex v afte edistibuting is ch (v). If at most k vetices finish with chage less than 0 (but all chages ae nonnegative), then e(g) = v V ch(v) = v V ch (v) ( 0(n k)) = 5n 5k. Choose V (G) such that d() =. Fo each positive intege i, let N i denote the set of vetices at distance i fom. Also, let N + = i N i. By the Small Neighbohood Lemma with u =, if v N +, then d(v). We edistibute chage accoding to the following two dischaging ules.. Each vetex v N takes chage fom some neighbo in N. If d(v) =, then v also takes chage fom its othe neighbo.. Each vetex v N + with d(v) = takes chage d(u). fom each neighbo u with We show that nealy all vetices finish with chage at least 0. Conside a vetex v V (G) \ N[]. If d(v) 5, then ch (v) d(v) d(v) = d(v) 0. Now suppose v N and d(v). In this case, ch (v) d(v) + (d(v) ) = d(v) + 0. Suppose instead that v N, d(v) =, and eithe v has both neighbos in N o the neighbo of v outside of N has degee at least. Now ch (v) = d(v) + = 0. We show that G has at most two -vetices in N with -neighbos in N. Suppose, to the contay, that u, u, and u ae -vetices in N, each with a -neighbo in N ; by symmety, assume u u E(G). By Lemma 5, u and u cannot have a common neighbo v N, since then v would be a cut-vetex. Thus, u and u have distinct neighbos in N. Howeve, now u is distance thee fom eithe u o u ; by symmety, say u. Now u and u contadict the Small Neighbohood Lemma. So indeed N has at most two -vetices with -neighbos in N. Now we conside -vetices in N +. Rathe than compute the chages of these -vetices individually, we goup them togethe as follows. Let H be the subgaph induced by -vetices in N +, and let H be a component of H with k vetices. If H contains a cycle, then H contains at least k edges, so vetices of H give chage to at most k (k) = k vetices outside H. Thus, ch (H ) ch(h ) k( ) = k k = 0k. Similaly, if H has some adjacent vetex that is not a -vetex, then ch (H ) ch(h ) (k + )( ) + = 0k. Instead, assume that H is a tee and evey vetex adjacent to H is a -vetex. Recall that each such -vetex is in N. If evey -neighbo of H is adjacent to the same vetex of N, call it v, then v is a cut-vetex. Thus, H has -neighbos that ae adjacent to both vetices of N ; call these -neighbos u and u. By the Small Neighbohood Lemma, evey pai of -vetices in N ae adjacent o have a common neighbo. Since u and u ae both adjacent to H, they can t be adjacent to each othe; thus, they must have a common

13 neighbo, u. Futhe, evey -vetex in N must be adjacent to u. Since u V (H ), u is a -vetex, so N has at most fou -vetices. Thus, H is the only component of H with final chage less than 0 times its size. Futhemoe, H has only a single vetex, and ch (H ) = ( ) = 8. Now we compute the total final chage of V (G). Fo each vetex v not in H, the final excess of v is ch (v) 0. Fo each component H i of H with ode k, the final excess is ch (H i ) 0 k. We now show that the sum of all final excesses is geate than o equal to, which poves the lemma. If v N + and d(v) 5, then ch (v) 0, so v has nonnegative excess. Each component of H, othe than (possibly) H, has nonnegative excess. Futhe, H has excess geate than o equal to. Each v N with d(v) has nonnegative excess. Also, each v N with d(v) = has excess 0, except fo at most two adjacent -vetices, which each have excess. Finally, the sum of the final chages on N[] is at least (since takes no chage fom N()). Thus, the sum of excesses of N[] is at least ( 0 ) = 6. So the sum of excesses ove all vetices is at least ( ) + ( ) + ( 6) =. Thus v V (G) d(v) 0n, so e(g) 5n. Now we pove ou main theoem of this section. Theoem. If G is an n-vetex Class 0 gaph with diamete at least, then e(g) 5 n. Poof. Let G be Class 0 with diamete at least. By Lemma 5, δ(g). Lemma 6 poves the bound when δ(g) =. If δ(g), then e(g) δ(g)n n. Thus, we assume that δ(g) =. The poof is simila to that of Lemma 6, but easie. Recall that a k-vetex is a vetex of degee k. Similaly, a k + -vetex has degee at least k and a k-neighbo of a vetex v is a k-vetex adjacent to v. Choose to be a -vetex with as few vetices at distance as possible. Fo each intege i, let N i denote the set of vetices at distance i fom. Also, let N + = i N i. We fist handle the case N 8, which is shot. Claim. If N 8, then e(g) 5 n. Since was chosen among all -vetices to minimize N, each -vetex has eithe a 5 + -neighbo o at least two -neighbos. Thus, we let ch(v) = d(v) and use the following dischaging ule.. Each -vetex takes 6 fom each -neighbo and fom each 5+ -neighbo. If d(v) 5, then ch (v) d(v) d(v) = d(v) 0. If d(v) =, then ch (v) d(v) d(v) = 6 = 0. If d(v) =, then 6 ch (v) + = 0 o ch (v) + = 0. 6 Hence, e(g) = v V (G) ch(v) = v V (G) ch (v) 5 n. This poves the claim. Heeafte, we assume that N 7. Now a vaiation on the Small Neighbohood Lemma implies that d(v) fo each vetex v N +. Suppose instead that d(v) = fo some vetex v N +. Let p be the configuation with 5 pebbles on, 0

14 pebbles on each vetex in N N N[v], and pebble on each othe vetex. Since {} N N N[v] 5, the configuation has at least n pebbles, but no pebble can each v, since at most one pebble can leave N[] N. This contadicts that G is Class 0. Thus, d(v) fo each v N +. Now we again edistibute chage. We let ch(v) = d(v) and we use the following two dischaging ules.. Each vetex in N takes chage fom its neighbo in N.. Each vetex in N takes chage fom its neighbo in N. We show that each vetex in V (G) \ N[] finishes with chage at least 0. If v N +, then ch (v) = ch(v) = d(v). If v N, then ch (v) d(v) + 0. If v N, then ch (v) d(v) + (d(v) ) = d(v) + 0. The total chage on vetices of {} N is + () = 6. Thus, the sum of all final chages is at least 0 (n ) + 6 = 0n. Thus, e(g) 5n.. Diamete We now pove that evey n-vetex diamete Class 0 gaph G has at least n 5 edges. This bound is best possible. Befoe poving this esult, we descibe some gaphs whee equality holds. In what follows, we show that these ae the only gaphs whee equality holds. To begin, we need the following lemma. Lemma 7. Given a gaph G and a vetex v V (G), fom G fom G by adding a new vetex, v, with N(v ) = N(v). If G is Class 0, then G is also Class 0. Poof. Let G be Class 0, and fom G fom G as in the lemma. We show that G is Class 0. Let p be a configuation of size V (G ) on G and be a taget vetex in G. Fist suppose that / {v, v }. We fom configuation p fo G as follows. Let p(w) = p (w) fo all w V (G) \ {v}, and let p(v) = max(p (v) + p (v ), 0). Now p V (G), so is eachable fom p in G; let σ be a pebbling sequence that eaches in G. If σ eaches fom p in G, then we ae done. Othewise, v must make moe moves in σ in G fom p than ae possible in G fom p. Now all of these exta moves fom v can be made instead fom v (pecisely because p(v) = p (v) + p (v ) ). Thus is eachable in G fom p. Suppose instead that {v, v }; by symmety, assume that = v. We may assume that p (v) = 0 and p (v ) <. If p (v ), then we can poceed as in the pevious paagaph. So assume that p (v ) {, }. Since G is Class 0, Lemma 5 implies that d(v). Choose u, u N(v). Since p(v ) {, }, we can assume that p(u ) = p(u ) = 0. We fom p fo G as follows. Let p(w) = p (w) fo all w V (G) \ {u, u } and p(u ) = p(u ) =. Now p V (G), so v is eachable fom p in G; let σ be a pebbling sequence that eaches v in G. If σ makes no moves fom u o u, then σ also eaches v fom p in G. So assume that σ makes a move fom u o u. Fom σ fom σ by tuncating σ just befoe the fist time that it moves fom

15 8 8 8 Figue 8: The tee stategies fo C 5 and fo K with the edges of a K, subdivided. u o u, say u, and then appending a move fom v to u and a move fom u to v. Now σ eaches v fom p in G. Thus, G is Class 0. Now we use Lemma 7 to show that two infinite families of gaphs ae all Class 0. Example. The following ae two infinite families of Class 0 gaphs. Each n-vetex 5

16 gaph has exactly n 5 edges. To fom an instance of F p,q, begin with K and eplace the two edges incident to some vetex v with p paallel edges and q paallel edges (whee p and q ae positive); finally, subdivide each of these p + q new edges. To fom an instance of G p,q,, begin with K and eplace the thee edges incident to some vetex v with p paallel edges, q paallel edges, and paallel edges (whee p, q, and ae positive); finally, subdivide each of these p + q + new edges. It is easy to see that each n-vetex gaph in F p,q has n 5 edges, since the - vetices induce an independent set (when p and q ), and the thee high-degee vetices have among them a single edge. Similaly, G p,q, has n 5 edges, since the -vetices induce an independent set and the fou high-degee vetices have among them edges. We pove that all of F p,q is Class 0, by induction on p+q; the induction step follows immediately fom Lemma 7. The base case is F,, which is the 5-cycle. To show that it is Class 0, we use the tee stategies shown in the fist ow of Figue 8. Since C 5 is vetex-tansitive, we can pick the oot abitaily. Let w be the sum of the weights in the two tee stategies fo C 5. Note that w(v) fo evey vetex v V (G) \ {} and v V (G)\{} w(v) = < ( + ). Thus, by the Coveing Lemma, C 5 is Class 0. We pove that all of G p,q, is Class 0, by induction on p + q + ; the induction step follows immediately fom Lemma 7. The base case is G,,. To show that G,, is Class 0, we use the tee stategies shown in Figue 8. Up to symmety, G,, has thee types of vetices: a degee vetex, the cente degee vetex, and a peipheal degee vetex. The tee stategies fo these cases ae given in the fist, second, and thid ow below the stategies fo C 5. Let be a degee vetex, and let w(v) be the sum of the two weight functions in the second ow of Figue 8. Note that w(v) fo all v V (G) \ {}. Futhe, v V (G)\{} w(v) = 0 < (6 + ). Thus, the Coveing Lemma implies that π(g,,, ) V (G,, ). Now let be the cente vetex, and let w(v) be the sum of the thee weight functions in the thid ow of Figue 8. Note that w(v) = fo all v V (G) \ {}. Thus, v V (G)\{} w(v) = < (6 + ). Thus, the Coveing Lemma implies that π(g,,, ) V (G,, ). Finally, let be a peipheal vetex, and let w(v) be the sum of the thee weight functions in the fouth ow of Figue 8. Note that w(v) 7 fo all v V (G)\{}. Futhe, v V (G)\{} w(v) = 6 < 7(6+). Thus, the Coveing Lemma implies that π(g,,, ) V (G,, ). Since π(g,,, ) V (G,, ) fo each oot, we conclude that π(g,, ) V (G,, ). So, G,, is Class 0. Now we show that evey diamete Class 0 gaph G has e(g) n 5 and chaacteize when equality holds. Clake et al. [5, Theoem.] chaacteized diamete gaphs that ae not Class 0. It seems likely that we could deive ou esult fom theis. Howeve, we pefe the poof below, since it seems simple and moe staightfowad. Futhe, the poof below genealizes to diamete gaphs with no cut-vetices. Theoem 5. Let G be an n-vetex gaph with diamete. If G has no cut-vetex (in paticula, if G is Class 0) then e(g) n 5. Futhe, equality holds if and only if G is the Petesen gaph o one of the gaphs in Example. 6

17 Poof. If δ(g), then e(g) n = n and the theoem is tue. So we assume δ(g). Lemma 5 implies that δ(g), so e(g) nδ(g) n. If n 5, then e(g) n n 5, so the theoem is tue. Thus, we assume n 6. We conside two cases: (i) δ(g) = and (ii) δ(g) =. Case : δ(g) =. Choose V (G) with d() =, and let S = N(). Each vetex v V (G) \ S has a neighbo in S and has neighbos in S, so v S d(v) (n ) + = n. Also, v V (G)\S d(v) v V (G)\S = (n ). So e(g) = v V (G) d(v) ((n ) + (n )) = (n 0) = n 5. If equality holds in e(g) n 5, then each vetex in V \ S has degee and each vetex in V \ (S {}) has exactly one neighbo in S. Let {v, v, v } = S, let S i = N(v i ) fo each i {,, }, and let H = G[S S S ]. Note that H is a disjoint union of cycles, since each vetex has degee and has exactly one neighbo in S. Also S i fo each i, since δ(g) =. Suppose that S, and choose v S. Now S S + = 5, so v is distance at least fom some vetex of S S (pecisely because H is a disjoint union of cycles). Hence S = and, by symmety, S = S =. Similaly, if H consists of two -cycles, then some pai of its vetices is distance at least apat. Hence, H is a 6-cycle. Futhe, each pai of vetices in the same S i ae distance apat in H. Thus, if e(g) = n 5, then G is the Petesen gaph. Case : δ(g) =. Choose V (G) with d() =, and let {v, v } = N(). We patition V (G) \ N() into thee sets, S, S, and S,. (Note that S,.) Let S consist of all vetices adjacent only to v, S of all vetices adjacent only to v, and S, of all vetices adjacent to both v and v. Let H be the subgaph induced by S S, and let H,..., H t be the components of H. We fist show that e(g) n if evey H i eithe contains a cycle o has a vetex adjacent to some vetex of S,. We assign each edge to one of its endpoints as follows, so that each vetex othe than v and v has at least assigned edges. Each edge with exactly one endpoint in {v, v } is assigned to its othe endpoint. Thus, each vetex of S, has assigned edges and each vetex of S S has assigned edge. Suppose that T is some tee component of H and t is a vetex of T with a neighbo in S,. We can diect the edges of T so that t has outdegee 0 and each othe vetex has outdegee. Now we assign to t its edge to S, and assign to each othe vetex of T its out-edge. When H i is a component of H with a cycle, the pocess is simila. We choose some spanning tee T of H i and choose t to be some vetex incident to an edge of H i not in T. So assume that some component H is a tee and has no neighbo in S,. If V (H ) S, then v is a cut-vetex, which is fobidden. Hence, V (H ) S ; similaly, V (H ) S. Suppose that H has anothe component, with some vetex w. By symmety, assume that w S. Since H has vetices in both S and S, w is distance at least fom some vetex of H in S, a contadiction. Thus, H is the only component of H; so fom now on, we say H fo H. Choose t abitaily in H, and diect E(H) and assign edges as above. Now t has assigned edge and each othe vetex has assigned edges, so e(g) n 5. If v v E(G), then e(g) n, so we assume v v / E(G). Similaly, if any edge has both endpoints in S,, then 7

18 e(g) n, so we assume that S, induces an independent set. In what follows, we chaacteize when equality holds in e(g) n 5. Fist suppose that H has leaves in both S and S ; call these u and u, espectively. If u and u ae adjacent, then H is a single edge, which is possible; this is F,q, whee q = S,. Now suppose that u and u ae nonadjacent. Since G is diamete, u and u have some common neighbo, u. By symmety, assume that u S. Now u and v must have a common neighbo, so v v is an edge. Howeve, now e(g) n. So assume instead that H has leaves only in one of S and S ; by symmety, say S. Let S denote the vetices of S adjacent to a leaf. Now S induces a gaph with diamete at most (othewise some leaf is distance at least fom some vetex of S ). Since H is acyclic, S. Fist suppose that S =. Let {u } = S. Now all vetices in S ae leaves of H, since H is acyclic. Futhe, u is the only vetex in S. Thus, H is a sta centeed at u. This is possible; G = F p,q, whee q = S, and p is the numbe of leaves of H. Suppose instead that S =, and let {u, u } = S. Now u and u ae adjacent, since G has diamete ; othewise some leaf in S is distance at least fom u o u. Again, each vetex u S must be a leaf, since G is acyclic. Finally, S = {u, u }, again since G has diamete. Thus, H is a double sta, centeed at u and u, with all leaves in S. This is also possible; G = G p,q,, whee p = S, and q and ae (espectively) the numbes of leaves of H adjacent to u and u. Hence, e(g) = n 5 implies that H is (i) a single edge, which is F,q, (b) a sta with its cente in S (by symmety) and all of its leaves in S, which is F p,q, o (c) a double sta with both of its centes in S and all of its leaves in S, which is G p,q,. This finishes the chaacteization of when e(g) = n 5. Refeences [] A. Blasiak, A. Czyginow, A. Fu, D. Hescovici, G. Hulbet and J.R. Schmitt. Spase gaphs with small pebbling numbe. Manuscipt, (0). [] B. Bukh. Maximum pebbling numbe of gaphs of diamete thee. J. Gaph Theoy 5 (006), [] F. Chung. Pebbling in hypecubes. SIAM J. Discete Math. (989), [] B. Clak and K. Milans. The complexity of gaph pebbling, SIAM J. Discete Math. 0 (006), [5] T.A. Clake, R.A. Hochbeg, and G.H. Hulbet. Pebbling in diamete two gaphs and poducts of paths. J. Gaph Theoy 5 (997), 9 8. [6] P. Edős, A. Ginzbug, and A. Ziv. A theoem in additive numbe theoy. Bull. Res. Council Iseal 0F (96). 8

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